This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
2
(2)
(3)
etc. In 1995 Seiffert [6] considered another mean, namely T = T (x, y) =
x−y x − y (x 6= y), 2 arctan x+y
T (x, x) = x.
(4)
(Here T , as P in [1], is our notation for these means, see [2]). He proved that A < T < Q.
(5)
2. Our aim in what follows is to show that by a transformation of arguments, the mean T can be reduced to the mean P . Therefore, by using the known properties of P , these will be transformed into properties of T . Theorem 1. Let u, v > 0; and put p p 2(u2 + v 2 ) + u − v 2(u2 + v 2 ) + v − u , y= . x= 2 2 Then x, y > 0;p and T (u, v) = P (x, y). Proof. From p2(u2 + v 2 ) > |u − v| we get that x > 0, y > 0. Clearly one has x + y = 2(u2 + v 2 ), x − y = u − v. From the definitions (1) and u−v u−v = arcsin p . Let u > v, and put (4) we must prove arctan u+v 2(u2 + v 2 ) u−v α = arctan . By u+v tan α sin α = cos α tan α = √ 1 + tan2 α 266
and
u−v u−v s uµ+ v ¶ = p 2 2(u2 + v 2 ) u−v 1+ u+v
we get u−v u−v arcsin p = α = arctan , 2 2 u+v 2(u + v ) and the proof of the above relation is finished. It is interesting to remark that x+y = A(x, y) = 2 G(x, y) =
r
u2 + v 2 = Q(u, v) 2
u+v √ xy = = A(u, v). 2
Therefore, by using the transformations of Theorem 1, the following transformations of means will be true: G → A,
A → Q,
P → T.
Thus, the inequality G < P < A valid for P , will be transformed into A < T < Q, i.e. relation (5). By using our inequality (2), we get for T the following results: p A + 2Q 3 Q2 A < T < , (6) 3 while using (3), we get
µ 3
T >
Q+A 2
¶2 Q.
(7)
In fact, the following is true: Theorem 2. Let 0 < u < v. Then T = T (u, v) is the common limit of the sequences (un ) and (vn ) defined by u0 = A(u, v), v0 = Q(u, v), un+1 = For all n ≥ 0 one has un < T < vn and 267
u n + vn √ , vn+1 = un+1 vn . 2 p 3
vn2 un < T <
un + 2vn . 3
References [1] J. S´andor, On certain inequalities for means, III, Arch. Math., Basel, 76(2001), 34-40. ¨ [2] J. S´andor, Uber zwei Mittel von Seiffert, Wurzel, 36(2002), no.5, 104-107. [3] H.-J. Seiffert, Werte zwishen dem geometrischen und dem arithmetischen Mittel zweier Zahlen, Elem. Math., Basel, 42(1987), 105-107. [4] H.-J. Seiffert, Problem 887, Nieuw Arch. Wisk., (4)11(1993), 176. [5] H.-J. Seiffert, Ungleichungen f¨ ur einen bestimmten Mittelwert, Nieuw Arch. Wisk. (4)13(1995), 195-198. [6] H.-J. Seiffert, Aufgabe β16, Wurzel, 29(1995), no.3+4, 87.
9
Some new inequalities for means and convex functions 1. In what follows, for a, b > 0 let us denote A = A(a, b) =
a+b , 2
a2 + b2 W = W (a, b) = , a+b
G = G(a, b) =
√ ab,
H = H(a, b) = 2/
µ
1 1 + a b
¶ .
If f : [a, b] → R is an increasing (decreasing) function, then the following property is immediate: af (b) + bf (a) f (a) + f (b) af (a) + bf (b) Proposition 1. ≤ ≤ (1) a+b 2 a+b All inequalities in (1) are reversed, when f is decreasing. Proof. After simple computations, each parts of (1) become equivalent to (f (b) − f (a))(b − a) ≥ 0 (or (f (b) − f (a))(b − a) ≤ 0). For f (x) = x, relations (1) imply the classical inequality H ≤ A ≤ W. A more interesting example arises, when f (x) = ln x. Then we get (ab ba )1/(a+b) ≤ G ≤ (aa bb )1/(a+b) .
(2)
For the involved means in the extremal sides of (2), see e.g. the References [1]-[3]. 268
If f is convex, the following can be proved: Proposition 2. Let f be convex on [a, b]. Then f (W ) ≤
af (a) + bf (b) ; a+b
(3)
f (H) ≤
af (b) + bf (a) ; a+b
(4)
af (b) + bf (a) + f (W ) ≤ f (a) + f (b). a+b µ Proof. f (W ) = f
≤
a2 + b2 a+b
¶
(5)
µ ¶ a b =f a +b a+b a+b
a b af (a) + bf (b) f (a) + f (b) = , a+b a+b a+b
by the convexity of f (i.e. f (aλ + bµ) ≤ λf (a) + µf (b) for λ, µ > 0, λ + µ = 1). This proved (3). Now, µ ¶ µ ¶ 2ab a b f (H) = f =f b+ a a+b a+b a+b ≤
a b af (b) + bf (a) f (b) + f (a) = , a+b a+b a+b
yielding (4). Relation (5) follows by (3), since af (b) + bf (a) af (a) + bf (b) + = f (a) + f (b). a+b a+b 2. By taking into account of Propositions 1 and 2, one can ask the question of validity of relations of type af (b) + bf (a) af (a) + bf (b) ≤ f (W ) ≤ , a+b a+b or f (H) ≤
af (b) + bf (a) ≤ f (A), etc. a+b
We will prove the following results: 269
Theorem 1. ([4]) Let f : [a, b] → R be a differentiable, convex and increasf 0 (x) ing function. Suppose that the function g(x) = , x ∈ [a, b] is decreasing. x Then one has af (b) + bf (a) f (H) ≤ ≤ f (A). (6) a+b Proof. The left side of (6) is exactly relation (4). Let us write the righthand side of (6) in the form a[f (b) − f (A)] ≤ b[f (A) − f (a)]. By b − A = has
(∗)
b−a = A − a, and by the Lagrange mean value theorem one 2
b−a 0 b−a 0 f (ξ2 ), f (A) − f (a) = f (ξ1 ), 2 2 where ξ1 ∈ (a, A), ξ2 ∈ (A, b). Thus a < ξ1 < ξ2 < b. By f 0 (x) ≥ 0, and f 0 being increasing we get by the monotonicity of g: f (b) − f (A) =
f 0 (a) f 0 (b) ≤ , b a so af 0 (ξ2 ) ≤ af 0 (b) ≤ bf 0 (a) ≤ bf 0 (ξ1 ). This implies relation (∗), i.e. the proof of Theorem 1 is completed. The following theorem has a similar proof: Theorem 2. Let f : [a, b] → R be a differentiable, convex, and increasing √ function. Suppose that the function h(x) = f 0 (x)/ x is decreasing on [a, b]. Then af (b) + bf (a) ≤ f (G). (7) f (H) ≤ a+b For f (x) = x, (7) gives the classical inequality H ≤ G. Theorem 3. Let f : [a, b] → R be a differentiable, convex, and increasing f 0 (x) function. Suppose that the function g(x) = is decreasing on [a, b]. Then x f (A) ≤ f (W ) ≤
f (a) + f (b) . 2
(8)
Proof. The left side of (8) is trivial by A ≤ W and the monotonicity of f . The proof of right side is very similar to the proof of right side of (6). Indeed, W −a=
b(b − a) , a+b
b−W =
270
a(b − a) . a+b
By Lagrange’s mean value theorem one has f (W ) − f (a) =
b(b − a) 0 f (η1 ), a+b
f (b) − f (W ) =
a(b − a) 0 f (η2 ), a+b
where η1 ∈ (a, W ), η2 ∈ (W, b). Now, we can write that af 0 (η1 ) ≤ af 0 (b) ≤ bf 0 (a) ≤ bf 0 (η2 ), so f (W ) − f (a) ≤ f (b) − f (W ), and (8) follows. 3. Finally, we shall prove an integral inequality, which improves on certain known results. Theorem 4. ([4]) If f : [a, b] → R is convex and differentiable, then · ¸ Z b 1 1 af (b) + bf (a) f (a) + f (b) f (x)dx ≤ + f (W ) ≤ . (9) b−a a 2 a+b 2 Proof. Since f is convex, and differentiable, we can write that f (x) − f (y) ≤ (x − y)f 0 (x) for all x, y ∈ [a, b].
(∗∗)
Apply now (∗∗) for y = W and integrate the relation on x ∈ [a, b]: Z b Z b (x − W )f 0 (x)dx. f (x)dx ≤ (b − a)f (W ) + a
a
Here Z
b
Z 0
(x − W )f (x)dx =
b
xf 0 (x)dx − W [f (b) − f (a)]
a
a
Z = bf (b) − af (a) −
b
f (x)dx − W [f (b) − f (a)], a
by partial integration. Thus · ¸ Z b af (b) + bf (a) 2 f (x)dx ≤ (b − a) + (b − a)f (W ), a+b a and the left side of (9) follows. The right hand side inequality of (9) is a consequence of relation (5). Remarks. 1) Relation (9) improves the Hadamard inequality Z b f (a) + f (b) 1 f (x)dx ≤ . b−a a 2 271
2) If the conditions of Theorem 1 are satisfied, the following chain of inequalities holds true: Z b af (b) + bf (a) 1 f (H) ≤ ≤ f (A) ≤ f (x)dx a+b b−a a · ¸ f (a) + f (b) 1 af (b) + bf (a) + f (W ) ≤ . (10) ≤ 2 a+b 2 3) The methods of this paper show that the more general means Wk =
ak + bk ak−1 + bk−1
may be introduced.
References [1] J. S´andor, On the identric and logarithmic means, Aequations Math., 40(1990), 261-270. [2] J. S´andor, On certain integral inequalities, Octogon Math. Mag., 5 (1997), 29-35. [3] J. S´andor, Gh. Toader, On means generated by two positive functions, Octogon Math. Mag., 10(2002), no.1, 70-73. [4] J. S´andor, On certain new inequalities of convex functions (Hungarian), Erd´elyi Mat. Lapok, 5(2003), no.1, 29-34.
10
On an inequality of Sierpinski on the arithmetic, geometric and harmonic means
1. Let xi , i = 1, n, be strictly positive numbers and denote their usual arithmetic, geometric and harmonic mean by n X
An (x) =
Ã
xi
i=1
n
,
Gn (x) =
n Y i=1
!1
n
xi
,
Hn (x) =
n n X i=1
where x = (x1 , . . . , xn ). 272
1 xi
,
In 1909 W. Sierpinski ([5]) discovered the following double-inequality: (Hn (x))n−1 An (x) ≤ (Gn (x))n ≤ (An (x))n−1 Hn (x).
(1)
The aim of this note is to obtain a very short proof for (1) (in fact, a generalization), by using Maclaurin’s theorem for elementary symmetric functions. For another idea of proof for (1) (due to the present author), which leads also to a refinement of an inequality of Ky Fan, see [1]. For application of (1) see [4]. Now we state Maclaurin’s theorem as the following: Lemma. Let cr be the r-th elementary symmetric function of the x (i.e. the sum of the products, r at a time, of different xi ) and pr the average of these products, i.e. cr pr = µ ¶ . n r Then
1
1
1
1
n−k p1 ≥ p22 ≥ p33 ≥ · · · ≥ pn−k ≥ · · · ≥ pnn .
(2)
See [2], [3] for a proof and history of this result. 2. Our result is contained in the following Theorem. Let k = 1, 2, . . . and define the k-harmonic mean of x by µ ¶ n k Hn,k (x) = X . 1 x1 . . . xk (nk) Then one has the inequality (Gn (x))n ≤ (An (x))n−k Hn,k (x).
Remark. Letting x →
1 , we get x
(Gn (x))n ≥ (Hn,k (x))n−k An,k (x), where
X An,k (x) = pk = 273
x1 . . . xk Cnk
.
(3)
1 n−k Proof. Apply p1 ≥ pn−k from (2), where
pn−k
X x1 . . . xn−k µ ¶ = = n n−k
Ã
n Y i=1
! xi
X
1 x1 .¶. . xk , µ n k
and we easily get (3). For µ k = 1 one ¶reobtains the right side of inequality (1). By replacing x by 1 1 1 = ,..., , and remarking that x x1 xn µ ¶ µ ¶ µ ¶ 1 1 1 1 1 1 Gn = , An = , Hn = , x Gn (x) x Hn (x) x An (x) we immediately get the left side of (1) from the right side of this relation. This finishes the proof of (1).
References [1] H. Alzer, Versch¨ arfung einer Ungleichung vom Ky Fan, Aequationes Math., 36(1988), 246-250. [2] G.H. Hardy, J.E. Littlewood, G. P´olya, Inequalities, 2nd ed., Cambridge Univ. Press, Cambridge, 1959, Theorem 52. [3] C. Maclaurin, A second letter to Martin Folkes, Esq., concerning the roots of equations, with the demonstration of other rules in algebra, Phil. Transactions, 36(1729), 59-96. [4] J. S´andor, On an inequality of Ky Fan, Babe¸s-Bolyai Univ., Seminar on Math. Analysis, no.7, 1990, 29-34. [5] W. Sierpinski, Sur une in´egalit´e pour la moyenne arithm´etique, g´eometrique et harmonique, Warsch. Sitzunsber. 2(1909), 354-357.
11
An application of Rolle’s theorem
1. Rolle’s theorem from the differential calculus asserts that for a continuous function f : [a, b] → R which is differentiable on (a, b) and f (a) = f (b), there exists at least a θ ∈ (a, b) such that f 0 (θ) = 0. As it is well-known, this 274
theorem implies among others the classical mean-value theorems by Lagrange, Cauchy or Taylor. The aim of this note is to consider a less known application of this theorem, with interesting consequences. Two applications relating to the logarithmic mean of numbers, and the Euler gamma function will be given. 2. Let f : [0, 1] → R be a 3-times differentiable function and define Fk : [0, 1] → R by 1 Fk (x) = f (x) − f (0) − x[f 0 (x) + f 0 (0)] 2 ½ ¾ 1 k 0 0 −x f (1) − f (0) − [f (1) + f (0)] (1) 2 where k ∈ R is a fixed number. Clearly, one has Fk (0) = Fk (1) = 0, thus by Rolle’s theorem there exists θ1 ∈ (0, 1) with Fk0 (θ1 ) = 0. Since 1 1 Fk0 (x) = f 0 (x) − [f 0 (x) + f 0 (0)] − xf 00 (x) 2 2 ½ ¾ 1 0 k−1 0 −kx f (1) − f (0) − [f (0) + f (1)] 2
(2)
we have Fk0 (0) = 0. Applying once again the Rolle theorem, one finds a θ2 ∈ (0, θ1 ) with Fk00 (θ2 ) = 0. By ½ ¾ 1 1 Fk00 (x) = xf 00 (x) − k(k − 1)xk−2 f (1) − f (0) − [f 0 (0) + f 0 (1)] (3) 2 2 we get the following result: If f : [0, 1] → R satisfies the above conditions, and k ∈ R is given, then there exists θ = θ2 ∈ (0, 1) such that 1 f 00 (θ) f (1) = f (0) + [f 0 (0) + f 0 (1)] − . 2 2k(k − 1)θk−3
(4)
Let now f (x) = g[(b − a)x + a], where x ∈ [0, 1] and a < b, being 3-times differentiable etc. function. Applying (4) for this function, one can derive: If g : [a, b] → R is as above and k ∈ R is given, then there exists θ ∈ (0, 1) such that 1 g 000 (ξ)(b − a)3 g(b) = g(a) + [g 0 (a) + g 0 (b)](b − a) − , 2 2k(k − 1)θk−3
(5)
where ξ = (b − a)θ + a. For k = 3 one gets a beautiful result: 1 g 000 (ξ)(b − a)3 g(b) = g(a) + [g 0 (a) + g 0 (b)](b − a) − , 2 12 275
ξ ∈ (a, b).
(6)
If we select
Z
x
g(x) =
F (t)dt, a
where F : [a, b] → R has a continuous second order derivative, then from (6) we can obtain: Z b F 00 (ξ)(b − a)3 b−a F (t)dt = [F (a) + F (b)] − , ξ ∈ (a, b), (7) 2 12 a called also as the trapezium formula (see e.g. [3]). 3. As a first application, let F (t) = log t in (7) and 0 < a < b. After some elementary transformation, we can obtain: (b − a)2 (b − a)2 A(a, b) < + 1 < + 1, 12b2 L(a, b) 12a2 where A(a, b) =
a+b 2
and L(a, b) =
(8)
a−b log a − log b
denote the arithmetic and logarithmic mean, respectively, of a and b. We note that the left side of (8) refines the known relation L(a, b) < A(a, b) (see [2], [7]). For another application, let b = x+1, a = x, where x > 1 and F (t) = Ψ(t), where Ψ is the Euler ”digamma function”, i.e. Ψ(t) =
Γ0 (t) , Γ(t)
with Γ the Euler gamma function. Since 00
Ψ (t) = −2
∞ X n=0
1 (n + x)3
(see [1]), and from the inequalities Z
∞
h(t)dt < 0
∞ X
Z h(n) < h(0) +
∞
h(t)dt, 0
n=0
which are valid for all functions h : (0, ∞) → R+ which are strictly decreasing with lim h(x) = 0 (see [4]), one can derive that x→∞
log x −
1 1 1 1 − < Ψ(x) < log x − − , x > 1. 2 2x 12(x − 1) 2x 12(x + 1)2 276
(9)
which improve certain known relations for Ψ (see also [1], [4]). This application appears also in [5]. 4. Of course, there are various applications of (5), (6), (7). By the same method, the following general result of form (4) can be proved: Let f : [a, b] → R be a 3-times differentiable function, and let α, β, γ, δ, k, γ 6= 0, be given real numbers. Assume that the following conditions hold true: (i) (α − 2β)f 0 (0) = 0; (ii) [f (1) − f (0)](α − γ) + [f 0 (1) + f 0 (0)] + [f 0 (1) + f 0 (0)](γδ − β) = 0. Then there exists θ ∈ (0, 1) such that f (1) = f (0) + δ[f 0 (0) + f 0 (1)] +
α − 2β f 00 (θ) β f 000 (θ) · k−2 − · k−3 . k(k − 1)δ θ k(k − 1)γ θ
(10)
Proof. Consider the function F : [0, 1] → R defined by F (x) = α[f (x) − f (0)] − βx[f 0 (x) + f 0 (0)] −γxk {[f (1) − f (0)] − δ[f 0 (1) + f 0 (0)]}
(11)
and apply the same argument as in the proof of (4) (we omit the details). For various selections of α, β, γ, δ, k and f we can obtain different meanvalue theorems with many possible applications.
References [1] E. Artin, The Gamma function, Holt, Einchart, Winston, New York, 1964. [2] B.C. Carlson, The logarithmic mean, Amer. Math. Monthly, 79 (1972), 615-618. [3] B.P. Demidovich, I.A. Maron, Computational Mathematics, Mir Publishers, Moscow, 1981. [4] D.S. Mitrinovi´c, Analytic Inequalities, Springer Verlag, 1970. [5] J. S´andor, Remark on a function which generalizes the harmonic series, C.R. Acad. Buld. Sci., 41(1988), 19-21. [6] J. S´andor, Sur la function Gamma, Publ. Centre Rech. Math. Pures, Neuchˆatel, S´erie I, 21, 1989, 4-7. [7] J. S´andor, Some integral inequalities, Elem. Math., 43(1988), 177-180. 277
12
An application of Lagrange’s mean value theorem for the computation of a limit
We must determine f : R∗+ → R∗+ so that f (x) → ∞ and x[f (x + 1) − f (x)] → a ∈ R (x → ∞). For a = 0 this is OQ.1074 ([1]). We note here that when f is differentiable, the following is true: If lim xf 0 (x) = a, (1) x→∞
then lim x[f (x + 1) − f (x)] = a.
x→∞
(2)
Indeed, by the Lagrange mean-value theorem one can write f (x + 1) − f (x) = (x + 1 − x)f 0 (ξ) for certain x < ξ < x + 1. x x x < 1 < + 1 it follows →1 ξ ξ ξ x as x → ∞. By x[f (x + 1) − f (x)] = f 0 (ξ) = [ξf 0 (ξ)] and the assumption ξ ξf 0 (ξ) → a as x → ∞ (i.e. ξ → ∞), we get the result (2). Now, as x → ∞, clearly ξ → 0, and by
References [1] M. Bencze, OQ.1074, Octogon Math. Mag., 10(2002), no.2, 1064.
13
An inequality of Alzer, as an application of Cauchy’s mean value theorem The inequality of Alzer [1] is the following: n ≤ n+1
Ã
n
n+1
i=1
i=1
1X r 1 X r i / i n n+1
!1/r (1)
where r ≥ 0 is a real number, while n is a positive integer. In the first part [3] we have obtained an easy proof based on mathematical induction and Cauchy’s mean value theorem. Recently, Chen and Qi [2] discovered the interesting fact that (1) is true also for r < 0. They use a complicated function and Jensen’s inequality (and, of course, mathematical induction). We now simply prove that here one can apply Cauchy’s mean value theorem, again. 278
Indeed, the mathematical induction process (see [3], [2]) leads to the inequality (k + 1)s [(k + 1)1−s − k 1−s ] > (k + 2)s [(k + 2)1−s − (k + 1)1−s ].
(2)
Now, let f, g : [k, k + 1] → R be given by f (x) = (x + 1)1−s ,
g(x) = x1−s
where 0 < s < 1. Remark that by the Cauchy mean value theorem one can write: f 0 (ξ) f (k + 1) − f (k) = 0 with ξ ∈ (k, k + 1). g(k + 1) − g(k) g (ξ) Since
f 0 (ξ) = g 0 (ξ)
µ
ξ ξ+1
¶s
µ <
k+1 k+2
¶s
we immediately get (k + 2)1−s − (k + 1)1−s (k + 1)s < , (k + 1)1−s − k 1−s (k + 2)s implying relation (2).
References [1] H. Alzer, On an inequality of Minc and Sathre, J. Math. Anal. Appl., 179(1993), 396-402. [2] C.-P. Chen, F. Qi, The inequality of Alzer for negative powers, Octogon Math. Mag., 11(2003), no.2, 442-445. [3] J. S´andor, On an inequality of Alzer, J. Math. Anal. Appl., 192(1995), 1034-1035.
14
A new mean value theorem
A new mean value theorem for differentiable functions with applications is given. 279
Introduction A function f : [a, b] → R is called a Rolle function when f is continuous on [a, b] and differentiable on its interior (a, b). The well known Cauchy mean value theorem from the differential calculus asserts that if f and g are Rolle functions on [a, b] then there exists ξ ∈ (a, b) with f (b) − f (a) f 0 (ξ) = 0 g(b) − g(a) g (ξ)
(1)
where the involved expressions are well defined (thus g is strictly monotonic). This mean value theorem has many remarkable applications in different branches of Mathematics and it is one of the powerful methods of Analysis. As a consequence for (1) it is sufficient to consider Lagrange’s and Taylor’s mean value theorems, respectively (see [2], [3], [4], [5]). The aim of this note is to prove an extension of a new type for (1) and to deduce certain applications.
Main results Theorem. Let f, g : [a, b] → R be Rolle functions and suppose that g(b) − g(a) 6= (b − a)g 0 (a),
g(b) − g(a) 6= (b − a)g 0 (b)
and g 0 (x) 6= g 0 (a), g 0 (b) for x ∈ (a, b). Then there exist ξ, η ∈ (a, b) such that f (b) − f (a) − (b − a)f 0 (a) f 0 (ξ) − f 0 (a) = 0 0 g(b) − g(a) − (b − a)g (a) g (ξ) − g 0 (a) and
(3)
f (b) − f (a) − (b − a)f 0 (b) f 0 (η) − f 0 (b) = . g(b) − g(a) − (b − a)g 0 (b) g 0 (η) − g 0 (b)
Proof. Let
f (b) − f (a) − (b − a)f 0 (c) = k, g(b) − g(a) − (b − a)g 0 (c)
where c ∈ {a, b}. Then after certain simple transformation we get f (b) − kg(b) − bf 0 (c) + kbg 0 (c) = f (a) − kg(a) − af 0 (c) + kag 0 (c). Let us introduce the auxiliar function F (x) = f (x) − kg(x) − xf 0 (c) + kxg 0 (c). 280
(5)
This is a Rolle function and by (5) one has F (b) = F (a), so by Rolle’s theorem there exists θ ∈ (a, b) with F 0 (θ) = 0. Since F 0 (x) = f 0 (x) − f 0 (c) − k[g 0 (x) − g 0 (c)] for c = a we get (3) and, similarly for c = b the relation (4). We note that condition (2) are necessary for the expressions in (3) and (4) to have sense. Remarks. 1) If f 0 and g 0 are Rolle functions too, and g 0 (x) 6= g 0 (a), g 0 (b) with g 00 (x) 6= 0 on (a, b), then there exists θ1 , θ2 ∈ (a, b) with f (b) − f (a) − (b − a)f 0 (a) f 00 (θ1 ) = g(b) − g(a) − (b − a)g 0 (a) g 00 (θ1 )
(6)
f 00 (θ2 ) f (b) − f (a) − (b − a)f 0 (b) = . g(b) − g(a) − (b − a)g 0 (b) g 00 (θ2 )
(7)
This follows by Cauchy’s mean value theorem (1) applied to (3) and (4). 2) If f 0 (a) = g 0 (a) = 0 (or f 0 (b) = g 0 (b) = 0) then from (3) (or (4)) we recapture Cauchy’s formula (1).
Applications 1) As a first application, we show that (b − a)2 eb (b − a)2 ea < eb − ea − (b − a)ea < 2 2
(8)
and
(b − a)2 ea (b − a)2 eb < (b − a)eb + ea − eb < . (9) 2 2 These follow from (6) and (7) applied to the functions f (x) = ex and g(x) = x2 . Then (2) and the conditions of (6) and (7) are satisfied with g(b) − g(a) − (b − a)g 0 (a) = (b − a)2 and g(b) − g(a) − (b − a)g 0 (b) = −(b − a)2 . 2) For a second application, let us suppose that F : [a, b] → R is a Rolle function with F 0 a Lipschitz function. Then ¯Z b ¯ ¯ ¯ (b − a)3 b−a ¯ ¯< F (t)dt − [F (a) + F (b)] K (10) ¯ ¯ 2 2 a 281
where K is the Lipschitz constant of F 0 . Set Z x f (x) = F (t)dt, g(x) = x2 a
in (6) and (7). By adding the two obtained relations, one obtains Z 2
b
F (t)dt − (b − a)[F (a) + F (b)] = a
(b − a)2 0 [F (θ1 ) − F 0 (θ2 )]. 2
(11)
By |F 0 (θ1 )−F 0 (θ2 )| < K|θ1 −θ2 | < K(b−a), and by taking absolute values in (11), we get (10). Corollary. Let I = I(a, b) and G = G(a, b) be the identric and geometric means of the positive numbers a and b, where 1 log I = b−a
Z
b
log xdx
and
G=
√ ab
a
(see [1], [6], [7]). Suppose that 1 ≤ a < b. Then 1<
b−a I <e 2 . G
(12)
The left side of (12) is well known (for all a, b > 0), for the right side apply (10) with F (t) = log t. Then ¯ ¯ ¯ ¯ ¯1 1¯ ¯ x − y¯ 1 0 ¯ ≤ x − y for xy ≥ 1. and ¯¯ − ¯¯ = ¯¯− F (t) = t x y xy ¯ Thus K = 1 and the right side of (12) follows by simple computations.
References [1] H. Alzer, Ungleichungen f¨ ur Mittelwerte, Arch. Math., 47(1986), 422426. ´ ements d’analyse, Tome I, Gauthier-Villars, Paris, 1969. [2] J. Dieudonn´e, El´ [3] B. Gelbaum, J.M.H. Olmsted, Conterexamples in analysis, Holden Day, San Francisco, London, 1964. [4] G.H. Hardy, J.E. Littlewood, G. P´olya, Inequalities, 2nd ed., Cambridge Univ. Press, 1952. 282
[5] W. Rudin, Principles of mathematical analysis, 2nd ed., McGraw-Hill Company, New York, 1964. [6] J. S´andor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [7] J. S´andor, On certain inequalities for means, J. Math. Anal. Appl., 189(1995), 602-606.
15
Some mean value theorems and consequences
1. Cauchy’s mean value theorem of differential calculus (for function of a single real variable) asserts that for two functions f, g : [a, b] → R which are continuous on [a, b], differentiable on (a, b), with g 0 (x) 6= 0 for all x ∈ (a, b) there exists at least a number ξ ∈ (a, b) with the property f (b) − f (a) f 0 (ξ) = 0 . g(b) − g(a) g (ξ)
(1)
This mean value theorem has remarkable applications in different branches of Mathematics and it is one of the powerful methods of Mathematical Analysis. For two notable applications in Number Theory, and the Theory of Inequalities, respectively, we quote the papers [2] and [3] of the second author. In a recent note [4] J. S´andor has proved the following mean value theorem: Let f, g : [a, b] → R differentiable functions satisfying g(a) − g(b) 6= (b − a)g 0 (a),
g(b) − g(a) 6= (b − a)g 0 (b)
g 0 (x) 6= g 0 (b) for all x ∈ (a, b).
(2) (3)
Then there exist ξ, η ∈ (a, b) such that
and
f (b) − f (a) − (b − a)f 0 (a) f 0 (ξ) − f 0 (a) = g(b) − g(a) − (b − a)g 0 (a) g 0 (ξ) − g 0 (a)
(4)
f (b) − f (a) − (b − a)f 0 (b) f 0 (η) − f 0 (b) = . g(b) − g(a) − (b − a)g 0 (b) g 0 (η) − g 0 (b)
(5)
The proof is based on Rolle’s theorem relating to the existence of certain θ with f 0 (θ) = 0 for Rolle functions f with f (a) = f (b). We will show that (4) and (5) follows by Cauchy’s mean value theorem, and in fact, a more general result is obtainable by this way. The aim of this note however is to give certain 283
new applications of Cauchy’s theorem, besides applications for the proof of (4) and (5). 2. First of all remark that conditions (3) imply relations (2), since the applications Ga , Gb : [a, b] → R, G(a) = g(x) − g(a) − (x − a)g 0 (a) and Gb (x) = g(x) − g(b) − (x − b)g 0 (b) respectively, satisfy the conditions of Rolle’s theorem, this implying the existence of certain c, d ∈ (a, b) with G0a (c) = 0, G0b (d) = 0 (if one admits conditions (2)). These are impossible by (3). Consider now the functions Fa , Ga : [a, b] → R defined by Fa (x) = f (x) − f (a) − (x − a)f 0 (a), Since
Ga (x) = g(x) − g(a) − (x − a)g 0 (a).
f (b) − f (a) − (b − a)f 0 (a) Fa (b) − Fa (a) = Ga (b) − Ga (a) g(b) − g(a) − (b − a)g 0 (a)
and
Fa0 (ξ) f 0 (ξ) − f 0 (a) , = G0a (ξ) g 0 (ξ) − g 0 (a)
from (1) we get relation (4). Similarly, by considering Fb (x) = f (x) − f (b) − (x − b)f 0 (b),
Gb (x) = g(x) − g(b) − (x − b)g 0 (b)
and remarking that Fb (b) − Fb (a) f (b) − f (a) − (b − a)f 0 (b) = , Gb (b) − Gb (a) g(b) − g(a) − (b − a)g 0 (b) by the same procedure we recapture equality (5). 3. In order to obtain a generalization of (4) and (5), define a Taylor polynomial function h : [a, b] → R which is n times differentiable, in point c, of order n. This polynomial will be (Tn,c h)(x) =
n X (x − c)k h(k) (c) k=0
k!
.
(6)
Let (Rn,c h)(x) = h(x) − (Tn,x h)(x)
(7)
the rest of order n in Taylor formula. Now we are able to formulate the following result: 284
Let us suppose that f, g : [a, b] → R have a derivative on [a, b], there exist the higher order derivatives f (k) , g (k) (2 ≤ k ≤ −1) in a neighbourhood of a, respectively b, and f (n) , g (n) in a, respectively b. If (8) g 0 (x) 6= (Tn−1,a g 0 )(b), (Tn−1,b g 0 )(a), ∀ x ∈ (a, b) g 0 (x) 6= g 0 (a), g 0 (b) then there exist ξ, η ∈ (a, b) such that
and
f (b) − (Tn,a f )(b) f 0 (ξ) − (Tn−1,a f 0 )(ξ) = 0 g(b) − (Tn,a g)(b) g (ξ) − (Tn−1,a g 0 )(ξ)
(9)
f 0 (η) − (Tn−1,b f 0 )(η) f (a) − (Tn,b f )(a) = 0 g(a) − (Tn,b g)(a) g (η) − (Tn−1,b g 0 )(η)
(10)
For the proof of this result, let us remark that one has (Tn,c h)(c) = h(c),
(Tn,c h)0 = (Tn−1,c h0 ),
(Tn,c h)(n) = h(n) (c),
and apply Cauchy’s mean value theorem to the functions (Rn,a f ), (Rn,a g) : [a, b] → R to obtain (10); and (Rn,b f ), (Rn,b g) : [a, b] → R to obtain (9) respectively. As a Corollary of this theorem one can derive: If f, g : [a, b] → R are n times differentiable on [a, b], with g (n+1) (x) 6= 0, ∀ x ∈ (a, b), then there exist θ1 , θ2 ∈ (a, b) such that f (b) − (Tn,a f )(b) f (n+1) (θ1 ) = (n+1) g(b) − (Tn,a g)(b) g (θ1 )
(11)
f (a) − (Tn,b f )(a) f (n+1) (θ2 ) = (n+1) . g(a) − (Tn,b g)(a) g (θ2 )
(12)
Relations (11), (12) are generalizations of a Corollary of (4) and (5) for n = 1 obtained in [4]. Of course, there are many particular applications for (11) and (12). 4. In what follows we shall obtain certain applications for Cauchy’s theorem, with nice consequences. Let F, G : [a, b] → R defined by F = f ◦ u + f ◦ v,
G=g◦u+g◦v
where f, g, u, v : [a, b] → R are Rolle functions satisfying the conditions: u(a) = a, u(b) = v(a) = c, c ∈ (a, b); 285
v(b) = b, u0 (x), v 0 (x) > 0, ∀ x ∈ (a, b).
(13)
Applying (1) and remarking that F (b) − F (a) = f (b) − f (a),
G(b) − G(a) = g(b) − g(a)
one gets the following result: Let f, g, u, v be Rolle functions on [a, b] with u(a) = a, u(b) = v(a) = c, c ∈ (a, b), v(b) = b and u0 (x), v 0 (x) > 0 and (g ◦ u)0 (x) + (g ◦ v)0 (x) 6= 0, ∀ x ∈ (a, b). Then there exists ξ ∈ (a, b) such that f (b) − f (a) u0 (ξ)f 0 (ξ1 ) + v 0 (ξ)f 0 (ξ2 ) = 0 g(b) − g(a) u (ξ)g 0 (ξ1 ) + v 0 (ξ)g 0 (ξ2 )
(14)
where ξ1 = u(ξ) ∈ (a, c), ξ2 = v(ξ) ∈ (c, b). For the particular case of u(x) =
x+a , 2
v(x) =
x+b 2
√ ax,
v(x) =
√ xb
respectively u(x) =
for 0 ≤ a < b one gets the corollaries: 1. Let f, g be Rolle functions on [a, b] with g 0 (x) + g 0 (y) 6= 0, ∀ x, y ∈ (a, b), µ Then there exist ξ1 ∈ such that
x−y =
b−a . 2
¶ µ ¶ a+b b−a a+b , b , ξ2 ∈ a, and ξ1 − ξ2 = , 2 2 2
f (b) − f (a) f 0 (ξ1 ) + f 0 (ξ2 ) = 0 . g(b) − g(a) g (ξ1 ) + g 0 (ξ2 )
(15)
2. Let f, g be Rolle functions on [a, b], 0 ≤ a < b with √ √ 0 ag (x) + bg 0 (y) 6= 0, ∀ x, y ∈ (a, b) r r √ √ x a ξ1 a with = . Then there exists ξ1 ∈ (a, ab), ξ2 ∈ ( ab, b), = , such y b ξ2 b that √ √ 0 f (b) − f (a) af (ξ1 ) + bf 0 (ξ2 ) √ = √ . (16) g(b) − g(a) ag 0 (ξ1 ) + bg 0 (ξ2 ) 286
For g(x) = x, x ∈ [a, b], and comparing (15) with Lagrange’s mean value theorem applied to the function f , one gets theµ ¶ ¶ µ a+b a+b Corollary. There exist θ ∈ (a, b), ξ1 ∈ , b , ξ2 ∈ a, , 2 2 b−a ξ1 − ξ2 = such that 2 f 0 (θ) =
f 0 (ξ1 ) + f 0 (ξ2 ) . 2
(17)
We note here that f 0 doesn’t need to be continuous (but clearly having Darboux’s property), in the case of continuous functions, relation (17) has been proved in [1]. Let now H, T : [a, b] → R be defined by H = f ◦ u − f ◦ v,
T = g ◦ u − g ◦ v,
where f, g, u, v are Rolle functions, satisfying the conditions (13) and (g ◦ u)0 (x) − (g ◦ v)0 (x) 6= 0 on (a, b). Then applying (1) to the functions H, T one easy obtain the following result: There exists ξ ∈ (a, b) such that u0 (ξ)f 0 (ξ1 ) − v 0 (ξ)f 0 (ξ2 ) f (b) − 2f (c) + f (a) = 0 g(b) − 2g(c) + g(a) u (ξ)g 0 (ξ1 ) − v 0 (ξ)g 0 (ξ2 )
(18)
where ξ1 = u(ξ) ∈ (a, c), ξ2 = v(ξ) ∈ (c, b). For the particular case of u(x) = x+a x+b , v(x) = and g 0 (x) injective (and f, g have a second order derivative 2 2 on (a, b)) one gets µ ¶ µ ¶ µ ¶ ξ+b ξ+a a+b 0 0 µ ¶ f (b) − 2f + f (a) f −f f 00 (θ) 2 2 2 µ ¶ µ ¶ µ ¶ = = 00 (19) a+b ξ+b ξ+a g (θ) g(b) − 2g + g(a) g0 − g0 2 2 2 where ξ ∈ (a, b), θ ∈ (a, b). For the particular case of g(x) = x2 , x ∈ [a, b], relation (19) yields that µ ¶ a+b (b − a)2 00 f (b) − 2f + f (a) = f (θ), θ ∈ (a, b) (20) 2 4 287
when f has a second order derivative on (a, b). To demonstrate the power of this simple relation, put f (x) = ln x, x ∈ [a, b], where 0 < a < b. Then simple calculations give the double inequality A (b − a)2 1 (b − a)2 1 · 2 < ln < · 2 8 b G 8 a
(21)
where
√ a+b , G = G(a, b) = ab 2 denote the arithmetic and geometric means of a and b, respectively. We note that for a = n, b = n + 1, n > 0 from (21) we obtain the logarithmic inequality A = A(a, b) =
1 2n + 1 1 < ln p < 2, 2 8(n + 1) 8n 2 n(n + 1)
n > 0.
(22)
5. Finally, we will consider certain mean value results and inequalities for Riemann integrals. For simplicity we apply (1) for g(x) = x, x ∈ [a, b]. Let Z u(x) Z v(x) F (x) = f (t)dt + f (t)dt a
b
where f : [a, b] → R is a continuous function and u, v : [a, b] → R are Rolle functions satisfying the conditions u(a) = a, u(b) = v(a) = c, c ∈ (a, b), v(b) = b. Since
Z F (b) − F (a) =
b
f (t)dt, a
one obtains that there exists ξ ∈ (a, b) such that Z b f (t)dt = (b − a)[u0 (ξ)f (u(ξ)) + v 0 (ξ)f (v(ξ))].
(23)
a
For u(x) =
a+x b+x , v(x) = one gets 2 2 · µ ¶ µ ¶¸ Z b b−a a+ξ b+ξ f (t)dt = f +f . 2 2 2 a
(24)
For particular functions f one can obtain interesting results from (24). A result of different type can be proved by considering the application Z x+b Z x+a 2 2 F (x) = f (t)dt − f (t)dt (25) a
a
288
where, as above, f is continuous on [a, b]. Since Z F (b) − F (a) =
Z
b a+b 2
f (t)dt −
a+b 2
f (t)dt,
a
we have · µ ¶ µ ¶¸ ξ+b ξ+a f (t)dt − f (t)dt = (b − a) f −f , ξ ∈ (a, b). a+b 2 2 a 2 (26) a+ξ Let us suppose that f is a convex function on [a, b]. Let ξ1 = , 2 µ µ ¶¶ a+b a+b b+ξ . Let A(a, f (a)), X(ξ1 , f (ξ1 )), M ,f , Y (ξ2 , f (ξ2 )), ξ2 = 2 2 2 B(b, f (b)). Since f is a convex function, it is well known that (see e.g. [5]) Z
b
Z
a+b 2
slopeAX ≤ slopeAM ≤ slopeXM ≤ slopeXY ≤ slopeY M ≤ slopeM B ≤ slopeY B thus in particular slopeAX ≤ slopeY M
(27)
and slopeXM ≤ slopeY B i.e.
µ
a+b f (ξ2 ) − f f (ξ1 − f (a) 2 ≤ a+b ξ1 − a ξ2 − 2
and
µ f
¶
¶ a+b − f (ξ1 ) f (b) − f (ξ2 ) 2 . ≤ a+b b − ξ2 − ξ1 2
a+b a+b and − ξ1 = b − ξ2 we get that 2 2 µ ¶ µ ¶ a+b a+b f − f (a) ≤ f (ξ2 ) − f (ξ1 ) ≤ f (b) − f . 2 2
Since ξ1 − a = ξ2 −
289
(28)
By comparing (26) and (28), we get the following unusual double inequality for a convex (continuous) function f : [a, b] → R: ¶ ¸ Z b · µ Z a+b 2 a+b − f (a) ≤ f (t)dt (b − a) f f (t)dt − a+b 2 a 2
· µ ¶¸ a+b ≤ (b − a) f (b) − f . 2
(29)
Clearly, when f is strictly convex, one has strict inequalities in (29). Remark that (29) is an improvement, involving integrals, for the classical inequality for convex functions µ ¶ a+b 2f ≤ f (a) + f (b). (30) 2
References [1] J. S´andor, Problem T36 (Hungarian), Mat. Lapok, Cluj, XCVII, 1992, no.6, 239. [2] J. S´andor, A note on the functions σk (n) and ϕk (n), Studia Univ. Babe¸sBolyai, Mathematica, XXXV, 2, 1990, 3-6. [3] J. S´andor, On an inequality of Alzer, J. Math. Anal. Appl., 192(1995), 1034-1035. [4] J. S´andor, On a mean value theorem, Octogon Math. Mag., 3(1995), no.2, 47-49. [5] A.W. Roberts, D.E. Varberg, Convex functions, Academic Press, 1973.
16
The second mean value theorem of integral calculus
1. The first mean value theorem for Riemann integrals can be stated as follows: Theorem 1. If f, g are integrable on [a, b], g having a constant sign, then there exist η ∈ [m, M ] (where m = inf f (x), M = sup f (x)) such that Z
Z
b
f (x)g(x)dx = η a
g(x)dx. a
290
b
(1)
If the function f is continuous, then there exists ξ ∈ [a, b] such that η = f (ξ). The second mean value theorem for integrals is less known, and as we shall see, it appears in various places under various conditions and different names. Theorem 2. If f, g are integrable on [a, b], and f is monotone, then there exists ξ ∈ [a, b] such that Z
Z
b
a
Z
ξ
f (x)g(x)dx = f (a)
b
g(x)dx + f (b) a
g(x)dx.
(2)
ξ
It will be sufficient to consider only the following theorem, known also as ”Bonnet’s formula” (P.O. Bonnet (1819-1892) French mathematician). Theorem 3. Let f, g be defined on [a, b], with integrable g, and f a positive, monotone decreasing function. Then there exists ξ ∈ [a, b] such that Z
Z
b
ξ
f (x)g(x)dx = f (a) a
g(x)dx.
(3)
a
Indeed, if (3) is true, let us apply it for the function F defined by F (x) = f (x) − f (b), f being decreasing function. One has F (x) ≥ 0, F decreasing, so by (3) we get Z b Z ξ F (x)g(x)dx = F (a) g(x)dx, a
i.e.
Z
a
Z
b
a
Z
b
f (x)g(x)dx −
f (b)g(x)dx = (f (a) − f (b)) a
Thus Z b
ξ
g(x)dx. a
Z
·Z
ξ
f (x)g(x)dx = f (a)
g(x)dx + f (b)
a
g(x)dx −
a
Z = f (a)
Z
b
a
Z
ξ
g(x)f x + f (b) a
ξ
¸ g(x)dx
a
b
g(x)dx, ξ
which was to be proved. Remark also that from the above it results also that the property is true also for f monotone increasing function, since we can apply the proved result for the function −f , which is decreasing. Therefore from (3) it results relation (2). 291
Before the demonstration of Bonnet’s formula, we will indicate some works where one can find particular cases or other names. In [1], Theorem 2 is proved for continuous f, g and increasing f ; and on page 341 it is stated that it holds true also for monotone f , and g as an integrable derivative. (Attention: a function which is a derivative, is not necessarily integrable!). In [2], p.337, the Theorem appears as (2), in which f (a) and f (b) are replaced with f (a + 0), resp. f (b − 0), with f, g bounded and integrable, f monotone. In [3], p.239, there is an incomplete proof. In [4], along with many interesting applications of the mean value theorems, appears also Bonnet’s formula, without a proof. In [7] relation (2) is called as ”Weierstrass formula”, while in [13] it is attributed to Du Bois-Reymond. In [8], the Bonnet formula is called as the ”second mean value theorem” with increasing f and integrable g. The proof given there is quite complicated. Finally, we mention [9], where the result (2) is proved for f, g continuous functions, f having a continuous derivative, and a constant sign. The book [6] (p.155) contains some interesting applications. See also [4], [5], [6], [11]-[14]. 2. The proof of Bonnet’s formula (Relation (3)) If f (a) = 0, then from 0 ≤ f (x) ≤ f (a) we get f (x) = 0, thus any ξ ∈ [a, b] is acceptable. Let us suppose thus f (a) > 0, and consider a division ∆ = {x0 , . . . , xn+1 } of [a, b] with x0 = a, xn+1 = b. Put S=
n X (xk+1 − xk )f (xk )g(xk ),
S0 =
n X
(xk+1 − xk )f (xk )λk ,
k=0
k=0
where λk ∈ [mk , Mk ], with mk , Mk being the extreme values of g on [xk , xk+1 ]. Suppose that λk is exactly the intermediate point which appears in the first mean value theorem (relation (1)) of g on [xk , xk+1 ]: Z xk+1 g(x)dx = (xk+1 − xk )λk . xk
Put
Z
x
G(x) =
g(t)dt. a
The function G being continuous, attains its margins m1 and M1 . From the equality G(xk+1 ) − G(xk ) = (xk+1 − xk )λk , we get 0
S =
n X
f (xk )[g(xk+1 ) − G(xk )] = −f (a)G(a) + G(x1 )[f (a) − f (x1 )] + · · · +
k=0
292
+G(xn )[f (xn−1 ) − f (xn )] + G(b)f (xn ). By G(a) = 0 and f decreasing, we get S 0 ≥ m1 [f (a) − f (x1 ) + · · · + f (xn−1 − f (xn ) + f (xn )] = m1 f (a), S 0 ≤ M1 [f (a) − f (x1 ) + · · · + f (xn−1 − f (xn ) + f (xn )] = M1 f (a) Let
Z
(4)
b
A=
f (x)g(x)dx. a
In what follows we shall prove that m1 f (a) ≤ A ≤ M1 f (a).
(5)
Indeed, if we would have e.g. A > M1 f (a), from the definition of Riemann’s integral and Darboux’s theorem (applied for the function g) we could write: ∀ ε > 0, ∃ δ > 0 such that k∆k < δ ⇒ |A − S| < ε and D =
n X
(xk+1 − xk )(Mk − mk ) < ε/f (a).
k=0
On the other hand, n X (xk+1 − xk )f (xk )|g(xk ) − λk | ≤ Df (a) < f (a)ε/f (a) = ε. |S − S | ≤ 0
k=0
Now, |A − S| < ε, |S − S 0 | < ε and A > M1 f (a), so 2 − ε < S 0 − A < 2ε. In other words S 0 → A > M1 f (a), thus for sufficiently large n we have S 0 > M1 f (a), a contradiction to (4). Let now α = A/f (a). On base of (5) we have m1 ≤ α ≤ M1 , and G being continuous, by the intermediate value property (i.e. ”Darboux property”) there is a ξ ∈ [a, b] with G(ξ) = α. This implies the equality Z b Z ξ f (x)g(x)dx = f (a) g(x)dx. a
a
See also [10] (where in Part II a proof based on Abel’s inequality is given). 3. Now, applying (2), and using a method by E.L. Stark [11], a rather elementary proof of Euler’s sum ∞ X 1 π2 = k2 6 k=1
293
(6)
will be given. The following identity is well known: n
Dn (x) =
1 X sin(2n + 1)x/2 + cos kx = 2 2 sin x/2
(x ∈ R, n ∈ N).
(7)
k=1
Set
Z Mn =
π
0
tDn (t)dt.
By employing the polynomial representation from (7), by partial integration we get ( ) m π2 X 1 M2m−1 = 2 − (m ∈ N). (8) 8 (2k − 1)2 k=1
On the other hand, by using the closed representation of (7) and applying x x the second mean value theorem (2) for f (x) = sin , x ∈ (0, π); f (0) = 1, 2 2 ³ π x´ f (π) = and g(x) = sin (4m − 1) we get 2 2 ¶ Z πµ x x/2 sin(4m − 1) dx M2m−1 = sin x/2 2 0 ½ ¾ ³π ´ ξ 1 =2 1+ − 1 cos(4m − 1) 2 2 4m − 1 µ ¶ 1 =O as m → ∞ (0 ≤ ξ ≤ π). (9) m By combining (8) and (9), we obtain ∞ X k=1
1 π2 , = (2k − 1)2 8
which immediately implies relation (6). 4. We note that the second mean value theorem of integral calculus (2) has many other applications in mathematics. For example in [5] it is given an application to the proof of Taylor’s formula. It is also important in the theory of trigonometric series (see e.g. [14]), or Laplace transforms (see e.g. [12]).
References [1] M. Bal´azs, J. Kolumb´an, Mathematical analysis (Hungarian), Ed. Dacia, Cluj, 1978. 294
[2] G. Chilov, Analyse math´ematique, Fonctions d’une variable, Ed. Mir, Moscou, 1978. [3] M. Craiu, M.N. Ro¸sculet¸, Problems of mathematical analysis (Romanian), Ed. Did. Ped., Bucure¸sti, 1976. [4] C.V. Cr˘aciun, Mean value theorems of mathematical analysis (Romanian), Univ. Bucure¸sti, 1986. [5] R.R. Goldberg, Methods of real analysis, Wiley, 1964. [6] N.M. G¨ unther, R.O. Cuzmin, Problems of higher mathematics (Romanian), vol.II, Ed. Tehnic˘ a, Bucure¸sti, 1950, 155-157. [7] C. Meghea, The basics of mathematical analysis (Romanian), Ed. S¸tiint¸. Enc., Bucure¸sti, 1977. [8] S.M. Nikolsky, A course of mathematical analysis, vol.I, Mir Publ., Moscou, 1977. [9] C. Popa, V. Hiri¸s, M. Megan, Introduction to mathematical analysis via exercises and problems (Romanian), Ed. Facla, 1976. [10] J. S´andor, On the second mean value theorem for integrals, Lucr. Semin. Didactica Mat., 5(1989), 275-280; II ibid., 2005, to appear. [11] E.L. Stark, Application of a mean value theorem for integrals to series summation, Amer. Math. Monthly, 1978, 481-483. [12] D. Widder, The Laplace transform, Princeton, 1941. [13] E.T. Whittaker, G.N. Watson, A course of modern analysis, Cambridge, 1969. [14] A. Zygmund, Trigonometric series, 2nd ed., vol.I, (p.58), Cambridge, 1959.
295
296
Chapter 7
Functional equations and inequalities ”... If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy.” (Alfred R´enyi)
”... The authors have chosen to emphasize applications, though not at the expense of theory.” (J. Acz´el and J. Dhombres)
297
1
The Bohr-Mollerup theorem
1. In 1922 H. Bohr and J. Mollerup [3] discovered the following surprising fact: Theorem 1. The only log-convex solution f : (0, ∞) → (0, ∞) to the functional equation f (x + 1) = xf (x), x > 0, satisfying f (1) = 1 is f (x) = Γ(x), the Euler Gamma function. Proof. This proof is based on ideas by E. Artin [1]. It is easy to see that Z ∞ Γ(x) = e−t tx−1 dt, x > 0 0
satisfies these conditions. Clearly Γ(1) = 1. By partial integration it follows that Γ(x + 1) = xΓ(x). Finally, by H¨older’s inequality one can prove that Γ is log-convex. 1 1 Indeed, let p > 1 and suppose that + = 1. Then p q ¶ Z ∞ µ Z ∞³ ´ 1 ³ y 1 t´ x y x y − t x −p − − −t p + q −1 + = e t Γ dt = e ptp e q q e q dt p q 0 0 µZ
∞
≤
−t x−1
e t
¶1/p µZ dt
0
∞
−t y−1
e t
¶1/q dt
0
by the H¨older inequality ¯Z ¯ ¯ ¯
0
∞
¯ µZ ¯ f (t)g(t)dt¯¯ ≤
∞
p
¶1/p µZ
|f (t)|
0
∞
q
¶1/q
|g(t)|
,
0
1 =λ∈ p (0, 1). Then Γ(λx + (1 − λ)y) ≤ (Γ(x))λ (Γ(y))1−λ , which shows that log Γ is a convex function on (0, ∞). We now prove that the given conditions uniquely determines the function f . Clearly, it is sufficient to consider x ∈ (0, 1]. Since f (1) = 1, we get f (2) = 1, f (3) = 1 · 2, . . . , f (n) = (n − 1)! (by mathematical induction) for any positive integer n ≥ 1. Let now n ≥ 2 be any integer and 0 < x ≤ 1. Since log f is convex, one has that x
applied to the functions f (t) = e−t/p t p
− p1
y
, g(t) = e−t/q t q
− 1q
. Let now
log f (−1 + n) − log f (n) log f (x + n) − log f (n) ≤ (−1 + n) − n (x + n) − n 298
≤ i.e.
log f (1 + n) − log f (n) (1 + n) − n
log f (x + n) − log(n − 1)! ≤ log n. x Writing these inequalities without logarithms, we get log(n − 1) ≤
(n − 1)x (n − 1)! ≤ f (x + n) ≤ nx (n − 1)!
(1)
Since f (x + 1) = xf (x), we have f (x + 2) = x(x + 1)f (x), . . . , f (x + n) = x(x + 1) . . . (x + n − 1)f (x), (1) can be written also as (n − 1)x (n − 1)! nx (n − 1)! ≤ f (x) ≤ . x(x + 1) . . . (x + n − 1) x(x + 1) . . . (x + n − 1)
(2)
By writing n + 1 in place of n in the left side of (2), we can write: nx n! nx n! x+n ≤ f (x) ≤ · , x(x + 1) . . . (x + n) x(x + 1) . . . (x + n) n or f (x)
n nx n! ≤ ≤ f (x). x+n x(x + 1) . . . (x + n)
(3)
(4)
Relation (4) shows that f (x) is uniquely determined by nx n! n→∞ x(x + 1) . . . (x + n)
f (x) = lim
(5)
This finishes the proof of the theorem, which shows also that one has the equality nx n! Γ(x) = lim (6) n→∞ x(x + 1) . . . (x + n) Remark. Equality (6) can be proved also by e.g. the Lebesgue (domination) criterion: Let (fn ) be a measurable sequence of functions on (a, b) and a.e. fn −→ f . Suppose there exists function g such that |fn (u)| ≤ g(u) for all u ∈ (a, b). Then f is measurable, and one has Z b Z b lim fn (u)du = f (u)du. n→∞ a
Z Let an (x) =
1
a
(1 − t)n tx−1 dt. By partial integration we get
0
n an (x) = x
Z
1
(1 − t)n−1 tx ,
0
299
and repeating the argument n times, we get an (x) = By letting t =
u , we obtain n 1 an (x) = x n
i.e.
n! . x(x + 1) . . . (x + n)
Z
n³
1− 0
n!xn = x(x + 1) . . . (x + n) Put now
³ Since 1 −
Z
u ´n x−1 u du, n
n³
1− 0
u ´n x−1 u du n
(∗)
( ³ u ´n x−1 u , 0
u ´n n
u ∈ (a, b) = (0, +∞).
< e−u , by the above criterion by Lebesgue Z
lim
n³
n→∞ 0
1−
u ´n x−1 u du = n
Z
∞
e−u ux−1 du,
0
and by (∗) the result follows. 2. Let 0 < q < 1. In 1869 J. Thomae [5], and independently in 1904 F.H. Jackson [4] introduced the so-called q-Gamma function: Γq : (0, ∞) → (0, ∞) by (1 − q)(1 − q 2 ) . . . (1 − q n+1 ) , n→∞ (1 − q x )(1 − q x+1 ) . . . (1 − q x+n )
Γq (x) = (1 − q)1−x lim
x > 0.
(7)
In 1978 R. Askey [2] proved the following: Theorem 2. The only log-convex solution f : (0, ∞) → (0, ∞) to the functional equation f (x + 1) =
1 − qx f (x), 1−q
x>0
satisfying f (1) = 1 is f (x) = Γq (x). In 1997 R. Webster [6] obtained a common generalization to the above theorems. In what follows we will obtain a simplified proof. 300
Theorem 3. Let g : (0, ∞) → (0, ∞) be a log-concave function which has the property that for all w > 0 one has g(x + w)/g(x) → 1 as x → ∞. Then the only log-convex solution to the functional equation f (x + 1) = g(x)f (x),
x > 0,
satisfying f (1) = 1 is g(n) . . . g(1)(g(n))x . n→∞ g(n + x) . . . g(x)
f (x) = lim
Proof. The uniqueness of f , i.e. equality (8), follows on the same lines as in the proof of Theorem 1, by using the log-convexity of f and the property g(x + w)/g(x) → 1 as x → ∞ for all w > 0. We must prove also the existence of such a solution. For each n ∈ N define a function fn : (0, ∞) → (0, ∞) by fn (x) = Then
g(n) . . . g(1)(g(n))x , g(n + x) . . . g(x)
x > 0.
(9)
(g(n + 1))x+1 f (x) = fn (x) n+1 g(n + x + 1)(g(n))x fn+1 (x + 1) =
(10) g(n) g(x)fn (x). g(n + x + 1)
Let 0 < x ≤ 1. By the log-concavity of g one can write (g(n + 1))x+1 ≥ (g(n))x g(n + x + 1), which by the first relation of (10) shows that fn+1 (x) ≥ fn (x), i.e. the sequence (fn (x))n is decreasing. Writing n times the log-concavity of g, we get: g(x + 1) ≥ (g(1))1−x (g(2))x , . . . , g(x + n) ≥ (g(n))1−x (g(n + 1))x , and their product, together with (9) shows that µ ¶x (g(1))x g(n) fn (x) ≤ g(x) g(n + 1) Now, it can be shown that g(n) ≤ g(n + 1), so (11) implies µ ¶ g(1) x fn (x) ≤ , g(x) 301
(11)
i.e. (fn (x))n is bounded above. Thus (fn (x))n being bounded, and monotone, converges to a function fe : (0, 1] → (0, ∞) by g(n) . . . g(1)(g(n))x fe(x) = lim for 0 < x ≤ 1. n→∞ g(n + x) . . . g(x)
(12)
By (10) it follows that (12) may be extended to all x > 0 defining a function f : (0, ∞) → (0, ∞) by g(n) . . . g(1)(g(n))x , n→∞ g(n + x) . . . g(x)
f (x) = lim
x > 0.
We have to prove only that g(n) ≤ g(n + 1), or more generally that g is an increasing function. Indeed, let 0 < a < b. Since log g is concave on (0, ∞), log g(a + x) − log g(a) ≥ log g(b + x) − log g(b),
x > 0,
so
g(a) g(a + x) ≤ . g(b) g(b + x) g(a + x) → 1, thus implying g(a)/g(b) ≤ Now, by assumption, as x → ∞, g(b + x) 1. Since a < b, this proves that g is an increasing function. 1 − qx Remarks. g(x) = x and g(x) = (0 < q < 1) satisfy the conditions 1−q of Theorem 3.
References [1] E. Artin, Einf¨ uhrung in die Theorie der Gammafunktion, Teubner, Leipzig, 1931. [2] R. Askey, The q-gamma and q-beta functions, Appl. Anal. 8(1978), 125141. [3] H. Bohr and J. Mollerup, Laerebog i matematisk Analyse, vol.III, 149164. [4] F.H. Jackson, A generalization of the functions Γ(n) and xn , Proc. Roy. Soc. London, 74(1904), 64-72. [5] J. Thomae, Beitr¨ age zur Theorie der durch der Heinesche Reihe..., J. Reine Angew. Math. 70(1869), 258-281. [6] R. Webster, Log-convex solutions to the functional equation f (x + 1) = g(x)f (x): Γ-type functions, J. Math. Anal. Appl. 209(1997), 605-623. 302
2
On certain functional equations containing more unknown functions
1. Problem C:564 proposed by V. B˘andil˘ a [2] asks for the determination of three continuous functions f, g, h : R → (0, ∞) such that µ ¶ x+y f (x)g(y) = h , x, y ∈ R. (1) 2 This problem can be reduced, as we shall see, to the solution of Cauchy’s functional equation F (x + y) = F (x) + F (y),
x, y ∈ R.
(2)
On the other hand, we will show that, equation (1) is a particular case of a ”Popoviciu type equation”, studied and generalized by Stamate [3], Iani´c and Peˇcari´c [4], and Acu [5]. See also [6]. In the second part, we shall study a functional equation suggested by the equation by Popoviciu. As a particular case, we will obtain the solutions to a problem by Vlada [7]. Then we will suggest some generalizations which are analogous to the ones studied by Stamate. 2. All continuous solutions of equation (2) are of the form F (x) = ax,
x ∈ R,
where a is fixed. Indeed, by letting successively y = x, y = 2x, . . . we get F (2x) = 2F (x), 1 F (3x) = 3F (x), . . . , F (nx) = nF (x) (by mathematical induction). For x = n µ ¶ µ ¶ ³m´ F (1) 1 m 1 1 = = a. Similarly, F = a. From the we get F = mF n n n n n n observation F (0) = 0 one gets F (−x) = −F (x), so we get finally F (r) = ar for all rational numbers r. Let now x ∈ R \ Q, and consider a sequence (xn ) of rational numbers with xn → x as n → ∞. By F (xn ) = axn , and the continuity of F we can deduce F (x) = ax (where in fact, a = F (1)). 3. We now consider equation (1). Letting y = x, it results h(x) = f (x)g(x). So h(y) = f (y)g(y), and by substitution, µ ¶ x+y h(x)h(y) = h2 . 2 The function h being strictly positive, we can define H by h(x) = eH(x) . This yields the equation (known also as ”Jensen’s functional equation”) µ ¶ x+y H(x) + H(y) = 2H . 2 303
³x´ ³x´ For y = 0 we obtain H(x) = 2H − H(0), thus 2H − H(0) + 2 2 µ ¶ ³y ´ ³x´ x+y − H(0) = 2H − H(0). 2H . Let us introduce h1 (x) = H 2 2 2 Then we easily arrive at Cauchy’s functional equation h1 (x + y) = h1 (x) + h1 (y). By 2., the most general continuous solutions to this equation are h1 (x) = ax, thus H(x) = 2ax + b (where b = H(0)). Therefore h(x) + e2ax+b = cekx (where c > 0, k ∈ R). By applying (1) to y = 0, we get finally: f (x) = h2
³x´ 2
/g(0) =
c2 kx e , g(0)
g(y) =
c2 ky e , f (0)
where c2 = f (0)g(0). 4. A more general equation like (1) is h(x + y) = F (x)G(y),
(3)
where h, F, G : R → R are continuous functions. Indeed, f (2x) = F (x), g(2y) = G(y) imply to the above equation, with the assumption that F (x), G(x) ∈ (0, ∞). This equation is a particular case of a ”Popoviciu type equation” (see [5]): h(x + y + z) = F (x)G(y, z),
(4)
where G is a continuous function of two variables. Let us remark that, for z = 0, and with the notation G(y, 0) = G(y), from (4) we arrive at (3). It is possible to prove that the solutions of (4) are given by: h(x) = abecx , F (x) = aecx , G(y, z) = bec(y+z) where a, b, c are certain constants. From this we are able to reobtain the solutions obtained in 3. 5. Let us consider in what follows the functional equation f (x + y + z) = g(x) + h(y, z)
(5)
suggested by the equation (4), of Popoviciu type. We will obtain all continuous solutions f, g : R → R, h : R2 → R. Putting x = 0, we get f (y + z) = g(0) + h(y, z); thus f (x + y + z) = g(x) + f (y + z) − g(0). 304
Applying (5) for y = z = 0, it results f (x) = g(x) + h(0, 0) = g(x) + f (0) − g(0). Thus, by substitution, f (x + y + z) = f (x) − f (0) + g(0) + f (y + z) − g(0) = f (x) + f (y + z) − f (0). For x = u, y + z = v and the notation f (x) − f (0) = f1 (x), we get f1 (u + v) = f1 (u) + f1 (v), i.e. the Cauchy functional equation (2). Since f1 (x) = Ax, we get f (x) = f (0) + Ax = Ax + B. The function g can be found by g(x) = Ax + B − (f (0) − g(0)) = Ax + C; and similarly h by h(y, z) = f (y + z) − g(0) = A(y + z) + D (A, B, C, D ∈ R). Remark. This functional equation contains the special case h(y, z) = h(y) + k(z) (see [7]). From A(y + z) + D = h(y) + k(z) with z = 0 it immediately results that h(y) = Ay + A1 , k(z) = Ax + A2 (for y = 0). 6. From the generalizations of the functional equation (5) we may consider the followings: 1) f (x1 + x2 + · · · + xn ) = f1 (x1 ) + f2 (x2 , x3 , . . . , xn ), where f1 : R → R, f : R → R, f2 : Rn−1 → R are continuous functions; 2) f (x1 +x2 +· · ·+xn ) = f1 (x1 )+f2 (x2 )+· · ·+fn−2 (xn−2 )+fn−1 (xn−1 xn ), where f, f1 , . . . , fn−2 : R → R, fn−1 : R2 → R are continuous. These equations can be studied with the methods shown before. For example, for eq. 1) we proceed as follows: for x1 = 0 we get f (x2 + · · · + xn ) = f1 (0) + f2 (x2 , x3 , . . . , xn ), thus f (x1 + · · · + xn ) = f1 (x1 ) + f (x2 + · · · + xn ) − f1 (0). For x2 = · · · = xn = 0 we get f1 (x1 ) = f (x1 ) − f (0) − f1 (0), thus by substitution, f (x1 + · · · + xn ) = f (x1 ) + f (x2 + · · · + xn ) − k (k = constant). Put F (x) = f (x) − k, x1 = u, x2 + · · · + xn = v. Then we get F (u + v) = F (u) + F (v), etc. 7. As a final remark, we notice that the above equations may be further generalized, obtaining the so-called ”Pexider type” equations (see [1]). 305
References [1] J. Acz´el, Lectures on functional equations and their applications, Academic Press, 1966. [2] V. B˘andil˘a, Problem C:564, Gazeta Mat. Seria B, 2/1986. [3] I. Stamate, Equations fonctionnelles contenant plusieurs fonctions inconnues, Univ. Beograd Publ. Elektr. Fak. Ser. Mat. Fiz., nr.354-356, 1971. [4] R. Iani´c, J.E. Peˇcari´c, On some functional equations for functions of several variables, Ibid., nr.716-734, 1981. [5] D. Acu, On a functional equation of Popoviciu type (Romanian), Gaz. Mat. Seria A, 3-4/1984. [6] J. S´andor, On a functional equation (Romanian), Lucr. Sem. Did. Mat., 6(1990), 293-296. [7] M. Vlada, Problem 13259, Gaz. Mat. 6/1973.
3
Locally integrable solutions of Cauchy’s functional equation
A function f : R → R is called locally integrable, if it is integrable over every finite interval of R. Theorem. The locally integrable solutions of the equation f (x + y) = f (x) + f (y),
x, y ∈ R
(1)
are f (x) = cx, c = constant. Proof. On the basis of (1), and the hypothesis of local integrability one easily verifies the following identity: Z x+y Z x Z y yf (x) = f (t)dt − f (t)dt − f (t)dt. (2) 0
0
0
Relation (2) implies at once xf (y) = yf (x),
x, y ∈ R.
Letting x = x0 6= 0, from (3) we get f (y) = cy, where c = finishes the proof of the theorem. 306
(3) f (x0 ) ; and this x0
Remark. W. Sierpinski [1] proved that the above result remains valid under the weaker hypothesis that f (x) is measurable. The given simple proof of the above theorem is due to H.N. Shapiro [2].
References [1] W. Sierpinski, Sur l’´equation fonctionnelle f (x + y) = f (x) + f (y), Fundamenta Math., 1(1920), 116-122. [2] H.N. Shapiro, A micronote on a functional equation, American Math. Monthly, 80(1973), no.9, 1041.
4
Generalizations of Haruki’s and Cior˘ anescu’s functional equations 1. In 1979 S. Haruki [5] proved that the functional equation f (x) − g(y) = h(x + y), x−y
x 6= y,
(1)
where f, g, h : R → R, has f (x) = g(x) = ax2 + bx + c and h(x) = ax + b as its only solutions, without any regularity condition. His proof went by reduction to the well known Jensen functional equation (see e.g. [1]). In 1985 J. Acz´el [2] used an interesting elementary argument to prove the above considered result. In what follows we shall apply Acz´el’s method in order to solve some functional equations, one of which generalizes Haruki’s equation, while the others extend certain results by N. Cior˘anescu [3] (see also [5]). 2. Let g : R → ∞R be a given additive function, and let us consider the equation f (x) − F (y) = h(x + y), g(x) 6= g(y), (2) g(x) − g(y) where f, F, h : R → R are the unknown functions. Theorem 1. Without any regularity conditions, all solutions of equation (2) are given by ³x´ f (x) = F (x) = ag 2 (x) + bg(x) + c, h(x) = ag + b, 2 where a, A, b, c are certain real constants. Proof. By interchanging x and y we obtain easily f = F , consequently the equation becomes f (x) − f (y) = (g(x) − g(y))h(x + y). 307
(3)
If f satisfies this equation, so does f + b (b = constant), so we may suppose that f (0) = 0. Put y = 0 in (3) in order to get f (x) = g(x)h(x)
(4)
(because of g(0) = 0, g being an additive function, i.e. g(x + y) = g(x) + g(y) for all x, y ∈ R). This transforms (3) into g(x)h(x) − g(y)h(y) = [g(x) − g(y)]h(x + y).
(5)
Again, we may suppose h(0) = 0. Therefore, putting x = −y into (5), we get g(−y)h(−y) = g(y)h(y).
(6)
We take this into consideration when replacing y by −y in (5) getting [g(x) − g(y)]h(x + y) = [g(x) − g(−y)]g(x − y). Substituting here x + y = u, x − y = v, we have: µ µ ¶ µ ¶¶ µ µ ¶ µ ¶¶ u+v u−v u+v v−u g −g h(u) = g −g h(v). 2 2 2 2
(7)
(8)
The function g being additive, one has: ¶ µ ¶ µ ³v ´ u−v u+v −g = 2g g , 2 2 2 and similarly
µ g
u+v 2
so one arrives to g
¶
³v ´
µ −g
v−u 2
= 2g
³u´
³u´
2
,
h(v). (9) ³u´ Let v = v0 in (9). Thus we get h(u) = ag . If we do not assume 2 ³u´ + b. By (4) this gives h(0) = 0, we have in general h(u) = ag 2 ³ ³x´ ´ f (x) = g(x) ag + b = Ag 2 (x) + bg(x); 2 2
h(u) = g
¶
2
and if we do not assume f (0) = 0, we have: f (x) = Ag 2 (x) + bg(x) + c. 308
3. N. Cior˘anescu [3] considered the following equations: f (x) − f (y) 1 = [f 0 (x) + f 0 (y)]; x−y 2 µ
f (x) − f (y) x−y
¶2
= f 0 (x)f 0 (y),
(10)
(11)
where f is differentiable. Equations (10) and (11) were introduced in connection with some properties of conics. In what follows we will study the more general equations f (v) − f (u) = h(u) + h(v), u 6= v, v−u ¶ µ f (v) − f (u) 2 = h(u)h(v), u = 6 v, v−u 1 f (v) − f (u) = , v−u h(u) + h(v) Equation (14) arises from
u 6= v.
(12) (13) (14)
f (v) − f (u) 2f 0 (u)f 0 (v) = 0 , not studied by v−u f (u) + f 0 (v)
Cior˘anescu. In what follows, we shall apply the method to equations (12) and (14). Theorem 2. Let f, h : R → R satisfy equation (12). If h ≡ 0, then f = constant. If h 6≡ 0, then h(u) = au + b, and satisfies f (u) = au2 + b1 u + c, where a, b1 , c are constants. Proof. Assume as above f (0) = 0. Put u = 0 in order to deduce f (v) = v[h(v) + h(0)],
v 6= 0.
(15)
Therefore, we have vh(v) − uh(u) + vh(0) − uh(0) = (h(u) + h(v))(v − u).
(16)
This is equivalent to vh(u) − uh(v) = (v − u)h(0).
(17)
Let v = v0 (6= 0) in (17). This implies h(u) =
1 uh(v0 ) − h(0) [uh(v0 ) + vh(0) − uh(0)] = + h(0). v0 v0 309
(18)
We have to distinguish here two cases: Case 1. h ≡ 0. Then (15) gives f ≡ 0. if we do not assume f (0) = 0, we get f = constant. Case 2. h 6≡ 0, i.e. there exists v0 ∈ R such that h(v0 ) 6= 0. Then obviously v0 6= 0 (see (18)), so we can write: h(u) = au + b, implying f (u) = au2 + b1 u. Finally, it results f (u) = au2 + b1 u + c. Remark. As an application we can find immediately the solutions of the Cior˘anescu equation (10): One finds f (x) = Ax or f (x) = Ax2 + Bx + C. Theorem 3. Suppose that f, h : R → R satisfy the equation (14), where h(u) + h(v) 6= 0 for u 6= v. If h = constant, then f (u) = Au + B (A, B constants). If h 6= constant, then h2 (u) = au + b (a 6= 0), and u + f (0). f (u) = A + h(u) Proof. Assuming f (0) = 0, and letting u = 0, we get 1 f (v) = , v h(0) + h(v) so that we obtain µ
u v − a + h(v) a + h(u)
v 6= 0
(19)
(h(u) + h(v)) = v − u.
(20)
¶
Here h(u) 6= −a, with a = h(0) for all u ∈ R. After some elementary operations, (20) becomes equivalent to vh2 (u) − uh2 (v) = a2 v − a2 u. Letting v = v0 in (21), where v0 6= 0, we obtain: · 2 ¸ h (v) − h2 (0) + h2 (0). h2 (u) = u v0 Case 1. h(u) = h(0) (since h(u) 6= −h(0) by assumption). This gives f (u) =
u u + f (0) = , a + h(u) 2a
and if we do not assume f (0) = 0, then f (u) =
u + f (0). 2a 310
(21)
(22)
Case 2. h(u) 6= h(0) for some u 6= 0, i.e. h(u0 ) 6= h(0) for some u0 6= 0. Then we can easily deduce h2 (u) = Au + B (A 6= 0), B = a2 , and f (u) =
u + f (0), a + h(u)
where in fact A = (h2 (u0 ) − h2 (0))/u0 .
References [1] J. Acz´el, Lectures in functional equations and their applications, Academic Press, 1966. [2] J. Acz´el, A mean value property of the derivative of quadratic polynomials - without mean values and derivatives, Math. Mag. 586(1985), 42-45. [3] N. Cior˘anescu, Deux ´equations fonctionnelles et quelques propri´et´es de ´ certaines coniques, Bull. Math. Phys. Pures Appl. Ecole Polyt. Bucarest, 9(1937-1938), 52-53. [4] S. Haruki, A property of quadratic polynomials, Amer. Math. Monthly, 86(1979), 577-579. [5] J. S´andor, On certain functional equations, Itinerant Sem. Funct. Eq. Approx. Convexity 1988, Cluj, 285-288.
5
The equation à nfunctional ! n X X f xk = (f (xk ))k k=1
k=1
In this note we obtain a solution of OQ.32 (Octogon Math. Mag., 4(1996), no.1, p.85) proposed by M. Bencze and F. Popovici. This problem asks for the determination of all functions f : R → R having the property f (x1 + · · · + xn ) = f (x1 ) + f 2 (x2 ) + · · · + f n (xn ),
(1)
for all x1 , . . . , xn ∈ R (n fixed). Here f n (x) := (f (x))n . When f is continuous, a more general equation has been studied by the author [2]. For functional equations without any regularity we quote [3]. Certain (quite difficult) open problems were stated in [4]. 311
1) First assume that n ≥ 4. We shall prove that equation (1) is equivalent with the following system ½ 2 f (x) = f (x), x∈R (2) f (x + y) = f (x) + f (y), x, y ∈ R. By putting x1 = x2 = · · · = xn = 0 in (1) we get f 2 (0)[1 + f (0) + · · · + f n−2 (0)] = 0.
(3)
Since the equation 1+z+· · ·+z n−2 = 0 has only complex roots for n−2 ≥ 2, from (3) we obtain that f (0) = 0. Letting x1 = x3 = · · · = xn = 0, x2 = x in (1) we can deduce that f 2 (x) = f (x). Similarly f 3 (x) = · · · = f n (x) = f (x). Thus f (x1 + · · · + xn ) = f (x1 ) + · · · + f (xn ), and taking x1 = x, x2 = y, x3 = 0, . . . , xn = 0 by f (0) = 0 we get the second equation of (2). Reciprocally, each solution of the system (2) is a solution of equation (1). This follows easily by induction. Now, if we admit that f satisfies a regularity condition (for example, f is continuous at a point, is monotone, or is measureable, etc.), then it is well-known that Cauchy’s function equation (i.e. the second equation of (2)) has the general solution f (x) = cx, c ∈ R (fixed). Since c2 x2 = cx, ∀ x ∈ R, clearly c = 0 (e.g. x = 1 and x = 2). Thus the general solution of equation (1) is f ≡ 0 (obviously, if one of the above stated regularity conditions is satisfied). Remark 1. For the above stated results on Cauchy’s functional equation, see [1]. 2) For n = 1 the equation (1) becomes f (x1 ) = f (x1 ), x1 ∈ R, and all functions f : R → R satisfy this property. 3) Let n = 2. Then, as in (3), we get f (0) = 0, and we obtain the same solutions as in case 1). 4) Let n = 3. Then the equation becomes f (x + y + z) = f (x) + f 2 (y) + f 3 (z),
x, y, z ∈ R.
(4)
From x = y = z = 0 we have f (0)[1 + f (0)] = 0. Thus, we have to consider two cases: a) f (0) = 0, b) f (0) = −1. In case a) we get, if we put x = z = 0 that f (y) = f 2 (y) and the same solutions as in 1) are obtained. When f (0) = −1, by substituting x = z = 0, we can deduce that f (y) = f 2 (y) − 2. (5) Let y = 0 in (4). Then f (x + z) = f (x) + 1 + f 3 (z). Now if x = 0 in this equation, clearly f (z) = f 3 (z). On the other hand, from (5) we have 312
f 3 (z) = (f (z) + 2)f (z) = f 2 (z) + 2f (z) = 3f (z) + 2. From 3f (z) + 2 = f (z) we obtain that f (z) = −1, z ∈ R. This is a most general solution of equation (4), when f (0) = −1. Remark 2. The discontinuous general solutions of Cauchy’s functional equation can be obtained via the so called ”Hamel basis”. This is in fact a base of Q-vectorial space R. The interested reader can deduce that the most general solution of system (2) if f ≡ 0. Thus, this is the general solution of equation (1) in case n ≥ 4 and n = 2 or n = 3, f (0) = 0.
References [1] J. Acz´el, Lectures on functional equations and their applications, Academic Press, New York, San Francisco, London, 1966. [2] J. S´andor, Asupra unei ecuat¸ii funct¸ionale, Lucr. Semin. Did. Mat., 6(1990), 293-296. [3] J. S´andor, On certain functional equations, Itinerant Seminar Funct. Eq. Approx. and Convexity, Babe¸s-Bolyai Univ., 1988, 285-288. [4] J. S´andor, On certain open problems in the theory of functional equations, Proc. Symp. Appl. of Funct. Eq. in Ed., Science and Prod., 4th iune 1988, Odorheiu Secuiesc, Romania.
6
Two functional equations
Problem OQ.57 (M. Bencze and F. Popovici) from Octogon Math. Mag., 4(1996), no.2, p.78 asks for the determination of positive functions f with the property ! Ã n n q X X f xk = f (xk ), ∀ x ∈ R (n fixed). k=1
k=1
The misprint in f (xk ) can be interpreted in two ways, namely, as f (xk ) or as f (xkk ). These two situations lead to two distinct functional equations; and our aim is to study both cases in this note. 1) The functional equation à n ! n p X X k f xk = f (xk ). (1) k=1
k=1
313
In this case, the method is very similar to that shown inpour Note [1]. Namely, let x1 = · · · = xn = 0 in (1). We get f (0) = f (0) + f (0)p + ··· + p n f (0), p and this implies f (0) = 0 (since on the right-hand side of 0 = f (0)+ · · ·+ n f (0) are only nonnegative members. Letting x3 = · · · = xn = 0, x1 = x, x2 = y we get f (x p+ y) = f (x) + f (y). But from x1 = 0, x3 = 0, . . . , xnp= 0 we obtain f (x2 ) = f (x2 ), and in the same manner, we have f (xn ) = n f (xn ). Thus f (x) = f 2 (x) = · · · = f n (x) (where f n (x) := (f (x))n ). Thus we get the system ½ f (x + y) = f (x) + f (y), x, y ∈ R, (2) f (x) = f 2 (x), x ∈ R. Reciprocally, each solution of system (2) is a solution of (1). We have to solve the system (2) for arbitrary nonnegative functions. This can be made by using the theory of Cauchy’s equation (i.e. the first equation of (2)) but we shall follow here a very simple other argument. Clearly one has f 2 (x + y) = f 2 (x) + f 2 (y) + 2f (x)f (y) or f (x + y) = f (x) + f (y) + 2f (x)f (y). This gives f (x)f (y) = 0, yielding f (x) = 0, ∀ x ∈ R. This is the most general solution of the functional equation (1). 2) The functional equation ! Ã n n q X X k xk = f f (xkk ). (3) k=1
k=1
By putting x1 = · · · = xn =p0 in (3) we obtain f (0) = 0. From x1 = x3 = · · · = xn = 0, we get f (x2 ) = f (x22 ), i.e. f (x2 ) = (f (x))2 for all x ∈ R. In the same manner we can deduce that (f (x))2 = f (x2 ), (f (x))3 = f (x3 ), . . . , (f (x))n = f (xn ).
(4)
We will show that the functional equation (3) is equivalent to the system f (x + y) = f (x) + f (y), f (x2 ) = (f (x))2 .
(5)
By f (x2 + y 2 ) = f (x2 ) + f (y 2 ) = f 2 (x) + f 2 (y) and f ((x + y)2 ) = f (x2 + 2xy + y 2 ) = f (x2 + y 2 ) + f (2xy) = (f (x))2 + (f (y))2 + f (2xy) = f 2 (x + y) = f 2 (x) + f 2 (y) + 2f (x)f (y) we get f (2xy) = 2f (x)f (y). 314
(6)
We shall prove that f (xy) = f (x)f (y). (7) µ ¶ 1 1 Indeed, let y = in (6). Then f (x) = 2f f (x) and if we suppose that 2 2 µ ¶ 1 = 0, then f (0) = 0, ∀ x ∈ R, and then clearly (7) is satisfied. If one f 2 µ ¶ µ ¶ 1 1 1 admits that f 6= 0, then clearly f = and from the first equation 2 2 2 of (5) we get f (1) = 1. Letting y = 1 in (6), we can deduce f (2x) = 2f (x), thus f (2xy) = 2f (xy). This relation combined with (6) gives the equation (7). Now, from (7) easily follow all relations (4). Thus we have proved that equation (3) is equivalent with the system (5), or, on base of (7), with the system ½ f (x + y) = f (x) + f (y) (8) f (xy) = f (x)f (y) By f (1) = f 2 (1) we have two possibilities: a) f (1) = 0, in which case f (x) = 0 is the most general solution of (8); and b) f (1) = 1. Since f (0) = 0, (8) represents nothing else than all automorphisms of the field of real numbers R. When f is continuous at a point, is monotonic, or measurable, etc., all solutions of (8) are f (x) = x, ∀ x ∈ R, and there are no other solutions.
7
A functional equation satisfied by the sin and identity functions Let f : R → R such that µ
(n + 1)x f (nx)f n X 2 ³x´ f (kx) = f k=1 2
¶ for all x ∈ R and all n ∈ N.
(1)
Clearly, f (x) = x and f (x) = sin x are examples. By writing (1) for n + 1 and subtracting, one obtains the equation ·µ ¶ ¸ n+1 ¶ f x · µ ³ nx ´¸ (n + 2)x 2 ³ ´ f [(n + 1)x] = −f , (2) f x 2 2 f 2 315
∀ x ∈ R, ∀ n ∈ N. In fact, (1) and (2) are equivalent. Indeed, by writing (2) for n = 1, 2, . . . , (n − 1), after addition one gets: µ ¶ µ ¶ 3x 3x f (x)f f f (2x) 2 2 ³x´ ³x´ f (2x) + f (3x) + · · · + f (nx) = − f (x) + f f 2 2 µ ¶ µ ¶ µ ¶ µ ¶ 3x 3x 5x 3x f (x) f (2x)f f (2x)f f f (3x) f 2 2 2 2 ³x´ ³x´ ³x´ ³x´ − + − + − ···+ f f f f 2 2 2 2 ¶ ³ ´ ¶ µ ¶ µ µ nx (n − 2)x (n − 1)x (n − 1)x f f f f 2 2 2 2 ³x´ ³x´ + − f f 2 2 ¶ µ ³ nx ´ ³ nx ´ µ (n − 1)x ¶ (n + 1)x f f f f 2 2 2 2 ³x´ ³x´ + − , f f 2 2 giving (1). Therefore we have to consider equation (2). Without any auxiliary assumption on f it seems difficult to deduce any significant result on f . Let us assume that f (0) = 0 and that there exists a function g : R → R such that ¶ µ ¶ µ a+b a−b g , ∀ a, b ∈ R. f (a) − f (b) = 2f 2 2 (For example f (x) = sin x and f (x) = x are such functions, with the selections g(x) = cos x or g(x) = 1). Then, by putting b = 0 one gets ³a´ ³a´ f (a) = 2f g . 2 2 Since µ ¶ ³ nx ´ ³ x ´ µ (n + 1)x ¶ (n + 2)x f −f = 2f g , 2 2 2 2 ³a´ ³a´ by the above equality f (a) = 2f g applied to a := (n + 1)x, we can 2 2 deduce that (2) holds true. All in all, we have proved that all functions f with the property µ ¶ µ ¶ a−b a+b f (0) = 0, f (a) − f (b) = 2f g , a, b ∈ R 2 2 316
satisfy relation (1).
8
On certain functional equations by Rassias 1. Solve the equation: µ µ ¶ ¶ µ ¶ x x + f (xy) = f f + f (y) . f y y µ ¶ x Let f + f (x, y) = t. Then (1) can be written as y f (t) = t.
(1)
(2)
This means that if for each t ∈ A there exist x, y ∈ R, y 6= 0 such that µ ¶ x f + f (x, y) = t (3) y then the general solution of equation (1) is the identity function. For example, x when f : A → B is surjective (A ⊆ R such that x, y ∈ A ⇒ xy, ∈ A) then y (3) is satisfied. Indeed, let y = 1. Then 2f (x) = f
³x´ 1
t + f (x · 1) = t ⇔ f (x) = . 2
t t Let ∈ B. Since f is surjective, there exists x ∈ A with f (x) = . 2 2 2. Solve the equation: √ √ √ f (x xy + y x + y y) = f (f (x) + f (y)).
(4)
Put x → x2 , y → y 2 in (4) (where x, y ≥ 0). Then we get f (x3 y + xy 2 + y 3 ) = f [f (x2 ) + f (y 2 )].
(5)
Let x = 0 in (5). Then f (y 3 ) = f [f (0) + f (y 2 )]. For y = 0 in (5), we get f (0) = f [f (x2 ) + f (0)]. Since f [f (0) + f (x2 )] = f (x3 ), (put y → x in the above equality), we obtain f (0) = f (x3 ). (6) √ By putting x → 3 x, this implies f (x) = f (0) = constant. 317
3. Find the general solution of the equation f (xf (y) + yf (x) −
√ √ xf (y) − yf (x)) = f (x) + f (y).
(7)
Clearly x, y ≥ 0 in (1). Put y = 0 in (7). Then f ((x −
√ x)f (0)) = f (x) + f (0)
(8)
√ which for x = 0 gives f (0) = 0. Therefore (x − x)f (0) = 0, ∀ x ∈ A (A ⊂ [0, +∞)) implying f (0) = f (x) + f (0), i.e. 0 = f (x), ∀ x ∈ A. Reciprocally, f (x) = 0, ∀ x ∈ A satisfies (7). 4. What is the general solution of the equation: f (x)f (y) + f (y)f (x) = xf (x) + y f (x) .
(9)
We must have x, y > 0, f (x) > 0, f (y) > 0. Let x = 1 in (9). Then f (1)f (y) + f (y)f (1) = 1 + y f (1) .
(10)
Putting y = 1 in (10), we deduce 2f (1)f (1) = 2, i.e. f (1)f (1) = 1. Therefore f (1) = 1 and (10) implies f (y) = y. (11) By concluding, f (x) = x, ∀ x > 0. 5. We must solve the functional equation: f (xy) + f (x + y) = f (xy + x) + f (x − y).
(12)
Let f : A → R (A ⊆ R). Since x − y ∈ A, clearly 0 ∈ A. Put y = 0 in (12). Then we get f (0) + f (x) = f (x) + f (x), giving f (x) = f (0), ∀ x ∈ A. Therefore, f = constant. Reciprocally, f = constant satisfies equation (12).
References [1] Th.M. Rassias, OQ.738, Octogon Math. Mag., 9(2001), no.1, 1086. [2] Th.M. Rassias, OQ.739, Octogon Math. Mag., 9(2001), no.2, 1086. [3] Th.M. Rassias, OQ.735, Octogon Math. Mag., 9(2001), no.2, 1085. [4] Th.M. Rassias, OQ.736, Octogon Math. Mag., 9(2001), no.2, 1085. [5] Th.M. Rassias, OQ.737, Octogon Math. Mag., 9(2001), no.2, 1085. 318
9
Bartha’s functional equation
Let n ≥ 1 be fixed. We have to determine all functions f : R → R such that f 2 (x1 + x2 + · · · + xn ) = f 3 (x1 ) + f 3 (x2 ) + · · · + f 3 (xn ) (1) for all xk ∈ R (k = 1, n). Let n = 1. Then (1) gives f 2 (x1 ) = f 3 (x1 ), ∀ x1 ∈ R, i.e. f 2 (x1 )[f (x1 )−1] = 0. Let A = {x ∈ R : f (x) = 0}, B = {x ∈ R : f (x) = 1}. Then any function with the property A ∪ B = R satisfies the equality. Suppose n ≥ 2. By putting x1 = x2 = · · · = xn = 0, we get f 2 (0) = nf 3 (0), i.e. f 2 (0)[nf (0) − 1] = 0. i) f (0) = 0. By putting x2 = · · · = xn = 0 in (1) we get f 2 (x1 ) = f 3 (x1 ), ∀ x1 ∈ R, therefore f 2 (x1 + · · · + xn ) = f 2 (x1 ) + · · · + f 2 (xn ). Let g(x) = f 2 (x). Then g(x1 + · · · + xn ) = g(x1 ) + · · · + g(xn ) and letting x3 = · · · = xn = 0 we obtain g(x1 + x2 ) = g(x1 ) + g(x2 ).
(1)
Let x2 = −x1 in (1). Then 0 = g(x1 ) + g(x2 ), i.e. g(−x1 ) = −g(x1 ) giving 0 ≤ g(x1 ) = −g(−x1 ) ≤ 0 since g(x) = f 2 (x) ≥ 0 for all x ∈ R (particularly for x := x1 and x := −x1 ). This implies g(x1 ) = 0, and since x is arbitrary, g ≡ 0. Therefore f ≡ 0 is the general solution in case i). 1 ii) f (0) = . By letting x2 = · · · = xn = 0 in (1) we obtain n f 2 (x1 ) = f 3 (x1 ) +
n−1 n3
and this implies f 2 (x1 + · · · + xn ) = f 2 (x1 ) + · · · + f 2 (xn ) −
depending only on n). Then, by (2)
i.e. k =
(2)
n−1 = y 2 gives y = k (constant, n3
Let f (x1 ) = y. The equation y 3 +
k 2 = nk 2 −
(n − 1)n . n3
(n − 1)n n3
1 1 . Therefore f (x) = in case ii). n n 319
References [1] B. Bartha, OQ.742, Octogon Math. Mag., 9(2001), no.2, 1086.
10
An application of Chebyshev’s inequality We have to determine all functions f : R → (0, ∞) such that n n X 0 f (xk ) n Y f 0 (xk ) k=1 ≤ n X f (xk ) k=1 f (xk ) k=1
for all xk ∈ R (k = 1, n).
f 0 (xi ) = ai (i = 1, n) and f (xi ) suppose that ai > 0. Then by the arithmetic-geometric inequality one can write v u n uY a1 + · · · + an n t ak ≤ . n We will present here a partial solution. Put
k=1
Now, we will study an inequality of type a1 + · · · + an a1 f (x1 ) + · · · + an f (xn ) ≤ n f (x1 ) + · · · + f (xn ) ! ! à n à n X X 0 f (xk ) / f (xk ) . =
(1)
k=1
k=1
Remark that, when f (x1 ) + · · · + f (xn ) > 0, (1) is equivalent to (a1 + · · · + an )[f (x1 ) + · · · + f (xn )] ≤ n[a1 f (x1 ) + · · · + an f (xn )].
(2)
When the sequences (ai ) and (f (xi )) have the same type of monotony, by Chebysev’s inequality, this is true. Therefore, the following has been proved. Theorem. Suppose that f 0 (xi ) i) > 0, i = 1, n f (xi ) ii) f·(x1 ) + · · · + f (xn¸) > 0 f 0 (xi ) f 0 (xj ) iii) − [f (xi ) − f (xj )] > 0 for all i, j. f (xi ) f (xj ) Then inequality (1) is true. 320
11
On a functional inequality We will describe solutions of the following inequality: µ ¶ x+y 2 x+y f (x)f (y) ≤ f , ∀ x, y, ∈A⊂R 2 2
(1)
(see [1]). Clearly A must be a mid (or Jensen) convex set. Let X = {x ∈ A : f (x) = 0}, Y = {x ∈ A : f (x) > 0}, Z = {x ∈ A : f (x) < 0}. If X 6= ∅, Y = ∅, Z = ∅, then f ≡ 0. If X 6= ∅, Y 6= ∅, Z 6= ∅, then f (x) ≥ 0 and any x ∈ X, y ∈ Y or x, y ∈ X satisfy (1) (so if at least x+y one of x and y is in X). If x, y ∈ Y and Y is mid-convex (i.e. ∈ Y, 2 ¶ µ g(x) + g(y) x+y ≤ , i.e. g too), then g(t) = − ln f (t), t ∈ Y satisfies g 2 2 is Jensen-convex function on Y . For References to these functions, see¶[1]. In µ x+y < 0, case, if Y is not mid-convex, i.e. f (x) > 0, f (y) > 0 but f 2 remark that still (1) can be written as ¯ µ ¶¯ p ¯ x + y ¯¯ ¯ f (x)f (y) ≤ ¯f ¯, 2 so we may take h(t) = − ln |f (t)|, t ∈ Y ∪ Z (since remark that for such x, y one has ¯ µ ¶¯ ¶ µ ¯ 2 x+y ¯ x + y 2 ¯=f |f (x)f (y)| ≤ ¯¯f ) ¯ 2 2 which is J-convex on Y ∪ Z (we have used f (x)f (y) > 0, ∀ x, y ∈ Y ∪ Z). The other possibilities (e.g. X = ∅) are even more simple to handle.
References [1] J. S´andor, On the Open Problem OQ.573, Octogon Math. Mag., 9(2001), 943-944.
12
An unsolved functional equation ”What is the general solution of the functional equation: f {xf (y) + yf (z) + zf (x) − xyz} = f (x)f (y)f (z) + xyz?” Put x = 0. Then f [yf (z) + zf (0)] = f (0)f (y)f (z). 321
Let z = 0 here; implying f [yf (0)] = f 2 (0)f (y). For y = 0 this gives f (0) = f 3 (0), so three possibilities arose: i) f (0) = 0, ii) f (0) = 1, iii) f (0) = −1. In case i) the first equation above yields f (yf (z)) = 0.
(1)
z0 , z = z0 in (1). Then f (z0 ) f (z0 ) = 0, contradiction. Thus f ≡ 0, which is in fact impossible, since then the initial equation would imply 0 = xyz, which is not true for x 6= 0, y 6= 0, z 6= 0. So, in case i) there is no solution. Cases ii) and iii) remain momentary unsolved. Suppose there exists z0 with f (z0 ) 6= 0. Put y =
References [1] Th.M. Rassias, OQ.982, Octogon Math. Mag., 10(2002), no.2, 10411042.
322
Chapter 8
Diophantine equations ”... Wherever there is number, there is beauty.” (Proclus Diadochus)
”... An equation means nothing to me unless it expresses a thought of God.” (Srinivasa Ramanujan)
323
1
Variations on a problem with factorials
1. In the recent issue of Erd´elyi Matematikai Lapok [1] F. Smarandache has proposed the following problem: Solve in positive integers the equation: (x!)2 + (y!)2 = (z!)2 .
(1)
Since this is a particular case of the equation X2 + Y 2 = Z2
(2)
where solutions are known to be given by X = d(a2 − b2 ),
Y = 2dab,
Z = d(a2 + b2 )
(3)
where d ≥ 1 is arbitrary, a > b are of opposite parity, and (a, b) = 1; we can use (2) to solve (1). Clearly Z > X and Z > Y , so in (1) z! > y!. This implies z > y, since otherwise z ≤ y would imply z! = 1 · 2 . . . z ≤ 1 · 2 . . . y = y! But then z! = 1 · 2 . . . y(y + 1) · · · = y!K, so y! is a divisor of z!. By (3) we must have that 2dab divides d(a2 + b2 ), i.e. 2ab|(a2 + b2 ), which is impossible, since 2ab is even, while a2 + b2 is odd. Therefore, equation (1) is not solvable in positive integers. 2. There are many possible variations on equation (1). Namely: (x2 )! + (y 2 )! = (z 2 )!
(4)
(x!)2 + (y!)2 = (z 2 )!
(5)
(x2 )! + (y 2 )! = (z!)2
(6)
(x!)2 + (y 2 )! = (z!)2
(7)
(x!)2 + (y 2 )! = (z 2 )!
(8)
In fact, these are the special cases of the following equation: a! + b! = c!
(40 )
a2 + b2 = c!
(50 )
a! + b! = c2
(60 )
324
a2 + b! = c2
(70 )
a2 + b! = c!
(80 )
Remark that in (40 ) a = x2 , b = y 2 , c = z 2 , in (50 ) a = x!, b = y!, c = z 2 , in (60 ) a = x2 , b = y 2 , c = z!, in (70 ) a = x!, b = y 2 , c = z!, in (80 ) a = x!, b = y 2 , c = z 2 . Though, after the study of equations (40 )−(80 ), these particular cases need a special treatment, in what follows, we will restrict our attention to these equations (40 ) − (80 ) only. 3. The equation a! + b! = c! (i.e. (40 )). This is well-known, but we will give for the sake of completeness the solution (see also [3] for the more general equation). Let a, b, c ≥ 1. As in the case of equation (1), here in the same manner, as c > a, c > b, one has a!|c!, b!|c!; let c! = a!M , c! = b!K. Then b! = a!(M − 1), and a! = b!(K − 1), so a!|b! and b!|a!, implying a! = b!. This gives a = b, and 1 = M − 1, 1 = K − 1, i.e. M = 2, K = 2. On base of c! = 2a! c! = (a + 1) . . . c = 2, which is possible only if a = 1, since for a ≥ 2 one has a! clearly (a + 1) . . . c ≥ 3. Thus a = 1, b = 1, c = 2 is the only solution to (40 ). (By the way, since c = 2 = z 2 is not solvable in integers, equation (4) doesn’t have any solution). The equation a2 + b2 = c! (i.e. (50 )). This is a particular case of the equation a2 + b2 = n. It is well-known (see e.g. that [2]) this is possible only if in the prime factorization of n each prime of the form p ≡ 3 (mod 4) appears to an even power (here ”even” includes zero, too). There are many cases when c! satisfies this property, e.g. z! = 24 ·32 ·5·7 (each prime factor is of type p ≡ 1 (mod 4)) 10! = 28 · 34 · 52 · 7. For 12! = 210 · 35 · 52 · 7 · 11 this is not satisfied, since 2 ≡ 3 (mod 4) has an odd power 5. On the other hand, equation (50 ) cannot be solved for infinitely many c. Let c ≡ 3 (mod 4) be a prime. Then clearly in c! this prime appears to an odd power. The equation a! + b! = c2 (i.e. (60 )). First remark that for a = 1 one gets the equation b! + 1 = c2 (9) which is a famous unsolved equation (see e.g. [4]). It is easy to verify that 4! + 1 = 52 , 5! + 1 = 112 , 7! + 1 = 712 . It is conjectured that these are the only solutions to (9) but this seems hopeless at present. Let now 2 ≤ a ≤ b. Though the general case seems difficult, we can completely settle the case b = a + 1. We will show that the only solution of the equation a! + (a + 1)! = c2
(10)
is a = 4, c = 12, i.e. 4! + 5! = 122 . Equation (10) can be written also as 1 · 2 · 3 . . . a(a + 2) = c2 . 325
(∗)
If a + 2 = prime, clearly this is impossible (right side a square, left side a not). Let p the greatest prime between and a + 2 (by Chebyshev’s theorem, 2 a there is such a prime). If a+1 is not a prime, then < p < a, then a < 2p, so p 2 will appear at power 1. This means that (∗) is again impossible. Let therefore a + 1 = q prime, when the equation becomes (q − 1)!(q + 1) = c2 .
(∗∗)
We will show that this is possible only for q = 5. Let p < q − 1 be the greatest prime. If p - (q + 1), clearly (∗∗) will be impossible, as above. Let q − 1 = 2n; then m < p < 2m. By p ≥ m + 1 and q + 1 = 2(m + 1), one can have p = 2 or p = m + 1. Now, for m ≥ 4 it is known that there are at least two prime numbers between m and 2m. Thus p 6= m + 1, by the definition of p. Thus m ≤ 3. The case m = 3 is impossible (p = m + 1 being even), so q−1 = 2, i.e. q = 5. m = 2, when 2 2 Equation a + b! = c2 (i.e. (70 )). This is a particular case of an equation of type c2 − a2 = n (11) (where n = b! in our case). We will show that (11) is solvable if and only if n > 1 is an odd integer, or divisible by 4. Indeed, if n = c2 − a2 , and c, a have the same parity, then m is a multiple of 4. If c and a are of opposite parity, then clearly, n will be odd. Reciprocally, if n = 4m, then n can be written as n = (m + 1)2 − (m − 1)2 . If n = 2m + 1 (odd), then n = (m + 1)2 − m2 . In other words n cannot have the form n ≡ 2 (mod 4). Equation a2 + b! = c! (i.e. (80 )). We will show that this equation can have at most a finite number of solutions. For example, when b = 1, written in the form a2 + 1 = c! (12) has the only solution a = 1, b = 1, c = 2. Indeed, if c ≥ 4, then 4|c!, so a must be odd. Then it is well-known that a2 ≡ 1 (mod 8), so a2 + 1 ≡ 2 (mod 8), which means that 4 - (a2 + 1). Therefore c ≤ 3, and by verifying these cases one easily finds the solutions. Let p be a prime factor of b! such that p2n+1 kb! (m ≥ 0). Such a prime does exist since otherwise each prime would appear to an even power and b! would be a perfect square, which is impossible. Since c > b, then b!|c!, so p2n+1 |c. Clearly one can select c ≥ K (e.g. K ≥ p2m+2 ), such that p2m+2 |c. Now, since p2m+2 |a2 , one has a2 = p2m · p · a0 , so p2m |a2 , giving pm |a, or a = pm · A. This implies p2m A = p2m · p · a0 , thus A2 = pa0 . Since p|A2 , and p is prime, p|A, so A = pA0 , giving a = pm+1 A0 , thus a2 = 326
p2m+2 A0 2 = M p2m+2 . Now a2 + b! = M p2m+2 + p2m+1 B = p2m+1 (M p + B), where p - B, so p2m+1 ka2 + b!. Since p2m+2 |c!, the equation is impossible for c ≥ K. Therefore, we must have a finite number of solutions.
References [1] F. Smarandache, Problem 11136, Erd´elyi Matematikai Lapok 5 (2004), no.2, 108. [2] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford, 1960. [3] J. S´andor, On the Diophantine equation x1 ! + · · · + xn ! = xn+1 !, Octogon Math. Mag., 9(2001), no.2, 963-965. [4] R.K. Guy, Unsolved problems in Number Theory, Springer Verlag, 1994.
2
On the Diophantine equation x1 x2 + x2 x3 + · · · + xn x1 = n + x1 + · · · + xn
This equation has been proposed by Th.M. Rassias in [1]. Remark first that, this equation always has at least a solution in integers, since x1 = 0, x2 = 0, . . . , xn = −n provide a solution. In nonnegative integers a solution is x1 = 0, x2 = 0, . . . , xn−2 = 0, xn−1 = 2, xn = x2 (n ≥ 3); while a solution with all components ≥ 1 is x1 = 1, x2 = 1, . . . , xn−1 = 1, xn = n + 1. Now, we shall prove that equation (1) has at most a finite number of solutions in positive integers; while in the set of all integers, it has infinitely many solutions. By writing (1) for the equation, it can be written as x1 (x2 − 1) + x2 (x3 − 1) + · · · + xn−1 (xn − 1) + xn (x1 − 1) = n
(2)
Clearly for xi ≥ 1, i = 1, n, one must have 0 ≤ x1 (x2 − 1) ≤ n, . . . , 0 ≤ xn (x1 − 1) ≤ n, which imply surely 1 ≤ xi ≤ n + 1, i = 1, n. When, e.g. x1 = 0 and xi ≥ 1, i = 2, n, then by the obvious relation xy ≥ x + y − 1, x, y ≥ 1, one can write successively x2 x3 ≥ x2 + x3 − 1, . . . , xn−1 xn ≥ xn−1 + xn − 1, 327
so by supposing x2 ≤ x3 ≤ · · · ≤ xn−1 ≤ xn , by n + x2 + · · · + xn ≥ x2 + 2x3 + · · · + 2xn−1 + xn − (n − 1) we get x3 + · · · + xn−1 ≤ 2n − 1, i.e. indeed one can have a finite number of x2 , x3 , . . . , xn−1 . For these finite values each time we get an equation axn = b, giving also a finite number of xn ’s. We now show, that in the set of all integers, the number of solutions is infinite. Let, for simplicity n = 3. When x, y, z < 0 in the equation xy + yz + xz = 3 + x + y + z,
(3)
then by putting x = −X, y = −Y , z = −Z one has XY + Y Z + XZ = 3 − (X + Y + Z) < 3
(4)
(where X, Y, Z > 0) having a finite number of solutions. The same happens when x = 0, y < 0, z < 0. Let us now consider the two remaining cases, namely: (i) x < 0, y < 0, z > 0 (ii) x > 0, y > 0, z < 0. Case (ii) is very similar to (i), so we shall study only this latest possibility. Let x = −a, y = −b, z = c. Then a, b, c ≥ 1 and by (3), one gets ab − bc − ac = 3 − a − b + c.
(5)
Let a = λ be arbitrary. Then (5) gives b(λ + 1) + λ − 3 = c(b + λ + 3), i.e. (b + λ + 3)(λ + 1) − [(λ + 1)(λ + 3) − λ + 3] = c(b + λ + 3). Therefore b + λ + 3 must divide (λ + 1)(λ + 3) − λ + 3 = λ2 + 3λ + 6. Since λ2 + 3λ + 6 has always as the divisor λ2 + 3λ + 6, we get for b the value given by b + λ + 3 = λ2 + 3λ + 6, i.e. b = λ2 + 2λ + 3. Thus a = λ, c = λ, b = λ2 + 2λ + 3 give a solution to (5). But all solutions may be determined in this way! Indeed, b + λ + 3 is a divisor of λ2 + 3λ + 6
(6)
so for fixed λ we can obtain a finite number of values of b. The value of c is given by λ2 + 3λ + 6 c=λ+1− , b+λ+3 for b + λ + 3 dividing λ2 + 3λ + 6. Acknowledgement. The author thanks Mih´aly Bencze for pointing out a fallacy in the first draft of this note. 328
References [1] Th.M. Rassias, OQ.725, Octogon Math. Mag., 9(2001), no.2, 1084.
3
On the equation n + (n + 1) + · · · + (n + k) = n(n + k) In what follows we will determine all n, k positive integers such that n + (n + 1) + (n + 2) + · · · + (n + k) = n(n + k)
(1)
(the examples 3 + 4 + 5 + 6 = 3 · 6 and 15 + 16 + 17 + · · · + 35 = 15 · 35 are given in [2]). By simple formulae, (1) can be transformed into 2n2 − 2n − k(k + 1) = 0. Resolving this equation of second degree, one gets p 1 + 1 + 2k(k + 1) n= 2
(2)
(3)
where 1 + 2k(k + 1) = u2 and u is odd. This gives the equation 2k 2 + 2k + 1 − u2 = 0, having the positive solution p −1 + 1 − 2(1 − u2 ) , where 2u2 − 1 = v 2 . k= 2
(4)
(5)
The equation 2u2 − v 2 = 1
(6)
is known as a ”conjugate” Pell equation; namely the conjugate of the classical Pell equation ([3]) u2 − 2v 2 = 1. (7) √ It is well-known that (since 2 is irrational), this equation has infinitely many solutions, and all solutions are determined by the recurrence ½ u1 = u0 um+1 = u0 um + 2v0 vm , vm+1 = v0 um + u0 vm , (8) v1 = v0 where (u0 , v0 ) is the smallest solution of (7). Thus (u0 , v0 ) = (3, 2) give um+1 = 3um + 4vm ,
vm+1 = 2um + 3vm .
329
This gives a sequence of solutions (3, 2); (17, 12); (99, 70), . . . On the other hand, the general solutions of equation (6) are given by u = xm , v = ym , where ½ xm = Aum + Bvm ym = Bum + 2Avm where (un , vm ) are the solutions given in (8), while (A, B) is the smallest solution of (6); i.e. (A, B) = (1, 1) (see e.g. [4], [1]). Thus the solutions of (6) are xm = um + vm , ym = um + 2vm . (9) By (3) and (5), we get n=
1 + xm , 2
k=
ym − 1 , 2
m≥0
(10)
where xm and ym are given by (9). Remark that, by induction it follows from (8) that um is odd and vm is even for all m ≥ 0. By (9) this means that xm is odd and ym is odd for all m ≥ 0, so in (10) n and k are indeed positive integers. Equation (10) gives all possible solutions of (1). For example, x1 = u1 +v1 = 5, y1 = u1 + 2v1 = 7, giving n = 3, k = 3, whence x2 = u2 + v2 = 29, y2 = 41, so n = 15, k = 20. Now x3 = u3 + v3 = 99 + 70 = 169, y3 = 99 + 270 = 239; so n = 85, k = 119. This last example gives the third solution of (1); however all are given by (8), (9) and (10).
References [1] T. Andreescu, D. Andrica, On the resolvation of the equation ax2 −by 2 = 1 in natural numbers (Romanian), Gaz. Mat. 4(1980), 146-148. [2] M. Bencze, OQ.756, Octogon Math. Mag., 9(2001), no.2, 1088. [3] I.M. Vinogradov, Bazele teoriei numerelor, Bucure¸sti, 1954. [4] D.T. Walker, On the Diophantine equation mx2 − ny 2 = ±1, Amer. Math. Monthly, 74(1966), no.5, 504-513.
4
On a problem of Subramanian, and Pell equations
”Find all positive integers x such that both 2x2 + 1 and 3x2 + 1 are simultaneously perfect squares”. Let 2x2 + 1 = X 2 , 3x2 + 1 = Y 2 . This gives 2Y 2 − 3X 2 = −1, so 3X 2 − 2Y 2 = 1 (1) 330
X2 − 1 = 4k, so x is even. 2 Since Y 2 = 3x2 + 1, Y must be odd. Therefore we must find all such solutions of equation (1), where X is odd and Y is odd. Now, the general theory of equations aX 2 − bY 2 = c says (see e.g. [2], [3]) that all its solutions can be written as X = s0 u + bt0 v, Y = t0 u + as0 v, where (u, v) is any solution to X must be odd, so X 2 = 8k + 1, implying x2 =
u2 − abv 2 = c
(∗)
while (s0 , t0 ) is the minimal solution to as2 − bt2 = 1.
(∗∗)
In our case c = 1, a = 3, b = 2, s0 = 1, t0 = 1, so we get X = u + 2v,
Y = u + 3v
(2)
where u2 − 6v 2 = 1.
(∗ ∗ ∗)
Since the minimal solution√to (∗ ∗ ∗) is √ u0 = 5, v0 = 2, the general solution will be obtained from un + √ vn 6 = (5 +√2 6)n (see e.g. [2],√[3]). √ By writing un+1 + vn+1 6 = (5 + 2 6)n+1 = (un + vn 6)(5 + 2 6), we get the recurrence ½ un+1 = 5un + 12vn (3) vn+1 = 2un + 5vn Since u0 = 5, v0 = 2 we get u1 = 49, v1 = 20, etc., and by induction immediately follows un = odd, vn = even for all n ≥ 0. Thus, by (2) we get Xn = un + 2vn = odd, Yn = un + 3vn = odd, so all solutions of the initial problem are found.
References [1] K.B. Subramanian, OQ.1241, Octogon Math. Mag., 11(2003), no.1, 395. [2] D.T. Walker, On the Diophantine equation mx2 − ny 2 = ±1, Amer. Math. Monthly, 74(1966), 504-513. [3] T. Nagell, Introduction to Number Theory, John Wiley and Inc., 1951. 331
5
On the Diophantine equation x3 + y 3 + z 3 = a This note contains certain informations on equations of type: x3 + y 3 + z 3 = a
(1)
particularly when a = 30, we obtain the Open Problem OQ.277 proposed by Th.M. Rassias (Octogon). From the considerations which follow, we can state that when a has the form a = 9m + 4 or a = 9m − 4, the equation (1) have no solutions in integers (m ∈ Z). Thus the equations x3 +y 3 +z 3 = 31 (31 = 9·3 = 4), x3 +y 3 +z 3 = 32 (32 = 9 · 4 − 4), x3 + y 3 + z 3 = 22 (22 = 9 · 2 + 4), etc., none have solutions. When a = 30, unfortunately we can prove only a partial result. 1. For x, y, z < 0, clearly there are no solutions. 2. When x, y, z > 0, there are no solutions, since x, y, z ≤ 3 and x = 0: 13 + 13 = 2, 13 + 23 = 9, 13 + 33 = 28, 23 + 33 > 30; for x = 1 ⇒ y 3 + z 3 6= 29, . . . , x = 3 ⇒ y 3 + z 3 = 3, impossible. 3. Let x < 0, y > 0, z > 0. Put x = −X, when the equation becomes y 3 + z 3 = X 3 + 30,
X, y, z > 0.
(2)
We will prove that this equation has no solutions. Let X = 3k + r, y = 3m + p, z = 3n + t, where r, p, t ∈ {0, 1, 2}. Then we obtain from (2): 27k 3 + 27k 2 r + 9kr2 + r3 + 27m3 + 27m2 p + 9mp2 + p3 = 27n3 + 27n2 t + 9nt2 + t3 + 30.
(3)
This implies: r3 + p3 ≡ t3
(mod 3).
(∗)
i) r = 0. Then from (∗) p = t, and after reducing each term divisible by 3, excepting 10. ii) r = 1. Then 1 + p3 ≡ t3 (mod 3). For p = 0 we have t = 1. Then r = t, and we proceed as above. For p = 1, 2 ≡ t3 (mod 3) is impossible, for p = 3 we can have t = 0, when each term is ≡ (mod 9), excepting 30. iii) r = 2. Then p 6= 0; for p = 1 we have t = 0, for p = 2, t = 1, when 31 − 16 = 15 is not divisible by 9. Thus, in all cases the equation (3) (thus (2)) are not solvable. 332
4. x < 0, y < 0, z > 0. Put x = −X, y = −Y , when we obtain: z 3 = X 3 + Y 3 + 30,
X, Y, z > 0.
(4)
Let z = 3k + r, X = 3m + p, Y = 3n + t, r, p, t ∈ {0, 1, 2}. Then we get from (4): 27k 3 + 27k 2 r + 9kr2 + r3 = 27m3 + 27m2 p + 9mp2 + p3 + 27n3 + 27n2 t + 9nt2 + t3 .
(5)
This implies: r3 ≡ p3 + t3
(mod 3).
(∗∗)
By analyzing all cases as above, in the case r = 1, p = 2 we obtain t = 2. This case is possible, since in (5) we deduce 30 + 16 − 1 = 45, which is ≡ 0 (mod 9). By dividing with 9 each term in (5), we can deduce: 3k 3 + 3k 2 + k = 3m3 + 6m2 + 4m + 3n3 + 6n2 + 4n + 5.
(6)
Thus, we have obtained that when (4) is solvable, then necessarily z = 3k +1, X = 3m+2, Y = 3n+2, with k, m, n satisfying (6). We have considered many particular cases, and many computations, but were unable to prove that (6) has no solutions. The author feels that the equation has no solutions at all. Thus we make the conjecture that the equation x3 + y 3 + z 3 = 30 has no solutions in integers. (This is equivalent that (6) is not solvable in positive integers k, m, n).
References [1] Th.M. Rassias, OQ.277, Octogon Math. Mag., 7(1999), no.2, 195.
6
On the Diophantine equation x3 + y 3 + z 3 = a, II In the first part [12], we have considered equations of type x3 + y 3 + z 3 = a
(1)
and a special study was denoted to the particular case a = 30. This was proposed as an OQ.277 [13]. We have reduced the study of equation to x < 0, y < 0, z < 0. We note have that this Open Problem (when a = 30) is included also in W. Sierpinski’s famous book [11] or L.J. Mordell’s monograph [10]. In [14] Z. Tuzson published an erroneous proof that this equation has no solution (the fallacy in the proof follows at one by remarking that on 333
p.219 of [14] in relation (1) we must have 30 − 3xyz (and not 30 + 3xyz)) and as a result, 10 − S3 in relation (2) (not 10 + S3 )! This invalidates the argument what follows. Equations of type (1) (and particularly, with a = 30) have a long history. Practically, today theoretical methods are combined with computer search algorithms. In finding all solutions for a range of values of a with max{|x|, |y|, |z|} ≤ A, a straightforward two-dimensional algorithm [3], [6], [9] was applied in O(A2 ) steps. In [6], a computer search based on this algorithm for max{|x|, |y|, |z|} ≤ 221 − 1, 1 ≤ n ≤ 999 was discussed. All 5418 solutions were deposited into the UMT file of American Math. Society. Particularly, the search gives solutions for 17 values of a for which no solutions has been found before: a ∈ {39, 143, 180, 231, 312, 321, 367, 438, 462, 516, 542, 556, 660, 663, 754, 777, 870}. Recently (see [7]) for a = 439 the solution (−869418, −2281057, 2322404) and for a = 462 the solution (1612555, 2598019, −2790488) were obtained. For a = 478 (see [7]) the solution (−1368722, −13434503, 13439237) √ was found. In [5] a new algorithm based on the class number of Q( 3 a) was applied. On a CYBER 205 vector computer for many valued of a in the range max{|x|, |y|, |z|} ≤ A no solutions has been found. Particular values of a include: a = 30, 33, 42, 52, 74, 75, 110, 114, 156, 165, 195, 290, 318, . . . , so for all these equations we could conjecture that they have no solutions at all. In a recent paper new solutions for a = 75, 435, 444, 501, 600, 618, 912, 969 in the range |x| = min{|x|, |y|, |z|} ≤ 2 · 107 were found. For history or other aspects of these equations, see other titles in the References. Equations of such type appear also in [15]. 334
References [1] A. Bremner, On sums of three cubs, Canadian Math. Soc. Conf. Proc., 15(1995), 87-91. [2] B. Conn, L. Vaserstein, On sums of three integral cubs, Contemp. Math., 166(1994), 285-294. [3] V.L. Gardiner, R.B. Lazarus, P.R. Stein, Solutions of the Diophantine equation x3 + y 3 = z 3 − d, Math. Comp., 18(1964), 408-413. [4] R.K. Guy, Unsolved problems in Number Theory, Springer, New York, 1994 (second edition). [5] D.R. Heath-Brown, W.M. Lioen, H.J.J. teRiele, On solving the Diophantine equation x3 + y 3 + z 3 = k on a vector processor, Math. Comp., 61(1993), 235-244. [6] K. Koyama, Tables of solutions of the Diophantine equation x3 +y 3 +z 3 = n, Math. Comp., 62(1994), 941-942. [7] K. Koyama, Y. Tsuruoka, H. Sekigawa, On searching for solutions of the Diophantine equation x3 +y 3 +z 3 = n, Math. Comp., 66(1997), 841-851. [8] R.F. Lukes, A very fast electronic number sieve, Ph.D. Thesis, Univ. of Manitoba, 1995. [9] J.C.P. Miller, M.F.C. Woollett, Solutions of the Diophantine equation x3 + y 3 + z 3 = k, J. London Math. Soc., 30(1955), 101-110. [10] L.J. Mordell, Diophantine equations, Academic Press, 1969. [11] W. Sierpinski, A selection of problems in the theory of numbers, Pergamon Press, 1964. [12] J. S´andor, On a Diophantine equation, Octogon Math. Mag., 8(2000), no.1, 221-222. [13] Th.M. Rassias, OQ.277, Octogon Math. Mag., 7(1999), no.2. [14] Z. Tuzson, A solution to OQ.277, Octogon Math. Mag., 8(2000), no.1, 219-220. [15] M. Bencze, OQ.340, Octogon Math. Mag., 8(2000), no.1, 273. 335
7
n dimensional cuboids with integer sides and diagonals
In what follows we will determine all n dimensional cuboids with integer sides a1 , a2 , . . . , an , having integer diagonals. Clearly, this leads to the Diophantine equation a21 + a22 + · · · + a2n = a2 .
(1)
Let d = (a1 , a2 , . . . , an ), i.e. ai = dxi (i = 1, n), where (x1 , xn ) = 1. Then from (1) we get that d2 |a2 . It is well-known, that then this implies d|a. Let a = dx. Then (1) becomes: x21 + x22 + · · · + x2n = x2
(2)
where (x1 , x2 , . . . , xn ) = 1. This implies (x1 , x2 , . . . , xn , x) = 1, too. Let x1 = ky1 , . . . , xn−1 = kyn−1 , xn = kyn − x, for integers y1 , . . . , yn , and certain rational k. Then (2) becomes: k(y12 + y22 + · · · + yn2 ) = 2xyn .
(3)
This gives 2xn yn = 2(kyn − x)yn = 2kyn2 − 2xyn 2 = 2kyn2 − k(y12 + · · · + yn2 ) = k(yn2 − y12 − · · · − yn−1 ).
Thus: 2 k(yn2 − y12 − · · · − yn−1 ) = 2xn yn .
(4)
(3) and (4) imply x 1 = y1
yn2
2xn yn 2xn yn , . . . , xn−1 = yn−1 2 2 2 2 2 yn − y1 − · · · − yn−1 − y1 − . . . yn−1
and xn = kyn − i.e.
2 ) k(yn2 − y12 − · · · − yn−1 k(y12 + · · · + yn2 ) = , 2yn 2yn
x1 xn−1 xn = ··· = = 2 2 2 2y1 yn 2yn−1 yn yn − y1 − · · · − yn−1 =
y12
x k . = 2 2yn + · · · + yn 336
(5)
Let us suppose that
p k is the reduced form of (i.e. (p, q) = 1). Then q 2yn
q q q 2 2y1 yn = x1 , . . . , 2yn−1 yn = xn−1 , yn2 − y12 − · · · − yn−1 = xn , p p p q y12 + · · · + yn2 = x. p Since on the left sides we have integers and (q, p) = 1, thus p|x, p|xn , . . . , p|xn−1 , p|x1 , which is possible only when p = 1, since (x, xn , . . . , xn−1 , x1 ) = 1. Thus we have obtained that if (x1 , . . . , xn , x) is a solution of equation (2), then there exist integers y1 , . . . , yn and q > 0 such that qx1 = 2y1 yn , qx2 = 2y2 yn , . . . , qxn−1 = 2yn−1 yn , qxn = yn2 − y12 − · · · − yn2 , (6) qx = y12 + · · · + yn2 . Here q > 0 is taken so that (x1 , . . . , xn ) = 1. Conversely it is easy to see that if x1 , . . . , xn , x satisfy the equations (6) with positive integers y1 , . . . , yn and q > 0, integer, then we have obtained a solution for (2), if (y1 , . . . , yn ) = 1. Thus, the general solution of equation (1) is given by a1 = dx1 = d
2y2 yn 2yn−1 yn 2y1 yn , a2 = dx2 = d , . . . , an−1 = d , q q q
an = d
2 yn2 − y12 − · · · − yn−1 y 2 + · · · + yn2 , a=d 1 q q
(7)
where yi (i = 1, n) and q are as above.
8
The equation {x2 } + {y 2 } = {z 2 }
Let {x} denote the fractional part of the real number x. We will solve the equation {x2 } + {y 2 } = {z 2 }. (1) Since x2 = a has solutions only when a ≥ 0, and x = ±a, clearly it will be sufficient to solve the equation {a} + {b} = {c}
(2)
in nonnegative real numbers. Then if (a, b, c) is a solution of (2), this will generate eight solutions of (1) (i.e. (a, b, c); (−a, b, c); (a, −b, c); (a, b, −c); 337
(−a, −b, c); (a, −b, −c); (−a, b, −c); (−a, −b, −c)). We prove that (2) has solutions only if {a} + {b} < 1. Indeed, if (2) is true, then {c} = c − [c] < 1, so {a} + {b} < 1. Reciprocally, if θ = {a} + {b} < 1 (clearly θ ≥ 0), then the equation θ = {c} is solvable in c (in fact in each interval [m, m + 1) (m ≥ 0, integer), there is a single solution c). Therefore, we must study the inequality {a} + {b} < 1.
(∗)
6
y = {x}
1 y=0 O
-
1
2
3
n n+1
Let a ∈ [m, m + 1), b ∈ [n, n + 1). If a ∈ [m, m + 1/2), b ∈ [n, n + 1/2), then 1 1 {a} + {b} = (a − m) + (b − n) < + = 1, so (∗) is true. But this condition 2 2 is only sufficient. One may happen that (∗) is true for a ∈ [m, m + 1/2), b ∈ [n + 1/2, n + 1) or a ∈ [m + 1/2, m + 1), b ∈ [n, n + 1/2) (e.g. m = n = 0, 3 1 a = , b = . Then a < 1/2, b > 1/2 and a + b < 1). In fact, (∗) can be 5 5 written equivalently as a + b < m + n + 1. (3) Equation (2) may be generalized for three or more variables, but their study follows the same line as above.
9
On the Diophantine equation
1 1 1 a + +· · ·+ = x1 x2 xn b
1. We will study the equation in the title, where the unknowns xi (i = 1, n) are positive integers, while a, b are given positive integer numbers. For a = 6, 338
b = n2 −1, this contains the OQ.1119 [1] (by putting x1 = y1 , x2 = y1 +y2 , . . . , xn = y1 + y2 + · · · + yn ). The particular case n = 3 of this Open Problem will b be studied in detail. Clearly, for n = 1 one has x1 = , which is integer only a if a|b. Let n = 2. The equation 1 a 1 + = x1 x2 b
(1)
has been studied in detail in [2]. See also [3]. We will show that the resolvation of the general equation reduces inductively to the resolvation of equations of type (1). Let n = 3. We may suppose that x1 ≤ x2 ≤ x3 . Thus, if 1 1 a 1 + + = , x1 x2 x3 b
(2)
a 3 3b ≤ , implying x1 ≤ . This means that x1 can take a finite number b x1 a · ¸ 3b values). Let x0 be such a value. Then of values (at most a
then
1 a 1 ax0 − b A 1 + = − = = . x2 x3 b x0 bx0 B Thus
1 1 A + = x2 x3 B
(3)
which is an equation of type (1). Thus, in order to solve equation (2) we must solve a finite number of equations of type (3). For general n we can assume n nb a , which gives x1 ≤ , i.e. a finite number of x1 ≤ x2 ≤ · · · ≤ xn . Thus ≤ b x1 a values. For each fixed value of x1 one obtains an equation of n − 1 unknown. 1 1 A + ··· + = , x2 xn B where x2 ≤
(4)
(n − 1)B , etc.; inductively all will be reduced to equations of type A
(1). 2. We will take as an example the case n = 3 of OQ.1119, i.e. the equation 1 1 3 1 + + = . x x+y x+y+z 4 339
(5)
Clearly x ≥ 2; while for x ≥ 4 one has x + y ≤ 5, x + y + z ≥ 6, so the 1 1 1 37 3 37 . Now, ≤ , i.e. 180 ≤ 148 is sum of the left side is ≤ + + = 4 5 6 60 4 60 impossible. Therefore, we must have x > 1, x < 4, i.e. x ∈ {2, 3}. 3 1 1 a) Let x = 2. By − = one obtains the equation 4 2 4 1 1 1 + = (6) t t+z 4 where y + 2 = t. By elementary transformations from (6), we get the equation t2 + t(z − 8) − 4z = 0. (7) √ 8 − z ± z 2 + 64 This gives t1,2 = , and since t > 0, 2 √ 8 − z + z 2 + 64 t= . (8) 2 Here z 2 + 64 = u2 , n ∈ N, otherwise t is irrational. Writing 64 = (u − z)(u + z), and considering the divisors 1, 2, 4, 8, 16, 32, 64 of 64, by an easy verification follows u = 10, z = 6; respectively u = 17, z = 15. By (8) these give t = 6, respectively t = 5. Thus all solutions of equation (6) are: z = 6, y = 4 and z = 15, y = 3. (9) 3 1 5 b) x = 3. Now, by − = we get the equation 4 3 12 1 1 5 + = , (10) t t+z 13 where t = y + 3. We could use the method of a) however we will follow another way, i.e. by solving an equation of type (see [2]) 1 1 5 + = . (11) a b 12 Since 12(a + b) = 5ab, put (a, b) = d, i.e. a = dA, b = dB, with (A, B) = 1. Then 12(A + B) = 5dAB, and by the known fact that (AB, A + B) = 1, it follows that AB|12, i.e. AB ∈ {1, 2, 3, 4, 6, 12}. But (5, 12) = 1 implies 5|(A + B), so we immediately get the solutions A = 1, B = 4; A = 2, B = 3. Then d = 3 or d = 2 and we obtain a = 3 · 1 = 3, b = 3 · 4 = 12; respectively a = 2 · 2 = 4, b = 3 · 2 = 6. From this we can deduce the single solution of (10), namely y = 1, z = 2. (12) By concluding, all solutions of the equation (5) are the following: x = 2, y = 4, z = 6;
x = 2, y = 3, z = 15; 340
x = 3, y = 1, z = 2.
References [1] M. Bencze, OQ.1119, Octogon Math. Mag., 10(2002), no.2, 1075. [2] J. S´andor, On two Diophantine equations (Hungarian), Mat. Lapok, Cluj, 8/2001, 285-286. [3] J. S´andor, Geometric theorems, Diophantine equations and arithmetic functions, American Research Press, New Mexico, 2002.
10
Harmonic triangles
1. Let ABC be a triangle with side lengths a, b, c. If C = 60◦ , problem US4 presented to the XXth IMO asked for a proof of relation c c + ≥ 2. a b
(1)
For (1) quite complicated proofs were given. We note here that, by using elementary means, in fact an easier proof of a stronger relation can be deduced. Indeed, (1) can be written also as H(a, b) ≤ c,
(2)
where H(a, b) denotes the harmonic mean of a and b. Remark that, the stronger inequality a+b ≤c (3) A(a, b) = 2 is valid. Indeed, a + b ≤ 2c ⇔ (a + b)2 ⇔ 4c2 , or a2 + 2ab + b2 ≤ 4c2 = 4(a2 +b2 −ab), since c2 = a2 +b2 −2ab cos 60◦ = a2 +b2 −ab. The last inequality becomes 3(a − b)2 ≥ 0, which is obvious. Therefore H(a, b) ≤ G(a, b) ≤ A(a, b) ≤ c
(4)
is valid in any triangle ABC having C = 60◦ . Since a = 2R sin A, etc., and sin A + sin B = 2 sin
A−B A+B cos , 2 2
sin C = 2 sin
C C cos , 2 2
we can remark that, in any triangle ABC, relation (3) is valid iff cos
A−B C ≤ 2 sin . 2 2 341
(5)
C 1 The case sin = is a particular case of (5). 2 2 2. Now, let us suppose that ABC has integral sides. We say that the triangle ABC is harmonic relatively to the vertex C, if H(a, b) = c.
(6)
We say that the triangle ABC is harmonic, if H(a, b, c) = integer, (7) ¶ µ 1 1 1 denotes the harmonic mean of a, b, c. In where H(a, b, c) = 3/ + + a b c what follows we shall study these two types of harmonic triangles. Relation (6) can be written also as 2ab = (a + b)c
(8)
Let gcd(a, b) = d, i.e. a = da1 , b = db1 , with (a1 , b1 ) = 1. Then (8) becomes 2da1 b1 = (a1 +b1 )c. This implies a1 b1 |(a1 +b1 )c, and since it is well-known that for (a1 , b1 ) = 1 one has also (a1 + b1 , a1 b1 ) = 1, we get a1 b1 |c, so c = ka1 b1 . k(a1 + b1 ) This in turn implies 2d = k(a1 + b1 ), so d = , where at least one of 2 k and a1 + b1 is an even number. From this we get a=
ka1 (a1 + b1 ) , 2
b=
kb1 (a1 + b1 ) , 2
c = ka1 b1
(9)
with (a1 , b1 ) = 1 (and at least one of k and a1 + b1 = even). Now a + b > c, i.e. k(a1 + b1 ) (a1 + b1 ) > ka1 b1 ⇔ (a1 + b1 )2 > 2a1 b1 is always true. The other 2 inequalities b + c > a and a + c > b lead to relations of type a21 + 2a1 b1 > b21 and b21 + 2a1 b1 > a21 so −2a1 b1 < a21 − b21 < 2a1 b1 , i.e. |a21 − b21 | < 2a1 b1 . All in all, the triangle ABC is harmonic relatively to C, if its sides are given by (9), where (a1 , b1 ) = 1; k or a1 + b1 are even; and |a21 − b21 | < 2a1 b1
(∗)
For example, when a1 = b1 + 1 and k = even, the inequality (∗) is true, since 2b1 + 1 < 2b1 (b1 + 1) ⇔ 1 < 2b21 ≥ 2. 3. Now, we shall determine all harmonic triangles of type 2, i.e. for which (7) is true, i.e. X ab|3abc. (10) 342
Let d = gcd(a, b, c), i.e. a = da1 , b³= db1 , c = dc1 ´where (a1 , b1 , c1 ) = 1. X X Then a1 b1 |3da1 b1 c1 . Let D = gcd a1 b1 , a1 b1 c1 . Then a1 b1 c1 = Dk, X a1 b1 = Dk 0 , where (k, k 0 ) = 1 so Dk 0 |3dDk, i.e. k 0 |3dk. This implies k 0 |3d, X X X m a1 b1 so a1 b1 = Dk 0 |3dD. Let 3dD = m a1 b1 , i.e. d = . This gives 3D X X m a1 b1 m a1 b1 a = da1 = a1 , b = db1 = b1 , 3D 3D X m a1 b1 . c = cc1 = c1 3D X Here D| a1 b1 , so for a, b, c to be integers, we have two possibilities: 1) 3|m. Let m = 3s. Then a=
sa1
X D
a1 b1
,
b=
sb1
X D
a1 b1
,
c=
sc1
X
a1 b1
D
.
Clearly a + b > c if a1 + b1 > c1 . Therefore (a1 , b1 , c1 ) = 1 must satisfy also the inequalities a1 + b1 > c1 , b1 + c1 > a1 , a1 + cX 1 > b1 . 2) 3 - m. Remark that we cannot have 3 a1 b1 , since then we must X have 3|a1 , 3|b1 , 3|c1 , contradiction to (a1 , b1 , c1 ) = 1. Thus we have 3| a1 b1 (and again, of course a1 + b1 > c1 , etc.). These two cases determine all harmonic triangles.
11
A Diophantine equation involving Euler’s totient The equation in the title is ϕ(x1 x2 . . . xn ) = ϕ(x1 ) + ϕ(x2 ) + · · · + ϕ(xn ).
(1)
The aim of this Note is to prove that equation (1) has a finite number of solutions. For particular n, all solutions can be obtained by verifications. The method is based on two simple lemmas. Lemma 1. For all positive integers x and y one has ϕ(xy) ≥ ϕ(x)ϕ(y) 343
(2)
with equality only when x and y are coprime. Proof. This inequality is well known, see e.g. [3]. In fact, if one denotes by P = product of all common prime divisors of x and y, then immediately follows the identity P ϕ(x)ϕ(y), ϕ(xy) = ϕ(P ) where P ≥ ϕ(P ). This may be proved e.g. by ¶ Y µ 1 ϕ(x) = x 1− . q q/x,prime
Lemma 2. For n ≥ 2 and ui > 2 positive integers (i = 1, 2, . . . , n), one has u1 u2 . . . un > u1 + u2 + · · · + un .
(3)
Proof. For n = 2 this follows from (u1 −1)(u2 −1) > 1 (i.e. u1 u2 > u1 +u2 ), valid for u1 > 2, u2 > 2. Now, by admitting that (3) is true for n, by (u1 . . . un )un+1 > (u1 + · · · + un )un+1 > u1 + · · · + un + un+1 since this is equivalent to (u1 + · · · + un − 1)(un+1 − 1) > 1. So, by induction, relation (3) is valid for all n ≥ 2. Now, first consider the case n = 2. By (2) and the notations ϕ(x1 ) = u, ϕ(x2 ) = v one can write: u + v ≥ uv, i.e. (u − 1)(v − 1) ≤ 1. We have two cases a) (u − 1)(v − 1) = 0 b) (u − 1)(v − 1) = 1. In case a) we have u = 1 or v = 1. Thus ϕ(x1 ) = 1, i.e. x1 = 1 or x1 = 2, in which case ϕ(x2 ) = 1 + ϕ(x2 ) (which is impossible), or ϕ(2x2 ) = 1 + ϕ(x2 ), for x2 = odd this cannot be true, since then ϕ(2x2 ) = ϕ(2)ϕ(x2 ) = ϕ(x2 ) 6= 1 + ϕ(x2 ). If x2 is even, let x2 = 2k m, with m = odd. Then ϕ(2x2 ) = ϕ(2k+1 m) = k+1 ϕ(2 )ϕ(m) = 2k ϕ(m) = 1 + ϕ(2k m), only if 2k ϕ(m) = 1 + 2k−1 ϕ(m), or 2k−1 ϕ(m) = 1. Then k = 1 and ϕ(m) = 1. Thus m = 1 or 2, and clearly we have determined all solutions. In case b) we have u = 2 and v = 2. Thus ϕ(x1 ) = 2 and ϕ(x2 ) = 2. Then ϕ(x1 x2 ) = 4 etc. 344
Now, let n = 3, i.e. consider the equation ϕ(x1 x2 x3 ) = ϕ(x1 ) + ϕ(x2 ) + ϕ(x3 ). By (2) one has clearly ϕ(x1 x2 x3 ) ≥ ϕ(x1 )ϕ(x2 )ϕ(x3 ), and by (3) we cannot have ϕ(xi ) > 2 for all i = 1, 2, 3. Let ϕ(xi ) = ui . Then, let u1 ∈ {1, 2}. For u1 = 1 one gets 1 + u2 + u3 ≥ u2 u3 , i.e. (u2 − 1)(u3 − 1) ≤ 2. Thus one have to distinguish the cases: a) (u2 − 1)(u3 − 1) = 1 b) (u2 − 1)(u3 − 1) = 2 and in each case one arrives to a finite number of solutions, and these solutions may be obtained by direct verifications. For u1 = 2, one gets 2 + u2 + u3 ≥ 2u2 u3 , implying u2 + u3 ≥ 2(u2 u3 − 1) > u2 u3 if u2 u3 > 2. Thus (u2 − 1)(u3 − 1) < 1. Here it is easy to study all possibilities to deduce the finite number of possible solutions. In the general case one can follow the same argument. Namely, let ϕ(xi ) = ui . By the inequality ϕ(x1 . . . x2 ) ≥ ϕ(x1 ) . . . ϕ(xn ) (which is a consequence of Lemma 1 with induction), and by relation (3) one can deduce that there exists i0 ∈ {1, . . . , n}, with xi0 ∈ {1, 2}. Let i0 = 1 (for simplification of notation), when ϕ(x1 ) = 1, then x1 ∈ {1, 2} and one obtains ϕ(x2 . . . xn ) = 1+ϕ(x2 )+· · ·+ϕ(xn ) or ϕ(2x2 . . . xn ) = 1+ϕ(x2 )+· · ·+ϕ(xn ). From 1+ϕ(x2 )+ · · · + ϕ(xn ) ≥ ϕ(x2 ) . . . ϕ(xn ) we have that there exists j0 ∈ {2, . . . , n} with ϕ(xj0 ) ∈ {1, 2, 3}. (Since we can prove, completely analogously with Lemma 2, that for ui > 3, 1 + u2 + · · · + un < u2 . . . un for n ≥ 2). When ϕ(x1 ) = 2, we have 2+ϕ(x2 )+· · ·+ϕ(xn ) ≥ 2ϕ(x2 ) . . . ϕ(xn ), implying ϕ(x2 )+· · ·+ϕ(xn ) ≥ 2[ϕ(x2 ) + · · · + ϕ(xn ) − 1] > ϕ(x2 ) . . . ϕ(xn ), if ϕ(x2 ) . . . ϕ(xn ) > 2. By Lemma 2, this is impossible if all ϕ(xi ) > 2 for i ∈ {2, . . . , n}. Thus there exists i1 ∈ {2, . . . , n} with ϕ(xi1 ) ∈ {1, 2}. In all cases, we obtain an equation with n − 1 arguments, then two (or three) equations with n − 2 arguments, and so on. Since the equation ϕ(x) = k can have only a finite number of solutions √ (since, e.g. for x > 6, by the known inequality ϕ(x) > x it follows x < k 2 ), finally all equations can have a finite number of solutions. Indeed, one can prove: Theorem. The equation (1) has at most a finite number of solutions. Proof. By (2), equation (1) implies u1 + u2 + · · · + un ≥ u1 u2 . . . un
(4)
where ui ≥ 1 (i = 1, 2, . . . , n) are positive integers, ui = ϕ(xi ). First we prove that inequality (4) has a finite number of solutions (u1 , . . . , un ). Let u1 ≤ u2 ≤ · · · ≤ un . Then (4) yields u1 u2 . . . un ≤ nun or u1 u2 . . . un−1 ≤ n. Thus 1 ≤ u1 ≤ u2 ≤ · · · ≤ un−1 ≤ n, which means that the numbers u1 , . . . , un−1 can take at most the values 1, 2, . . . , n. For un we 345
u1 + · · · + un−1 , u1 u2 . . . un−1 − 1 if u1 u2 . . . un−1 > 1 (if not, then u1 = · · · = un−1 = 1, and we obtain the equation ϕ(x1 . . . xn ) = n − 1 + ϕ(xn ) which we study separately. Thus un can take a finite number of values, too. Since the equation ϕ(x) = k (k fixed) can have at most a finite number of solutions, the proof (in this case) is completed. For the equation ϕ(x1 . . . xn ) = n−1+ϕ(xn ) we note that by ϕ(x1 . . . xn ) ≥ ϕ(x1 . . . xn−1 )ϕ(xn ) we can deduce ϕ(xn )[ϕ(x1 . . . xn−1 ) − 1] ≤ n − 1 thus, if ϕ(x1 . . . xn−1 ) 6= 1, ϕ(xn ) ≤ n − 1. If ϕ(x1 . . . xn−1 ) = 1, then x1 . . . xn−1 ∈ {1, 2}, and the proof is finished. For Diophantine equations involving various arithmetical functions we quote [1], [2]. have un (u1 u2 . . . un−1 − 1) ≤ u1 + u2 + · · · + un−1 , i.e. un ≤
References [1] C.A. Nicol, Some Diophantine equations involving arithmetic functions, J. Math. Anal. Appl., 15(1966), 154-161. [2] J. S´andor, Some Diophantine equations for particular arithmetic functions, Seminarul de teoria structurilor, no.53, Univ. Timi¸soara, Romania, 1989, 1-10. [3] J. S´andor, Some arithmetic inequalities, Bulletin Number Theory Rel. Topics, 11(1987), 149-161.
12
On f (n) + f (n + 1) + · · · + f (n + k) = f (n)f (n + k) for f ∈ {ϕ, ψ, σ}
Our aim is to study the equations in the title in positive integers n, and particular values of k for the arithmetical functions ϕ (Euler’s totient), ψ (Dedekind’s function) and σ (sum of divisors). Theorem 1. The only solution in positive integers of the equation ϕ(n) + ϕ(n + 1) + ϕ(n + 2) = ϕ(n)ϕ(n + 2)
(1)
is n = 3. Proof. An easy computation shows that for n < 19, the only solution of (1) is n = 3: ϕ(3) + ϕ(4) + ϕ(5) = ϕ(3)ϕ(5), i.e. 2 + 2 + 4 = 2 · 4. Let now n ≥ 19. We note that for such n one has ϕ(n) >
√ n + 2.
346
(2)
This can be proved by many arguments. For example, in [2] it is proved that ϕ(n) > 6n/(12 + 5 log n). √ Now, the inequality 6n/(12 + 5 log n) > n + 2 becomes ¶ µ √ 24 10 log n 2 (12 + 5 log n) = 12 + √ + 5 log n + √ 6 n> 1+ √ n n n √ and this is true for n ≥ 19. Here one has n > log n for n ≥ 4, but we need a √ 24 24 slightly stronger 6 n > 15 log n (n ≥ 11), and since 12 + √ < 12 + = 20, 3 n the above inequality follows. This proves (2). Now, (1) can be written as [ϕ(n) − 1][ϕ(n + 2) − 1] = ϕ(n + 1) + 1.
(3)
On the right side of (3) one has ϕ(n + 1) + 1 ≤ n + 1, but on the left side, for n ≥ 19 one can write that √ √ [ϕ(n) − 1][ϕ(n + 2) − 1] > ( n + 1)( n + 2 + 1) p √ √ = n(n + 2) + n + n + 2 + 1 > n + 1, i.e. n2 + 2n + 1 + a > n2 + 2n + 1 for a > 0. This finishes the proof of Theorem 1. Theorem 2. None of the equations ψ(n) + ψ(n + 1) + ψ(n + 2) = ψ(n)ψ(n + 2)
(2)
σ(n) + σ(n + 1) + σ(n + 2) = σ(n)σ(n + 2)
(3)
and has solutions in positive integers. Proof. Write (2) in the form [ψ(n) − 1][ψ(n + 2) − 1] = ψ(n + 1) + 1.
(4)
k(k + 1) 2 (see e.g. [3]) on the left side of (4) we have ≥ n(n + 2), while on the right side of (4) one has (n + 1)(n + 2) ψ(n + 1) + 1 ≤ + 1. 2 Since ψ(n) ≥ n + 1, n ≥ 2, and ψ(k) ≤ σ(k) ≤ 1 + 2 + · · · + k =
347
Therefore, if (4) is true, then n(n + 2) − 1 ≤
(n + 1)(n + 2) , 2
(5)
i.e. n2 + n ≤ 4. Since for n ≥ 2, n(n + 1) ≥ 6, this is impossible. For n = 1, (4) doesn’t give a solution. The proof of (3) runs on the same lines. Remark 1. The proof shows that for any arithmetical function f such that n(n + 1) f (1) = 1, f (n) ≥ n + 1, f (n) ≤ , n≥2 2 the equation f (n) + f (n + 1) + f (n + 2) = f (n)f (n + 2) doesn’t have a solution in positive integers. Remark 2. Equations (1) and (2) are particular cases of OQ.753 and OQ.754 (t = 1, k = 2) see [1]. Theorem 3. The equations ψ(n) + ψ(n + 1) + ψ(n + 2) + ψ(n + 3) = ψ(n)ψ(n + 3)
(6)
σ(n) + σ(n + 1) + σ(n + 2) + σ(n + 3) = σ(n)σ(n + 3)
(7)
and do not have solutions in positive integers. Proof. Now the method of proof of Theorem 2 doesn’t work. We need the following auxiliary results: ψ(n) ≤ σ(n) < 2n log n for n ≥ 3.
(8)
The result is well-known, but we give here the simple proof: σ(n) = n
X1 d|n
d
≤n
X1 d≤n
s
< 2n log n,
since it is well-known that 1+
1 1 + · · · + < 2 log n for n ≥ 3. 2 n
Now, (6) can be written equivalently as: [ψ(n) − 1][ψ(n + 3) − 1] = ψ(n + 1) + ψ(n + 2) + 1. 348
(8)
The left side of (8) is ≥ n(n + 3) for n ≥ 2, but the right side is < 2(n + 1) log(n + 1) + 2(n + 2) log(n + 2) + 1, so we must have µ ¶ µ ¶ 1 2 1 n+3<2 1+ log(n + 1) + 1 + log(n + 2) + . n n n Since 1+
1 1 4 ≤1+ = , n 3 3
1+
2 2 5 ≤1+ = n 3 3
we must have 3n + 8 ≤ 6 log(n + 2).
(9)
By considering the application f (x) = 6 log(x + 2) − 3x − 8 it is easy to see that (9) is impossible. Clearly, for equation (7) the same argument applies. Theorem 4. For fixed k ≥ 2 and t ≥ 1, the equations ψ t (n) + ψ t (n + 1) + · · · + ψ t (n + k) = ψ t (n)ψ t (n + k)
(10)
σ t (n) + σ t (n + 1) + · · · + σ t (n + k) = σ t (n)σ t (n + k)
(11)
can have at most a finite number of solutions. Proof. The same argument as in the proof of Theorem 3 can be used. Indeed, the left side of (10) is ≥ [(n + 1)t − 1][(n + k)t − 1] while the right side is < 2t [(n + 1)t + · · · + (n + k − 1)t ] logt (n + k − 1) or ((n + 1)t − 1)((n + k)t − 1) k2t (n + k − 1)t logt (n + k − 1) < . n2t n2t Now, if k, t are fixed and n → ∞ this would give 1 ≤ 0, which is impossible. Therefore n ≤ n0 (k, t).
References [1] M. Bencze, OQ.753, Octogon Math. Mag., 9(2001), no.2, 1088. [2] J. S´andor, On certain Diophantine equations for particular arithmetic functions (Romanian), Seminarul de teoria structurilor, no.53, 1989, Univ. of Timi¸soara. [3] J. S´andor, On Dedekind’s arithmetical function, Seminarul de teoria structurilor, no.51, 1988, Univ. of Timi¸soara, 1-15. 349
13
On certain equations for the Euler, Dedekind, and Smarandache functions Here appear equations of type: f (n)f (n + 1) . . . f (n + k) = (f (n))t + (f (n + k))t
(1)
with f ∈ {ϕ, ψ, S}, where ϕ, ψ, S are the Euler totient, Dedekind’s functions and Smarandache function, respectively. Let f be arbitrary and suppose f (n) ≥ 2, f (n + 1) ≥ 2. Then (1) cannot have solutions for t = 1 since f (n)f (n + 1) . . . f (n + k) ≥ 2f (n)f (n + k) > f (n) + f (n + k). Indeed, (f (n) − 1)(f (n + k) − 1) ≥ 0 ⇒ f (n)f (n + k) ≥ f (n) + f (n + k) − 1 ⇒ 2f (n)f (n + k) ≥ 2f (n) + 2f (n + k) − 2 > f (n) + f (n + k) since f (n) + f (n + k) ≥ 3, by assumption. It is easy to see that for n ≥ 3 one has ϕ(n) ≥ 2, ϕ(n + 1) ≥ 2, ψ(n) ≥ 2, ψ(n + 1) ≥ 2, S(n) ≥ 2, S(n + 1) ≥ 2, so only the cases n ∈ {1, 2} should be considered. Since (1) implies f (n + k)|f (n) (for t = 1) and ϕ(1) = ϕ(2) = 1; ψ(1) = 1, ψ(2) = 3, S(1) = 1, S(2) = 2 we should have ϕ(1 + k) = 1, ϕ(2 + k) = 1, ψ(1 + k) = 1, ψ(2 + k) = 1, ψ(2 + k) = 3, S(1 + k) = 1, S(2 + k) = 2. Only for ϕ the k = 1 case is acceptable, but then ϕ(1)ϕ(2) = 1 6= ϕ(1) + ϕ(2). Therefore, for t = 1 the proposed equations does not have solutions.
14
Some equations involving the arithmetical functions ϕ, σ, ψ WeX will consider the following equations: X n 1) ϕ(d ) = ϕ(nd ) d|n
d|n
350
2)
X
(σ(d))d =
d|n,d6=n
3)
X
X
(σ(d))n−d
d|n,d6=n X ψ(d) = ψ(n − d).
d|n,d6=n
d|n,d6=n
In what follows all equations (1)-(3) will be solved. 1) Let d1 = 1 < d2 < · · · < dr = n be all divisors of n. Then ϕ(1n ) + ϕ(dn2 ) + · · · + ϕ(dnr ) = ϕ(nd1 ) + · · · + ϕ(ndr )
(1)
is impossible for n ≥ 3, since it is well known that for x ≥ 3, ϕ(x) is an even number. Now di ≥ 1, i = 1, r, and on the right side of (1) one has an even number. On the other hand, the left side of (1) is odd, since dn2 ≥ 2n ≥ 8, so ϕ(dn2 ), . . . , ϕ(dnr ) all are even, while ϕ(1n ) = 1 is odd. For n = 1 or n = 2, by ϕ(12 ) + ϕ(22 ) = ϕ(21 ) + ϕ(22 ) one has equality, by ϕ(2) = 1. Thus all solutions of equation 1) are n = 1 and n = 2. 2) Let d be a divisor of n. Then n = kd, so n − d = (k − 1)d ≥ d if k ≥ 2. This is true, if n ≥ 2, d 6= n. For such d’s one has (σ(d))n−d ≥ (σ(d))d . On the other hand, if n ≥ 3, there is at least a divisor d such that n − d ≥ 2d (since n ≥ 3d, i.e. X Xk ≥ 3 e.g. take d = 1, when n ≥ 3 = 1 · 3). Then clearly n−d (σ(d)) > (σ(d))d in such a case. For n = 1, 2 there is no solution, too. 3) We shall use the following property of the function ψ (see e.g. [1]): ψ(ab) ≥ aψ(b), (∀) a, b ≥ 1.
(2)
Now, if n = dk, then n − d = d(k − 1), so ψ(n − d) ≥ (k − 1)ψ(d) > ψ(d), if k − 1 > 1 for at least a divisor d. Clearly, for n ≥ 3 one has such a divisor, namely d = 1. Thus the equality ψ(d1 ) + · · · + ψ(dr ) = ψ(n − d1 ) + · · · + ψ(n − dr )
(3)
cannot be true. Clearly n ≥ 2, and we do not have solutions.
References [1] J. S´andor, On Dedekind’s arithmetical function, Seminarul de teoria structurilor, no.51, 1988, Univ. of Timi¸soara, 1-15.
15
Equations with composite functions
Let d(n), σ(n), ϕ(n) denote the number of divisors, the sum of divisors, and Euler’s totient, respectively. Our aim is to solve completely the four equations stated in the title. 351
The following well-known inequalities will be used (see e.g. [3]): √ Lemma 1. d(n) < 2 n for all n ≥ 1. (1) Lemma 2. ϕ(n) ≤ n − 1 for all n ≥ 2. (2) σ(n) n+1 Lemma 3. ≤ . (3) d(n) √2 Lemma 4. σ(n) < n n for all n ≥ 2. (4) Theorem 1. All solutions of equation σ(d(n)) = n are n = 1, 3, 4, 12. In fact, σ(d(n)) < n for all n > 12. (5) Proof. By (3), √ p d(n) + 1 2 n+1 σ(d(n)) ≤ d(d(n)) < 2 d(n) 2 2 √ q √ √ 2 n+1 √ √ <2 2 n· = 2 · 4 n(2 n + 1) < n ⇔ 2 √ √ 2(2 n + 1) < n3/4 . √ √ Put n = x. Then 2(2x + 1) < x3/2 ⇔ 2(2x + 1)2 < x3 ⇔ x3 > 8x2 + 8x + 2. Let f (x) = x3 − 8x2 − 8x − 2. Since f 0 (x) = 3x2 − 16x − 8, f 00 (x) = 6x − 16, clearly for all x ≥ 9 we have f 00 (x) > 0 so f 0 (x) > f 0 (9) > 0, f (x) ≥ f (9) = 7 > 0. Thus for n ≥ 81, the inequality σ(d(n)) < n is proved. A computer search shows that this is true for all n > 12. For n = 1, 3, 4, 12 there is equality, finishing the proof of Theorem 1. Remark. For Theorem 1 see also [2]. Theorem 2. All solutions of equation d(σ(n)) = n are n = 1, 2, 3. In fact, d(σ(n)) < n for all n ≥ 4. (6) p √ p 3/4 Proof. By (1) and (4), d(σ(n)) < 2 σ(n) < 2 n n = 2n < n ⇔ 16n3 < n4 ⇔ n > 16. For n ≥ 4, n ≤ 16 a direct computation applies. It is immediate that n = 1, 2, 3 are solutions. Theorem 3. The single solution of ϕ(d(n)) = n is n = 1. In fact, for any n ≥ 2 one has ϕ(d(n)) < n. (7) Proof. By (2) and (1) one can write successively for n ≥ 2, ϕ(d(n)) ≤ √ √ √ n+1 (i.e. ( n − 1)2 > 0). Clearly n = 1 d(n) − 1 < 2 n − 1 < n since n < 2 is the single solution. Theorem 4. The single solution of equation d(ϕ(n)) = n is n = 1. In fact, for all n ≥ 2 one has d(ϕ(n)) < n. (8) p Proof. Let n ≥ 2. Then by (1) and (2) one can write d(ϕ(n)) < 2 ϕ(n) < √ 2 n ≤ n ⇔ n ≥ 4. For n = 2, 3 one has ϕ(2) = 1, ϕ(3) = 2, d(1) = 1 < 2, d(2) = 2 < 3. Remark. For difficulties regarding the equations σ(ϕ(n)) = n and ϕ(σ(n)) = n, see [1], [4]. 352
References [1] R. K. Guy, Unsolved problems in number theory, Third Edition, 2004, Springer Verlag. [2] J. S´andor, On a note by Amarnath Murthy, Octogon Math. Mag. 9(2001), no. 2, 836-838. [3] J. S´andor, Geometric theorems, Diophantine equations, and arithmetic functions, American Research Press, Rehoboth, 2002. [4] J. S´andor, Handbook of number theory, II, Kluwer Acad. Publ., to appear.
16
On d(n) + σ(n) = 2n
In a recent paper, Jason Earls [1] raised some problems and conjectures on abundant and deficient numbers. Particularly, in Problem 17 he denotes the deficiency of n by α(n) = 2n − σ(n), and notes that n = 1, 3, 14, 52, 130, 184, 656, 8648, 12008, 34688, 2118656 are solutions to α(n) = d(n). Here, as usual, d(n) and σ(n) denote the number, resp., sum- of divisors of n. The above equation is clearly equivalent with d(n) + σ(n) = 2n
(1)
Particularly, in [1] it is asked if (1) has infinitely many solutions. Our aim in what follows is to investigate certain properties of solutions of equation (1). Theorem 1. Let p be a prime and α ≥ 1 a positive integer. Then n = pα is a solution to (1) only if p = 2 and α = 1 (i.e. n = 3). Proof. (1) is equivalent to α+1+
pα+1 − 1 = 2pα , p−1
(2)
or after certain elementary transformations to pα (p − 2) = p(α + 1) − (α + 2)
(3)
For p = 2 we get α = 0; contradiction. Thus p ≥ 3. For α = 1 we get p = 3, as a solution to (3). Now, suppose p ≥ 3 and α ≥ 2. Then, by induction upon α it follows immediately that pα > p(α + 1) − (α + 2), 353
(4)
contradicting (3), since p − 2 ≥ 1. Indeed, for α = 2 one gets p2 > 3p − 4, i.e. p(p − 3) > −4, which is trivial. Now, assuming (4), one can write pα+1 > p[p(α + 1) − (α + 2)] > p(α + 2) − (α + 3) iff p2 (α + 1) > 2p(α + 2) − (α + 3), i.e. p[p(α+1)−2(α+2)] > −(α+3). Here p(α+1)−2(α+2) ≥ 3(α+1)−2(α+2) = α − 1 ≥ 1, and all is clear, since −(α + 3) < 0. Theorem 2. Let p be an odd prime. Then n = 2k p is a solution to (1) only if p = 2k+1 + 2k + 1. Proof. Since d(n) = 2(k+1), σ(n) = (2k+1 −1)(p+1), an easy computation yields that (1) is equivalent to 2k+1 + 2k + 2 − p − 1 = 0, i.e. p = 2k+1 + 2k + 1. Remarks. Therefore, for primes of the form p = 2k+1 + 2k + 1
(4)
we get a solution. For k = 1, p = 7 = prime, so n = 14. For k = 2, p = 13 = prime, so n = 52; for k = 3, p = 23 = prime, so n = 184; for k = 4, p = 41 = prime, so n = 656; for k = 7, p = 271 = prime, so n = 27 · 271 = 34688; for k = 10, p = 2069 = prime, so n = 210 · 2069 = 2118656. For k = 13 we get the prime p = 214 + 27 = 16411, which gives the number n = 213 · 1641 = 134448912, not listed by Earls. It seems very likely that there are infinitely many primes of the form (4). If this is true, then clearly an infinite number of solutions to (1) will be provided. Theorem 3. Let p < q be odd primes. Then n = 2k · p · q is a solution to (1) only if (p + q)(2k+1 − 1) + 2k+1 + 4k + 3 = p · q (5) There are no solutions for p = 3. The only solution for p = 5 is n = 2 · 5 · 13 = 130. We cannot have k = 2. For k = 3 the only solutions are p = 19, q = 79 and p = 23, q = 47. Proof. d(n) = 4(k + 1), σ(n) = (2k+1 − 1)(p + 1)(q + 1), and after some elementary computations, (1) can be transformed into (5). For p = 3 remark, that since 2k+1 − 1 ≥ 3, the left side of (5) becomes > 3q, while the right side is 3q; a contradiction. For p = 5 we cannot have k ≥ 2, since then 2k+1 −1 ≥ 7, so the left side of (5) will be greater than the right side. For k = 1, however we can obtain the solution q = 13, yielding n = 130. For k = 2 we get 7(p + q) + 19 = pq, i.e. (p − 7)(q − 7) = 68 (6) Here p − 7 and q − 7 are both even, and (68) is impossible, since 68 = 2 · 34, where p = 9 is not a prime. For k = 3, however we get the equation 15(p + q) + 31 = pq, i.e. (p − 15)(q − 15) = 256 354
(7)
Here p − 15, q − 15 are even, and writing 256 as 256 = 2 · 128 = 4 · 64 = 8 · 32 = 16 · 16, we get the only solutions p − 15 = 4, q − 15 = 64; p − 15 = 8, q − 15 = 32; i.e. p = 19, q = 79, and p = 23, q = 47 (p − 15 = 2, q − 15 = 128 do not give solution, since q = 128 + 15 = 143 = 11 · 13 is not prime). Remarks. Thus for k = 3 the only solutions are n = 23 · 23 · 47 = 8648 and n = 23 · 19 · 79 = 12008, which are listed by J. Earls in [1]. Relation (5) can be written also as (p − 2k+1 + 1)(q − 2k+1 + 1) = 22k+2 − 2k+1 + 4k + 4
(8)
Thus for k = 4 one obtains: (p − 31)(q − 31) = 1012
(9)
Since 1012 = 2 · 506 = 2 · 2 · 253, we have the only possibility p − 31 = 2, q−31 = 506, which do not give a solution, since p = 33 is not prime! Therefore, we cannot have k = 4. Similarly for k = 5 one has (p − 63)(q − 63) = 4056,
(10)
and since 4056 = 2 · 2028 = 4 · 1014 = 8 · 507, we cannot have again solution by p = 63 + 4 = 67, q = 63 + 1014 = 1077, which is not prime.
References [1] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250.
355
356
Chapter 9
Arithmetic functions ”... There still remain three studies suitable for free man. Arithmetic is one of them.” (Plato)
”... The story was told that the young Dirichlet had as a constant companion all his travels, like a devout man with his prayer book, an old, worn copy of the Disquisitiones Arithmetica of Gauss.” (H. Tietze)
357
1
The non-Lipschitz property of certain arithmetic functions
Let d, σ, ϕ, π be the usual arithmetic functions denoting respectively the number of divisors, sum of divisors, Euler’s totient and the counting function of primes, respectively. We note here that none of these functions has the Lipschitz property, i.e. |f (x) − f (y)| ≤ L|x − y| (L > 0 constant), x, y ∈ N.
(1)
First remark that by Dirichlet’s theorem, for all m ≥ 1 there are infinitely many primes n of the form n = k · 2m − 1. Now d(n + 1) − d(n) = d(k2m ) − 2 ≥ d(2m ) − 2 = m + 1 − 2 = m − 1, so for m → ∞ we get d(n + 1) − d(n) → ∞. Therefore lim sup[d(n + 1) − d(n)] = +∞ n→∞
(2)
which means that d cannot have the Lipschitz property (1). We have used also the property d(ab) ≥ d(a). For the function σ remark that: 1 1 − σ(n!) − 1 σ(n!) σ(n!) − σ(1) σ(n!) , = = 1 n! − 1 n! − 1 n! 1− n! and by
σ(n!) X 1 1 1 = ≥ 1 + + ··· + n! d 2 n d|n!
one has
σ(n!) → ∞, n → ∞, giving n! lim
n→∞
σ(n!) − σ(1) = +∞. n! − 1
(3)
This contradicts (1) for x = n!, y = 1, f = σ. For the function ϕ it is well-known that (see e.g. [1]) lim sup[ϕ(n + 1) − ϕ(n)] = +∞ n→∞
so ϕ doesn’t have as well, the Lipschitz property. 358
(4)
Finally, since π(m) > m log m and π(m) < m log m − m log log m for sufficiently large m, one has π(3n) − π(2n) 3n log 3n − 2n log 2n + 2n log log 2n > → ∞ as n → ∞, n n so
· lim
n→∞
¸ π(3n) − π(2n) = +∞. 3n − 2n
(5)
Thus π cannot have the property (1).
References [1] J. S´andor, D.S. Mitrinovi´c (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1995.
2
On certain open problems considered by Murthy In a recent note A. Murthy [1] has considered the indices Id = Id (n) =
n , d(n)
IS = IS (n) =
n , S(n)
Iσ = Iσ (n) =
n , σ(n)
Iϕ = Iϕ (n) =
n , ϕ(n)
where d(n), S(n), ϕ(n), σ(n) are respectively the number of divisors of n, the Smarandache function, the Euler totient, and the sum of divisors of n. He conjectures that the functions Id , IS , Iϕ , Iσ : N∗ → N∗ are all surjective functions. For example, in case of Iσ this means that every positive integer k is the abundancy index of some positive integer µn, i.e. ¶ there exists n ≥ 1 with σ(n) σ(n) = k. It is well-known that the sequence is dense in (1, ∞), n n n≥2 so that the above question is natural even in the weaker form: is every rational number k the abundancy index Iσ (n) of some integer n? We note that for a fixed k, the solutions of (1) are called k-perfect numbers; the numbers n with n(σ(n)) (i.e. k ∈ N) are called multiperfect numbers. These are extremely difficult problems, and only partial solutions are known, see e.g. R. Laatsch [2] for a presentation of the abundancy index. (See also the recent book by the author [7].) H.J. Kanold showed first that the set of all integers n with n|σ(n) has asymptotic density zero. For similar results on the set of integers with d(n)|n, 359
see [5]. For example, in 1985 C.A. Spiro proved that if S(x) = card{n ≤ x : d(n)|n}, then S(x) = (1 + θ(1))x(log x)−1/2 (log log x)−1
(2)
which improves S(x)/x → 0 as x → ∞. P. Erd¨os [6] proved that for n 6= 3, 5 one has d(n!)|n! and this yields a weaker version of (2). We now show that Murthy’s conjecture for the surjectivity of Iϕ is not true. Thus the proposition: for all k one can find n such that n =k ϕ(n)
(3)
is not true! Writing (3) as n = ϕ(n)k, and remarking that for n ≥ 3, ϕ(n) is even, this equality implies that for any k, n ≥ 3 is even. When n = 1, then clearly k = 1; for n = 2 one has k = 2. Let n = 2m N (N odd) be even. Then (3) implies 2m N = 2m−1 ϕ(N )k, i.e. 2N = ϕ(N )k. (4) If N = 1, this is possible, namely for k = 2 (i.e. n = 2m are the solution of (3)); let N ≥ 3, and suppose k is even. Then the left side of (4) is divisible only by 2, but the right side by 22 , a contradiction. But the real surprise is that one can prove that all integer solutions (n, k) or (3) are given by n = 2a , a ≥ 0, n = 2a 3b , a, b ≥ 1, when k ∈ {1, 2, 3}. Thus Iϕ (N∗ ) ∩ N∗ = {1, 2, 3}.
(5)
Result (5) is essentially due to W. Sierpinski [3]. (j) For the complete solutions of ϕn |n (ϕ(j) is the j-th iteration) see M. Hausman [4]. For example, when j = 2, all solutions are given by n = 1, 2a , 3, 2a · 3b , 2a · 5, 2a · 7,
a, b ≥ 1
(6)
so one gets easily that Iϕ◦ϕ is not surjective. We now prove that Murthy’s conjecture on the surjectivity of IS , where S is the Smarandache function, is correct. Let k be a given positive integer, and p > k a prime. Remark that S(kp) = p, since p! = 1 · 2 · 3 . . . k . . . p is the least kp kp n factorial such that kp|p!. Now, by = = k one has = k with S(kp) p S(n) n = kp. The fact that Id is not surjective (e.g. 18 6∈ Id (N∗ )) has been proved recently by M. Lee, and independently, by J. S´andor (see [8]). 360
References [1] A. Murthy, A conjecture on d(n) and the divisor function itself as divisor with required justification, Octogon Math. Mag., 11(2003), no.2, 647-650. [2] R. Laatsch, Measuring the abundancy of integers, Math. Mag., 59(1986), no.2, 84-92. [3] W. Sierpinski, Elementary theory of numbers, Warsaw, 1964. [4] M. Hausman, The solution of a special arithmetic equation, Canad. Math. Bull. 25(1982), 114-117. [5] J. S´andor, D.S. Mitrinovi´c (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1996. [6] P. Erd¨os, Problem 3 of M. Schweitzer Math. Competition, Mat. Lapok (Budapest), 25(1974), no.3-4(1976), 353-357. [7] J. S´andor (in coop. with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005. [8] J. S´andor, On certain open problems considered by A. Murthy, Octogon Math. Mag., 13(2005), no.1B, 894-896.
3
On an inequality of Moree on d(n)
Let d(n) denote the number of all distinct divisors of n. Then it is easy to see that d(nm) ≥ max{d(n), d(m)} (1) for all n, m ∈ N. This relation has been applied by us in certain problems related to a factorial of a number ([2]). Inequality (1) may be defined as follows: d(nm) ≥ d(n) + d(m) − 1
(2)
for all m, n ∈ N. This is due to P. Moree [1]. Clearly, (2) improves (1) since d(m) − 1 ≥ 0, d(n) − 1 ≥ 0. On the other hand, (2) may be even refined, as for all m > 1, n > 1 one has d(nm) > d(n) + d(m).
(3)
Therefore, for such values of n, m one has in fact d(nm) ≥ d(n) + d(m) + 1. 361
(4)
Let k ≥ 0 be a nonnegative integer. Let σk (n) be the sum of kth powers of divisors of n, i.e. X σk = dk . d|n
We shall extend Moree’s inequality as follows: σk (mn) ≥ σk (m) + σk (n) − 1
(5)
for all m ≥ 1, n ≥ 1, k ≥ 0. We note that for k = 0 this gives relation (2). First remark that if at least one of m, n is = 1, one has equality in (5). So we may suppose m, n > 1. Then the following stronger result will be true: σk (mn) > σk (m) + σk (n) for all m, n > 1.
(6)
If (m, n) = 1 this is trivial, since by the multiplicativity of σk one can write σk (mn) = σk (m)σk (n). But ab > a + b for a, b ≥ 2 since (a − 1)(b − 1) > 1 implies ab > a + b. Let us suppose that (m, n) > 1; let Y Y Y 0Y m= px qy , n = px rz , where (p, q) = (p, q) = (q, r) = 1 and x, x0 , y, z are positive integers (y or z may be zero). We do not use indices for simplicity. Now, ³Y ´ ³Y ´ ³Y ´ 0 σk (mn) = σk px+x σk q y σk rz ³Y ´ ³ ³Y ´ ³Y ´´ 0 ≥ σk px+x σk q y + σk rz ³Y ´ ³Y ´ ³Y ´ 0 0 = σk px+x σk (q y ) + σk px+x σk rz ³Y ´ ³Y ´ Y ³Y 0 ´ ³Y ´ Y 0 > pkx σk px σk qy + pkx σk px σk rz ³Y ´ ³Y ´ ³Y 0 ´ ³Y ´ ≥ σk px σk q y + σk px σk rz = σk (m) + σk (n). have ´used the fact that (6) is valid for (m, n) = 1 (in case of ³YWe Y y q , rz = 1) and the following known fact: σk (AB) ≥ Ak σk (B) for all A, B ≥ 1, k ≥ 0 (with strict inequality for A, B > 1). Corollary. For all m, n > 1 one has σ(mn) > σ(m) + σ(n)
(7)
d(mn) > d(m) + d(n).
(8)
and
Proof. Select k = 1, respectively k = 0 in relation (6). 362
References andor, dated 29th April 2003. [1] P. Moree, e-mail to J. S´ [2] J. S´andor, On values of arithmetical functions at factorials, I, Smarandache Notions J., 10(1999), 87-94.
4
On duals of the Smarandache simple function 1. The Smarandache simple function is defined by Sp (n) = min{k ≥ 1 : pn |k!}
(1)
where p is a fixed prime number. The multiplicative ”dual” of this function has been defined by us (see [1]) as: Sp∗ (n) = max{k ≥ 1 : k!|pn }.
(2)
The additive variants of these functions are Sp (x) = min{k ≥ 1 : px ≤ k!}
(3)
Sp∗ (n) = max{k ≥ 1 : k! ≤ pn }
(4)
where p > 1 is a fixed real number, and x > 0 is a real number. These functions have been studied in [1]. For example, we have proved that:
and that the series
log Sp∗ (x) ∼ log x as x → ∞
(5)
µ ¶ ∞ X 1 log log n α n log Sp∗ (n)
(6)
n=1
is convergent for α > 1, and divergent for α ≤ 1. The additive variants of the Smarandache function S(n) have been introduced in [2], and further generalized by C. Adiga and T. Kim [3] and C. Adiga, T. Kim, D.D. Somashekara and A.N. Fathima [4]. 2. The aim of this note is to consider the function Sp∗ (n) given by (2), which (though introduced in [10] has not been studied up to now). First we note that the following simple result is true: Theorem 1. ½ 1, if p ≥ 3 Sp∗ (n) = = L(p). 2, if p = 2 363
Proof. If p = 2, then we cannot have k ≥ 3, since then 3|k!, but 3 - 2n . Clearly kmax = 2, since 2! = 2|2n for all n ≥ 1. For p ≥ 3 remark that we cannot have k ≥ 2, since then 2|k!|pn would imply 2|pn , impossible since pn is odd. Therefore (7) is proved. Remarks. Therefore, Sp∗ (n) is independent of n, i.e. Sp∗ (1) = Sp∗ (2) = · · · = Sp∗ (n) = constant, when p is fixed. The things are different when one considers Sp∗ (n) given by (2) not only for primes p, but arbitrary positive integers a ≥ 1. Sa∗ (n) = max{k ≥ 1 : k!|an }. For example,
½ S6∗ (n) =
3, if n = 1, 2 4, if n ≥ 3
(8)
(9)
Indeed, remark that k!|6 and k!|36 are valid for k ≤ 3 only, while k!|63 = · 33 is true for k ≤ 4. Now, for k ≥ 5, k! ≡ 0 (mod 10), while clearly 6n 6≡ 0 (mod 10). This proves (9). Remark also that when a is a prime-power, a = ps , then Sps ∗ (n) = So∗ (n) = L(p) for any s, with the same proof as (7). Hence the function f (a) = Sa∗ (n) (n = fixed) is a prime-independent function. Also, for a = odd, f (a) = 1 for any n ≥ 1 (since for k ≥ 2, k! is even). Generally, for all a, a similar situation to (9) is true; namely Sa∗ (n) can take a finite number of values. Theorem 2. Let a = pα1 1 pα2 2 . . . pαr r be the prime factorization of a > 1 and suppose p1 < p2 < · · · < pr . Then if p 6∈ {p1 , . . . , pr } is any prime, then 23
Sa∗ (n) ≤ p − 1.
(10)
Proof. Suppose that k = Sa∗ (n) ≥ p. Then p!|k!|an implies that p!|an , so p|an , impossible by the definition of p. Clearly, this is true for any prime p 6∈ {p1 , . . . , pr }. For example, for a = 6 = 2 · 3 one can select p = 5 and (10) gives S6∗ (n) ≤ 4, in concordance with (9). For a = 30 = 2 · 3 · 5 one has S30∗ (n) ≤ 6.
(11)
Remark that for a = odd one can select p = 2, so Sa∗ (n) ≤ 1, which by Sa∗ (n) ≥ 1 gives the relation Sa∗ (n) = 1, i.e. f (a) = 1. Corollary. If (a, 3) = 1, then Sa∗ (n) ≤ 2. 364
(12)
For example, S10∗ (n) = 2 for any n ≥ 1. 3. As we have seen, the function f (a) = Sa∗ (n) for any fixed n ≥ 1, has the property that f (ps ) = f (p) = L(p), for any s ≥ 1, i.e. f is primeindependent. Now, one can ask what are the functions f , if we suppose that f is multiplicative? Theorem 3. Let f be multiplicative, prime-independent, and f (p) = L(p) for primes p (where L is given by (7)), f (1) = 1. Then one has ½ 1, if n is odd f (n) = (13) 2, if n is even Y Proof. Let 1 < n = pα be the prime factorization of n. If n is odd, then clearly Y Y Y f (n) = f (pα ) = f (p) = L(p) = 1 for p ≥ 3. Now, if n is even, n = 2a
Y
pα , then
f (n) = f (2a )
Y
f (pα ) = L(2) · 1 = 2
and (13) is proved. Remark. (13) gives an example of a strongly-multiplicative function (i.e. multiplicative and prime-independent). Now, search for additive functions, which are prime-independent (so, they will be strongly-additive) and which satisfy (7). Theorem 4. Let g be an additive, prime independent function such that g(p) = L(p) for primes p, and let g(1) = 0. Then 1, if n is an odd prime power 2, if n is a power of 2 ω(n), if n is odd, and ω(n) ≥ 2, (14) g(n) = ω(n) + 1, if n is even, but it has at least an odd divisor, where ω(n) denotes the number of distinct prime divisors of n. Proof. Assume g(ab) = g(a)+g(b) for any (a, b) = 1, g(pα ) = g(p) = L(p). Then if nY = pα (p odd), g(n) = L(p) = 1 for p ≥ 3. If n = 2α , clearly g(n) = 2. Let n = pα with π(n) ≥ 2, p ≥ 3. Then g(n) =
X
g(pα ) =
X
g(p) = 365
X
L(p) =
X
1 = ω(n),
if all p is odd. If n = 2a
Y
pα , then
g(n) = g(2a ) + = L(2) +
X
X
g(pα ) = g(2) +
X
g(p)
L(p) = 2(ω(n) − 1) = ω(n) + 1.
Together with g(1) = 0, (14) gives an example for a strongly-additive function.
References [1] J. S´andor, On additive analogues of certain arithmetic functions, Smarandache Notions J., 14(2004), 128-133. [2] J. S´andor, On an additive analogues of the function S, Notes Number Theory Discr. Math., 7(2001), 91-95. andor’s function, Proc. Jang[3] C. Adiga, T. Kim, On a generalization of S´ jeon Math. Soc., 5(2002), 121-124. [4] C. Adiga, T. Kim, D.D. Somashekara, A.N. Fathima, On a q-analogue of S´ andor’s functions, JIPAM, 4(2003), no.5, article 84.
5
A modification of the Smarandache function
1. For a given function f : N∗ → N∗ and a given set A ⊂ N∗ , in papers [1], [2] we have introduced the arithmetical function FfA (n) = min{k ∈ A : n|f (k)}
(1)
if this is well defined. We have also considered the ”dual” function GA g (n) = max{k ∈ A : f (k)|n},
(2)
if this is again, well defined. In various papers (see [3] and [4]), which have been recently published, or are under publication, we have studied certain properties of these functions for A = N ∗ and f (k) = g(k) = k!,
f (k) = g(k) = ϕ(k),
f (k) = g(k) = σ(k), f (k) = g(k) = S(k),
f (k) = d(k),
f (k) = g(k) = T (k),
366
etc., where ϕ, σ, d, S, T represents Euler’s totient, the sum of divisors, number of divisors, Smarandache’s function, and the product of divisors functions, respectively. For additive analogs, see e.g. [5], [6], [7]. 2. The aim of this note is the introductions and preliminary study of a particular case of (1) when A = N ∗ and f (k) = k! + (k − 1)! = (k − 1)!(k + 1). Here 0! = 1. Let us denote this function by F (n) = min{k ≥ 1 : n|(k − 1)!(k + 1)}.
(3)
This seems to be closely analogous to the Smarandache function S(n) = min{k ≥ 1 : n|k!},
(4)
and we shall call it as ”a modification of the Smarandache function”. 3. In spite of the similarity between (3) and (4), the two functions S(n) and F (n) have quite distinct properties. For example, it is well-known, that S(p) = p for all primes p. For the function F (p) we have: Theorem 1. F (p) = p − 1 for all primes p.
(5)
(6)
Proof. Clearly, p|(p − 2)!p for all primes p, so F (p) ≤ p − 1. Incidentally, for general n ≥ 2 we have n|(n − 2)!n, so by (3) we have: F (n) ≤ n − 1 for all n ≥ 2.
(7)
On the other hand, if k ≤ p − 2, then p - (k − 1)!(k + 1), since k − 1 ≤ p − 3 and k + 1 ≤ p − 1, so the prime factors of (k − 1)! and k + 1 are less than p. This proves relation (6). It is well-known and obvious that: S(k!) = k, for all k.
(8)
Theorem 2. F (k!) = k + 1 for all k ≥ 3,
367
F (1!) = F (2!) = 1.
(9)
Proof. Clearly F (1) = F (2) = 1 since 1|0! · 2, 2|0! · 2, where 0! = 1 by the known convention. Let now k ≥ 3. Clearly k!|k!(k + 2), so F (k!) ≤ k + 1. Let us suppose that there is an m ≤ k with k!|(m − 1)!(m + 1). Since m ≤ k, so m − 1 ≤ k − 1, thus (k − 1)! = (n − 1)!l, l ≥ 1, and on the other hand, one has (m − 1)!|(m + 1) = k!A. Since k! = (k − 1)k, one gets m + 1 = klA. Thus m + 1 ≥ k and by m + 1 ≤ k + 1, we can have only m + 1 = k or m + 1 = k + 1. Since (k, k + 1) = 1, this last possibility cannot hold. But for m + 1 = k we get l = A = 1, so k − 1 = m − 1, i.e. k = m, impossible by m!|(m − 1)!(m + 1). This proves relation (9). Corollary 1. For infinitely many n one has S(n) < F (n).
(10)
Corollary 2. For infinitely many m one has S(m) > F (m).
(11)
Proof. Put n = p in (11). Then S(p) = p > p − 1 = F (p) by (5) and (6). Now, let n = k!, k ≥ 3 in (10). Then S(k!) = k < k + 1 = F (k!) by (8) and (9). Remark. Since F (n) ≥ 1, by (7) we get the limit p lim n F (n) = 1. (12) n→∞
References [1] J. S´andor, On certain generalizations of the Smarandache functions, Notes Numb. Th. Discr. Math., 5(1999), no.2, 41-51. [2] J. S´andor, On certain generalizations of the Smarandache functions, Smarandache Notions J., 11(2000), 202-212. [3] J. S´andor, The product of divisors minimum and maximum functions, RGMIA Research Report collection, 7(2004), no.2, art.18, 11. [4] J. S´andor, A note on the divisor minimum function, Octogon Math. Mag., 12(2004), no.1, 273-275. [5] J. S´andor, On an additive analogue of the function S, Notes Number Theory Discr. Math., 7(2001), 91-95. 368
[6] J. S´andor, On additive analogues of the function S, Smarandache Notions J., 14(2004), 128-133. [7] J. S´andor, An additive analogue of the Euler minimum function, Adv. Stud. Contemp. Math., 10(2005), no.1, 53-62.
6
The star function of an arithmetic function
Let d(n) be the number of distinct positive divisors of n. Recently, in an interesting note, Murthy and Bencze [1] have considered a ”star function” X d∗ (n) = d(k), k|n
and studied some of its properties, as well as analogous notions. We must note here that the ”star” notation is already a standard notation when considering ”unitary divisors” and associated arithmetic functions³ (see´e.g. [2], [3]). An n integer k is called a unitary divisor of n if k|n and k, = 1. Then the k number of unitary divisors of n is denoted in the literature as X 1. d∗ (n) = n k|n,(k, k )=1 Therefore, when using this ”star” notation, we always must note that this is not the unitary divisor theory notation. The star function of an arbitrary function f : N → R can be defined similarly as X f ∗ (n) = f (k). k|n
Thus we get: Theorem 1. If f is multiplicative (i.e. f (mn) = f (m)f (n), for all (m, n) = 1 and f (n) 6= 0), then f ∗ is multiplicative, too, and for the prime factorization n = pα1 1 . . . pαr r one has r Y f (n) = (1 + f (pi ) + · · · + f (pαi i )) ∗
(1)
i=1
Particularly, d∗ (n) =
r Y (α + 1)(α + 2) Y (α1 + 1)(α1 + 2) = . 2 2 α i=1
p kn
369
(2)
Proof. The first part is a classical result, see e.g. [4]. Clearly f (1) = 1. Since d(1) + d(p) + · · · + d(pα ) = 1 + 2 + · · · + (α + 1) =
(α + 1)(α + 2) , 2
the second part also follows. Remark 1. Let ϕ be Euler’s totient. By Gauss theorem (which easily follows from (1), too) ϕ∗ (n) = n, so the star function of Euler’s totient is the identity function. If σ denotes the sum of divisors function, then, by (1) σ ∗ (n) =
r Y [1 + (pi + 1) + · · · + (pαi i + · · · + 1)]. i=1
This must not be confused with the sum of unitary divisors function, for which r Y σ ∗ (n) = (1 + pαi i ) i=1
(see e.g. [3], [5]). Theorem 2. Let ω(n), resp. Ω(n) denote the number of distinct, respectively total-number of prime factors of n. Then 1 · ¸ i · d∗ (n) ¸ ω(n) 1 1h 1 Ω(n) 3 ω(n) ≤ ≤ 2+ . (3) 1 + d(n) ≤ 2 2 d(n) 2 ω(n) Proof. By (2) one can write d∗ (n) =
d(n) (α1 + 2) . . . (αr + 2). 2ω(n)
Now, by the arithmetic-geometric inequality one has µ ¶ µ ¶ α1 + 2 + · · · + αr + 2 r Ω(n) ω(n) (α1 + 2) . . . (αr + 2) ≤ = 2+ , r ω(n) since ω(n) = r and Ω(n) = α1 + · · · + αr . This gives the right side of (3). For the second part of left side apply Chrystal’s inequality p √ r (1 + x1 ) . . . (1 + xr ) ≥ r x1 . . . xr + 1 with xi = αi + 1, i = 1, r. Since d(n) = (α1 + 1) . . . (αr + 1) ≥ 2r , 370
the first inequality of (3) follows. Corollary 1. The normal order of magnitude of the arithmetical function µ F =
d∗ d
¶1
ω
3 . 2 Proof. This follows by (3) and the classical result due to Hardy and RaΩ(n) manujan that the normal order of magnitude of is 1 (see e.g. [6]). ω(n) Theorem 3. Let d∗∗ be the star function of d∗ . Then is
d∗∗ (n) ≤ d∗ (n)d(n). Proof. FirstY we note the property m|n ⇒ d∗ (m) ≤ d∗ (n). Indeed if Y n= pa , m = pb with b ≤ a, then d∗ (m) =
Y (b + 1)(b + 2) 2
≤
Y (a + 1)(a + 2) 2
= d∗ (n).
Now, by definition, X X X d∗∗ (n) = d∗ (k) ≤ d∗ (n) = d∗ (n) 1 = d∗ (n)d(n) k|n
k|n
k|n
by the above proved property. Remark 2. The notation d∗∗ (n) should not be confused with the number of ”bi-unitary” divisors of n (see [4], [6]), with the same notation. Theorem 4. σ ∗ (n) ≤ σ(n)d(n). Proof. Similarly as above m|n ⇒ σ(n) ≤ σ(n), so X σ ∗ (n) = σ(k) ≤ σ(n)d(n). k|n
References [1] A. Murthy, M. Bencze, Extending the scope of some number theoretic functions, Octogon Math. Mag., 11(2003), no.1, 110-113. [2] E. Cohen, The number of unitary divisors of an integer, Amer. Math. Monthly, 67(1960), 879-880. 371
[3] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z., 74(1960), 66-80. [4] J.M. DeKoninck, A. Ivi´c, Topics in arithmetical functions, North Holland Math. Studies, 72(1980). [5] J. S´andor, L. T´oth, On certain number-theoretic inequalities, Fib. Quart., 28(1990), 255-258. [6] J. S´andor, D.S. Mitrinovi´c, Handbook of number theory, Kluwer Acad. Publ., 1995. [7] J. S´andor, On the arithmetical functions dk (n) and d∗k (n), Portugaliae Math., 53(1996), 107-115.
7
On Jordan’s arithmetical function
Prelimiaries Jordan’s arithmetical function is X a generalization of Euler’s totient func1 = the number of all k-tuples tion ϕ. By definition, ϕk (n) = (a1 ,...,ak ,n)=1 1≤a1 ,...,ak ≤n
(a1 , . . . , ak ) with all components between 1 and n such that (a1 , . . . , ak , n) = 1. We note that this function has some applications in the theory of linear groups. For k = 1 we have ϕ1 = ϕ ([4]). Furthermore, let σk (n) be the sum of kth powers of divisiors of n, i.e. X k σk (n) = d . For k = 1 and k = 0 we reobtain the well-known arithmetical d|n
functions σ1 (n) = σ(n), the sum of divisors of n, and σ0 (n) = d(n), the number of divisors of n. The aim of this note is to prove some interesting relations for the above mentioned arithmetical functions. First we need some lemmas. Lemma 1. X ϕk (d) = nk d|n
Proof. Let us consider the set S = {(a1 , . . . , ak ) : a1 , . . . , ak ∈ N∗ , 1 ≤ a1 , . . . , ak ≤ n} =
[ d|n
where (a1 , . . . , ak , n) = d iff d|n. 372
Sd ,
³ n´ n From a1 = b1 d, . . . , ak = bk d, b1 , . . . , bk , = 1 we get b1 ≤ , . . . , bk ≤ d d ³n´ n and so Sd has ϕk distinct elements, the set S trivially has a number of d d k n elements, hence we obtain X ³n´ = nk . ϕk d d|n
n Since |n iff d|n, we have the result. d Lemma 2. X µ(d) ϕk (n) = nk dk d|n
where µ denotes the M¨ obius arithmetical function. Proof. Apply Lemma 1 and the M¨obius inversion formula ([2], [3]). ϕr (m) ϕr (n) Lemma 3. If m|n (m divides n) then ≥ . r m nr Proof. It is immediate from the definition (and Lemma 2) that ϕr (ab) ≤ (ϕr (b))a, so writing n = qm, and applying this inequality the result follows. Theorem 1.³ ´ X n ϕk (i)d a) = σk (n), n ≥ 1 i i|n ³n´ X b) ϕk (i)σk = nk d(n), n ≥ 1. i i|n
Proof. It is easy to see that ϕk and σk , d are multiplicative functions and by a known result ([2], [3]) the left sides in a) and b) are also multiplicative. Thus it is sufficient to prove these relations for n = pa (prime powers). However we shall use another argument, which is based on Dirichlet series. Denote D(f, s) =
∞ X f (n) n=1
ns
the Dirichlet series of the arithmetical function X f . It ³isn ´well-known that D(f ∗ g, s) = D(f, s)D(g, s), where (f ∗ g)(n) = f (i)g is the Dirichlet i i|n
product of f and g, by supposing that the considered series are absolute convergent ([3]). Let Ek be the function Ek (n) = nk and U (n) ≡ 1 the identity function. Then evidently, ³n´ X σk (n) = Ek (i)U . d i|n
373
Since D(Ek , s) =
∞ X n=1
1 ns−k
= ζ(s − k) (which is true for Re s > k + 1) and
D(U, s) = ζ(s), the Riemann zeta function, hence we have D(σk , s) = ζ(s − k)ζ(s). For k = 0 one has D(d, s) = ζ 2 (s). Similarly, from Lemma 2, we obtain D(ϕk , s) =
ζ(s − k) . ζ(s)
We now consider the identity ζ(s − k) 2 ζ (s) = ζ(s − k)ζ(s) ζ(s) or D(ϕk , s)D(d, s) = D(σk , s), i.e. D(ϕk ∗ d, s) = D(σk , s). But it is well-known by the uniqueness theorem of Dirichlet series ([8]) that this implies ϕk ∗ d = σk , which is exactly a). In order to prove the second relation, we may consider the identity ζ(s − k) ζ(s − k)ζ(s) = ζ 2 (s − k) ζ(s) or D(ϕk ∗ σk , s) = D(nk d, s), yielding relation b). Next we shall prove: Theorem 2. ∞ X n=1
∞
X xn xn nk for |x| < 1 = ϕk (n) 1 − xn n=1
Proof. We need the following result: X If g(n) = f (d), then d|n ∞ X n=1
n
g(n)x =
∞ X i=1
374
f (i)
xi , 1 − xi
if the involved series are convergent (see [5]). To prove this, let us observe that ∞ ∞ ∞ X ∞ X X X X f (i)xij g(n)xn = f (i) xn = n=1
n=1
=
∞ X
i=1 j=1
i|n
f (i)
i=1
xi , for |x| < 1. 1 − xi
Apply now this theorem for f (i) = ϕk (i) and use Lemma 1. One obtains ∞
∞ X
ϕk (i)
X xi = nk xn , i 1−x n=1
i=1
which for k = 1 gives ∞ X
∞
ϕ(i)
X xi x = nxx = . 1−x (1 − x)2 n=1
i=1
By differentiation we have ∞ X
n2 xn =
n=1
hence
∞ X
ϕ1 (x)
i=1
x(1 + x) , (1 − x)3
xi x(1 + x) = . i 1−x (1 − x)3
Remark. The above proved identities imply the interesting relations µ ¶ 1 x= : 2
∞ X ϕ(n) =2 2n − 1
and
n=1
∞ X ϕ1 (n) = 6. 2n − 1
n=1
We conclude with a result on the composite function ϕ(σk (n)). Theorem 3. Let k be an odd natural number. Then we have lim inf n→∞
ϕ(σk (n)) = 0. n
Proof. Let p be a prime number of the form p ≡ −1 (mod p1 . . . ps ), where pi denotes the ith prime number. (By the well-known Dirichlet theorem 375
on arithmetical progressions, there exist such primes ([2], [3]). Then σk (p) = pk + 1 ≡ 0 (mod p1 . . . ps ) and Lemma 3 (r = 1) implies the inequality ¶ s µ ϕ(σk (p)) ϕ(p1 . . . ps ) Y 1 . ≤ = 1− p p1 . . . ps pi i=1
This last product tends to 0 as s → ∞ (see [2], [3]), so the theorem is proved. Remark. For k = 1 Theorem 3 gives a result of L. Alaoglu and P. Erd¨os [1]. See also [6], [7]. Our argument is completely different. By more complicated argument we can prove that lim inf ϕ((σk (n)) log log log n)/n < ∞. n→∞
References [1] L. Alaoglu, P. Erd¨os, A conjecture in elementary theory of numbers, Bull. Amer. Math. Soc., 50(1944), 881-882. [2] I. Creang˘a ¸si colectiv, Introducere ˆın teoria numerelor, Ed. Did. Ped., Bucure¸sti, 1965. [3] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford, 1938. [4] C. Jordan, Trait´e de substitutions et des ´equations alg´ebriques, Paris, 1957. [5] G. P´olya, G. Szeg¨o, Aufgaben und Lehrs¨ atze aus der Analysis, Springer Verlag, 1924. [6] J. S´andor, On the arithmetical functions ϕk and σk , Math. Student, 58(1990), 49-54. [7] J. S´andor, On Euler’s arithmetical function, Proc. VIIth National Conf. on Algebra, Bra¸sov, Romania, 1988. [8] E.C. Titchmarsch, The theory of functions, Oxford Univ. Press, 2nd ed., 1978, Theorem 9.6. 376
8
Generalization of a theorem of Lucas on Euler’s totient
Let ϕ be the Euler totient function. In 1845 E. Prouchet [1] proved that for all m, n ≥ 1 one has ϕ(mn) = ϕ(m)ϕ(n)
d , ϕ(d)
(1)
where d = (m, n) = gcd of m and n. Clearly (1) implies the multiplicative property of ϕ, namely ϕ(mn) = ϕ(m)ϕ(n) (2) if (m, n) = 1 and reciprocally, (2) holds only if (m, n) = 1 (since (1) implies ϕ(d) = d, which is true for d = 1). In 1891 E. Lucas [2] proved that for all m, n ≥ 1 ϕ(mn) = (m, n)ϕ([m, n])
(3)
where [m, n] = lcm of m and n. Let f : N∗ → R be an arithmetical function, such that f (n) 6= 0 for all n ≥ 1. Then that relation ϕ(mn) = (m, n)f (|m, n|)
(4)
is satisfied by the following two functions: f1 (n) = n, n ≥ 1,
f2 (n) = ϕ(n).
Indeed, (4) for f2 is exactly Lucas’ theorem (4), while for f1 , (4) is exactly the known relation mn = (m, n)[m, n],
(m, n ≥ 1).
(5)
In 1950 H.N. Shapiro [3], as an extension of Prouchet’s theorem (1) introduced the so-called over-multiplicative functions as follows: f : N∗ → R is called over-multiplicative, if there exists a function g : N∗ → R, g(n) 6= 0 for all n, such that f (mn) = f (m)f (n)g(d), m, n ≥ 1 (6) where d = (m, n). Now, we shall prove the following generalization of the Lucas theorem: 377
Theorem 1. If f is over-multiplicative, then there exists g : N∗ → R, g(n) 6= 0 for all n ≥ 1 such that f (mn) = f (d)g(d)f [m, n],
m, n ≥ 1.
(7)
Proof. Let m = dm0 , n = dn0 , where d = (m, n), so (m0 , n0 ) = 1. Then since (d, dm0 n0 ) = d, by (6) one can write f (d2 m0 n0 ) = f (d)f (dm0 n0 )g(d). But mn = d2 m0 n0 and [m, n] = dm0 n0 by (5). Thus (7) follows. Remarks. If f is over-multiplicative and g(1) = 1, then f is multiplicative. Indeed, letting d = 1 in (6), it follows: f (mn) = f (m)f (m) for (m, n) = 1
(8)
i.e. f is a multiplicative function. Generally speaking, (6) implies only that f is quasi-multiplicative, i.e. f (mn) = kf (m)f (n) for (m, n) = 1
(9)
where k = g(1). Now, relation (7) shows that for over-multiplicative functions f , (4) is true if and only if f (d)g(d) = d, i.e. g(n) =
n , f (n)
n ≥ 1.
(10)
In this case, (6) becomes f (mn) = f (m)f (n)
d f (d)
(11)
generalizing Prouchet’s relation (1). If f (1) = 1, then (11) clearly implies that f is multiplicative, i.e. has the property (8). But not all multiplicative functions satisfy (11)! Indeed, letting m = n in (11) it follows f (m2 ) = mf (m).
(12)
For example, the σ-function (sum of divisors) is multiplicative, but doesn’t satisfy (12). Indeed, e.g. σ(4) = 7 6= 2σ(2) = 6. By induction easily follows that if f (1) = 1, then f (mα ) = mf (mα−1 ) for all m ≥ 1, α ≥ 2. 378
(13)
Now, we prove that, reciprocally, if f is multiplicative and has property (13), then it has also the property (11). Even a stronger result is true: Theorem 2. Suppose that f (1) = 1, f is multiplicative, and f (pα ) = pf (pα−1 ) for all primes Yp≥ Y2, α ≥ 2. Then Y (11) is true. α β α Proof. Let m = p q ,n=p rγ be the canonical factorizations Y Y Y 0 of m and n. Then nm = pα+α qβ rγ , where α+α0 ≥ 2 if (m, n) 6= 1 (if 0
0
(m, n) = 1, then (11) is trivially satisfied). We have f (pα+α ) = pf (pα+α −1 ) = 0 pα+α −1 f (p), since from f (pn ) = pf (pn−1 ) it follows by induction that f (pn ) = pn−1 f (p). Similarly, for all α0 ≥ 2, β0 ≥ 1, γ0 ≥ 2, f (pα0 ) = pα0 −1 f (p), fY (q β0 ) =Yq β0 −1 f (q), f (rγ0 ) = rγ0 −1 f (r). When all α, β = 1, then f (m) = f (p) f (q), f being multiplicative. Therefore, we may assume that all 0 α, β, α , γ ≥ 2. Then f (mn) = f (m)f (n) Y Y Y Y Y Y 0 pα+α −1 q β−1 rγ−1 f (p) f (q) f (r) Y Y 0 Y Y Y Y Y =Y pα−1 q β−1 pα −1 rγ−1 f (p) f (q) f (p) f (r) =
Y p d = , f (p) f (d)
Y 0 as now d = pmin{α,α } and in case min{α, α0 } ≥ 2 one can again apply f (pn ) = pn−1 f (p).
References [1] E. Prouchet, Nouv. Ann. Math., 4(1845), 75-80. [2] E. Lucas, Th´eorie des nombres I, Gauthier-Villars, Paris, 1891. [3] H.N. Shapiro, On the iterates of certain class of arithmetic functions, Comm. Pure Applied Math., 3(1950), 259-272.
9
Some arithmetic inequalities connected with the divisors of an integer
1. Let 1 = d1 < d2 < · · · < dk = n be the consecutive divisors of n > 1. n n n Then, since di |n we have also |n, and in fact, if di < dj , then > . For di di dj 379
example, let k = 2n (even). Then the divisors in increasing order are 1 = d1 < d2 < · · · < dm < thus dm+1 =
n n n n < < ··· < < =n dm dm−1 d2 d1
(1)
n n n n , dm+2 = , . . . , dk−1 = , dk = d2n = . dm dm−1 d2 d1
In fact, this happens when n is not a square, since it is well-known that if n = pα1 1 . . . pαr r is the prime factorization of n, then k = d(n) = (1 + α1 ) . . . (1 + αr ) and this is even only if all αi are not even, so n is not a square. When n is a square, i.e. n = s2 , then k is odd, k = 2m + 1, and one can write 1 = d1 < d2 < · · · < dm < s < where dm+1 = s, dm+2 =
n n n n < < ··· < < =n dm dm−1 d2 d1
(2)
n n n , . . . , d2m = , d2m+1 = . dm d2 d1
From (1) it is clear that dm ≥ m, so by dm dm+1 = n we have n ≥ m(m + √ 1) > m2 , i.e. m < n. Since k = 2m, one gets √ (3) k = d(n) < 2 n for n 6= square. √ √ In (2) dm+1 = s = n ≥ n + 1, so m ≤ n − 1, implying k = 2m + 1 ≤ √ 2 n − 1, i.e. √ k = d(n) ≤ 2 n − 1 for n = square. (4) We note that actually from m(m + 1) ≤ n one has the slightly stronger relation, improving (3): √ 1 + 4n − 1 k = d(n) ≤ , for n 6= square. (30 ) 2 2. From (1) it follows that dm dm+1 = n, so one can write di di+1 < n for 1 ≤ i ≤ m − 1. Therefore, S = d1 d2 + d2 d3 + · · · + dm−1 dm + dm dm+1 + dm+1 dm+2 + · · · + dk−1 dk =n+
m−1 Xµ i=1
n2 di di+1 + di di+1 380
¶ .
Remark that i(i + 1) ≤ di di+1 < n for 1 ≤ i ≤ m − 1, so S < n + (m − 1)n +
m−1 X i=1
µ ¶ n2 1 2 = mn + n 1 − . i(i + 1) n
Thus S < n2 + mn −
n2 , m
n 6= square.
(5)
n2 Since, by (3) m2 < n, mn − < 0 relation (5) implies the weaker inm equality S < n2 . (6) For n = square (i.e. case (2)) a similar argument gives (6), so the p = 2 case of OQ.1284 [1] is settled. The case p = 3 can be studied in a similar manner, with the difference that e.g. in case (1). S 0 = d1 d2 d3 + d2 d3 d4 + · · · + dm−2 dm−1 dm + dm−1 dm dm+1 +dm dm+1 dm+2 + · · · + dk−2 dk−1 dk µ ¶ µ ¶ m−2 X n3 n2 di di+1 di+2 + = + ndm−1 + , di di+1 di+2 dm−1 i=1
since here dm−1 dm dm+1 = dm−1 dm and dm dm+1 dm+2 = dm
n = ndm−1 ≤ n2 dm
n n n2 · = ≤ n2 . dm dm−1 dm−1
Now di di+1 di+2 ≤ dm−1 dm dm+1 ≤ n2 for 1 ≤ i ≤ m − 2 and di di+1 di+2 ≥ i(i + 1)(i + 2). Using
1 1 1 1 1 1 = · − + · , i(i + 1)(i + 2) 2 i i+1 2 i+2
by addition one gets the well-known identity m−1 X i=1
1 1 1 1 1 = + − < , i(i + 1)(i + 2) 4 2m 2m − 2 4 381
so S 0 < (m − 2)n2 + 2n2 + n3 · i.e.
√ √ 1 n3 3 ≤ n2 n + < n3 ⇔ n2 3 < n3 , 4 4 4
√ √ 3 n < n, i.e. 3 n > 4, which is true for all n ≥ 2, finishing the proof of 4 S 0 < mn2 +
√ n3 n3 ≤ n2 n + < n3 , 4 4
n 6= square
(7)
which improves the case p = 3 of [1]. When n is a square, a similar study applies. I think that the most general case with d1 d2 . . . dp . . . can be performed in the same lines, and I so propose for the patient reader to carry out the calculations. √ We finish with the remark that the inequality d(n) ≤ 2 n sometimes is called as Sierpinski’s inequality [2]. (3), (4), (3’) give improvements of this relation.
References [1] M. Bencze, OQ.1284, Octogon Math. Mag., 11(2003), no.2, 851. [2] W. Sierpinski, Elementary theory of numbers, Warsawa, 1964.
10
A new arithmetic function 1. Let n =
r Y
pai i ≥ 2 be the prime factorization of the positive integer n
i=1
(pi primes, ai ≥ 1). Let us define fk (n) =
¶a r µ k Y p +1 i i
i=1
2
,
k ≥ fixed.
(1)
Let fk (1) = 1 by definition. This arithmetic function appears naturally from a result in [10], where it is proved that ¶a r µ k Y p +1 i i
i=1
2
n
≤
i +1 σk (n) Y pka i ≤ . d(n) 2
i=1
See also [11] for similar relations. 382
(2)
Here σk (n) and d(n) = σ0 (n) stand for the sum of kth powers of divisors of n, and the number of divisors of n, respectively. Let σk∗ (n) and d∗ (n) denote the sum of kth powers of unitary divisors of n, and the number of unitary divisors of n, resp. (see e.g. [2], [6], [9]). Then (2) can be rewritten as fk (n) ≤
σ ∗ (n) σk (n) ≤ ∗k . d(n) d (n)
(3)
There is equality at each side only when n is squarefree. For k = 1 we get the arithmetical function ¶ r µ Y pi + 1 a i f (n) = , r ≥ 1, (4) 2 i=1
with f (1) = 1. Clearly fk (n) and f (n) are examples of multiplicative functions, i.e. they satisfy the functional equation g(nm) = g(n)g(m) for (n, m) = 1. 2. Results on composite functions such as σ(ϕ(n)), d(ϕ(n)), etc., where ϕ(n) is Euler’s totient, are very difficult to obtain. For example, it is conjectured (by Makowski and Schinzel, see e.g. [12] for recent and/or related results) that σ(ϕ(n)) ≥ n/2 (5) for all n. We now prove the following simple result: Theorem 1. For all n ≥ 1 one has f (ϕ(n)) ≤ n/2,
(6)
with equality only for n = 1, 2, 3. Proof. It is known (see [9]) that σk∗ (n) nk + 1 ≤ . d∗ (n) 2
(7)
Now, by (3) and (7) we obtain f (n) ≤
n+1 , implying f (ϕ(n)) ≤ 2
ϕ(n) + 1 n ≤ , since ϕ(n) ≤ n − 1 for all n ≥ 2. An immediate verifica2 2 tion shows that for n = 1, 2, 3 there is equality, and that there are no other such values of n. Remarks. 1) More generally, if an arithmetic function k(n) satisfies n σ(k(n)) ≤ , d(k(n)) 2 383
n ∈ S ⊂ N,
then f (k(n)) ≤ 2) Since for n even ϕ(n) ≤
n for all n ∈ S. 2
n , by the proof of Theorem 1 one can see that 2 n+2 n < for n = even. 4 2
f (ϕ(n)) ≤
(8)
µ
¶ x+1 a xa + 1 ≥ , x > 0, a ≥ 1, applied to x = pki , 2 2a a = ai , after term-by-term multiplication implies 3. The inequality
fk (n) ≥
σk∗ (n) , D∗ (n)
(9)
where D∗ (n) = 2Ω(n) , defined in analogy with d∗ (n) = 2ω(n) , where ω(n), Ω(n) denote the distinct, resp. total number of prime factors of n. Relation (3) and (9) give at once σ ∗ (n) ω(n) log 2 ≤ log k ≤ Ω(n) log 2. (10) fk (n) Theorem 2. The normal order of magnitude of the arithmetical function σ ∗ (n) log k is (log 2) log log n. fk (n) Proof. This follows by the double-inequality (10) and the well-known fact, due to Hardy and Ramanujan, that the normal order of magnitude of ω(n) and Ω(n) is log log n (see e.g. [1], [3]). Theorem 3. σ ∗ (n) log log n lim sup log k ≥ log 2, (11) log n fk (n) n→∞ lim sup m→∞
σ ∗ (n) 1 log k ≤ 1. log n fk (n)
(12)
Proof. This follows by (10) and the known facts lim sup n→∞
ω(n) log log n = 1, log n
lim sup n→∞
Ω(n) log 2 = 1. log n
Theorem 4. Y σ ∗ (n) k = 2x log log x+O(x) as x → ∞, fk (n)
n≤x
384
(13)
X
1 x 1 ∼ · , ∗ σ (n) log 2 log log x 2≤n≤x log k fk (n)
x → ∞.
(14)
Proof. The proof of (13) is based on inequalities (10) and the known facts that (see e.g. [1], [3]) X X ω(n) ∼ x log log x, Ω(n) ∼ x log log x as x → ∞, n≤x
n≤x
while (14) is a consequence of (10) and X 2≤n≤x
1 x ∼ ω(n) log log x
X
and
2≤n≤x
1 x ∼ , Ω(n) log log x
x → ∞,
see e.g. [4]. 4. By (3) and (7) we can write fk (n)/nk ≤ (nk + 1)/2nk , and since for a prime n = p we have fk (p)/pk ≤ (pk + 1)/2pk , clearly lim sup n→∞
1 fk (n) = . 2 nk
(15)
On the other hand, fk (2m )/2mk = (1 + 2−k )m /2m → 0 as m → ∞, thus fk (n) = 0. nk
(16)
σ ∗ (n) σ ∗ (n) 1 1 log k = lim sup log k = log 2 ω(n) fk (n) fk (n) n→∞ Ω(n)
(17)
lim inf n→∞
Theorem 5. lim inf n→∞
σk∗ (p) = 2 for a prime p, and use inequality (10). fk (p) For the function D∗ (n) = 2Ω(n) remark that Proof. Remark that
lim sup log(f (n)D∗ (n)/n) = +∞.
(18)
n→∞
Indeed, by log(f (n)D∗ (n)) =
r X
ai log(pi +1) and by the known inequality
i=1
log(pi + 1) > log pi + 2/(2pi + 1) it follows ∗
log(f (n)D (n)/n) >
r X i=1
r
1X 2ai /(pi + 1) > 1/pi . 2
385
i=1
(19)
r X i=1
If one selects for pi , i = 1, r, the sequence of consecutive primes, then as 1/pi → ∞ as r → ∞, by (19) we get the weaker result (18). Theorem 6. There exists a sequence (nk ) such that log f (nk )D∗ (nk )/nk À log nk .
(20)
Proof. Let nk = 2k , in which case we have µ ¶k µ ¶k µ ¶log nk / log 2 3 1 3 k f (nk )D (nk )/nk = 2 = , 2 2 2 ∗
hence log f (nk )D∗ (nk )/nk = (log nk )
log 3/2 À log nk . log 2
Inequality (19) suggests the introduction of the arithmetic function h(n) =
n X ai i=1
pi
.
(21)
Theorem 7. For all n ≥ 2 one has 4 h(n) < log f (n)D∗ (n)/n < h(n). 5
(22)
p Proof. By the known inequality log(x + 1) < log x + 1/ x(x + 1), we get log f (n)D∗ (n)/n <
r X
p ai / pi (pi + 1).
(23)
i=1
2ai 4 ai ai ai ≥ · and p < , by (23) we get relation (22). 2ai + 1 5 pi pi pi (pi + 1) Corollary. f (n)D∗ (n) lim inf = 0. (24) n→∞ n 1 Proof. h(p) = for a prime p, and apply the right side of Theorem 7. p 5. For the sequence (fk (n))k≥1 we have: Theorem 8. The sequence (fk (n))k≥1 is supermultiplicative; (25) the sequence (fk (n)/nk )k≥1 is strictly decreasing. (26) Since
386
Proof. It is easy to prove that xk+m + 1 xm + 1 ≥ , xk + 1 2
x > 0, m, k ≥ 0
(27)
where the inequality is strict, when at least one of k and m is ≥ 1. Apply (27) to x = pi , in order to obtain fk+n (n) ≥ fk (n)fm (n),
k, m ≥ 1,
which shows that (fk (n))k is supermultiplicative. The algebraic inequality xk+m + 1 ≤ xm , xk + 1
x > 0, m, k ≥ 0
(28)
and relation (1) give at once that fk+m (n) ≤ nm fk (n),
k, m ≥ 1.
(29)
For m = 1 inequality (29) gives fk+1 (n)/nk+1 ≤ fk (n)/nk , so assertion (26) is proved, too. 6. Finally, we consider certain divisibility problems. An important property for multiplicative functions, which is valid for f (n), too, is the following Theorem 9. If m|n (m divides n), then f (m) ≤ f (n). (30) Remark. The values of f (n) are not integers for all n. For example, f (2) = 3 . If n is odd, then all prime factors of n are odd, thus n is an integer. Let 2 n = 29 m, with (m, 2) = 1. Then by multiplicative property of f one has µ ¶9 3 f (n) = f (m), so 2a must divide f (m). Let p1 , . . . , pt , 1 ≤ t ≤ r, be all 2 prime factors of n which are of the form pi = 2mi +1 Mi − 1 (Mi odd, mi ≥ 0, 1 ≤ i ≤ t). Theorem 10. If m =
r Y
pbi i , pi odd, then the necessary and sufficient
i=1
condition for f (n) to be an integer is that r X
bi mi ≥ a,
i=1
where 2a kn. 387
(31)
References [1] T. Apostol, Introduction to analytic number theory, Springer Verlag, 1976. [2] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z., 34(1960), 66-80. [3] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [4] M.J. deKoninck, On a class of arithmetical functions, Duke Math. J., 39(1972), 807-818. [5] A. Makowski, A. Schinzel, On the functions σ(n) and ϕ(n), Colloq. Math., 8(1964), 95-99. [6] J. Morgado, On the arithmetical function σk∗ , Portugal Math., 23(1964), 35-40. [7] C. Pomerance, On the composition of the arithmetic functions σ and ϕ, Colloq. Math., 58(1989), 11-15. [8] J. S´andor, Remarks on the functions σ(n) and ϕ(n), Babe¸s-Bolyai Univ. Preprint Nr.7, 1989, 7-12. [9] J. S´andor, L. T´oth, On certain number-theoretic inequalities, Fib. Quart. 28(1990), 255-258. [10] J. S´andor, An application of the Jensen-Hadamard inequality, Nieuw Arch. Wiskunde, 8(1990), no.4, 63-66. [11] J. S´andor, On the arithmetical functions dk (n) and d∗k (n), Portugal Math., 53(1996), no.1, 107-115. [12] J. S´andor, Handbook of number theory, II, Springer Verlag, 2005.
11
A generalization of Ruzsa’s theorem
Recently [1] we have published a new proof of a theorem of Ruzsa, which states that all even numbers a and b satisfying σ(a) = 2b, σ(b) = 2a (the so-called ”lovely pairs”) are given by a = 2k (2q+1 − 1),
b = 2q (2k+1 − 1),
388
where 2k+1 − 1 and 2q+1 − 1 are both prime numbers. The aim of this note is to show, that the methods of [1] enable us to give the following generalization. Theorem. Let f : N∗ → N∗ be a given multiplicative arithmetic function which satisfies the following properties: 1) f (2k ) is odd; 2) f (mn) ≥ mf (n) for all m, n ≥ 1, with equality only for m = 1; 3) f [f (2k )] ≥ 2k+1 for all k ≥ 1, with equality only for k ∈ A where A ⊂ N∗ . Then all even solutions of the system ½ f (a) = 2b f (b) = 2a in positive integers are given by a = 2k f (2a ), b = 2q f (2k ), where k, q ∈ A. If there is strict inequality in at least one of 2) and 3), then the system doesn’t have even solutions. Proof. Let a = 2k A, b = 2q B, where A and B are odd numbers. Since f is multiplicative, f (a) = f (2k )f (A), f (b) = f (2k )f (B), so f (2k )f (A) = 2q+1 B, f (2q )f (B) = 2k+1 A. By 1) f (2k ) is odd, so f (2k ) must divide B, i.e. B = f (2k )B 0 . Similarly, A = f (2q )A0 , where A0 , B 0 are odd integers. These give f (A) = 2q+1 B 0 , f (B) = 2k+1 A0 . By using relations 2) and 3) one can deduce: 2q+1 B = f (2k )f (A) = f (2k )f [A0 f (2q )] ≥ f (2k )A0 f [f (2q )] ≥ f (2k )A0 2q+1 . Since B = f (2k )B 0 , this gives B 0 ≥ A0 . In a completely analogous way one can obtain A0 ≥ B 0 . Thus A0 = B 0 , so A = f (2q )S, B = f (2k )S. Now, f (A) = 2q+1 S, f (B) = 2k+1 S give f [f (2q S)] = 2q+1 S, f [f (2k )S] = 2k+1 S. But f [f (2q )S] ≥ Sf [f (2q )] ≥ S · 2q+1 . Therefore, we must have equality, so S = 1 and q ∈ A. Similarly, k ∈ A, and the theorem is proved. When f (n) = σ(n), Ruzsa’s theorem is reobtained (select A = set of primes). When f (n) = nσ(n), all conditions are satisfied, with strict inequality in 3), so in this case the system is not solvable.
References [1] J. S´andor, On Ruzsa’s lovely pairs, Octogon Math. Mag., 12(2004), no.1, 287-289.
12
On the monotonicity of the sequence (σk /σk∗ )
1. Introduction Let n > 1 be a positive integer and k ≥ 0 a nonnegative integer. A divisor d of n is called a unitary divisor of n, if (d, n/d) = 1. Let σ ∗ (n) be the sum of 389
unitary divisors of n, i.e. X
σ ∗ (n) =
d.
(1)
r Y σ (n) = (pai i + 1),
(2)
d|n,(d,n/d)=1
Then it is well-known (see e.g. [1], [8]) that ∗
i=1
where n =
n Y
pai i is the prime factorization of n > 1 (pi distinct primes, ai ≥ 1
i=1
positive integers). More generally, if σk∗ (n) is the sum of kth powers of unitary divisors of n (i.e. (1) generalized to dk in place of d in the sum), then, similarly to (2), one has r Y ∗ (3) σk (n) = (pikai + 1). i=1
We note that, for k = 0 we get the number d∗ (n) = σ0∗ (n) of unitary divisors of n, when (3) gives d∗ (n) = 2r = 2ω(n) ,
(4)
where ω(n) = r denotes the number of unitary divisors of n. The similar formulae for the (classical) sum of divisors of n are the well-known (see e.g. [2], [9], [7]) r Y σ(n) = (pai i +1 − 1)/(pi − 1), (5) i=1
resp.
r Y k(a +1) σk (n) = (pi i − 1)/(pki − 1).
(6)
i=1
For k = 0, (6) provides the number d(n) of classical divisors of n: r Y (ai + 1). d(n) =
(7)
i=1
There are many results involving inequalities on these arithmetical functions. See e.g. [3]-[6]. For surveys of results, see e.g. [9], [8]. 390
2. Main results Langford ([9]) proved that µ σk (n) ≤ d(n)
nk + 1 2
¶ ,
(8)
while we proved ([10], [4], [5]) the stronger relation d(n)σk∗ (n) σk (n) ≤ ≤ d(n) 2ω(n)
µ
nk + 1 2
¶ .
(9)
The second inequality of (9) is a consequence of the elementary inequality r Y (xi + 1) ≤ 2r−1 i=1
à r Y
! xi + 1
(xi ≥ 1, r ≥ 1)
(10)
i=1
i applied to xi = pka i , r = ω(n), and using relation (3). Remark that the first inequality of (9) may be written also as
σ0 (n) σk (n) ≤ ∗ . ∗ σk (n) σ0 (n)
(11)
Our aim is to give a generalization of (11) as follows: µ ¶ σk (n) Theorem. For all fixed n ≥ 1, the sequence is monotone σk∗ (n) k≥0 decreasing. Proof. We have to prove that σk (n) σl (n) ≤ ∗ for all k ≥ l ≥ 0. ∗ σk (n) σl (n)
(12)
By (3) and (6), fk (n) =
σk (n)/σk∗ (n)
r Y k(a +1) i (pi i − 1)/(pki − 1)(pka + 1), = i i=1
so to prove that fk (n) ≤ fl (n) for k ≥ l, it will be sufficient to show that pl(a+1) − 1 pk(a+1) − 1 ≤ , (pk − 1)(pka + 1) (pl − 1)(pla + 1) 391
k ≥ l ≥ 0, p ≥ 2.
(13)
Put pk = x, pl = y, where x > y ≥ 1. After some elementary transformations (which we omit here) it can be shown that (13) becomes equivalent to xa − y a (xy)a − 1 ≤ (x > y ≥ 1). (14) x−y xy − 1 For y = 1, relation (14) is trivial, so we may suppose y ≥ 2. Now, remark that xa − y a = xa−1 + xa−2 y + · · · + xy a−2 + y a−1 ≤ axa−1 , x−y by y < x and a ≥ 1. On the other hand, we will prove that (xy)a − 1 ≥ axa−1 . xy − 1
(15)
This is equivalent to (xy)a − 1 ≥ axa y − axa−1 , or xa y(y a−1 − a) + axa−1 − 1 ≥ 0. Here axa−1 − 1 ≥ a − 1 ≥ 0, and y a−1 − a ≥ 2a−1 − a ≥ 0 for all a ≥ 1, so the result follows. By (15), and the above remark, inequality (14) is established. By (13), the inequality (12) follows, so the theorem is proved. Remarks. 1) For l = 0, k ≥ 0 arbitrary, we reobtain relation (11). 2) For l = 1, k ≥ 1 we get σ(n) σk (n) ≤ ∗ , ∗ σk (n) σ (n)
(16)
which offers an improvement of (11) for k ≥ 1, since by (11) applied to k = 1, and by (16), one has σ(n) σ0 (n) d(n) σk (n) ≤ ∗ ≤ ∗ = ∗ . ∗ σk (n) σ (n) σ0 (n) d (n)
(17)
For other improvements of the right side of (17), see [5].
References [1] E. Cohen, Arithmetical functions associated with the unitary divisors of an integer, Math. Z. 74(1960), 66-80. 392
[2] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [3] J. S´andor, An application of the Jensen-Hadamard inequality, Nieuw Arch. Wiskunde, Serie 4, 8(1990), no. 1, 43-66. [4] J. S´andor, On an inequality of Klamkin with arithmetical applications, Int. J. Math. Ed. Sci. Technol. 25(1994), 157-158. [5] J. S´andor, On certain inequalities for arithmetic functions, Notes Numb. Th. Discr. Math. 1(1995), 27-32. [6] J. S´andor, On the arithmetical functions dk (n) and d∗k (n), Portugaliae Math. 53(1996), no. 1, 107-115. [7] J. S´andor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth, New Mexico, 2002. [8] J. S´andor, Handbook of number theory II, Springer-Verlag, 2004. [9] J. S´andor, D. S. Mitrinovi´c, Handbook of number theory, Kluwer Acad. Publ., 1996. [10] J. S´andor, L. T´oth, On certain number-theoretic inequalities, Fib. Quart. 28(1990), no. 3, 255-258.
13
A note on exponential divisors and related arithmetic functions
1. Introduction Let n > 1 be a positive integer, and n = pa11 . . . par r its prime factorization. A number d|n is called an exponential divisor (or e-divisor, for short) of n if d = pb11 . . . pbrr with bi |ai (i = 1, r). This notion has been introduced by E. G. Straus and M. V. Subbarao [1]. Let σe (n), resp. de (n) denote the sum, resp. number of e-divisors of n, and let σe (1) = de (1) = 1, by convention. A number n is called e-perfect, if σe (n) = 2n. For results and References involving e-perfect numbers, and the arithmetical functions σe (n) and de (n), see [4]. For example, it is well-known that de (n) is multiplicative, and de (n) = d(a1 ) . . . d(ar ), 393
(1)
where n = pa11 . . . par r is the canonical form of n, and d(a) denotes the number of (ordinary) divisors of a. The e-totient function ϕe (n), introduced and studied in [4] is multiplicative, and one has ϕe (n) = ϕ(a1 ) . . . ϕ(ar ), (2) where ϕ is the classical Euler totient function. Let σ(a) denote the sum of (ordinary) divisors of a. The product of edivisors of n, denoted by Te (n) has the following expression (see [9]): σ(a1 )d(a2 )...d(ar )
Te (n) = p1
r )d(a1 )...d(ar−1 ) . . . pσ(a r
(3)
A number n is called multiplicatively e-perfect if Te (n) = n2 . Based on (3), in [9] we have proved that n is multiplicatively e-perfect iff n can be written as n = pm , where σ(m) = 2m, and p is a prime. Two notions of exponentially-harmonic numbers have been recently introduced by the author in [11]. Finally, we note that for a given arithmetic function f : N∗ → N∗ , in [5], [6] we have introduced the minimum function of f by Ff (n) = min{k ≥ 1 : n|f (k)}
(4)
Various particular cases, including f (k) = ϕ(k), f (k) = σ(k), f (k) = d(k), f (k) = S(k) (Smarandache function), f (k) = T (k) (product of ordinary divisors), have been studied recently by the present author. He also studied the duals of these functions (when these have sense) defined by Ff∗ (n) = max{k ≥ 1 : f (k)|n}
(5)
See e.g. [10] and the References therein.
2. Main notions and results The aim of this note is to introduce certain new arithmetic functions, related to the above considered notions. Since for the product of ordinary divisors of n one can write T (n) = nd(n)/2 ,
(6)
trying to obtain a similar expression for Te (n) of the product of e-divisors of n, by (3) the following can be written: Theorem 1. Te (n) = (t(n))de (n)/2 , (7) 394
where de (n) is the number of exponential divisors of n, given by (1); while the arithmetical function t(n) is given by t(1) = 1 σ(a )
2 d(a 1)
t(n) = p1
1
2
σ(ar )
. . . pr d(ar )
(8)
n = pa11 . . . par r being the prime factorization of n > 1. Proof. This follows easily by relation (3), and the definition of t(n) given by (8). Remark. For multiplicatively perfect numbers given by T (n) = n2 , see [7]. For multiplicatively deficient numbers, see [8]. Remark that de (n) ≤ d(n) (9) for all n, with equality only for n = 1. Indeed, by d(a) < a + 1 for a ≥ 2, via (1) this is trivial. On the other hand, the inequality t(n) ≤ n
(10)
is not generally valid. Let e.g. n = pq11 . . . pqrr , where all qi (i = 1, r) are primes. Then, by (8) t(n) = pq11 +1 . . . pqrr +1 = (p1 . . . pr )n > n. However, there is a particular case, when (10) is always true, namely suppose that ω(ai ) ≥ 2 for all i = 1, r (where ω(a) denotes the number of distinct prime factors of a). In σ(a) a [3] it is proved that if ω(a) ≥ 2, then < . This gives (10) with strict d(a) 2 inequality, if the above conditions are valid. Without any condition one can prove: Theorem 2. For all n ≥ 1 Te (n) ≤ T (n), with equality only for n = 1 and n = prime. Proof. The inequality to be proved becomes ³ ´d(a1 )...d(ar ) σ(a )/d(a1 ) r )/d(ar ) p1 1 . . . pσ(a ≤ (pa11 . . . par r )(a1 +1)...(ar +1)/2 r
(11)
(12)
We will prove that σ(a1 ) a1 (a1 + 1) . . . (ar + 1) d(a1 ) . . . d(ar ) ≤ d(a1 ) 2 with equality only if r = 1 and a1 = 1. Indeed, it is known that (see [2]) a1 + 1 σ(a1 ) ≤ , with equality only for a1 = 1 and a1 = prime. On the d(a1 ) 2 395
other hand, d(a1 ) . . . d(ar ) ≤ a1 (a2 + 1) . . . (ar + 1) is trivial by d(a1 ) ≤ a1 , d(a2 ) < a2 + 1, . . . , d(ar ) < ar + 1, with equality only for a1 = 1 and r = 1. Thus (12) follows, with equality for r = 1, a1 = 1, so n = p1 = prime for n > 1. Remark. In [4] it is proved that ϕe (n)de (n) ≥ a1 . . . ar
(13)
a1 ar ... ≥ 2r if all ai (i = 1, r) are even, since ϕ(a1 ) ϕ(ar ) a it is well-known that ϕ(a) ≤ for a = even. Since d(n) = (a1 +1) . . . (ar +1) ≤ 2 2a1 . . . 2ar = 2a1 +···+ar = 2Ω(n) (where Ω(n) denotes the total number of prime divisors of n), by (9) one can write: Now, by (2), de (n) ≥
2ω(n) ≤ de (n) ≤ 2Ω(n) ,
(14)
if all ai are even, i.e. when n is a perfect square (right side always). Similarly, in [4] it is proved that ϕe (n)de (n) ≥ σ(a1 ) . . . σ(ar )
(15)
when all ai (i = 1, r) are odd. Let all ai ≥ 3 be odd. Then, since σ(ai ) ≥ ai + 1 (with equality only if ai = prime), (15) implies ϕe (n)de (n) ≥ d(n),
(16)
which is a converse to inequality (9). √ √ 2 a 2 a Let now introduce the arithmetical function t1 (n) = p1 1 . . . pr r , t1 (1) = 1 and let γ(n) = p1 . . . pr denote the ”core” of n (see [2]). Then: Theorem 3. t1 (n) ≤ t(n) ≤ nγ(n) for all n ≥ 1.
(17)
Proof. This follows at once by the known double-inequality √ a+1 σ(a) ≤ , a≤ d(a) 2
(18)
with equality for a = 1 on the left side, and for a = 1 and a = prime on the right side. Therefore, in (17) one has equality when n is squarefree, while on the right side if n is squarefree, or n = pq11 . . . pqrr with all qi (i = 1, r) primes. Clearly, the functions t1 (n), t(n) and γ(n) are all multiplicative. 396
Finally, we introduce the minimum exponential totient function by (4) for f (k) = ϕe (k): Ee (n) = min{k ≥ 1 : n|ϕe (k)}, (19) where ϕe (k) is the e-totient function given by (2). Let E(n) = min{k ≥ 1 : n|ϕ(k)} be the Euler minimum function (see [10]). The following result is true: Theorem 4. Ee (n) = 2E(n) for n > 1.
(20)
(21)
Proof. Let k = pα1 1 . . . pαs s . Then k ≥ 2α1 +···+αs ≥ 2s . Let s be the least integer with n|ϕ(s) (i.e. s = E(n) by (20)). Clearly ϕe (2s ) = ϕ(s), so k = 2s is the least k ≥ 1 with property n|ϕe (k). This finishes the proof of (21). For properties of E(n), see [10]. Remark. It is interesting to note that the ”maximum e-totient”, i.e. Ee∗ (n) = max{k ≥ 1 : ϕe (k)|n}
(22)
is not well defined. Indeed, e.g. for all primes p one has ϕe (p) = 1|n, and Ee∗ (p) = +∞, so Ee∗ (n) given by (22) is not an arithmetic function.
References [1] E. G. Straus and M. V. Subbarao, On exponential divisors, Duke Math. J. 41(1974), 465-471. [2] D. S. Mitronovi´c and J. S´andor, Handbook of number theory, Kluwer Acad. Publ., 1995. [3] J. S´andor, On the Jensen-Hadamard inequality, Nieuw Arch Wiskunde (4)8(1990), 63-66. [4] J. S´andor, On an exponential totient function, Studia Univ. Babe¸sBolyai, Math. 41(1996), no. 3, 91-94. [5] J. S´andor, On certain generalizations of the Smarandache functions, Notes Number Th. Discr. Math. 5(1999), no. 2, 41-51. [6] J. S´andor, On certain generalizations of the Smarandache function, Smarandache Notion J. 11(2000), no. 1-3, 202-212. 397
[7] J. S´andor, On multiplicatively perfect numbers, J. Ineq. Pure Appl. Math. 2(2001), no. 1, Article 3, 6 pp. (electronic). [8] J. S´andor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth, 2002. [9] J. S´andor, On multiplicatively e-perfect numbers, to appear. [10] J. S´andor, On the Euler minimum and maximum functions, to appear. [11] J. S´andor, On exponentially harmonic numbers, to appear.
14
The Euler minimum and maximum functions 1. The Euler minimum function is defined by E(n) = min{k ≥ 1 : n|ϕ(k)}
(1)
It was introduced by P. Moree and H. Roskam [5]; and independently by J. S´andor [7], as a particular case of the more general function FfA (n) = min{k ∈ A : n|f (k)} (A ⊂ N∗ ),
(2)
where f : N∗ → N∗ is a given function, and A is a given set of positive integers. For A = N∗ , f = ϕ (Euler’s totient), one obtains the function E given by (1) (denoted also as Fϕ in [7]). Since by Dirichlet’s theorem on arithmetical progression, there exists a ≥ 1 such that k = an + 1 = prime, by ϕ(k) = an˙:n, so E(n) is well defined. We note that for A = N∗ , f (k) = k! one reobtains the Smarandache function S(n) = min{k ≥ 1 : n|k!}, (3) while for A = P = {2, 3, 5, . . . } = set of all primes, f (k) = k!, (2) gives a new function, denoted by us as P (n): P (n) = min{k ∈ P : n|k!}
(4)
We note that this function should not be confused with the greatest prime divisor of n (denoted also sometimes by P (n)). For properties of this function, see [6], [7]. There is a dual of (2) (see [7]), namely GA g (n) = max{k ∈ A : g(k)|n}, 398
(5)
where g : N∗ → N∗ , A ⊂ N∗ are given, if this is well defined. For A = N∗ , g(k) = k!, this has been denoted by us by S∗ (n), and called as the dual of the Smarandache function: S∗ (n) = max{k ≥ 1 : k!|n}
(6)
For properties of this function, see [6]. See also F. Luca [4], where a conjecture of the author has been proved, and M. Le [3] for a recent new proof. See also K. Atanassov [1]. For A = N∗ , g(k) = ϕ(k) one obtains the dual E∗ (n) of the Euler minimum function, which we shall call as the Euler maximum function: E∗ (n) = max{k ≥ 1 : ϕ(k)|n} (7) √ Since for k > 6, ϕ(k) > k, clearly k < n2 , so E∗ (n) ≤ n2 < ∞. Generally, for A = N∗ , let us write simply FfA (n) = Ff (n), GA g (n) = Gf (n). 2. First we prove the following property of the Euler minimum function: Theorem 1. If pi (i = 1, r) are distinct primes, and αi ≥ 1 are integers, then ! Ã r Y α αi (8) max{E(pi ) : i = 1, r} ≤ E pi i ≤ [E(pα1 1 ), . . . , E(pαr r )], i=1
where [, . . . , ] denotes l.c.m. Proof. For simplicity we shall prove (8) for r = 2. Let pα , q β be distinct prime powers. Then E(pα q β ) = min{k ≥ 1 : pα q β |ϕ(k)} = k0 , so pα q β |ϕ(k0 ), which is equivalent to pα |ϕ(k0 ), q β |ϕ(k0 ), thus k0 ≥ E(pα ), k0 ≥ E(q β ), imβ ) ≥ max{E(pα ), E(q β )}. It is immediate that the same proof plying E(pα q³ Y ´ applies to E pα ≥ max{E(pα )}, where pα are distinct prime powers. Therefore, the left side of (8) follows. Now, let E(pα ) = k1 , E(q β ) = k2 , implying pα |ϕ(k1 ), q β |ϕ(k2 ). Let [k1 , k2 ] = g. Since k1 |g, one has ϕ(k1 )|ϕ(g) (by a known property of the function ϕ). Similarly, since k2 |g, one can write ϕ(k2 )|ϕ(g). Thus pα |ϕ(k1 )|ϕ(g) and q β |ϕ(k2 )|ϕ(g), yielding pα q β |ϕ(g). By the definition (1) this gives g ≥ E(pα q β ), i.e. [E(pα ), E(q β )] ≥ E(pα q β ), so the right side of (8) for r = 2 is proved. The general case follows exactly the same lines. Remark 1. The above proof shows that the left side of (8) holds true for any function f (for which Ff is well defined), so we get à r ! Y max{Ff (pαi i ) : i = 1, r} ≤ Ff pαi i (9) i=1
399
For the right side of (8), with the same proof the following is valid: if f has the property a|b ⇒ f (a)|f (b) (a, b ≥ 1), (10) then Ff
à r Y
! pαi i
≤ [Ff (pα1 1 ), . . . , Ff (pαr r )]
(11)
i=1
Now, if one replaces (10) with a stronger property, then a better result will be true: Theorem 2. Assume that f satisfies the following property a ≤ b ⇒ f (a)|f (b) Then Ff
à r Y
(a, b ≥ 1)
(12)
! pαi i
= max{Ff (pαi i ) : i = 1, r}
(13)
i=1
Proof. By taking into account of (9), one needs only to show that the reverse inequality is true. For simplicity, let us take again r = 2. Let Ff (pα ) = m, Ff (q β ) = n, with m ≤ n. Then the definition (2) of Ff implies that pα |f (m), q β |f (n). By (12) one has f (m)|f (n), so pα |f (m)|f (n). We have pα |f (n), q β |f (n), so pα q β |f (n). But this implies n ≥ Ff (pα q β ), i.e. max{Ff (pα ), Ff (q β )} ≥ Ff (pα q β ). The general case follows exactly the same lines. s r Y Y β qj j , (pi , qj ) = 1, Remark 2. If (a, b) = 1, by writing a = pαi i , b = i=1
j=1
it follows that β
Ff (ab) = max{E(pαi i ), E(qj j ) : i = 1, r, j = 1, s} = β
= max{max{E(pαi i ) : i = 1, r}, max{E(qj j ) : j = 1, s}} = = max{Ff (a), Ff (b)}, so: Ff (ab) = max{Ff (a), Ff (b)} for (a, b) = 1
(14)
When f (n) = n!, then clearly (12) is true, so (14) gives: S(ab) = max{S(a), S(b)} for (a, b) = 1, discovered by F. Smarandache [9]. 400
(15)
3. The Euler minimum function must be studied essentially (by Theorem 1) for prime powers pα . For values of E(p), E(p2 ), etc., see [5]. On the other hand, for each prime p ≥ 3 one has E(p − 1) = p
(16)
Indeed, if (p − 1)|ϕ(k), then p − 1 ≤ ϕ(k). Since ϕ(k) ≤ k − 1 for k ≥ 3, one has p ≤ k. Now k = p gives ϕ(p) = p − 1, giving (16). The values of the Euler maximum function E∗ given by (7) however are even difficult to calculate in some cases. This function doesn’t seem to have been studied up to now. Clearly E∗ (1) = 2 since ϕ(1) = 1, ϕ(2) = 1. One has E∗ (2) = 6 since ϕ(6) = 2, 2|2, and it is well-known that ϕ(n) ≥ 4 for n ≥ 7. Now let p ≥ 3 be a prime. Since ϕ(k)|p implies ϕ(k) = 1 or ϕ(k) = p, for p ≥ 3 the last equality is impossible for ϕ(k) is even for all k ≥ 3, we can have only ϕ(k) = 1 and kmax = 2. Actually since for k ≥ 3, ϕ(k) is even, ϕ(k)|n is impossible for n = odd, so remains k ≤ 2, and kmax = 2. We have½proved: 6, if p = 2 Theorem 3. One has E∗ (1) = 2, E∗ (p) = for all primes 2, if p ≥ 3 p; and E∗ (n) = 2 for all n ≥ 1 odd. For all n ≥ 2 one has E∗ (n) ≥ 2. (17) The last inequality is a consequence of ϕ(2) = 1 and the definition (7). The value 2 is taken infinitely often, but the same is true for the value 6: Theorem 4. For all α ≥ 1 one has E∗ (2 · 7α ) = 6
(18)
Proof. If ϕ(k)|(2 · 7α ), then assuming k ≥ 3, as ϕ(k) is even, we can only have ϕ(k) = 2 or ϕ(k) = 2 · 7a where 1 ≤ a ≤ α. Now, A. Schinzel [8] has shown that the equation ϕ(x) = 2 · 7a is not solvable for any a ≥ 1. Thus, it remains ϕ(k) = 2 and the maximal value of k ≥ 3 is k = 6. This finishes the proof of (18). Remark. One has similarly E∗ (2 · 52α ) = 6 for any α ≥ 1. (19) The function E∗ can take greater values, too; the values at powers of 2 is shown by the following theorem: Theorem 5. E∗ (2m ) = k, where k is the greatest number which can be α αr written as k = 2α p1 . . . pr , with p1 = 22 1 + 1, . . . , pr = 22 + 1 distinct Fermat primes, and where α = a + 1 − (2k1 + · · · + 2kr ), with k1 , . . . , kr ≥ 0, 0 ≤ a ≤ m. (20) m a Proof. Since ϕ(k)|2 , clearly ϕ(k) = 2 , where 0 ≤ a ≤ m. Now let k = 2α pα1 1 . . . pαr r with p1 , . . . , pr distinct odd primes. Since ϕ(k) = 401
2α−1 pα1 1 −1 . . . pαr r −1 (p1 − 1) . . . (pr − 1) = 2a , we must have α1 − 1 = · · · = αr − 1 = 0 and p1 − 1 = 2a1 , . . . , pr − 1 = 2ar with α − 1 + a1 + · · · + ar = a. This gives p1 = 2a1 + 1, . . . , pr = 2ar + 1. Since p1 is prime, it is well-known that it is a Fermat prime, so a1 = 2k1 , etc., and the theorem follows. Remark 3. For m = 2 we get α ≤ 3 − (2k1 + · · · + 2kr ), so with k1 = 0 (when p1 = 3), we get k = 22 · 3 = 12. Another value would be k = 2 · 5 = 10, so we get E∗ (4) = 12 Similarly, for m = 3, E∗ (8) = 30 If can be shown also that E∗ (16) = 60,
E∗ (32) = 120,
E∗ (64) = 240,
E∗ (128) = 510, etc.
However, since the structure (or the cardinality) of the Fermat primes is not well-known, there are problems also with the calculation of great values of E∗ (2m ). The function E∗ (n) can take arbitrarily large values, since one has: Theorem 6. For all m ≥ 1 the following inequality is true: E∗ (m!) ≥
(m!)2 ϕ(m!)
(21)
Proof. It is known (see e.g. [2]) that the equation ϕ(x) = m!
(22)
admits the solution x = (m!)2 /ϕ(m!). Now, since ϕ(x) = m!|m!, clearly E∗ (m!) ≥ x, giving inequality (21). E∗ (m!) = +∞ (23) Corollary. lim m→∞ m! m! → ∞ as m → ∞. Proof. Indeed, it is well-known (see e.g. [10]) that ϕ(m!) By (21), this implies (23).
References [1] K. T. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] P. Erd¨os, Amer. Math. Monthly 58(1951), p. 98. 402
[3] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notion J. 14(2004), 153-155. [4] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [5] P. Moree, H. Roskam, On an arithmetical function related to Euler’s totient and the discriminantor, Fib. Quart. 33(1995), 332-340. [6] J. S´andor, On certain generalizations of the Smarandache function, Smarandache Notions J. 11(2000), no. 1-3, 202-212. [7] J. S´andor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [8] A. Schinzel, Sur l’´equation ϕ(x) = m, Elem. Math. 11(1956), 75-78. [9] F. Smarandache, A function in the number theory, An. Univ. Timi¸soara, Ser. Mat., 38(1980), 79-88. [10] J. S´andor, On values of arithmetical functions at factorials, I, Smarandache Notions J., 10(1999), 87-94.
15
The Smarandache minimum and maximum functions
1. Let f : N∗ → N be a given arithmetic function and A ⊂ N a given set. The arithmetical function FfA (n) = min{k ∈ A : n|f (k)}
(1)
has been introduced in [4] and [5]. For A = N, f (k) = k! one obtains the Smarandache function; for A = N∗ , A = P = {2, 3, 5, . . . } = set of all primes, one obtains a function P (n) = min{k ∈ P : n|k!}
(2)
For properties of this function, see [4], [5]. The ”dual” function of (1) has been defined by GA g (n) = max{k ∈ A : g(k)|n}, 403
(3)
where g : N∗ → N is a given function, and A ⊂ N is a given set. Particularly, for A = N∗ , g(k) = k! one obtains the dual of the Smarandache function, S∗ (n) = max{k ≥ 1 : k!|n}
(4)
For properties of this function, see [4], [5]. F. Luca [3], K. Atanassov [1] and L. Le [2] have proved in the affirmative a conjecture of the author. For A = N∗ and f (k) = g(k) = ϕ(k) in (1), resp. (3) one obtains the Euler minimum, resp. maximum-functions, defined by E(n) = min{k ≥ 1 : n|ϕ(k)},
(5)
E∗ (n) = max{k ≥ 1 : ϕ(k)|n}
(6)
For properties of these functions, see [6]. When A = N∗ , f (k) = d(k) = number of divisors of k, one obtains the divisor minimum function (see [4], [5], [7]) D(n) = min{k ≥ 1 : n|d(k)}
(7)
It is interesting to note that the divisor maximum function (i.e. the ”dual” of D(n)) given by D∗ (n) = max{k ≥ 1 : d(k)|n} (8) is not well-defined! Indeed, for any prime p one has d(pn−1 ) = n|n and pn−1 is unbounded as p → ∞. For a finite set A, however D∗A (n) does exist. On the other hand, it has been shown in [4], [5] that Σ(n) = min{k ≥ 1 : n|σ(k)}
(9)
(denoted there by Fσ (n)) is well defined. (Here σ(k) denotes the sum of all divisors of k). The dual of the sum-of-divisors minimum function is Σ∗ (n) = max{k ≥ 1 : σ(k)|n}
(10)
Since σ(1) = 1|n and σ(k) ≥ k, clearly Σ∗ (n) ≤ n, so this function is well-defined (see [8]). 2. The Smarandache minimum function will be defined for A = N∗ , f (k) = S(k) in (1). Let us denote this function by Smin : Smin (n) = min{k ≥ 1 : n|S(k)}
(11)
Let us assume that S(1) = 1, i.e. S(n) is defined by (1) for A = N∗ , f (k) = k!: S(n) = min{k ≥ 1 : n|k!} (12) 404
Otherwise (i.e. when S(1) = 0) by n|0 for all n, by (11) one gets the trivial function Smin (n) ≡ 0. By this assumption, however, one obtains a very interesting (and difficult) function Smin given by (11). Since n|S(n!) = n, this function is correctly defined. The Smarandache maximum function will be defined as the dual of Smin : Smax (n) = max{k ≥ 1 : S(k)|n} (13) We prove that this is well-defined. Indeed, for a fixed n, there are a finite number of divisors of n, let i|n be one of them. The equation S(k) = i
(14)
is well-known to have a number of d(i!) − d((i − 1)!) solutions, i.e. in a finite number. This implies that for a given n there are at most finitely many k with S(k)|n, so the maximum in (13) is attained. Clearly Smin (1) = 1, Smin (2) = 2, Smin (3) = 3, Smin (4) = 4, Smin (5) = 5, Smin (6) = 9, Smin (7) = 7, Smin (8) = 32, Smin (9) = 27, Smin (10) = 25, Smin (11) = 11, etc., which can be determined from a table of Smarandache numbers: n 1 2 3 4 5 6 7 8 9 10 11 12 13 S(n) 1 2 3 4 5 3 7 4 6 5 11 4 13 n 14 15 16 17 18 19 20 21 22 23 24 25 S(n) 7 5 6 7 6 19 5 7 11 23 4 10 We first prove that: Theorem 1. Smin (n) ≥ n for all n ≥ 1, with equality only for n = 1, 4, p (p = prime). (15) Proof. Let n|S(k). If we would have k < n, then since S(k) ≤ k < n we would get S(k) < n, in contradiction with n|S(k). Thus k ≥ n, and taking minimum, the inequality follows. There is equality for n = 1 and n = 4. Let now n > 4. If n = p = prime, then p|S(p) = p, but for k < p, p - S(k). Indeed, by S(k) ≤ k < p this is impossible. Reciprocally, if min{k ≥ 1 : n|S(k)} = n, then n|S(n), and by S(n) ≤ n this is possible only when S(n) = n, i.e. when n = 1, 4, p (p = prime). Theorem 2. For all n ≥ 1, Smin (n) ≤ n! ≤ Smax (n).
405
(16)
Proof. Since S(n!) = n, definition (11) gives the left side of (16), while definition (13) gives the right side inequality. X 1 Corollary. The series is divergent, while the series Smin (n) n≥1 X 1 is convergent. Smax (n) n≥1 X X 1 1 Proof. Since ≤ = e − 1 by (16), this series is converSmax (n) n! n≥1 n≥1 X X X1 1 1 gent. On the other hand, ≥ = = +∞, so Smin (n) Smin (p) p p p prime
n≥1
the first series is divergent. Theorem 3. For all primes p one has Smax (p) = p!
(17)
Proof. Let S(k)|p. Then S(k) = 1 or S(k) = p. We prove that if S(k) = p, then k ≤ p!. Indeed, this follows from the definition (12), since S(k) = min{m ≥ 1 : k|m!} = p implies k|p!, so k ≤ p!. Therefore the greatest value of k is k = p!, when S(k) = p|p. This proves relation (17). Since S(p2 ) = 2p, for all primes p, from (11) and (13) one can deduce the first part of the following theorem: Theorem 4. For all primes p, Smin (2p) ≤ p2 ≤ Smax (2p),
(18)
and more generally; for all m ≤ p, Smin (mp) ≤ pm ≤ Smax (mp)
(19)
Proof. (19) follows by the known relation S(pm ) = mp if m ≤ p and the definitions (11), (13). Particularly, for m = 2, (19) reduces to (18). For m = p, (19) gives Smin (p2 ) ≤ pp ≤ Smax (p2 ) (20) The case when m is also an arbitrary prime is given in Theorem 5. For all odd primes p and q, p < q one has Smin (pq) ≤ q p ≤ pq ≤ Smax (pq) (21) holds also when p = 2 and q ≥ 5. 406
(21)
Proof. Since S(q p ) = pq and S(pq ) = qp for primes p and q, the extreme inequalities of (21) follow from the definitions (11) and (13). For the inequality ln x q p < pq remark that this is equivalent to f (p) > f (q), where f (x) = x 1 − ln x 0 (x ≥ 3). Since f (x) = = 0 ⇔ x = e immediately follows that f is x2 strictly decreasing for x ≥ e = 2.71 . . . From the graph of this function, since ln 2 ln 4 ln 2 ln 3 ln 2 ln q = we get that < , but > for q ≥ 5. Therefore (21) 2 4 2 3 2 q holds also when p = 2 and q ≥ 5. Indeed, f (q) ≤ f (5) < f (4) = f (2). Remark. For all primes p, q Smin (pq) ≤ min{pq , q p } and Smax (pq) ≥ max{pq , q p }.
(22)
For p = q this implies relation (21). Proof. Since S(q p ) = S(pq ) = pq, one has Smin (pq) ≤ pq ,
Smin (pq) ≤ q p ,
Smax (pq) ≥ pq ,
Smax (pq) ≥ q p .
References [1] K. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notions J. 14(2004), 153-155. [3] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [4] J. S´andor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [5] J. S´andor, On certain generalizations of the Smarandache function, Smarandache Notions J., 11(2000), no. 1-3, 202-212. [6] J. S´andor, On the Euler minimum and maximum functions, RGMIA Research Report Collection 8(1), Article 1, 2005. [7] J. S´andor, A note on the divisor minimum function, Octogon Math. Mag., 12(2004), no.2A, 273-275. [8] J. S´andor, The sum-of-divisors minimum and maximum functions, RGMIA Research Report Collection, 8(1), Article 20, 2005. 407
16
The pseudo-Smarandache minimum and maximum functions
1. Introduction K. Kashihara [2] defined the pseudo-Smarandache function Z by ½ ¾ m(m + 1) Z(n) = min m ≥ 1 : n| 2
(1)
For properties of this function see e.g. [2], [1], [6]. The dual of the pseudoSmarandache function has been introduced and studied by the author (see [5], [6): ½ ¾ m(m + 1) Z∗ (n) = max m ≥ 1 : |n (2) 2 More generally, for functions f, g : N∗ → N∗ , the author [3], [4] (see also [6]) has defined the functions Ff (n) = min{k ≥ 1 : n|f (k)},
(3)
Gg (n) = max{k ≥ 1 : g(k)|n},
(4)
if these functions are well-defined. Various particular cases, including f (k) = g(k) = ϕ(k), f (k) = g(k) = σ(k), f (k) = d(k), f (k) = S(k) (Smarandache function), f (k) = T (k) (product of ordinary divisors) have been studied by the author. See e.g. [7] and the References therein. The aim of this note is the initial study of the functions Z(n) and Z ∗ (n) given by (3), resp. (4) for the particular cases f (k) = g(k) = Z(k).
2. The pseudo-Smarandache minimum and maximum functions The pseudo-Smarandache minimum function will be defined by Z(n) = min{k ≥ 1 : n|Z(k)},
(5)
while the pseudo-Smarandache maximum function will be the dual of Z(n): Z ∗ (n) = max{k ≥ 1 : Z(k)|n} (6) The particular cases of (3) and (4) for f (k) = g(k) = Z∗ (k), given by (2) (i.e. the pseudo-Smarandache dual minimum and maximum functions) will be studied in another paper. 408
Recall the following known properties of Z(n) (see e.g. [6]) ½ 2n − 1, if n is even, Z(n) ≤ n − 1, if n is odd,
(7)
Z(2k ) = 2k+1 − 1,
(8)
Z(p) = p − 1 for p ≥ 3 prime,
(9)
Z(2p) = p − 1 for p ≡ 1 (mod 4) a prime, µ ¶ n(n + 1) Z = n for all n ≥ 1, 2 r 1 1 Z(n) ≥ − + 2n + for all n ≥ 1 2 4 Theorem 1. For all n ≥ 1, n+1 n(n + 1) ≤ Z(n) ≤ 2 2
(10) (11) (12)
(13)
µ
¶ n(n + 1) Proof. Since by (11), n|Z for all n, by (5) we get the right side 2 inequality of (13). On the other hand, remark that if n|Z(k), then n ≤ Z(k). n+1 . This implies From (7) we get Z(k) ≤ 2k − 1, so n ≤ 2k − 1, giving k ≥ 2 the left side inequality of (13). p n Corollary. lim Z(n) = 1. n→∞ Theorem 2. For all primes p, Z(p − 1) ≤ p
(14)
Z(p − 1) = p
(15)
If p ≡ 3 (mod 4), then
Proof. For p = 2, Z(1) = 1 < 2; let p ≥ 3. By (9) one has p − 1|Z(p) so (14) follows from definition (5). Now let p − 1|Z(k). If k is odd, then by p − 1 ≤ Z(k) ≤ k − 1 (see (7)) one gets p ≤ k. But Z(p) = p − 1. Further, if k < p is odd, then p ≤ k < p, which is impossible. If k is even, then Z(k) ≤ 2k − 1 and p − 1|Z(k) gives Z(k) = (p − 1)a ≤ 2k − 1 < 2p − 1. For a = 2 one can write 2(p − 1) = 2p − 2 ≤ 2k − 1 so 2k ≥ 2p − 1, 1 i.e. k ≥ p − , and k being integer, k ≥ p. Since p is odd, k ≥ p + 1, so for 2 409
k = p (in the odd case) we get a smaller value. For a = 1, Z(k) = p − 1. (n − 1)p We search for even k < p for which this is true. Since by (1), k| , we 2 get (p − 1)p = 2k · A (A = integer), so k|(p − 1)p. But p being prime, and k < p, clearly (k, p) = 1, so we must have k|(p − 1), i.e. p − 1 = km. Thus km(kn + 1) = 2kA, i.e. m(km + 1) = 2A. Since k is even, km + 1 is odd, so m must be even. From p − 1 = km it follows p ≡ 1 (mod 4). This contradicts the made assumption X on p. Corollary. 1/Z(p − 1) is divergent. Theorem 3. Z(2n − 1) = 2n−1 for all n ≥ 1 (16) Proof. The left side of (13) gives Z(2n − 1) ≥
2n − 1 + 1 = 2n−1 2
(∗)
On the other hand, by (8), Z(2n−1 ) = 2n − 1, so (2n − 1)|Z(2n−1 ), implying by (5) that Z(2n − 1) ≤ 2n−1 (∗∗) Relations (∗)X and (∗∗) imply (16). Corollary. 1/Z(2n − 1) is convergent. Theorem 4. n(n + 1) (17) 1 ≤ Z∗ (n) ≤ 2 Proof. n. Since by (12) this yields n ≥ r By (6), if Z(k)|n, thenµZ(k) ≤ ¶2 1 1 1 1 1 n(n + 1) − + 2k + , we get 2k + ≤ n + = n2 + n + , so k ≤ , 2 4 4 2 4 2 giving the right side of (17). The left side of (17) is obvious. p(p + 1) Theorem 5. Z∗ (p) = for all primes p. (18) 2 Proof. Let Z(k)|p. Since p is prime, we must have Z(k) = 1 (i.e. k = 1) or p(p + 1) is a solution. Z(k) = p. This equation is always solvable, since k = 2 p(p + 1) p(p + 1) So Z∗ (p) ≥ . On the other hand, (17) reads Z∗ (p) ≤ , so (18) 2 2 follows. Remark. The calculation of Z(p) seems much difficult. Theorem 6. If p ≡ 1 (mod 4) is a prime, then Z∗ (p − 1) ≥ 2p 410
(19)
Proof. By (10), Z(2p)|(p − 1), so (19) is a consequence of this fact and definition (6). Theorem 7. Z(Z∗ (n))|n|Z(Z(n)) for all n ≥ 1. (20) Proof. Let Z∗ (n) = k. Then by (6), Z(k)|n, so the left side of (20) follows. The right side follows similarly by (5). Corollary. Z(Z∗ (n)) ≤ n ≤ Z(Z(n)) for all n ≥ 1. (21)
References [1] C. Ashbacher, The pseudo-Smarandache function and the classical functions of Number Theory, Smarandache Notions J., 9(1998), No. 1-2, 7881. [2] K. Kashihara, Comments and Topics on Smarandache Notions and Problems, Erhus U.P., AZ, 1996. [3] J. S´andor, On certain generalizations of the Smarandache function, Notes Number Th. Discr. Math. 5(1999), no. 2, 41-51. [4] J. S´andor, On certain generalizations of the Smarandache function, Smarandache Notions J., 11(2000), no. 1-3, 202-212. [5] J. S´andor, On a dual of the pseudo-Smarandache function, Smarandache Notions J., 13(2002), no. 1-2-3, pp. 18-23. [6] J. S´andor, Geometric theorems, diophantine equations, and arithmetic functions, American Research Press, Rehoboth 2002. [7] J. S´andor, On the Euler minimum and maximum functions, RGMIA Research Report Collection, 8(1), Article 1, 2005.
411
412
Chapter 10
Miscellaneous themes ”... Say what you know, do what you must, come what may.” (Sofia Kovalevskaya)
”... The real purpose of mathematics is to be the means to illuminate reason and to exercise spiritual forces.” (August Crelle)
413
1
On a divisibility problem
1. The problem in the title is: Determine all positive integers x and y such that: xy−1 + y x−1 ≡ 0 (mod (x + y)). (1) In our opinion the most general solution of this divisibility cannot be never obtained, and only particular (special) cases are treatable. In what follows we shall indicate certain cases when (1) is soluble, or not. 2. First remark that (1) is always solvable when x = 2m, y = 2n with m > n; m + n = 2s , where s + 1 ≤ 2n − 1. Indeed, by substituting x = 2m, y = 2n in (1) we get xy−1 + y x−1 = (2m)2n−1 + (2n)2m−1 = 22n−1 (m2n−1 + 22m−2n n2m+1 ) which is divisible by 2(m + n) = 2 · 2s = 2s+1 where m > n and s + 1 ≤ 2n − 1. 3. Now, we will prove that (1) is not solvable if x = 2m + 1, y = 2n + 1 and m + n 6= 0 (mod 2). We shall use the known fact that the square of an odd number a is the form a2 ≡ 1 (mod 8). By this result xy−1 + y x−1 = (2m + 1)2n + (2n + 1)2m = (8M + 1) + (8N + 1) = 8K + 2 = 2(4K + 1). Since x + y = 2(m + n) + 2 = 2(m + n + 1); if m + n is odd, then m + n + 1 is even, which cannot divide an odd number (of the form 4K + 1). 4. Let x > y, y even and (x, y) = 1. Then (1) is true if and only if y x−y − 1 ≡ 0 (mod (x + y)).
(2)
Indeed, xy−1 + y x−1 = xy−1 + y y−1 + y x−1 − y y−1 = (xy−1 + y y−1 ) + y y−1 (y x−y − 1). Now, since y − 1 is odd, it is well-known that xy−1 + y y−1 is divisible by x + y. Since (xy−1 , x + y) = 1, by the above equality, (2) follows.
2
A generalization of Fermat’s little theorem
Let n, k ≥ 1 be positive integers, and put Sk (n) = 1k + 2k + · · · + nk . By writing the binomial theorem for the expressions (0 + 1)k = 1k 414
µ
¶ k (1 + 1) = 1 + · 1k−1 + . . . k−1 µ ¶ k k k (2 + 1) = 2 + · 2k−1 + . . . k−1 ... µ ¶ k k k (n + 1) = n + nk−1 + . . . , k−1 k
k
after addition, and reducing the common terms, we get the following well known formula: Lemma 1. One has the identity µ ¶ µ ¶ µ ¶ k k k k (n + 1) − (n + 1) = S1 (n) + S2 (n) + · · · + Sk−1 (n). 1 2 k−1 In what follows, we shall need also the following simple (but not so wellknown) lemma: µ ¶ µ ¶ m n−1 Lemma 2. If (m, k) = 1, then m| and k| . k k−1 Proof. The trivial identity m (m − 1) . . . (m − k + 1) m(m − 1) . . . (m − k + 1) = · 1 · 2...k k 1 · 2 . . . (k − 1) gives at once the combinatorial relation µ ¶ µ ¶ m m m−1 = . k k−1 k µ ¶ µ ¶ m m−1 This implies k =m . By assuming (m, k) = 1, since m dik k−1 vides the right µ hand ¶ side, by Euclid’s theorem (or Gauss’ µ lemma) ¶ it follows m m−1 that m divides on the left hand side. Similarly, k| . k k−1 µ ¶ p Corollary. If p is a prime number, then p| for all 1 ≤ k < p. k One has (p, k) = 1 and Lemma 2 applies. We now are able to prove the main result of this note: Theorem. Let (a, b) denote the g.c.d. of a and b. The following congruence is true: (n + 1)k − (n + 1) ≡ A(k, n) (mod k), 415
where A(k, n) =
X 1≤l≤k−1, (l,k)≥2
µ ¶ k Sl (n) l
and an empty sum is considered to µ be ¶ 0. k Proof. By Lemma 2 one has k| for all (k, l) = 1. By writing the sum l of Lemma 1 in two sums, namely the sum of terms with (k, l) = 1, and the otherone with (k, l) ≥ 2, we can write µ ¶ X k k (n + 1) − (n + 1) = M k + Sl (n) , M = integer, l 1≤l≤k−1, (l,k)≥2
and this implies the stated congruence. Remarks. 1) If k is prime, then by the Corollary one has A(k, n) = 0, so we can deduce the (”little”) Fermat theorem: (n + 1)k − (n + 1) ≡ 0
(mod k).
2) We have obtained also a new proof (without mathematical induction) of Fermat’s theorem. We note that this theorem has important applications e.g. in Cryotography (see e.g. [1]).
References [1] J. S´andor (in coop with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005.
3
On an inequality of Klamkin
1. Introduction In 1974 M. S. Klamkin [3] proved the following result: Let x be a nonnegative real number, and m, n integers with m ≥ n ≥ 1. Then (m + n)(1 + xm ) ≥ 2n
1 − xm+n . 1 − xn
(1)
We note that for x = 1, the right side of (1) is understood as lim , when x→1
the inequality becomes an equality. Also, for x = 0 (1) becomes m + n ≥ 2n, which is true. For m = n there is equality in (1). In fact, it can be shown 416
that for all real numbers m > n > 0, and all x > 0, (1) holds true with strict inequality (see the solutions of (1) in [1]). Assume now that x = a ≥ 1, m = p, n = q, where p ≥ q ≥ 0 are real numbers. Then, since (1 + ap )(1 − aq ) = ap − aq + 1 − ap+q , after some transformations, (1) becomes equivalent to (p − q)(ap+q − 1) ≥ (p + q)(ap − aq ).
(2)
In the case of p − q ≤ 1, a weaker result than (2) appears in the famous monograph by D. S. Mitrinovi´c [4] (3.6.26, page 276). For certain arithmetical applications of Klamkin’s inequality, see [5]. In what follows we will point out some surprising connections of inequality (2) (i.e., in fact (1)) with certain special means of two arguments. Also, a new application of (1) will be given.
2. Stolarsky means Let m, n > 0 and put p + q = m, p − q = n. Then p = and (2) gives
m+n m−n ,q= 2 2
am − 1 m > a(m−n)/2 . an − 1 n
By letting a =
(3)
x (x > y > 0), relation (3) may be written also as y µ
xm − y n n · xn − yn m
¶1/(m−n) >
√ xy.
(4)
If n = 1, the expression on the left side of (4) is called as the Stolarsky mean of x and y. Put µ S(m) = S(x, y, m) =
xm − y m 1 · x−y m
¶1/(m−1) .
It is not difficult to see that S can be defined also for all real numbers m 6∈ {0, 1}, while for m = 0, and m = 1, by the limits lim S(x, y, m) =
m→0
417
x−y ln x − ln y
and
1 lim S(x, y, m) = (y y /xx )1/(y−x) (y 6= x), m→1 e the definition of S can be extended to all real numbers m. Let x−y 1 L(x, y) = , I(x, y) = (y y /xx )1/(y−x) (x 6= y), ln x − ln y e L(x, x) = I(x, x) = x. These means are known as the logarithmic and identric means of x and y (see e.g. [8] for their properties). Stolarsky [10] has proved that S is a strictly increasing function of m. Therefore S(−1) < S(0) < S(1) < S(2), giving x+y √ . (5) xy < L(x, y) < I(x, y) < 2 Since S(−1) < S(0) < S(m) for m > 0, we get ¶ µ m x − y m 1/(m−1) √ xy < L(x, y) < , (6) m(x − y) which is an improvement of (4), when n = 1.
3. Main results We shall prove that the following refinement of (4) holds true: Theorem 1. µ m ¶ x − y m n 1/(m−n) √ m−n m−n 1/(m−n) xy < (L(x ,y )) < · (m > n). (7) xn − y n m ax − 1 f (p + q) (x > 0), where a > 1; and let ϕ(p) = x f (p − q) (p > q > 0), where q is fixed. We first show that ϕ is strictly increasing function. Since Proof. Put f (x) =
ϕ0 (p) =
f 0 (p + q)f (p − q) − f 0 (p − q)f (p + q) , f 2 (p − q)
f 0 (p + q) f 0 (p − q) > . Since p + q > p − q, f (p + q) f (p − q) this will follow, if f 0 /f = g is an increasing function. By
it will be sufficient to prove that
g 0 (t) = (f 0 (t)/f (t))0 =
f 00 (t)f (t) − (f 0 (t))2 , f 2 (t)
418
it will be sufficient to show that f is strictly log-convex (i.e. ln f is strictly convex). Lemma. The function f is strictly log-convex. Proof. After certain simple computations (which we omit here), it follows that tat ln t − (at − 1) , f 0 (t) = t2 f 00 (t) =
t2 at ln2 a − 2tat ln a + 2at − 2 , t3
and a2t − 2at − t2 at ln2 a + 1 t4 √ √ (at − 10 at ln at )(at − 1 + at ln at ) = . t4 √ √ h−1 √ Put at = h. Then h − 1 − h ln h > 0, since > h by L(h, 1) > h ln h (left side of (5)). This proves the log-convexity property of f for a > 1. Since ϕ is strictly increasing, one can write f 00 (t)f (t) − (f 0 (t))2 =
ϕ(p) >
lim
p→q,p>q
f (p + q)/f (p − q) =
a2q − 1 . 2q ln a
x , and the right side of (7) follows. y For the left side of (7) remark that again by the left side of (5) one has Write p + q = m, p − q = n, a =
L(xm−n , y m−n ) >
p xm−n y m−n = (xy)(m−n)/2 ,
which implies the desired inequality. Remark. ϕ being strictly increasing, it follows also that ap+q − 1 p − q · = a2q , p→∞ ap−q − 1 p + q
ϕ(p) < lim i.e.
(p − q)(ap+q − 1) ≤ (p + q)a2q (ap−q − 1), which is complementary to (2). 419
(8)
4. Arithmetical applications A divisor d of N is called unitary divisor of the positive integer N > 1, if (d, N/d) = 1. For k ≥ 0, let σk (N ) resp. σk∗ (N ) denote the sum of kth powers of divisors, resp. unitary divisors of N . Remark that σ0 (N ) = d(N ), σ0∗ (N ) = d∗ (N ) are the number of these divisors of N . It is well-known that (see e.g. [3], [9]) if the prime factorization of N is N=
r Y
pai i
i=1
(pi distinct primes, ai ≥ 1 integers), then σk (N ) =
r r Y Y k(a +1) (pi i − 1)/(pki − 1), d(N ) = (ai + 1), i=1
i=1
r Y ∗ i σk (N ) = (pka + 1), d∗ (N ) = 2r (= 2ω(N ) ), i
(9)
i=1
where ω(r) = r denotes the number of distinct prime divisors of N . Write now (1), and a reverse of it (see [1]) in the form 2n
xm+n − 1 xm+n − 1 m ≤ (m + n)(1 + x ) ≤ 2m , xn − 1 xn − 1
(10)
where x > 1, m ≥ n ≥ 1. Put n = k, m = kai , x = pi (i = 1, 2, . . . , r). Writing (10), after term-by-term multiplication, we get 2ω(N ) σk (N ) ≤ d(N )σk∗ (N ) ≤ 2ω(N ) β(N )σk (N ), where β(N ) =
r Y
(11)
ai (for this, and the other functions, too, see e.g. [6], [9]).
i=1
The left side of (11) appears also in [5]. Now, remarking that 2
ω(N )
r r Y Y β(N ) = (2ai ) ≤ 2ai = 2Ω(N ) , i=1
i=1
where Ω(N ) denotes the total number of prime factors of N (we have used the classical inequality 2a−1 ≥ a for all a ≥ 1), relation (11) implies also 2ω(N ) ≤
d(N )σk∗ (N ) ≤ 2Ω(N ) . σk (N ) 420
(12)
Theorem 2. The normal order of magnitude of log(d(N )σk∗ (N )/σk (N )) is (log 2)(log log N ). Proof. Let P be a property in the set of positive integers and set X ap (n) = 1 if n has the property P ; ap (n) = 0, otherwise. Let Ap (x) = ap (n). n≤x
If Ap (x) ∼ x (x → ∞) we say that the property P holds for almost all natural numbers. We say that the normal order of magnitude of the arithmetical function f (n) is the function g(n), if for each ε > 0, the inequality |f (n) − g(n)| < εg(n) holds true for almost all positive integers n. By a well-known result of Hardy and Ramanujan (see e.g. [2], [4], [6]), the normal order of magnitude of ω(N ) and Ω(N ) is log log N . By (12) we can write 1 (1 − ε)(log log N ) < ω(N ) ≤ log d(N )σk∗ (N )/σk (N ) log 2 ≤ Ω(N ) < (1 + ε) lg lg N for almost all N , so Theorem 2 follows. Acknowledgements. The author thanks Professor Klamkin for sending him a copy of [1] and for his interest in applications of his inequality.
References [1] J. M. Brown, M. S. Klamkin, B. Lepson, R. K. Meany, A. Stenger, P. Zwier, Solutions to Problem E2483, Amer. Math. Monthly, 82(1975), 758-760. [2] G. H. Hardy, E. M. Wright, An introduction to the theory of numbers, Oxford Univ. Press, 1960. [3] M. S. Klamkin, Problem E2483, Amer. Math. Monthly, 81(1974), 660. [4] E. Kr¨atzel, Zahlentheorie, Berlin, 1981. [5] D. S. Mitrinovi´c, Analytic inequalities, Springer Verlag, 1970. [6] D. S. Mitrinovi´c, J. S´andor (in coop. with B. Crstici), Handbook of number theory, Kluwer Acad. Publ., 1995. [7] J. S´andor, On an inequality of Klamkin with arithmetical applications, Int. J. Math. E. Sci. Techn., 25(1994), 157-158. 421
[8] J. S´andor, On the identric and logarithmic means, Aequationes Math., 40(1990), 261-270. [9] J. S´andor (in coop. with B. Crstici), Handbook of number theory, II, Springer Verlag, 2005. [10] K. B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545-548.
4
Euler-pretty numbers
Let ϕ be the Euler totient. For a fixed positive rational number k, we say that the pair (a, b) is k-Euler-pretty, if ϕ(a) =
b k
and ϕ(b) =
a . k
(1)
Our aim is to solve the system (1) for k = 2, i.e. to find all 2-Euler-pretty pairs: Theorem. All 2-Euler-pretty numbers are given by a = 2m , b = 2m , where m is an arbitrary positive integer. The proof of this result is based on the following auxiliary result. Lemma. For all positive integers u, v one has ϕ(uv) ≤ uϕ(v).
(2)
¶ Yµ ¶ Yµ ϕ(uv) 1 1 ϕ(v) Proof. = 1− ≤ 1− = , since for any prime uv p p v p|uv
p|v
1 divisor q - v one has 1 − ≤ 1. Thus (2) follows. There is equality only if each q prime divisor of u is a prime divisor of v, too. To prove the theorem, first remark that for k = 2, a and b must be even, so a = 2A, b = 2B. Then the system (1) becomes ϕ(2A) = B, ϕ(2B) = A.
(3)
Applying two times relation (2), one can write: B ≤ Aϕ(2) = A and A ≤ Bϕ(2) = B, since ϕ(2) = 1. Therefore B = A = x, and (3) becomes ϕ(2x) = x. 422
(4)
Apply again (2) for (4), one has: ϕ(2x) ≤ xϕ(2) = x, with equality only if each prime factor of x is also a prime factor of 2. In other words, x = 2s . Since ϕ(2) = 1, one may assume s ≥ 0. Thus a = 2s+1 = 2m , b = 2s+1 = 2m , where m ≥ 1. This finishes the proof of the theorem. Remark. By (3) we have solved in fact the following equation: ϕ(2ϕ(2A)) = A.
(5)
The more general system (1) can be reduced (for positive integers k), to a similar equation, too, namely ϕ(kϕ(kA)) = A.
(6)
What are the most general solutions of this equation? An equation of type (4) has been studied, by other arguments, in [1].
References [1] J. S´andor, On the Open Problem PD.90 (Romanian), Gamma, 11(1988), no.1-2, 26-27.
5
Abundant numbers involving the smallest and largest prime factors
Problem 11 of [1] asks for abundant numbers n such that p(n) + P (n) is also abundant. Here p(n), resp. P (n) denote the smallest, resp. largest prime factors of n. Recall that n is called abundant, if σ(n) > 2n, where σ(n) denotes the sum of divisors of n. Our aim is to prove the following results: Theorem 1. Let n = 51·5b ·7c , where b ≥ 2 and c ≥ 1 are positive integers. Then p(n) + P (n) and n are abundant at the same time. Proof. Since 51 = 3 · 17, clearly p(n) + P (n) = 3 + 17 = 20, an abundant number: σ(20) = σ(4) · σ(5) = 42 > 40. On the other hand, by the multiplicatively property of σ one can write σ(n) = 4 · 18
5b+1 − 1 7c+1 − 1 · > 2 · 51 · 5b · 7c 4 6
iff (5b+1 − 1)(7c+1 − 1) > 34 · 5b · 7c , or 5b · 7c − 5b+1 − 7c+1 + 1 > 0. Equivalently, (5b − 7)(7c − 5) > 34 423
(1)
Now, in (1) 7c − 5 ≥ 2 for any c ≥ 1 and 5b − 7 ≥ 18 for any b ≥ 2. Since 2 · 18 = 36 > 34, relation (1) is true. Remark 1. Therefore for b = 2, 3, . . . , c = 1, 2, . . . we get such numbers as n = 8925, 44625, 62475, 223125, . . . Problem 12 of [1] asks for abundant numbers n such that the sum of all composite numbers strictly between p(n) and P (n) is also abundant. The following is true: Theorem 2. For any a ≥ 1, b ≥ 1, the number n = 19 · 2a · 3b satisfies the required property. Proof. Since p(n) = 2, P (n) = 19, the sum of the composite numbers between 3 and 18 is 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 112. Since σ(112) = 248 > 224, this is abundant number. Similarly, µ a+1
σ(n) = (2
− 1)
3b+1 − 1 2
¶ · 20 > 2 · 2a · 3b · 19
iff 5(2a+1 ·3b+1 −2a+1 −3b+1 +1) > 19·2a ·3b , or 11·2a ·3b −10·2a −15·3b +5 > 0. This can be written also as 2a ·(11·3b −10) > 15·3b −5. Now 2a ·(11·3b −10) ≥ 2 · (11 · 3b − 10) = 22 · 3b − 20 > 15 · 3b − 5 by 7 · 3b > 15, which is true for any b ≥ 1, since 7 · 3b ≥ 7 · 3 = 21. Remark 2. For a = 1, 2, . . . , b = 1, 2, . . . we get various numbers n = 114, 228, 342, 456, 684, . . .
References [1] J. Earls, Some Smarandache-type sequences and problems concerning abundant and deficient numbers, Smarandache Notions J., 14(2004), 243-250.
6
An inequality with sh x and ch x We will determine all α, β, γ > 0 such that µ
sh x x
¶β
µ +
Here sh x =
sh x x
¶γ
≤ (ch x + 1)α for all x ∈ R.
ex − e−x x x3 x5 = + + + ... 2 1! 3! 5! 424
(1)
and
ex + e−x x2 x4 =1+ + + ..., 2 2! 4! where we have used the classical formula ch s =
et = 1 +
t t2 + + ... 1! 2!
Now (1) for β = γ = α = 1 can be written as sh x x2 x4 ch x + 1 x2 x4 =1+ + + ··· ≤ =1+ + + ... x 3! 5! 2 2 · 2! 2 · 4!
(2)
sh 0 = 1 since lim shxx = 1). Now (2) trivially holds true for x = 0, while for x→0 0 x 6= 0 remark that 1 1 1 1 < , < ,... 3! 2 · 2! 5! 2 · 4!
(
sh x ≥ 1, for all x ∈ R, b = ch x + 1 ≥ 2. For x = 0 relation x α (2) gives 2 ≤ 2 , so α ≥ 1. By (2) one has b ≥ 2a. Clearly bα ≥ (2a)α so if β ≤ α, γ ≤ α, then aβ−α + aγ−α ≤ 2 ≤ 2α by α ≥ 1. Therefore, inequality (1) holds true for all α ≥ 1 and all 0 < β ≤ α, 0 < γ ≤ α. If, for example, β > α, then one can obtain a > 1, b > 2a so that aβ + aγ > bα (indeed, for ln b , one has β ln a > α ln b, so for β > α(1 + ε), ε > 0 the inequality β >α ln a is valid). In the same manner, γ > α implies the reverse inequality for some sh t , t > 0 and g(t) = ch t + 1, t > 0, f : (0, ∞) → (1, ∞), a, b. Since f (t) = t g : (0, ∞) → (2, ∞) are surjective functions, all solutions of (1) are Now, let a =
α ≥ 1,
7
0 < β ≤ α,
0 < γ ≤ α.
(3)
On geometric numbers
1. Introduction
³ n´ Let d be a positive divisor of the integer n > 1. If d, = 1, then d is d called a unitary divisor of n. If the greatest common unitary divisor of d and n/d is 1, then d is called a bi-unitary divisor of n. If n = pa11 . . . par r > 1 is the prime factorization of n, a divisor d of n is called an exponential divisor (or e-divisor, for short), if d = pb11 . . . pbrr , with bi |ai (i = 1, r). For the history 425
of these notions, as well as the connected arithmetical functions, see e.g. [2], [7]. Let T (n), T ∗ (n), T ∗∗ , resp. Te (n) denote the product of divisors, unitary divisors, bi-unitary divisors, resp. e-divisors of n. It½ is easily seen ¾ that, if n n ,..., , implying d1 , . . . , ds are all divisors of n, then {d1 , . . . , ds } = d1 ds n n that d1 . . . ds = . . . , giving d1 ds T (n) = nd(n)/2 ,
(1)
where s = d(n) is the number of divisors of n. If {d∗1 , . .½. , d∗m } are all¾unitary n n , . . . , ∗ , so, in divisors of n, then one has similarly, {d∗1 , . . . , d∗m } = ∗ d1 dm analogy with (1), one has ∗ T ∗ (n) = nd (n)/2 , (2) where m = d∗ (n) denotes the number of unitary divisors of n. In a completely similar manner, one can write ∗∗ (n)/2
T ∗∗ (n) = nd
,
(3)
where d∗∗ (n) denotes the number of bi-unitary divisors of n. It is well known, that if p = paa1 . . . par r > 1 is the prime representation of n, then d(n) = (a1 + 1) . . . (ar + 1), d∗ (n) = 2r , d∗∗ (n) =
Y ai even
ai
Y
(aj + 1). (4)
aj odd
For Te (n), the things are slightly more complicated, for the following formula holds true (see [6]): σ(a1 )d(a2 )...d(ar )
Te (n) = p1 where σ(k) =
X
r )d(a1 )...d(ar−1 ) . . . pσ(a , r
(5)
d is the sum of (ordinary) divisors of k.
d|k
Let σ ∗ (k), resp. σ ∗∗ (k) denote the sum of unitary, resp. bi-unitary divisors of k. In 1948 O. Ore [3] has studied the classical means related to the divisors of an integer. Let A(n), G(n), H(n) denote the arithmetic, geometric, and harmonic means of all ordinary divisors of n, i.e. A(n) =
d1 + · · · + ds , s 426
G(n) = (d1 . . . ds )1/s ,
µ H(n) = s/
¶
1 1 + ··· + d1 ds
.
It is immediate from the above notations, that A(n) =
σ(n) , d(n)
G(n) = (T (n))1/d(n) ,
where the last formula is a consequence of
H(n) =
X1 d|n
=
d
nd(n) , σ(n)
(6)
1X d. n d|n
Ore called a number n arithmetic, geometric, resp. harmonic, if A(n), G(n), resp. H(n) is integer. Similarly, other authors have studied the unitary analogues A∗ (n) =
σ ∗ (n) , d∗ (n)
∗ (n)
G∗ (n) = (T ∗ (n))1/d
,
H ∗ (n) =
nd∗ (n) . σ ∗ (n)
(7)
Recently, the present author [5] has studied the bi-unitary harmonic numbers n, i.e. with H ∗∗ (n) an integer; where H ∗∗ (n) is the third of the following means: A∗∗ (n) =
σ ∗∗ (n) , d∗∗ (n)
∗∗ (n)
G∗∗ (n) = (T ∗∗ (n))1/d
,
H ∗∗ (n) =
nd∗∗ (n) . σ ∗∗ (n)
(8)
He also introduced some variants of these bi-unitary harmonic numbers, e.g. H1 (n) =
nd(n) , σ ∗ (n)
H2 (n) =
nd∗ (n) , etc. σ(n)
(four other related fractions, involving also d∗∗ (n), σ ∗∗ (n)). It is immediate that, by (1)-(3), one has G(n) = G∗ (n) = G∗∗ (n) =
√ n,
(9)
so a geometric number, etc., is in fact a perfect square. We note that actually the relationships of arithmetic, geometric and harmonic numbers is not perfectly known, for example, it is conjectured (but not proved up to now, see e.g. [1]) that no harmonic number is geometric (i.e., a perfect square). Many harmonic numbers seem to be arithmetic, too; but their proportion is not clarified (see e.g. [2], [7]). Our aim in what follows is to introduce and study certain notions of geometric numbers, which are not so obvious than the ones given by (9). 427
2. Main notions and results The geometric mean of e-divisors is clearly given by Ge (n) = (Te (n))1/de (n) , and since it is known that de (n) = d(a1 ) . . . d(ar ) (see e.g. [7]), then by (5) one can write: σ(a )/d(a1 ) r )/d(ar ) Ge (n) = p1 1 . . . pσ(a . (10) r Let us now introduce the following expressions: ∗ (n)
G1 (n) = (T ∗ (n))1/d(n) ,
G2 (n) = (T (n))1/d
G3 (n) = (T ∗∗ (n))1/d(n) ,
G4 (n) = (T (n))1/d
∗∗ (n)
G5 (n) = (T ∗ (n))1/d
,
,
∗∗ (n)
,
∗ (n)
G6 (n) = (T ∗∗ (n))1/d
.
(11)
We note that, we could introduce also G1e (n) = (T ∗ (n))1/de (n) , etc., but these expressions will not be considered here. In what follows ω(n) = r will denote the number of distinct prime factors of n. We say that n is G1 -geometric (or G1 -number), if G1 (n) is an integer. The G2 -, etc. numbers are defined in a similar manner. We say that n is e-geometric, if Ge (n) is an integer. Theorem 1. Let n = pa11 . . . par r > 1 be the canonical factorization of n. Then n is e-geometric number if and only if all of a1 , . . . , ar are (ordinary) arithmetic numbers. Proof. The proof is a consequence of (10); the definition of arithmetic numbers, and the following auxiliary result: Lemma 1. Let b = pα1 1 . . . pαr r > 1, where pi are distinct primes, and αi are positive rational numbers. Then n is an integer if and only if all of α1 , . . . , αr are integers. b1 br Proof of the Lemma. Let us write α1 = , . . . , αr = , with common k k nominator k, and b1 , . . . , br positive integers. Then nk = pb11 . . . pbrr , so n cannot have other prime divisors, then p1 , . . . , pr . Indeed, if p|nk , (i.e. p|n), then p mr be 1 divides one of pb11 , . . . , pbrr , so p is one of p1 , . . . , pr . Let n = pm 1 . . . pr the prime factorization of n. Then by the unique factorization theorem one b1 br must have km1 = b1 , . . . , kmr = br , so = α1 , . . . , = αr are the integers k k m1 , . . . , mr . Lemma 1 clearly implies Theorem 1. 428
Remark. Two notions of e-harmonic numbers have been recently introduced by the present author in [5]. Lemma 2. One has, for all n ≥ 1: d∗ (n)|d∗∗ (n)
and
d∗∗ (n) ≤ d(n).
(12)
Proof. Let us suppose that there are t even numbers, and q odd num∗∗ bers Y between Y a1 , . . . , ar (t, q ≥ 0, t + q = r). Then, clearly d (n) = t q t+q r ∗ ai (aj +1) is divisible by 2 2 = 2 = 2 = d (n). This proved the ai even
aj odd
divisibility relation of (12). The second relation is trivial, by remaking that for ai even one has ai < ai + 1. It is immediate that in the first relation one has equality only if n has the form n = pε11 . . . pεrr , where εi ∈ {1, 2} for all i = 1, r. There is equality in the second relation of (12) only if all of ai are odd (in which case, one has also σ ∗∗ (n) = σ(n), so A∗∗ (n) = A(n), H ∗∗ (n) = H(n), too). Corollary. d∗ (n) ≤ d∗∗ (n) ≤ d(n) for all n ≥ 1. Theorem 2. For n > 1, there are no G1 -numbers n. The general form of G5 -numbers n is n = p21 . . . p2r , where pi , i = 1, r, are distinct primes. ∗ α d∗ (n)/2d(n) α d∗ (n)/2d(n) Proof. Since G1 (n) = nd (n)/2d(n) = p1 1 . . . pr r , by Lemma 1, one must have 2d(n)|α1 d∗ (n), etc., i.e. 2(α1 + 1) . . . (αr + 1)|α1 · 2r . But (α2 + 1) . . . (αr + 1) ≥ 2r−1 , and α1 + 1 > α1 , thus 2(α1 + 1) . . . (αr + 1) > α1 · 2r , in contradiction with the above divisibility relation. This proves the first part of Theorem 2. For the second part, note that, since G5 (n) = nd
∗ (n)/2d∗∗ (n)
α d∗ (n)/2d∗∗ (n)
= p1 1
. . . pαr r d
∗ (n)/2d∗∗ (n)
,
we must have Y Y Y Y 2 αi (αj + 1)|α1 · 2r , . . . , 2 αi (αj + 1)|αr · 2r , where αi are the even, while αj the odd exponents. Now, remark that if α1 is odd, then the left side is divisible by 2r+1 , but the right side only by 2r , a contradiction. So α1 , . . . , αr are all even. Since 2α1 . . . αr |α1 2r implies α2 . . . αr |2r−1 , and since 2r−1 |α2 . . . αr , we must have α2 = · · · = αr = 1. Similarly, α1 = 1, and this finishes the proof of the second part of Theorem 2. Theorem 3. n > 1 is a G6 -number if and only if a prime power pα kn satisfies one of the following conditions: a) α is even; 429
b) α ≡ −1 (mod 4); c) α is odd, α 6≡ −1 (mod 4), d∗∗ (n/(α + 1)) ≡ 0 (mod 2ω(n) ). The number n > 1 is a G2 -number, if and only if αd(n) ≡ 0
(mod 2ω(n)+1 ).
∗∗ (n)/2d∗ (n)
Proof. G6 (n) = nd
, so by Lemma 1 we must have Y Y 2r+1 |α αi (αj + 1),
where αi are the even, and αj are the odd exponents. In the cases a) and b) these are always true, by Lemma 2. Now, case c) follows by a direct verification. ∗ Since G2 (n) = nd(n)/2d (n) , by Lemma 1 we must have 2r+1 |αd(n), where r = ω(n). Corollary. Any ordinary geometric number if a G6 number. Any number n = pα1 1 . . . pαr r with αi ≡ −1 (mod 4), i = 1, r, is a G2 -number. Indeed, if n is perfect square, then all α are even, and case a) of Theorem 3 applies. If 4|(α + 1), then clearly 4r |d(n), and 2r+1 |4r for any r ≥ 1 (r = ω(n)) implies the result. Theorem 4. There are no G4 -numbers n > 1. ∗∗ Proof. Let n = pα1 1 . . . pαr r > 1. Then G4 (n) = nd(n)/2d (n) , so by Lemma 1, n is a G4 -number iff 2d∗∗ (n)|αd(n) for all α. Remark that if all α are odd, then d∗∗ (n) = d(n), so 2|α, and this is impossible. Therefore, there exists αi = even. By relation (4), one obtains Y Y 2 αi |α (αi + 1) for all α. Particularly, when Y α is one Yof αi , this is impossible. Indeed, by reducing with α, one obtains 2 αi | (αi + 1), where the left side is even, while the αi 6=α
right side odd. This gives a desired contradiction. Theorem 5. Let n = pα1 1 . . . pαr r > 1. Then n is a G3 -number iff for any prime power pα kn one has Y Y 2 (αi + 1)|α αi , (∗) where αi denote the even exponents (here an empty product is taken to be 1). ∗∗ Proof. G3 (n) = nd (n)/2d(n) , so 2d(n)|αd∗∗ (n). By using relation (4), this reduces to (∗). 430
Remarks. If all α are odd, then (∗) reduces to 2|α, which is impossible. If r = 1, then n = pα with α even, so 2(α+1)|α2 , which cannot be true for any α, 2b+1 2b since (α+1, α2 ) = 1. If r = 2, then n = p2a or n = p2a 1 p2 1 p2 . In the first case (∗) becomes 2(2a + 1)|α(2a) for α = 2a and α = 2b + 1. But 2(2a + 1)|(2a)2 is not true by (2a + 1, (2a)2 ) = 1. In the second case, 2(2a + 1)(2b + 1)|α(2a)(2b) for α = 2a and α = 2b. Here 2(2a + 1)(2b + 1)|2a(2a)(2b) by (2a + 1, 4a2 ) = 1 implies (2a+1)|b; and similarly (2b+1, 4b2 ) = 1 gives (2b+1)|a. Since b ≥ 2a+1 and a ≥ 2b + 1 ≥ 4a + 3, this is an obvious contradiction. Therefore, for all G3 -numbers n one has ω(n) ≥ 3.
References [1] G.L. Cohen, R.M. Sordy, Harmonic seeds, Fib. Quart. 36(1998), 386390. [2] R.K. Guy, Unsolved problems in number theory, Third ed., Springer Verlag, 2004. [3] O. Ore, On the averages of the divisors of a number, Amer. Math. Monthly, 55(1948), 615-619. [4] J. S´andor, On bi-unitary harmonic numbers, submitted. [5] J. S´andor, On exponentially harmonic numbers, submitted. [6] J. S´andor, On multiplicatively e-perfect numbers. J. Ineq. Pure Appl. Math., 5(2004), no.4, article 114 (electronic). [7] J. S´andor, Handbook of number theory, II, Springer Verlag, 2004.
8
The sum-of-divisors minimum and maximum functions
1. Let f : N∗ → N be a given arithmetic function, and A ⊂ N∗ a given set. The arithmetic function FfA (n) = min{k ∈ A : n|f (k)}
(1)
has been introduced in [7] and [6]. For A = N∗ , f (k) = k! one obtains the Smarandache function; for A = N∗ , A = P = {2, 3, 5, . . . } = set of all primes, one obtains a function P (n) = min{k ∈ P : n|k!} 431
(2)
For properties of this function, see [7], [6]. For A = {k 2 : k ∈ N∗ } = set of perfect squares, and f (k) = k! one obtains the function Q(n) = min{m2 ≥ 1 : n|(m2 )!}, (3) while for A = set of squarefree numbers ≥ 1, f (k) = k! we get Q1 (n) = min{m ≥ 1 squarefree: n|m!}
(4)
For properties of Q(n) and Q1 (n), see [11]. The ”dual” function of (1) has been defined by GA g (n) = max{k ∈ A : g(k)|n}
(5)
where g : N∗ → N is a given function. Particularly for A = N∗ , g(k) = k! one obtains the dual of the Smarandache function S∗ (n) = max{k ≥ 1 : k!|n}
(6)
For properties of this function, see [7], [6]. F. Luca [4], K. Atanassov [1] and M. Le [2] have proved in the affirmative a conjecture of the author stated in [7] and [6]. For A = N∗ , f (k) = g(k) = ϕ(k) (where ϕ is Euler’s totient) in (1), resp. (5) one obtains the Euler minimum, resp. maximum-functions, defined by E(n) = min{k ≥ 1 : n|ϕ(k)},
(7)
E∗ (n) = max{k ≥ 1 : ϕ(k)|n}
(8)
For properties of these functions, see [5], [8]. When A = N∗ , f (k) = d(k) = number of divisors of k, one has the divisor minimum function (see [7], [6], [9]): D(n) = min{k ≥ 1 : n|d(k)}
(9)
It is interesting to note that the divisor maximum function (i.e. the ”dual” of D(n)) given by D∗ (n) = max{k ≥ 1 : d(k)|n} (10) is not well-defined! Indeed, for any prime p we have d(pn−1 ) = n, and pn−1 is unbounded as p → ∞. When A is a finite set, however, D∗A (n) = max{k ∈ A : d(k)|n} 432
(11)
does exist. When A = N∗ , f (k) = g(k) = S(k) = min{m ≥ 1 : k|m!} (Smarandache function) one obtains the Smarandache minimum and maximum functions, given by Smin (n) = min{k ≥ 1 : n|S(k)}, (12) Smax (n) = max{k ≥ 1 : S(k)|n}.
(13)
These functions have been introduced and studied recently in [10]. 2. Let σ(n) be the sum of divisors of n. The function Σ(n) = min{k ≥ 1 : n|σ(k)}
(14)
has been introduced in [7], [6] (denoted there by Fσ ). Let k be a prime of the form k = an − 1, where n ≥ 1 is given. By Dirichlet’s theorem on arithmetical progressions, such a prime does exist. Then clearly σ(k) = an, so n|σ(k), and Σ(n) is well defined. The dual of Σ(n) is Σ∗ (n) = max{k ≥ 1 : σ(k)|n}
(15)
Since σ(1) = 1|n and σ(k) ≥ k, clearly Σ∗ (n) ≤ n, so this function is correctly defined. The aim of this note is the initial study of these functions Σ(n) and Σ∗ (n). Some values of Σ(n) are: Σ(1) = 1, Σ(2) = 3, Σ(3) = 2, Σ(4) = 3, Σ(5) = 8, Σ(6) = 5, Σ(7) = 4, Σ(8) = 7, Σ(9) = 10, Σ(11) = 43, Σ(12) = 6, Σ(13) = 9, Σ(14) = 12, Σ(15) = 8, Σ(16) = 21, Σ(17) = 67, Σ(18) = 10, Σ(19) = 37, Σ(20) = 19, Σ(21) = 20, Σ(22) = 43, Σ(23) = 137, Σ(24) = 14, Σ(25) = 149, Σ(26) = 45, Σ(27) = 34, Σ(28) = 12, Σ∗ (1) = 1, Σ∗ (2) = 1, Σ∗ (3) = 2, Σ∗ (4) = 3, Σ∗ (5) = 1, Σ∗ (6) = 5, Σ∗ (7) = 4, Σ∗ (8) = 7, Σ∗ (9) = 2, Σ∗ (10) = 1, Σ∗ (11) = 1, Σ∗ (12) = 11, Σ∗ (13) = 9, Σ∗ (14) = 13, Σ∗ (15) = 8, Σ∗ (16) = 7, Σ∗ (17) = 1, Σ∗ (18) = 17, Σ∗ (19) = 1, Σ∗ (20) = 19, Σ∗ (21) = 4, Σ∗ (22) = 1, Σ∗ (23) = 1, Σ∗ (24) = 23, Σ∗ (25) = 1, Σ∗ (26) = 9, Σ∗ (27) = 2, Σ∗ (28) = 12. 3. The first theoretical result gives informations on values of these functions at n = p + 1, where p is a prime: Theorem 1. If p is a prime, then Σ(p + 1) ≤ p ≤ Σ∗ (p + 1)
(16)
Proof. Since (p+1)|σ(p) = p+1, by definition (14) one can write Σ(p+1) ≤ p. Similarly, definition (15) gives (by σ(p) = (p + 1)|(p + 1)) Σ∗ (p + 1) ≥ p. 433
Remark. On the left side of (16) one can have equality, e.g. Σ(3) = 2, Σ(6) = 5, Σ(8) = 7. But the inequality can be strict, as Σ(12) = 6 < 11, Σ(18) = 10 < 17. For the right side of (16) however, one can prove the more precise result: Theorem 2. For all primes p, one has Σ∗ (p + 1) = p
(17)
Proof. First we prove that for all n ≥ 2 we have Σ∗ (n) ≤ n − 1
(18)
Indeed, since σ(k)|n, clearly we must have σ(k) ≤ n. On the other hand, for all k ≥ 2 we have σ(k) ≥ k + 1 (with equality only for k = prime), so k ≤ n − 1, and this is true for all k, so (18) follows. Let now n = p + 1 ≥ 3 in (18). Then Σ∗ (p + 1) ≤ p, which combined with (16) implies relation (17). Theorem 3. Let p be a prime and suppose that (p + 1)|n
(19)
Σ∗ (n) ≥ p
(20)
Then
Proof. Indeed, by σ(p) = (p + 1)|n, and definition (15), relation (20) follows. By letting p = 2, 3, 5, 7, 11 one gets: Corollary. If 3|n, then Σ∗ (n) ≥ 2. (21) If 4|n, then Σ∗ (n) ≥ 3. (22) If 6|n, then Σ∗ (n) ≥ 5. (23) If 8|n, then Σ∗ (n) ≥ 7. (24) If 12|n, then Σ∗ (n) ≥ 11. (25) Remark. If 7|n, then Σ∗ (n) ≥ 4. (26) Indeed, σ(4) = 7|n. If 15|n, then Σ∗ (n) ≥ 8. (27) Indeed, σ(8) = 15|n. It is immediate that Σ(n) = 1 only for n = 1. On the other hand, there exist many integers m with Σ∗ (m) = 1. Theorem 4. Let p be a prime such that p 6∈ σ(N∗ ) 434
(28)
Then Σ∗ (p) = 1
(29)
Proof. Remark that σ(k)|p ⇔ σ(k) = 1 or σ(k) = p. Now, if (28) is true, then the equation σ(k) = p is impossible for all k ≥ 1, so σ(k) = 1, i.e. k = 1, giving relation (29). For example, p = 17, 19, 23 satisfy relation (28). Theorem 5. If for all d > 1, d|n one has d 6∈ σ(N∗ ),
(30)
Σ∗ (n) = 1
(31)
then
Proof. Let d > 1, d|n. If d 6∈ σ(N∗ ), then the equation σ(k) = d is impossible. But then σ(k)|n is also impossible for σ(k) > 1, yielding (31). For example, n = 10, 22, 25 satisfy relation (30). Theorem 6. Let n be odd and suppose that Σ∗ (n) 6= 1, 2. Then µ Σ∗ (n) ≤
−1 +
√ ¶2 −3 + 4n 2
(32)
Proof. We use the following well-known results: Lemma 1. σ(k) is odd iff k = m2 or k = 2α m2 , where α ≥ 1 and m is an odd integer. (33) α1 α r Proof. Let k = p1 . . . pr . Then σ(k) = (1 + p1 + · · · + pα1 1 ) . . . (1 + pr + · · · + pαr r ). If k is odd, the σ(k) is odd if each term 1 + p1 + · · · + pα1 1 , . . . , 1 + pr + · · · + pαr r is odd, and since pi (1 = 1, r) are all odd numbers, we must have α1 = even, . . . , αr = even. This gives k = m2 , with m = odd. When k is even, then k = 2α pα1 1 . . . pαr r , and since σ(2α ) = 2α + 1 = odd, by the same argument as above, k = 2α m2 , with m = odd. Lemma 2. If k is composite, then √ σ(k) ≥ k + k + 1 (34)
435
√ Proof. Write k = ab, where√1 < a ≤ b < k. Then k ≤ b2 , so b ≥ k, implying σ(k) ≥ 1 + b + k ≥ 1 + k + k, i.e. relation (34). When k = p2 , with p an odd prime, one has equality since σ(p2 ) = p2 + p + 1. Now, if σ(k)|n and n is odd, then clearly σ(k) must be odd, too. Now, by (33) this is possible only when k = m2 or k = 2α m2 , with m ≥ 1 odd. If m > 1, then k = m2 is composite, while if m = 1 in k = 2α m2 , then k = 2α is prime only if α = 1, i.e. √ if k = 2. Supposing k 6= 1, 2 then √ k is always composite, so σ(k) √ ≥ k + k + 1. Since σ(1) ≤ n, we get k + k + 1 − n ≤ 0 √ −1 + −3 + 4n so k ≤ , and this gives (32). 2 Remark. For example, by (26), for 7|n, n odd, (32) is true. Theorem 7. If n ≥ 4, then Σ(n) ≥ 3. For all n ≥ 4, Σ(n) > n2/3
(35)
Proof. Σ(n) = 1 iff n|1, when n = 1. For Σ(n) = 2 we have σ(2) = 3 so n|3 ⇔ n = 1, 3. Thus for n ≥ 4, we have k = Σ(n) ≥ 3. Now, if n|σ(k), then clearly n ≤ σ(k). Let k ≥ 3. Then, it is known (see [3]) that √ (36) σ(k) < k k √ By n < k k = k 3/2 , inequality (35) follows. Corollary. For all m ≥ 2 (left side), and m ≥ 1 (right side): (2m+1 − 1)2/3 < Σ(2m+1 − 1) ≤ 2m
(37)
Proof. 2m+1 − 1 > 4 for m ≥ 2, and the left side is a consequence of (35). Now, the right side follows by (2m+1 − 1)|σ(2m ), since σ(2m ) = 2m+1 − 1, and apply definition (14). Theorem 8. Let f : [1, ∞) → [1, ∞) be given by f (x) = x + x log x. Then for all n ≥ 1, Σ(n) ≥ f −1 (n), (38) where f −1 is the inverse function of f . X Xn X1 X 1 Proof. σ(n) = d= =n ≤n ≤ n(1 + log n) as it is d d d d|n
d|n
d|n
1≤d≤n
1 1 well known that 1 + + · · · + ≤ 1 + log n for all n ≥ 1. Thus if n|σ(k), then 2 n n ≤ σ(k) ≤ f (k), so (38) follows. The function f is strictly increasing and continuous, so it is bijective, having an inverse function f −1 : [1, ∞) → [1, ∞). 436
√ √ Remark. The inequality f (x) < x x, i.e. log x < x − 1 is true for √ x sufficiently large (e.g. x ≥ e3 ). Indeed, let g(x) = x −√lg x − 1, when x−2 g(e3 ) = e3/2 − 4 > 0 by e3 ≈ 19.6 > 42 = 16, and g 0 (x) = > 0 for 2x √ x > 4. So g(x) ≥ g(e3 ) > 0 for x ≥ e3 . Thus x + x lg x < x x. By putting x = n2/3 we get f (n2/3 ) < n, i.e. for n2/3 ≥ e3 (m ≥ e9/2 ) we get: f −1 (n) > n2/3 for n ≥ e9/2
(39)
which improves, by (38), inequality (35). For values of Σ(n) and Σ∗ (n) at primes n = p the following is true: Theorem 9. For all primes p ≥ 5, 1 ≤ Σ∗ (p) ≤ p − 2 and
µ Σ∗ (p) ≤
−1 +
¶2 √ −3 + 4p 2
(40)
(41)
Proof. The inequality Σ∗ (n) ≥ 1 is true for all n (but remains an Open Problem the determination of all n with equality). Now, remark that σ(k)|p iff σ(k) = 1 or σ(k) = p. If σ(k) > 1, then by σ(k) ≥ k+1 we get k ≤ p−1. But we cannot have equality, since then k = q = prime, when σ(q) = q+1 = p ≥ 5 and this is impossible, since q + 1 is even for q ≥ 3, while for q = 2, q + 1 =√3 < 5. Thus k ≤ p − 2, so (40) follows. By applying the inequality σ(k) ≥ k + k + 1 (see (34)) then one arrives at (41), which is sharp, since e.g. Σ∗ (7) = 4 ≤ 4. Theorem 10. For all Mersenne primes p one has Σ(p) ≤
p+1 2
(42)
Proof. This follows from the right side of (37), by remarking that when p+1 p = 2m+1 − 1 is a prime, by Σ(2m+1 − 1) ≤ 2m = we get (42). 2
References [1] K. T. Atanassov, Remark on J´ ozsef S´ andor and Florian Luca’s theorem, C. R. Acad. Bulg. Sci. 55(2002), no. 10, 9-14. [2] M. Le, A conjecture concerning the Smarandache dual function, Smarandache Notion J. 14(2004), 153-155. 437
[3] C. C. Lindner, Problem E1888, Amer. Math. Monthly 73(1966), Solution by A. Bager and S. Russ, same journal 74(1967), 1143. [4] F. Luca, On a divisibility property involving factorials, C. R. Acad. Bulg. Sci. 53(2000), no. 6, 35-38. [5] P. Moree, H. Roskam, On an arithmetical function related to Euler’s totient and the discriminator, Fib. Quart. 33(1995), 332-340. [6] J. S´andor, On certain generalizations of the Smarandache function, Smarandache Notions J. 11(2000), no. 1-3, 202-212. [7] J. S´andor, On certain generalizations of the Smarandache function, Notes Number Theory Discr. Math. 5(1999), no. 2, 41-51. [8] J. S´andor, On the Euler minimum and maximum functions, (to appear). [9] J. S´andor, A note on the divisor minimum function, (to appear). [10] J. S´andor, The Smarandache minimum and maximum functions, (to appear). [11] J. S´andor, The Smarandache function of a set, Octogon Math. Mag. 9(2001), No. 1B, 369-371.
9
On certain new means and their Ky Fan type inequalities
1. Introduction Let x = (x1 , . . . , xn ) be an n-tuple of positive numbers. The unweighted arithmetic, geometric and harmonic means of x, denoted by A = An , G = Gn , H = Hn , respectively, are defined as follows n
1X xi , A= n i=1
G=
à n Y
!1/n xi
i=1
Ã
,
n X 1 H = n/ xi
! .
i=1
Assume 0 < xi < 1, 1 ≤ i ≤ n and define x0 := 1 − x = (1 − x1 , . . . , 1 − xn ). Throughout the sequel the symbols A0 = A0n , G0 = G0n and H 0 = Hn0 will stand for the unweighted arithmetic, geometric and harmonic means of x0 . The arithmetic-geometric mean inequality Gn ≤ An (and its weighted variant) played an important role in the development of the theory of inequalities. 438
Because of its importance, many proofs and refinements have been published. The following inequality is due to Ky Fan: ¸ µ remarkable 1 If xi ∈ 0, (1 ≤ i ≤ n), then 2 G A ≤ 0 0 G A
(1)
with equality only if x1 = · · · = xn . The paper by H. Alzer [1] (who obtained many results related to (1)) contains a very good account up to 1995 of the Ky Fan type results (1). For example, in 1984 Wang and Wang [11] established the following counterpart of (1): H G ≤ 0 H0 G
(2)
1 x2 x1 1/(x2 −x1 ) (x /x1 ) (x1 6= x2 ), I(x, x) = x denote e 2 the so-called identric mean of x1 , x2 > 0. In 1990 J. S´andor [8] proved the following refinement of (1) in the case of two arguments (i.e. n = 2): Let I = I(x1 , x2 ) =
G I A ≤ 0 ≤ 0, 0 G I A
(3)
where I 0 = I 0 (x1 , x2 ) = I(1 − x1 , 1 − x2 ). We note that, inequality (14) in Rooin’s paper [6] is exactly (3). In 1999 S´andor and Trif [10] have introduced an extension of the identric mean to n arguments, as follows. For n ≥ 2, let En−1 = {(λ1 , . . . , λn−1 ) : λi ≥ 0, 1 ≤ i ≤ n − 1, λ1 + · · · + λn−1 ≤ 1} be the Euclidean simplex. Given any probability measure µ on En−1 , for a continuous strictly monotone function f : (0, ∞) → R, the following functional means of n arguments can be introduced: ÃZ Mf (x; µ) = f −1
where xλ =
n X
! f (xλ)dµ(λ) ,
(4)
En−1
xi λi denotes the scalar product, λ = (λ1 , . . . , λn−1 ) ∈ En−1 ,
i=1
and λn = 1 − λ1 − · · · − λn−1 . 439
For µ = (n − 1)! and f (t) = 1/t, one obtains the unweighted logarithmic mean, studied by A.O. Pittenger [5]. For f (t) = ln t, however we obtain a mean ÃZ ! I = I(x) = exp
ln(xλ)dµ(λ)
(5)
En−1
which may be considered as a generalization of the identric mean. Indeed, it is immediately seen that µZ 1 ¶ I(x1 , x2 ) = exp ln(tx1 + (1 − t)x2 )dt , 0
in concordance with (5), which for µ = (n − 1)! gives the unweighted (and symmetric) identric mean of n arguments: Ã ! Z I = In = In (x1 , . . . , xn ) = exp (n − 1)!
En−1
ln(xλ)dλ1 . . . dλn−1
(6)
Let I 0 = In0 = In (1 − x) in (5) for µ = (n − 1)!. Then ndor ¸¶ and Trif µ S´aµ 1 . The [10] proved that relation (3) holds true for any n ≥ 2 xi ∈ 0, 2 weighted versions hold also true. In 1990 J. S´andor [7] discovered ¸ the following additive analogue of the Ky µ 1 (1 ≤ i ≤ n), then Fan inequality (1): If xi ∈ 0, 2 1 1 1 1 − ≤ 0− 0 H H A A
(7)
In 2002, E. Neuman and J. S´andor [2] proved the following refinement of (7): 1 1 1 1 1 1 − ≤ 0− ≤ 0− , (8) 0 H H L L A A where L is the (unweighted) logarithmic mean, obtained from (4) for f (t) = 1/t, i.e. Ã !−1 Z 1 L = Ln = Ln (x1 , . . . , xn ) = (n − 1)! dλ1 . . . dλn−1 , (9) En−1 xλ and L0 = L(1 − x). For n = 2 this gives the logarithmic mean of two arguments, L(x1 , x2 ) =
x2 − x1 ln x2 − ln x1
(x1 6= x2 ),
440
L(x, x) = 1.
We note that for n = 2, relation (8) is exactly inequality (27) in Rooin’s paper [6]. Alzer ([1]) proved another refinement of S´andor inequality, as follows: 1 1 1 1 1 1 − ≤ 0− ≤ 0− 0 H H G G A A
(10)
In [2] we have introduced a new mean J = Jn and deduced a new refinement of the Wang-Wang inequality: H J G ≤ 0 ≤ 0 0 H J G
(11)
We note that in a recent paper, Neuman and S´andor [4] have proved the following strong improvements of Alzer’s inequality (10): 1 1 1 1 1 1 1 1 1 1 ≤ 0− ≤ 0− ≤ 0− ≤ 0− − 0 H H J J G G I I A A
(12)
(where J 0 = J(1 − x) etc.).
2. New means and Ky Fan type inequalities 2.1. The results obtained by J. Rooin [6] are based essentially on the following Lemma 1. Let f be a convex function defined on a convex set C, and let xi ∈ C, 1 ≤ i ≤ n. Define F : [0, 1] → R by n
1X F (t) = f [(1 − t)xi + txn+1−i ], n
t ∈ [0, 1].
i=1
Then
µ
¶
f (x1 ) + · · · + f (xn ) , n Z 1 and the similar double inequality holds for F (t)dt. f
x1 + · · · + xn n
≤ F (t) ≤
0
Proof. By the definition of convexity, one has f [(1 − t)xi + txn+1−i ] ≤ (1 − t)f (xi ) + tf (xn+1−i ), and after summation, remarking that n X
[f (xn+1−i ) − f (xi )] = 0,
i=1
441
we get the right-side inequality. On the other hand, by Jensen’s discrete inequality for convex functions, ! Ã n µ ¶ x1 + · · · + xn 1X [(1 − t)xi + txn+1−i ] = f F (t) ≥ f , n n i=1
giving the left-side inequality. By integrating on [0, 1], clearly the same result holds true. 2.2. Now define the following mean of n arguments: Ã n !1/n Y K = Kn = Kn (x1 , . . . , xn ) = I(xi , xn+1−i ) (13) i=1
Letting f (x) = − ln x for x ∈ (0, +∞), and remarking that Z 1 ln[(1 − t)a + tb]dt = ln I(a, b), 0
Lemma 1 gives the following new refinement of the arithmetic-geometric inequality: G ≤ K ≤ A, (14) which holds true for any xi > 0 (i = 1,µn). ¸ 1−x 1 Selecting f (x) = ln for C = 0, , and remarking that x 2 Z 1 ln{1 − [(1 − t)a + tb]}dt = ln I(x01 , x02 ) = ln I 0 (x1 , x2 ), 0
we get the following Ky Fan-type inequality: G K A ≤ 0 ≤ 0 0 G K A
(15)
This is essentially inequality (13) in [6] (discovered independently by the author). 1 2.3. Let now f (x) = for x ∈ (0, ∞). Since f is convex, and x Z 1 1 1 dt = , (1 − t)a + tb L(a, b) 0 Lemma 1 gives H ≤ R ≤ A, 442
(16)
where R = Rn = Rn (x1 , . . . , xn ) = n/
n X i=1
1 L(xi , xn+1−i )
(17)
This is a refinement - involving the new mean R - of the harmonicarithmetic inequality. µ ¸ 1 1 1 Letting f (x) = − for x ∈ 0, , the above arguments imply the x 1−x 2 relations 1 1 1 1 1 1 − 0 ≤ − 0 ≤ − 0, (18) A A R R H H µ ¸ 1 where xi ∈ 0, , and R0 = Rn0 = Rn (1 − x1 , . . . , 1 − xn ). Relation (18) 2 coincides essentially with (26) of Rooin’s paper [6]. 2.4. Let S = Sn (x1 , . . . , xn ) = (xx1 1 . . . xxnn )1/(x1 +···+xn )
(19)
For n = 2, this mean has been extensively studied e.g. in [8], [9], [3]. Applying the Jensen inequality for the convex function f (x) = x ln x (x > 0), we get A ≤ S. On the other hand, remarking that S is a weighted geometric mean of x1 , . . . , xn with weights α1 = x1 /(x1 + · · · + xn ), . . . , αn = xn /(x1 + · · · + xn ), by applying the weighted geometric-arithmetic inequality xα1 1 . . . xαnn ≤ α1 x1 + · · · + αn xn , we can deduce S ≤ Q, where Q = Qn (x1 , . . . , xn ) =
x21 + · · · + x2n . x1 + · · · + xn
Therefore, we have proved that A≤S≤Q
(20)
In [8] it is shown that Z
b
x ln xdx = a
b2 − a2 ln I(a2 , b2 ) 4 443
(21)
Denote J(a, b) = (I(a2 , b2 ))1/2 and put J 0 (a, b) = J(1 − a, 1 − b). By applying Lemma 1, we get A ≤ T ≤ S, (22) where the mean T is defined by #1/A " n Y A(xi ,xn+1−i ) n (J(xi , xn+1−i )) T = Tn (x1 , . . . , xn ) =
(23)
i=1
µ ¸ 1 Letting now f (x) = x ln x − (1 − x) ln(1 − x), x ∈ 0, , by f 00 (x) = 2 1 − 2x ≥ 0 we can state that f is convex, so by Lemma 1 and by (21) we x(1 − x) ¸ µ 1 can write, for xi ∈ 0, : 2 AA /A0
A0
≤ T A /T 0
A0
A0
≤ S A /S 0 ,
(24)
where the mean T is defined by (23), while T 0 = T (1−x). Since for n = 2, T ≡ J, for means of two arguments (24) gives a Ky Fan-type inequality involving A, I, S. 2.5. Relation (23) shows that T is a generalization of the mean J to n arguments. In what follows we shall introduce another generalization, provided by the formula U = Un (x1 , . . . , xn ) ( Ã !)1/A Z = exp (n − 1)! (xλ) ln(xλ)dλ1 . . . dλn−1 (25) En−1
Here the notations are as in the Introduction. Since, by (21), Z b Z 1 1 [(1 − t)a + tb] ln[(1 − t)a + tb]dt = x ln xdx b−a a 0 A ln I(a2 , b2 ) = ln J A , 2 for n = 2, we have U ≡ J, thus U is indeed another generalization of the mean J. Now, the following result is due to E. Neuman (see e.g. [2]). Lemma 2. Let K be an interval containing x1 , . . . , xn , and suppose that f : K → R is convex. Then ¶ µ Z x1 + · · · + xn ≤ (n − 1)! f (λx)dλ1 . . . dλn−1 f n En−1 =
444
≤
f (x1 ) + · · · + f (xn ) . n
µ ¸ 1 Letting K = 0, , and f (x) = x ln x − (1 − x) ln(1 − x) in Lemma 2, we can µ2 ¸ 1 deduce for xi ∈ 0, 2 AA /A0
A0
≤ U A /U 0
A0
≤ S A /S 0
A0
(26)
Remark that for n = 2, inequalities (24) and (26) reduce to the same inequality, as in that case one has T = J = U . The mean U separates also A and S, since applying Lemma 2 for f (x) = x ln x (x > 0), we have A ≤ U ≤ S.
(27)
There remains an Open Problem, namely the comparability of the above defined means T and U for n > 2. Also, the connections of these means to K and R, introduced in the preceding sections.
References [1] H. Alzer, The inequality of Ky Fan and related results, Acta Appl. Math. 38, 305-354(1995). [2] E. Neuman, J. S´andor, On the Ky Fan inequality and related inequalities, I, Math. Ineq. Appl. 5, 49-56(2002). [3] E. Neuman, J. S´andor, Inequalities involving Stolarsky and Gini means, Math. Pannonica 14, 29-44(2003). [4] E. Neuman, J. S´andor, On the Ky Fan inequality and related inequalities, II, submitted. [5] A.O. Pittenger, The logarithmic mean in n variables, Amer. Math. Monthly 92, 99-104(1985). [6] J. Rooin, Some refinements of Ky Fan’s and S´ andor’s inequalities, Southest Asian Bull. Math. 27, 1101-1109(2004). [7] J. S´andor, On an inequality of Ky Fan, Babe¸s-Bolyai Univ., Fac. Math. Phys., Res. Semin. 7, 29-34(1990). 445
[8] J. S´andor, On the identric and logarithmic means, Aequationes Math. 40, 261-270(1990). [9] J. S´andor, I. Ra¸sa, Inequalities for certain means of two arguments, Nieuw Arch. Wiskunde 15, 51-55(1997). [10] J. S´andor, T. Trif, A new refinement of the Ky Fan inequality, Math. Ineq. Appl. 4, 529-533(1999). [11] W.-L. Wang, P.-F. Wang, A class of inequalities for symmetric functions (in Chinese), Acta Math. Sinica 27, 485-497(1984).
10
On Lehman’s inequality and electrical networks
1. Introduction A. Lehman’s inequality (see [6], [2]) (and also SIAM Review 4(1962), 150155), states that if A, B, C, D are positive numbers, then (A + B)(C + D) AC BD ≥ + . (1) A+B+C +D A+C B+D This was discovered as follows: interpret A, B, C, D as resistances of an electrical network. It is well-known that if two resistances R1 and R2 are serially connected, then their compound resistance is R = R1 + R2 , while in parallel connecting one has 1/R = 1/R1 + 1/R2 . Now consider two networks, as given in the following two figures: (A + B)(C + D) AC BD R0 = + A+B+C +D A+C B+D By Maxwell’s principle, the current chooses a distribution such as to minimize the energy (or power), so clearly R0 ≤ R, i.e. Lehman’s inequality (1). In fact, the above construction may be repeated with 2n resistances, in order to obtain: Theorem 1. If ai , bi (i = 1, n) are positive numbers, then R=
a1 b1 an bn (a1 + · · · + an )(b1 + · · · + bn ) ≥ + ··· + a1 + · · · + an + b1 + · · · + bn a1 + b1 an + bn
(2)
for any n ≥ 2.
2ab = H(a, b) is in fact the harmonic mean of two a+b positive numbers, Lehman’s inequality (2) can be written also as Remark. Since
H(a1 + · · · + an , b1 + · · · + bn ) ≥ H(a1 , b1 ) + · · · + H(an , bn ) 446
(3)
A
C
A
C
B
D
B
D
2. Two variable generalizations In what follows, by using convexity methods, we shall extend (3) in various ways. First we introduce certain definitions. Let f : A ⊂ R2 → R be a function with two arguments, where A is a cone (e.g. A = R2+ ). Let k ∈ R be a real number. Then we say that f is k-homogeneous, if f (rx, ry) = rk f (x, y)
(4)
for any r > 0 and x, y ∈ A. When k = 1, we simply say that f is homogeneous. Let F : I ⊂ R → R be a function of an argument defined on an interval I. We say that F is k-convex (k-concave), if F (λa + µb) ≤ λk F (a) + µk f (b),
(5)
(≥)
for any a, b ∈ I, and any λ, µ > 0, λ + µ = 1. We note, that if k = 1, then F will be called simply convex. For example, F (t) = |t|k , t ∈ R is k-convex, for k ≥ 1, since |λa+µb|k ≤ λk |a|k +µk |b|k by (u+v)k ≤ uk +v k (u, v > 0), k ≥ 1, which is well-known. On the other hand, the function F (t) = |t|, though is convex, is not 2-convex on R. The k-convex functions have been introduced for the first time by W. W. Breckner [4]. See also [5] for other examples and results. A similar convexity notion, when in (5) one replaces λ + µ = 1 by λk + µk = 1, was introduced by W. Orlicz [12] (see also [8] for these convexities). 447
Now, let A = (0, +∞) × (0, +∞) = R2+ and I = (0, +∞). Define F (t) = f (1, t) for t ∈ I. Theorem 2. If f is k-homogeneous, and F is k-convex (k-concave) then f (a1 + · · · + an , b1 + · · · + bn ) ≤ f (a1 , b1 ) + · · · + f (an , bn )
(6)
(≥)
for any ai , bi ∈ A (i = 1, 2, . . . , n). Proof. First remark, that by (4) and the definition of F , one has µ ¶ µ ¶ b b ak F = ak f 1, = f (a, b) a a
(7)
On the other hand, by induction it can be proved the following Jensen-type inequality: F (λ1 x1 + λ2 x2 + · · · + λn xn ) ≤ λk1 F (x1 ) + λk2 F (x2 ) + · · · + λkn F (xn ),
(8)
(≥)
for any xi ∈ I, λi > 0 (i = 1, n), λ1 + · · · + λn = 1. E.g. for n = 3, relation (8) can be proved as follows: λ1 λ2 Put a = x1 + x2 , b = x3 , λ = λ1 + λ2 , µ = λ3 in (5). Then, λ1 + λ2 λ1 + λ2 as λ1 x1 + λ2 x2 + λ3 x3 = λa + µb, we have F (λ1 x1 + λ2 x2 + λ3 x3 ) ≤ λk F (a) + µk F (b) ≤ · k
≤ (λ1 + λ2 )
¸ λk1 λk2 F (x1 ) + F (x2 ) + λk3 F (x3 ) = (λ1 + λ2 )k (λ1 + λ2 )k = λk1 F (x1 ) + λk2 F (x2 ) + λk3 F (x3 ).
Put now in (8) x1 = λ1 =
b1 b2 bn , x2 = , . . . , xn = , a1 a2 an
a1 a2 an , λ2 = , . . . , λn = a1 + · · · + an a1 + · · · + an a1 + · · · + an
in order to obtain µ µ F
b1 + · · · + bn a1 + · · · + an
ak1 F
¶ ≤ (≥)
b1 a1
¶
µ
µ ¶ ¶ bn b2 k + + · · · + an F a2 an (a1 + · · · + an )k ak2 F
448
(9)
Now, by (7) this gives f (a1 + · · · + an , b1 + · · · + bn ) ≤ f (a1 , b1 ) + · · · + f (an , bn ), i.e. relation (6). a+b Remark. Let f (a, b) = . Then f is homogeneous (i.e. k = 1), and ab t+1 F (t) = f (1, t) = is 1-convex (i.e., convex), since F 00 (t) = 2/t3 > 0. Then t relation (6) gives the following inequality: 1 1 1 ≤ + ··· + . H(a1 + · · · + an , b1 + · · · + bn ) H(a1 , b1 ) H(an , bn )
(10)
ab t . Then f is homogeneous, with F (t) = , which a+b t+1 is concave. From (6) (with ≥ inequality), we recapture Lehman’s inequality (3). The following theorem has a similar proof: Theorem 3. Let f be k-homogeneous, and suppose that F is l-convex (l-concave) (k, l ∈ R). Then Let now f (a, b) =
(a1 + · · · + an )l−k f (a1 + · · · + an , b1 + · · · + bn ) ≤ (≥) l−k al−k 1 f (a1 , b1 ) + · · · + an f (an , bn ).
(11)
Remarks. For k = l, (11) gives (9). a For example, let f (a, b) = , where a, b ∈ (0, ∞) × (0, ∞). Then k = 0 b 1 (i.e. f is homogeneous of order 0), and F (t) = , which is 1-convex, since t 2 00 F (t) = 3 > 0. Thus l = 1, and relation (11) gives the inequality t (a1 + · · · + an )2 a2 a2 ≤ 1 + ··· + n b1 + · · · + bn b1 bn Finally, we given another example of this type. Put f (a, b) =
(12) a2 + b2 . a+b
t2 + 1 , after elementary computations, F 00 (t) = t+1 4/(t + 1)3 > 0, so l = 1, and (11) (or (9)) gives the relation Then k = 1. Since F (t) =
a2 + b21 a2 + b2n (a1 + · · · + an )2 + (b1 + · · · + bn )2 ≤ 1 + ··· + n a1 + · · · + an + b1 + · · · + bn a1 + b1 an + bn 449
(13)
a2 + b2 ap+1 + bp+1 (more generally, Lp (a, b) = ) are the a+b ap + bp so-called ”Lehmer means” [9], [7], [1] of a, b > 0, (13) can be written also as Since L1 (a, b) =
L1 (a1 + · · · + an , b1 + · · · + bn ) ≤ L1 (a1 , b1 ) + · · · + L1 (an , b1 ).
(14)
Clearly, one can obtain more general forms for Lp . For inequalities on more general means (e.g. Gini means), see [10], [11].
3. H¨ older’s inequality As we have seen, there are many applications to Theorems 2 and 3. Here we wish to give an important application; namely a new proof of H¨older’s inequality (one of the most important inequalities in Mathematics). Let f (a, b) = a1/p b1/q , where 1/p + 1/q = 1 (p > 1). Then clearly f 1 is homogeneous (k = 1), with F (t) = t1/q . Since F 0 (t) = t−1/p , F 00 (t) = q 1 −(1/p)−1 − t < 0, so by Theorem 2 one gets pq 1/p 1/q
(a1 + · · · + an )1/p (b1 + · · · + bn )1/q ≥ a1 b1
1/q + · · · + a1/p n bn
(15)
Replace now ai = Api , bi = Biq (i = 1, n) in order to get n X i=1
Ai Bi ≤
à n X
Api
!1/p à n X
!1/q Biq
,
(16)
i=1
i=1
which is the classical H¨older inequality.
4. Many variables generalization Let f : A ⊂ Rn+ → R be of n arguments (n ≥ 2). For simplicity, put p = (x1 , . . . , xn ), p0 = (x01 , . . . , x0n ), when p + p0 = (x1 + x01 , . . . , xn + x0n ) and rp = (rx1 , . . . , rxn ) for r ∈ R. Then the definitions of k-homogeneity and k-convexity can be extended to this case, similarly to paragraph 2. If A is a cone, then f is k-homogeneous, if f (rp) = rk f (p) (r > 0) and if A is convex set then f is k-convex, if f (λp + µp0 ) ≤ λk f (p) + µk f (p0 ) for 0 any µ p, p 0 ∈ ¶ A, λ, µ > 0, 0λ + µ = 1. We say that f is k-Jensen convex, if p+p f (p) + f (p ) f ≤ . We say that f is r-subhomogeneous of order 2 2k k, if f (rp) ≤ rk f (p). Particularly, if k = 1 (i.e. f (rp) ≤ rf (p)), we say that f is r-subhomogeneous (see e.g. [14], [15]). If f is r-subhomogeneous of order 450
k for any r > 1, we say that f is subhomogeneous of order k. For k = 1, see [13]. We say that f is subadditive on A, if f (p + p0 ) ≤ f (p) + f (p0 )
(17)
We note that in the particular case of n = 2, inequality (6) with ”≤” says exactly that f (a, b) of two arguments is subadditive. Theorem 4. If f is homogeneous of order k, then f is subadditive if and only if it is k-Jensen convex. Proof. If f is subadditive, i.e. f (p + p0 ) ≤ f (p) + f (p0 ) for any p, p0 ∈ A, then µ ¶ p + p0 1 f (p) + f (p0 ) f = k f (p + p0 ) ≤ , 2 2 2k so f is k-Jensen convex. Reciprocally, if f is k-Jensen convex, then µ ¶ p + p0 f (p) + f (p0 ) f ≤ , 2 2k so
· µ ¶¸ µ ¶ p + p0 p + p0 f (p + p0 ) = f 2 = 2k f ≤ f (p) + f (p0 ), 2 2
i.e. (17) follows. Remark. Particularly, a homogeneous subadditive function is convex, a simple, but very useful result in the theory of convex bodies (e.g. ”distance function”, ”supporting function”, see e.g. [3], [16]). Theorem 5. If f is 2-subhomogeneous of order k, and is k-Jensen convex, then it is subadditive. Proof. Since µ µ ¶¶ µ ¶ p + p0 p + p0 f (p + p0 ) = f 2 ≤ 2k f , 2 2 and
µ f
p + p0 2
¶ ≤
f (p) + f (p0 ) , 2k
we get f (p + p0 ) ≤ f (p) + f (p0 ), so (17) follows. Remark. Particularly, if f is 2-subhomogeneous, and Jensen convex, then it is subadditive. (18) It is well-known that a continuous Jensen convex function (defined on an open convex set A ⊂ Rn ) is convex. Similarly, for continuous k-Jensen convex functions, see [4]. 451
To give an interesting example, connected with Lehman’s inequality, let us ¶ µ 1 1 + ··· + . consider A = Rn+ , f (p) = H(p) = n/ x1 xn 1 1 1 Let = + ··· + . Then g(p) x1 xn n
dg X dxi = , g2 x2i i=1
so 1 d2 g = 2 g3
Ã
n X dxi i=1
x2i
n
X dx2 dg 2 d2 g i − 2 = −2 , g2 g3 x3i i=1
!2
à −
n X 1 xi i=1
!Ã
n X dx2 i
i=1
x3i
! .
(Here d denotes a differential.) Now apply H¨older’s inequality (16) for p = q = √ √ 2 (i.e. Cauchy-Bunjakovski inequality), Ai = 1/ xi , Bi = (1/xi xi )dxi . Then d2 g one obtains 3 ≤ 0, and since g > 0, we get d2 g ≤ 0. It is well-known ([16]) g that this implies the concavity of function g(p) = H(p)/n, so −H(p) will be a convex function. By consequence (17) of Theorem 5, H(p) is subadditive, i.e. H(x1 + x01 , x2 + x02 , . . . , xn + x0n ) ≥ H(x1 , x2 , . . . , xn )+ +H(x01 , x02 , . . . , x0n ),
(xi , x0i > 0).
(19)
For n = 2 this coincides with (3), i.e. Lehman’s inequality (1). Finally, we prove a result, which is a sort of reciprocal to Theorem 5: Theorem 6. Let us suppose that f is subadditive, and k-convex, where k ≥ 1. Then f is subhomogeneous of order k. Proof. For any r > 1 one can find a positive integer n such that r ∈ [n, n + 1]. Then r can be written as a convex combination of n and n + 1: r = nλ + (n + 1)µ. By the k-convexity of f one has f (rp) = f (nλp + (n + 1)µp) ≤ λk f (np) + µk f [(n + 1)p]. Since f is subadditive, from (17) it follows by induction that f (np) ≤ nf (p), so we get f (rp) ≤ nλk f (p) + (n + 1)µk f (p) = [nλk + (n + 1)µk ]f (p). Now, since k ≥ 1, it is well-known that [λn + (n + 1)µ]k ≥ (λn)k + ((n + 1)µ)k . 452
But (λn)k ≥ nλk and ((n + 1)µ)k ≥ (n + 1)µk , so finally we can write f (rp) ≤ [λn + (n + 1)µ]k f (p) = rk f (p), which means that f is subhomogeneous of order k. Remark. For k = 1 Theorem 6 contains a result by R. A. Rosenbaum [13]. Acknowledgments. The author thanks Professors J. Peetre and H. Alzer for providing their reprints [2], resp. [1]. He is indebted to Professor F. A. Valentine for a copy of his book [16], and also to Professors W. W. Breckner of Cluj, and V. E. Szab´o of Budapest for helpful discussions. NOTE ADDED IN PROOF. Recently (23th February, 2005) we have discovered that Lehman’s inequality (2) (or (3)) appears also as Theorem 67 in G. H. Hardy, J. E. Littlewood and G. Polya [Inequalities, Cambridge Univ. Press, 1964; see p.61], and is due to E. A. Milne [Note on Rosseland’s integral for the stellar absorption coefficient, Monthly Notices, R.A.S. 85(1925), 979-984]. Though we are unable to read Milne’s paper, perhaps we should call Lehman’s inequality as the ”Milne-Lehman inequality”. We note also that the Milne-Lehman inequality is published as a Proposed Problem 2113 (by M.E. Kuczma), as well as Problem 2392 (by G. Tsintsifas) in the Canadian journal Crux Mathematicorum.
References ¨ Lehmers Mittelwertfamilie, Elem. Math., 43(1988), no. [1] H. Alzer, Uber 2, 50-54. [2] J. Arazy, T. Claesson, S. Janson and J. Peetre, Means and their iteration, Proc. 19th Nordic Congr. Math., Reykjavik 1984, 191-212. [3] T. Bonnesen and W. Fenchel, Theorie der konvexen K¨ orper, Berlin, 1934. [4] W. W. Breckner, Stetigkeitsaussagen f¨ ur eine Klasse verallgemeinerter konvexer Funktionen in topologischen linearen R¨ aume, Publ. Inst. Math. (Beograd), 23(1978), 13-20. [5] W. W. Breckner and Gh. Orb´an, Continuity properties of rationally sconvex mappings with values in an ordered topological linear space, 92 pp., Babe¸s-Bolyai Univ., Cluj, Romania, 1978. [6] R. J. Duffin, Network models, in SIAM-AMS Proceedings vol. III, pp. 65-91, AMS, Providence, 1971. 453
[7] H. W. Gould and M. E. Mays, Series expansions of means, J. Math. Anal. Appl., 101(1984), 611-621. [8] H. Hudzik and L. Maligranda, Some remarks on s-convex functions, Aequationes Math., 48(1994), 100-111. [9] D. H. Lehmer, On the compounding of certain means, J. Math. Anal. Appl., 36(1971), 183-200. [10] E. Neuman and J. S´andor, Inequalities involving Stolarsky and Gini means, Math. Pannonica 14(2003), 29-44. [11] E. Neuman and J. S´andor, On certain new means of two arguments and their extensions, Int. J. Math. Math. Sci., 16(2003), 981-993. [12] W. Orlich, A note on modular spaces, I, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys., 9(1961), 157-162. [13] R. A. Rosenbaum, Subadditive functions, Duke Math. J., 17(1950), 227247. [14] J. S´andor and Gh. Toader, On some exponential means, Babe¸s-Bolyai Univ., Preprint No. 7, 1990, 35-40. [15] J. S´andor, On certain subhomogeneous means, Octogon Math. Mag., 8(2000), 156-160. [16] F. A. Valentine, Convex sets, Mc Graw-Hill Inc., New York, 1964.
454
Author Index A ˇ Arslanagi´c 1.4, 4.4 S. H. Alzer 2.1, 2.10, 4.9, 4.10, 4.16, 5.12, 6.3, 6.4, 6.10, 6.13, 6.14, 6.15, 10.9, 10.10 J.L. d’Alambert 2.5 T.M. Apostol 3.7, 9.10 Ch. Ashbacher 3.11, 3.12, 3.17, 9.16 W. Aiello 3.19 E. Artin 5.9, 5.11, 5.13, 5.15, 6.11, 7.1 R.P. Agarwal 5.13 M. Abramowitz 5.13 C. Alsina 5.15 J. Arazy 6.3, 10.10 R. Askey 7.1 J. Acz´el 7.2, 7.4, 7.5 D. Acu 7.2 T. Andreescu 8.3 D. Andrica 8.3 C. Adiga 9.4 L. Alaoglu 9.7 K.T. Atanassov 9.14, 10.8
O. Bottema 1.5, 1.6 H. Brocard 1.6 D.M. B˘atinet¸u-Giurgiu 1.8, 2.2A, 2.2L, 2.2M, 2.2N, 2,2O, 2.2P, 2.2S, 2.2T, 5.5, 5.6, 5.7 H.J. Brothers 2.1, 2.2F, 2.2P J. Bernoulli 2.2I, 2.2J, 3.7, 5.3 V. Berinde 2.5 T.J. Bromwich 2.5 Gy. Bereznai 2.6 N.G. de Bruijn 2.7 F. Bencherif 2.7 D. Bradley 2.9 G.M. Bell 2.9 T. Bonnesen 10.10 G. Bennett 2.10, 5.12 A. Balog 3.2 D. Bode 3.9 R. Bojani´c 3.16 A. Bege 3.19 V. Bunjakovski (Bunyakovsky) 4.2, 4.4, 6.6, 10.10 E.F. Beckenbach 4.14 R. Bellman 4.14 B H. Bohr 5.9, 5.13, 7.1 E. B´ezout 1.2 J. Bass 5.11 M. Bencze 1.5, 1.8, 1.11, 1.12, 2.2A, J. Bursuc 5.11 2.2B, 2.2C, 2.2G, 2.2H, 2.2J, 2.2S, N.S. Barnett 5.13 2.2T, 2.2U, 2.3, 2.9, 3.9, 3.10, 3.14, C.W. Borchardt 6.7 4.13, 5.5, 5.6, 5.14, 8.3, 8.9, 9.6, 9.9 455
O. Bonnett 6.16 M. Bal´azs 6.16 V. B˘andil˘a 7.2 B. Bartha 7.9 A. Bremner 8.6 J.M. Brown 10.3 W.W. Breckner 10.10
L. Denbath 2.2P G.L. Dirichlet 2.9, 9.7, 9.14 S.S. Dragomir 2.10, 4.9, 4.10, 5.13 P. Dusart 3.2 F.W. Dodd 3.8 R. Dedekind 3.13, 8.12, 8.13, 8.14 L.E. Dickson 3.13 M. Deng 3.19 C B.P. Demidovich 5.11, 6.11 R. Cotes 1.2 B.C. Drachman 5.16 L. Cauchy 1.5, 2.2H, 2.10, 4.2, 4.4, J. Dieudonn´e 6.14 4.16, 6.6, 6.10, 6.13, 6.15, 7.1, 7.3, 7.4, G. Darboux 6.15, 6.16 7.5, 7.6, 10.10 R.J. Duffin 10.10 L. Carlitz 1.6 F.T. Cˆampan 1.14 E N. de Cusa 1.14 P. Erd¨os 1.3, 1.9. 9.3, 9.7, 9.14 E. Cesar`o 2.2I, 2.2O, 2.2R, 2.3A, 2.3B L. Euler 1.5, 2.2J, 2.2K, 2.2M, 2.2O, J.H. Conway 2.2U 2.2Q, 2.3, 2.6, 2.9, 2.10, 3.7, 3.11, 3.12, J. Choi 2.9 3.13, 5.1, 5.2, 5.4, 5.5, 5.7, 5.11, 6.11, E. Catalan 2.9 6.16, 8.11, 8.12, 8.13, 9.7, 9.8, 9.14, B. Crstici 3.1, 3.2, 5.15, 9.1, 9.2, 10.2 10.4 W. Chen 3.8 A. Erd´elyi 2.4 H.Z. Cao 3.8 Euclid of Alexandria 3.12, 10.2 G.L. Cohen 3.19, 10.7 J. Earls 3.15, 3.16, 3.17, 3.18, 8.16, G. Chrystal 4.3, 4.4 10.5 P. Chebyshev 4.10, 4.11, 7.10 F M. Craiu 5.11, 6.16 P. Fermat 1.2, 3.17, 10.2 M.J. Cloud 5.16 B. Finta 1.3 B.C. Carlson 6.3, 6.7, 6.11 J.B. Fourier 1.11 T. Claeson 6.7, 10.10 S. Finch 2.1 C.-P. Chen 6.13 L. Filep 2.6 G. Chilov 6.16 K. Ford 3.15 C.V. Cr˘aciun 6.16 J. Findley 3.15 R.O. Cuzmin 6.16 J. Fabrykowski 3.19 N. Cior˘anescu 7.4 A.M. Fink 4.12 B. Conn 8.6 K. Fan 4.13, 4.14, 6.10, 10.9 E. Cohen 9.6, 9.10, 9.11 J. Frauenthal 5.16 I. Creang˘a 9.7 A.N. Fathima 9.4 D W. Fenchel 10.10 456
G R. Iani´c 7.2 M. Gherm˘anescu 2.1, 2.2O, 2.2P J G. Garnir 2.2I, 2.2R C. Jordan 1.4, 1.11, 6.7, 9.7 R.K. Guy 3.4, 8.1, 8.6, 8.15, 10.7 J.L. Jensen 1.11, 4.7, 4.9, 5.13, 7.4, P. Gronas 3.12 K.F. Gauss 3.12, 3.13, 5.4, 6.3, 6.7, 9.11, 9.13, 10.9, 10.10 W. Janous 1.12, 3.1 9.6, 10.2 S. Janson 6.7, 10.10 S.W. Golomb 3.13 F.H. Jackson 7.1 J. Galambos 3.14 W. Gautschi 5.13 K B. Gelbaum 6.14 N.D. Kazarinoff 1.9 N.M. G¨ unther 6.16 J.A. Knox 2.1, 2.2F, 2.2O R.R. Goldberg 6.16 K. Knopp 2.8 V.L. Gardiner 8.6 H.M. Korobov 3.4 C. Gini 10.9, 10.10 H.J. Kanold 3.9, 3.10 H.W. Gould 10.10 I. K´atai 3.14 J.M. DeKoninck 3.19, 9.6, 9.10 H N. Keyfitz 5.16 Heron of Alexandria J. Hadamard 1.11, 2.1, 4.7, 4.8, 4.9, F. Klein 6.4 J. Karamata 6.5 4.10, 5.11, 5.13, 6.9, 9.11, 9.13 J. Kolumb´ an 6.16 Ch. Hermite 1.11, 4.7, 5.13 K. Koyama 8.6 G. de l’Hˆopital 1.15, 2.1, 2.2R T. Kim 9.4 D.R. Hofstadter 2.2U G.H. Hardy 2.9, 2.10, 3.9, 4.3, 4.11, M.S. Klamkin 9.11, 10.3 4.14, 5.13, 6.10, 6.14, 8.1, 9.6, 9.7, K. Kashihara 9.16 E. Kr¨atzel 10.3 9.10, 9.11, 10.3 J. Van der Hoek 2.10 L M. Hazewinkel 3.14 P.S. Laplace 6.16 P. Hagis 3.19 T. Lalescu 1.6, 5.1, 5.12 J. Hanumanthachari 3.19 F. Leuenberger 1.6 O. H¨older 5.13, 5.14, 7.1, 10.10 E. Lemoine 1.6 V. Hiri¸s 6.16 J. Lagrange 2.2Q, 2.10, 4.9, 5.2, 6.7, S. Haruki 7.4 6.9, 6.11 D.R. Heath-Brown 8.6 D.A. Lavis 2.9 M. Hausman 9.2 J.E. Littlewood 2.10, 3.4, 4.3, 4.11, H. Hudzik 10.10 4.14, 5.13, 6.10, 6.14 I G.G. Lorentz 2.10 A. Ivi´c 3.19, 9.6 G. Lord 3.19 457
L. Lucht 3.19 T.P. Lin 4.7, 6.3 A.M. Legendre 5.4 A. Lupa¸s 5.11 J. Lew 5.16 A. Lehman 10.10 D.H. Lehmer 10.10 E.B. Leach 6.2, 6.3, 6.5, 6.7 H. Lebesgue 7.1 F. Luca 9.14, 10.8 C.C. Lindner 10.8 E.S. Langford 9.11 B. Lepson 10.3 W.M. Lioen 8.6 R.F. Lukes 8.6 R.B. Lazarus 8.6 R. Lipschitz 9.1 M. Lee 9.2, 9.14, 10.8 R. Laatsch 9.2 E. Lucas 9.8
L. Maligranda 10.10 I.A. Maron 5.11, 6.11 M. Megan 6.16 J.C.P. Miller 8.6 L.J. Mordell 8.6 P. Moree 9.3, 9.14, 9.15, 10.8 A. M¨obius 9.7 A. Makowski 9.10 J. Morgado 9.10 M.E. Mays 10.10 J.C. Maxwell 10.10
M D.S. Mitrinovi´c 1.4, 1.5, 1.11, 1.13, 1.16, 2.10, 3.1, 3.2, 3.4, 3.8, 4.10, 4.12, 5.8, 5.10, 5.11, 5.15, 6.11, 7.1, 9.1, 9.2, 9.6, 9.11, 10.3 D.M. Milosevi´c 1.4, 1.5, 4.10 H. Minc 2.10, 5.11 R.K. Meany 10.3 Gy. Maurer 1.2 K.B. Mollweide 1.7 L.J. Mordell 1.9 A.W. Marshall 1.12 C. Maclaurin 2.2H, 2.3A, 6.10 A. Murthy 2.8, 3.15, 9.2, 9.3, 9.6 J.S. Martins 2.10 L.E. Matties 3.8 A.P. Minin 3.13 P. Mersenne 3.15 J. Mollerup 4.9, 5.13, 7.1
N C. Nedelcu 1.4 T. Negoi 2.2F J. Nagura 3.5 C.A. Nicol 3.13, 8.11 K. Nageswara Rao 3.19 E. Neuman 4.13, 10.9, 10.10 S.M. Nikolsky 6.16 T. Nagell 8.4 A. Nikiforov 5.15 O M. Olteanu 1.10 I. Olkin 1.12 L. Olivier 2.8 O. Ore 3.19, 10.7 J.M.H. Olmsted 6.14 W. Orlicz 10.10 Gh. Orb´an 10.10 V. Ouvarov 5.15 P J.E. Peˇcari´c 1.5, 4.9, 4.10, 7.2 G. P´olya 1.6, 2.10, 4.3, 4.11, 4.14, 5.13, 6.10, 6.14, 9.7 D. Pedoe 1.6 A.O. Pittenger 2.1, 6.3, 10.9 T. Popoviciu 2.1
458
ˇ I.I. Pjateckii-Sapiro 3.2 F. Popovici 3.9 J.L. Pe 3.12, 3.13 M. Petrovi´c 5.9 J. Peetre 6.7, 10.10 C. Popa 6.16 E. Prouchet 9.8 C. Pomerance 9.10 Q F. Qi 6.13 R R. Redheffer 1.4, 1.11, 1.13 G. Robin 2.7 S. Ramanujan 2.9, 6.5, 9.6 J.B. Rosser 3.1, 3.2, 3.6 B. Riemann 3.4, 3.7, 5.5, 6.16, 9.7 I.Z. Ruzsa 3.10, 9.11 A.W. Roberts 4.3, 4.15, 6.15 J.L. Raabe 5.11 I. Ra¸sa 6.3, 6.6, 10.9 M. Rolle 6.11, 6.14, 6.15 W. Rudin 6.14 M.N. Ro¸sculet¸ 5.11, 6.16 Th. M. Rassias 7.8, 7.12, 8.2, 8.5, 8.6 H.J.J. te Riele 8.6 H. Roskam 9.14, 9.15, 10.8 J. Rooin 10.9 R.A. Rosenbaum 10.10 P. du Bois-Reymond 6.16 S J. S´andor 1.3, 1.4, 1.5, 1.6, 1.8, 1.9, 1.10, 1.11, 1.13, 1.14, 2.1, 2.2A, 2.2B, 2.2C, 2.2D, 2.2F, 2.2H, 2.2J, 2.2O, 2.2P, 2.2S, 2.5, 2.9, 2.10, 3.1, 3.2, 3.4, 3.6, 3.7, 3.8, 3.9, 3.10, 3.11, 3.12, 3.13, 3.14, 3.16, 3.19, 4.3, 4.7, 4.8, 4.9, 4.10, 4.11, 4.15, 4.16, 5.5, 5.7, 5.8, 5.10,
5.11, 5.13, 5.14, 5.15, 6.1, 6.2, 6.3, 6.4, 6.6, 6.7, 6.8, 6.9, 6.10, 6.11, 6.13, 6.14, 6.15, 6.16, 7.2, 7.4, 7.5, 7.11, 8.1, 8.6, 8.9, 8.11, 8.12, 8.14, 8.15, 9.1, 9.3, 9.4, 9.5, 9.7, 9.10, 9.11, 9.13, 9.14, 9.15, 9.16, 10.2, 10.3, 10.4, 10.7 G. Szeg¨o 1.6, 9.7 S.B. Steˇckin 1.11 W. Snellius 1.14 K.B. Stolarsky 2.1, 4.9, 6.3, 6.7, 10.3, 10.10 Gh. Stoica 2.1 O. Stolz 2.2I, 2.2O, 2.2Q, 2.3A, 2.3B I. Sathre 2.10, 5.11 I. Schoenfeld 3.1, 3.2, 3.6 J. Stirling 3.7, 5.5 D. Suryanarayana 3.9, 3.10 I. Smarandache 3.9, 3.11, 3.12, 3.15, 3.17, 8.1, 8.13, 8.16, 9.4, 9.5, 9.13, 9.14, 9.15, 9.16, 10.5 I. A. Stegun 5.13 R.M. Sorli 3.19, 10.7 E.G. Straus 3.19, 9.13 M.V. Subbarao 3.19, 9.13 H.J. Seiffert 6.1, 6.3, 6.4, 6.8 V.E.S. Szab´o 6.1, 6.2, 6.3 M.C. Sholander 6.2, 6.3, 6.5, 6.7 J. Steiner 6.4 H. Shniad 6.5 W. Sierpinski 6.10, 7.3, 9.3, 9.9 E.L. Stark 6.16 I. Stamate 7.2 H.N. Shapiro 7.3, 9.8 K.B. Subramanian 8.4 P.R. Stein 8.6 H. Sekigawa 8.6 D.D. Somashekara 9.4 A. Schinzel 9.10, 9.14 A. Stenger 10.3
459
T F.A. Valentine 10.10 D.W. De Tempe 2.2F S.R. Tims 2.2F W J.A. Tyrrel 2.2F R. Weitzenb¨ ock F. Tricomi 2.4 J. Wallis 2.4 L. T´oth 2.4, 5.12, 9.10, 9.11 E.M. Wright 3.9, 8.1, 9.7, 9.10, 9.11, B. Taylor 3.7, 6.11, 6.15, 6.16 10.3 Gh. Toader 4.9, 4.10, 4.11, 6.3, 6.6, Ch. Wall 3.19 6.9, 10.10 E.T. Whittaker 5.3, 5.4, 5.5, 5.6, 5.11, T. Trif 4.13, 10.9 5.14, 5.15, 5.16, 6.16 M.S. Tom´as 5.15 G.N. Watson 5.3, 5.4, 5.5, 5.6, 5.11, J. Thomae 7.1 5.14, 5.15, 5.16, 6.16 Y. Tsuruoka 8.6 K. Weierstrass 5.11, 6.16 Z. Tuzson 8.6 J. Wendel 5.13, 5.14, 5.16 E.C. Titchmarsh 9.7 D. Widder 6.16 R. Webster 7.1 V D.T.Walker 8.3, 8.4 I. Vir´ag 1.2 M.F.C. Woollett 8.6 V. Volenec 1.5 W.-L. Wang 10.9 A. Vernescu 2.4 P.-F. Wang 10.9 Ch.J. de la Vall´ee-Poussin 3.4 P. Vlamos 3.6 Y D.E. Varberg 4.3, 4.15, 6.15 W.H. Young 5.10 M. Vuorinen 5.14, 6.2, 6.3, 6.7 M.K. Vamanamurthy 6.2, 6.3, 6.7 Z M. Vlada 7.2 A. Zygmund 6.16 I.M. Vinogradov 8.3 P. Zwier 10.3 L. Vaserstein 8.6
460
