Cheat_Sheet

f(n) =O(g(n))f(n) =Ω(g(n))Definitionsiff ∃ positive c, n 0 such that0 ≤ f(n) ≤ cg(n) ∀n ≥ n 0 .iff ∃ positive c, n 0 such thatf(n) ≥ cg(n) ≥ 0 ∀n ≥ n 0 .f(n) =Θ(g(n)) iff f(n) = O(g(n)) andf(n) =Ω(g(n)).f(n) =o(g(n)) iff lim n→∞ f(n)/g(n) =0.lim a n = a iff ∀ɛ > 0, ∃n 0 such thatn→∞|a n − a|

Theoretical Computer Science Cheat Sheetπ ≈ 3.14159, e ≈ 2.71828, γ ≈ 0.57721, φ = 1+√ 52≈ 1.61803, ˆφ =1− √ 52≈−.61803i 2 i p i General Probability123248235Bernoulli Numbers (B i =0,oddi ≠ 1):B 0 =1,B 1 = − 1 2 , B 2 = 1 6 , B 4 = − 130 ,B 6 = 142 , B 8 = − 1 30 , B 10 = 5 66 .4 16 7 Change of base, quadratic formula:5 32 11log b x = log a xlog a b , −b ± √ b 2 − 4ac.6 64 132a7 128 17 Euler’s number e:8 256 19e =1+ 1 2 + 1 6 + 124 + 1120(+ ···lim 1+ x n9 512 23= en→∞ n) x .10 1,024 29( )1+1 n (n

Theoretical Computer Science Cheat SheetTrigonometry Matrices More Trig.Multiplication:C(0,1)n∑C = A · B, c i,j = a i,k b k,j .bb a(cos θ, sin θ)k=1hCAθDeterminants: det A ≠0iffA is non-singular.(-1,0) (1,0)A c Bdet A · B = det A · det B,Law of cosines:c a(0,-1)Bdet A = ∑ n∏c 2 = a 2 +b 2 −2ab cos C.sign(π)a i,π(i) .Pythagorean theorem:πArea:i=1C 2 = A 2 + B 2 .2 × 2 and 3 × 3 determinant:Definitions:∣ a bA = 1 2 hc,c d∣ = ad − bc,= 1 2ab sin C,sin a = A/C, cos a = B/C,csc a = C/A, sec a = C/B,a b c∣ ∣ ∣ ∣∣∣ tan a = sin acos a = A B , cos acot a =sin a = B d e fA . ∣ g h i ∣ = g b c∣∣∣ e f ∣ − h a c∣∣∣ d f ∣ + i a b= c2 sin A sin B.d e∣2 sin CHeron’s formula:aei + bfg + cdhArea, radius of inscribed circle:=− ceg − fha− ibd.A = √ s · s a · s b · s c ,12 AB, ABA + B + C .Permanents:s = 1Identities:perm A = ∑ 2(a + b + c),n∏s a = s − a,a i,π(i) .sin x = 1csc x , cos x = 1sπsec x ,i=1b = s − b,Hyperbolic Functionss c = s − c.tan x = 1cot x , sin2 x + cos 2 x =1,Definitions:More identities:√1 − cos x1 + tan 2 x = sec 2 x, 1 + cot 2 x = csc 2 sinh x = ex − e −x, cosh x = ex + e −x, sin xx,2 = ,222sin x = cos ( π2 − x) tanh x = ex − e −x1, sin x = sin(π − x),e x , csch x =+ e−x sinh x ,cos x = − cos(π − x),tan x = cot ( π2 − x) ,cot x = − cot(π − x), csc x = cot x 2− cot x,sin(x ± y) = sin x cos y ± cos x sin y,cos(x ± y) = cos x cos y ∓ sin x sin y,tan x ± tan ytan(x ± y) =1 ∓ tan x tan y ,cot x cot y ∓ 1cot(x ± y) =cot x ± cot y ,sin 2x = 2 sin x cos x, sin 2x = 2 tan x1 + tan 2 x ,cos 2x = cos 2 x − sin 2 x, cos 2x = 2 cos 2 x − 1,cos 2x =1− 2 sin 2 x,cos 2x = 1 − tan2 x1 + tan 2 x ,tan 2x =2 tan x1 − tan 2 x , cot 2x = cot2 x − 12 cot x ,sin(x + y) sin(x − y) = sin 2 x − sin 2 y,cos(x + y) cos(x − y) = cos 2 x − sin 2 y.Euler’s equation:e ix = cos x + i sin x, e iπ = −1.v2.02 c○1994 by Steve Seidensseiden@acm.orghttp://www.csc.lsu.edu/~seidensech x = 1cosh x , coth x = 1tanh x .Identities:cosh 2 x − sinh 2 x =1, tanh 2 x + sech 2 x =1,coth 2 x − csch 2 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 2 x + sinh 2 x,cosh x + sinh x = e x , cosh x − sinh x = e −x ,(cosh x + sinh x) n = cosh nx + sinh nx, n ∈ Z,2 sinh 2 x 2 = cosh x − 1, 2 cosh2 x 2= cosh x +1.θ sin θ cos θ tan θ0 0 1 0π6π4π3π12√22√32√ √3 32 3√221√12 321 0 ∞...in mathematicsyou don’t understandthings, youjust get used tothem.– J. von Neumann√1 + cos xcos x 2 = ,2√1 − cos xtan x 2 = 1 + cos x ,= 1 − cos xsin x ,= sin x1 + cos x ,√1 + cos xcot x 2 = 1 − cos x ,= 1 + cos xsin x ,= sin x1 − cos x ,sin x = eix − e −ix,2icos x = eix + e −ix,2tan x = −i eix − e −ixe ix + e −ix ,= −i e2ix − 1e 2ix +1 ,sinh ixsin x = ,icos x = cosh ix,tanh ixtan x = .i

Number TheoryThe Chinese remainder theorem: There existsa number C such that:C ≡ r 1 mod m 1...C ≡ r n mod m nif m i and m j are relatively prime for i ≠ j.Euler’s function: φ(x) is the number ofpositive integers less than x relativelyprime to x. If ∏ ni=1 pei i is the prime factorizationof x thenn∏φ(x) = p ei−1i (p i − 1).i=1Euler’s theorem: If a and b are relativelyprime then1 ≡ a φ(b) mod b.Fermat’s theorem:1 ≡ a p−1 mod p.The Euclidean algorithm: if a>bare integersthengcd(a, b) = gcd(a mod b, b).If ∏ nis the prime factorization of xtheni=1 pei iS(x) = ∑ d =d|xn∏i=1p ei+1i − 1p i − 1 .Perfect Numbers: x is an even perfect numberiff x =2 n−1 (2 n −1) and 2 n −1 is prime.Wilson’s theorem: n is a prime iff(n − 1)! ≡−1modn.Möbius ⎧inversion:1 if i =1.⎪⎨0 if i is not square-free.µ(i) =⎪⎩ (−1) r if i is the product ofr distinct primes.IfG(a) = ∑ F (d),d|athenF (a) = ∑ ( a)µ(d)G .dd|aPrime numbers:ln ln np n = n ln n + n ln ln n − n + n( )ln nn+ O ,ln nπ(n) =nln n + n(ln n) 2 + 2!n(ln n)( )3n+ O(ln n) 4 .Theoretical Computer Science Cheat SheetGraph TheoryDefinitions:Loop An edge connecting a vertexto itself.Directed Each edge has a direction.Simple Graph with no loops ormulti-edges.Walk A sequence v 0 e 1 v 1 ...e l v l .Trail A walk with distinct edges.Path A trail with distinctvertices.Connected A graph where there existsa path between any twovertices.Component A maximal connectedsubgraph.Tree A connected acyclic graph.Free tree A tree with no root.DAG Directed acyclic graph.Eulerian Graph with a trail visitingeach edge exactly once.Hamiltonian Graph with a cycle visitingeach vertex exactly once.Cut A set of edges whose removalincreases the numberof components.Cut-set A minimal cut.Cut edge A size 1 cut.k-Connected A graph connected withthe removal of any k − 1vertices.k-Tough ∀S ⊆ V,S ≠ ∅ we havek · c(G − S) ≤|S|.k-Regular A graph where all verticeshave degree k.k-Factor A k-regular spanningsubgraph.Matching A set of edges, no two ofwhich are adjacent.Clique A set of vertices, all ofwhich are adjacent.Ind. set A set of vertices, none ofwhich are adjacent.Vertex cover A set of vertices whichcover all edges.Planar graph A graph which can be embededin the plane.Plane graph An embedding of a planargraph.∑deg(v) =2m.v∈VIf G is planar then n − m + f =2,sof ≤ 2n − 4, m ≤ 3n − 6.Any planar graph has a vertex with degree≤ 5.Notation:E(G) Edge setV (G) Vertex setc(G) Number of componentsG[S] Induced subgraphdeg(v) Degree of v∆(G) Maximum degreeδ(G) Minimum degreeχ(G) Chromatic numberχ E (G) Edge chromatic numberG c Complement graphK n Complete graphK n1,n 2Complete bipartite graphr(k, l) Ramsey numberGeometryProjective coordinates: triples(x, y, z), not all x, y and z zero.(x, y, z) =(cx, cy, cz) ∀c ≠0.Cartesian Projective(x, y) (x, y, 1)y = mx + b (m, −1,b)x = c (1, 0, −c)Distance formula, L p and L ∞metric:√(x1 − x 0 ) 2 +(y 1 − y 0 ) 2 ,[|x1 − x 0 | p + |y 1 − y 0 | p] 1/p,[lim |x1 − x 0 | p + |y 1 − y 0 | p] 1/p.p→∞Area of triangle (x 0 ,y 0 ), (x 1 ,y 1 )and (x 2 ,y 2 ):∣ ∣1 ∣∣∣2 abs x 1 − x 0 y 1 − y 0 ∣∣∣.x 2 − x 0 y 2 − y 0Angle formed by three points:(x 2 ,y 2 )l 2θ(0, 0) l 1 (x 1 ,y 1 )cos θ = (x 1,y 1 ) · (x 2 ,y 2 ).l 1 l 2Line through two points (x 0 ,y 0 )and (x 1 ,y 1 ):x y 1x 0 y 0 1∣ x 1 y 1 1 ∣ =0.Area of circle, volume of sphere:A = πr 2 , V = 4 3 πr3 .If I have seen farther than others,it is because I have stood on theshoulders of giants.– Issac Newton

Theoretical Computer Science Cheat SheetπCalculusWallis’ identity:π =2· 2 · 2 · 4 · 4 · 6 · 6 ···Derivatives:1 · 3 · 3 · 5 · 5 · 7 ···1. d(cu)dx= cdud(u + v), 2. = dudx dx dx + dvdx ,Brouncker’s continued fraction expansion:π4 =1+ 1 24. d(un ) du= nun−132+ 2dx dx , d(u/v)5. = v( ) (dudx − u dv)dxdxv 2 , 6. d(ecu )dx2+ 522+2+···727. d(cu ) du= (ln c)cudxdxGregrory’s series:π4 =1− 1 3 + 1 5 − 1 7 + 1 9 −···Newton’s series:π6 = 1 2 + 12 · 3 · 2 3 + 1 · 32 · 4 · 5 · 2 5 + ···Sharp’s series:π6 = √ 1 (1 − 13 3 1 · 3 + 13 2 · 5 − 1 ···)3 3 · 7 +Euler’s series:π 26 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + ···π 28 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + 1 9 2 + ···π 212 = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + 1 5 2 −···Partial FractionsLet N(x) and D(x) be polynomial functionsof x. We can break downN(x)/D(x) using partial fraction expansion.First, if the degree of N is greaterthan or equal to the degree of D, divideN by D, obtainingN(x)D(x) = (x)Q(x)+N′ D(x) ,where the degree of N ′ is less than that ofD. Second, factor D(x). Use the followingrules: For a non-repeated factor:N(x)(x − a)D(x) = Ax − a + N ′ (x)D(x) ,where [ ] N(x)A =.D(x)x=aFor a repeated factor:m−1N(x)(x − a) m D(x) = ∑ A k (x)(x − a) m−k +N′ D(x) ,whereA k = 1 k!k=0[ ( )]dk N(x)dx k .D(x)x=aThe reasonable man adapts himself to theworld; the unreasonable persists in tryingto adapt the world to himself. Thereforeall progress depends on the unreasonable.– George Bernard Shaw9.11.13.15.17.19.21.23.25.27.29.d(sin u)dxd(tan u)dxd(sec u)dxd(arcsin u)dxd(arctan u)dxd(arcsec u)dxd(sinh u)dxd(tanh u)dxd(sech u)dxd(arcsinh u)dxd(arctanh u)dxd(uv)3.dx= u dvdx + v dudx ,, 8.d(ln u)dxcu du= cedx ,= 1 duu dx ,= cos u dud(cos u), 10. = − sin u dudx dxdx ,= sec 2 u dud(cot u), 12. = csc 2 u dudx dxdx ,= tan u sec u dud(csc u), 14. = − cot u csc u dudx dxdx ,=d(arcsech u)31.dxIntegrals:∫ ∫1. cu dx = c3.6.8.10.12.14.∫∫∫1 dud(arccos u)√ , 16. = −1 du√1 − u2 dx dx 1 − u2 dx ,= 1 du1+u 2 dx1=u √ du1 − u 2 dx, 18.d(arccot u)dx, 20.d(arccsc u)dx== −11+u 2 dudx ,−1u √ 1 − u 2 dudx ,= cosh u dud(cosh u), 22. = sinh u dudx dxdx ,= sech 2 u dud(coth u), 24. = − csch 2 u dudx dxdx ,= − sech u tanh u dud(csch u), 26. = − csch u coth u dudx dxdx ,1 dud(arccosh u) 1 du= √ , 28. = √1+u2 dx dx u2 − 1 dx ,= 1 dud(arccoth u)1 − u 2 , 30. = 1 dudx dx u 2 − 1 dx ,−1=u √ du1 − u 2 dx, 32.d(arccsch u)dx=−1|u| √ 1+u 2 dudx .∫∫ ∫u dx, 2. (u + v) dx = udx+ v dx,x n dx = 1∫ ∫1n +1 xn+1 , n ≠ −1, 4.x dx =lnx, 5. e x dx = e x ,∫dx1+x 2 = arctan x, 7. u dv ∫dx dx = uv − v dudx dx,∫∫∫sin xdx= − cos x, 9.tan xdx= − ln | cos x|, 11.sec xdx=ln| sec x + tan x|, 13.arcsin x a dx = arcsin x a + √ a 2 − x 2 , a > 0,∫∫∫cos xdx= sin x,cot xdx=ln| cos x|,csc xdx=ln| csc x + cot x|,

15.17.19.21.23.25.26.29.33.36.∫∫∫∫∫∫∫∫∫∫Theoretical Computer Science Cheat SheetCalculus Cont.arccos x a dx = arccos x a − √ ∫a 2 − x 2 , a > 0, 16. arctan x a dx = x arctan x a − a 2 ln(a2 + x 2 ), a > 0,∫sin 2 ( )(ax)dx = 12a ax − sin(ax) cos(ax) , 18. cos 2 )(ax)dx =2a( 1 ax + sin(ax) cos(ax) ,∫sec 2 xdx= tan x, 20. csc 2 xdx= − cot x,sin n xdx= − sinn−1 x cos x+ n − 1 ∫∫sin n−2 x dx, 22. cos n xdx= cosn−1 x sin x+ n − 1 ∫cos n−2 x dx,n nn n∫∫∫tan n xdx= tann−1 x− tan n−2 x dx, n ≠1, 24. cot n xdx= − cotn−1 x− cot n−2 x dx, n ≠1,n − 1n − 1sec n xdx= tan x secn−1 xn − 1+ n − 2 ∫n − 1sec n−2 x dx, n ≠1,csc n xdx= − cot x cscn−1 x+ n − 2 ∫∫∫csc n−2 x dx, n ≠1, 27. sinh xdx= cosh x, 28. cosh xdx= sinh x,n − 1 n − 1∫∫∫tanh xdx=ln| cosh x|, 30. coth xdx=ln| sinh x|, 31. sech xdx= arctan sinh x, 32. csch xdx=ln ∣ tanhx∣ ,sinh 2 xdx= 1 4 sinh(2x) − 1 2 x,34. ∫cosh 2 xdx= 1 4 sinh(2x)+ 1 2 x,arcsinh x a dx = x arcsinh x a − √ x 2 + a 2 , a > 0, 37.∫35. ∫sech 2 xdx= tanh x,arctanh x a dx = x arctanh x a + a 2 ln |a2 − x 2 |,238.39.40.42.43.46.48.50.52.54.56.58.60.⎧∫⎨ x arccosh xarccosh x a dx = a − √ x 2 + a 2 ,⎩x arccosh x a + √ x 2 + a 2 ,∫dx(√a2 + x =ln x + √ )a 2 + x 2 , a > 0,2∫∫if arccosh x a> 0 and a>0,if arccosh x a< 0 and a>0,∫dx√a2a 2 + x 2 = 1 a arctan x a , a > 0, 41. √− x 2 dx = x 2 a2 − x 2 + a22 arcsin x a , a > 0,(a 2 − x 2 ) 3/2 dx = x 8 (5a2 − 2x 2 ) √ a 2 − x 2 + 3a48 arcsin x a , a > 0,∫∫dx√a2 − x = arcsin x 2 a , a > 0, 44. dxa 2 − x 2 = 1 ∣ ∣∣∣2a ln a + xa − x∣ ,∫ √a2 √ ∣± x 2 dx = x ∣∣x2 a2 ± x 2 ± a22 ln √ ∣ ∫∣∣+ a2 ± x 2 , 47.∫dxax 2 + bx = 1 ∣ ∣∣∣a ln xa + bx∣ ,∫45. dx(a 2 − x 2 ) = x3/2 a 2√ a 2 − x , 2dx∣ ∣∣x√x2 − a =ln √ ∣ ∣∣ + x2 − a 2 , a > 0,2∫49. x √ 2(3bx − 2a)(a + bx)3/2a + bx dx =15b 2 ,∫ √a + bxdx =2 √ ∫∫1a + bx + axx √ a + bx dx,51. x√ dx = 1√ √ ∣ √ lna + bx − a∣∣∣a + bx 2∣√ √ , a > 0,a + bx + a∫ √a2 − x 2dx = √ ax2 − x 2 a + √ a− a ln2 − x 2∫∣ x ∣ ,53. x √ a 2 − x 2 dx = − 1 3 (a2 − x 2 ) 3/2 ,∫x 2√ a 2 − x 2 dx = x 8 (2x2 − a 2 ) √ ∫∣a 2 − x 2 + a48 arcsin x a , a > 0, 55. dx∣∣∣∣√a2 − x = − 1 2 a ln a + √ a 2 − x 2x ∣ ,∫∫xdx√a2 − x = −√ a 2 − x 2 x 2 dx, 57. √ 2 a2 − x = − √ x2 2 a2 − x 2 + a22 arcsin x a,a>0,∫ √a2 + x 2dx = √ ax2 + x 2 a + √ a− a ln2 + x 2∫ √ ∣ x ∣ , 59. x2 − a 2dx = √ xx2 − a 2 − a arccos a|x| , a > 0,∫x √ ∫∣ ∣x 2 ± a 2 dx = 1 3 (x2 ± a 2 ) 3/2 dx∣∣∣, 61.x √ x 2 + a = 1 2 a ln x ∣∣∣a + √ ,a 2 + x 2

Theoretical Computer Science Cheat SheetCalculus Cont.Finite Calculus∫∫√dx62.x √ x 2 − a = 1 2 a arccos a|x| , a > 0, 63. dxx 2√ x 2 ± a = ∓ x2 ± a 2 Difference, shift operators:2 a 2 ,x∆f(x) =f(x +1)− f(x),∫xdx64. √x2 ± a = √ ∫ √x2x 2 ± a 2 ± a, 65.22 x 4 dx = ∓ (x2 + a 2 ) 3/2E f(x) =f(x +1).3a 2 x 3 ,∣1 ∣∣∣∣∫⎧⎪dx ⎨ √66.ax 2 + bx + c = b2 − 4ac ln 2ax + b − √ Fundamental Theorem:b 2 − 4ac2ax + b + √ b 2 − 4ac∣ , if f(x) =∆F (x) ⇔ ∑ f(x)δx = F (x)+C.b2 > 4ac,b∑ ∑b−1⎪ 2 ⎩ √ arctan √ 2ax + b , if f(x)δx = f(i).4ac − b2 4ac − b2 b2 < 4ac,ai=a1∣∫⎧⎪dx ⎨ √a ln ∣2ax + b +2 √ a √ Differences:ax 2 ∆(cu) =c∆u, ∆(u + v) =∆u +∆v,+ bx + c∣ , if a>0,67. √ax2 + bx + c = ⎪ 1 ⎩ √ arcsin √ −2ax − b∆(uv) =u∆v + E v∆u,−a b2 − 4ac ,if a0,if c 0,x 0 =1,x n 1=(x +1)···(x + |n|) , n < 0,x n+m = x m (x − m) n .Rising Factorial Powers:x n = x(x +1)···(x + n − 1), n > 0,x 0 =1,x n 1=(x − 1) ···(x −|n|) , n < 0,x n+m = x m (x + m) n .Conversion:x n =(−1) n (−x) n =(x − n +1) n=1/(x +1) −n ,x n =(−1) n (−x) n =(x + n − 1) n=1/(x − 1) −n ,n∑{ } nn∑{ } nx n = x k = (−1) n−k x k ,k kx n =x n =k=1n∑k=1n∑k=1k=1[ nk](−1) n−k x k ,[ nk]x k .

Taylor’s series:Expansions:f(x) =f(a)+(x − a)f ′ (a)+Theoretical Computer Science Cheat Sheet(x − a)2f ′′ (a)+···=2Series∞∑ (x − a) if (i) (a).i!i=011 − x=1+x + x 2 + x 3 + x 4 + ··· =11 − cx=1+cx + c 2 x 2 + c 3 x 3 + ··· =11 − x n =1+x n + x 2n + x 3n + ··· =x(1 − x) 2 = x +2x 2 +3x 3 +4x 4 + ··· =x k dndx n ( 11 − x)= x +2 n x 2 +3 n x 3 +4 n x 4 + ··· =∞∑x i ,i=0∞∑c i x i ,i=0∞∑x ni ,i=0∞∑ix i ,i=0∞∑i n x i ,∞∑e x =1+x + 1 2 x2 + 1 x i6 x3 + ··· =i! ,i=0∞∑ln(1 + x) = x − 1 2 x2 + 1 3 x3 − 1 4 x4 i+1 xi−··· = (−1)i ,i=11∞∑ln= x + 11 − x2 x2 + 1 3 x3 + 1 x i4 x4 + ··· =i ,i=1∞∑sin x = x − 1 3! x3 + 1 5! x5 − 1 7! x7 + ··· = (−1) i x 2i+1(2i + 1)! ,cos x =1− 1 2! x2 + 1 4! x4 − 1 6! x6 + ··· =tan −1 x = x − 1 3 x3 + 1 5 x5 − 1 7 x7 + ··· =(1 + x) n =1+nx + n(n−1)2x 2 + ··· =1(1 − x) n+1 =1+(n +1)x + ( )n+22 x 2 + ··· =xe x − 1=1− 1 2 x + 112 x2 − 1720 x4 + ··· =12x (1 − √ 1 − 4x) = 1+x +2x 2 +5x 3 + ··· =121√ 1 − 4x=1+x +2x 2 +6x 3 + ··· =( √1 1 − 1 − 4x√ 1 − 4x 2x11 − x ln 11 − x(ln11 − x) n=1+(2+n)x + ( )4+n2 x 2 + ··· == x + 3 2 x2 + 11 6 x3 + 2512 x4 + ··· =) 2= 1 2 x2 + 3 4 x3 + 1124 x4 + ··· =x1 − x − x 2 = x + x 2 +2x 3 +3x 4 + ··· =F n x1 − (F n−1 + F n+1 )x − (−1) n x 2 = F n x + F 2n x 2 + F 3n x 3 + ··· =i=0i=0∞∑i=0(−1) i x2i(2i)! ,∞∑(−1) i x 2i+1(2i +1) ,∞∑( ) nx i ,ii=0i=0∞∑( ) i + nx i ,i∞∑ B i x i,i!∞∑( )1 2ix i ,i +1 i∞∑( ) 2ix i ,i∞∑( ) 2i + nx i ,i∞∑H i x i ,i=0i=0i=0i=0i=0i=1∞∑ H i−1 x i,i∞∑F i x i ,i=2i=0∞∑F ni x i .i=0Ordinary power series:∞∑A(x) = a i x i .i=0Exponential power series:∞∑ x iA(x) = a ii! .i=0Dirichlet power series:∞∑ a iA(x) =i x .i=1Binomial theorem:n∑( n(x + y) n = xk)n−k y k .k=0Difference of like powers:n−1∑x n − y n =(x − y) x n−1−k y k .k=0For ordinary power series:∞∑αA(x)+βB(x) = (αa i + βb i )x i ,x k A(x) =i=0∞∑a i−k x i ,i=kA(x) − ∑ k−1i=0 a ix ix k =A(cx) =A ′ (x) =∞∑a i+k x i ,i=0∞∑c i a i x i ,i=0∞∑(i +1)a i+1 x i ,i=0∞∑xA ′ (x) = ia i x i ,i=1∫∞∑ a i−1A(x) dx = x i ,ii=1A(x)+A(−x)∞∑= a 2i x 2i ,2i=0A(x) − A(−x)∞∑= a 2i+1 x 2i+1 .2i=0Summation: If b i = ∑ ij=0 a i thenB(x) = 11 − x A(x).Convolution:⎛∞∑ i∑A(x)B(x) = ⎝i=0j=0a j b i−j⎞⎠ x i .God made the natural numbers;all the rest is the work of man.– Leopold Kronecker

Theoretical Computer Science Cheat SheetSeriesExpansions:1(1 − x) n+1 ln 1∞∑( ) ( ) −n n + i1= (H n+i − H n ) x i ,=1 − xixi=0∞∑[ ] nx n = x i , (e x − 1) n =i∞∑i=0{ in}x i ,∞∑{ i n!xi,n}i!i=0i=0( ) n1∞∑[ i n!xi∞∑ (−4)ln=, xcot x =1 − xn] i B 2i x 2i,i!(2i)!i=0i=0∞∑tan x = (−1) i−1 22i (2 2i − 1)B 2i x 2i−1∞∑ 1, ζ(x) =(2i)!i x ,i=1i=11∞∑ µ(i)=ζ(x)i x , ζ(x − 1)∞∑=ζ(x)i=1ζ(x) = ∏ p∞∑ζ 2 (x) =ζ(x)ζ(x − 1) =i=1∞∑i=111 − p −x ,d(i)x i where d(n) = ∑ d|n 1,S(i)x i where S(n) = ∑ d|n d,ζ(2n) = 22n−1 |B 2n |π 2n , n ∈ N,(2n)!x∞∑= (−1) i−1 (4i − 2)B 2i x 2i,sin x(2i)!i=0( √ ) n1 − 1 − 4x∞∑ n(2i + n − 1)!=x i ,2xi!(n + i)!i=0∞∑e x sin x =i=1√1 − √ 1 − x∞∑=xi=0( ) 2 arcsin x∞∑=xi=02 i/2 sin iπ 4i!x i ,(4i)!16 i√ 2(2i)!(2i + 1)! xi ,4 i i! 2(i + 1)(2i + 1)! x2i .i=1φ(i)i x , Stieltjes IntegrationEscher’s KnotIf G is continuous in the interval [a, b] and F is nondecreasing thenexists. If a ≤ b ≤ c then∫ caG(x) dF (x) =∫ ba∫ baG(x) dF (x)G(x) dF (x)+∫ cIf the integrals involved exist∫ b∫( ) bG(x)+H(x) dF (x) = G(x) dF (x)+a∫ ba∫ ba∫ bG(x) d ( F (x)+H(x) ) =ac · G(x) dF (x) =∫ baa∫ babG(x) dF (x)+G(x) d ( c · F (x) ) = cG(x) dF (x) =G(b)F (b) − G(a)F (a) −G(x) dF (x).∫ ba∫ ba∫ ba∫ baH(x) dF (x),G(x) dH(x),G(x) dF (x),F (x) dG(x).If the integrals involved exist, and F possesses a derivative F ′ at everypoint in [a, b] then∫ b∫ bG(x) dF (x) = G(x)F ′ (x) dx.Cramer’s RuleFibonacci Numbers00 47 18 76 29 93 85 34 61 52If we have equations:86 11 57 28 70 39 94 45 02 63a 1,1 x 1 + a 1,2 x 2 + ···+ a 1,n x n = b 11, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,...95 80 22 67 38 71 49 56 13 04a 2,1 x 1 + a 2,2 x 2 + ···+ a 2,n x n = b 2Definitions:59 96 81 33 07 48 72 60 24 1573 69 90 82 44 17 58 01 35 26.F i = F i−1 +F i−2 , F 0 = F 1 =1,..68 74 09 91 83 55 27 12 46 30F −i =(−1) i−1 F i ,a n,1 x 1 + a n,2 x 2 + ···+ a n,n x n = b n37 08 75 19 92 84 66 23 50 41F14 25 36 40 51 62 03 77 88 99i = √ 15(φ i − ˆφ i) ,Let A =(a i,j ) and B be the column matrix (b i ). Then21 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 i be A42 53 64 05 16 20 31 98 79 87with column i replaced by B. ThenF i+1 F i−1 − Fi 2 =(−1) i .x i = det A idet A . The Fibonacci number system:Additive rule:Every integer n has a unique F n+k = F k F n+1 + F k−1 F n ,representationF 2n = F n F n+1 + F n−1 F n .Improvement makes strait roads, but the crooked n = F k1 + F k2 + ···+ F km , Calculation by matrices:roads without Improvement, are roads of Genius.( ) ( )where k ni ≥ k i+1 + 2 for all i, Fn−2 F n−1 0 1– William Blake (The Marriage of Heaven and Hell) 1 ≤ i