Matematika/Matematinės eilutės

Tai yra sąrašas matematinių eilučių talpinančios formulę baigtinėms ir begalinėms sumoms. Tai gali būti naudota konjunkcijoje su kitais įrankiais evaluacinėms sumoms.

Galios sumos

• $\sum _{m=1}^{n}m={\frac {n(n+1)}{2}}\,\!$

• $\sum _{m=1}^{n}m^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\,\!$

• $\sum _{m=1}^{n}m^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left(\sum _{m=1}^{n}m\right)^{2}\,\!$

• $\sum _{m=1}^{n}m^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {6n^{5}+15n^{4}+10n^{3}-n}{30}}\,\!$

• $\sum _{m=0}^{n}m^{s}={\frac {(n+1)^{s+1}}{s+1}}+\sum _{k=1}^{s}{\frac {B_{k}}{s-k+1}}{s \choose k}(n+1)^{s-k+1}\,\!$

where $B_{k}\,$ is the $k\,$th Bernoulli number, and $B_{1}\,$ is negative.

• $\sum _{m=1}^{\infty }{\frac {1}{m^{2}}}={\frac {\pi ^{2}}{6}}\,\!$
• $\sum _{m=1}^{\infty }{\frac {1}{m^{4}}}={\frac {\pi ^{4}}{90}}\,\!$
• $\sum _{m=1}^{\infty }{\frac {1}{m^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$

• $\sum _{m=1}^{\infty }m^{-s}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}=\zeta (s)\,\!$

where $\zeta (s)\,$ is the Riemann zeta function.

Galios eilutės

begalinė suma (for $|x|<1$)
$\sum _{m=0}^{\infty }x^{m}={\frac {1}{1-x}}\,\!$	$\sum _{m=0}^{n}x^{m}={\frac {1-x^{n+1}}{1-x}}=1+{\frac {1}{r}}\left(1-{\frac {1}{(1+r)^{n}}}\right)$ where $r>0$ and $x={\frac {1}{1+r}}.\,\!$
$\sum _{m=0}^{\infty }x^{2m}={\frac {1}{1-x^{2}}}\,\!$
$\sum _{m=1}^{\infty }mx^{m}={\frac {x}{(1-x)^{2}}}\,\!$	$\sum _{m=1}^{n}mx^{m}=x{\frac {1-x^{n}}{(1-x)^{2}}}-{\frac {nx^{n+1}}{1-x}}\,\!$
$\sum _{m=1}^{\infty }m^{2}x^{m}={\frac {x(1+x)}{(1-x)^{3}}}\,\!$	$\sum _{m=1}^{n}m^{2}x^{m}={\frac {x(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}{(1-x)^{3}}}\,\!$
$\sum _{m=1}^{\infty }m^{3}x^{m}={\frac {x(1+4x+x^{2})}{(1-x)^{4}}}\,\!$
$\sum _{m=1}^{\infty }m^{4}x^{m}={\frac {x(1+x)(1+10x+x^{2})}{(1-x)^{5}}}\,\!$
$\sum _{m=1}^{\infty }m^{k}x^{m}=\operatorname {Li} _{-k}(x),\,\!$ where Lis(x) is the polylogarithm of x.

• Įrodysime, kad

$\arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...\quad (|x|<1).$

Kaip yra žinoma,

${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1.$

Dabar vietoje x įstatome $-x^{2}$ ir gauname lygybę:

${\frac {1}{1-(-x^{2})}}={\frac {1}{1+x^{2}}}=1-x^{2}+x^{4}-x^{6}+x^{8}-x^{10}+...+(-1)^{n}x^{2n}+...,$

Teisginai kai $|-x^{2}|<1.$

Integruodami gauname:

$\arctan x=\int _{0}^{x}{\frac {dx}{1+x^{2}}}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...+(-1)^{n+1}{\frac {x^{2n-1}}{2n-1}}+...,$

Iš kur

$\arctan x=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+{\frac {x^{9}}{9}}-{\frac {x^{11}}{11}}+...+(-1)^{n+1}{\frac {x^{2n-1}}{2n-1}}+...\quad (|x|<1).$

Iš to galime įrodyti Leibnico formulę:

${\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}.$

Kadangi ${\frac {\sin {\frac {\pi }{4}}}{\cos {\frac {\pi }{4}}}}=1=\tan {\frac {\pi }{4}},$ tai $\arctan 1={\frac {\pi }{4}}.$ Todėl:

$\arctan 1=\int _{0}^{1}{\frac {dx}{1+x^{2}}}=1-{\frac {1^{3}}{3}}+{\frac {1^{5}}{5}}-{\frac {1^{7}}{7}}+{\frac {1^{9}}{9}}-{\frac {1^{11}}{11}}+...+(-1)^{n+1}{\frac {1^{2n-1}}{2n-1}}+...=$

$={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}\approx 0.785398163.$

• Į eilutę

${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1,$

įstatome$-x$ vietoje x ir gauname:

${\frac {1}{1-(-x)}}={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+...+(-1)^{n}x^{n}+...,$

kai $|-x|<1,$ tada abi puses integruojame:

$\int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}x=\int _{0}^{x}(1-x+x^{2}-x^{3}+x^{4}-x^{5}+...+(-1)^{n}x^{n}+...){\mathsf {d}}x,$

$\ln(1+x)=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-{\frac {x^{6}}{6}}+...+(-1)^{n}{\frac {x^{n+1}}{n+1}}+...,\quad kai\;\;|x|<1,$

nes $\int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}x=\int _{0}^{x}{\frac {1}{1+x}}\;{\mathsf {d}}(1+x)=\ln(1+x)|_{0}^{x}=\ln(1+x)-\ln(1+0)=\ln(1+x)-0=\ln(1+x).$

• Pasirenkame $x=0.3$ ir įstatome:

${\frac {1}{1-x}}={\frac {1}{1-0.3}}={\frac {1}{0.7}}=1.428571429.$

Prasumuojant gauname:

$1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}=1+0.3+0.3^{2}+0.3^{3}+0.3^{4}+0.3^{5}+0.3^{6}+0.3^{7}+0.3^{8}+0.3^{9}+0.3^{10}=1.428568898.$

• Pasirenkame $x=-0.3$ ir įstatome:

${\frac {1}{1-x}}={\frac {1}{1-(-0.3)}}={\frac {1}{1.3}}=0.769230769.$

Prasumuojant gauname:

$1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}=1-0.3+0.3^{2}-0.3^{3}+0.3^{4}-0.3^{5}+0.3^{6}-0.3^{7}+0.3^{8}-0.3^{9}+0.3^{10}=0.769232131.$

• Įrodymas, kad

${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^{n}+...,\quad kai\;\;|x|<1.$

${\begin{aligned}&{\text{Tegu }}\;s=1+x+x^{2}+x^{3}+\cdots .\\[4pt]&{\text{Tada }}\;x\cdot s=x+x^{2}+x^{3}+x^{4}+\cdots .\\[4pt]&{\text{Tada }}\;x\cdot s-s=x^{\infty }-1=-1,{\text{ nes}}\;|x|<1,\;\;{\text{ taigi }}\;s(x-1)=-1,{\text{ ir todel }}\;s={\frac {-1}{x-1}}={\frac {-1}{-(1-x)}}={\frac {1}{1-x}}.\end{aligned}}$

Išraišką $x^{\infty }-1$ galima apibūdinti tiksliau, vietoje $x^{\infty },$ parašydami $x^{n}$, kur n koks nors labai didelis sveikasis skaičius, pavyzdžiui, $n=1000$. Pavyzdžiui, kai $x=0.9$, tai $0.9^{100}=0.000026561,$ o $0.9^{1000}={\frac {1.7478712}{10^{46}}},$ kas beveik lygu nuliui. Todėl $\lim _{n\to \infty }0.9^{n}=0$ arba $\lim _{n\to \infty }x^{n}=0,$ kai -1<x<1.

Paprasti denominatoriai

• $\sum _{m=1}^{\infty }{\frac {x^{m}}{m}}=\ln \left({\frac {1}{1-x}}\right)\quad {\mbox{ for }}|x|<1\!$

• $\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{2m+1}}x^{2m+1}=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots =\arctan(x)\,\!$

• $\sum _{m=0}^{\infty }{\frac {x^{2m+1}}{2m+1}}=\mathrm {arctanh} (x)\quad {\mbox{ for }}|x|<1\,\!$

Faktorialiniai denominatoriai

Daug geometrinių eilučių kurios kyla iš Tailoro theoremos turi koficientą turintį faktorialą.

• $\sum _{m=0}^{\infty }{\frac {x^{m}}{m!}}=e^{x}$
• $\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!}}x^{m}={\frac {1}{e^{x}}}$

• $\sum _{m=0}^{\infty }m{\frac {x^{m}}{m!}}=xe^{x}$ (c.f. mean of Poisson distribution)
• $\sum _{m=0}^{\infty }m^{2}{\frac {x^{m}}{m!}}=(x+x^{2})e^{x}$ (c.f. second moment of Poisson distribution)
• $\sum _{m=0}^{\infty }m^{3}{\frac {x^{m}}{m!}}=(x+3x^{2}+x^{3})e^{x}$
• $\sum _{m=0}^{\infty }m^{4}{\frac {x^{m}}{m!}}=(x+7x^{2}+6x^{3}+x^{4})e^{x}$
• $\sum _{m=0}^{\infty }m^{n}{\frac {x^{m}}{m!}}=x{\frac {d}{dx}}\sum _{m=0}^{\infty }m^{n-1}{\frac {x^{m}}{m!}}$

• $\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m+1)!}}x^{2m+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots =\sin x$

• $\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{(2m)!}}x^{2m}=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-\cdots =\cos x$

• $\sum _{m=0}^{\infty }{\frac {x^{2m+1}}{(2m+1)!}}=\sinh x$

• $\sum _{m=0}^{\infty }{\frac {x^{2m}}{(2m)!}}=\cosh x$

Modifikuoti-faktorialo denominatoriai

• $\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}=\arcsin x\quad {\mbox{ for }}|x|<1\!$

• $\sum _{m=0}^{\infty }{\frac {(-1)^{m}(2m)!}{4^{m}(m!)^{2}(2m+1)}}x^{2m+1}=\mathrm {arcsinh} (x)\quad {\mbox{ for }}|x|<1\!$

• $\sum _{m=0}^{\infty }{\frac {(4m)!}{16^{m}{\sqrt {2}}(2m)!(2m+1)!}}x^{m}={\sqrt {\frac {1-{\sqrt {1-x}}}{x}}}$

• $\sum _{m=0}^{\infty }{\frac {4^{m}(m)!^{2}}{(m+1)(2m+1)!}}x^{2m}=\left({\frac {\arcsin {x}}{x}}\right)^{2}$

• $\sum _{m=0}^{\infty }{\frac {\prod _{n=0}^{m-1}(4n^{2}+1)}{(2m)!}}x^{2m}+\sum _{m=0}^{\infty }{\frac {4^{m}\prod _{n=1}^{m}({\frac {1}{2}}-n+n^{2})}{(2m+1)!}}x^{2m+1}=e^{\arcsin {x}}$

Binominės eilutės

Geometrinės eilutės:

• $(1+x)^{-1}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }(-x)^{m}&|x|<1\\\displaystyle \sum _{m=1}^{\infty }-(x)^{-m}&|x|>1\\\end{cases}}$

Binominė Teorema:

• $(a+x)^{n}={\begin{cases}\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{n-m}x^{m}&|x|\!<\!|a|\\\displaystyle \sum _{m=0}^{\infty }{\binom {n}{m}}a^{m}x^{n-m}&|x|\!>\!|a|\\\end{cases}}$

• $(1+x)^{\alpha }=\sum _{m=0}^{\infty }{\alpha \choose m}x^{m}\quad {\mbox{ for all }}|x|<1{\mbox{ and all complex }}\alpha \!$

su apibendrintais binominiais koficientais

${\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}\!$

Šaknis:

• ${\sqrt {1+x}}=\sum _{m=0}^{\infty }{\frac {(-1)^{m}(2m)!}{(1-2m)m!^{2}4^{m}}}x^{m}\quad {\mbox{ for }}|x|<1\!$

Maišyta, įvairu:

• $\sum _{m=0}^{\infty }{m+n \choose m}x^{m}={\frac {1}{(1-x)^{n+1}}}$
• $\sum _{m=0}^{\infty }{\frac {1}{m+1}}{2m \choose m}x^{m}={\frac {1}{2x}}(1-{\sqrt {1-4x}})$
• $\sum _{m=0}^{\infty }{2m \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}$
• $\sum _{m=0}^{\infty }{2m+n \choose m}x^{m}={\frac {1}{\sqrt {1-4x}}}\left({\frac {1-{\sqrt {1-4x}}}{2x}}\right)^{n}$

Bernulio Skaičiai

• $\sum _{m=0}^{\infty }{\frac {B_{m}}{m!}}x^{m}={\frac {x}{e^{x}-1}}$
• $\sum _{m=0}^{\infty }{\frac {(-4)^{m}B_{2m}}{(2m)!}}x^{2m}=x\cot {x}$
• $\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}2^{2m}(2^{2m}-1)B_{2m}}{(2m)!}}x^{2m-1}=\tan {x}$
• $\sum _{m=0}^{\infty }{\frac {(-1)^{m-1}(4^{m}-2)B_{2m}}{(2m)!}}x^{2m}={\frac {x}{\sin {x}}}$

Harmoniniai Skaičiai

• $\sum _{m=1}^{\infty }H_{m}x^{m}={\frac {\log {\frac {1}{1-x}}}{1-x}}$
• $\sum _{m=2}^{\infty }{\frac {H_{2m-1}}{m}}x^{m}={\frac {1}{2}}\left(\log {\frac {1}{1-x}}\right)^{2}$
• $\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}H_{2m}}{2m+1}}x^{2m+1}={\frac {1}{2}}\arctan {x}\log {(1+x^{2})}$
• $\sum _{m=0}^{\infty }{\frac {\sum _{n=0}^{2m}{\frac {(-1)^{n}}{2n+1}}}{4m+2}}x^{4m+2}={\frac {1}{4}}\arctan {x}\log {\frac {1+x}{1-x}}$

Binominiai koeficientai

• $\sum _{m=0}^{n}{n \choose m}=2^{n}$
• $\sum _{m=0}^{n}{n \choose m}a^{(n-m)}b^{m}=(a+b)^{n}$
• $\sum _{m=0}^{n}(-1)^{i}{n \choose m}=0$
• $\sum _{m=0}^{n}{m \choose k}={n+1 \choose k+1}$
• $\sum _{m=0}^{n}{k+m \choose m}={k+n+1 \choose n}$
• $\sum _{m=0}^{r}{r \choose m}{s \choose n-m}={r+s \choose n}$

Trigonometrinės funkcijos

Sumos sinuso ir kosinuso iškyla Furje eilutėse.

• $\sum _{m=1}^{n}\sin \left({\frac {m\pi }{n}}\right)=0$
• $\sum _{m=1}^{n}\cos \left({\frac {m\pi }{n}}\right)=0$

Neklasifikuota

• $\sum _{m=b+1}^{\infty }{\frac {b}{m^{2}-b^{2}}}={\frac {1}{2}}H_{2b}$

• $\sum _{m=1}^{\infty }{\frac {y}{m^{2}+y^{2}}}=-{\frac {1}{2y}}+{\frac {\pi }{2}}\coth(\pi y)$

Integralinės eilutės

• ${\frac {x^{n+1}}{n+1}}\approx 1^{n}+2^{n}+3^{n}+4^{n}+5^{n}+...+x^{n},$ kai x artėja į begalybę.
• ${\frac {(ax)^{n+1}}{n+1}}=a((1a)^{n}+(2a)^{n}+(3a)^{n}+(4a)^{n}+(5a)^{n}+...+(xa)^{n}),$ 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 $y=x^{7}$ (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:

${\frac {(0.001\cdot 10000)^{7+1}}{7+1}}=0.001((1\cdot 0.001)^{7}+(2\cdot 0.001)^{7}+(3\cdot 0.001)^{7}+(4\cdot 0.001)^{7}+(5\cdot 0.001)^{7}+...+(10000\cdot 0.001)^{7}),$

${\frac {10^{8}}{8}}=0.001((1\cdot 0.001)^{7}+(2\cdot 0.001)^{7}+(3\cdot 0.001)^{7}+(4\cdot 0.001)^{7}+(5\cdot 0.001)^{7}+...+(9999\cdot 0.001)^{7}+10^{7}).$

Nuorodos

• http://en.wikipedia.org/wiki/Geometric_series
• https://en.wikipedia.org/wiki/Series_(mathematics)
• https://en.wikipedia.org/wiki/List_of_mathematical_series