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Theoretical Computer Science Cheat Sheet De(cid:12)nitions Series f(n)=O(g(n)) i(cid:11) 9 positive c;n0 such that Xn n(n+1) Xn n(n+1)(2n+1) Xn n2(n+1)2 0(cid:20)f(n)(cid:20)cg(n) 8n(cid:21)n0. i= ; i2 = ; i3 = : 2 6 4 f(n)=Ω(g(n)) i(cid:11) 9 positive c;n such that i=1 i=1 i=1 0 In general: f(n)(cid:21)cg(n)(cid:21)0 8n(cid:21)n . (cid:20) (cid:21) 0 Xn Xn (cid:0) (cid:1) 1 f(n)=(cid:2)(g(n)) i(cid:11) f(n) = O(g(n)) and im = (n+1)m+1−1− (i+1)m+1−im+1−(m+1)im m+1 f(n)=Ω(g(n)). i=1 i=1 (cid:18) (cid:19) nX−1 Xm 1 m+1 f(n)=o(g(n)) i(cid:11) limn!1f(n)=g(n)=0. im = Bknm+1−k: m+1 k nl!im1an =a jia(cid:11)n8−(cid:15)a>j<0(cid:15),, 89nn0(cid:21)snu0c.h that GiX=en1ometriccns+er1ie−s:k1=0 X1 1 X1 c supS least b 2 R such that b (cid:21) s, ci = c−1 ; c6=1; ci = 1−c; ci = 1−c; jcj<1; 8s2S. i=0 i=0 i=1 Xn ncn+2−(n+1)cn+1+c X1 c infS greatest b2R such that b(cid:20) ici = ; c6=1; ici = ; jcj<1: s, 8s2S. i=0 (c−1)2 i=0 (1−c)2 Harmonic series: linm!i1nfan nl!im1inffai ji(cid:21)n;i2Ng. Xn 1 Xn n(n+1) n(n−1) Hn = ; iHi = Hn− : linm!s1upan nl!im1supfai ji(cid:21)n;i2Ng. i=1 i i=1 (cid:18) (cid:19) 2(cid:18) (cid:19)(cid:18) 4 (cid:19) (cid:0) (cid:1) Xn Xn nk Combinations: Size k sub- Hi =(n+1)Hn−n; mi Hi = mn++11 Hn+1− m1+1 : sets of a size n set. i=1 i=1 (cid:2) (cid:3) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) n Xn Stirling numbers (1st kind): n n! n n n k 1. = ; 2. =2n; 3. = ; Arrangements of an n ele- k (n−k)!k! k k n−k ment set into k cycles. (cid:18) (cid:19) (cid:18) (cid:19) k=0 (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:8)nk(cid:9) Stirlingnumbers(2ndkind): 4. nk = nk nk−−11 ; 5. nk = n−k 1 + nk−−11 ; (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) Partitions of an n element n m n n−k Xn r+k r+n+1 set into k non-empty sets. 6. = ; 7. = ; (cid:10) (cid:11) m k k m−k k n n (cid:18) (cid:19) (cid:18) (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19) 1st order Eulerian numbers: Xn Xn k k n+1 r s r+s Permutations (cid:25)1(cid:25)2:::(cid:25)n on 8. m = m+1 ; 9. k n−k = n ; f1;2;:::;ng with k ascents. k=(cid:18)0 (cid:19) (cid:18) (cid:19) k=0 (cid:26) (cid:27) (cid:26) (cid:27) (cid:10)(cid:10) (cid:11)(cid:11) n k−n−1 n n n 2ndorderEuleriannumbers. 10. =(−1)k ; 11. = =1; k k k 1 n (cid:26) (cid:27) (cid:26) (cid:27) (cid:26) (cid:27) (cid:26) (cid:27) Cn Catalan Numbers: Binary n n n−1 n−1 trees with n+1 vertices. 12. =2n−1−1; 13. =k + ; 2 k k k−1 (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:26) (cid:27) n n n n n 14. 1 =(n−1)!; 15. 2 =(n−1)!Hn−1; 16. n =1; 17. k (cid:21) k ; (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:26) (cid:27) (cid:20) (cid:21) (cid:18) (cid:19) (cid:20) (cid:21) (cid:18) (cid:19) n n−1 n−1 n n n Xn n 1 2n 18. k =(n−1) k + k−1 ; 19. n−1 = n−1 = 2 ; 20. k =n!; 21. Cn = n+1 n ; (cid:28) (cid:29) (cid:28) (cid:29) (cid:28) (cid:29) (cid:28) (cid:29) (cid:28) (cid:29)k=0 (cid:28) (cid:29) (cid:28) (cid:29) n n n n n n−1 n−1 22. = =1; 23. = ; 24. =(k+1) +(n−k) ; 0 n−1 k n−1−k k k k−1 (cid:28) (cid:29) n (cid:28) (cid:29) (cid:28) (cid:29) (cid:18) (cid:19) 25. 0 = 1 if k =0, 26. n =2n−n−1; 27. n =3n−(n+1)2n+ n+1 ; k 0 otherwise 1 2 2 (cid:28) (cid:29)(cid:18) (cid:19) (cid:28) (cid:29) (cid:18) (cid:19) (cid:26) (cid:27) (cid:28) (cid:29)(cid:18) (cid:19) Xn Xm Xn 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 (cid:28) (cid:29)k=0 (cid:26) (cid:27)(cid:18) (cid:19) k=0 (cid:28)(cid:28) (cid:29)(cid:29) (cid:28)(cid:28) k(cid:29)=(cid:29)0 n Xn n n−k n n 31. = (−1)n−k−mk!; 32. =1; 33. =0 for n6=0; m k m 0 n (cid:28)(cid:28) (cid:29)(cid:29) k=0 (cid:28)(cid:28) (cid:29)(cid:29) (cid:28)(cid:28) (cid:29)(cid:29) (cid:28)(cid:28) (cid:29)(cid:29) n n−1 n−1 Xn n (2n)n 34. =(k+1) +(2n−1−k) ; 35. = ; k k k−1 k 2n (cid:26) (cid:27) (cid:28)(cid:28) (cid:29)(cid:29)(cid:18) (cid:19) (cid:26) (cid:27) (cid:18) (cid:19)(cid:26) (cid:27) k=(cid:26)0 (cid:27) x Xn n x+n−1−k n+1 X n k Xn k 36. = ; 37. = = (m+1)n−k; x−n k 2n m+1 k m m k=0 k k=0 Theoretical Computer Science Cheat Sheet Identities Cont. Trees (cid:20) (cid:21) (cid:20) (cid:21)(cid:18) (cid:19) (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:28)(cid:28) (cid:29)(cid:29)(cid:18) (cid:19) X Xn Xn Xn n+1 n k k 1 k x n x+k Every tree with n 38. m+1 = k m = m nn−k =n! k! m ; 39. x−n = k 2n ; vertices has n−1 (cid:26) (cid:27) X(cid:18)k (cid:19)(cid:26) (cid:27) k=0 k=0 (cid:20) (cid:21) X(cid:20) k=0(cid:21)(cid:18) (cid:19) edges. n n k+1 n n+1 k 40. = (−1)n−k; 41. = (−1)m−k; Kraft inequal- m k m+1 m k+1 m (cid:26) k (cid:27) (cid:26) (cid:27) (cid:20) k(cid:21) (cid:20) (cid:21) ity: If the depths Xm Xm 42. m+n+1 = k n+k ; 43. m+n+1 = k(n+k) n+k ; of the leaves of m k m k a binary tree are (cid:18) (cid:19) (cid:26) k=0(cid:27)(cid:20) (cid:21) (cid:18) (cid:19) (cid:20) (cid:21)(cid:26) (cid:27) k=0 44. n =X n+1 k (−1)m−k; 45. (n−m)! n =X n+1 k (−1)m−k; for n(cid:21)m, d1;X:n::;dn: (cid:26)m (cid:27)k k+(cid:18)1 m(cid:19)(cid:18) (cid:19)(cid:20) (cid:21) m(cid:20) k (cid:21) k+1(cid:18) m (cid:19)(cid:18) (cid:19)(cid:26) (cid:27) 2−di (cid:20)1; 46. n =X m−n m+n m+k ; 47. n =X m−n m+n m+k ; i=1 n−m m+k n+k k n−m m+k n+k k andequalityholds (cid:26) (cid:27)(cid:18) k (cid:19) (cid:26) (cid:27)(cid:26) (cid:27)(cid:18) (cid:19) (cid:20) (cid:21)(cid:18)k (cid:19) (cid:20) (cid:21)(cid:20) (cid:21)(cid:18) (cid:19) n ‘+m X k n−k n n ‘+m X k n−k n only if every in- 48. = ; 49. = : ‘+m ‘ ‘ m k ‘+m ‘ ‘ m k ternal node has 2 k k sons. Recurrences Master method: (cid:0) (cid:1) Generating functions: T(n)=aT(n=b)+f(n); a(cid:21)1;b>1 1 T(n)−3T(n=2)=n 1. Multiply both sides of the equa- (cid:0) (cid:1) If 9(cid:15)>0 such that f(n)=O(nlogba−(cid:15)) 3 T(n=2)−3T(n=4)=n=2 tion by xi. 2. Sum both sides over all i for then .. .. .. T(n)=(cid:2)(nlogba): .(cid:0) . . (cid:1) which the equation is valid. 3log2n−1 T(2)−3T(1)=2 3. Choose a generatingPfunction If f(n)=(cid:2)(nlogba) then G(x). UsuallyG(x)= 1i=0xigi. T(n)=(cid:2)(nlogbalog2n): wLeetgmet =T(nlo)g2−n3.mSTu(m1)m=ingTt(hne) −lef3tmsid=e 3. Rewrite the equation in terms of the generating function G(x). If9(cid:15)>0suchthatf(n)=Ω(nlogba+(cid:15)), T(n) − nk where k = log 3 (cid:25) 1:58496. afonrdla9rcge<n1, tshuecnh that af(n=b) (cid:20) cf(n) SummingmXt−h1e right sidemXw−1e2(cid:0)ge(cid:1)t 54.. TSohlevecofoe(cid:14)r Gci(exn)t.ofxi inG(x)isgi. T(n)=(cid:2)(f(n)): n3i =n 3 i: Example: i=0 2i i=0 2 gi+1 =2gi+1; g0 =0: Substitution (example): Consider the fNLoeoltltoetwiti=hnTagil+tor1geTc2=iuTirs2ir.i2eainTlw(cid:1)chTeaeiy2n;swaeTp1hoaw=vee2r:of two. Let c=nmXi32=−.01Tcihe==n2nwn(cid:18)e(cchlcmoag−v2−en11−(cid:19)1) WMinuetlXeict(cid:21)rhimp0ologsyis+oeaf1nGxGd(ix(s=x)u)m=X:i(cid:21):P02ig(cid:21)ix0ix+igiXi.(cid:21)R0xewi:rite ti+1 =2 +2ti; t1 =1: =2n(c(k−1)logcn−1) G(x)−g0 =2G(x)+Xxi: Ltheetpuriev=iotui=s2eiq.uDatiivoindibnyg2bio+t1hwseidgeestof =2nk−2n; x i(cid:21)0 2tii++11 = 2i2+i1 + 2tii: caunrdresnoceTs(nca)n=of3tnenk −be2cnh.anFguelldhtiostloimryitreed- SimplifyG:(x) =2G(x)+ 1 : history ones (example): Consider x 1−x Substituting we (cid:12)nd Xi−1 ui+1 = 12 +ui; u1 = 12; Ti =1+ Tj; T0 =1: Solve for G(x): x G(x)= : which is simply ui = i=2. So we (cid:12)nd j=0 (1−x)(1−2x) thatTi hastheclosedformTi =2i2i−1. Note that Summing factors (example): Consider Xi Expand this(cid:18)using partial fract(cid:19)ions: 2 1 the following recurrence Ti+1 =1+ Tj: G(x)=x 1−2x − 1−x T(n)=3T(n=2)+n; T(1)=1: j=0 0 1 X X Rewrite so that all terms involving T Subtracting we (cid:12)ndXi Xi−1 =x@2 2ixi− xiA are on thTe(lnef)t−sid3Te(n=2)=n: Ti+1−Ti =1+ Tj −1− Tj X i(cid:21)0 i(cid:21)0 j=0 j=0 = (2i+1−1)xi+1: Nowexpandtherecurrence,andchoose =Ti: i(cid:21)0 afactorwhichmakestheleftside\tele- scope" And so Ti+1 =2Ti =2i+1. So gi =2i−1. Theoretical Computer Science Cheat Sheet p p (cid:25) (cid:25)3:14159, e(cid:25)2:71828, γ (cid:25)0:57721, (cid:30)= 1+ 5 (cid:25)1:61803, (cid:30)^= 1− 5 (cid:25)−:61803 2 2 i 2i pi General Probability 1 2 2 Bernoulli Numbers (Bi =0, odd i6=1): Continuous distributionsZ: If b 2 4 3 B =1, B =−1, B = 1, B =− 1 , 0 1 2 2 6 4 30 Pr[a<X <b]= p(x)dx; 3 8 5 B = 1 , B =− 1 , B = 5 . a 6 42 8 30 10 66 thenpistheprobabilitydensityfunctionof 4 16 7 Change of base, quadratic formula: p X. If 5 32 11 log x= logax; −b(cid:6) b2−4ac: Pr[X <a]=P(a); b 6 64 13 logab 2a then P is the distribution function of X. If Euler’s number e: 7 128 17 P and p both exist then Z e=1+ 1 + 1 + 1 + 1 +(cid:1)(cid:1)(cid:1) a 8 256 19 2 (cid:16) 6 2(cid:17)4 120 x n P(a)= p(x)dx: 9 512 23 lim 1+ =ex: −1 n!1 n 10 1,024 29 (cid:0)1+ 1(cid:1)n <e<(cid:0)1+ 1(cid:1)n+1: Expectation: If XXis discrete 11 2,048 31 n n (cid:18) (cid:19) E[g(X)]= g(x)Pr[X =x]: (cid:0) (cid:1) 12 4,096 37 1+ 1 n =e− e + 11e −O 1 : x n 2n 24n2 n3 If X continZuous then Z 13 8,192 41 1 1 Harmonic numbers: E[g(X)]= g(x)p(x)dx= g(x)dP(x): 14 16,384 43 1, 3, 11, 25, 137, 49, 363, 761, 7129;::: −1 −1 2 6 12 60 20 140 280 2520 15 32,768 47 Variance, standard deviation: 16 65,536 53 lnn<Hn <lnn(cid:18)+1(cid:19); VAR[X]=pE[X2]−E[X]2; 17 131,072 59 1 (cid:27) = VAR[X]: Hn =lnn+γ+O : 18 262,144 61 n For events A and B: 19 524,288 67 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](cid:1)Pr[B]; 21 2,097,152 73 p (cid:18) (cid:19)n(cid:18) (cid:18) (cid:19)(cid:19) i(cid:11) A and B are independent. n 1 22 4,194,304 79 n!= 2(cid:25)n 1+(cid:2) : Pr[A^B] e n Pr[AjB]= 23 8,388,608 83 Pr[B] Ackerman8n’s function and inverse: 24 16,777,216 89 <2j i=1 For random variables X and Y: 25 33,554,432 97 a(i;j)= a(i−1;2) j =1 E[X (cid:1)Y]=E[X](cid:1)E[Y]; : a(i−1;a(i;j−1)) i;j (cid:21)2 if X and Y are independent. 26 67,108,864 101 (cid:11)(i)=minfj ja(j;j)(cid:21)ig: E[X +Y]=E[X]+E[Y]; 27 134,217,728 103 E[cX]=cE[X]: 28 268,435,456 107 Binomial distr(cid:18)ibu(cid:19)tion: Bayes’ theorem: n 2390 15,03763,8,77401,9,81224 110193 Pr[X =k]=Xnk p(cid:18)knq(cid:19)n−k; q =1−p; Pr[AijB]= PnjP=1r[PBrj[AAij]]PPrr[[ABij]Aj]: 31 2,147,483,648 127 k n−k E[X]= k p q =np: Inclusion-exclusion: 32 4,294,967,296 131 k=1 k h_n i Xn Pascal’s Triangle Poisson distribution: Pr Xi = Pr[Xi]+ e−(cid:21)(cid:21)k i=1 i=1 1 Pr[X =k]= ; E[X]=(cid:21): Xn X h ^k i k! 1 1 Normal (Gaussian) distribution: (−1)k+1 Pr Xij : 1 2 1 p(x)= p1 e−(x−(cid:22))2=2(cid:27)2; E[X]=(cid:22): Momentkin=e2qualities:ii<(cid:1)(cid:1)(cid:1)<ik j=1 1 3 3 1 2(cid:25)(cid:27) (cid:2) (cid:3) 1 1 4 6 4 1 The \coupon collector": We are given a Pr jXj(cid:21)(cid:21)E[X] (cid:20) ; (cid:21) 1 5 10 10 5 1 rdai(cid:11)nedroemntctoyuppeosnoefaccohudpaoyn,sa.nTdhtehedriestarribeun- Prh(cid:12)(cid:12)X −E[X](cid:12)(cid:12)(cid:21)(cid:21)(cid:1)(cid:27)i(cid:20) 1 : 1 6 15 20 15 6 1 (cid:21)2 tion of coupons is uniform. The expected Geometric distribution: 1 7 21 35 35 21 7 1 number of days to pass before we to col- Pr[X =k]=pqk−1; q =1−p; 1 8 28 56 70 56 28 8 1 lect all n types is X1 1 9 36 84 126 126 84 36 9 1 nHn: E[X]= kpqk−1 = 1: p 1 10 45 120 210 252 210 120 45 10 1 k=1 Theoretical Computer Science Cheat Sheet Trigonometry Matrices More Trig. Multiplication: C Xn (0,1) C =A(cid:1)B; ci;j = ai;kbk;j: a b b (cos(cid:18);sin(cid:18)) k=1 h A C (cid:18) Determinants: detA6=0 i(cid:11) A is non-singular. (-1,0) (1,0) A c B detA(cid:1)B =detA(cid:1)detB; Law of cosines: c a (0,-1) XYn c2 =a2+b2−2abcosC: B detA= sign((cid:25))ai;(cid:25)(i): Area: Pythagorean theorem: (cid:25) i=1 C2 =A2+B2: 2(cid:2)2 and 3(cid:2)3 d(cid:12)eterm(cid:12)inant: (cid:12) (cid:12) A= 1hc; De(cid:12)nitions: (cid:12)(cid:12)ac db(cid:12)(cid:12)=ad−bc; = 21absinC; sina=A=C; cosa=B=C; (cid:12)(cid:12)(cid:12)a b c(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) 2c2sinAsinB csca=C=A; seca=C=B; (cid:12)(cid:12)d e f(cid:12)(cid:12)=g(cid:12)(cid:12)b c(cid:12)(cid:12)−h(cid:12)(cid:12)a c(cid:12)(cid:12)+i(cid:12)(cid:12)a b(cid:12)(cid:12) = 2sinC : sina A cosa B (cid:12)g h i(cid:12) e f d f d e Heron’s formula: tana= = ; cota= = : cosa B sina A aei+bfg+cdh p Area, radius of inscribed circle: = −ceg−fha−ibd: A= s(cid:1)sa(cid:1)sb(cid:1)sc; 1AB; AB : Permanents: s= 12(a+b+c); 2 A+B+C XYn sa =s−a; Identities:1 1 permA= (cid:25) i=1ai;(cid:25)(i): sb =s−b; sinx= ; cosx= ; Hyperbolic Functions sc =s−c: cscx secx tanx= 1 ; sin2x+cos2x=1; De(cid:12)nitions: More iderntities: cotx ex−e−x ex+e−x 1−cosx sinhx = ; coshx= ; sinx = ; 1+tan2x=sec2x; 1+cot2x=csc2x; 2 2 2 2 ex−e−x 1 r (cid:0) (cid:1) tanhx= ; cschx = ; 1+cosx sinx=cos (cid:25) −x ; sinx=sin((cid:25)−x); ex+e−x sinhx cosx = ; 2 (cid:0) (cid:1) 1 1 2 r 2 sechx = ; cothx= : cosx=−cos((cid:25)−x); tanx=cot (cid:25) −x ; coshx tanhx 1−cosx 2 tanx = ; Identities: 2 1+cosx cotx=−cot((cid:25)−x); cscx=cotx2 −cotx; 1−cosx cosh2x−sinh2x=1; tanh2x+sech2x=1; = ; sinx sin(x(cid:6)y)=sinxcosy(cid:6)cosxsiny; sinx coth2x−csch2x=1; sinh(−x)=−sinhx; = ; cos(x(cid:6)y)=cosxcosy(cid:7)sinxsiny; r1+cosx tan(x(cid:6)y)= 1ta(cid:7)ntxan(cid:6)xttaannyy; csionshh((x−+x)y=)=cossihnxh;xcoshy+tacnohs(h−xxs)in=h−y;tanhx; cotx2 = 11+−ccoossxx; 1+cosx cotxcoty(cid:7)1 = ; cot(x(cid:6)y)= ; cosh(x+y)=coshxcoshy+sinhxsinhy; sinx cotx(cid:6)coty sinx 2tanx sinh2x=2sinhxcoshx; = ; sin2x=2sinxcosx; sin2x= ; 1−cosx 1+tan2x cosh2x=cosh2x+sinh2x; eix−e−ix cos2x=cos2x−sin2x; cos2x=2cos2x−1; sinx= ; 2i cos2x=1−2sin2x; cos2x= 1−tan2x; coshx+sinhx=ex; coshx−sinhx=e−x; eix+e−ix 1+tan2x (coshx+sinhx)n =coshnx+sinhnx; n2Z; cosx= 2 ; 2tanx cot2x−1 eix−e−ix tan2x= 1−tan2x; cot2x= 2cotx ; 2sinh2 x2 =coshx−1; 2cosh2 x2 =coshx+1: tanx=−ieix+e−ix; sin(x+y)sin(x−y)=sin2x−sin2y; e2ix−1 =−i ; (cid:18) sin(cid:18) cos(cid:18) tan(cid:18) :::inmathematics e2ix+1 cos(x+y)cos(x−y)=cos2x−sin2y: 0 0 1 0 you don’t under- sinhix p p stand things, you sinx= ; Euler’s equation: (cid:25) 1 3 3 i eix =cosx+isinx; ei(cid:25) =−1: (cid:25)6 p22 p22 13 jtuhsetm.get used to cosx=coshix; v2.02 (cid:13)c1994 by Steve Seiden (cid:25)4 p23 21 p3 { J. von Neumann tanx= tanhix: [email protected] 3 2 2 i (cid:25) 1 0 1 http://www.csc.lsu.edu/~seiden 2 Theoretical Computer Science Cheat Sheet Number Theory Graph Theory TheChineseremaindertheorem: Thereex- De(cid:12)nitions: Notation: ists a number C such that: Loop An edge connecting a ver- E(G) Edge set tex to itself. V(G) Vertex set C (cid:17)r modm 1 1 Directed Each edge has a direction. c(G) Number of components .. .. .. Simple Graph with no loops or G[S] Induced subgraph . . . multi-edges. deg(v) Degree of v C (cid:17)rn modmn (cid:1)(G) Maximum degree Walk A sequence v0e1v1:::e‘v‘. ifmi andmj arerelativelyprimefori6=j. Trail Awalkwithdistinctedges. (cid:14)(G) Minimum degree (cid:31)(G) Chromatic number Path A trail with distinct Euler’s function: (cid:30)(x) is the number of vertices. (cid:31)E(G) Edge chromatic number pproismiteivetoixn.tegIferQs nlesspeithisanthexprreimlaetivfaelcy- Connected Agraphwherethereexists Gc Complement graph torization of x thein=1 i a path between any two Kn Complete graph Yn vertices. Kn1;n2 Complete bipartite graph (cid:30)(x)= piei−1(pi−1): Component A maximal connected r(k;‘) Ramsey number i=1 subgraph. Geometry Euler’s theorem: If a and b are relatively Tree Aconnectedacyclicgraph. Projective coordinates: triples prime then Free tree A tree with no root. 1(cid:17)a(cid:30)(b) modb: DAG Directed acyclic graph. (x;y;z), not all x, y and z zero. (x;y;z)=(cx;cy;cz) 8c6=0: Eulerian Graph with a trail visiting Fermat’s theorem: 1(cid:17)ap−1 modp: each edge exactly once. Cartesian Projective Hamiltonian Graphwithacyclevisiting (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(amodb;b): moval increases the num- Distance formula, Lp and L1 Q If n pei is the prime factorization of x ber of components. metpric: theni=1 i Cut-set A minimal cut. (x −x )2+(y −y )2; S(x)=Xd=Yn piepi+i−1−11: Ck-uCtoendngeected AA sgizreap1hcucto.nnected with (cid:2)jx(cid:2)1−1x0jp0+jy1−1y0jp(cid:3)01=(cid:3)p; djx i=1 the removal of any k − 1 lim jx −x jp+jy −y jp 1=p: 1 0 1 0 vertices. p!1 PerfectNumbers: xisanevenperfectnum- beri(cid:11)x=2n−1(2n−1)and2n−1isprime. k-Tough 8S (cid:18) V;S 6= ; we have Area of triangle (x0;y0), (x1;y1) Wilson’s theorem: n is a prime i(cid:11) k(cid:1)c(G−S)(cid:20)jSj. and (x2;y(cid:12)2): (cid:12) (n−1)!(cid:17)−1modn: k-Regular Ahavgeradpehgrweehekr.e all vertices 12abs(cid:12)(cid:12)(cid:12)xx1−−xx0 yy1−−yy0(cid:12)(cid:12)(cid:12): 2 0 2 0 M¨obius 8inversion: ><1 if i=1. k-Factor A k-regular spanning Angle formed by three points: subgraph. 0 if i is not square-free. (cid:22)(i)= >:(−1)r if i is the product of Matching A set of edges, no two of (x2;y2) r distinct primes. which are adjacent. ‘ Clique A set of vertices, all of 2 If X (cid:18) which are adjacent. G(a)= F(d); (0;0) ‘ (x ;y ) Ind. set A set of vertices, none of 1 1 1 dja which are adjacent. cos(cid:18) = (x1;y1)(cid:1)(x2;y2): then X (cid:16)a(cid:17) Vertex cover A set of vertices which ‘1‘2 F(a)= (cid:22)(d)G : cover all edges. Line through two points (x ;y ) d 0 0 dja Planar graph A graph which can be em- and (x1;(cid:12)y1): (cid:12) (cid:12) (cid:12) Prime numbers: beded in the plane. (cid:12) x y 1(cid:12) pn =nlnn+nlnlnn−n+nlnlnn Plane graph An embedding of a planar (cid:12)(cid:12)(cid:12)x0 y0 1(cid:12)(cid:12)(cid:12)=0: (cid:18) (cid:19) lnn graph. x1 y1 1 n X Area of circle, volume of sphere: +O ; deg(v)=2m: lnn A=(cid:25)r2; V = 4(cid:25)r3: v2V 3 n n 2!n (cid:25)(n)= + + If G is planar then n−m+f =2, so lnn (lnn)2 (lnn)3 IfIhaveseenfartherthanothers, (cid:18) (cid:19) f (cid:20)2n−4; m(cid:20)3n−6: n it is because I have stood on the +O : Any planar graph has a vertex with de- shoulders of giants. (lnn)4 gree (cid:20)5. { Issac Newton Theoretical Computer Science Cheat Sheet (cid:25) Calculus Wallis’ identity: Derivatives: 2(cid:1)2(cid:1)4(cid:1)4(cid:1)6(cid:1)6(cid:1)(cid:1)(cid:1) (cid:25) =2(cid:1) d(cu) du d(u+v) du dv d(uv) dv du 1(cid:1)3(cid:1)3(cid:1)5(cid:1)5(cid:1)7(cid:1)(cid:1)(cid:1) 1. =c ; 2. = + ; 3. =u +v ; dx dx dx dx dx dx dx dx (cid:0) (cid:1) (cid:0) (cid:1) Brouncker’scontinuedfractionexpansion: (cid:25) 12 4. d(un) =nun−1du; 5. d(u=v) = v ddux −u ddxv ; 6. d(ecu) =cecudu; 4 =1+ 2+ 32 dx dx dx v2 dx dx 2+2+522+72(cid:1)(cid:1)(cid:1) 7. d(cu) =(lnc)cudu; 8. d(lnu) = 1du; dx dx dx udx Gregrory’s series: (cid:25)4 =1− 13 + 15 − 17 + 19 −(cid:1)(cid:1)(cid:1) 9. d(sinu) =cosudu; 10. d(cosu) =−sinudu; dx dx dx dx Newton’s series: d(tanu) du d(cotu) du (cid:25) = 1 + 1 + 1(cid:1)3 +(cid:1)(cid:1)(cid:1) 11. dx =sec2udx; 12. dx =csc2udx; 6 2 2(cid:1)3(cid:1)23 2(cid:1)4(cid:1)5(cid:1)25 d(secu) du d(cscu) du 13. =tanu secu ; 14. =−cotu cscu ; Sharp’s series: dx dx dx dx (cid:25) = p1 (cid:16)1− 1 + 1 − 1 +(cid:1)(cid:1)(cid:1)(cid:17) 15. d(arcsinu) = p 1 du; 16. d(arccosu) = p −1 du; 6 3 31(cid:1)3 32(cid:1)5 33(cid:1)7 dx 1−u2dx dx 1−u2dx d(arctanu) 1 du d(arccotu) −1 du Euler’s series: 17. = ; 18. = ; dx 1+u2dx dx 1+u2dx (cid:25)62 = 112 + 212 + 312 + 412 + 512 +(cid:1)(cid:1)(cid:1) 19. d(arcsecu) = p 1 du; 20. d(arccscu) = p−1 du; (cid:25)2 = 1 + 1 + 1 + 1 + 1 +(cid:1)(cid:1)(cid:1) dx u 1−u2dx dx u 1−u2dx 8 12 32 52 72 92 (cid:25)122 = 112 − 212 + 312 − 412 + 512 −(cid:1)(cid:1)(cid:1) 21. d(sinhu) =coshudu; 22. d(coshu) =sinhudu; dx dx dx dx Partial Fractions d(tanhu) du d(cothu) du 23. =sech2u ; 24. =−csch2u ; Let N(x) and D(x) be polynomial func- dx dx dx dx tions of x. We can break down d(sechu) du d(cschu) du N(x)=D(x) using partial fraction expan- 25. =−sechu tanhu ; 26. =−cschu cothu ; dx dx dx dx sion. First, if the degree of N is greater d(arcsinhu) 1 du d(arccoshu) 1 du than or equal to the degree of D, divide 27. = p ; 28. = p ; N by D, obtaining dx 1+u2dx dx u2−1dx N(x) N0(x) d(arctanhu) 1 du d(arccothu) 1 du =Q(x)+ ; 29. = ; 30. = ; D(x) D(x) dx 1−u2dx dx u2−1dx wherethedegreeofN0 islessthanthatof d(arcsechu) −1 du d(arccschu) −1 du 31. = p ; 32. = p : D. Second, factor D(x). Use the follow- dx u 1−u2dx dx juj 1+u2dx ing rules: For a non-repeated factor: Integrals: N(x) A N0(x) Z Z Z Z Z = + ; (x−a)D(x) x−a D(x) 1. cudx=c udx; 2. (u+v)dx= udx+ vdx; where (cid:20) (cid:21) Z Z Z A= N(x) : 3. xndx= 1 xn+1; n6=−1; 4. 1dx=lnx; 5. exdx=ex; D(x) n+1 x x=a Z Z Z dx dv du For a repeated factor: 6. =arctanx; 7. u dx=uv− v dx; N(x) mX−1 Ak N0(x) Z 1+x2 dx Z dx = + ; (x−a)mD(x) k=0 (x−a)m−k D(x) 8. sinxdx=−cosx; 9. cosxdx=sinx; Z Z where (cid:20) (cid:18) (cid:19)(cid:21) 1 dk N(x) 10. tanxdx=−lnjcosxj; 11. cotxdx=lnjcosxj; Ak = k! dxk D(x) x=a: Z Z 12. secxdx=lnjsecx+tanxj; 13. cscxdx=lnjcscx+cotxj; Thereasonablemanadaptshimselftothe Z world; the unreasonable persists in trying p to adapt the world to himself. Therefore 14. arcsinxdx=arcsinx + a2−x2; a>0; a a all progress dependson the unreasonable. { George Bernard Shaw Theoretical Computer Science Cheat Sheet Calculus Cont. Z Z p 15. arccosxdx=arccosx − a2−x2; a>0; 16. arctanxdx=xarctanx − aln(a2+x2); a>0; a a a a 2 Z Z (cid:0) (cid:1) (cid:0) (cid:1) 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. sec2xdx=tanx; 20. csc2xdx=−cotx; Z Z Z Z sinn−1xcosx n−1 cosn−1xsinx n−1 21. sinnxdx=− + sinn−2xdx; 22. cosnxdx= + cosn−2xdx; n n n n Z Z Z Z tann−1x cotn−1x 23. tannxdx= − tann−2xdx; n6=1; 24. cotnxdx=− − cotn−2xdx; n6=1; n−1 n−1 Z Z tanxsecn−1x n−2 25. secnxdx= + secn−2xdx; n6=1; n−1 n−1 Z Z Z Z cotxcscn−1x n−2 26. cscnxdx=− + cscn−2xdx; n6=1; 27. sinhxdx=coshx; 28. coshxdx=sinhx; n−1 n−1 Z Z Z Z (cid:12) (cid:12) 29. tanhxdx=lnjcoshxj; 30. cothxdx=lnjsinhxj; 31. sechxdx=arctansinhx; 32. cschxdx=ln(cid:12)tanhx(cid:12); 2 Z Z Z 33. sinh2xdx= 1sinh(2x)− 1x; 34. cosh2xdx= 1sinh(2x)+ 1x; 35. sech2xdx=tanhx; 4 2 4 2 Z Z p 36. arcsinhxdx=xarcsinhx − x2+a2; a>0; 37. arctanhxdx=xarctanhx + alnja2−x2j; a a a a 2 8 p Z <xarccoshx − x2+a2; if arccoshx >0 and a>0, 38. arccoshxdx= a p a a : x xarccosh + x2+a2; if arccoshx <0 and a>0, a a Z (cid:16) p (cid:17) dx 39. p =ln x+ a2+x2 ; a>0; a2+x2 Z Z p p 40. dx = 1 arctanx; a>0; 41. a2−x2dx= x a2−x2+ a2 arcsinx; a>0; a2+x2 a a 2 2 a Z p 42. (a2−x2)3=2dx= x(5a2−2x2) a2−x2+ 3a4 arcsinx; a>0; 8 8 a Z Z (cid:12) (cid:12) Z (cid:12) (cid:12) 43. pa2dx−x2 =arcsinxa; a>0; 44. a2d−xx2 = 21aln(cid:12)(cid:12)aa−+xx(cid:12)(cid:12); 45. (a2−dxx2)3=2 = a2pax2−x2; Z p p (cid:12) p (cid:12) Z (cid:12) p (cid:12) 46. a2(cid:6)x2dx= x a2(cid:6)x2(cid:6) a2 ln(cid:12)(cid:12)x+ a2(cid:6)x2(cid:12)(cid:12); 47. p dx =ln(cid:12)(cid:12)x+ x2−a2(cid:12)(cid:12); a>0; 2 2 x2−a2 Z (cid:12) (cid:12) Z 48. dx = 1ln(cid:12)(cid:12)(cid:12) x (cid:12)(cid:12)(cid:12); 49. xpa+bxdx= 2(3bx−2a)(a+bx)3=2; ax2+bx a a+bx 15b2 Z p Z Z (cid:12)p p (cid:12) 50. a+bxdx=2pa+bx+a p 1 dx; 51. p x dx= p1 ln(cid:12)(cid:12)(cid:12)pa+bx−pa(cid:12)(cid:12)(cid:12); a>0; Z p x p x(cid:12)(cid:12) a+pbx (cid:12)(cid:12) a+bx Z p2 a+bx+ a a2−x2 (cid:12)a+ a2−x2(cid:12) 52. dx= a2−x2−aln(cid:12) (cid:12); 53. x a2−x2dx=−1(a2−x2)3=2; x (cid:12) x (cid:12) 3 Z p p Z (cid:12)(cid:12) p (cid:12)(cid:12) 54. x2 a2−x2dx= x(2x2−a2) a2−x2+ a4 arcsinx; a>0; 55. p dx =−1 ln(cid:12)(cid:12)a+ a2−x2(cid:12)(cid:12); 8 8 a a2−x2 a (cid:12) x (cid:12) Z Z p p 56. pxdx =− a2−x2; 57. px2dx =−x a2−x2+ a2 arcsin x a>0; Z pa2−x2 p (cid:12)(cid:12) p (cid:12)(cid:12) Z pa2−x2 2p 2 a; a2+x2 (cid:12)a+ a2+x2(cid:12) x2−a2 58. dx= a2+x2−aln(cid:12) (cid:12); 59. dx= x2−a2−aarccos a ; a>0; x (cid:12) x (cid:12) x jxj Z Z (cid:12) (cid:12) p (cid:12) (cid:12) 60. x x2(cid:6)a2dx= 13(x2(cid:6)a2)3=2; 61. xpxd2x+a2 = a1 ln(cid:12)(cid:12)a+pax2+x2(cid:12)(cid:12); Theoretical Computer Science Cheat Sheet Calculus Cont. Finite Calculus Z Z p dx dx x2(cid:6)a2 Di(cid:11)erence, shift operators: 62. p = 1 arccos a ; a>0; 63. p =(cid:7) ; x x2−a2 a jxj x2 x2(cid:6)a2 a2x (cid:1)f(x)=f(x+1)−f(x); Z Z p xdx p x2(cid:6)a2 (x2+a2)3=2 Ef(x)=f(x+1): 64. p = x2(cid:6)a2; 65. dx=(cid:7) ; x2(cid:6)a2 8 (cid:12) p x4 (cid:12) 3a2x3 Fundamental TheoXrem: (cid:12) (cid:12) Z >>><p 1 ln(cid:12)(cid:12)2ax+b−pb2−4ac(cid:12)(cid:12); if b2 >4ac, f(x)=(cid:1)F(x), f(x)(cid:14)x=F(x)+C: 66. dx = b2−4ac (cid:12)2ax+b+ b2−4ac(cid:12) Xb Xb−1 ax2+bx+c >>>:p 2 arctanp2ax+b ; if b2 <4ac, f(x)(cid:14)x= f(i): 4ac−b2 4ac−b2 a i=a Di(cid:11)erences: Z 8>><p1 ln(cid:12)(cid:12)(cid:12)2ax+b+2papax2+bx+c(cid:12)(cid:12)(cid:12); if a>0, (cid:1)(cu)=c(cid:1)u; (cid:1)(u+v)=(cid:1)u+(cid:1)v; dx a 67. p = (cid:1)(uv)=u(cid:1)v+Ev(cid:1)u; ax2+bx+c >>:p1 arcsinp−2ax−b ; if a<0, (cid:1)(xn)=nxn−1; −a b2−4ac Z p 2ax+bp 4ax−b2 Z dx (cid:1)(Hx)=x−1; (cid:0)(cid:1)(cid:1)(2x)(cid:0)=2x(cid:1); 68. ax2+bx+cdx= ax2+bx+c+ p ; (cid:1)(cx)=(c−1)cx; (cid:1) x = x : 4a 8a ax2+bx+c m m−1 Sums: Z p Z P P xdx ax2+bx+c b dx cu(cid:14)x=c u(cid:14)x; 69. p = − p ; ax2+bx+c a 2a ax2+bx+c P P P (u+v)(cid:14)x= u(cid:14)x+ v(cid:14)x; 8 (cid:12)(cid:12) p p (cid:12)(cid:12) P P Z >>><−p1ln(cid:12)(cid:12)2 c ax2+bx+c+bx+2c(cid:12)(cid:12); if c>0, u(cid:1)v(cid:14)x=uv− Ev(cid:1)u(cid:14)x; dx c (cid:12) x (cid:12) P P 70. Z xpax2+bx+c =>>>:p1−carcsinjxjpbxb2+−2c4ac; if c<0, Pcxxn(cid:14)(cid:14)xx==cxmc−nx+1+;11; P(cid:0)mx(cid:1)x(cid:14)−x1(cid:14)=x(cid:0)=mx+H1x(cid:1);: p Falling Factorial Powers: 71. x3 x2+a2dx=(13x2− 125a2)(x2+a2)3=2; xn =x(x−1)(cid:1)(cid:1)(cid:1)(x−n+1); n>0; Z Z x0 =1; 72. xnsin(ax)dx=−1xncos(ax)+ n xn−1cos(ax)dx; a a 1 n Z Z x = ; n<0; (x+1)(cid:1)(cid:1)(cid:1)(x+jnj) 73. xncos(ax)dx= a1xnsin(ax)− na xn−1sin(ax)dx; xn+m =xm(x−m)n: Z Z xneax Rising Factorial Powers: 74. xneaxdx= − n xn−1eaxdx; a a xn =x(x+1)(cid:1)(cid:1)(cid:1)(x+n−1); n>0; Z (cid:18) (cid:19) 75. xnln(ax)dx=xn+1 ln(ax) − 1 ; x0 =1; n+1 (n+1)2 1 Z Z xn = ; n<0; 76. xn(lnax)mdx= xn+1(lnax)m− m xn(lnax)m−1dx: (x−1)(cid:1)(cid:1)(cid:1)(x−jnj) n+1 n+1 xn+m =xm(x+m)n: Conversion: x1 = x1 = x1 xn =(−1)n(−x)n =(x−n+1)n x2 = x2+x1 = x2−x1 =1=(x+1)−n; x3 = x3+3x2+x1 = x3−3x2+x1 xn =(−1)n(−x)n =(x+n−1)n x4 = x4+6x3+7x2+x1 = x4−6x3+7x2−x1 =1=(x−1)−n; (cid:26) (cid:27) (cid:26) (cid:27) x5 = x5+15x4+25x3+10x2+x1 = x5−15x4+25x3−10x2+x1 Xn n Xn n xn = xk = (−1)n−kxk; k k x1 = x1 x1 = x1 k=1(cid:20) (cid:21) k=1 Xn x2 = x2+x1 x2 = x2−x1 xn = n (−1)n−kxk; x3 = x3+3x2+2x1 x3 = x3−3x2+2x1 k=1(cid:20)k(cid:21) x4 = x4+6x3+11x2+6x1 x4 = x4−6x3+11x2−6x1 Xn n n k x = x : x5 = x5+10x4+35x3+50x2+24x1 x5 = x5−10x4+35x3−50x2+24x1 k k=1 Theoretical Computer Science Cheat Sheet Series Taylor’s series: Ordinary power series: (x−a)2 X1 (x−a)i X1 f(x)=f(a)+(x−a)f0(a)+ 2 f00(a)+(cid:1)(cid:1)(cid:1)= i! f(i)(a): A(x)= aixi: Expansions: i=0 i=0 X1 Exponential power series: 1 1−x =1+x+x2+x3+x4+(cid:1)(cid:1)(cid:1) = xi; X1 xi i=0 A(x)= ai : X1 i! 1 =1+cx+c2x2+c3x3+(cid:1)(cid:1)(cid:1) = cixi; i=0 1−cx Dirichlet power series: i=0 X1 1 X1 ai 1−xn =1+xn+x2n+x3n+(cid:1)(cid:1)(cid:1) = i=0xni; A(x)= i=1 ix: X1 Binomial theorem: x =x+2x2+3x3+4x4+(cid:1)(cid:1)(cid:1) = ixi; Xn (cid:18) (cid:19) (1−x)2 n n n−k k (cid:18) (cid:19) i=0 (x+y) = x y : xk dn 1 =x+2nx2+3nx3+4nx4+(cid:1)(cid:1)(cid:1) =X1 inxi; k=0 k dxn 1−x Di(cid:11)erence of like powers: ex =1+x+ 1x2+ 1x3+(cid:1)(cid:1)(cid:1) =Xi1=0 xi; xn−yn =(x−y)nX−1xn−1−kyk: 2 6 i! i=0 k=0 X1 xi For ordinary power series: ln(1+x) =x− 1x2+ 1x3− 1x4−(cid:1)(cid:1)(cid:1) = (−1)i+1 ; X1 2 3 4 i i=1 (cid:11)A(x)+(cid:12)B(x)= ((cid:11)ai+(cid:12)bi)xi; ln 1 =x+ 1x2+ 1x3+ 1x4+(cid:1)(cid:1)(cid:1) =X1 xi; i=0 1−x 2 3 4 i X1 Xi1=1 x2i+1 xkA(x)= ai−kxi; sinx =x− 1x3+ 1x5− 1x7+(cid:1)(cid:1)(cid:1) = (−1)i ; i=k cosx =1− 231!!x2+ 451!!x4− 671!!x6+(cid:1)(cid:1)(cid:1) =Xi1=0(−1)i((x222ii)i+!;1)! A(x)−Pxkki=−01aixXi1=Xi1=0ai+kxi; i=0 X1 x2i+1 A(cx)= ciaixi; tan−1x =x− 13x3+ 15x5− 17x7+(cid:1)(cid:1)(cid:1) = (−1)i(2i+1); i=0 i=0(cid:18) (cid:19) X1 X1 0 i (1+x)n =1+nx+ n(n−1)x2+(cid:1)(cid:1)(cid:1) = n xi; A(x)= (i+1)ai+1x ; 2 i i=0 i=0(cid:18) (cid:19) X1 (cid:0) (cid:1) X1 1 =1+(n+1)x+ n+2 x2+(cid:1)(cid:1)(cid:1) = i+n xi; xA0(x)= iaixi; (1−x)n+1 2 i=0 i Z i=1 x =1− 1x+ 1 x2− 1 x4+(cid:1)(cid:1)(cid:1) =X1 Bixi; A(x)dx=X1 ai−1xi; ex−1 2 12 720 i! i i=0 (cid:18) (cid:19) i=1 21x(1−p1−4x) =1+x+2x2+5x3+(cid:1)(cid:1)(cid:1) =X1 i+11 2ii xi; A(x)+2A(−x) =X1 a2ix2i; Xi1=0(cid:18) (cid:19) i=0 p(cid:18)1−1 4px (cid:19) =1+x+2x2+6x3+(cid:1)(cid:1)(cid:1) = i=0(cid:18)2ii xi;(cid:19) A(x)−2A(−x) =X1 a2i+1x2i+1: 1 1− 1−4x n (cid:0) (cid:1) X1 2i+n i=P0 p1−4x 2x =1+(2+n)x+ 4+2n x2+(cid:1)(cid:1)(cid:1) = i xi; Summation: If bi = ij=0ai then Xi1=0 1 1 1 B(x)= A(x): 1−xln1−x =x+ 32x2+ 161x3+ 2152x4+(cid:1)(cid:1)(cid:1) = Hixi; 1−x (cid:18) (cid:19) i=1 Convolution: 0 1 12 ln1−1 x 2 = 12x2+ 34x3+ 1214x4+(cid:1)(cid:1)(cid:1) =X1 Hi−i1xi; A(x)B(x)=X1 @Xi ajbi−jAxi: i=2 X1 i=0 j=0 x 1−x−x2 =x+x2+2x3+3x4+(cid:1)(cid:1)(cid:1) = Fixi; i=0 God made the natural numbers; X1 1−(Fn−1+FFnn+x1)x−(−1)nx2 =Fnx+F2nx2+F3nx3+(cid:1)(cid:1)(cid:1) = i=0Fnixi: a{llLtehoeporledstKisrotnheeckweorrk of man. Theoretical Computer Science Cheat Sheet Series Escher’s Knot Expansions: (cid:18) (cid:19) (cid:18) (cid:19) (cid:26) (cid:27) X1 −n X1 1 1 n+i 1 i (1−x)n+1 ln1−x = (Hn+i−Hn) i xi; x = n xi; i=0(cid:20) (cid:21) i=0(cid:26) (cid:27) X1 n X1 i n!xi xn = xi; (ex−1)n = ; i n i! (cid:18) (cid:19) i=0(cid:20) (cid:21) i=0 ln 1 n =X1 i n!xi; xcotx =X1 (−4)iB2ix2i; 1−x n i! (2i)! i=0 i=0 tanx =X1 (−1)i−122i(22i−1)B2ix2i−1; (cid:16)(x) =X1 1; (2i)! ix i=1 i=1 1 X1 (cid:22)(i) (cid:16)(x−1) X1 (cid:30)(i) = ; = ; (cid:16)(x) ix (cid:16)(x) ix Yi=1 i=1 1 (cid:16)(x) = 1−p−x; Stieltjes Integration p If G is continuous in the interval [a;b] and F is nondecreasing then (cid:16)2(x) =X1 d(i) where d(n)=P 1; Z b xi djn G(x)dF(x) i=1 a X1 S(i) P exists. If a(cid:20)b(cid:20)c then (cid:16)(x)(cid:16)(x−1) = where S(n)= d; Z Z Z xi djn c b c i=1 G(x)dF(x)= G(x)dF(x)+ G(x)dF(x): (cid:16)(2n) = 22n−1jB2nj(cid:25)2n; n2N; If the integarals involved exisat b (2n)! Z Z Z b(cid:0) (cid:1) b b x =X1 (−1)i−1(4i−2)B2ix2i; G(x)+H(x) dF(x)= G(x)dF(x)+ H(x)dF(x); sinx (2i)! Za Za Za (cid:18) p (cid:19) i=0 b (cid:0) (cid:1) b b 1− 1−4x n X1 n(2i+n−1)! G(x)d F(x)+H(x) = G(x)dF(x)+ G(x)dH(x); i = x ; a a a 2x i!(n+i)! Z b Z b (cid:0) (cid:1) Z b i=0 X1 2i=2sini(cid:25) c(cid:1)G(x)dF(x)= G(x)d c(cid:1)F(x) =c G(x)dF(x); exsinx = 4 xi; Za a Z a i! b b s i=1 G(x)dF(x)=G(b)F(b)−G(a)F(a)− F(x)dG(x): p 1− 1−x =X1 p (4i)! xi; Iftheinategralsinvolvedexist,andF possessesadearivativeF0 atevery (cid:18) x (cid:19) i=0 16i 2(2i)!(2i+1)! point in [a;b] theZn Z arcsinx 2 X1 4ii!2 b b = x2i: G(x)dF(x)= G(x)F0(x)dx: x i=0 (i+1)(2i+1)! a a Cramer’s Rule Fibonacci Numbers 00 47 18 76 29 93 85 34 61 52 If we have equations: a1;1x1+a1;2x2+(cid:1)(cid:1)(cid:1)+a1;nxn =b1 8965 1810 5272 2687 7308 3791 9449 4556 0123 6034 1;1;2;3;5;8;13;21;34;55;89;::: a2;1x1+a2;2x2+(cid:1)(cid:1)(cid:1)+a2;nxn =b2 59 96 81 33 07 48 72 60 24 15 De(cid:12)nitions: . . . 73 69 90 82 44 17 58 01 35 26 Fi =Fi−1+Fi−2; F0 =F1 =1; . . . . . . 68 74 09 91 83 55 27 12 46 30 F−i =(−(cid:16)1)i−1Fi(cid:17); an;1x1+an;2x2+(cid:1)(cid:1)(cid:1)+an;nxn =bn 37 08 75 19 92 84 66 23 50 41 Fi = p1 (cid:30)i−(cid:30)^i ; 14 25 36 40 51 62 03 77 88 99 5 LetA=(ai;j)andB bethecolumnmatrix(bi). Then wthiethreciosluamunniiqureepslaocluedtiobnyiB(cid:11).dTethAen6= 0. Let Ai be A 2412 3523 4634 5045 6156 0260 1301 8998 9779 7887 CasFsiin+i1’sFii−de1n−titFyi2: =for(−i>1)i0:: detAi Additive rule: xi = : TheFibonaccinumbersystem: detA 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 +(cid:1)(cid:1)(cid:1)+Fkm; C(cid:18)alculation by(cid:19)matr(cid:18)ices: (cid:19) r{oWadisllwiaimthoBulatkIem(pTrhoveeMmaenrrti,aagreeorfoHadesavoefnGaenndiuHs.ell) w1h(cid:20)erie<kim(cid:21)ankdi+k1m+(cid:21)22f.or all i, FFnn−−21 FFn−n1 = 01 11 n:

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