多重超幾何級数のFourier-Legendre展開

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この著者は初心者として投稿しています。間違いや考慮が足りていない点が含まれている可能性が高いです。見つけたらコメント欄で優しく指摘してあげましょう。

$\underset{―}{記法}$

$β_{n}^{} = 2^{- 2 n} (\binom{2 n}{n}), K (x) = \frac{π}{2} \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n}, A = \sum_{n = 0}^{\infty} β_{n}^{3}, B = \sum_{n = 1}^{\infty} {(2 n β_{n}^{})}^{- 3} C = \sum_{n = 0}^{\infty} β_{n}^{4}, D = \sum_{n = 1}^{\infty} {(2 n β_{n}^{})}^{- 4}$
$E = \sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} (2 m β_{m}^{})^{- 3}, F = \sum_{n = 1}^{\infty} {(2 n β_{n}^{})}^{- 4} \sum_{m = 0}^{n} β_{m}^{3}, G = \sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} (2 m β_{m}^{})^{- 4}, H = \sum_{n = 1}^{\infty} {(2 n β_{n}^{})}^{- 4} \sum_{m = 0}^{n} β_{m}^{4}$
$γ_{n}^{(1)} = \sum_{m = 0}^{n} β_{m}^{2} β_{n - m}^{}, γ_{n}^{(2)} = \sum_{m = 1}^{n} \frac{β_{n - m}^{}}{(2 m β_{m}^{})^{2}}, J = \sum_{n = 0}^{\infty} β_{n}^{2} γ_{n}^{(2)}$

$L e g e n d r e 多項式$

$定義$ Legendre多項式 $P_{n}^{} (x)$ は次式で定義されます．

\begin{array}{r} \frac{1}{\sqrt{1 - 2 t x + t^{2}}} = \sum_{n = 0}^{\infty} t^{n} P_{n}^{} (x) \end{array}

　
$漸化式・微分方程式$ $P_{n}^{} (x)$ は次の漸化式・微分方程式を満たします．

\begin{aligned} (2 n + 1) x P_{n}^{} (x) = (n + 1) P_{n + 1}^{} (x) + n P_{n - 1}^{} (x) \\ (2 n + 1) (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) = n (n + 1) (P_{n - 1}^{} (x) - P_{n + 1}^{} (x)) \\ \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) + n (n + 1) P_{n}^{} (x) = 0 \end{aligned}

また，これらを用いて次の等式を証明することができます．

\begin{aligned} (\frac{d}{d x} x (1 - x^{2}) \frac{d}{d x} - x) f (x) \cdot P_{n}^{} (x) = \frac{d}{d x} x (1 - x^{2}) (P_{n}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{n}^{} (x)) - \frac{n^{3} P_{n - 1}^{} (x) + (n + 1)^{3} P_{n + 1}^{} (x)}{2 n + 1} f (x) \\ (1) & ⟺ & {\begin{cases} P_{2 n}^{} (x) f (x) = {(- 1)}^{n} β_{n}^{3} (f (x) - \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{d}{d x} x (1 - x^{2}) (P_{2 m - 1}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m - 1}^{} (x)) - (\frac{d}{d x} x (1 - x^{2}) \frac{d}{d x} - x) f (x) \cdot P_{2 m - 1}^{} (x))) \\ P_{2 n - 1}^{} (x) f (x) = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{k}^{3} (\frac{d}{d x} x (1 - x^{2}) (P_{2 m}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m}^{} (x)) - (\frac{d}{d x} x (1 - x^{2}) \frac{d}{d x} - x) f (x) \cdot P_{2 m}^{} (x)) \end{cases} \end{aligned}

Fourier - Legendre 展開

f (x)

のFourier-Legendre展開は次式で定義されます．

\begin{array}{r} f (x) = \frac{1}{2} \sum_{n = 0}^{\infty} (2 n + 1) P_{n}^{} (x) \int_{- 1}^{1} f (t) P_{n}^{} (t) d t \end{array}

また，本稿では， $f (x) P_{n}^{} (x)$ を $(0, 1)$ で積分したものをFL係数と呼ぶことにします． $n$ の偶奇で $P_{n}^{} (x)$ の偶奇が異なり， $f (x)$ の偶奇も分けて考えるのが妥当と思います．

$m u l t i p l e h y p e r g e o m e t r i c s e r i e s$

次の微分方程式について考えてみます．

\begin{array}{r} (\frac{d}{d x} x (1 - x^{2}) \frac{d}{d x} - x) f (x) = g (x) \end{array}

$f (x), g (x)$ を偶奇で分けて考えると，次のように表現できます．

\begin{array}{r} {\begin{cases} g (x) = \sum_{n = 1}^{\infty} g_{n}^{} x^{2 n - 1}, & f (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{g_{m}^{}}{(2 m β_{m}^{})^{2}} (A) \\ g (x) = \sum_{n = 0}^{\infty} g_{n}^{} x^{2 n}, & f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{2} g_{m}^{} (B) \end{cases} \end{array}

$(1)$ 式より， $g (x)$ のFL係数が計算できるならば， $f (x)$ のFL係数が計算でき，ひいては多重級数のFL係数も計算できることになると思います．
以降，上記のようなHypergeometric termの多重和のみを扱うこととします． $(1)$ 式の両辺を積分すると，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} f (x) d x - \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (lim_{x \to 1} [x (1 - x^{2}) (P_{2 m - 1}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m - 1}^{} (x))] - lim_{x \to 0} [x (1 - x^{2}) (P_{2 m - 1}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m - 1}^{} (x))] - \int_{0}^{1} P_{2 m - 1}^{} (x) g (x) d x)) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (lim_{x \to 1} [x (1 - x^{2}) (P_{2 m}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m}^{} (x))] - lim_{x \to 0} [x (1 - x^{2}) (P_{2 m}^{} (x) \frac{d}{d x} f (x) - f (x) \frac{d}{d x} P_{2 m}^{} (x))] - \int_{0}^{1} P_{2 m}^{} (x) g (x) d x) \end{cases} \end{array}

ここで， $lim_{x \to 1} x (1 - x^{2}) f (x) \frac{d}{d x} P_{k}^{} (x) = lim_{x \to 0} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} f (x) = lim_{x \to 0} x (1 - x^{2}) f (x) \frac{d}{d x} P_{k}^{} (x) = 0$ であり，

\begin{aligned} (A) の 場 合 ： lim_{x \to 1} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} f (x) = lim_{x \to 1} (1 - x^{2}) f^{'} (x) = \frac{2}{π} \sum_{n = 1}^{\infty} \frac{g_{n}^{}}{(2 n β_{n}^{})^{2}} \\ (B) の 場 合 ： lim_{x \to 1} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} f (x) = lim_{x \to 1} (1 - x^{2}) f^{'} (x) = \frac{π}{2} \sum_{n = 0}^{\infty} β_{n}^{2} g_{n}^{} \end{aligned}

となります．

\begin{array}{r} (2) & {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} f (x) d x - \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (lim_{x \to 1} (1 - x^{2}) f^{'} (x) - \int_{0}^{1} P_{2 m - 1}^{} (x) g (x) d x)) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (lim_{x \to 1} (1 - x^{2}) f^{'} (x) - \int_{0}^{1} P_{2 m}^{} (x) g (x) d x) \end{cases} \end{array}

$Fourier-Legendre 展開$

具体的な $g_{n}^{}$ について考えていきます．

$\underset{―}{1. g_{n}^{} = β_{n}^{} の場合}$

\begin{array}{r} g (x) = \sum_{n = 0}^{\infty} β_{n}^{} x^{2 n} = \frac{1}{\sqrt{1 - x^{2}}}, f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{3} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} f (x) d x - \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{π}{2} A - \frac{1}{(2 m β_{m})^{2}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3})) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = \frac{π}{2} \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (A - β_{m}^{2}) \end{cases} \end{array}

\begin{array}{r} f (x) = \frac{π}{2} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1) P_{2 n - 1}^{} (x)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (A - β_{m}^{2}) \end{array}

と書けることになります．いま，右辺の $m$ に関する和は $n \to \infty$ で $0$ に収束することより $- \sum_{n \leq m}$ と書き換えることができ， $\sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}}$ のFL展開を思い出すと， $\hat{f} (x) = \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{m = 1}^{n} \frac{x^{2 m - 1}}{(2 m β_{m}^{})^{2}}$ と書くことで

\begin{array}{r} \hat{f} (x) = - \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{5} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1) P_{2 m - 1}^{} (x)}{(2 m β_{m}^{})^{3}} \end{array}

となります．

$\underset{―}{2. g_{n}^{} = \frac{1}{2 n β_{n}^{}} の場合}$

\begin{array}{r} g (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n β_{n}^{}} = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}}, f (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{3}}, \hat{f} (x) = \sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} β_{m}^{2} x^{2 m} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} B - \frac{1}{(2 m β_{m})^{2}}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{2}{π} B - β_{m}^{2} (\frac{π^{2}}{8} + \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l)^{2} (2 l - 1) β_{l}^{}})) \end{cases} \end{array}

\begin{aligned} f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} B - \frac{1}{(2 m β_{m})^{2}}) \\ \hat{f} (x) = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{5}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} P_{2 m}^{} (x) \end{aligned}

$\underset{―}{3. g_{n}^{} = β_{n}^{2} の場合}$

\begin{array}{r} g (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} = \frac{2}{π} K (x), f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{4}, \hat{f} (x) = \sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} \frac{x^{2 m - 1}}{(2 m β_{m}^{})^{2}} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{π}{2} C - \frac{2}{π} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} C - \frac{2}{π} {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}}) \end{cases} \end{array}

\begin{aligned} f (x) = - \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1) P_{2 n - 1}^{} (x)}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} C - \frac{2}{π} {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}}) \\ \hat{f} (x) = - \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1) P_{2 l - 1}^{} (x)}{(2 l β_{l}^{})^{3}} \end{aligned}

$\underset{―}{4. g_{n}^{} = \frac{1}{(2 n β_{n}^{})^{2}} の場合}$

\begin{array}{r} g (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}}, f (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}}, \hat{f} (x) = \sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{4}} \sum_{m = 0}^{n - 1} β_{m}^{2} x^{2 m} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} D + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{2}{π} D - {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}} (\frac{π}{2} - \frac{(- 1)^{l - 1} β_{l}^{}}{2 l - 1})) \end{cases} \end{array}

\begin{aligned} f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} D + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \\ \hat{f} (x) = - \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3} P_{2 l}^{} (x) \end{aligned}

$\underset{―}{5. g_{n}^{} = β_{n}^{2} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{3}} の場合}$

\begin{array}{r} g (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{3}}, f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{n}^{4} \sum_{l = 1}^{m} \frac{1}{(2 l β_{l}^{})^{3}}, \hat{f} (x) = \sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} β_{m}^{4} \sum_{l = 1}^{m} \frac{x^{2 l - 1}}{(2 l β_{l}^{})^{2}} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{π}{2} E + \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3} (\frac{2}{π} B - β_{l}^{2} (\frac{π^{2}}{8} + \sum_{k = 1}^{l} \frac{(- 1)^{k - 1} (4 k - 1)}{(2 k)^{2} (2 k - 1) β_{k}^{}}))) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} E - {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}} (\frac{2}{π} B - \frac{1}{(2 l β_{l})^{2}})) \end{cases} \end{array}

\begin{aligned} f (x) = - \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1) P_{2 n - 1}^{} (x)}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} E - {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}} (\frac{2}{π} B - \frac{1}{(2 l β_{l})^{2}})) \\ \hat{f} (x) = - \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{5}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1) P_{2 l - 1}^{} (x)}{(2 l β_{l}^{})^{3}} \end{aligned}

$\underset{―}{6. g_{n}^{} = \frac{1}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{3} の場合}$

\begin{array}{r} g (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{3}, f (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{3}, \hat{f} (x) = \sum_{n = 0}^{\infty} β_{n}^{3} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{2} x^{2 l} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} F + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3} (A - β_{l}^{2})) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{2}{π} F - {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}} (\frac{π}{2} A - \frac{1}{(2 l β_{l})^{2}} \sum_{k = 0}^{l - 1} {(- 1)}^{k} (4 k + 1) β_{k}^{3})) \end{cases} \end{array}

\begin{aligned} f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} F + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3} (A - β_{l}^{2})) \\ \hat{f} (x) = - \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{5} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3} P_{2 l}^{} (x) \end{aligned}

$\underset{―}{7. g_{n}^{} = β_{n}^{2} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} の場合}$

\begin{array}{r} g (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}}, f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{n}^{4} \sum_{l = 1}^{m} \frac{1}{(2 l β_{l}^{})^{4}}, \hat{f} (x) = \sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{4}} \sum_{m = 0}^{n - 1} β_{m}^{4} \sum_{l = 1}^{m} \frac{x^{2 l - 1}}{(2 l β_{l}^{})^{2}} \end{array}

であり，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{π}{2} G - \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3} (\frac{2}{π} D - {(- 1)}^{l} β_{l}^{3} \sum_{l < k} \frac{(- 1)^{k - 1} (4 k - 1)}{(2 k β_{k}^{})^{3}} (\frac{π}{2} - \frac{(- 1)^{k - 1} β_{k}^{}}{2 k - 1}))) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} G - {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} D + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3})) \end{cases} \end{array}

\begin{aligned} f (x) = - \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1) P_{2 n - 1}^{} (x)}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} G - {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} D + \frac{π}{2} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3})) \\ \hat{f} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} (4 l + 1) β_{l}^{6} \sum_{k = 1}^{l} \frac{(- 1)^{k - 1} (4 k - 1) P_{2 k - 1} (x)}{(2 k β_{k}^{})^{3}} \end{aligned}

$\underset{―}{8. g_{n}^{} = \frac{1}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{4} の場合}$

\begin{array}{r} g (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{4}, f (x) = \sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{4}, \hat{f} (x) = \frac{π}{2} \sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{2} x^{2 l} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) f (x) d x = {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} H + \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} C - \frac{2}{π} {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}})) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) f (x) d x = - \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{2}{π} H - {(- 1)}^{n} β_{n}^{3} \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{π}{2} C - \frac{2}{π} \frac{(- 1)^{m - 1}}{(2 m β_{m}^{})^{3}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3})) \end{cases} \end{array}

\begin{aligned} f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) \sum_{n < m} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} (\frac{2}{π} H + \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{n \leq m} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (\frac{π}{2} C - \frac{2}{π} {(- 1)}^{m} β_{m}^{3} \sum_{m < l} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l β_{l}^{})^{3}})) \\ \hat{f} (x) = - \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{4 l - 1}{(2 l β_{l}^{})^{6}} \sum_{k = 0}^{l - 1} {(- 1)}^{k} (4 k + 1) β_{k}^{3} P_{2 k}^{} (x) \end{aligned}

となります．
次に， $x \to \sqrt{1 - x^{2}}$ とした場合のFL展開を計算します．
まず， $h (x) = f (\sqrt{1 - x^{2}}), j (x) = g (\sqrt{1 - x^{2}})$ とし，微分方程式は

\begin{array}{r} (\frac{d}{d x} x (1 - x^{2}) \frac{d}{d x} - x) h (x) = \frac{x}{\sqrt{1 - x^{2}}} j (x) \end{array}

であり，以下の場合分けもします．

\begin{array}{r} {\begin{cases} j (x) = \sum_{n = 1}^{\infty} g_{n}^{} {(\sqrt{1 - x^{2}})}^{2 n - 1}, & h (x) = \sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n} \sum_{m = 1}^{n} \frac{g_{m}^{}}{(2 m β_{m}^{})^{2}} & (C) \\ j (x) = \sum_{n = 0}^{\infty} g_{n}^{} {(1 - x^{2})}^{n}, & h (x) = \sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{2} g_{m}^{} & (D) \end{cases} \end{array}

また， $lim_{x \to 1} x (1 - x^{2}) h (x) \frac{d}{d x} P_{k}^{} (x) = lim_{x \to 1} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} h (x) = lim_{x \to 0} x (1 - x^{2}) h (x) \frac{d}{d x} P_{k}^{} (x) = 0$ ，

\begin{array}{r} {\begin{cases} (C) の 場 合 ： lim_{x \to 0} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} h (x) = P_{k}^{} (0) lim_{x \to 0} x h^{'} (x) = - \frac{2}{π} P_{k}^{} (0) \sum_{n = 1}^{\infty} \frac{g_{n}^{}}{(2 n β_{n}^{})^{2}} \\ (D) の 場 合 ： lim_{x \to 0} x (1 - x^{2}) P_{k}^{} (x) \frac{d}{d x} h (x) = P_{k}^{} (0) lim_{x \to 0} x h^{'} (x) = - \frac{π}{2} P_{k}^{} (0) \sum_{n = 0}^{\infty} β_{n}^{2} g_{n}^{} \end{cases} \end{array}

であるので，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) h (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} h (x) d x + \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} \int_{0}^{1} \frac{x P_{2 m - 1}^{} (x) j (x)}{\sqrt{1 - x^{2}}} d x) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) h (x) d x = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} (- {(- 1)}^{m} β_{m}^{} lim_{x \to 0} x h^{'} (x) - \int_{0}^{1} \frac{x P_{2 m}^{} (x) j (x)}{\sqrt{1 - x^{2}}} d x) \end{cases} \end{array}

となります． $\int_{0}^{1} \frac{x P_{n}^{} (x) j (x)}{\sqrt{1 - x^{2}}} d x$ をうまく計算するために制限がかかりそうです．

$\underset{―}{1. g_{n}^{} = β_{n}^{} の場合}$

\begin{array}{r} j (x) = \sum_{n = 0}^{\infty} β_{n}^{} {(1 - x^{2})}^{n} = \frac{1}{x}, h (x) = \sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{3} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) h (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} h (x) d x + \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) h (x) d x = \frac{π}{2} \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (A - {(- 1)}^{m} β_{m}) \end{cases} \end{array}

\begin{array}{r} h (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (B + \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \end{array}

$\underset{―}{2. g_{n}^{} = \frac{1}{2 n β_{n}^{}} の場合}$

\begin{array}{r} j (x) = \sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{2 n β_{n}^{}} = \frac{1}{x} (\frac{π}{2} - \sin^{- 1} x), h (x) = \sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{3}}, \hat{h} (x) = \sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} β_{m}^{2} {(1 - x^{2})}^{m} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) h (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} h (x) d x + \frac{π}{2} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) h (x) d x = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (\frac{2}{π} B - {(- 1)}^{m} β_{m} \sum_{m < l} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}^{}}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) \hat{h} (x) d x = \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{5} \sum_{m < l} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m)^{2} (2 m - 1) β_{m}^{}} \end{cases} \end{array}

\begin{array}{r} h (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (\int_{0}^{1} h (x) d x + \frac{π}{2} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3}) \end{array}

$\underset{―}{3. g_{n}^{} = γ_{n - 1}^{(1)} の場合}$

\begin{array}{r} j (x) = \sum_{n = 1}^{\infty} γ_{n - 1}^{(1)} {(\sqrt{1 - x^{2}})}^{2 n - 1} = \frac{\sqrt{1 - x^{2}}}{x} \sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n}, h (x) = \sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n} \sum_{m = 1}^{n} \frac{γ_{m - 1}^{(1)}}{(2 m β_{m}^{})^{2}}, \hat{h} (x) = \sum_{n = 1}^{\infty} \frac{γ_{n - 1}^{(1)}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{2} {(1 - x^{2})}^{m} \end{array}

とすれば，

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) h (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} h (x) d x + \frac{2}{π} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} (4 l + 1) β_{l}^{4}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) h (x) d x = \frac{π}{2} \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (C - β_{m}^{2}) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) \hat{h} (x) d x = \frac{π}{2} \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \end{cases} \end{array}

\begin{array}{r} h (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (\int_{0}^{1} h (x) d x + \frac{2}{π} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} (4 l + 1) β_{l}^{4}) \end{array}

$\underset{―}{4. g_{n}^{} = γ_{n}^{(2)} の場合}$

\begin{array}{r} j (x) = \sum_{n = 0}^{\infty} γ_{n}^{(2)} {(1 - x^{2})}^{n} = \frac{\sqrt{1 - x^{2}}}{x} \sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{(2 n β_{n}^{})^{2}}, h (x) = \sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{2} γ_{m}^{(2)} \end{array}

\begin{array}{r} {\begin{cases} \int_{0}^{1} P_{2 n}^{} (x) h (x) d x = {(- 1)}^{n} β_{n}^{3} (\int_{0}^{1} h (x) d x + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} (\frac{π}{2} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} - \sum_{m = 0}^{n - 1} \frac{(- 1)^{m} β_{m}^{}}{2 m + 1} \sum_{l = 0}^{m} {(- 1)}^{l} (4 l + 1) β_{l}^{3} + \sum_{m = 1}^{n - 1} \frac{(- 1)^{m - 1} β_{m}^{}}{2 m} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3})) \\ \int_{0}^{1} P_{2 n - 1}^{} (x) h (x) d x = \frac{π}{2} \frac{(- 1)^{n - 1}}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (J - β_{m}^{2} (\frac{π^{2}}{8} + \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1)}{(2 l)^{2} (2 l - 1) β_{l}^{}})) \end{cases} \end{array}

となります．
ここまでをまとめると

$f (x)$	FL展開
$\sum_{n = 0}^{\infty} β_{n}^{} x^{2 n}$	$\frac{π}{2} \sum_{n = 0}^{\infty} (4 n + 1) β_{n}^{2} P_{2 n}^{} (x)$
$\sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n β_{n}^{}}$	$\sum_{n = 1}^{\infty} \frac{4 n - 1}{(2 n β_{n}^{})^{2}} P_{2 n - 1}^{} (x)$
$\sum_{n = 0}^{\infty} β_{n}^{2} x^{2 n}$	$\frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} P_{2 m}^{} (x)$
$\sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{(2 n β_{n}^{})^{2}}$	$- \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{3}} P_{2 m - 1}^{} (x)$
$\sum_{n = 0}^{\infty} β_{n}^{3} \sum_{m = 1}^{n} \frac{x^{2 m - 1}}{(2 m β_{m}^{})^{2}}$	$- \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{5} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1) P_{2 m - 1}^{} (x)}{(2 m β_{m}^{})^{3}}$
$\sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} β_{m}^{2} x^{2 m}$	$\sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{5}} \sum_{m = 0}^{n - 1} {(- 1)}^{m} (4 m + 1) β_{m}^{3} P_{2 m}^{} (x)$
$\sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} \frac{x^{2 m - 1}}{(2 m β_{m}^{})^{2}}$	$- \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1) P_{2 l - 1}^{} (x)}{(2 l β_{l}^{})^{3}}$
$\sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{4}} \sum_{m = 0}^{n - 1} β_{m}^{2} x^{2 m}$	$- \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3} P_{2 l}^{} (x)$
$\sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} β_{m}^{4} \sum_{l = 1}^{m} \frac{x^{2 l - 1}}{(2 l β_{l}^{})^{2}}$	$- \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{5}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{(- 1)^{l - 1} (4 l - 1) P_{2 l - 1}^{} (x)}{(2 l β_{l}^{})^{3}}$
$\sum_{n = 0}^{\infty} β_{n}^{3} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{2} x^{2 l}$	$- \frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{5} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3} P_{2 l}^{} (x)$
$\sum_{n = 1}^{\infty} \frac{1}{(2 n β_{n}^{})^{4}} \sum_{m = 0}^{n - 1} β_{m}^{4} \sum_{l = 1}^{m} \frac{x^{2 l - 1}}{(2 l β_{l}^{})^{2}}$	$\frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} (4 l + 1) β_{l}^{6} \sum_{k = 1}^{l} \frac{(- 1)^{k - 1} (4 k - 1) P_{2 k - 1} (x)}{(2 k β_{k}^{})^{3}}$
$\sum_{n = 0}^{\infty} β_{n}^{4} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{4}} \sum_{l = 0}^{m - 1} β_{l}^{2} x^{2 l}$	$- \frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} (4 n - 1)}{(2 n β_{n}^{})^{3}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{6} \sum_{l = 1}^{m} \frac{4 l - 1}{(2 l β_{l}^{})^{6}} \sum_{k = 0}^{l - 1} {(- 1)}^{k} (4 k + 1) β_{k}^{3} P_{2 k}^{} (x)$
$\sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n}$	$\frac{π}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x)$
$\sum_{n = 1}^{\infty} \frac{{(\sqrt{1 - x^{2}})}^{2 n - 1}}{(2 n β_{n}^{})^{2}} \sum_{m = 0}^{n - 1} β_{m}^{3}$	$\sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (B + \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{l = 0}^{m - 1} {(- 1)}^{l} (4 l + 1) β_{l}^{3})$
$\sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n} \sum_{m = 1}^{n} \frac{1}{(2 m β_{m}^{})^{3}}$	$\sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (\int_{0}^{1} f (x) d x + \frac{π}{2} \sum_{m = 1}^{n} \frac{(- 1)^{m - 1} (4 m - 1)}{(2 m β_{m}^{})^{5}} \sum_{m \leq l} {(- 1)}^{l} (4 l + 1) β_{l}^{3})$
$\sum_{n = 0}^{\infty} β_{n}^{2} {(1 - x^{2})}^{n} \sum_{m = 1}^{n} \frac{γ_{m - 1}^{(1)}}{(2 m β_{m}^{})^{2}}$	$\sum_{n = 0}^{\infty} {(- 1)}^{n} (4 n + 1) β_{n}^{3} P_{2 n}^{} (x) (\int_{0}^{1} f (x) d x + \frac{2}{π} \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{6}} \sum_{l = 0}^{m - 1} (4 l + 1) β_{l}^{4})$

となります．

$4乗のFourier-Legendre 展開$

$(1)$ 式を受けて， $4$ 乗となる式もどうなるか気になりましたので計算してみました．

\begin{aligned} (2 n + 1) x P_{n}^{} (x) = (n + 1) P_{n + 1}^{} (x) + n P_{n - 1}^{} (x) & [1] \\ (2 n + 1) (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) = n (n + 1) (P_{n - 1}^{} (x) - P_{n + 1}^{} (x)) & [2] \\ \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) = - n (n + 1) P_{n}^{} (x) & [3] \end{aligned}

$[2], [3]$ 式より

\begin{array}{r} (2 n + 1) (1 - x^{2}) \frac{d^{2}}{d x^{2}} (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) = - (n^{4} + 2 n^{3} + n^{2}) P_{n - 1}^{} (x) + (n^{4} + 2 n^{3} + n^{2}) P_{n + 1}^{} (x) \end{array}

$[1], [3]$ 式より

\begin{array}{r} (2 n + 1) x \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) = - (n^{3} + n^{2}) P_{n - 1}^{} (x) - (n^{3} + 2 n^{2} + n) P_{n + 1}^{} (x) \end{array}

以上の式を適当に足し引きして整理すれば

\begin{array}{r} (\frac{d}{d x} (1 - x^{2}) \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} - (1 - x^{2}) \frac{d}{d x} + x) P_{n}^{} (x) = \frac{1}{2 n + 1} ({(n + 1)}^{4} P_{n + 1}^{} (x) - n^{4} P_{n - 1}^{} (x)) \end{array}

となりました．また，これにより

\begin{array}{r} {\begin{cases} \int_{0}^{1} f (x) P_{2 n - 1}^{} (x) d x = \frac{1}{(2 n β_{n}^{})^{4}} \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (lim_{x \to 1} D_{2 m}^{} (x) - lim_{x \to 0} D_{2 m}^{} (x) - \int_{0}^{1} g (x) P_{2 m}^{} (x) d x) \\ \int_{0}^{1} f (x) P_{2 n}^{} (x) d x = β_{n}^{4} (\int_{0}^{1} f (x) d x + \sum_{m = 1}^{n} \frac{4 m - 1}{(2 m β_{m}^{})^{4}} (lim_{x \to 1} D_{2 m - 1}^{} (x) - lim_{x \to 0} D_{2 m - 1}^{} (x) - \int_{0}^{1} g (x) P_{2 m - 1}^{} (x) d x)) \end{cases} \end{array}

\begin{array}{r} D_{n}^{} (x) : = (1 - x^{2}) \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) \cdot f (x) - (1 - x^{2}) \frac{d}{d x} P_{n}^{} (x) \cdot (1 - x^{2}) \frac{d}{d x} f (x) + P_{n}^{} (x) \cdot (1 - x^{2}) \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} f (x) - (1 - x^{2}) P_{n}^{} (x) f (x) \end{array}

\begin{array}{r} g (x) : = (\frac{d}{d x} (1 - x^{2}) \frac{d}{d x} (1 - x^{2}) \frac{d}{d x} - (1 - x^{2}) \frac{d}{d x} + x) f (x) [♠] \end{array}

となりました．また， $[♠]$ 式の解として

\begin{array}{r} f (x) = \frac{2}{π^{2}} \int_{0}^{x} {(K (\sqrt{\frac{1 - x}{2}}) K (\sqrt{\frac{1 + t}{2}}) - K (\sqrt{\frac{1 + x}{2}}) K (\sqrt{\frac{1 - t}{2}}))}^{2} g (t) d t \end{array}

があることがわかりました．これをみると

\begin{array}{r} lim_{x \to 1} D_{n}^{} (x) = \frac{4}{π^{2}} \int_{0}^{1} K {(\sqrt{\frac{1 - t}{2}})}^{2} g (t) d t, lim_{x \to 0} D_{n}^{} (x) = 0 \end{array}

となりそうです． $f (x)$ は奇関数なので

\begin{array}{r} f (x) = \sum_{n = 1}^{\infty} \frac{4 n - 1}{(2 n β_{n}^{})^{4}} P_{2 n - 1}^{} (x) \sum_{m = 0}^{n - 1} (4 m + 1) β_{m}^{4} (\frac{4}{π^{2}} \int_{0}^{1} K {(\sqrt{\frac{1 - t}{2}})}^{2} g (t) d t - \int_{0}^{1} g (t) P_{2 m}^{} (t) d t) \end{array}

となります．

$\bar{f} (x) = s i g n (x) f (x)$

ところで， $\bar{f} (x) = s i g n (x) f (x)$ と定義するならば， $f (x)$ と $\bar{f} (x)$ の偶奇は逆になり，また $0 < x < 1$ において両者は一致することにより

\begin{array}{r} f (x) = \sum_{n = 0}^{\infty} (4 n + 1) P_{2 n}^{} (x) \int_{0}^{1} f (t) P_{2 n}^{} (t) d t = \sum_{n = 1}^{\infty} (4 n - 1) P_{2 n - 1}^{} (x) \int_{0}^{1} f (t) P_{2 n - 1}^{} (t) d t = \bar{f} (x) (0 < x < 1) \end{array}

と書けるのではないでしょうか．

\begin{array}{r}  \end{array}

投稿日：2024年9月23日

更新日：2024年10月9日

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$L e g e n d r e 多項式$

$m u l t i p l e h y p e r g e o m e t r i c s e r i e s$

$Fourier-Legendre 展開$

$4乗のFourier-Legendre 展開$

$\bar{f} (x) = s i g n (x) f (x)$