Tai yra sąrašas matematinių eilučių talpinančios formulę baigtinėms ir begalinėms sumoms. Tai gali būti naudota konjunkcijoje su kitais įrankiais evaluacinėms sumoms.
∑
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}\,}
k
{\displaystyle k\,}
Bernoulli number , and
B
1
{\displaystyle B_{1}\,}
∑
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)\,}
Riemann zeta function .
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)}
r
>
0
{\displaystyle r>0}
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),\,\!}
s 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}}
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}},}
arctan
1
=
π
4
.
{\displaystyle \arctan 1={\frac {\pi }{4}}.}
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}
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,}
∫
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}
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}
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}
x
∞
,
{\displaystyle x^{\infty },}
x
n
{\displaystyle x^{n}}
n koks nors labai didelis sveikasis skaičius, pavyzdžiui,
n
=
1000
{\displaystyle n=1000}
x
=
0.9
{\displaystyle x=0.9}
0.9
100
=
0.000026561
,
{\displaystyle 0.9^{100}=0.000026561,}
0.9
1000
=
1.7478712
10
46
,
{\displaystyle 0.9^{1000}={\frac {1.7478712}{10^{46}}},}
lim
n
→
∞
0.9
n
=
0
{\displaystyle \lim _{n\to \infty }0.9^{n}=0}
lim
n
→
∞
x
n
=
0
,
{\displaystyle \lim _{n\to \infty }x^{n}=0,}
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 [ keisti ] 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}}
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}}
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}}}
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}}
∑
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]
∑
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}}
[ 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}
∑
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)}
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},}
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}),}
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}}
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> žymė; nebuvo pateiktas tekstas nuorodoms su pavadinimu gfo
![{\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}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b401073f6581dbeae3739e0167704d5749a5917)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/141178384151305b532fc88af3485a05a5c01e6c)
