Theoretical Computer Science Cheat Sheet

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 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 + 1
3 +


=1+ 2 2
3
2 2
4
5
2 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 Dierence, 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

Dierences:
(cu) = c
u;
(u + v) =
u +
v;
(uv) = u
v + E v
u;
(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 u
v x = uv , P E v
u 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 +

, n
x +


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

Dierence 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