Cheat_Sheet

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