Theoretical Computer Science Cheat Sheet Definitions Series f(n)= O(g(n)) iff positive c,n0 such that n n n 2 2 ∃ n(n + 1) 2 n(n + 1)(2n + 1) 3 n (n + 1) 0 f(n) cg(n) n n0. i = , i = , i = . ≤ ≤ ∀ ≥ 2 6 4 i=1 i=1 i=1 f(n)=Ω(g(n)) iff positive c,n such that X X X 0 In general: f(n∃) cg(n) 0 n n . ≥ ≥ ∀ ≥ 0 n n m 1 m+1 m+1 m+1 m f(n)=Θ(g(n)) iff f(n) = O(g(n)) and i = (n + 1) 1 (i + 1) i (m + 1)i m +1 − − − − i=1 i=1 f(n)=Ω(g(n)). X  X  n 1 m
− m 1 m +1 m+1 k f(n)= o(g(n)) iff limn f(n)/g(n) = 0. i = B n − . →∞ m +1 k k i=1 k=0   lim an = a iff ǫ > 0, n0 such that X X n →∞ ∀ ∃ Geometric series: an a <ǫ, n n0. n | − | ∀ ≥ cn+1 1 ∞ 1 ∞ c R ci = − , c =1, ci = , ci = , c < 1, sup S least b such that b s, c 1 6 1 c 1 c | | s S.∈ ≥ i=0 − i=0 − i=1 − Xn X X ∀ ∈ ncn+2 (n + 1)cn+1 + c ∞ c inf S greatest b R such that b ici = − , c =1, ici = , c < 1. ∈ ≤ (c 1)2 6 (1 c)2 | | s, s S. i=0 i=0 ∀ ∈ X − X − N Harmonic series: lim inf an lim inf ai i n,i . n n n n { | ≥ ∈ } 1 n(n + 1) n(n 1) →∞ →∞ H = , iH = H − . N n i i 2 n − 4 lim sup an lim sup ai i n,i . i=1 i=1 n n { | ≥ ∈ } →∞ n X Xn →∞ i n +1 1 n Combinations: Size k sub- H = (n + 1)H n, H = H . k i n − m i m +1 n+1 − m +1 sets of a size n set. i=1 i=1
X X       n n k Stirling numbers (1st kind): n n! n n n n Arrangements of an n ele- 1. = , 2. =2 , 3. = ,   k (n k)!k! k k n k ment set into k cycles.   − k=0      −  n n n 1 X n n 1 n 1 n 4. = − , 5. = − + − , Stirling numbers (2nd kind): k k k 1 k k k 1 k    −       −  Partitions of an n element n m n n k n r + k r + n +1  set into k non-empty sets. 6. = − , 7. = , m k k m k k n n      −  k=0     1st order Eulerian numbers: n Xn k k n +1 r s r + s Permutations π π ...π on 8. = , 9. = , 1 2 n m m +1 k n k n 1, 2,...,n with k ascents. k=0     k=0   −    { } X X n k k n 1 n n n 2nd order Eulerian numbers. 10. = ( 1) − − , 11. = =1, k k − k 1 n         Cn Catalan Numbers: Binary n n 1 n n 1 n 1 trees with n + 1 vertices. 12. =2 − 1, 13. = k − + − , 2 − k k k 1        −  n n n n n 14. = (n 1)!, 15. = (n 1)!Hn 1, 16. =1, 17. , 1 − 2 − − n k ≥ k           n n 1 n 1 n n n n n 1 2n 18. = (n 1) − + − , 19. = = , 20. = n!, 21. C = , k − k k 1 n 1 n 1 2 k n n +1 n      −   −   −    k=0     n n n n n X n 1 n 1 22. = =1, 23. = , 24. = (k + 1) − + (n k) − , 0 n 1 k n 1 k k k − k 1    −     − −       −  0 1 if k = 0, n n n +1 25. = 26. =2n n 1, 27. =3n (n + 1)2n + , k 0 otherwise 1 − − 2 − 2   n n m    n   n x + k n n +1 n n k 28. xn = , 29. = (m +1 k)n( 1)k, 30. m! = , k n m k − − m k n m k=0      k=0     k=0    X n X X − n n n k n k m n n 31. = − ( 1) − − k!, 32. =1, 33. =0 for n =0, m k m − 0 n 6   k=0        X n n n 1 n 1 n (2n)n 34. = (k + 1) − + (2n 1 k) − , 35. = , k k − − k 1 k 2n       k=0   n − nX x n x + n 1 k n +1 n k k n k 36. = − − , 37. = = (m + 1) − , x n k 2n m +1 k m m  −  Xk=0      Xk    kX=0   Theoretical Computer Science Cheat Sheet Identities Cont. Trees n n n n +1 n k k n k 1 k x n x + k Every tree with n 38. = = n − = n! , 39. = , m +1 k m m k! m x n k 2n vertices has n 1 k k=0 k=0 k=0 −   X    X   X    −  X    edges. n n k +1 n k n n +1 k m k 40. = ( 1) − , 41. = ( 1) − , m k m +1 − m k +1 m − Kraft inequal-   k      k    X m X m ity: If the depths m + n +1 n + k m + n +1 n + k 42. = k , 43. = k(n + k) , of the leaves of m k m k a binary tree are   k=0     k=0   X X d ,...,d : n n +1 k m k n n +1 k m k 1 n 44. = ( 1) − , 45. (n m)! = ( 1) − , for n m, n m k +1 m − − m k +1 m − ≥ di k k 2− 1,   X      X    ≤ n m n m + n m + k n m n m + n m + k i=1 46. = − , 47. = − , X n m m + k n + k k n m m + k n + k k  −  k      −  k     and equality holds n ℓ +Xm k n k n n Xℓ + m k n k n only if every in- 48. = − , 49. = − . ℓ + m ℓ ℓ m k ℓ + m ℓ ℓ m k ternal node has 2    Xk        Xk     sons. Recurrences Master method: Generating functions: T (n)= aT (n/b)+ f(n), a 1,b> 1 1 T (n) 3T (n/2) = n 1. Multiply both sides of the equa- ≥ − 3 T (n/2) 3T (n/4) = n/2 tion by xi. If ǫ> 0 such that f(n)= O(nlogb a ǫ)
− − 2. Sum both sides over all i for then∃ . . . . . .
which the equation is valid. T (n)=Θ(nlogb a). log n 1 3 2 − T (2) 3T (1) = 2 3. Choose a generating function logb a − i If f(n)=Θ(n ) then G(x). Usually G(x)= i∞=0 x gi. T (n)=Θ(nlogb a log n). Let m = log 2 n. Summing the
left side 3. Rewrite the equation in terms of 2 m m P we get T (n) 3 T (1) = T (n) 3 = the generating function G(x). If ǫ> 0 such that f(n)=Ω(nlogb a+ǫ), T (n) nk where− k = log 3 −1.58496. ∃ − 2 ≈ 4. Solve for G(x). and c < 1 such that af(n/b) cf(n) Summing the right side we get i ∃ ≤ m 1 m 1 5. The coefficient of x in G(x) is gi. for large n, then − − n i 3 i Example: T (n)=Θ(f(n)). 3 = n . 2i 2 i=0 i=0 gi+1 =2gi +1, g0 =0. Substitution (example): Consider the X X
3 Multiply and sum: following recurrence Let c = 2 . Then we have i m 1 i i i 2 2 − m gi+1x = 2gix + x . Ti+1 =2 Ti ,T1 =2. i c 1 · n c = n − i 0 i 0 i 0 c 1 X≥ X≥ X≥ Note that Ti is always a power of two. i=0  −  X We choose G(x)= xig . Rewrite log2 n i 0 i Let ti = log2 Ti. Then we have =2n(c 1) ≥ i − in terms of G(x): ti+1 =2 +2ti, t1 =1. (k 1) logc n P =2n(c − 1) G(x) g0 i i − − =2G(x)+ x . Let ui = ti/2 . Dividing both sides of k x =2n 2n, i 0 the previous equation by 2i+1 we get − X≥ i k ti+1 2 ti and so T (n)=3n 2n. Full history re- Simplify: i+1 = i+1 + i . − G(x) 1 2 2 2 currences can often be changed to limited =2G(x)+ . x 1 x Substituting we find history ones (example): Consider i 1 − 1 1 − ui+1 = + ui, u1 = , Solve for G(x): 2 2 Ti =1+ Tj,T0 =1. x G(x)= . which is simply ui = i/2. So we find j=0 (1 x)(1 2x) − X i2i 1 − − that Ti has the closed form Ti =2 . Note that i Expand this using partial fractions: Summing factors (example): Consider 2 1 the following recurrence Ti+1 =1+ Tj. G(x)= x j=0 1 2x − 1 x T (n)=3T (n/2)+ n, T (1) = 1. X  − −  Subtracting we find i i i Rewrite so that all terms involving T i i 1 = x 2 2 x x −  −  are on the left side T T =1+ T 1 T i 0 i 0 i+1 − i j − − j X≥ X≥ T (n) 3T (n/2) = n. j=0 j=0  i+1 i+1  − X X = (2 1)x . − Now expand the recurrence, and choose = Ti. i 0 X≥ a factor which makes the left side “tele- And so T =2T =2i+1. So g =2i 1. scope” i+1 i i − Theoretical Computer Science Cheat Sheet

1+√5 1 √5 π 3.14159, e 2.71828, γ 0.57721, φ = 1.61803, φˆ = − .61803 ≈ ≈ ≈ 2 ≈ 2 ≈− i i 2 pi General Probability

1 2 2 Bernoulli Numbers (Bi = 0, odd i = 1): Continuous distributions: If 6 b 2 4 3 B = 1, B = 1 , B = 1 , B = 1 , 0 1 − 2 2 6 4 − 30 Pr[anumber e: P and p both exist then e =1+ 1 + 1 + 1 + 1 + a 8 256 19 2 6 24 120 ··· x n P (a)= p(x) dx. 9 512 23 lim 1+ = ex. n n Z−∞ →∞ Expectation: If X is discrete 10 1,024 29 n  n+1 1+ 1 Factorial, Stirling’s approximation: Pr[A B] = Pr[A] + Pr[B] Pr[A B] ∨ − ∧ 20 1,048,576 71 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, ... Pr[A B] = Pr[A] Pr[B], ∧ · 21 2,097,152 73 iff A and B are independent. n n 1 22 4,194,304 79 n!= √2πn 1+Θ . Pr[A B] e n Pr[A B]= ∧      | Pr[B] 23 8,388,608 83 Ackermann’s function and inverse: For random variables X and Y : 24 16,777,216 89 2j i =1 [X Y ]= [X] [Y ], 25 33,554,432 97 a(i, j)= a(i 1, 2) j =1 E E E  · · a(i − 1,a(i, j 1)) i, j 2 if X and Y are independent. 26 67,108,864 101  − − ≥ α(i) = min j a(j, j) i . E[X + Y ]= E[X]+ E[Y ], 27 134,217,728 103  { | ≥ } E[cX]= c E[X]. 28 268,435,456 107 Binomial distribution: Bayes’ theorem: n k n k 29 536,870,912 109 Pr[X = k]= p q − , q =1 p, k − Pr[B Ai] Pr[Ai] 30 1,073,741,824 113   Pr[Ai B]= n | . n | j=1 Pr[Aj ] Pr[B Aj ] 31 2,147,483,648 127 n k n k | E[X]= k p q − = np. Inclusion-exclusion: k P 32 4,294,967,296 131 k=1   n n X Pr X = Pr[X ]+ Pascal’s Triangle Poisson distribution: i i λ k i=1 i=1 e− λ h i 1 Pr[X = k]= , E[X]= λ. _ X k! n k 1 1 ( 1)k+1 Pr X . Normal (Gaussian) distribution: − ij k=2 ii< Trigonometry Matrices More Trig. Multiplication: C n (0,1) C = A B, c = a b . · i,j i,k k,j b a b (cos θ, sin θ) k=1 h C X A θ Determinants: det A = 0 iff A is non-singular. (-1,0) (1,0) 6 A c B det A B = det A det B, Law of cosines: · · c a n 2 2 2 (0,-1) c = a +b 2ab cos C. B det A = sign(π)ai,π(i). − π i=1 Area: Pythagorean theorem: X Y C2 = A2 + B2. 2 2 and 3 3 determinant: × × A = 1 hc, a b 2 Definitions: = ad bc, 1 c d − = 2 ab sin C, sin a = A/C, cos a = B/C, a b c c2 sin A sin B b c a c a b = . csc a = C/A, sec a = C/B, d e f = g h + i 2 sin C e f − d f d e sin a A cos a B g h i Heron’s formula: tan a = = , cot a = = . cos a B sin a A aei + bfg + cdh = Area, radius of inscribed circle: A = √s sa sb sc, ceg fha ibd. · · · 1 AB − − − 1 AB, . Permanents: s = 2 (a + b + c), 2 A + B + C n sa = s a, perm A = a . − Identities: i,π(i) s = s b, π i=1 b 1 1 X Y − sin x = , cos x = , Hyperbolic Functions sc = s c. csc x sec x − 1 2 2 Definitions: More identities: tan x = , sin x + cos x =1, x x x x cot x e e− e + e− 1 cos x sinh x = − , cosh x = , sin x = − , 1 + tan2 x = sec2 x, 1+cot2 x = csc2 x, 2 2 2 2 x x r e e− 1 π tanh x = x − x , csch x = , x 1+cos x sin x = cos x , sin x = sin(π x), e + e− sinh x cos = , 2 − − 1 1 2 2 sech x = , coth x = . r cos x = cos( π
x), tan x = cot π x , cosh x tanh x 1 cos x − − 2 − tan x = − , 2 1+cos x x
Identities: r cot x = cot(π x), csc x = cot 2 cot x, 1 cos x − − − 2 2 2 2 = − , cosh x sinh x =1, tanh x + sech x =1, sin x sin(x y) = sin x cos y cos x sin y, − sin x ± ± coth2 x csch2 x =1, sinh( x)= sinh x, = , cos(x y) = cos x cos y sin x sin y, − − − 1+cos x ± ∓ cosh( x) = cosh x, tanh( x)= tanh x, 1+cos x tan x tan y − − − cot x = , tan(x y)= ± , 2 ± 1 tan x tan y sinh(x + y) = sinh x cosh y + cosh x sinh y, r1 cos x ∓ 1+cos− x cot x cot y 1 = , cot(x y)= ∓ , cosh(x + y) = cosh x cosh y + sinh x sinh y, sin x ± cot x cot y ± 2 tan x sin x sin 2x = 2 sin x cos x, sin 2x = , sinh 2x = 2 sinh x cosh x, = , 1 + tan2 x 1 cos x 2 2 ix− ix cosh2x = cosh x + sinh x, e e− cos2x = cos2 x sin2 x, cos2x = 2cos2 x 1, sin x = − , − − 2i 2 x x 1 tan x cosh x + sinh x = e , cosh x sinh x = e− , ix ix cos2x =1 2 sin2 x, cos2x = − , − e + e− − 1 + tan2 x cos x = , (cosh x + sinh x)n = cosh nx + sinh nx, n Z, 2 2 ∈ ix ix 2 tan x cot x 1 e e− 2 2 tan2x = 2 , cot2x = − , 2 sinh x = cosh x 1, 2 cosh x = cosh x +1. tan x = i − , 1 tan x 2cot x 2 2 − eix + e ix − − − sin(x + y) sin(x y) = sin2 x sin2 y, e2ix 1 − − = i − , θ sin θ cos θ tan θ ... in mathematics − e2ix +1 2 2 cos(x + y) cos(x y) = cos x sin y. you don’t under- sinh ix − − 00 1 0 sin x = , π 1 √3 √3 stand things, you i Euler’s equation: 6 2 2 3 ix iπ just get used to e = cos x + i sin x, e = 1. √ √ cos x = cosh ix, − π 2 2 1 them. 4 2 2 tanh ix c √ – J. von Neumann v2.02 1994 by Steve Seiden π 3 1 √3 tan x = . [email protected] 3 2 2 i π 1 0 http://www.csc.lsu.edu/~seiden 2 ∞ Theoretical Computer Science Cheat Sheet Number Theory Graph Theory The Chinese remainder theorem: There ex- Definitions: Notation: ists a number C such that: Loop An edge connecting a ver- E(G) Edgeset tex to itself. V (G) Vertex set C r mod m ≡ 1 1 Directed Each edge has a direction. c(G) Number of components . . . G[S] Induced subgraph . . . Simple Graph with no loops or multi-edges. deg(v) Degree of v C rn mod mn ∆(G) Maximum degree ≡ Walk A sequence v0e1v1 ...eℓvℓ. δ(G) Minimum degree if mi and mj are relatively prime for i = j. Trail A walk with distinct edges. 6 χ(G) Chromatic number Euler’s function: φ(x) is the number of Path A trail with distinct χ (G) Edge chromatic number positive integers less than x relatively vertices. E Gc Complement graph prime to x. If n pei is the prime fac- Connected A graph where there exists i=1 i K Complete graph torization of x then a path between any two n n Q vertices. Kn1,n2 Complete bipartite graph ei 1 φ(x)= p − (p 1). r(k,ℓ) Ramsey number i i − Component A maximal connected i=1 Y subgraph. Geometry Euler’s theorem: If a and b are relatively Tree A connected acyclic graph. Projective coordinates: triples prime then Free tree A tree with no root. φ(b) (x,y,z), not all x, y and z zero. 1 a mod b. DAG ≡ Directed acyclic graph. Eulerian (x,y,z) = (cx,cy,cz) c =0. Fermat’s theorem: Graph with a trail visiting ∀ 6 p 1 each edge exactly once. Cartesian Projective 1 a − mod p. ≡ Hamiltonian Graph with a cycle visiting (x, y) (x, y, 1) The Euclidean algorithm: if a>b are in- each vertex exactly once. y = mx + b (m, 1,b) tegers then Cut A set of edges whose re- x = c (1, 0−, c) gcd(a,b) = gcd(a mod b,b). − moval increases the num- Distance formula, Lp and L ∞ n ei ber of components. metric: If i=1 pi is the prime factorization of x Cut-set A minimal cut. 2 2 then (x1 x0) + (y1 y0) , n Q ei+1 Cut edge A size 1 cut. − − pi 1 p p 1/p S(x)= d = − . k-Connected A graph connected with px1 x0 + y1 y0 , pi 1 | − | | − | d x i=1 − the removal of any k 1 p p 1/p X| Y lim x1 x0 + y1 y0 . − p | − | | − | Perfect Numbers: x is an even perfect num- vertices. →∞ n 1 n n k-Tough S V,S = we have Area of triangle (x0,y0), (x1,y1) ber iff x =2 − (2 1) and 2 1 is prime. ∀ ⊆ 6 ∅ − − k c(G S) S . and (x2,y2): Wilson’s theorem: n is a prime iff · − ≤ | | (n 1)! 1 mod n. k-Regular A graph where all vertices x1 x0 y1 y0 1 abs . − ≡− have degree k. 2 x − x y − y 2 − 0 2 − 0 M¨obius inversion: k-Factor A k-regular spanning 1 if i = 1. Angle formed by three points: 0 if i is not square-free. subgraph. µ(i)= r Matching  ( 1) if i is the product of A set of edges, no two of (x2,y2)  − which are adjacent. r distinct primes. ℓ Clique A set of vertices, all of 2 If   which are adjacent. θ G(a)= F (d), (0, 0) ℓ (x ,y ) Ind. set A set of vertices, none of 1 1 1 d a X| (x1,y1) (x2,y2) which are adjacent. cos θ = · . then ℓ ℓ a Vertex cover A set of vertices which 1 2 F (a)= µ(d)G . cover all edges. Line through two points (x ,y ) d 0 0 d a and (x ,y ): X|   Planar graph A graph which can be em- 1 1 Prime numbers: beded in the plane. x y 1 ln ln n Plane graph An embedding of a planar x0 y0 1 =0. pn = n ln n + n ln ln n n + n − ln n graph. x y 1 1 1 n + O , Area of circle, volume of sphere: ln n deg(v)=2m. 2 4 3   v V A = πr , V = 3 πr . n n 2!n X∈ π(n)= + + If G is planar then n m + f =2, so ln n (ln n)2 (ln n)3 − If I have seen farther than others, f 2n 4, m 3n 6. n ≤ − ≤ − it is because I have stood on the + O . (ln n)4 Any planar graph has a vertex with de- shoulders of giants.   gree 5. – Issac Newton ≤ Theoretical Computer Science Cheat Sheet π Calculus Wallis’ identity: Derivatives: 2 2 4 4 6 6 π =2 · · · · · ··· d(cu) du d(u + v) du dv d(uv) dv du · 1 3 3 5 5 7 1. = c , 2. = + , 3. = u + v , · · · · · ··· dx dx dx dx dx dx dx dx Brouncker’s continued fraction expansion: n du dv cu 2 d(u ) n 1 du d(u/v) v dx u dx d(e ) cu du π 1 4. 5. 6. = nu − , = −2 , = ce , 4 =1+ 32 dx dx dx v dx dx 2+ 52

2+ 2 u 2+ 7 d(c ) du d(ln u) 1 du 2+··· 7. = (ln c)cu , 8. = , dx dx dx u dx Gregrory’s series: π =1 1 + 1 1 + 1 d(sin u) du d(cos u) du 4 3 5 7 9 9. = cos u , 10. = sin u , − − − · · · dx dx dx − dx Newton’s series: d(tan u) du d(cot u) du 11. = sec2 u , 12. = csc2 u , π 1 1 1 3 dx dx dx dx 6 = + 3 + · 5 + 2 2 3 2 2 4 5 2 ··· d(sec u) du d(csc u) du · · · · · 13. = tan u sec u , 14. = cot u csc u , Sharp’s series: dx dx dx − dx 1 1 1 1 d(arcsin u) 1 du d(arccos u) 1 du π = 1 + + 15. = , 16. = − , 6 √3 − 31 3 32 5 − 33 7 ··· dx √1 u2 dx dx √1 u2 dx · · · − −   d(arctan u) 1 du d(arccot u) 1 du Euler’s series: 17. = , 18. = − , dx 1+ u2 dx dx 1+ u2 dx π2 1 1 1 1 1 6 = 12 + 22 + 32 + 42 + 52 + d(arcsec u) 1 du d(arccsc u) 1 du 2 ··· 19. = , 20. = − , π 1 1 1 1 1 2 2 = 2 + 2 + 2 + 2 + 2 + dx u√1 u dx dx u√1 u dx 8 1 3 5 7 9 ··· − − π2 1 1 1 1 1 d(sinh u) du d(cosh u) du 12 = 12 22 + 32 42 + 52 21. = cosh u , 22. = sinh u , − − − · · · dx dx dx dx Partial Fractions d(tanh u) du d(coth u) du 23. = sech2 u , 24. = csch2 u , Let N(x) and D(x) be polynomial func- dx dx dx − dx tions of x. We can break down d(sech u) du d(csch u) du N(x)/D(x) using partial fraction expan- 25. = sech u tanh u , 26. = csch u coth u , dx − dx dx − dx sion. First, if the degree of N is greater d(arcsinh u) 1 du d(arccosh u) 1 du than or equal to the degree of D, divide 27. = , 28. = , N by D, obtaining dx √1+ u2 dx dx √u2 1 dx − N(x) N ′(x) d(arctanh u) 1 du d(arccoth u) 1 du = Q(x)+ , 29. = , 30. = , D(x) D(x) dx 1 u2 dx dx u2 1 dx − − where the degree of N ′ is less than that of d(arcsech u) 1 du d(arccsch u) 1 du 31. = − , 32. = − . D. Second, factor D(x). Use the follow- dx u√1 u2 dx dx u √1+ u2 dx − | | ing rules: For a non-repeated factor: Integrals: N(x) A N (x) = + ′ , (x a)D(x) x a D(x) − − 1. cu dx = c u dx, 2. (u + v) dx = u dx + v dx, where Z Z Z Z Z N(x) 1 1 A = . 3. xn dx = xn+1, n = 1, 4. dx = ln x, 5. ex dx = ex, D(x) n +1 6 − x  x=a Z Z Z dx dv du For a repeated factor: 6. = arctan x, 7. u dx = uv v dx, m 1 2 N(x) − A N (x) 1+ x dx − dx = k + ′ , Z Z Z (x a)mD(x) (x a)m k D(x) k=0 − 8. sin x dx = cos x, 9. cos x dx = sin x, − X − − where Z Z 1 dk N(x) A = . 10. tan x dx = ln cos x , 11. cot x dx = ln cos x , k k! dxk D(x) − | | | |   x=a Z Z 12. sec x dx = ln sec x + tan x , 13. csc x dx = ln csc x + cot x , The reasonable man adapts himself to the | | | | world; the unreasonable persists in trying Z Z to adapt the world to himself. Therefore 14. arcsin x dx = arcsin x + a2 x2, a> 0, a a − all progress depends on the unreasonable. Z p – George Bernard Shaw Theoretical Computer Science Cheat Sheet Calculus Cont.

15. arccos x dx = arccos x a2 x2, a> 0, 16. arctan x dx = x arctan x a ln(a2 + x2), a> 0, a a − − a a − 2 Z p Z 17. sin2(ax)dx = 1 ax sin(ax)cos(ax) , 18. cos2(ax)dx = 1 ax + sin(ax)cos(ax) , 2a − 2a Z Z

19. sec2 x dx = tan x, 20. csc2 x dx = cot x, − Z Z n 1 n 1 n sin − x cos x n 1 n 2 n cos − x sin x n 1 n 2 21. sin x dx = + − sin − x dx, 22. cos x dx = + − cos − x dx, − n n n n Z Z Z Z n 1 n 1 n tan − x n 2 n cot − x n 2 23. tan x dx = tan − x dx, n =1, 24. cot x dx = cot − x dx, n =1, n 1 − 6 − n 1 − 6 Z − Z Z − Z n 1 n tan x sec − x n 2 n 2 25. sec x dx = + − sec − x dx, n =1, n 1 n 1 6 Z − − Z n 1 n cot x csc − x n 2 n 2 26. csc x dx = + − csc − x dx, n =1, 27. sinh x dx = cosh x, 28. cosh x dx = sinh x, − n 1 n 1 6 Z − − Z Z Z 29. tanh x dx = ln cosh x , 30. coth x dx = ln sinh x , 31. sech x dx = arctansinh x, 32. csch x dx = ln tanh x , | | | | 2 Z Z Z Z

33. sinh2 x dx = 1 sinh(2x) 1 x, 34. cosh2 x dx = 1 sinh(2x)+ 1 x, 35. sech2 x dx = tanh x, 4 − 2 4 2 Z Z Z 36. arcsinh x dx = x arcsinh x x2 + a2, a> 0, 37. arctanh x dx = x arctanh x + a ln a2 x2 , a a − a a 2 | − | Z p Z x 2 2 x x arccosh x + a , if arccosh a > 0 and a> 0, 38. x a − arccosh a dx = x  x arccosh + px2 + a2, if arccosh x < 0 and a> 0, Z  a a dx p 39. = lnx + a2 + x2 , a> 0, √a2 + x2 Z  p  dx 2 40. = 1 arctan x , a> 0, 41. a2 x2 dx = x a2 x2 + a arcsin x , a> 0, a2 + x2 a a − 2 − 2 a Z Z p p 4 42. (a2 x2)3/2dx = x (5a2 2x2) a2 x2 + 3a arcsin x , a> 0, − 8 − − 8 a Z dx p dx 1 a + x dx x 43. = arcsin x , a> 0, 44. = ln , 45. = , √a2 x2 a a2 x2 2a a x (a2 x2)3/2 a2√a2 x2 Z Z Z − − − − − x a2 dx 46. a2 x2 dx = a2 x2 ln x + a2 x2 , 47. = ln x + x2 a2 , a> 0, ± 2 ± ± 2 ± √x2 a2 − Z Z − p dx 1 p x p 2(3bxp 2a)(a + bx)3/2 48. = ln , 49. x√a + bx dx = − , ax2 + bx a a + bx 15b2 Z Z

√a + bx 1 x 1 √a + bx √a 50. dx =2√ a + bx + a dx, 51. dx = ln − , a> 0, x x√a + bx √a + bx √2 √a + bx + √a Z Z Z √ 2 2 √ 2 2 a x 2 2 a + a x 2 2 1 2 2 3/2 52. − dx = a x a ln − , 53. x a x dx = 3 (a x ) , x − − x − − − Z p Z p 4 √ 2 2 2 2 2 x 2 2 2 2 a x dx 1 a + a x 54. x a x dx = (2x a ) a x + arcsin , a> 0, 55. = ln − , − 8 − − 8 a √a2 x2 − a x Z Z p p 2 − x dx x dx 2 56. = a2 x2, 57. = x a2 x2 + a arcsin x a> 0, √a2 x2 − − √a2 x2 − 2 − 2 a, Z − Z − √ 2 2 p √ 2 2 √ 2 2 p a + x 2 2 a + a + x x a 2 2 a 58. dx = a + x a ln , 59. − dx = x a a arccos x , a> 0, x − x x − − | | Z Z p pdx x 60. x x2 a2 dx = 1 (x2 a2)3/2, 61. = 1 ln , ± 3 ± x√x2 + a2 a a + √a2 + x2 Z Z p

Theoretical Computer Science Cheat Sheet Calculus Cont. Finite Calculus dx dx √x2 a2 Difference, shift operators: 62. 1 a 63. = a arccos x , a> 0, = 2± , x√x2 a2 | | x2√x2 a2 ∓ a x ∆f(x)= f(x + 1) f(x), Z − Z ± − x dx √x2 a2 (x2 + a2)3/2 E f(x)= f(x + 1). 64. = x2 a2, 65. ± dx = , √x2 a2 ± x4 ∓ 3a2x3 Fundamental Theorem: Z ± Z p 1 2ax + b √b2 4ac f(x) = ∆F (x) f(x)δx = F (x)+ C. ln − − , if b2 > 4ac, ⇔ dx √ 2 √ 2 66.  b 4ac 2ax + b + b 4ac b X b 1 2 = − − − ax + bx + c  2 2ax + b f(x)δx = f(i). Z  2 arctan , if b < 4ac, a i=a √4ac b2 √4ac b2 X X  − − Differences:   1 ∆(cu)= c∆u, ∆(u + v) = ∆u + ∆v, ln 2ax + b +2√a ax2 + bx + c , if a> 0, dx √a 67. =  ∆(uv)= u∆v + E v∆u, √ 2 1 2ax bp ax + bx + c  n n 1 Z  arcsin − − , if a< 0, ∆(x )= nx − , √ a √b2 4ac − − 1 x x  ∆(Hx)= x− , ∆(2 )=2 ,  2ax + b 4ax b2 dx 68. ax2 + bx + c dx = ax2 + bx + c + − , ∆(cx) = (c 1)cx, ∆ x = x . 4a 8a √ax2 + bx + c − m m 1 Z Z − p p Sums:

x dx √ax2 + bx + c b dx cuδx = c u δx, 69. = , √ax2 + bx + c a − 2a √ax2 + bx + c Z Z P(u + v) δxP= u δx + v δx, 2 1 2√c√ax + bx + c + bx +2c P u∆v δx = uvP E v∆Pu δx, − ln , if c> 0, − dx √c x n+1 70. =  P xn δx = x , P x 1 δx = H , √ 2  m+1 − x x ax + bx + c  1 bx +2c Z  arcsin , if c< 0, x x x 2 x c √ c x √b 4ac P c δx = c 1 , Pm δx = m+1 . − | | − −  Falling Factorial Powers: 3 2 2 1 2 2 2 2 2 3/2 P P

71. x x + a dx = ( x a )(x + a ) , n 3 − 15 x = x(x 1) (x n + 1), n> 0, − ··· − Z p 0 n 1 n n n 1 x =1, 72. x sin(ax) dx = x cos(ax)+ x − cos(ax) dx, − a a 1 Z Z xn = , n< 0, (x + 1) (x + n ) n 1 n n n 1 73. x cos(ax) dx = x sin(ax) x − sin(ax) dx, ··· | | a − a xn+m = xm(x m)n. Z Z − n ax Rising Factorial Powers: n ax x e n n 1 ax 74. x e dx = x − e dx, a − a xn = x(x + 1) (x + n 1), n> 0, Z Z ··· − ln(ax) 1 0 75. xn ln(ax) dx = xn+1 , x =1, n +1 − (n + 1)2 Z   n 1 n+1 x = , n< 0, n m x m m n m 1 (x 1) (x n ) 76. x (ln ax) dx = (ln ax) x (ln ax) − dx. − ··· − | | n +1 − n +1 n+m m n Z Z x = x (x + m) . Conversion: n n n n 1 1 1 x = ( 1) ( x) = (x n + 1) x = x = x − − − n x2 = x2 + x1 = x2 x1 =1/(x + 1)− , − x3 = x3 +3x2 + x1 = x3 3x2 + x1 xn = ( 1)n( x)n = (x + n 1)n 4 4 3 2 1 4 −3 2 1 − − n − x = x +6x +7x + x = x 6x +7x x =1/(x 1)− , − − − x5 = x5 + 15x4 + 25x3 + 10x2 + x1 = x5 15x4 + 25x3 10x2 + x1 n n n n k n n k k − − x = x = ( 1) − x , k k − x1 = x1 x1 = x1 k=1   k=1   Xn X 2 2 1 2 2 1 x = x + x x = x x n n n k k − x = ( 1) − x , 3 3 2 1 3 3 2 1 k − x = x +3x +2x x = x 3x +2x k=1   − Xn x4 = x4 +6x3 + 11x2 +6x1 x4 = x4 6x3 + 11x2 6x1 n − − xn = xk. x5 = x5 + 10x4 + 35x3 + 50x2 + 24x1 x5 = x5 10x4 + 35x3 50x2 + 24x1 k − − Xk=1   Theoretical Computer Science Cheat Sheet Series Taylor’s series: Ordinary power series: 2 i (x a) ∞ (x a) (i) ∞ f(x)= f(a) + (x a)f ′(a)+ − f ′′(a)+ = − f (a). i − 2 ··· i! A(x)= aix . i=0 i=0 Expansions: X X ∞ Exponential power series: 1 2 3 4 i =1+ x + x + x + x + = x , i 1 x ··· ∞ x i=0 A(x)= a . − X i i! 1 ∞ i=0 =1+ cx + c2x2 + c3x3 + = cixi, X 1 cx ··· Dirichlet power series: i=0 − X ∞ ∞ ai 1 n 2n 3n ni =1+ x + x + x + = x , A(x)= x . 1 xn ··· i i=0 i=1 − X X x ∞ Binomial theorem: = x +2x2 +3x3 +4x4 + = ixi, n (1 x)2 ··· n n n k k i=0 (x + y) = x − y . n − X k n k!zk ∞ k=0   = x +2nx2 +3nx3 +4nx4 + = inxi, X k (1 z)k+1 ··· Difference of like powers: k=0 i=0   − n 1 X X i ∞ x n n − n 1 k k ex =1+ x + 1 x2 + 1 x3 + = , x y = (x y) x − − y . 2 6 ··· i! − − i=0 k=0 X X ∞ xi For ordinary power series: ln(1 + x) = x 1 x2 + 1 x3 1 x4 = ( 1)i+1 , − 2 3 − 4 − · · · − i ∞ i=1 i X αA(x)+ βB(x)= (αai + βbi)x , ∞ i 1 1 2 1 3 1 4 x i=0 ln = x + 2 x + 3 x + 4 x + = , X 1 x ··· i ∞ i=1 k i − X x A(x)= ai kx , ∞ 2i+1 − 1 3 1 5 1 7 i x i=k sin x = x 3! x + 5! x 7! x + = ( 1) , X − − ··· − (2i + 1)! k 1 i=0 i ∞ X A(x) i=0− aix i ∞ 2i − = ai+kx , 1 2 1 4 1 6 i x xk cos x =1 x + x x + = ( 1) , i=0 − 2! 4! − 6! ··· − (2i)! P X i=0 ∞ X i i ∞ 2i+1 A(cx)= c aix , 1 1 3 1 5 1 7 i x tan− x = x x + x x + = ( 1) , i=0 − 3 5 − 7 ··· − (2i + 1) X i=0 X ∞ ∞ i n n(n 1) 2 n i A′(x)= (i + 1)ai+1x , (1 + x) =1+ nx + − x + = x , 2 ··· i i=0 i=0   X X ∞ ∞ i 1 n+2 2 i + n i xA′(x)= iaix , n+1 =1+(n + 1)x + 2 x + = x , (1 x) ··· i i=1 − i=0   X
X i ∞ x ∞ Bix ai 1 i =1 1 x + 1 x2 1 x4 + = , A(x) dx = − x , ex 1 − 2 12 − 720 ··· i! i i=0 Z i=1 − X X ∞ 1 1 2i i A(x)+ A( x) ∞ (1 √1 4x) =1+ x +2x2 +5x3 + = x , = a x2i, 2x − − ··· i +1 i − 2i i=0 2 X   i=0 ∞ X 1 2 3 2i i ∞ =1+2x +6x + 20x + = x , A(x) A( x) 2i+1 √1 4x ··· i − − = a2i+1x . i=0   2 − n X i=0 1 1 √1 4x ∞ 2i + n X − − =1+(2+ n)x + 4+n x2 + = xi, Summation: If b = i a then √1 4x 2x 2 ··· i i j=0 i   i=0   −
X 1 1 1 ∞ B(x)= PA(x). ln = x + 3 x2 + 11 x3 + 25 x4 + = H xi, 1 x 1 x 1 x 2 6 12 ··· i − − − i=1 Convolution: 2 X 1 1 ∞ H xi i 1 2 3 3 11 4 i 1 ∞ ln = x + x + x + = − , i 2 1 x 2 4 24 ··· i A(x)B(x)= aj bi j x .   i=2  −  − X i=0 j=0 x ∞ X X = x + x2 +2x3 +3x4 + = F xi,   1 x x2 ··· i i=0 God made the natural numbers; − − X ∞ Fnx 2 3 i all the rest is the work of man. n 2 = Fnx + F2nx + F3nx + = Fnix . 1 (F 1 + F +1)x ( 1) x ··· – Leopold Kronecker n n i=0 − − − − X Theoretical Computer Science Cheat Sheet Series Escher’s Knot Expansions: n 1 1 ∞ n + i 1 − ∞ i ln = (H H ) xi, = xi, (1 x)n+1 1 x n+i − n i x n i=0 i=0 − − X     X   ∞ n ∞ i n!xi xn = xi, (ex 1)n = , i − n i! i=0   i=0   n X X 1 ∞ i n!xi ∞ ( 4)iB x2i ln = , x cot x = − 2i , 1 x n i! (2i)! i=0 i=0  −  X   X ∞ 2i 2i 2i 1 ∞ i 1 2 (2 1)B2ix − 1 tan x = ( 1) − − , ζ(x) = , − (2i)! ix i=1 i=1 X X 1 ∞ µ(i) ζ(x 1) ∞ φ(i) = , − = , ζ(x) ix ζ(x) ix i=1 i=1 X 1 X ζ(x) = , 1 p x Stieltjes Integration p − Y − If G is continuous in the interval [a,b] and F is nondecreasing then ∞ 2 d(i) b ζ (x) = i where d(n)= d n 1, x | G(x) dF (x) i=1 X P Za ∞ S(i) exists. If a b c then ζ(x)ζ(x 1) = i where S(n)= d n d, c≤ ≤ b c − x | i=1 G(x) dF (x)= G(x) dF (x)+ G(x) dF (x). X2n 1 P 2 − B a a b ζ(2n) = | 2n|π2n, n N, Z Z Z (2n)! ∈ If the integrals involved exist b b b ∞ i 2i x i 1 (4 2)B2ix G(x)+ H(x) dF (x)= G(x) dF (x)+ H(x) dF (x), = ( 1) − − , sin x − (2i)! a a a i=0 Z Z Z X b
b b n G(x) d F (x)+ H(x) = G(x) dF (x)+ G(x) dH(x), 1 √1 4x ∞ n(2i + n 1)! i − − = − x , a a a 2x i!(n + i)! Z Z Z i=0 b b
b   X iπ c G(x) dF (x)= G(x) d c F (x) = c G(x) dF (x), ∞ 2i/2 sin · · ex sin x = 4 xi, Za Za Za i! b
b i=1 G(x) dF (x)= G(b)F (b) G(a)F (a) F (x) dG(x). X − − Za Za 1 √1 x ∞ (4i)! i − − = x , If the integrals involved exist, and F possesses a derivative F ′ at every s x i√ i=0 16 2(2i)!(2i + 1)! point in [a,b] then X 2 i 2 b b arcsin x ∞ 4 i! 2i = x . G(x) dF (x)= G(x)F ′(x) dx. x (i + 1)(2i + 1)! i=0 a a   X Z Z Cramer’s Rule Fibonacci Numbers 00 47 18 76 29 93 85 34 61 52 If we have equations: 86 11 57 28 70 39 94 45 02 63 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... a1,1x1 + a1,2x2 + + a1,nxn = b1 95 80 22 67 38 71 49 56 13 04 ··· Definitions: a x + a x + + a x = b 59 96 81 33 07 48 72 60 24 15 2,1 1 2,2 2 ··· 2,n n 2 73 69 90 82 44 17 58 01 35 26 Fi = Fi 1+Fi 2, F0 = F1 =1, . . . − − . . . i 1 68 74 09 91 83 55 27 12 46 30 F i = ( 1) − Fi, − a x + a x + + a x = b 37 08 75 19 92 84 66 23 50 41 − n,1 1 n,2 2 ··· n,n n n F = 1 φi φˆi , 14 25 36 40 51 62 03 77 88 99 i √5 Let A = (a ) and B be the column matrix (b ). Then − i,j i 21 32 43 54 65 06 10 89 97 78 Cassini’s identity: for i> 0: there is a unique solution iff det A = 0. Let A be A i 42 53 64 05 16 20 31 98 79 87 2 i 6 Fi+1Fi 1 Fi = ( 1) . with column i replaced by B. Then − − − det Ai Additive rule: x = . The Fibonacci number system: i det A Every integer n has a unique Fn+k = FkFn+1 + Fk 1Fn, − representation F2n = FnFn+1 + Fn 1Fn. − Improvement makes strait roads, but the crooked n = Fk1 + Fk2 + + Fkm , Calculation by matrices: ··· roads without Improvement, are roads of Genius. n where ki ki+1 + 2 for all i, Fn 2 Fn 1 0 1 – William Blake (The Marriage of Heaven and Hell) ≥ − − = . 1 i