아래는 킨친 상수 (Khinchin constant)에 대한 설명이다.
수 이론 에서 알렉산드르 야코블레비치 킨친 (Aleksandr Yakovlevich Khinchin) 은 거의 모든 실수
x
{\displaystyle x}
x
{\displaystyle x}
연분수 ) 확장의 계수
a
n
{\displaystyle a_{n}}
a
i
{\displaystyle a_{i}}
x
{\displaystyle x}
x
=
a
0
+
1
a
1
+
1
a
2
+
1
a
3
+
1
⋱
{\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+{\cfrac {1}{\ddots }}}}}}}}\;}
lim
n
→
∞
(
a
1
,
a
2
,
.
.
.
,
a
n
)
1
n
=
K
0
{\displaystyle \lim _{n\rightarrow \infty }\left(a_{1},a_{2},...,a_{n}\right)^{1 \over n}=K_{0}}
킨친 상수(Khinchin constant)[ 1]
K
=
K
0
=
∏
k
=
1
∞
(
1
+
1
k
(
k
+
2
)
)
log
2
k
≈
2.6854520010
…
(
O
E
I
S
A
002210
)
{\displaystyle K=K_{0}=\prod _{k=1}^{\infty }{\left(1+{1 \over k(k+2)}\right)}^{\log _{2}k}\approx 2.6854520010\dots (OEISA002210)}
연속적인 분수 확장이 이 속성을 갖지 않는 것으로 알려진
x
{\displaystyle x}
ϕ
{\displaystyle \phi }
e
{\displaystyle e}
e
{\displaystyle e}
"Khinchin"은 때때로 오래된 수학 문헌에서 "Khintchine" (러시아어 Хинчин의 프랑스어 음역)으로 표기된다.
K
=
∏
k
=
1
∞
(
1
+
1
k
2
+
2
k
)
log
2
k
=
∏
k
=
1
∞
k
log
2
(
1
+
1
k
2
+
2
k
)
{\displaystyle K=\prod _{k=1}^{\infty }{\left(1+{1 \over k^{2}+2k}\right)}^{\log _{2}k}=\prod _{k=1}^{\infty }k^{\log _{2}^{}\left(1+{1 \over k^{2}+2k}\right)}}
K
=
∏
k
=
1
∞
(
1
+
1
k
(
k
+
2
)
)
log
2
k
=
∏
k
=
1
∞
(
1
+
1
k
(
k
+
2
)
)
l
n
k
l
n
2
{\displaystyle K=\prod _{k=1}^{\infty }{\left(1+{1 \over k(k+2)}\right)}^{\log _{2}k}=\prod _{k=1}^{\infty }{\left(1+{1 \over k(k+2)}\right)}^{{lnk} \over {ln2}}}
K
=
1
log
(
2
)
∑
s
=
1
∞
ζ
(
2
s
)
−
1
s
∑
k
=
1
2
s
−
1
−
1
(
k
+
1
)
k
{\displaystyle K={{1} \over {\log(2)}}\sum _{s=1}^{\infty }{{\zeta (2s)-1} \over {s}}\sum _{k=1}^{2s-1}{{-1^{(k+1)}} \over {k}}}
K
=
e
x
p
(
1
l
n
2
∑
k
=
1
∞
H
2
k
−
1
′
(
ζ
(
2
k
)
−
1
)
k
)
{\displaystyle K=exp\left({{1} \over {ln2}}\sum _{k=1}^{\infty }{{H_{2k-1}^{'}(\zeta (2k)-1)} \over {k}}\right)}
H
{\displaystyle H}
조화수 ,
ζ
{\displaystyle \zeta }
리만 제타 함수
log
(
K
0
)
=
1
log
(
2
)
∑
s
=
1
∞
ζ
(
2
s
)
−
1
s
∑
k
=
1
2
s
−
1
−
1
(
k
+
1
)
k
{\displaystyle \log(K_{0})={{1} \over {\log(2)}}\sum _{s=1}^{\infty }{{\zeta (2s)-1} \over {s}}\sum _{k=1}^{2s-1}{{-1^{(k+1)}} \over {k}}}
log
(
K
0
)
log
(
2
)
=
∑
s
=
1
∞
ζ
(
2
s
)
−
1
s
∑
k
=
1
2
s
−
1
−
1
(
k
+
1
)
k
{\displaystyle \log(K_{0}){\log(2)}=\sum _{s=1}^{\infty }{{\zeta (2s)-1} \over {s}}\sum _{k=1}^{2s-1}{{-1^{(k+1)}} \over {k}}}
log
(
K
0
)
log
(
2
)
=
∑
s
=
1
∞
ζ
(
2
s
)
−
1
s
(
1
1
−
1
2
+
1
3
−
⋯
+
1
(
2
s
−
1
)
)
{\displaystyle \log(K_{0}){\log(2)}=\sum _{s=1}^{\infty }{{\zeta (2s)-1} \over {s}}\left({1 \over 1}-{1 \over 2}+{1 \over 3}-\cdots +{1 \over (2s-1)}\right)}
K
p
=
[
∑
k
=
1
∞
−
k
p
log
2
(
1
−
1
(
k
+
1
)
2
)
]
1
p
{\displaystyle K_{p}=\left[\sum _{k=1}^{\infty }-k^{p}\log _{2}\left(1-{{1} \over {(k+1)^{2}}}\right)\right]^{1 \over p}}
=
[
1
l
n
2
∑
k
=
1
∞
k
p
l
n
(
1
+
1
k
(
k
+
2
)
)
]
1
p
{\displaystyle \;\;\;=\left[{{1} \over {ln2}}\sum _{k=1}^{\infty }k^{p}ln\left(1+{{1} \over {k(k+2)}}\right)\right]^{1 \over p}}
K
−
1
=
1.74540566240
…
(
O
E
I
S
A
087491
)
{\displaystyle K_{-1}=1.74540566240\dots (OEISA087491)}
K
−
2
=
1.45034032849563
…
(
O
E
I
S
A
087492
)
{\displaystyle K_{-2}=1.45034032849563\dots (OEISA087492)}
K
−
3
=
1.3135070786879
…
(
O
E
I
S
A
087493
)
{\displaystyle K_{-3}=1.3135070786879\dots (OEISA087493)}
![{\displaystyle K_{p}=\left[\sum _{k=1}^{\infty }-k^{p}\log _{2}\left(1-{{1} \over {(k+1)^{2}}}\right)\right]^{1 \over p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad1e90934f2ce04bf62283477804060122fd8060)
![{\displaystyle \;\;\;=\left[{{1} \over {ln2}}\sum _{k=1}^{\infty }k^{p}ln\left(1+{{1} \over {k(k+2)}}\right)\right]^{1 \over p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f154659c68260b1c87fd22ba6253a40040fa1409)
