arXiv:math.HO/0412154 v1 8 Dec 2004
An analytical exercise∗ Leonhard Euler†
1. In considering the infinite product for...

Author:
Euler L.

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arXiv:math.HO/0412154 v1 8 Dec 2004

An analytical exercise∗ Leonhard Euler†

1. In considering the infinite product for expressing cosine in terms of any angle, which is cos

1 1 1 1 π = (1 − )(1 − )(1 − )(1 − ) etc., 2n nn 9nn 25nn 49nn

it comes to mind to investigate a method which itself works by the nature of the value of this product, which we in fact already know to be equal to π cos 2n . Indeed, it is possible, though with many labors which were brought about by artifice, for a geometric demonstration of this to be given which I do not consider at all unpleasant. 2. Therefore I set, S = (1 −

1 1 1 )(1 − )(1 − ) etc., nn 9nn 25nn

and then the logarithms produced by me are summed: lS = l(1 −

1 1 1 ) + l(1 − ) + l(1 − ) + etc., nn 9nn 25nn

∗

Delivered October 3, 1776. Originally published as Exercitatio analytica, Nova Acta Academiae Scientarum Imperialis Petropolitinae 8 (1794), 69-72, and republished in Leonhard Euler, Opera Omnia, Series 1: Opera mathematica, Volume 16, Birkh¨auser, 1992. A copy of the original text is available electronically at the Euler Archive, at www.eulerarchive.org. This paper is E664 in the Enestr¨om index. † Date of translation: December 8, 2004. Translated from the Latin by Jordan Bell, 2nd year undergraduate in Honours Mathematics, School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada. Email: [email protected]. Part of this translation was written during an NSERC USRA supervised by Dr. B. Stevens.

1

and since

1 1 1 1 1 l(1 − ) = − − − 3 − 4 − etc., x x 2xx 3x 4x the successive series set out with their signs changed are: −lS =

1 1 1 1 + 4 + 6 + 8 + etc. nn 2n 3n 4n 1 1 1 1 + + + + etc. + 2 4 3 6 9nn 2 · 9 n 3·9 n 4 · 94 n8 1 1 1 1 + + + + etc. + 2 4 3 6 25nn 2 · 25 n 3 · 25 n 4 · 254 n8 1 1 1 1 + + + + + etc. 2 4 3 6 49nn 2 · 49 n 3 · 49 n 4 · 494 n8 etc.

3. But now if we arrange each of the rows vertically in columns, we will obtain the following series for −lS: −lS =

1 1 1 1 1 (1 + 2 + 2 + 2 + 2 + etc.) nn 3 5 7 9 1 1 1 1 1 + 4 (1 + 4 + 4 + 4 + 4 + etc.) 2n 3 5 7 9 1 1 1 1 1 + 6 (1 + 6 + 6 + 6 + 6 + etc. 3n 3 5 7 9 1 1 1 1 1 + 8 (1 + 8 + 8 + 8 + 8 + etc. 4n 3 5 7 9 etc.

Thus in this way, the investigation is led to the summation of series of even powers of the harmonic progression, 1, 13 , 51 , 17 , etc. 4. By putting π2 = ρ for the sake of brevity, it can then be seen that the sums of powers can be represented in the following manner: 1 1 1 + 2 + 2 + etc. = Aρ2 , 2 3 5 7 1 1 1 1 + 4 + 4 + 4 + etc. = Bρ4 , 3 5 7 1 1 1 1 + 6 + 6 + 6 + etc. = Cρ6 , 3 5 7 etc. 1+

2

with the first one taken as A = 21 , and then the other letters determined by the preceding ones in the following way: 2 2 2 B = A2 , C = · 2AB, D = (2AC + BB), 3 5 7 2 2 E = (2AD + 2BC), F = (2AE + 2BD + CC), etc. 9 11 the truth of which will soon shine forth from a most beautiful conjunction of analysis. 5. With the substituted values, this series is obtained: −lS = If we put

ρ n

aρ2 1 Bρ4 1 Cρ6 1 Dρ8 + · 4 + · 6 + · 8 + etc. nn 2 n 3 n 4 n

= x, so that x =

π , 2n

then this series is obtained:

1 1 1 −lS = Axx + Bx4 + Cx6 + Dx8 + etc. 2 3 4 So that we can get rid of the fractions 21 , 31 , 14 , etc., we differentiate, and then after dividing the result by 2∂x it follows, −

∂S = Ax + Bx3 + Cx5 + Dx7 + etc. 2S∂x

∂S = t, so that we have 6. Then for the sake of brevity we set − 2S∂x

t = Ax + Bx3 + Cx5 + Dx7 + etc., for which the square is seen to be this series: tt =A2 xx

+2ABx4 + 2ACx6 + BB

+2ADx8 + 2AEx10 +2BC + 2BD + CC

+etc. +etc. +etc.

and in this, each of the powers of x are the very same as those found in the formulas which express the determination of the letters A, B, C, D: the only thing missing are the coefficients 32 , 52 , 72 , etc. 3

7. But we can indeed introduce these coefficients by integrating after we have multiplied through by 2∂x. So it will be obtained that Z 2 2 2 2 tt∂x = A2 x3 + · 2ABx5 + (2AC + BB)x7 + 3 3 7 9 2 (2AD + 2BC)x9 + (2AE + 2BD + CC)x11 + etc. 9 11 Since at present, 2 2 2 2 A = B, · 2AB = C, (2AC + BB) = D, etc., 3 3 7 with these values substituted in, we then attain this series: Z 2 tt∂x = Bx3 + Cx5 + Dx7 + Ex9 + etc. 8. Therefore with the series we had from before: t = Ax + Bx3 + Cx5 + Dx7 + etc., this equation clearly follows: t = Ax + 2

Z

tt∂x,

which when differentiated yields 1 1 ∂t = A∂x + 2tt∂x = ∂x + 2tt∂x, because A = . 2 2 2∂t , From this we therefore have 2∂t = ∂x(1 + 4tt), from which it is ∂x = 1+4tt 1 whose integral is at hand, namely x = A tang.2t, for which the addition of a constant is not desired, since by setting x = 0 then t should simultaneously vanish. Hence with this equation that has been discovered, if the quantity x ∂S is seen as an angle then it will be 2t = tang.x. Since it was already t = − 2S∂x , from this are deduced these equations:

−

∂S ∂S ∂x sin x = tang.x, and − = . S∂x S cos x

1

Translator: Euler uses here A tang. 2t to express the arctangent of 2t, and later uses tang. x to stand for the tangent of x.

4

cos x 9. Therefore, since it is ∂x sin x = −∂ cos x, it will be ∂S = ∂cos , and S x then by integrating, lS = l cos x+C, which to be consistent should be defined such that setting x = 0 makes lS = 0. Thus it will be C = 0, so lS = l cos x, and then so that the numbers can come forth, S = cos x. π , from this it is clear 10. Moreover, since we already had that x = 2n π that the value being sought out for S is revealed to be S = cos 2n , which agrees totally with what was known from before. Therefore, analysis most excellently confirms the former relation between the letters A, B, C, D, which I introduced in earlier calculations.

5

An analytical exercise∗ Leonhard Euler†

1. In considering the infinite product for expressing cosine in terms of any angle, which is cos

1 1 1 1 π = (1 − )(1 − )(1 − )(1 − ) etc., 2n nn 9nn 25nn 49nn

it comes to mind to investigate a method which itself works by the nature of the value of this product, which we in fact already know to be equal to π cos 2n . Indeed, it is possible, though with many labors which were brought about by artifice, for a geometric demonstration of this to be given which I do not consider at all unpleasant. 2. Therefore I set, S = (1 −

1 1 1 )(1 − )(1 − ) etc., nn 9nn 25nn

and then the logarithms produced by me are summed: lS = l(1 −

1 1 1 ) + l(1 − ) + l(1 − ) + etc., nn 9nn 25nn

∗

Delivered October 3, 1776. Originally published as Exercitatio analytica, Nova Acta Academiae Scientarum Imperialis Petropolitinae 8 (1794), 69-72, and republished in Leonhard Euler, Opera Omnia, Series 1: Opera mathematica, Volume 16, Birkh¨auser, 1992. A copy of the original text is available electronically at the Euler Archive, at www.eulerarchive.org. This paper is E664 in the Enestr¨om index. † Date of translation: December 8, 2004. Translated from the Latin by Jordan Bell, 2nd year undergraduate in Honours Mathematics, School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada. Email: [email protected]. Part of this translation was written during an NSERC USRA supervised by Dr. B. Stevens.

1

and since

1 1 1 1 1 l(1 − ) = − − − 3 − 4 − etc., x x 2xx 3x 4x the successive series set out with their signs changed are: −lS =

1 1 1 1 + 4 + 6 + 8 + etc. nn 2n 3n 4n 1 1 1 1 + + + + etc. + 2 4 3 6 9nn 2 · 9 n 3·9 n 4 · 94 n8 1 1 1 1 + + + + etc. + 2 4 3 6 25nn 2 · 25 n 3 · 25 n 4 · 254 n8 1 1 1 1 + + + + + etc. 2 4 3 6 49nn 2 · 49 n 3 · 49 n 4 · 494 n8 etc.

3. But now if we arrange each of the rows vertically in columns, we will obtain the following series for −lS: −lS =

1 1 1 1 1 (1 + 2 + 2 + 2 + 2 + etc.) nn 3 5 7 9 1 1 1 1 1 + 4 (1 + 4 + 4 + 4 + 4 + etc.) 2n 3 5 7 9 1 1 1 1 1 + 6 (1 + 6 + 6 + 6 + 6 + etc. 3n 3 5 7 9 1 1 1 1 1 + 8 (1 + 8 + 8 + 8 + 8 + etc. 4n 3 5 7 9 etc.

Thus in this way, the investigation is led to the summation of series of even powers of the harmonic progression, 1, 13 , 51 , 17 , etc. 4. By putting π2 = ρ for the sake of brevity, it can then be seen that the sums of powers can be represented in the following manner: 1 1 1 + 2 + 2 + etc. = Aρ2 , 2 3 5 7 1 1 1 1 + 4 + 4 + 4 + etc. = Bρ4 , 3 5 7 1 1 1 1 + 6 + 6 + 6 + etc. = Cρ6 , 3 5 7 etc. 1+

2

with the first one taken as A = 21 , and then the other letters determined by the preceding ones in the following way: 2 2 2 B = A2 , C = · 2AB, D = (2AC + BB), 3 5 7 2 2 E = (2AD + 2BC), F = (2AE + 2BD + CC), etc. 9 11 the truth of which will soon shine forth from a most beautiful conjunction of analysis. 5. With the substituted values, this series is obtained: −lS = If we put

ρ n

aρ2 1 Bρ4 1 Cρ6 1 Dρ8 + · 4 + · 6 + · 8 + etc. nn 2 n 3 n 4 n

= x, so that x =

π , 2n

then this series is obtained:

1 1 1 −lS = Axx + Bx4 + Cx6 + Dx8 + etc. 2 3 4 So that we can get rid of the fractions 21 , 31 , 14 , etc., we differentiate, and then after dividing the result by 2∂x it follows, −

∂S = Ax + Bx3 + Cx5 + Dx7 + etc. 2S∂x

∂S = t, so that we have 6. Then for the sake of brevity we set − 2S∂x

t = Ax + Bx3 + Cx5 + Dx7 + etc., for which the square is seen to be this series: tt =A2 xx

+2ABx4 + 2ACx6 + BB

+2ADx8 + 2AEx10 +2BC + 2BD + CC

+etc. +etc. +etc.

and in this, each of the powers of x are the very same as those found in the formulas which express the determination of the letters A, B, C, D: the only thing missing are the coefficients 32 , 52 , 72 , etc. 3

7. But we can indeed introduce these coefficients by integrating after we have multiplied through by 2∂x. So it will be obtained that Z 2 2 2 2 tt∂x = A2 x3 + · 2ABx5 + (2AC + BB)x7 + 3 3 7 9 2 (2AD + 2BC)x9 + (2AE + 2BD + CC)x11 + etc. 9 11 Since at present, 2 2 2 2 A = B, · 2AB = C, (2AC + BB) = D, etc., 3 3 7 with these values substituted in, we then attain this series: Z 2 tt∂x = Bx3 + Cx5 + Dx7 + Ex9 + etc. 8. Therefore with the series we had from before: t = Ax + Bx3 + Cx5 + Dx7 + etc., this equation clearly follows: t = Ax + 2

Z

tt∂x,

which when differentiated yields 1 1 ∂t = A∂x + 2tt∂x = ∂x + 2tt∂x, because A = . 2 2 2∂t , From this we therefore have 2∂t = ∂x(1 + 4tt), from which it is ∂x = 1+4tt 1 whose integral is at hand, namely x = A tang.2t, for which the addition of a constant is not desired, since by setting x = 0 then t should simultaneously vanish. Hence with this equation that has been discovered, if the quantity x ∂S is seen as an angle then it will be 2t = tang.x. Since it was already t = − 2S∂x , from this are deduced these equations:

−

∂S ∂S ∂x sin x = tang.x, and − = . S∂x S cos x

1

Translator: Euler uses here A tang. 2t to express the arctangent of 2t, and later uses tang. x to stand for the tangent of x.

4

cos x 9. Therefore, since it is ∂x sin x = −∂ cos x, it will be ∂S = ∂cos , and S x then by integrating, lS = l cos x+C, which to be consistent should be defined such that setting x = 0 makes lS = 0. Thus it will be C = 0, so lS = l cos x, and then so that the numbers can come forth, S = cos x. π , from this it is clear 10. Moreover, since we already had that x = 2n π that the value being sought out for S is revealed to be S = cos 2n , which agrees totally with what was known from before. Therefore, analysis most excellently confirms the former relation between the letters A, B, C, D, which I introduced in earlier calculations.

5

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