Sei
X
1
,
X
2
,
X
3
,
⋯
{\displaystyle X_{1},X_{2},X_{3},\cdots }
bernoulli-verteilter Zufallsvariablen und deren Summe
S
n
{\displaystyle S_{n}}
binomialverteilt mit Parametern
n
∈
N
{\displaystyle n\in \mathbb {N} }
p
∈
]
0
,
1
[
{\displaystyle p\in ]0,1[}
σ
2
=
p
(
1
−
p
)
{\displaystyle \sigma ^{2}=p(1-p)}
(1)
P
(
S
n
=
k
)
=
B
(
k
∣
p
,
n
)
≈
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
{\displaystyle \quad \operatorname {P} \left(S_{n}=k\right)\;=\;B(k\mid p,n)\;\approx \;{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
lim
n
→
∞
P
(
x
1
≤
S
n
−
n
p
n
≤
x
2
)
=
1
2
π
σ
2
∫
x
1
x
2
exp
(
−
x
2
2
σ
2
)
d
x
{\displaystyle \quad \lim _{n\to \infty }\operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x}
x
1
,
x
2
∈
R
{\displaystyle x_{1},x_{2}\in \mathbb {R} }
x
1
<
x
2
{\displaystyle x_{1}<x_{2}}
Nach Gauß lässt sich die Gammafunktion
Γ
(
x
)
{\displaystyle \Gamma (x)}
x
∈
C
∖
{
0
,
−
1
,
−
2
,
…
}
{\displaystyle x\in \mathbb {C} \setminus \{0,-1,-2,\dotsc \}}
n
∈
N
0
{\displaystyle n\in \mathbb {N} _{0}}
Γ
(
x
)
=
lim
n
→
∞
n
!
n
x
x
(
x
+
1
)
(
x
+
2
)
⋯
(
x
+
n
)
{\displaystyle \Gamma (x)=\lim _{n\to \infty }{\frac {n!\,n^{x}}{x(x+1)(x+2)\dotsm (x+n)}}}
Bemerkungen:
Es gilt
n
!
=
Γ
(
n
+
1
)
=
n
Γ
(
n
)
{\displaystyle n!=\Gamma (n+1)=n\Gamma (n)}
Die Stirlingformel lautet
n
!
=
lim
n
→
∞
(
2
π
n
(
n
e
)
n
+
ϵ
n
)
{\displaystyle n!=\lim _{n\to \infty }\left({\sqrt {2\pi n}}\;\left({\tfrac {n}{\mathrm {e} }}\right)^{n}+\epsilon _{n}\right)}
Nachfolgend wird das Näherungszeichen "
≈
{\displaystyle \approx }
=
{\displaystyle =}
In der Halbebene
Re
(
x
)
>
0
{\displaystyle \operatorname {Re} (x)>0}
log
Γ
(
x
)
=
(
x
−
1
2
)
log
x
−
x
+
log
2
π
+
O
(
1
|
x
|
)
{\displaystyle \log \Gamma (x)=(x-{\tfrac {1}{2}})\log x-x+\log {\sqrt {2\pi }}+{\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}
Dabei ist
log
x
{\displaystyle \log x}
x
{\displaystyle x}
log
Γ
(
x
)
{\displaystyle \log \Gamma (x)}
x
>
0
{\displaystyle x>0}
Nach Gauß ist
Γ
(
x
)
=
lim
n
→
∞
n
!
n
x
x
(
x
+
1
)
(
x
+
2
)
⋯
(
x
+
n
)
{\displaystyle \Gamma (x)=\lim _{n\to \infty }{\frac {n!\,n^{x}}{x(x+1)(x+2)\dotsm (x+n)}}}
also
log
Γ
(
x
)
=
lim
n
→
∞
(
log
n
!
+
x
log
n
−
∑
k
=
0
n
log
(
x
+
k
)
)
{\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\sum _{k=0}^{n}\log(x+k){\bigg )}}
x
{\displaystyle x}
d
2
f
(
t
)
d
t
2
|
t
=
k
=
d
2
log
(
x
+
t
)
d
t
2
|
t
=
k
=
1
(
x
+
t
)
2
{\displaystyle {\bigg .}{\tfrac {\mathrm {d} ^{2}f(t)}{\mathrm {d} t^{2}}}{\bigg |}_{t=k}={\bigg .}{\tfrac {\mathrm {d} ^{2}\log(x+t)}{\mathrm {d} t^{2}}}{\bigg |}_{t=k}={\tfrac {1}{(x+t)^{2}}}}
∑
k
=
0
n
log
(
x
+
k
)
=
∫
0
n
log
(
x
+
t
)
d
t
+
1
2
(
log
x
+
log
(
x
+
n
)
)
+
∫
0
n
φ
(
t
)
1
(
x
+
t
)
2
d
t
{\displaystyle \sum _{k=0}^{n}\log(x+k)=\int _{0}^{n}\log(x+t)\mathrm {d} t+{\frac {1}{2}}{\bigg (}\log x+\log(x+n){\bigg )}\;+\;\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t}
log
Γ
(
x
)
=
lim
n
→
∞
(
log
n
!
+
x
log
n
−
∫
0
n
log
(
x
+
t
)
d
t
−
1
2
(
log
x
+
log
(
x
+
n
)
)
−
∫
0
n
φ
(
t
)
1
(
x
+
t
)
2
d
t
)
{\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\int \limits _{0}^{n}\log(x+t)\mathrm {d} t-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}-\int \limits _{0}^{n}\varphi (t){\frac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg )}}
|
φ
(
t
)
|
≤
1
8
{\displaystyle {\big |}\varphi (t){\big |}\leq {\tfrac {1}{8}}}
|
∫
0
n
φ
(
t
)
1
(
x
+
t
)
2
d
t
|
≤
1
8
|
∫
0
n
1
(
x
+
t
)
2
d
t
|
{\displaystyle {\bigg |}\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}\leq {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}}
1
8
|
∫
0
n
1
(
x
+
t
)
2
d
t
|
=
1
8
|
[
−
1
x
+
t
]
0
n
|
=
1
8
|
−
1
x
+
n
−
−
1
x
|
=
1
8
|
−
x
−
n
+
x
−
x
(
x
+
n
)
|
=
1
8
|
1
x
(
x
n
+
1
)
|
≤
1
8
1
|
x
|
≈
O
(
1
|
x
|
)
{\displaystyle {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}={\tfrac {1}{8}}{\bigg |}{\bigg [}{\tfrac {-1}{x+t}}{\bigg ]}_{0}^{n}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-1}{x+n}}-{\tfrac {-1}{x}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-x-n+x}{-x(x+n)}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {1}{x({\tfrac {x}{n}}+1)}}{\bigg |}\leq {\tfrac {1}{8}}{\tfrac {1}{|x|}}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}
|
∫
0
n
φ
(
t
)
1
(
x
+
t
)
2
d
t
|
≈
O
(
1
|
x
|
)
{\displaystyle {\bigg |}\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}
O
(
1
|
x
|
)
{\displaystyle {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}
log
Γ
(
x
)
=
lim
n
→
∞
(
log
n
!
+
x
log
n
−
∫
0
n
log
(
x
+
t
)
d
t
−
1
2
(
log
x
+
log
(
x
+
n
)
)
)
{\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\int \limits _{0}^{n}\log(x+t)\mathrm {d} t-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}{\bigg )}}
∫
0
n
log
(
x
+
t
)
d
t
=
[
(
x
+
t
)
log
(
x
+
t
)
−
1
]
0
n
=
(
x
+
n
)
log
(
x
+
n
)
−
x
log
x
+
C
{\displaystyle \int _{0}^{n}\log(x+t)\mathrm {d} t={\Big [}(x+t)\log(x+t)-1{\Big ]}_{0}^{n}=(x+n)\log(x+n)-x\log x+C}
o.B.d.A. gewählt werden kann, sei
C
=
−
n
{\displaystyle C=-n}
log
Γ
(
x
)
=
lim
n
→
∞
(
log
n
!
+
x
log
n
−
(
x
+
n
)
log
(
x
+
n
)
+
x
log
x
+
n
−
1
2
(
log
x
+
log
(
x
+
n
)
)
)
=
lim
n
→
∞
(
log
n
!
+
x
log
n
−
(
x
+
n
+
1
2
)
log
(
x
+
n
)
+
(
x
−
1
2
)
log
x
+
n
)
{\displaystyle {\begin{aligned}\log \Gamma (x)&=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-{\Big (}x+n{\Big )}\log(x+n)+x\log x+n-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}{\bigg )}\\&=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+n{\bigg )}\end{aligned}}}
n
!
=
lim
n
→
∞
(
2
π
n
(
n
e
)
n
+
ϵ
n
)
{\displaystyle n!=\lim _{n\to \infty }\left({\sqrt {2\pi n}}\;\left({\tfrac {n}{\mathrm {e} }}\right)^{n}+\epsilon _{n}\right)}
log
e
=
1
{\displaystyle \log e=1}
log
ϵ
n
=
ϵ
n
′
{\displaystyle \log \epsilon _{n}=\epsilon _{n}'}
log
n
!
=
lim
n
→
∞
(
(
n
+
1
2
)
log
n
−
n
+
log
2
π
+
ϵ
n
′
)
{\displaystyle \log n!=\lim _{n\to \infty }\left((n+{\tfrac {1}{2}})\log n-n+\log {\sqrt {2\pi }}+\epsilon _{n}'\right)}
log
Γ
(
x
)
=
lim
n
→
∞
(
(
n
+
1
2
)
log
n
−
n
+
x
log
n
−
(
x
+
n
+
1
2
)
log
(
x
+
n
)
+
(
x
−
1
2
)
log
x
+
n
+
ϵ
n
′
)
+
log
2
π
=
lim
n
→
∞
(
(
x
+
n
+
1
2
)
log
n
−
(
x
+
n
+
1
2
)
log
(
x
+
n
)
+
ϵ
n
′
)
+
(
x
−
1
2
)
log
x
+
log
2
π
{\displaystyle {\begin{aligned}\log \Gamma (x)&=\lim _{n\to \infty }{\bigg (}{\Big (}n+{\frac {1}{2}}{\Big )}\log n-n+x\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+n+\epsilon _{n}'{\bigg )}+\log {\sqrt {2\pi }}\\&=\lim _{n\to \infty }{\bigg (}{\Big (}x+n+{\frac {1}{2}}{\Big )}\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+\epsilon _{n}'{\bigg )}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\log {\sqrt {2\pi }}\end{aligned}}}
Nun ist
log
(
x
+
n
)
=
log
(
n
(
1
+
x
n
)
)
=
log
n
+
log
(
1
+
x
n
)
{\displaystyle \log(x+n)=\log(n(1+{\tfrac {x}{n}}))=\log n+\log(1+{\tfrac {x}{n}})}
x
{\displaystyle x}
n
{\displaystyle n}
log
(
1
+
x
n
)
≈
x
n
+
O
(
1
n
2
)
{\displaystyle \log(1+{\tfrac {x}{n}})\approx {\tfrac {x}{n}}+{\mathcal {O}}{\Big (}{\tfrac {1}{n^{2}}}{\Big )}}
O
(
1
n
2
)
{\displaystyle {\mathcal {O}}{\Big (}{\tfrac {1}{n^{2}}}{\Big )}}
log
(
x
+
n
)
=
log
n
+
x
n
{\displaystyle \log(x+n)=\log n+{\tfrac {x}{n}}}
log
Γ
(
x
)
=
log
2
π
+
(
x
−
1
2
)
log
x
+
lim
n
→
∞
(
(
x
+
n
+
1
2
)
log
n
−
(
x
+
n
+
1
2
)
(
log
n
+
x
n
)
+
ϵ
n
′
)
=
log
2
π
+
(
x
−
1
2
)
log
x
+
lim
n
→
∞
(
−
x
n
(
x
+
n
+
1
2
)
+
ϵ
n
′
)
=
log
2
π
+
(
x
−
1
2
)
log
x
−
x
{\displaystyle {\begin{aligned}\log \Gamma (x)&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\lim _{n\to \infty }{\bigg (}{\Big (}x+n+{\frac {1}{2}}{\Big )}\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}{\Big (}\log n+{\frac {x}{n}}{\Big )}+\epsilon _{n}'{\bigg )}\\&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\lim _{n\to \infty }{\bigg (}-{\frac {x}{n}}(x+n+{\frac {1}{2}})+\epsilon _{n}'{\bigg )}\\&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x-x\end{aligned}}}
◻
{\displaystyle \Box }
An dieser Stelle sei erwähnt, dass diese Näherung für
x
≥
9
{\displaystyle x\geq 9}
1
%
{\displaystyle 1\%}
Für die weitere Beweisführung seien folgende Umformungen angegeben:
Γ
(
x
)
≈
2
π
x
(
x
−
1
2
)
e
−
x
{\displaystyle \Gamma (x)\approx {\sqrt {2\pi }}x^{{\Big (}x-{\frac {1}{2}}{\Big )}}e^{-x}}
n
!
=
Γ
(
n
+
1
)
=
n
Γ
(
n
)
{\displaystyle n!=\Gamma (n+1)=n\Gamma (n)}
n
!
=
n
Γ
(
n
)
≈
n
2
π
n
(
n
−
1
2
)
e
−
n
=
2
π
n
(
n
e
)
n
=
n
n
e
−
n
2
π
n
{\displaystyle n!=n\Gamma (n)\approx n{\sqrt {2\pi }}n^{{\Big (}n-{\tfrac {1}{2}}{\Big )}}e^{-n}={\sqrt {2\pi n}}{\bigg (}{\frac {n}{e}}{\bigg )}^{n}=n^{n}e^{-n}{\sqrt {2\pi n}}}
Der Beweis wird in zwei Schritten durchgeführt.
Zunächst wird gezeigt:
B
(
k
∣
p
,
n
)
≈
n
n
e
−
n
2
π
n
k
k
e
−
k
2
π
k
(
n
−
k
)
n
−
k
e
−
(
n
−
k
)
2
π
(
n
−
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle B(k\mid p,n)\approx {\frac {n^{n}e^{-n}{\sqrt {2\pi n}}}{k^{k}e^{-k}{\sqrt {2\pi k}}(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}}p^{k}(1-p)^{n-k}}
Dazu werden mit Hilfe der Stirlingformel die Fakultäten ersetzt, also folgende Näherungen vorgenommen:
n
!
≈
2
π
n
(
n
e
)
n
=
n
n
e
−
n
2
π
n
{\displaystyle n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}=n^{n}e^{-n}{\sqrt {2\pi n}}}
k
!
≈
2
π
k
(
k
e
)
k
=
k
k
e
−
k
2
π
k
{\displaystyle k!\approx {\sqrt {2\pi k}}\left({\frac {k}{e}}\right)^{k}=k^{k}e^{-k}{\sqrt {2\pi k}}}
(
n
−
k
)
!
≈
2
π
(
n
−
k
)
(
n
−
k
e
)
n
−
k
=
(
n
−
k
)
n
−
k
e
−
(
n
−
k
)
2
π
(
n
−
k
)
{\displaystyle (n-k)!\approx {\sqrt {2\pi (n-k)}}\left({\frac {n-k}{e}}\right)^{n-k}=(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}
Damit lässt sich die Binomialverteilung wie folgt ausdrücken:
B
(
k
∣
p
,
n
)
=
(
n
k
)
p
k
(
1
−
p
)
n
−
k
=
n
!
k
!
(
n
−
k
)
!
p
k
(
1
−
p
)
n
−
k
≈
n
n
e
−
n
2
π
n
k
k
e
−
k
2
π
k
(
n
−
k
)
n
−
k
e
−
(
n
−
k
)
2
π
(
n
−
k
)
p
k
(
1
−
p
)
n
−
k
=
n
2
π
k
(
n
−
k
)
n
n
k
k
(
n
−
k
)
n
−
k
p
k
(
1
−
p
)
n
−
k
=
n
2
π
k
(
n
−
k
)
(
n
p
k
)
k
(
n
(
1
−
p
)
n
−
k
)
n
−
k
=
1
2
π
n
k
n
(
1
−
k
n
)
(
n
p
k
)
k
(
n
(
1
−
p
)
n
−
k
)
n
−
k
{\displaystyle {\begin{aligned}B(k\mid p,n)&={n \choose k}p^{k}(1-p)^{n-k}\\&={\frac {n!}{k!(n-k)!}}p^{k}(1-p)^{n-k}\\&\approx {\frac {n^{n}e^{-n}{\sqrt {2\pi n}}}{k^{k}e^{-k}{\sqrt {2\pi k}}(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}}p^{k}(1-p)^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}{\frac {n^{n}}{k^{k}\left(n-k\right)^{n-k}}}p^{k}(1-p)^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\end{aligned}}}
Mit der Näherung
k
n
≈
p
{\displaystyle {\frac {k}{n}}\approx p}
Taylorapproximation wird gezeigt:
B
(
k
∣
p
,
n
)
≈
1
2
π
n
k
n
(
1
−
k
n
)
(
n
p
k
)
k
(
n
(
1
−
p
)
n
−
k
)
n
−
k
≈
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
{\displaystyle B(k\mid p,n)\approx {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\approx {\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
n
≫
k
{\displaystyle n\gg k}
k
n
≈
p
{\displaystyle {\frac {k}{n}}\approx p}
1
2
π
n
k
n
(
1
−
k
n
)
≈
1
2
π
n
p
(
1
−
p
)
=
1
2
π
n
σ
2
{\displaystyle {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\approx {\frac {1}{\sqrt {2\pi np(1-p)}}}={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}}
B
(
k
∣
p
,
n
)
≈
1
2
π
n
σ
2
(
n
p
k
)
k
(
n
(
1
−
p
)
n
−
k
)
n
−
k
=
1
2
π
n
σ
2
(
p
k
/
n
)
k
(
(
1
−
p
)
1
−
k
/
n
)
n
−
k
=
1
2
π
n
σ
2
exp
(
n
n
log
(
(
p
k
/
n
)
k
(
1
−
p
1
−
k
/
n
)
n
−
k
)
)
=
1
2
π
n
σ
2
exp
(
n
n
log
(
(
k
/
n
p
)
−
k
(
1
−
k
/
n
1
−
p
)
−
(
n
−
k
)
)
)
=
1
2
π
n
σ
2
exp
(
n
(
−
k
n
log
(
k
/
n
p
)
⏟
f
(
x
)
−
(
1
−
k
n
)
log
(
1
−
k
/
n
1
−
p
)
⏟
g
(
x
)
)
)
{\displaystyle {\begin{aligned}B(k\mid p,n)&\approx {\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\left({\frac {p}{k/n}}\right)^{k}\left({\frac {(1-p)}{1-k/n}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left({\frac {n}{n}}\log \left(\left({\frac {p}{k/n}}\right)^{k}\left({\frac {1-p}{1-k/n}}\right)^{n-k}\right)\right)\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left({\frac {n}{n}}\log \left(\left({\frac {k/n}{p}}\right)^{-k}\left({\frac {1-k/n}{1-p}}\right)^{-(n-k)}\right)\right)\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp {\Bigg (}n{\bigg (}-\underbrace {{\frac {k}{n}}\log \left({\frac {k/n}{p}}\right)} _{f(x)}-\underbrace {\left(1-{\frac {k}{n}}\right)\log \left({\frac {1-k/n}{1-p}}\right)} _{g(x)}{\bigg )}{\Bigg )}\end{aligned}}}
Taylorapproximation in der Annäherung durch die Schmiegparabel . Wir erhalten mit
x
=
k
n
{\displaystyle x={\frac {k}{n}}}
f
(
x
)
=
x
ln
x
p
{\displaystyle f(x)=x\ln {\frac {x}{p}}}
g
(
x
)
=
(
1
−
x
)
ln
1
−
x
1
−
p
{\displaystyle g(x)=(1-x)\ln {\frac {1-x}{1-p}}}
x
=
p
{\displaystyle x=p}
T
2
f
(
x
,
p
)
=
(
x
ln
x
p
)
|
x
=
p
+
d
d
x
(
x
ln
x
p
)
|
x
=
p
⋅
(
x
−
p
)
+
1
2
d
2
d
x
2
(
x
ln
x
p
)
|
x
=
p
⋅
(
x
−
p
)
2
=
(
ln
x
p
+
x
⋅
p
x
⋅
1
p
)
|
x
=
p
⋅
(
x
−
p
)
+
1
2
d
d
x
(
ln
x
p
+
1
)
|
x
=
p
⋅
(
x
−
p
)
2
=
x
−
p
+
1
2
(
p
x
⋅
1
p
)
|
x
=
p
⋅
(
x
−
p
)
2
=
x
−
p
+
1
2
p
⋅
(
x
−
p
)
2
{\displaystyle {\begin{aligned}T_{2}f(x,p)&=\left.\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}+\left.{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=\left.\left(\ln {\frac {x}{p}}+x\cdot {\frac {p}{x}}\cdot {\frac {1}{p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(\ln {\frac {x}{p}}+1\right)\right|_{x=p}\cdot (x-p)^{2}\\&=x-p+\left.{\frac {1}{2}}\left({\frac {p}{x}}\cdot {\frac {1}{p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=x-p+{\frac {1}{2p}}\cdot (x-p)^{2}\end{aligned}}}
und
T
2
g
(
x
,
p
)
=
(
(
1
−
x
)
ln
1
−
x
1
−
p
)
|
x
=
p
+
d
d
x
(
(
1
−
x
)
ln
1
−
x
1
−
p
)
|
x
=
p
⋅
(
x
−
p
)
+
1
2
d
2
d
x
2
(
(
1
−
x
)
ln
1
−
x
1
−
p
)
|
x
=
p
⋅
(
x
−
p
)
2
=
(
−
ln
1
−
x
1
−
p
+
(
1
−
x
)
⋅
1
−
p
1
−
x
⋅
−
1
1
−
p
)
|
x
=
p
⋅
(
x
−
p
)
+
1
2
d
d
x
(
−
ln
1
−
x
1
−
p
−
1
)
|
x
=
p
⋅
(
x
−
p
)
2
=
−
(
x
−
p
)
−
1
2
(
1
−
p
1
−
x
⋅
−
1
1
−
p
)
|
x
=
p
⋅
(
x
−
p
)
2
=
−
(
x
−
p
)
+
1
2
(
1
−
p
)
⋅
(
x
−
p
)
2
{\displaystyle {\begin{aligned}T_{2}g(x,p)&=\left.\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}+\left.{\frac {\mathrm {d} }{\mathrm {d} x}}\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=\left.\left(-\ln {\frac {1-x}{1-p}}+(1-x)\cdot {\frac {1-p}{1-x}}\cdot {\frac {-1}{1-p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(-\ln {\frac {1-x}{1-p}}-1\right)\right|_{x=p}\cdot (x-p)^{2}\\&=-(x-p)-\left.{\frac {1}{2}}\left({\frac {1-p}{1-x}}\cdot {\frac {-1}{1-p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=-(x-p)+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\end{aligned}}}
Bemerkungen: zu beachten ist
d
d
x
ln
h
(
x
)
=
h
′
(
x
)
h
(
x
)
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\ln h(x)={\frac {h'(x)}{h(x)}}}
Integralrestglied
R
2
f
(
x
,
p
)
{\displaystyle R_{2}f(x,p)}
R
2
g
(
x
,
p
)
{\displaystyle R_{2}g(x,p)}
Die Zusammenfassung beider Taylorapproximationen liefert dann mit
σ
2
=
p
(
1
−
p
)
{\displaystyle \sigma ^{2}=p(1-p)}
T
2
f
(
x
,
p
)
+
T
2
g
(
x
,
p
)
=
x
−
p
+
1
2
p
⋅
(
x
−
p
)
2
−
(
x
−
p
)
+
1
2
(
1
−
p
)
⋅
(
x
−
p
)
2
=
1
2
p
⋅
(
x
−
p
)
2
+
1
2
(
1
−
p
)
⋅
(
x
−
p
)
2
=
1
−
p
+
p
2
p
(
1
−
p
)
⋅
(
x
−
p
)
2
=
1
2
p
(
1
−
p
)
⋅
(
x
−
p
)
2
=
1
2
σ
2
⋅
(
x
−
p
)
2
{\displaystyle {\begin{aligned}T_{2}f(x,p)+T_{2}g(x,p)&=x-p+{\frac {1}{2p}}\cdot (x-p)^{2}-(x-p)+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2p}}\cdot (x-p)^{2}+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\\&={\frac {1-p+p}{2p(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2p(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2\sigma ^{2}}}\cdot (x-p)^{2}\end{aligned}}}
B
(
k
∣
p
,
n
)
≈
1
2
π
n
k
n
(
1
−
k
n
)
(
n
p
k
)
k
(
n
(
1
−
p
)
n
−
k
)
n
−
k
≈
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
{\displaystyle B(k\mid p,n)\approx {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\;\approx \;{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
◻
{\displaystyle \Box }
Gezeigt wird, dass aus dem lokalen Grenzwertsatz im Limes
n
→
∞
{\displaystyle n\rightarrow \infty }
Riemann-Summe folgt:
P
(
x
1
≤
S
n
−
n
p
n
≤
x
2
)
=
1
2
π
σ
2
∫
x
1
x
2
exp
(
−
x
2
2
σ
2
)
d
x
{\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x}
x
1
,
x
2
∈
R
{\displaystyle x_{1},x_{2}\in \mathbb {R} }
x
1
<
x
2
{\displaystyle x_{1}<x_{2}}
Zunächst folgt mit dem lokalen Grenzwertsatz:
P
(
x
1
≤
S
n
−
n
p
n
≤
x
2
)
=
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
P
(
S
n
=
k
)
=
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
⋅
(
1
+
ϵ
n
,
p
(
k
)
)
{\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}\operatorname {P} \left(S_{n}=k\right)\;=\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\cdot (1+\epsilon _{n,p}(k))}
wobei
lim
n
→
∞
ϵ
n
,
p
(
k
)
=
0
{\displaystyle \lim _{n\to \infty }\epsilon _{n,p}(k)=\;0}
Setzen wir nun
ϵ
¯
n
,
p
:=
sup
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
|
ϵ
n
,
p
(
k
)
|
{\displaystyle {\bar {\epsilon }}_{n,p}:=\sup _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}|\epsilon _{n,p}(k)|}
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
⋅
ϵ
n
,
p
(
k
)
)
≤
ϵ
¯
n
,
p
⋅
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
{\displaystyle \sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\cdot \epsilon _{n,p}(k))\;\leq \;{\bar {\epsilon }}_{n,p}\cdot \sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
und im Limes
n
→
∞
{\displaystyle n\to \infty }
lim
n
→
∞
ϵ
n
,
p
⏟
→
0
⋅
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
⏟
→
∫
a
b
…
<
∞
=
0
{\displaystyle \lim _{n\to \infty }\;\underbrace {\quad \epsilon _{n,p}\quad } _{\to 0}\cdot \underbrace {\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} _{\to \int _{a}^{b}\ldots <\infty }=0}
Und daher gilt im Limes
n
→
∞
{\displaystyle n\to \infty }
P
(
x
1
≤
S
n
−
n
p
n
≤
x
2
)
=
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
{\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\;\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
Im Folgenden wird für die Riemann-Summe die Integraldarstellung gezeigt:
lim
n
→
∞
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
=
1
2
π
σ
2
∫
x
1
x
2
exp
(
−
x
2
2
σ
2
)
d
x
{\displaystyle \lim _{n\to \infty }\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\;=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x}
Wir bilden ein äquidistantes Gitter
Γ
n
:=
{
k
−
n
p
n
|
k
=
0
,
1
,
…
,
n
}
⊆
R
{\displaystyle \Gamma _{n}:=\left\{\left.{\tfrac {k-np}{\sqrt {n}}}\right|\,k=0,1,\ldots ,n\right\}\subseteq \mathrm {R} }
Maschenweite
1
n
{\displaystyle {\frac {1}{\sqrt {n}}}}
Riemannintegral überführen:
lim
n
→
∞
∑
k
∈
{
0
,
1
…
,
n
}
x
1
≤
k
−
n
p
n
≤
x
2
1
2
π
n
σ
2
exp
(
−
n
2
σ
2
(
k
n
−
p
)
2
)
=
lim
n
→
∞
∑
x
∈
Γ
n
x
1
≤
x
≤
x
2
1
2
π
σ
2
exp
(
−
n
2
σ
2
(
k
−
n
p
n
)
2
)
⋅
1
n
=
1
2
π
σ
2
∫
x
1
x
2
exp
(
−
x
2
2
σ
2
)
d
x
{\displaystyle {\begin{aligned}\lim _{n\to \infty }\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\;&=\;\lim _{n\to \infty }\sum _{x\in \Gamma _{n} \atop x_{1}\leq x\leq x_{2}}{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k-np}{n}}\right)^{2}\right)\cdot {\frac {1}{\sqrt {n}}}\;\\&=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x\end{aligned}}}
Unter Verwendung der Ergebnisse aus Schritt 1 und Schritt 2 folgt die eingangs formulierte Behauptung:
lim
n
→
∞
P
(
x
1
≤
S
n
−
n
p
n
≤
x
2
)
=
1
2
π
σ
2
∫
x
1
x
2
exp
(
−
x
2
2
σ
2
)
d
x
{\displaystyle \quad \lim _{n\to \infty }\operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x}
x
1
,
x
2
∈
R
{\displaystyle x_{1},x_{2}\in \mathbb {R} }
x
1
<
x
2
{\displaystyle x_{1}<x_{2}}
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![{\displaystyle p\in ]0,1[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cd07a5e96ded88e85782453bdae4353f369a648)
![{\displaystyle f:[n_{0},N]\to \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eddf7cfe6165a006c3b09737e53998a912ba8e63)
![{\displaystyle {\begin{aligned}\int \limits _{0}^{1}f(x)\mathrm {d} x&=-\int \limits _{0}^{1}\varphi ''(x)f(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg [}\varphi '(x)f(x){\bigg ]}_{0}^{1-\epsilon }+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\bigg .}\varphi '(x){\bigg |}_{x=1-\epsilon }f(1-\epsilon )-\varphi '(0)f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\frac {1}{2}}{\Big (}1-2(1-\epsilon )+0{\Big )}f(1-\epsilon )-{\frac {1}{2}}f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\frac {1}{2}}(2\epsilon -1)f(1-\epsilon )-{\frac {1}{2}}f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;+\;\lim _{\epsilon \to 0}{\Bigg (}{\bigg [}\varphi (x)f'(x){\bigg ]}_{0}^{1-\epsilon }\;-\;\int \limits _{0}^{1-\epsilon }\varphi (x)f''(x)\mathrm {d} x{\Bigg )}\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;-\;\lim _{\epsilon \to 0}\;\int \limits _{0}^{1-\epsilon }\varphi (x)f''(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;-\;\int \limits _{0}^{1}\varphi (x)f''(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5526fe44fb84afadd6d91789781dda89b1c94cc4)
![{\displaystyle {\begin{aligned}\int \limits _{n_{0}}^{N}f(x)\mathrm {d} x&=-\int \limits _{n_{0}}^{N}\varphi ''(x)f(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg [}\varphi '(x)f(x){\bigg ]}_{k}^{k+1-\epsilon }+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\bigg .}\varphi '(x){\bigg |}_{x=k+1-\epsilon }\cdot f(k+1-\epsilon )-\varphi '(k)f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}{\Big (}1-2(k+1-\epsilon )+2k{\Big )}f(k+1-\epsilon )-{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}(2\epsilon -1)f(k+1-\epsilon )-{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}f(k+1)+{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}-1}^{N-1}{\frac {1}{2}}f(k+1)-{\frac {1}{2}}f(n_{0})+\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(N)\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(n_{0})+\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(N)\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\Bigg (}{\bigg [}\varphi (x)f'(x){\bigg ]}_{0}^{1-\epsilon }\;-\;\int \limits _{k}^{k+1-\epsilon }\varphi (x)f''(x)\mathrm {d} x{\Bigg )}\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;-\;\lim _{\epsilon \to 0}\;\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi (x)f''(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;-\;\int \limits _{n_{0}}^{N}\varphi (x)f''(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1553de6000283dee5cc1aaf6d3ad2d86ef71d6a7)
![{\displaystyle {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}={\tfrac {1}{8}}{\bigg |}{\bigg [}{\tfrac {-1}{x+t}}{\bigg ]}_{0}^{n}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-1}{x+n}}-{\tfrac {-1}{x}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-x-n+x}{-x(x+n)}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {1}{x({\tfrac {x}{n}}+1)}}{\bigg |}\leq {\tfrac {1}{8}}{\tfrac {1}{|x|}}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b450c65c4798ff0daa9e89245dbcef2a2534b04a)
![{\displaystyle \int _{0}^{n}\log(x+t)\mathrm {d} t={\Big [}(x+t)\log(x+t)-1{\Big ]}_{0}^{n}=(x+n)\log(x+n)-x\log x+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cc1468841a4e5c953b4c6adf4e0f9b37916439e)
