Theoretical Computer Science Cheat Sheet
f(n) = O(g(n))
De nitions i 9 positive c; n such that 0 f(n) cg(n) 8n n . i 9 positive c; n such that f(n) cg(n) 0 8n n . i f(n) = O(g(n)) and f(n) = (g(n)). i limn!1 f(n)=g(n) = 0. i 8 2 R, 9n such that jan , aj < , 8n n . least b 2 R such that b s, 8s 2 S. greatest b 2 R such that b s, 8s 2 S. lim inf fai j i n; i 2 Ng. n!1
n X
0
0
f(n) = (g(n))
i
=1
In general:
0
f(n) = o(g(n)) lim a = a n!1 n sup S inf S lim inf a n!1 n
n X i
=1
n X
+ 1) ; i = n(n + 1)(2n 6
i = n (n4+ 1) : i
2
n , X
+1
=1
=1
+1
+1
+1
=1
0
=0
+1
i
n X
=0
i
ici =
ncn
+2
=0
+1
(c , 1)
2
=0
Harmonic series: n X Hn = 1i ;
n!1
i
, (n + 1)cn
i
+ c ; c 6= 1;
i
1 X
=1
2
i
=0
n X
ici = (1 ,c c) ; c < 1:
iHi = n(n2+ 1) Hn , n(n4, 1) :
i
=1
n i X
1
=1
+1
=1
2
=0
=0
1
25. 28.
1
=0
=0
31.
=0
=0
=0
34.
=0
36.
n + 1 Combinations: Size k subHi = (n + 1)Hn , n; Hi = m + 1 Hn , m 1+ 1 : m sets of a size n set. i i n n n n n X Stirling numbers (1st kind): n n! k n 1. k = (n , k)!k! ; 2. 3. k = n , k ; Arrangements of an n elek =2 ; k n , 1 ment set into k cycles. n , 1 n n , 1 n n n 4. k = k k , 1 ; 5. k = k + k , 1 ; Stirling numbers (2nd kind): k Partitions of an n element X r + k r + n + 1 6. mn mk = nk mn ,, kk ; 7. ; set into k non-empty sets. k = n
n k n 1st order Eulerian numbers: n k n + 1 n r s r + s k X X Permutations : : :n on 8. = ; 9. m+1 k n,k = n ; f1; 2; : : :; ng with k ascents. k m k n n
n n = (,1)k k , n , 1; 10. 11. 2nd order Eulerian numbers. k k k 1 = n = 1; n n n , 1 n , 1 Cn Catlan Numbers: Binary n , trees with n + 1 vertices. 12. 2 = 2 , 1; 13. k = k k + k , 1 ; n n n n n = (n , 1)!; 15. = (n , 1)!H ; 16. = 1; 17. k ; n, 1 2 n k n n , 1 n , 1 n n n n n X 1 2n ; = (n , 1) + ; 19. = = ; 20. = n!; 21. C = n n+1 n k,1 2 kn n k n n , 1 n n , 1 n k k n , 1 n , 1 = = 1; 23. = ; 24. = (k + 1) + (n , k) ; 0 n , 1 k n , 1 , k k k k , 1 0 n n n 1 if k = 0, n , n , 1; n , (n + 1)2n + n + 1 ; = 26. = 2 27. = 3 k 0 otherwise 1 2 2 n X n n x + k m n + 1 n n k X X n xn = ; 29. m = (m + 1 , k)n(,1)k ; 30. m! m = k n k k n,m ; k k k n X n n n n n , k n , k , m k!; 32. 0 = 1; 33. n = 0 for n 6= 0; m =k k m (,1) n n , 1 n , 1 n n (2n)n X = (k + 1) + (2n , 1 , k) ; 35. k k k,1 k = 2n ; k x X n + 1 X n k X n n x + n , 1 , k n k ; 37. m + 1 = = (m + 1)n,k ; x,n = k k 2n k m m k k n X
=0
22.
=1
=1
18.
2
1
0
n!1 ,n k
14.
2
3
i = m 1+ 1 (n + 1)m , 1 , (i + 1)m , im , (m + 1)im i i m nX , X m+1 Bk nm ,k : im = m 1+ 1 k i k Geometric series: 1 1 n n X X X ci = 1 ,1 c ; ci = 1 ,c c ; c < 1; ci = c c ,,1 1 ; c 6= 1;
lim supfai j i n; i 2 Ng.
lim sup an
n X m
0
f(n) = (g(n))
i = n(n2+ 1) ;
Series
=0
=0
Theoretical Computer Science Cheat Sheet
Identities Cont. n + 1 X n k X n k n n,k = n! X 1 k ; 38. m + 1 = = n k m k m k k! m n X k n k+1 =0
=0
40. m = (,1)n,k ; k m + 1 m + n +k1 X m n + k 42. = k k ; m
Trees x X n n x + k tree with n 39. x , n = ; Every vertices has n , 1 k 2n n X n +k 1 k edges. =0
41. m = (,1)m,k ; Kraft inequalk + 1 m m + n + 1k X n + k ity: If the depths m 43. = k(n + k) k ; of the leaves of m
a binary tree are n X n +k 1 k n X n + 1 k k :n: :; dn: m,k ; 45. (n , m)! m,k ; for n m, d ;X 44. m = ( , 1) = ( , 1) k+1 m m 2,di 1; n k X m , nm + n m + k n k k +X1 mm, nm + n m + k i =0
=0
1
46. n , m = +k k ; n ` +kmm +Xk kn n , k n 48. ` + m ` = ` m k ;
47. n , m = k ; and equality holds n k` + mm+ k X n k+k n , k only if every inn 49. ` + m ` = ` m k : ternal node has 2 =1
k
Master method: T(n) = aT(n=b) + f(n); a 1; b > 1 If 9 > 0 such that f(n) = O(n b a, ) then T (n) = (n b a ): If f(n) = (n b a ) then T(n) = (n b a log n): If 9 > 0 such that f(n) = (n b a ), and 9c < 1 such that af(n=b) cf(n) for large n, then T(n) = (f(n)): Substitution (example): Consider the following recurrence Ti = 2 i Ti ; T = 2: Note that Ti is always a power of two. Let ti = log Ti . Then we have ti = 2i + 2ti; t = 1: Let ui = ti =2i. Dividing both sides of the previous equation by 2i we get ti = 2i + ti : 2i 2i 2i log
log
log
log
2
log
2
+1
2
+
1
k
,
log
+1
+1
+1
Substituting we nd ui = + ui ; u = 12; which is simply ui = i=2. So we nd that Ti has the closed form Ti = 2i i,1 . Summing factors (example): Consider the following recurrence Ti = 3Tn= + n; T = n: Rewrite so that all terms involving T are on the left side Ti , 3Tn= = n: Now expand the recurrence, and choose a factor which makes the left side \telescope" +1
1
1
2
2
1
log
log
=0
Let c = andm = log n. Then we have m m , 1 X c n ci = n c , 1 3
2
2
+1
i
=0
= 2n(c c 2 n , 1) = 2n(c ck c n , 1) = 2nk , 2n 2n : log
+1
1 58496
1
1
j
Note that
0
=0
Ti = 1 + +1
1
2
2
, 2n;
where k = (log ), . Full history recurrences can often be changed to limited history ones (example): Consider the following recurrence i, X Ti = 1 + Tj ; T = 1: 3 2 2
Xi j
Tj :
=0
Subtracting we nd i, Xi X Ti , Ti = 1 + Tj , 1 , Tj 1
+1
j
=0
= Ti : And so Ti = 2Ti = 2i . +1
+1
Generating functions: 1. Multiply both sides of the equation by xi . 2. Sum both sides over all i for which the equation is valid. 3. Choose a generatingPfunction i G(x). Usually G(x) = 1 i x. 3. Rewrite the equation in terms of the generating function G(x). 4. Solve for G(x). 5. The coecient of xi in G(x) is gi . Example: gi = 2gi + 1; g = 0: Multiply X andi sum: X X gi x = 2gi xi + xi: =0
+1
i
0
+1
i
i
P We choose G(x) = i xi. Rewrite in terms of G(x): X i G(x) , g 0
0
0
0
log
1
+1
1 T(n) , 3T(n=2) = n , 3 T(n=2) , 3T(n=4) = n=2 .. .. .. . . . ,T(2) , 3T(1) = 2 n , 2 3 , 3 2 n T(1) , 0 = 1 Summing the left side we get T (n). Summing the right side we get X2 n n i i3 : i 2
2
+1
Recurrences
sons.
j
=0
0
x
= 2G(x) +
x:
i
0
Simplify: G(x) = 2G(x) + 1 : x 1,x Solve for G(x): x G(x) = (1 , x)(1 , 2x) : Expand this using partial fractions: G(x) = x 1 ,2 2x , 1 ,1 x
0 1 X X = x @2 2i xi , xi A i X ii i 0
=
(2
i
0
So gi = 2i , 1.
+1
0
, 1)x : +1
Theoretical Computer Science Cheat Sheet p e 2:71828,
0:57721, = 1:61803,
3:14159, i 2i pi General 1 2 2 Bernoulli Numbers (Bi = 0, odd i 6= 1): 2 4 3 B = 1, B = , , B = , B = , , B = ,B =, ,B = . 3 8 5 Change of base, quadratic formula: 4 16 7 p b , 4ac : log x , b 5 32 11 a logb x = log b ; 2a 6 64 13 a Euler's number e: 7 128 17 + e = 1 + 8 256 19 +x n+ x + 9 512 23 lim 1 + n = e : n!1 ,1 + n < e < ,1 + n : 10 1,024 29 n n 11 2,048 31 , e 11e n 1 + n = e , 2n + 24n , O n1 : 12 4,096 37 13 8,192 41 Harmonic numbers: 14 16,384 43 1, , , , , , , , ; : : : 15 32,768 47 16 65,536 53 lnn < Hn < ln n + 1; 17 131,072 59 Hn = ln n + + O n1 : 18 262,144 61 Factorial, Stirling's approximation: 19 524,288 67 ::: 20 1,048,576 71 1 21 2,097,152 73 p n n n! = 2n 1 + 22 4,194,304 79 e n : 23 8,388,608 83 Ackermann's 8 jfunction and inverse: 24 16,777,216 89 i=1 <2 a(i; j) = : a(i , 1; 2) j=1 25 33,554,432 97 a(i , 1; a(i; j , 1)) i; j 2 26 67,108,864 101 (i) = minfj j a(j; j) ig: 27 134,217,728 103 28 268,435,456 107 Binomial distribution: n 29 536,870,912 109 Pr[X = k] = k pk qn,k ; q = 1 , p; 30 1,073,741,824 113 n n X 31 2,147,483,648 127 [X] = k = 1k pk qn,k = np: E k 32 4,294,967,296 131 Poisson distribution: Pascal's Triangle , k Pr[X = k] = e k! ; E[X] = : 1 11 Normal (Gaussian) distribution: 121 p(x) = p 1 e, x, 2 = 2 ; E[X] = : 2 1331 The \coupon collector": We are given a 14641 random coupon each day, and there are n 1 5 10 10 5 1 di erent types of coupons. The distribu1 6 15 20 15 6 1 tion of coupons is uniform. The expected number of days to pass before we to col1 7 21 35 35 21 7 1 lect all n types is 1 8 28 56 70 56 28 8 1 nHn: 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1+
2
0
1
1
6
1
30
5
10
30
1
4
6
1
8
42
1
2
2
66
2
1
1
1
1
2
6
24
120
1
1
+1
1
2
3
3
11
25
137
49
363
761
7129
2
6
12
60
20
140
280
2520
1, 2, 6, 24, 120, 720, 5040, 40320, 362880,
(
)
2
5
p
^ = , ,:61803 Probability Continuous distributions:Z If b Pr[a < X < b] = p(x) dx; a then p is the probability density function of X. If Pr[X < a] = P (a); then P is the distribution function of X. If P and p both existZthen a P(a) = p(x) dx: ,1 Expectation: If XX is discrete E[g(X)] = g(x) Pr[X = x]: 1
5
2
x
If X continuous Z 1 then Z1 E[g(X)] = g(x)p(x) dx = g(x) dP(x): ,1 ,1 Variance, standard deviation: VAR[X] = E[X ] , E[X] ; p = VAR[X]: Basics: Pr[X _ Y ] = Pr[X] + Pr[Y ] , Pr[X ^ Y ] Pr[X ^ Y ] = Pr[X] Pr[Y ]; i X and Y are independent. ^Y] Pr[X jY ] = Pr[X Pr[B] E[X Y ] = E[X] E[Y ]; i X and Y are independent. E[X + Y ] = E[X] + E[Y ]; E[cX] = c E[X]: Bayes' theorem: jAi] Pr[Ai ] Pr[AijB] = PnPr[B Pr[A ] Pr[B jA ] : 2
j
=1
2
j
j
Inclusion-exclusion: n h _n i X Pr Xi = Pr[Xi ] + i
=1
i n X
=1
k
(,1)k
+1
=1
X ii <
Pr
h ^k j
Moment inequalities: Pr jX j E[X] 1 ;
=1
h i Pr X , E[X] 1 : Geometric distribution: Pr[X = k] = pk, q; q = 1 , p; 1 X 1 E[X] = kpqk, = p : k 2
1
1
=1
i
Xij :
Theoretical Computer Science Cheat Sheet
Trigonometry
Multiplication:
C = A B; ci;j =
(0,1)
A
b
C
(-1,0)
c
a
B Pythagorean theorem: C =A De nitions: sin a = A=C; csc a = C=A; sin a = A ; tan a = cos a B 2
(cos ; sin ) (1,0)
det A =
= Permanents:
1
2
cos x = sec1 x ;
sin x + cos x = 1; 2
2
1 + cot x = csc x; 2
, sin x = cos , x ;
2
sin x = sin( , x);
cos x = , cos( , x);
, tan x = cot , x ;
cot x = , cot( , x);
csc x = cot x , cot x;
2
2
2
sin(x y) = sinx cos y cos x siny; cos(x y) = cos x cos y sin x sin y; x tan y ; tan(x y) = 1tan tan x tany cot y1 cot(x y) = cotxxcot cot y ;
sin 2x = 1 +2 tanx tan x ;
sin 2x = 2 sin x cos x;
2
cos 2x = cos x , sin x;
cos 2x = 2 cos x , 1; tan x cos 2x = 1 , 2 sin x; cos 2x = 11 , + tan x ; 2 tanx ; x , 1; tan 2x = 1 , cot 2x = cot2 cot x tan x sin(x + y) sin(x , y) = sin x , sin y; 2
2
2
2
2
2
2
2
2
2
cos(x + y) cos(x , y) = cos x , sin y: Euler's equation: eix = cos x + i sin x; ei = ,1:
c 1994 by Steve Seiden 2
n XY
sign()ai; i :
2
[email protected] http://www.ics.uci.edu/~sseiden
perm A =
b e
ai; i :
=1
x ,x cosh x = e +2e ; 1 ; csch x = sinh x 1 : coth x = tanh x
Identities: cosh x , sinh x = 1; tanh x + sech x = 1; coth x , csch x = 1; sinh(,x) = , sinh x; cosh(,x) = cosh x; tanh(,x) = , tanh x; sinh(x + y) = sinh x cosh y + cosh x sinh y; cosh(x + y) = cosh x cosh y + sinh x sinh y; sinh 2x = 2 sinh x cosh x; cosh 2x = cosh x + sinh x; cosh x + sinh x = ex ; cosh x , sinh x = e,x ; (cosh x + sinh x)n = cosh nx + sinh nx; n 2 Z; 2 sinh x = cosh x , 1; 2 cosh x = cosh x + 1: 2
2
2
2
2
2
2
2
2
sin cos tan 0 0 p1 p0 6
4
3 2
2
2
2
1
p p
2 2
2 3
2
1
p
3
2 2
2 1 2
0
3
3
1
p 3 1
2
A = hc; = ab sin C; A sin B : = c sin 2 sin C Heron's formula: 2 1 2
2
A = ps sa sb sc ; s = (a + b + c); sa = s , a; sb = s , b; sc = s , c: More identities: r x sin = 1 , 2cos x ; 2
Hyperbolic Functions
De nitions: x ,x sinh x = e ,2 e ; x e,x tanh x = eex , + e,x ; 1 ; sech x = cosh x
2
1
( )
i
A c B Law of cosines: c = a +b ,2ab cos C: Area: 1
aei + bfg + cdh , ceg , fha , ibd: n XY
a
b h
2
( )
i
a b c d e f = g b c , h a c + i a g h i e f d f d
cos a = B=C; sec a = C=B; a B cot a = cos sin a = A :
2
ai;k bk;j :
=1
2 2 and 3 3 determinant: a b c d = ad , bc;
2
Area, radius of inscribed circle: AB; A +AB B +C:
2
k
=1
+B :
Identities: sin x = csc1 x ; tan x = cot1 x ; 1 + tan x = sec x;
n X
More Trig. C
Determinants: det A = 0 i A is non-singular. det A B = det A det B;
(0,-1)
2
Matrices
: : : in mathematics you don't understand things, you just get used to them. { J. von Neumann
2
r
= 1 + 2cos x ; r 1 , cos x x tan = 1 + cos x ; x; = 1 ,sincos x sin = 1 + cosx x ; r 1 + cos x x cot = 1 , cos x ; = 1 +sincosx x ; x ; = 1 ,sincos x ix , e,ix e sin x = 2i ; ix ,ix cos x = e +2 e ; ix e,ix tan x = ,i eeix , + e,ix ; ix 1 = ,i ee ix , + 1; sin x = sinhi ix ; cos x = cosh ix; tan x = tanhi ix : cos x 2
2
2
2
2
Theoretical Computer Science Cheat Sheet
Number Theory The Chinese remainder theorem: There exists a number C such that: C r mod m .. .. .. . . . C rn mod mn if mi and mj are relatively prime for i 6= j. Euler's function: (x) is the number of positive integersQnless than x relatively prime to x. If i pei i is the prime factorization of x then n Y (x) = pei i , (pi , 1): 1
1
=1
1
i
=1
Euler's theorem: If a and b are relatively prime then 1 a b mod b: Fermat's theorem: 1 ap, mod p: The Euclidean algorithm: if a > b are integers then gcd(a; b) = gcd(a mod b; b): Q If ni pei i is the prime factorization of x then n ei X Y S(x) = d = pip ,,1 1 : ( )
1
=1
+1
djx
i
=1
i
Perfect Numbers: x is an even perfect number i x = 2n, (2n , 1) and 2n , 1 is prime. Wilson's theorem: n is a prime i (n , 1)! ,1 mod n: Mobius 8 inversion: if i = 1. > < 10 square-free. (i) = > (,1)r ifif ii isis not the product of : r distinct primes. If X G(a) = F(d); 1
then
F(a) =
X dja
dja
a
Prime numbers: n pn = n ln n + n lnln n , n + n lnln ln n n + O ln n ; 2!n (n) = lnnn + (lnnn) + (lnn) + O (lnnn) : 4
0
1
1
1
0
2
3
1
2
0
1
1
0
1
0
1
1
0
1
0
2
0
0
1
1
2
1
2
1
0
1
0
2
0
2
0
2
2
2
1
1
(d)G d :
2
Graph Theory Notation: De nitions: E(G) Edge set Loop An edge connecting a verV (G) Vertex set tex to itself. c(G) Number of components Directed Each edge has a direction. G[S] Induced subgraph Simple Graph with no loops or deg(v) Degree of v multi-edges. (G) Maximum degree Walk A sequence v e v : : :e` v` . (G) Minimum degree Trail A walk with distinct edges. (G) Chromatic number Path A trail with distinct E (G) Edge chromatic number vertices. Gc Complement graph Connected A graph where there exists K Complete graph n a path between any two K Complete bipartite graph n ;n 1 2 vertices. r(k; `) Ramsey number Component A maximal connected subgraph. Geometry Tree A connected acyclic graph. Projective coordinates: triples Free tree A tree with no root. (x; y; z), not all x, y and z zero. DAG Directed acyclic graph. (x; y; z) = (cx; cy; cz) 8c 6= 0: Eulerian Graph with a trail visiting Cartesian Projective each edge exactly once. Hamiltonian Graph with a path visiting (x; y) (x; y; 1) each vertex exactly once. y = mx + b (m; ,1; b) Cut A set of edges whose rex=c (1; 0; ,c) moval increases the numDistance formula, Lp and L1 ber of components. metric: p Cut-set A minimal cut. (x , x ) + (x , x ) ; Cut edge A size 1 cut. jx , x jp + jx , x jp =p; k-Connected A graph connected with jx , x jp + jx , x jp =p: the removal of any k , 1 lim p !1 vertices. k-Tough 8S V; S 6= ; we have Area of triangle (x ; y ), (x ; y ) and (x ; y ): k c(G , S) jS j. k-Regular A graph where all vertices x , x y , y abs x , x y , y : have degree k. k-Factor A k-regular spanning Angle formed by three points: subgraph. Matching A set of edges, no two of (x ; y ) which are adjacent. ` Clique A set of vertices, all of which are adjacent. (0; 0) ` (x ; y ) Ind. set A set of vertices, none of which are adjacent. cos = (x ; y `) `(x ; y ) : Vertex cover A set of vertices which cover all edges. Line through two points (x ; y ) Planar graph A graph which can be emand (x ; y ): beded in the plane. x y 1 Plane graph An embedding of a planar x y 1 = 0: graph. x y 1 X Area of circle, volume of sphere: deg(v) = 2m: v2V A = r ; V = r : If G is planar then n , m + f = 2, so If I have seen farther than others, f 2n , 4; m 3n , 6: it is because I have stood on the Any planar graph has a vertex with de- shoulders of giants. gree 5. { Issac Newton 1
1
2
1
1
2
2
0
1
1
0
0
1
1
2
4 3
3
0
Theoretical Computer Science Cheat Sheet
Wallis' identity: = 2 21 23 43 45 65 67
Brouncker's continued fraction expansion: 1 =1+ 2 2+ 52 2
4
3
2+
2+ 2+72
Gregrory's series: = 1, + , + , Newton's series: 1 + 13 + =1+ 2 232 2452 Sharp's series: = p1 1 , 1 + 1 , 1 + 3 3 3 5 3 7 3 Euler's series: 4
1
1
1
1
3
5
7
9
6
3
6
5
1
2
= = 2 = 6
+ 2+
2
+ 2+ 2, 2+ 2
1
2
1
8
1
12
1
1
1
2
1
2
1
3
1
2
+ 2+
3
+ 2+ 2, 2+ 2
1 3
1 5
1 3
2
1
4
1
7
1
4
+ 2 + 2
1 5
1 9
1
2
5
,
Partial Fractions Let N(x) and D(x) be polynomial functions of x. We can break down N(x)=D(x) using partial fraction expansion. First, if the degree of N is greater than or equal to the degree of D, divide N by D, obtaining N(x) = Q(x) + N 0(x) ; D(x) D(x)
where the degree of N 0 is less than that of D. Second, factor D(x). Use the following rules: For a non-repeated factor: N(x) = A + N 0 (x) ; (x , a)D(x) x , a D(x) where
N(x)
A = D(x) : x a For a repeated factor: mX , N(x) Ak + N 0 (x) ; = m (x , a) D(x) (x , a)m,k D(x)
Calculus
Derivatives: du dv d(uv) dv du 2. d(udx+ v) = du 1. d(cu) dx = c dx ; dx, + dx ; , 3. dx = u dx + v dx ; du dv n) d(ecu ) = cecu du ; n, du ; 5. d(u=v) = v dx , u dx ; 4. d(u = nu 6. dx dx dx v dx dx u u) 1 du 7. d(cdx ) = (ln c)cu du 8. d(ln dx ; dx = u dx ; d(cos u) = , sin u du ; u) = cos u du ; 10. 9. d(sin dx dx dx dx d(tan u) du d(cot u) 12. dx = csc u du 11. dx = sec u dx ; dx ; du u) du u) 14. d(csc 13. d(sec dx = tan u sec u dx ; dx = , cot u csc u dx ; u) 1 du u) = p ,1 du ; 15. d(arcsin 16. d(arccos dx = p1 , u dx ; dx 1 , u dx u) = 1 du ; d(arccot u) = ,1 du ; 17. d(arctan 18. dx 1 , u dx dx 1 , u dx 1
2
2
2
where
k
=0
dk N(x) Ak = k!1 dx : k D(x) x a =
The reasonable man adapts himself to the world; the unreasonable persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable. { George Bernard Shaw
2
2
2
u) = p,1 du ; 20. d(arccsc dx u 1 , u dx u) du 22. d(cosh dx = sinh u dx ;
u) = p 1 du ; 19. d(arcsec dx u 1 , u dx u) du 21. d(sinh dx = cosh u dx ; 2
2
u) du 23. d(tanh dx = sech u dx ;
u) du 24. d(coth dx = , csch u dx ;
2
2
u) du 25. d(sech dx = , sech u tanh u dx ; u) = p 1 27. d(arcsinh dx
u) du 26. d(csch dx = , csch u coth u dx ;
du ; 1 + u dx d(arctanh u) 1 du ; 29. = dx 1 , u dx u) = p,1 du ; 31. d(arcsech dx u 1 , u dx Integrals:
u) = p 1 28. d(arccosh dx
du ; u , 1 dx d(arccoth u) 1 du ; 30. = dx u , 1 dx u) = p,1 du : 32. d(arccsch dx juj 1 + u dx
2
2
2
2
2
1. 3.
=
1
2
6. 8.
Z
2
Z
Z
Z
xn dx =
Z 1 4. x dx = ln x; Z dv
1 n n + 1 x ; n 6= ,1; +1
Z dx 1 + x = arctan x; Z
10. 12. 14.
7.
2
Z Z Z
Z
Z
2. (u + v) dx = u dx + v dx;
cu dx = c u dx;
5.
9.
13.
sec x dx = ln j sec x + tan xj;
p
Z
arcsin xa dx = arcsin xa + a , x ; a > 0; 2
2
ex dx = ex ;
Z u dx dx = uv , v du dx dx;
sin x dx = , cos x;
tan x dx = , ln j cos xj;
Z
11.
Z
Z
cos x dx = sin x;
cot x dx = ln j cos xj;
csc x dx = ln j csc x + cot xj;
Theoretical Computer Science Cheat Sheet 15. 17. 19. 21. 23. 25. 26. 29. 33. 36.
Z Z Z Z Z Z Z Z Z Z
Calculus Cont.
p
arccos xa dx = arccos xa ,
16.
a , x ; a > 0;
,
2
2
Z
arctan xa dx = x arctan ax , a ln(a + x ); a > 0; 2
2
1
Z
18.
sin (ax)dx = a ax , sin(ax) cos(ax) ; 2
39.
40.
Z Z
43. 46. 48. 50. 52.
Z Z Z Z
1
Z
20.
2
csc x dx = , cot x; 2
Z n, n, x sinx n , 1 Z 1 Z sinn, x dx; n x dx = cos sinn x dx = , sin nx cos x + n , 22. cos + n cosn, x dx; n n Z n, x Z n, x Z tan cot n n , n 24. cot x dx = , n , 1 , cotn, x dx; n 6= 1; tan x dx = n , 1 , tan x dx; n 6= 1; n, x n , 2 Z n, sec n x dx = tan xnsec , 1 + n , 1 sec x dx; n 6= 1; Z Z n, x n , 2 Z n, x dx; n 6= 1; 27. sinh x dx = cosh x; 28. cosh x dx = sinh x; + csc csc n x dx = , cot xncsc ,1 n,1 1
1
2
2
1
1
2
2
1
2
1
2
tanh x dx = ln j cosh xj; 30.
Z
coth x dx = ln j sinh xj; 31.
1
1
4
2
arcsinh xa dx = x arcsinh xa ,
p
Z
34.
sinh x dx = sinh(2x) , x; 2
Z
sech x dx = arctan sinh x; 32. 1
1
4
37.
x + a ; a > 0; 2
2
Z
p
2
2
2
2
= ln x + a + x ; a > 0; a +x dx x a + x = a arctan a ; a > 0; 2
2
41.
1
2
2
2
2 3 2
2
p
2
2
2
2
2
2
a4 arcsin x ; a
3
8
2
2
2
2
a > 0;
Zp
2
2
2
2
2
2
2
2
2
2
2
2
51.
2
8
2
2
2
2
x x a dx = (x a ) ; 2
2
1 3
2
2
2 3 2
2
p
2
2
2
8
2
2
arctanh xa dx = x arctanh xa + a ln ja , x j;
2
2
2
2
sech x dx = tanh x;
2
2
2
2
2
2
Z
a , x dx = x a , x + a2 arcsin ax ; a > 0;
2
dx = 1 ln x ; ax + bx a a + bx p a + bx dx = 2pa + bx + a Z p 1 dx; x a +pbx p x a , x dx = pa , x , a ln a + a , x ; x x 2
2
2
Z 1 ln a + x ; p dx = arcsin ax ; a > 0; 44. a dx = , x 2a a , x a ,x p p p a x dx = x a x a2 ln x + a x ; 47. 8
2
csch x dx = ln tanh x ;
2
2
2
Z
35.
cosh x dx = sinh(2x) + x; 2
Z p p 54. x a , x dx = x (2x , a ) a , x + a4 arcsin xa ; a > 0; Z Z x dx p 56. pa , x = , a , x ; 57. p p Z a +x Z a + a + x p 58. dx = a + x , a ln ; 59. x Z px = 60.
2
sec x dx = tan x;
42. (a , x ) = dx = x (5a , 2x ) a , x +
Z
,
cos (ax)dx = a ax + sin(ax) cos(ax) ; 2
p 8 < x arccosh xa , x + a ; if arccosh xa > 0 and a > 0, 38. arccosh xa dx = : p x arccosh xa + x + a ; if arccosh xa < 0 and a > 0, Z dx p Z
2
2
Z
49.
2
Z
45.
Z
p dx Z xp , a 2
dx x p ; = = (a , x ) a a ,x p = ln x + x , a ; a > 0; 2
2
3 2
2
2
2
2
2
2
=
+ bx) ; x a + bxdx = 2(3bx , 2a)(a 15b 3 2
p
2
pa x 1 a + bx , ; a > 0; p dx = p ln p a + bx a + bx + pa 2 Z p 53. x a , x dx = , (a , x ) = ; 2
2
Z
1
2
3
p
2 3 2
55. p dx = , a ln a + ax , x ; a ,x p x dx p = , x a , x + a2 arcsin a;x a > 0; pa , x x , a dx = px , a , a arccos a ; a > 0; jxj x 2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
61.
Z
p dx
x x +a 2
2
1
2
= a ln
; a+ a +x px
2
2
Theoretical Computer Science Cheat Sheet
62. 64. 66.
Z Z
Calculus Cont.
p
p dx = a arccos jaxj ; a > 0; p dx 63. = xa x a ; x x ,a Z pxx ax a (x + a ) = p x dx p 65. dx = 3a x ; = x a ; x a 8 p x > + b , pb , 4ac ; if b > 4ac, > < pb ,1 4ac ln 2ax dx 2ax + b + b , 4ac = ax + bx + c > > : p 2 arctan p2ax + b ; if b < 4ac, 4ac , b 4ac , b 8 1 papax + bx + c ; if a > 0, > p ln 2ax + b + 2 < a p dx = ax + bx + c > : p1,a arcsin p,b2ax, ,4acb ; if a < 0, Z p p ax + bx + c dx = 2ax + b ax + bx + c + 4ax , b p dx ; 1
2
2
2
2
2
2
2
Z
Z
2
2
2
2
2
2
2
3 2
2
4
2
2
3
2
2
2
Finite Calculus Di erence, shift operators: f(x) = f(x + 1) , f(x); E f(x) = f(x + 1): Fundamental Theorem: X f(x) = F(x) , f(x)x = F (x) + C: b X
2
2
2
2
67. 68. 69. 70. 71. 72. 73. 74.
Z
2
2
2
Z
2
2
2
Z
2
4a
8a
2
Z x dx ax + bx + c b p = , 2a p dx ; a ax + bx + c ax + bx + c p
2
2
Z
ax + bx + c
2
8 ,1 2pcpax + bx + c + bx + 2c > ; if c > 0, > < pc ln x dx p => x ax + bx + c > 1 : p,c arcsin jxjpbxb+,2c4ac ; if c < 0, p x x + a dx = ( x , a )(x + a ) = ; Z 2
Z Z Z Z
2
2
1
2
2
3
2
2
2 3 2
xn sin(ax) dx = , a xn cos(ax) + na xn, cos(ax) dx; xneax dx =
Z
1
1
sin(ax) dx;
+1
ln(ax)
+1
1
= = = = =
x x x x x
= x = x +x = x + 3x + 2x = x + 6x + 11x + 6x = x + 10x + 35x + 50x + 24x
1
2 3 4 5
1 2 3 4 5
x x +x x + 3x + x x + 6x + 7x + x x + 15x + 25x + 10x + x
= = = = =
1
2
3
1
2
4
1
3
5
2
4
1
3
2
1
1
2
3
4
5
2
3
4
3
2
1
2
3
1
2
x x x x x
1
1
4
1
5
1
+1
1
x ,x x , 3x + x x , 6x + 7x , x x , 15x + 25x , 10x + x 2
3
1
2
4
1
3
5
2
4
1
3
2
1
= x = x ,x = x , 3x + 2x = x , 6x + 11x , 6x = x , 10x + 35x , 50x + 24x 1
2
3
4
=1
1
2
3
4
xn m = xm (x + m)n : Conversion: xn = (,1)n (,x)n = (x , m + 1)n = 1=(x + 1),n ; xn = (,1)n (,x)n = (x + m , 1)n = 1=(x , 1),n ; n n n X k = X n (,1)n,k xk ; xn = x k k k k n X n xn = (,1)n,k xk ; k k n n X n x = xk : k k +
x
5
1
0
2
x x x x x
1
+
1
xn ln(ax) dx = xn
Di erences: (cu) = cu; (u + v) = u + v; (uv) = uv + E vu; (xn) = nxn, ; (Hx ) = x, ; (2x ) = 2x ; , , (cx ) = (c , 1)cx ; mx = mx, : Sums: P cu x = c P u x; P(u + v) x = P u x + P v x; P uv x = uv , P E vu x; P x, x = H ; P xn x = xn+1 ; x m P, x x = , x : P cx x = cx ; c, m m Falling Factorial Powers: xn = x(x , 1) (x , m + 1); n > 0; x = 1; xn = (x + 1) 1 (x + jnj) ; n < 0; xn m = xm (x , m)n : Rising Factorial Powers: xn = x(x + 1) (x + m , 1); n > 0; x = 1; xn = (x , 1) 1 (x , jnj) ; n < 0;
1 , n + 1 (n + 1) ; Z Z n 76. xn(ln ax)m dx = nx + 1 (ln ax)m , n m+ 1 xn(ln ax)m, dx:
75.
f(i):
=
0
1
n n, n a x sin(ax) , a x xn eax , n Z xn, eax dx; a a
xn cos(ax) dx =
i a
1
15
1
1
1
2
3
b, X
+1
2
Z
a
f(x)x =
1
2
3
=1
1
2
1
=1
=1
Theoretical Computer Science Cheat Sheet Series
Taylor's series: 1 i X f(x) = f(a) + (x , a)f 0 (a) + (x ,2 a) f 00 (a) + = (x ,i! a) f i (a): i Expansions: 1 X 1 = 1 + x + x + x + x + = xi ; 1,x
Ordinary power series:
2
A(x) =
( )
=0
2
3
=0
4
i
1 X
= 1 + cx + c x + c x +
1 , cx 1 1 , xn x (1 , x) 1 dn xk dx n 1,x
2
2
3
3
=
i
ci xi ;
1 X
=0
=0
= 1 + xn + x n + x n + 2
3
= x + 2x + 3x + 4x + 2
2
3
4
= =
i
xni;
1 X
=1
=0
ixi ;
i
1 X n =0
= x + 2n x + 3nx + 4n x + = 2
3
4
k
i xi ;
i
= 1+x+ x + x + 1
1
2
2
3
6
=
1 xi X i
=0
= x, x + x , x ,
ln(1 + x)
1
1
2
2
1
3
3
4
4
ln 1 ,1 x
= x+ x + x + x +
sin x
= x, x + x , x +
cos x
= 1, x + x , x +
1
1
2
2
1
3
3
1
1
3
3!
4
4
1
5
5!
7
7!
= =
i
xn , yn = (x , y)
i! ;
1 X
(,1)i
+1
xi
i;
1 xi X i
=1
1 X =1
i;
1
2
2!
1
4
4!
6
6!
i = (,1)i (2ix + 1)! ; i 1 X xi; = (,1)i (2i)! i1 X xi ; = (,1)i (2i i + 1) 1 n X = xi ; i i 1 i + n X = xi ; i i 1 i X = Bi!i x ; i 1 1 2i X = i + 1 i xi ; i 1 2i X = xi ; i i 1 2i + n X = xi ; i i 2 +1
2
=0
tan, x
= x, x + x , x +
1
1
1
3
3
1
5
5
7
7
= 1 + nx + n n, x + (
, = 1 + (n + 1)x + n x + +2
2
2
+1
2 +1
A(x) + B(x) = xk A(x) =
= 1, x+ x , 1
1
2
12
1
2
720
x + 4
=0
= 1 + x + 2x + 5x + 2
3
=0
= 1 + x + 2x + 6x + 2
3
= 1 + (2 + n)x +
, nx + 4+
2
2
=0
1 X =0
= x+ x + x + x + = 3
11
2
2
25
3
6
4
12
i
Hi xi ;
1 H xi X i, =1
2
= x + x + x + 1
3
2
2
11
3
4
4
24
=
i
1 X =2
= x + x + 2x + 3x + 2
2
3
4
=
i
1 X
i
1
Fi xi;
=0
2
= Fnx + F nx + F nx + = 2
2
3
3
i
=0
k
xn, ,kyk : 1
=0
1 X
( ai + bi )xi ;
i
1 X
Fnixi :
;
ai,k xi;
i P 1 k , i A(x) , i aix = X ai,k xi ; xk 1 i X i i =0
1
=0
=0
A(cx) =
c ai x ;
i
1 X =0
A0 (x) =
(i + 1)ai xi; +1
i
1 X =0
=0
=0
1 (1 , p1 , 4x) 2x p 1 1 , 4x p n 1 1 , 1 , 4x p 2x 1 , 4x 1 1 1 , x ln 1 , x 1 ln 1 2 1,x x 1,x,x Fn x 1 , (Fn, + Fn )x , (,1)n x +1
2
2
1 (1 , x)n x ex , 1
1
1)
1
For ordinary power series:
=0
(1 + x)n
nX ,
=0
=0
1
=0
Di erence of like powers:
=0
ex
i
aixi :
Exponential power series: 1 i X A(x) = ai xi! : i Dirichlet power series: 1 X A(x) = iaxi : i Binomial theorem: n n X n,k k (x + y)n = k x y:
=0
1
1 X
Z
xA0 (x) =
i
iai xi;
1 a X i, =1
A(x) dx =
i i x; 1
i
1 A(x) + A(,x) = X a ix i; 2 i X A(x) , A(,x) 1 i =1
2
2
=0
= ai x : 2 i P Summation: If bi = ij ai then B(x) = 1 ,1 x A(x): Convolution: 0 1 2 +1
2 +1
=0
=0
A(x)B(x) =
1 X i X @ i
=0
j
aj bi,j A xi:
=0
God made the natural numbers; all the rest is the work of man. { Leopold Kronecker
Theoretical Computer Science Cheat Sheet Series
Expansions: n + i 1 X 1 1 i (1 , x)n ln 1 , x = (Hn i , Hn) i x ; +1
1 ,n
1 n X = xi ; i i 1 i n!xi X =0
xn
n 1 ln 1 , x
n i! ; 1 iB ix i X = (,4)(2i)! ; i 1 1 X = ix ; i 1 X = (i) ix ; i =
=0
=
2
=0
=0
2
2
2
1
=1
=1
2
i
(x)(x , 1)
=
=1
1 S(i) X =1
P where S(n) = djn d;
xi n, jB nj = 2 (2n)! n; n 2 N; 1 i 2)B ix i X = (,1)i, (4 ,(2i)! ; i 1 X = n(2ii!(n+ +n ,i)!1)! xi ; i 1 2i= sin i X = xi ; i! i i
=1 2
(2n) x sin x 1 , p1 , 4x n 2x
1
2
2
2
1
Z b,
Zb Zb G(x) + H(x) dF(x) = G(x) dF(x) + H(x) dF (x); a a Zb Zab , Zb
2
=0
Zb a
p
1 X
p (4i)! = xi ; i 16 2(2i)!(2i + 1)! i 1 X 4i i! = (i + 1)(2i x i: + 1)! i Crammer's Rule If we have equations: a ; x + a ; x + + a ;nxn = b a ; x + a ; x + + a ;nxn = b .. .. .. . . . an; x + an; x + + an;nxn = bn Let A = (ai;j ) and B be the column matrix (bi ). Then there is a unique solution i det A 6= 0. Let Ai be A with column i replaced by B. Then Ai xi = det det A : =0
2
Zb a
=0
1 2
2
1
1
2 1
1
2 2
2
2
2
2
2
Improvement makes strait roads, but the crooked roads without Improvement, are roads of Genius. { William Blake (The Marriage of Heaven and Hell)
a
G(x) dF(x) +
,
G(x) d c F(x) = c
Z ab a b
Z
G(x) dH(x);
G(x) dF(x);
G(x) dF(x) = G(b)F(b) , G(a)F (a) , F (x) dG(x): a If the integrals involved exist, and F possesses a derivative F 0 at every point in [a; b] then
2
1
Zb
a
a
2
1 1
1
c G(x) dF(x) =
Zb
4
1, 1,x x arcsin x x
G(x) d F(x) + H(x) =
a
=0
=1
1
a
exists. If a b c then Zc Zb Zc G(x) dF(x) = G(x) dF(x) + G(x) dF (x): a a b If the integrals involved exist
2
ex sin x
2
x cot x n i! ; i 1 i i 1)B i x i, X = (,1)i, 2 (2 ,(2i)! ; (x) i 1 X (x , 1) = (i) ; x i (x) i Y 1 = 1 , p,x ; Stieltjes Integration p If G is continuous in the interval [a; b] and F is nondecreasing then 1 d(i) X Zb P = G(x) dF(x) xi where d(n) = djn 1; 1
(x)
i
=0
=1
1 (x) (x)
s
=0
(ex , 1)n
2
tanx
1 i X = xi ; n i 1 i n!xi X
x
+
i
Escher's Knot
G(x) dF(x) =
0
47
18
76
29
93
85
34
61
52
86
11
57
28
70
39
94
95
80
22
67
38
71
49
45
2
63
56
13
4
59
96
81
33
7
48
73
69
90
82
44
17
72
60
24
15
58
1
35
68
74
9
91
83
55
26
27
12
46
30
37
8
75
19
92
84
66
23
50
41
14
25
36
40
51
62
3
77
88
99
21
32
43
54
65
6
10
89
97
78
42
53
64
5
16
20
31
98
79
87
The Fibonacci number system: Every integer n has a unique representation n = Fk1 + Fk2 + + Fkm ; where ki ki + 2 for all i, 1 i < m and km 2. +1
Zb a
G(x)F 0(x) dx: Fibonacci Numbers
1; 1; 2; 3; 5;8; 13; 21; 34; 55; 89; : : : De nitions: Fi = Fi, +Fi, ; F = F = 1; i, F,i = (, 1)i ^Fii; Fi = p , ; Cassini's identity: for i > 0: Fi Fi, , Fi = (,1)i : Additive rule: Fn k = Fk Fn + Fk, Fn; F n = FnFn + Fn, Fn: Calculation F F by matrices: n n, n, = 0 1 : F F 1 1 1
2
0
1
1
1
5
+1
2
1
+
+1
2
+1
2
n,
1
n
1
1
1
