科学文献

k −1 X `=1
1 (−1)k 0 (ζ (k) − 1) − 1 + ζ (k) , `2 k−1
!
µ
Ã
¶¸
!
#
ρ(ln(X ), ln(T X )) = −0.2275522084.... The cumulative distribution for T X can be expressed in terms of the digamma function: ¶ ∞ µ X 1 1 F (x) = P(T X ≤ x) = − = γ + ψ (x + 1), n+x n=1 n 1 = ψ 0 (x + 1). 2 ( n + x ) n=1
µ
µ
1 − n
!
¶
ρ(ln(X ), ln(T X )) = −0.1858801270... = r1 . √ The median of T X is 2 − 1 = 0.4142135623... for every j . We wish to understand the decay rate of ρ(X, T j X ) and ρ(ln(X ), ln(T j X )) as j increases, but this appears to be a difficult problem.
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c 2007 by Steven R. Finch. All rights reserved. Copyright °
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Continued Fraction Transformation
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when x = π . The asymptotic distribution of denominators Qn , corresponding to uniformly distributed X as n → ∞, turns out to be related to our earlier work on ln(T j X ) statistics. 0.1. Uniform Distribution. Let γ denote the Euler-Mascheroni constant [2], ζ denote the Riemann zeta function and Lik denote the kth polylogarithm function [3]. If X is a random variable following the uniform distribution on [0, 1], then E(X ) =
Cov(X, T X ) = E(X · T X ) − E(X ) E(T X ) =
where ρ denotes cross-correlation. Likewise, E(ln(X )) = −1,
Z1
0
E(ln(X )2 ) = 2,
Var(ln(X )) = 1,
and, via the substitutions y = 1/x and z = y − n, E(ln(T X )) =
We will later study the partial convergents to x, for example, p1 3 p2 22 p3 333 p4 355 p5 103993 = , = , = = = , , , ... q1 1 q2 7 q3 106 q4 113 q5 33102
n+1 1 ln − = 1 − γ = 0.4227843351... n n+1
µ
1
¶
¶
(which is related to de la Vall´ ee Poussin’s theorem [2, 4]),
2 E((T X ) ) = ln(2π ) − γ − 1,
Var(T X ) = ln(2π) − γ 2 + γ − 2 = 0.0819148075... = (0.2862076300...)2 ,
Ã
1| 1| 1| 1| 1| 1| + + + ··· = + + + ···. |a1 |a2 |a3 |a2 |a3 |a4
!
What can be said about the moments of T j X and of ln(T j X ), where X is a random variable in [0, 1]? There are two cases: the first when X follows the uniform distribution, and the second when X follows the Gauss-Kuzmin distribution: P(X ≤ x) = ln(x + 1) . ln(2)
1 = 7, j π−3 k 1 = 15, j T (π−3) k 1 = 1, j T 2 (π−3) k 1 = 292, j T 3 (π−3) k 1 =1 T 4 (π −3)
k
1| 1| 1| 1| 1| 1| 1| 1| 1| 1| + + + + + + + + + + ··· |7 |15 |1 |292 |1 |1 |1 |2 |1 |3 is the regular continued fraction expansion for π . In words, T discards the first “digit” in any expansion, that is, π =3+ T
∞ X
and its density in terms of the trigamma function: f (x) =
For example, the median of T X is F −1 (1/2) = 0.3846747346.... The cumulative distribution for T 2 X is 1 1 G(x) = P(T X ≤ x) = F −F n n+x n=1 µ µ ¶ µ ¶¶ ∞ X 1 1 +1 −ψ +1 , = ψ n n+x n=1
2 ∞ µ X
µ ¶
µLeabharlann ¶¶its density is g(x) =
n=1
and its median is G−1 (1/2) = 0.42278.... It is certainly inconvenient that F 6= G !
∞ X
ψ0
µ
1 1 , +1 n+x (n + x)2
¶
Continued Fraction Transformation
(2 − γ ) ln(2) − 1 = −0.3474517057.... (3/2) ln(2) − 1
π2 = −1.1865691104... = E(ln(T X )), 12 ln(2) 3ζ (3) 2 = E(ln(T X )2 ), E(ln(X ) ) = 2 ln(2) 216 ln(2)ζ (3) − π 4 = 1.1933560457... 144 ln(2)2 = (1.0924083695...)2 = Var(ln(T X )),
´ 1 ³ 6 − π 2 + 6γ , 12 2 Cov(X, T X ) 6 − π + 6γ q ρ(X, T X ) = q =√ q = −0.4098133678... Var(X ) Var(T X ) 12 ln(2π) − γ 2 + γ − 2
E(X · T X ) = 1 −
π2 , 12
⎧ ½ ¾ ⎨ 1
0
x
if 0 < x ≤ 1, if x = 0
j
where {ξ } = ξ − bξ c denotes the fractional part of ξ . For example, π − 3 = 0.141592..., T (π − 3) = 0.062513..., T 2 (π − 3) = 0.996594..., T 3 (π − 3) = 0.003417..., T 4 (π − 3) = 0.634591..., and
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0.2. Gauss-Kuzmin Distribution. If X is a random variable following the Gauss-Kuzmin distribution on [0, 1], then E(X ) = 1 − 1 = 0.4426950408... = E(T X ), ln(2) 1 = E((T X )2 ), 2 ln(2)
" µ ¶2
Var(ln(X )) =
∞ 1 X 1 n+1 ln E(ln(X ) · ln(T X )) = ln(2) n=1 2 n
ln((n + 1)n) + ln(n) Li2
¶ Ã
1 1 − ln(n + 1) Li2 − + ln(n + 1) Li2 n+1 (n + 1)2 Ã !# µ ¶ µ ¶ 1 1 1 +2 Li3 − − 2 Li3 − + Li3 n n+1 (n + 1)2 " Ã !# ∞ ζ (2k) − 1 ζ 0 (2k) ζ 00 (2k ) 1 3ζ (3) X + = − − + , ln(2) 2 k3 k2 2k k=1
j
Continued Fraction Transformation
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0.3.
Variance of Sample Mean. Let us consider the sample mean μ ˆn (X ) = − 1 X ln(T j X ), n 0≤j<n
3
¶
=
n=1 0
= − ln(2) +
Ã
k=2
(−1)
k ζ (k )
−1 k−1
!
µ
= −1.2577468869...
(this constant appears elsewhere [5, 6]), E(ln(T X )2 ) = −2
∞ X 1 ζ (k + 1) − 1 1 Li2 − = ζ (2) − 2 (−1)k , n k2 n=1 n k=1 ∞ X
Z1
0
1 x dx = , 2
E(X ) =
2
Z1
0
1 x2 dx = , 3 1 12
Var(X ) = E(X 2 ) − E(X )2 = and, via the substitution y = 1/x, E(T X ) = =
Z1 ½ ¾
0 ∞ X
n=1
µ
∞ Z Z+1 ∞ n X 1 {y } y−n dx = dy = dy x y2 y2 n=1 n
µ
¶
Var(ln(T X )) = 1.2665694005... = (1.1254196552...)2 , E(ln(X ) · ln(T X )) = 1 n+1 1 ln (1 + ln(n)) − Li2 n n+1 n=1 n
k=2 ∞ X ∞ X
∙
µ
= −ζ (2) +
"Ã
¶
ζ (2) −
Continued Fraction Transformation Steven Finch May 23, 2007 We are interested in iterates of the continued fraction transformation T : [0, 1] → [0, 1] defined by [1] T (x) = ⎩
E(X 2 ) = 1 − Var(X ) =
(3/2) ln(2) − 1 = 0.0826735803... = (0.2875301381...)2 = Var(T X ) ln(2)2 by invariance under T , and E(X · T X ) = 1 − ρ(X, T X ) = Likewise, E(ln(X )) = − γ , ln(2) Cov(X, T X ) = (2 − γ ) ln(2) − 1 , ln(2)2
∞ Z Z+1 ∞ n X 1 ln {y } ln(y − n) ln dx = dy = dy x y2 y2 n=1 n 1
½ ¾
Continued Fraction Transformation
∞ Z X 1 ∞ X ln(z ) 1 n+1 ln dz = − 2 (z + n) n n=1 n ∞ X