Tai yra sąrašas matematinių eilučių talpinančios formulę baigtinėms ir begalinėms sumoms. Tai gali būti naudota konjunkcijoje su kitais įrankiais evaluacinėms sumoms.
Galios sumos
keisti
∑ m = 1 n m = n ( n + 1 ) 2 {\displaystyle \sum _{m=1}^{n}m={\frac {n(n+1)}{2}}\,\!} ∑ m = 1 n m 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = n 3 3 + n 2 2 + n 6 {\displaystyle \sum _{m=1}^{n}m^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\,\!} ∑ m = 1 n m 3 = [ n ( n + 1 ) 2 ] 2 = n 4 4 + n 3 2 + n 2 4 = ( ∑ m = 1 n m ) 2 {\displaystyle \sum _{m=1}^{n}m^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left(\sum _{m=1}^{n}m\right)^{2}\,\!} ∑ m = 1 n m 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 = 6 n 5 + 15 n 4 + 10 n 3 − n 30 {\displaystyle \sum _{m=1}^{n}m^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {6n^{5}+15n^{4}+10n^{3}-n}{30}}\,\!} ∑ m = 0 n m s = ( n + 1 ) s + 1 s + 1 + ∑ k = 1 s B k s − k + 1 ( s k ) ( n + 1 ) s − k + 1 {\displaystyle \sum _{m=0}^{n}m^{s}={\frac {(n+1)^{s+1}}{s+1}}+\sum _{k=1}^{s}{\frac {B_{k}}{s-k+1}}{s \choose k}(n+1)^{s-k+1}\,\!} where B k {\displaystyle B_{k}\,} is the k {\displaystyle k\,} th Bernoulli number , and B 1 {\displaystyle B_{1}\,} is negative. ∑ m = 1 ∞ 1 m 2 = π 2 6 {\displaystyle \sum _{m=1}^{\infty }{\frac {1}{m^{2}}}={\frac {\pi ^{2}}{6}}\,\!}
∑ m = 1 ∞ 1 m 4 = π 4 90 {\displaystyle \sum _{m=1}^{\infty }{\frac {1}{m^{4}}}={\frac {\pi ^{4}}{90}}\,\!}
∑ m = 1 ∞ 1 m 2 n = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \sum _{m=1}^{\infty }{\frac {1}{m^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} ∑ m = 1 ∞ m − s = ∏ p prime 1 1 − p − s = ζ ( s ) {\displaystyle \sum _{m=1}^{\infty }m^{-s}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}=\zeta (s)\,\!} where ζ ( s ) {\displaystyle \zeta (s)\,} is the Riemann zeta function . Galios eilutės
keisti
begalinė suma (for | x | < 1 {\displaystyle |x|<1} )
∑ m = 0 ∞ x m = 1 1 − x {\displaystyle \sum _{m=0}^{\infty }x^{m}={\frac {1}{1-x}}\,\!}
∑ m = 0 n x m = 1 − x n + 1 1 − x = 1 + 1 r ( 1 − 1 ( 1 + r ) n ) {\displaystyle \sum _{m=0}^{n}x^{m}={\frac {1-x^{n+1}}{1-x}}=1+{\frac {1}{r}}\left(1-{\frac {1}{(1+r)^{n}}}\right)} where r > 0 {\displaystyle r>0} and x = 1 1 + r . {\displaystyle x={\frac {1}{1+r}}.\,\!}
∑ m = 0 ∞ x 2 m = 1 1 − x 2 {\displaystyle \sum _{m=0}^{\infty }x^{2m}={\frac {1}{1-x^{2}}}\,\!}
∑ m = 1 ∞ m x m = x ( 1 − x ) 2 {\displaystyle \sum _{m=1}^{\infty }mx^{m}={\frac {x}{(1-x)^{2}}}\,\!}
∑ m = 1 n m x m = x 1 − x n ( 1 − x ) 2 − n x n + 1 1 − x {\displaystyle \sum _{m=1}^{n}mx^{m}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}\,\!}
∑ m = 1 ∞ m 2 x m = x ( 1 + x ) ( 1 − x ) 3 {\displaystyle \sum _{m=1}^{\infty }m^{2}x^{m}={\frac {x(1+x)}{(1-x)^{3}}}\,\!}
∑ m = 1 n m 2 x m = x ( 1 + x − ( n + 1 ) 2 x n + ( 2 n 2 + 2 n − 1 ) x n + 1 − n 2 x n + 2 ) ( 1 − x ) 3 {\displaystyle \sum _{m=1}^{n}m^{2}x^{m}={\frac {x(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}{(1-x)^{3}}}\,\!}
∑ m = 1 ∞ m 3 x m = x ( 1 + 4 x + x 2 ) ( 1 − x ) 4 {\displaystyle \sum _{m=1}^{\infty }m^{3}x^{m}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}\,\!}
∑ m = 1 ∞ m 4 x m = x ( 1 + x ) ( 1 + 10 x + x 2 ) ( 1 − x ) 5 {\displaystyle \sum _{m=1}^{\infty }m^{4}x^{m}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}\,\!}
∑ m = 1 ∞ m k x m = Li − k ( x ) , {\displaystyle \sum _{m=1}^{\infty }m^{k}x^{m}=\operatorname {Li} _{-k}(x),\,\!} where Lis (x ) is the polylogarithm of x .
arctan x = x − x 3 3 + x 5 5 − x 7 7 + x 9 9 − x 11 11 + . . . ( | x | < 1 ) . {\displaystyle \arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...\quad (|x|<1).}
Kaip yra žinoma,
1 1 − x = 1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n + . . . , k a i | x | < 1. {\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1.}
Dabar vietoje x įstatome − x 2 {\displaystyle -x^{2}} ir gauname lygybę:
1 1 − ( − x 2 ) = 1 1 + x 2 = 1 − x 2 + x 4 − x 6 + x 8 − x 10 + . . . + ( − 1 ) n x 2 n + . . . , {\displaystyle {\frac {1}{1-(-x^{2})}}={\frac {1}{1+x^{2}}}=1-x^{2}+x^{4}-x^{6}+x^{8}-x^{10}+...+(-1)^{n}x^{2n}+...,}
Teisginai kai | − x 2 | < 1. {\displaystyle |-x^{2}|<1.}
Integruodami gauname:
arctan x = ∫ 0 x d x 1 + x 2 = x − x 3 3 + x 5 5 − x 7 7 + x 9 9 − x 11 11 + . . . + ( − 1 ) n + 1 x 2 n − 1 2 n − 1 + . . . , {\displaystyle \arctan x=\int _{0}^{x}{\frac {dx}{1+x^{2}}}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...+(-1)^{n+1}{\frac {x^{2n-1}}{2n-1}}+...,}
Iš kur
arctan x = x − x 3 3 + x 5 5 − x 7 7 + x 9 9 − x 11 11 + . . . + ( − 1 ) n + 1 x 2 n − 1 2 n − 1 + . . . ( | x | < 1 ) . {\displaystyle \arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...+(-1)^{n+1}{\frac {x^{2n-1}}{2n-1}}+...\quad (|x|<1).}
Iš to galime įrodyti Leibnico formulę:
1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 . {\displaystyle {\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}.}
Kadangi sin π 4 cos π 4 = 1 = tan π 4 , {\displaystyle {\frac {\sin {\frac {\pi }{4}}}{\cos {\frac {\pi }{4}}}}=1=\tan {\frac {\pi }{4}},} tai arctan 1 = π 4 . {\displaystyle \arctan 1={\frac {\pi }{4}}.} Todėl:
arctan 1 = ∫ 0 1 d x 1 + x 2 = 1 − 1 3 3 + 1 5 5 − 1 7 7 + 1 9 9 − 1 11 11 + . . . + ( − 1 ) n + 1 1 2 n − 1 2 n − 1 + . . . = {\displaystyle \arctan 1=\int _{0}^{1}{\frac {dx}{1+x^{2}}}=1-{\frac {1^{3}}{3}}+{\frac {1^{5}}{5}}-{\frac {1^{7}}{7}}+{\frac {1^{9}}{9}}-{\frac {1^{11}}{11}}+...+(-1)^{n+1}{\frac {1^{2n-1}}{2n-1}}+...=}
= 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 ≈ 0.785398163. {\displaystyle ={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}\approx 0.785398163.}
1 1 − x = 1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n + . . . , k a i | x | < 1 , {\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1,}
įstatome− x {\displaystyle -x} vietoje x ir gauname:
1 1 − ( − x ) = 1 1 + x = 1 − x + x 2 − x 3 + x 4 − x 5 + . . . + ( − 1 ) n x n + . . . , {\displaystyle {\frac {1}{1-(-x)}}={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+...+(-1)^{n}x^{n}+...,}
kai | − x | < 1 , {\displaystyle |-x|<1,} tada abi puses integruojame:
∫ 0 x 1 1 + x d x = ∫ 0 x ( 1 − x + x 2 − x 3 + x 4 − x 5 + . . . + ( − 1 ) n x n + . . . ) d x , {\displaystyle \int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}x=\int _{0}^{x}(1-x+x^{2}-x^{3}+x^{4}-x^{5}+...+(-1)^{n}x^{n}+...){\mathsf {d}}x,}
ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + x 5 5 − x 6 6 + . . . + ( − 1 ) n x n + 1 n + 1 + . . . , k a i | x | < 1 , {\displaystyle \ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-{\frac {x^{6}}{6}}+...+(-1)^{n}{\frac {x^{n+1}}{n+1}}+...,\quad kai\;\;|x|<1,}
nes ∫ 0 x 1 1 + x d x = ∫ 0 x 1 1 + x d ( 1 + x ) = ln ( 1 + x ) | 0 x = ln ( 1 + x ) − ln ( 1 + 0 ) = ln ( 1 + x ) − 0 = ln ( 1 + x ) . {\displaystyle \int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}x=\int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}(1+x)=\ln(1+x)|_{0}^{x}=\ln(1+x)-\ln(1+0)=\ln(1+x)-0=\ln(1+x).} Pasirenkame x = 0.3 {\displaystyle x=0.3} ir įstatome: 1 1 − x = 1 1 − 0.3 = 1 0.7 = 1.428571429. {\displaystyle {\frac {1}{1-x}}={\frac {1}{1-0.3}}={\frac {1}{0.7}}=1.428571429.}
Prasumuojant gauname:
1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n = 1 + 0.3 + 0.3 2 + 0.3 3 + 0.3 4 + 0.3 5 + 0.3 6 + 0.3 7 + 0.3 8 + 0.3 9 + 0.3 10 = 1.428568898. {\displaystyle 1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}=1+0.3+0.3^{2}+0.3^{3}+0.3^{4}+0.3^{5}+0.3^{6}+0.3^{7}+0.3^{8}+0.3^{9}+0.3^{10}=1.428568898.} Pasirenkame x = − 0.3 {\displaystyle x=-0.3} ir įstatome: 1 1 − x = 1 1 − ( − 0.3 ) = 1 1.3 = 0.769230769. {\displaystyle {\frac {1}{1-x}}={\frac {1}{1-(-0.3)}}={\frac {1}{1.3}}=0.769230769.}
Prasumuojant gauname:
1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n = 1 − 0.3 + 0.3 2 − 0.3 3 + 0.3 4 − 0.3 5 + 0.3 6 − 0.3 7 + 0.3 8 − 0.3 9 + 0.3 10 = 0.769232131. {\displaystyle 1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}=1-0.3+0.3^{2}-0.3^{3}+0.3^{4}-0.3^{5}+0.3^{6}-0.3^{7}+0.3^{8}-0.3^{9}+0.3^{10}=0.769232131.}
1 1 − x = 1 + x + x 2 + x 3 + x 4 + x 5 + . . . + x n + . . . , k a i | x | < 1. {\displaystyle {\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1.}
Tegu s = 1 + x + x 2 + x 3 + ⋯ . Tada x ⋅ s = x + x 2 + x 3 + x 4 + ⋯ . Tada x ⋅ s − s = x ∞ − 1 = − 1 , nes | x | < 1 , taigi s ( x − 1 ) = − 1 , ir todel s = − 1 x − 1 = − 1 − ( 1 − x ) = 1 1 − x . {\displaystyle {\begin{aligned}&{\text{Tegu }}\;s=1+x+x^{2}+x^{3}+\cdots .\\[4pt]&{\text{Tada }}\;x\cdot s=x+x^{2}+x^{3}+x^{4}+\cdots .\\[4pt]&{\text{Tada }}\;x\cdot s-s=x^{\infty }-1=-1,{\text{ nes}}\;|x|<1,\;\;{\text{ taigi }}\;s(x-1)=-1,{\text{ ir todel }}\;s={\frac {-1}{x-1}}={\frac {-1}{-(1-x)}}={\frac {1}{1-x}}.\end{aligned}}}
Išraišką x ∞ − 1 {\displaystyle x^{\infty }-1} galima apibūdinti tiksliau, vietoje x ∞ , {\displaystyle x^{\infty },} parašydami x n {\displaystyle x^{n}} , kur n koks nors labai didelis sveikasis skaičius, pavyzdžiui, n = 1000 {\displaystyle n=1000} . Pavyzdžiui, kai x = 0.9 {\displaystyle x=0.9} , tai 0.9 100 = 0.000026561 , {\displaystyle 0.9^{100}=0.000026561,} o 0.9 1000 = 1.7478712 10 46 , {\displaystyle 0.9^{1000}={\frac {1.7478712}{10^{46}}},} kas beveik lygu nuliui. Todėl lim n → ∞ 0.9 n = 0 {\displaystyle \lim _{n\to \infty }0.9^{n}=0} arba lim n → ∞ x n = 0 , {\displaystyle \lim _{n\to \infty }x^{n}=0,} kai -1<x<1. Paprasti denominatoriai
keisti
∑ m = 1 ∞ x m m = ln ( 1 1 − x ) for | x | < 1 {\displaystyle \sum _{m=1}^{\infty }{\frac {x^{m}}{m}}=\ln \left({\frac {1}{1-x}}\right)\quad {\mbox{ for }}|x|<1\!} ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 x 2 m + 1 = x − x 3 3 + x 5 5 − ⋯ = arctan ( x ) {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{2m+1}}x^{2m+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)\,\!} ∑ m = 0 ∞ x 2 m + 1 2 m + 1 = a r c t a n h ( x ) for | x | < 1 {\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m+1}}{2m+1}}=\mathrm {arctanh} (x)\quad {\mbox{ for }}|x|<1\,\!} Faktorialiniai denominatoriai
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Daug geometrinių eilučių kurios kyla iš Tailoro theoremos turi koficientą turintį faktorialą.
∑ m = 0 ∞ x m m ! = e x {\displaystyle \sum _{m=0}^{\infty }{\frac {x^{m}}{m!}}=e^{x}}
∑ m = 0 ∞ ( − 1 ) m m ! x m = 1 e x {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!}}x^{m}={\frac {1}{e^{x}}}} ∑ m = 0 ∞ m x m m ! = x e x {\displaystyle \sum _{m=0}^{\infty }m{\frac {x^{m}}{m!}}=xe^{x}} (c.f. mean of Poisson distribution )
∑ m = 0 ∞ m 2 x m m ! = ( x + x 2 ) e x {\displaystyle \sum _{m=0}^{\infty }m^{2}{\frac {x^{m}}{m!}}=(x+x^{2})e^{x}} (c.f. second moment of Poisson distribution)
∑ m = 0 ∞ m 3 x m m ! = ( x + 3 x 2 + x 3 ) e x {\displaystyle \sum _{m=0}^{\infty }m^{3}{\frac {x^{m}}{m!}}=(x+3x^{2}+x^{3})e^{x}}
∑ m = 0 ∞ m 4 x m m ! = ( x + 7 x 2 + 6 x 3 + x 4 ) e x {\displaystyle \sum _{m=0}^{\infty }m^{4}{\frac {x^{m}}{m!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}}
∑ m = 0 ∞ m n x m m ! = x d d x ∑ m = 0 ∞ m n − 1 x m m ! {\displaystyle \sum _{m=0}^{\infty }m^{n}{\frac {x^{m}}{m!}}=x{\frac {d}{dx}}\sum _{m=0}^{\infty }m^{n-1}{\frac {x^{m}}{m!}}} ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) ! x 2 m + 1 = x − x 3 3 ! + x 5 5 ! − ⋯ = sin x {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m+1)!}}x^{2m+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x} ∑ m = 0 ∞ ( − 1 ) m ( 2 m ) ! x 2 m = 1 − x 2 2 ! + x 4 4 ! − ⋯ = cos x {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m)!}}x^{2m}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x} ∑ m = 0 ∞ x 2 m + 1 ( 2 m + 1 ) ! = sinh x {\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m+1}}{(2m+1)!}}=\sinh x} ∑ m = 0 ∞ x 2 m ( 2 m ) ! = cosh x {\displaystyle \sum _{m=0}^{\infty }{\frac {x^{2m}}{(2m)!}}=\cosh x} Modifikuoti-faktorialo denominatoriai
keisti
∑ n = 0 ∞ ( 2 n ) ! 4 n ( n ! ) 2 ( 2 n + 1 ) x 2 n + 1 = arcsin x for | x | < 1 {\displaystyle \sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}=\arcsin x\quad {\mbox{ for }}|x|<1\!} ∑ m = 0 ∞ ( − 1 ) m ( 2 m ) ! 4 m ( m ! ) 2 ( 2 m + 1 ) x 2 m + 1 = a r c s i n h ( x ) for | x | < 1 {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m}(2m)!}{4^{m}(m!)^{2}(2m+1)}}x^{2m+1}=\mathrm {arcsinh} (x)\quad {\mbox{ for }}|x|<1\!} ∑ m = 0 ∞ ( 4 m ) ! 16 m 2 ( 2 m ) ! ( 2 m + 1 ) ! x m = 1 − 1 − x x {\displaystyle \sum _{m=0}^{\infty }{\frac {(4m)!}{16^{m}{\sqrt {2}}(2m)!(2m+1)!}}x^{m}={\sqrt {\frac {1-{\sqrt {1-x}}}{x}}}} ∑ m = 0 ∞ 4 m ( m ) ! 2 ( m + 1 ) ( 2 m + 1 ) ! x 2 m = ( arcsin x x ) 2 {\displaystyle \sum _{m=0}^{\infty }{\frac {4^{m}(m)!^{2}}{(m+1)(2m+1)!}}x^{2m}=\left({\frac {\arcsin {x}}{x}}\right)^{2}} ∑ m = 0 ∞ ∏ n = 0 m − 1 ( 4 n 2 + 1 ) ( 2 m ) ! x 2 m + ∑ m = 0 ∞ 4 m ∏ n = 1 m ( 1 2 − n + n 2 ) ( 2 m + 1 ) ! x 2 m + 1 = e arcsin x {\displaystyle \sum _{m=0}^{\infty }{\frac {\prod _{n=0}^{m-1}(4n^{2}+1)}{(2m)!}}x^{2m}+\sum _{m=0}^{\infty }{\frac {4^{m}\prod _{n=1}^{m}({\frac {1}{2}}-n+n^{2})}{(2m+1)!}}x^{2m+1}=e^{\arcsin {x}}} Binominės eilutės
keisti
Geometrinės eilutės :
( 1 + x ) − 1 = { ∑ m = 0 ∞ ( − x ) m | x | < 1 ∑ m = 1 ∞ − ( x ) − m | x | > 1 {\displaystyle (1+x)^{-1}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }(-x)^{m}&|x|<1\\\displaystyle \sum _{m=1}^{\infty }-(x)^{-m}&|x|>1\\\end{cases}}} Binominė Teorema :
( a + x ) n = { ∑ m = 0 ∞ ( n m ) a n − m x m | x | < | a | ∑ m = 0 ∞ ( n m ) a m x n − m | x | > | a | {\displaystyle (a+x)^{n}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{n-m}x^{m}&|x|\!<\!|a|\\\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{m}x^{n-m}&|x|\!>\!|a|\\\end{cases}}}
( 1 + x ) α = ∑ m = 0 ∞ ( α m ) x m for all | x | < 1 and all complex α {\displaystyle (1+x)^{\alpha }=\sum _{m=0}^{\infty }{\alpha \choose m}x^{m}\quad {\mbox{ for all }}|x|<1{\mbox{ and all complex }}\alpha \!} su apibendrintais binominiais koficientais
( α n ) = ∏ k = 1 n α − k + 1 k = α ( α − 1 ) ⋯ ( α − n + 1 ) n ! {\displaystyle {\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}\!} Šaknis :
1 + x = ∑ m = 0 ∞ ( − 1 ) m ( 2 m ) ! ( 1 − 2 m ) m ! 2 4 m x m for | x | < 1 {\displaystyle {\sqrt {1+x}}=\sum _{m=0}^{\infty }{\frac {(-1)^{m}(2m)!}{(1-2m)m!^{2}4^{m}}}x^{m}\quad {\mbox{ for }}|x|<1\!} Maišyta, įvairu:
[1] ∑ m = 0 ∞ ( m + n m ) x m = 1 ( 1 − x ) n + 1 {\displaystyle \sum _{m=0}^{\infty }{m+n \choose m}x^{m}={\frac {1}{(1-x)^{n+1}}}}
[1] ∑ m = 0 ∞ 1 m + 1 ( 2 m m ) x m = 1 2 x ( 1 − 1 − 4 x ) {\displaystyle \sum _{m=0}^{\infty }{\frac {1}{m+1}}{2m \choose m}x^{m}={\frac {1}{2x}}(1-{\sqrt {1-4x}})}
[1] ∑ m = 0 ∞ ( 2 m m ) x m = 1 1 − 4 x {\displaystyle \sum _{m=0}^{\infty }{2m \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}}
[1] ∑ m = 0 ∞ ( 2 m + n m ) x m = 1 1 − 4 x ( 1 − 1 − 4 x 2 x ) n {\displaystyle \sum _{m=0}^{\infty }{2m+n \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}\left({\frac {1-{\sqrt {1-4x}}}{2x}}\right)^{n}} Bernulio Skaičiai
keisti
∑ m = 0 ∞ B m m ! x m = x e x − 1 {\displaystyle \sum _{m=0}^{\infty }{\frac {B_{m}}{m!}}x^{m}={\frac {x}{e^{x}-1}}} [2]
∑ m = 0 ∞ ( − 4 ) m B 2 m ( 2 m ) ! x 2 m = x cot x {\displaystyle \sum _{m=0}^{\infty }{\frac {(-4)^{m}B_{2m}}{(2m)!}}x^{2m}=x\cot {x}} [2]
∑ m = 1 ∞ ( − 1 ) m − 1 2 2 m ( 2 2 m − 1 ) B 2 m ( 2 m ) ! x 2 m − 1 = tan x {\displaystyle \sum _{m=1}^{\infty }{\frac {(-1)^{m-1}2^{2m}(2^{2m}-1)B_{2m}}{(2m)!}}x^{2m-1}=\tan {x}} [2]
∑ m = 0 ∞ ( − 1 ) m − 1 ( 4 m − 2 ) B 2 m ( 2 m ) ! x 2 m = x sin x {\displaystyle \sum _{m=0}^{\infty }{\frac {(-1)^{m-1}(4^{m}-2)B_{2m}}{(2m)!}}x^{2m}={\frac {x}{\sin {x}}}} [2] Harmoniniai Skaičiai
keisti
∑ m = 1 ∞ H m x m = log 1 1 − x 1 − x {\displaystyle \sum _{m=1}^{\infty }H_{m}x^{m}={\frac {\log {\frac {1}{1-x}}}{1-x}}}
∑ m = 2 ∞ H 2 m − 1 m x m = 1 2 ( log 1 1 − x ) 2 {\displaystyle \sum _{m=2}^{\infty }{\frac {H_{2m-1}}{m}}x^{m}={\frac {1}{2}}\left(\log {\frac {1}{1-x}}\right)^{2}} [2]
∑ m = 1 ∞ ( − 1 ) m − 1 H 2 m 2 m + 1 x 2 m + 1 = 1 2 arctan x log ( 1 + x 2 ) {\displaystyle \sum _{m=1}^{\infty }{\frac {(-1)^{m-1}H_{2m}}{2m+1}}x^{2m+1}={\frac {1}{2}}\arctan {x}\log {(1+x^{2})}} [2]
∑ m = 0 ∞ ∑ n = 0 2 m ( − 1 ) n 2 n + 1 4 m + 2 x 4 m + 2 = 1 4 arctan x log 1 + x 1 − x {\displaystyle \sum _{m=0}^{\infty }{\frac {\sum _{n=0}^{2m}{\frac {(-1)^{n}}{2n+1}}}{4m+2}}x^{4m+2}={\frac {1}{4}}\arctan {x}\log {\frac {1+x}{1-x}}} [2] Binominiai koeficientai
keisti
∑ m = 0 n ( n m ) = 2 n {\displaystyle \sum _{m=0}^{n}{n \choose m}=2^{n}}
∑ m = 0 n ( n m ) a ( n − m ) b m = ( a + b ) n {\displaystyle \sum _{m=0}^{n}{n \choose m}a^{(n-m)}b^{m}=(a+b)^{n}}
∑ m = 0 n ( − 1 ) i ( n m ) = 0 {\displaystyle \sum _{m=0}^{n}(-1)^{i}{n \choose m}=0}
∑ m = 0 n ( m k ) = ( n + 1 k + 1 ) {\displaystyle \sum _{m=0}^{n}{m \choose k}={n+1 \choose k+1}}
∑ m = 0 n ( k + m m ) = ( k + n + 1 n ) {\displaystyle \sum _{m=0}^{n}{k+m \choose m}={k+n+1 \choose n}}
∑ m = 0 r ( r m ) ( s n − m ) = ( r + s n ) {\displaystyle \sum _{m=0}^{r}{r \choose m}{s \choose n-m}={r+s \choose n}} Trigonometrinės funkcijos
keisti
Sumos sinuso ir kosinuso iškyla Furje eilutėse.
∑ m = 1 n sin ( m π n ) = 0 {\displaystyle \sum _{m=1}^{n}\sin \left({\frac {m\pi }{n}}\right)=0}
∑ m = 1 n cos ( m π n ) = 0 {\displaystyle \sum _{m=1}^{n}\cos \left({\frac {m\pi }{n}}\right)=0} Neklasifikuota
keisti
∑ m = b + 1 ∞ b m 2 − b 2 = 1 2 H 2 b {\displaystyle \sum _{m=b+1}^{\infty }{\frac {b}{m^{2}-b^{2}}}={\frac {1}{2}}H_{2b}} ∑ m = 1 ∞ y m 2 + y 2 = − 1 2 y + π 2 coth ( π y ) {\displaystyle \sum _{m=1}^{\infty }{\frac {y}{m^{2}+y^{2}}}=-{\frac {1}{2y}}+{\frac {\pi }{2}}\coth(\pi y)} Integralinės eilutės
keisti
x n + 1 n + 1 ≈ 1 n + 2 n + 3 n + 4 n + 5 n + . . . + x n , {\displaystyle {\frac {x^{n+1}}{n+1}}\approx 1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+...+x^{n},} kai x artėja į begalybę.
( a x ) n + 1 n + 1 = a ( ( 1 a ) n + ( 2 a ) n + ( 3 a ) n + ( 4 a ) n + ( 5 a ) n + . . . + ( x a ) n ) , {\displaystyle {\frac {(ax)^{n+1}}{n+1}}=a((1a)^{n}+(2a)^{n}+(3a)^{n}+(4a)^{n}+(5a)^{n}+...+(xa)^{n}),} kur a mažesnis už 1, ne neigiamas skaičius; kai x artėja į begalybę arba x tiesiog didelė reikšmė (o a tada artėja prie 0). Jei pavyzdžiui norima rasti plotą po šaka y = x 7 {\displaystyle y=x^{7}} (n=7, kai x yra nuo 0 iki 10 ant Ox ašies), tai a reikia parinkti a=1/1000, o x reikia parinkti x=10000, ir sudėti 10000 skaičių padaugintų iš a :( 0.001 ⋅ 10000 ) 7 + 1 7 + 1 = 0.001 ( ( 1 ⋅ 0.001 ) 7 + ( 2 ⋅ 0.001 ) 7 + ( 3 ⋅ 0.001 ) 7 + ( 4 ⋅ 0.001 ) 7 + ( 5 ⋅ 0.001 ) 7 + . . . + ( 10000 ⋅ 0.001 ) 7 ) , {\displaystyle {\frac {(0.001\cdot 10000)^{7+1}}{7+1}}=0.001((1\cdot 0.001)^{7}+(2\cdot 0.001)^{7}+(3\cdot 0.001)^{7}+(4\cdot 0.001)^{7}+(5\cdot 0.001)^{7}+...+(10000\cdot 0.001)^{7}),}
10 8 8 = 0.001 ( ( 1 ⋅ 0.001 ) 7 + ( 2 ⋅ 0.001 ) 7 + ( 3 ⋅ 0.001 ) 7 + ( 4 ⋅ 0.001 ) 7 + ( 5 ⋅ 0.001 ) 7 + . . . + ( 9999 ⋅ 0.001 ) 7 + 10 7 ) . {\displaystyle {\frac {10^{8}}{8}}=0.001((1\cdot 0.001)^{7}+(2\cdot 0.001)^{7}+(3\cdot 0.001)^{7}+(4\cdot 0.001)^{7}+(5\cdot 0.001)^{7}+...+(9999\cdot 0.001)^{7}+10^{7}).}
↑ 1,0 1,1 1,2 1,3 Theoretical computer science cheat sheet
↑ 2,0 2,1 2,2 2,3 2,4 2,5 2,6 Citavimo klaida: Netinkama <ref>
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