高校数学解説

Apery様の加速級数

級数,コネクター,ζ値

143

あいさつ

んちゃ！
今回はゼータ関数の加速級数を発見したので紹介いたします。
使用するのはコネクターだけです。
参考記事：記号や詳細についてはこちらの記事を参照(👈クリック)

Apery様のζ

コネクター

$C (n, m) = \frac{1}{{(\begin{matrix} n + m \\ m \end{matrix})}^{k}}$

差分を計算

\begin{array}{rcl} C (n, m - 1) - C (n, m) & = & \frac{1}{{(\begin{array}{c} n + m \\ m \end{array})}^{k}} {\frac{(n + m)^{k}}{m^{k}} - 1}) \\ = & \frac{\sum_{l = 1}^{k} (\begin{array}{c} k \\ l \end{array}) n^{l} m^{k - l}}{m^{k}} C (n, m) \end{array}

今回使用する級数

$f_{N} (n) = \sum_{n < m} \frac{C (n, m)}{m^{N}}$

$f_{N} (n - 1) = 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{matrix} 2 m \\ m \end{matrix})}^{k}} + \sum_{n - 1 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k}}{p^{N} {(\begin{matrix} m + p \\ p \end{matrix})}^{k}} - \sum_{n - 1 < m} \sum_{m < p} \frac{1}{p^{N} {(\begin{matrix} m + p \\ p \end{matrix})}^{k}}$

[1]
$\begin{array}{rcl} f_{N} (n - 1) - f_{N} (n) & = & \sum_{n - 1 < m} \frac{C (n - 1, m)}{m^{N}} - \sum_{n < m} \frac{C (n, m)}{m^{N}} \\ = & \frac{C (n - 1, n)}{n^{N}} + \sum_{n < m} \frac{C (n - 1, m) - C (n, m)}{m^{N}} \\ = & \frac{(n!)^{k} {(n - 1)!}^{k}}{n^{N} {(2 n - 1)!}^{k}} + \sum_{n < m} \frac{\sum_{l = 1}^{k} (\begin{array}{c} k \\ l \end{array}) n^{k - l} m^{l} C (n, m)}{n^{k} m^{N}} \\ = & \frac{2^{k}}{n^{N} {(\begin{array}{c} 2 n \\ n \end{array})}^{k}} + \frac{1}{n^{k}} \sum_{l = 1}^{k} (\begin{array}{c} k \\ l \end{array}) n^{k - l} \sum_{n < m} \frac{C (n, m)}{m^{N - l}} \\ = & \frac{2^{k}}{n^{N} {(\begin{array}{c} 2 n \\ n \end{array})}^{k}} + \frac{1}{n^{k}} \sum_{l = 1}^{k} (\begin{array}{c} k \\ l \end{array}) n^{k - l} f_{N - l} (n) \end{array}$
[3]
$\begin{array}{rcl} f_{N} (n - 1) & = & f_{N} (n) + \frac{2^{k}}{n^{N} {(\begin{array}{c} 2 n \\ n \end{array})}^{k}} + \sum_{l = 0}^{k} (\begin{array}{c} k \\ k - l \end{array}) n^{k - l - k} f_{N + k - l - k} (n) - f_{N} (n) \\ = & f_{N} (n) + \frac{2^{k}}{n^{N} {(\begin{array}{c} 2 n \\ n \end{array})}^{k}} + \sum_{l = 0}^{k} (\begin{array}{c} k \\ l \end{array}) n^{l - k} f_{N + l - k} (n) - f_{N} (n) \\ = & f_{N} (n) + \frac{2^{k}}{n^{N} {(\begin{array}{c} 2 n \\ n \end{array})}^{k}} + \sum_{l = 0}^{k - 1} (\begin{array}{c} k \\ l \end{array}) n^{l - k} f_{N + l - k} (n) \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \sum_{l = 0}^{k - 1} (\begin{array}{c} k \\ l \end{array}) \frac{1}{m^{k - l}} f_{N + l - k} (m) \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \sum_{l = 0}^{k - 1} (\begin{array}{c} k \\ l \end{array}) \frac{1}{m^{k - l}} \sum_{m < p} \frac{1}{p^{N + l - k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \sum_{m < p} \sum_{l = 0}^{k - 1} (\begin{array}{c} k \\ l \end{array}) \frac{1}{m^{k - l} p^{l}} \frac{1}{p^{N - k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \sum_{m < p} {\sum_{l = 0}^{k} (\begin{array}{c} k \\ l \end{array}) \frac{1}{m^{k - l} p^{l}} - \frac{1}{p^{k}}} \frac{1}{p^{N - k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \sum_{m < p} (\frac{1}{m} + \frac{1}{p})^{k} \frac{1}{p^{N - k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} - \sum_{n - 1 < m} \sum_{m < p} \frac{1}{p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & 2^{k} \sum_{n - 1 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{n - 1 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k}}{p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} - \sum_{n - 1 < m} \sum_{m < p} \frac{1}{p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \end{array}$

$ζ (N) = 2^{k} \sum_{0 < m} \frac{1}{m^{N} {(\begin{matrix} 2 m \\ m \end{matrix})}^{k}} + \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k}}{p^{N} {(\begin{matrix} m + p \\ p \end{matrix})}^{k}} - \sum_{0 < m} \sum_{m < p} \frac{1}{p^{N} {(\begin{matrix} m + p \\ p \end{matrix})}^{k}}$

直接的証明

[1]

\begin{array}{rcl} \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k}}{p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} - \sum_{0 < m} \sum_{m < p} \frac{1}{p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} & = & \sum_{0 < m} \sum_{m < p} \frac{(m + p)^{k} - m^{k}}{m^{k} p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & \sum_{1 < p} \sum_{0 < m < p} \frac{(m + p)^{k} - m^{k}}{m^{k} p^{N} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & \sum_{1 < p} \frac{1}{p^{N}} \sum_{0 < m < p} \frac{(m + p)^{k} - m^{k}}{m^{k} {(\begin{array}{c} m + p \\ m \end{array})}^{k}} \\ = & \sum_{1 < p} \frac{1}{p^{N}} \sum_{0 < m < p} {\frac{1}{{(\begin{array}{c} m + p - 1 \\ m - 1 \end{array})}^{k}} - \frac{1}{{(\begin{array}{c} m + p \\ m \end{array})}^{k}}} \\ = & \sum_{1 < p} \frac{1}{p^{N}} {1 - \frac{1}{{(\begin{array}{c} 2 p - 1 \\ p - 1 \end{array})}^{k}}} \\ = & \sum_{1 < p} \frac{1}{p^{N}} - 2^{k} \sum_{1 < p} \frac{1}{p^{N} {(\begin{array}{c} 2 p \\ p \end{array})}^{k}} \end{array}

[2]

\begin{array}{rcl} 2^{k} \sum_{0 < m} \frac{1}{m^{N} {(\begin{array}{c} 2 m \\ m \end{array})}^{k}} + \sum_{1 < p} \frac{1}{p^{N}} - 2^{k} \sum_{1 < p} \frac{1}{p^{N} {(\begin{array}{c} 2 p \\ p \end{array})}^{k}} & = & 1 + \sum_{1 < p} \frac{1}{p^{N}} \\ = & ζ (N) \end{array}

$\sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k} - m^{k}}{p^{k} {(\begin{matrix} m + p \\ p \end{matrix})}^{k}} = \sum_{0 < m} \frac{1}{m^{k} (\begin{matrix} 2 m \\ m \end{matrix})} + \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{p^{k} - m^{k}}{p^{k} (\begin{matrix} m + p \\ p \end{matrix})}$

$\begin{array}{rcl} \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k} - m^{k}}{p^{k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} & = & \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{(m + p)^{k} - p^{k}}{p^{k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} + \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{p^{k} - m^{k}}{p^{k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} {\frac{1}{(\begin{array}{c} m + p - 1 \\ p - 1 \end{array})} - \frac{1}{(\begin{array}{c} m + p \\ p \end{array})}} + \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{p^{k} - m^{k}}{p^{k} {(\begin{array}{c} m + p \\ p \end{array})}^{k}} \\ = & \sum_{0 < m} \frac{1}{m^{k} (\begin{array}{c} 2 m \\ m \end{array})} + \sum_{0 < m} \frac{1}{m^{k}} \sum_{m < p} \frac{p^{k} - m^{k}}{p^{k} (\begin{array}{c} m + p \\ p \end{array})} \end{array}$

任意の複素数 $s \in C (1 < Re (s))$ に対して以下の式が成り立つ。
$ζ (s) = (2^{s} + 1) \sum_{0 < m} \frac{1}{m^{s} {(\begin{matrix} 2 m \\ m \end{matrix})}^{s}} + \sum_{0 < m} \frac{1}{m^{s}} \sum_{m < p} \frac{p^{s} - m^{s}}{p^{s} {(\begin{matrix} m + p \\ p \end{matrix})}^{s}}$

謝罪

KBHMさんにより間違いに気づくことが出来ました。
一旦誤った記述部分は消しました。
まだ改善中です。あと少しのところまで来ました。

確認用コード

import sympy as sp

def binomial(n, k):
    return sp.binomial(n, k)

def compute_rhs(k, max_m=100, max_p=100):
    term1 = sum(1 / (sp.Pow(m, k) * sp.Pow(binomial(2 * m, m), k)) for m in range(1, max_m + 1)) * (sp.Pow(2, k) + 1)
   
    term2 = sum((sp.Pow(p, k) - sp.Pow(m, k)) / (sp.Pow(m, k) * sp.Pow(p, k) * sp.Pow(binomial(m + p, p), k))
                for m in range(1, max_m + 1) for p in range(m + 1, max_p + 1))
   
    return term1 + term2

def matching_digits(x, y):
    if x.is_real and y.is_real:
        x_val = float(x.evalf(50))
        y_val = float(y.evalf(50))
        x_str = f"{x_val:.50f}"
        y_str = f"{y_val:.50f}"
    else:
        x_real, x_imag = float(x.as_real_imag()[0].evalf(50)), float(x.as_real_imag()[1].evalf(50))
        y_real, y_imag = float(y.as_real_imag()[0].evalf(50)), float(y.as_real_imag()[1].evalf(50))
        x_str = f"{x_real:.50f}{x_imag:+.50f}j"
        y_str = f"{y_real:.50f}{y_imag:+.50f}j"

    count = 0
    for a, b in zip(x_str, y_str):
        if a == b:
            count += 1
        else:
            break
    return count

def main(k, max_m=10, max_p=10):
    lhs = sp.N(sp.zeta(k), 20)  # ζ(k) の高精度計算
    rhs = compute_rhs(k, max_m, max_p)

    matched_digits = matching_digits(lhs, rhs)
    print(f"LHS (Zeta Function): {lhs}")
    print(f"RHS (Computed Sum) : {rhs.evalf(20)}")
    print(f"Matching Digits    : {matched_digits}")

if name == "main":
    k = 2 + 0.5j  # 複素数の例
    main(k)

投稿日：2月24日

更新日：3月5日

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ずんだもん

Apery様の加速級数

あいさつ
Apery様のζ
確認用コード