べき級数の逆数を求る方法について

551

あいさつ

んちゃ！
今回は、与えられたべき級数 $f (z) : = \sum_{n = 0}^{\infty} a_{n} z^{n}$ に対する逆数 $\frac{1}{f (z)} = \sum_{n = 0}^{\infty} b_{n} z^{n}$ を求める方法を考えます。

導入
応用例
最後に

表記

$N_{0} : = {0} \cup N$
$C [x] : = {\sum_{k = 0}^{n} a_{k} x^{k} | a_{0}, a_{1}, . . ., a_{n} \in C}$
$C (x) : = {\frac{f (x)}{g (x)} | f (x) \in C [x], g (x) \in C [x] ∖ {0}}$

導入

同値関係

≃

$C [x]$ に下記の様な同値関係 $≃_{N}$ を定義する。
$\forall f, g \in C [x] : f ≃_{N} g \overset{d e f}{\Leftrightarrow} f (x) - g (x) = O (x^{N + 1})$

同値関係確認

反射律： $\begin{array}{rcl} \forall f (x) = \sum_{n = 0}^{\infty} f (x) \in C [x] : f (x) - f (x) & = & 0 \\ = & O (x^{N + 1}) \end{array}$
$f ≃_{N} f$
対称律：
$\begin{array}{r} \forall f (x) = \sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1}), g (x) = \sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1}) : f (x) - g (x) = O (x^{N + 1}) \end{array}$ $f ≃_{N} g \Leftrightarrow g ≃_{N} f$
推移律：
$\begin{array}{rcl} \forall & f, g, h \in C [x] : f ≃_{N} g \land g ≃_{N} : f (x) = \sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1}) \land g (x) = \sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1}) \land h (x) = \sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1}) \\ f & (x) - g (x) = O (x^{N + 1}) \land g (x) - h (x) = O (x^{N + 1}) \\ f & (x) - h (x) = O (x^{N + 1}) \end{array}$ $f ≃_{N} g \land g ≃_{N} h \Leftrightarrow f ≃ h$

和・積

以下、商集合 $C [x]_{n} : = C [x] / ≃$ に対して以下の和・積を考える。
$\begin{array}{r} {\begin{cases} \forall f, g \in C [x] : [f] + [g] = [f + g] \\ \forall f, g \in C [x] : [f] [g] = [f g] \\ \forall α \in C : f \in C [x] : α [f] = [α f] \end{cases} \end{array}$

well-defined

\begin{array}{r} \forall f, g \in C [x] : \exists f^{^{'}}, g^{^{'}} \in C [x] s . t . f ≃ f^{^{'}} \land g ≃ g^{^{'}} : {\begin{cases} f (x) = \sum_{n = 0}^{N} a_{n} + O (x^{N + 1}) \\ f^{^{'}} (x) = \sum_{n = 0}^{N} a_{n} + O (x^{N + 1}) \\ g (x) = \sum_{n = 0}^{N} b_{n} + O (x^{N + 1}) \\ g^{^{'}} (x) = \sum_{n = 0}^{N} b_{n} + O (x^{N + 1}) \end{cases} \end{array}

を用いると次の様に計算できる。
[1]和：

\begin{array}{rcl} f (x) + g (x) & = & \sum_{n = 0}^{N} a_{n} x^{N} + \sum_{n = 0}^{N} b_{n} x^{N} + O (x^{N + 1}) \\ = & \sum_{n = 0}^{N} (a_{n} + b_{n}) x^{n} + O (x^{N + 1}) \\ = & f^{^{'}} (x) + g^{^{'}} (x) + O (x^{N + 1}) \end{array}

[f + g] = [f^{^{'}} + g^{^{'}}]

[2]積：

\begin{array}{rcl} f (x) g (x) & = & \sum_{m = 0}^{N} a_{m} x^{m} \sum_{n = 0}^{N} b_{n} x^{n} + O (x^{N + 1}) \\ = & \sum_{n = 0}^{N} (\sum_{m = 0}^{n} a_{m} b_{n - m}) x^{n} + O (x^{N + 1}) \\ = & f^{^{'}} (x) g^{^{'}} (x) \end{array}

[f g] = [f^{^{'}} g^{^{'}}]

[3]スカラー積：

\begin{array}{rcl} α f (x) & = & α {\sum_{n = 0}^{N} a_{n} x^{n} + O (x^{N + 1})} \\ = & \sum_{n = 0}^{N} α a_{n} x^{n} + O (x^{N + 1}) \\ = & α f^{^{'}} (x) \end{array}

[α f] = [α f^{^{'}}]

同型射

$φ : C [x]_{N} ∋ [\sum_{n = 0}^{N} a_{n} x^{n}] \to (\begin{matrix} a_{0} \\ a_{1} \\ ⋮ \\ a_{N} \end{matrix}) \in C^{N + 1}$ とすると $φ$ は同型射。

同型射である事を確認

[1]

φ

の全単射性は明らかなので省略
[2]線形性：

\begin{array}{rcl} \forall α, β \in C : \forall [\sum_{n = 0}^{N} a_{n} x^{n}], [\sum_{n = 0}^{N} b_{n} x^{n}] \in C [x]_{n} : φ (α [\sum_{n = 0}^{N} a_{n} x^{n}] + β [\sum_{n = 0}^{N} b_{n} x^{n}]) & = & φ ([\sum_{n = 0}^{N} (α a_{n} + β b_{n}) x^{n}]) \\ = & (\begin{array}{c} α a_{0} + β b_{0} \\ α a_{1} + β b_{1} \\ ⋮ \\ α a_{N} + β b_{N} \end{array}) \\ = & α (\begin{array}{c} a_{0} \\ a_{1} \\ ⋮ \\ a_{N} \end{array}) + β (\begin{array}{c} b_{0} \\ b_{1} \\ ⋮ \\ b_{N} \end{array}) \\ = & α φ ([\sum_{n = 0}^{N} a_{n} x^{n}]) + β φ ([\sum_{n = 0}^{N} b_{n} x^{n}]) \end{array}

線形写像

$\forall f = \sum_{n = 0}^{N} a_{n} + O (x^{N + 1}) \in C [x]$ に対して以下の様な $(N + 1) \times (N + 1)$ 行列を対応付ける。
$f \mapsto F = (\begin{matrix} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{matrix})$
すると、以下の式が成り立つ。
$\forall [g] \in C [x]_{N} : φ ([f g]) = F φ ([g])$

$g (x) = \sum_{n = 0}^{N} b_{n} x^{n} + O (x^{N + 1})$ とする。
すると下記の様に計算できる。
$φ (g) = (\begin{matrix} b_{0} \\ b_{1} \\ ⋮ \\ b_{N} \end{matrix})$
$\begin{array}{rcl} f (x) g (x) & = & \sum_{m = 0}^{N} a_{m} \sum_{n = 0}^{N} b_{n} x^{m + n} + O (x^{N + 1}) \\ = & \sum_{n = 0}^{N} (\sum_{m = 0}^{n} a_{n - m} b_{m}) x^{n} + O (x^{N + 1}) \end{array}$
ゆえに、
$\begin{array}{rcl} φ ([f g]) & = & (\begin{array}{c} \sum_{m = 0}^{0} a_{0 - m} b_{m} \\ \sum_{m = 0}^{1} a_{1 - m} b_{m} \\ ⋮ \\ \sum_{m = 0}^{N} a_{N - m} b_{m} \end{array}) \\ = & (\begin{array}{c} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{array}) (\begin{array}{c} b_{0} \\ b_{1} \\ ⋮ \\ b_{N} \end{array}) \\ = & F φ ([g]) \end{array}$

可換性

$F = (\begin{matrix} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{matrix}), G = (\begin{matrix} b_{0} & 0 & 0 & 0 & \dots & 0 \\ b_{1} & b_{0} & 0 & 0 & \dots & 0 \\ b_{2} & b_{1} & b_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ b_{N} & b_{N - 1} & b_{N - 2} & b_{N - 3} & \dots & b_{0} \end{matrix})$ の様な $(N + 1) \times (N + 1)$ 行列を考えると $[F, G] : = F G - G F = 0$

[1] $N = 0$ の場合は明らか。
[2]そこで、 $0, 1, 2, . . ., N$ まで成り立つと仮定。
そして、次の様にブロック分割する。
$\begin{array}{r} {\begin{cases} F : = (\begin{array}{c} \overset{―}{F} & 0_{N + 1} \\ A_{N + 1} & a_{0} \end{array}) \\ \overset{―}{F} : = (\begin{array}{c} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{array}) \\ A_{N + 1} : = (\begin{array}{c} a_{N + 1} & a_{N} & \dots a_{1} \end{array}) \end{cases} \end{array}$
$\begin{array}{r} {\begin{cases} G : = (\begin{array}{c} \overset{―}{G} & 0_{N + 1} \\ B_{N + 1} & b_{0} \end{array}) \\ \overset{―}{F} : = (\begin{array}{c} b_{0} & 0 & 0 & 0 & \dots & 0 \\ b_{1} & b_{0} & 0 & 0 & \dots & 0 \\ b_{2} & b_{1} & b_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ b_{N} & b_{N - 1} & b_{N - 2} & b_{N - 3} & \dots & b_{0} \end{array}) \\ B_{N + 1} : = (\begin{array}{c} b_{N + 1} & b_{N} & \dots b_{1} \end{array}) \end{cases} \end{array}$
すると以下の式が得られる。
$\begin{array}{rcl} F G & = & (\begin{array}{c} \overset{―}{F} & 0_{N + 1} \\ A_{N + 1} & a_{0} \end{array}) (\begin{array}{c} \overset{―}{G} & 0_{N + 1} \\ B_{N + 1} & b_{0} \end{array}) \\ = & (\begin{array}{c} \overset{―}{F} \overset{―}{G} + 0_{N + 1} B_{N + 1} & \overset{―}{F} 0_{N + 1} + 0_{N + 1} b_{0} \\ A_{N + 1} \overset{―}{G} + a_{0} B_{N + 1} & A_{N + 1} 0_{N + 1} + a_{0} b_{0} \end{array}) \\ = & (\begin{array}{c} \overset{―}{F} \overset{―}{G} & 0_{N + 1} \\ A_{N + 1} \overset{―}{G} + a_{0} B_{N + 1} & a_{0} b_{0} \end{array}) \end{array}$
$\begin{array}{rcl} G F & = & (\begin{array}{c} \overset{―}{G} \overset{―}{F} & 0_{N + 1} \\ B_{N + 1} \overset{―}{F} + b_{0} A_{N + 1} & b_{0} a_{0} \end{array}) \end{array}$
ゆえに、以下の結果が得られる。
$\begin{array}{rcl} [F, G] & = & F G - G F \\ = & (\begin{array}{c} [\overset{―}{F}, \overset{―}{G}] & 0_{N + 1} \\ A_{N + 1} \overset{―}{G} + a_{0} B_{N + 1} - (B_{N + 1} \overset{―}{F} + b_{0} A_{N + 1}) & 0 \end{array}) \\ = & (\begin{array}{c} O_{N + 1, N + 1} & 0_{N + 1} \\ A_{N + 1} \overset{―}{G} + a_{0} B_{N + 1} - (B_{N + 1} \overset{―}{F} + b_{0} A_{N + 1}) & 0 \end{array}) \end{array}$
さらに次の様に計算できる。
$\begin{array}{rcl} [A_{N + 1} \overset{―}{G} + a_{0} B_{N + 1} - (B_{N + 1} \overset{―}{F} + b_{0} A_{N + 1})]_{k} & = & \sum_{l = 0}^{N - k} a_{N + 1 - k - l} b_{l} + a_{0} b_{N + 1 - k} - (\sum_{l = 0}^{N - k} b_{N + 1 - k - l} a_{l} + b_{0} a_{N + 1 - k}) \\ = & \sum_{l = 0}^{N - k + 1} a_{N + 1 - k - l} b_{l} - \sum_{l = 0}^{N - k + 1} b_{N + 1 - k - l} a_{l} \\ = & 0 \end{array}$
以上の計算により $[F, G] = 0$

$d e t F = a_{0}^{N + 1} \neq 0$ なので明らか。

$a_{0} \in C ∖ {0}$ に対して $F = (\begin{matrix} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{matrix})$ の逆行列 $G = (\begin{matrix} b_{0} & 0 & 0 & 0 & \dots & 0 \\ b_{1} & b_{0} & 0 & 0 & \dots & 0 \\ b_{2} & b_{1} & b_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ b_{N} & b_{N - 1} & b_{N - 2} & b_{N - 3} & \dots & b_{0} \end{matrix})$ は次の様に与えられる。
$\begin{array}{r} {\begin{cases} b_{0} = \frac{1}{a_{0}} \\ b_{n} = - b_{0} \sum_{k = 0}^{n - 1} a_{k - n} b_{k} (n = 1, 2, . . ., N) \end{cases} \end{array}$

$\begin{array}{rcl} F G & = & (\begin{array}{c} a_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} & a_{0} & 0 & 0 & \dots & 0 \\ a_{2} & a_{1} & a_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{N} & a_{N - 1} & a_{N - 2} & a_{N - 3} & \dots & a_{0} \end{array}) (\begin{array}{c} b_{0} & 0 & 0 & 0 & \dots & 0 \\ b_{1} & b_{0} & 0 & 0 & \dots & 0 \\ b_{2} & b_{1} & b_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ b_{N} & b_{N - 1} & b_{N - 2} & b_{N - 3} & \dots & b_{0} \end{array}) \\ = & (\begin{array}{c} a_{0} b_{0} & 0 & 0 & 0 & \dots & 0 \\ a_{1} b_{0} + a_{0} b_{1} & a_{0} b_{0} & 0 & 0 & \dots & 0 \\ a_{2} b_{0} + a_{1} b_{1} + a_{0} b_{2} & a_{1} b_{0} + a_{0} b_{1} & a_{0} b_{0} & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ \sum_{n = 0}^{N} a_{N - n} b_{n} & \sum_{n = 0}^{N - 1} a_{N - n - 1} b_{n} & \sum_{n = 0}^{N - 2} a_{N - n - 2} b_{n} & \sum_{n = 0}^{N - 3} a_{N - n - 3} b_{n} & \dots & a_{0} b_{0} \end{array}) \\ = & E_{N + 1} \end{array}$
よって
$\begin{array}{r} {\begin{cases} b_{0} = \frac{1}{a_{0}} \\ b_{n} = - b_{0} \sum_{k = 0}^{n - 1} a_{k - n} b_{k} \end{cases} \end{array}$

応用

では上記の理論を応用してみましょう。
下記の問題は書き手であるずんだもんの主(やなさん)が思い付きで書いたものになります。必ずしも正しいとは限りません。
必ず、自分で計算して正しいかどうか確認してください。

$f (x) = e^{x} = \sum_{n = 0}^{\infty} \frac{1}{n!} x^{n}$ の逆数 $g (x) = \sum_{n = 0}^{\infty} b_{n} x^{n}$ のテーラー展開を求めてください。

序盤の項の係数について計算しますと下記の様になる事が分かります。
$\begin{array}{r} {\begin{cases} b_{0} = 1 \\ b_{1} = - b_{0} a_{1} b_{0} = - 1 \\ b_{2} = - b_{0} (a_{2} b_{0} + a_{1} b_{1}) = \frac{1}{2} \\ b_{3} = - b_{0} (a_{3} b_{0} + a_{2} b_{1} + a_{1} b_{2}) = - \frac{1}{6} \end{cases} \end{array}$
ゆえに、次の様になると予測できます： $b_{n} = (- 1)^{n} \frac{1}{n!} (n \in N_{0})$
実際、次の様に計算出来きるのでこの予想は正しいです。
$\begin{array}{rcl} b_{n + 1} & = & - b_{0} \sum_{k = 0}^{n} a_{n + 1 - k} b_{k} \\ = & - \sum_{k = 0}^{n} (- 1)^{k} \frac{1}{(n + 1 - k)! k!} \\ = & - \sum_{k = 0}^{n} (- 1)^{k} \frac{1}{(n + 1 - k)! k!} \\ = & - \frac{1}{(n + 1)!} \sum_{k = 0}^{n} (- 1)^{k}_{n + 1} C_{k} \\ = & - \frac{1}{(n + 1)!} \sum_{k = 0}^{n + 1} (- 1)^{k}_{n + 1} C_{k} + (- 1)^{n + 1} \frac{1}{(n + 1)!}_{n + 1} C_{n + 1} \\ = & - \frac{1}{(n + 1)!} (1 - 1)^{n + 1} + (- 1)^{n + 1} \frac{1}{(n + 1)!} \\ = & (- 1)^{n + 1} \frac{1}{(n + 1)!} \end{array}$

Gaussの超幾何関数 $f (x) =_{2} F_{1} (a, b; c; z) = \sum_{n = 0}^{\infty} \frac{(a)_{n} (b)_{n}}{(c)_{n} n!} z^{n}$ の逆数 $g (x) = \sum_{n = 0}^{\infty} b_{n} x^{n}$ がどうなるか予測してください。

序盤の項の係数について計算します。すると下記の様な事が分かります。
$\begin{array}{r} {\begin{cases} b_{0} = 1 \\ b_{1} = - b_{0} a_{1} b_{0} = - \frac{a b}{c} \end{cases} \end{array}$
予想： $\forall n \in N_{0} : \exists {γ_{n}}_{n \in N_{0}} \subset C s . t . {\begin{cases} γ_{0} = 1 \\ b_{n} = (- 1)^{n} γ_{n} \frac{(a)_{n} (b)_{n}}{(c)_{n} n!} \end{cases}$ の形を取る。
$\begin{array}{rcl} b_{n + 1} & = & - b_{0} \sum_{k = 0}^{n} \frac{(a)_{n + 1 - k} (b)_{n + 1 - k}}{(c)_{n + 1 - k} (n + 1 - k)!} (- 1)^{k} γ_{k} \frac{(a)_{k} (b)_{k}}{(c)_{k} k!} \\ = & - \sum_{k = 0}^{n} (- 1)^{k} \frac{(a)_{n + 1 - k} (a)_{k} (b)_{n + 1 - k} (b)_{k}}{(c)_{n + 1 - k} (c)_{k} (n + 1 - k)! k!} γ_{k} \\ = & - \sum_{k = 0}^{n + 1} (- 1)^{k} \frac{(a)_{n + 1 - k} (a)_{k} (b)_{n + 1 - k} (b)_{k}}{(c)_{n + 1 - k} (c)_{k} (n + 1 - k)! k!} γ_{k} + (- 1)^{n + 1} \frac{(a)_{n + 1} (b)_{n + 1}}{(c)_{n + 1} (n + 1)!} γ_{n + 1} \end{array}$
ゆえに、以下の様に $γ_{n}$ を決めましょう。
$γ_{n + 1} = (- 1)^{n} \frac{(c)_{n + 1} (n + 1)!}{(a)_{n + 1} (b)_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a)_{n + 1 - k} (a)_{k} (b)_{n + 1 - k} (b)_{k}}{(c)_{n + 1 - k} (c)_{k} (n + 1 - k)! k!} γ_{k} (n \in N_{0})$
すると下記の式が得られます。
$\begin{array}{r} {\begin{cases} g (z) = 1 - \sum_{n = 1}^{\infty} \frac{z^{n}}{n!} {\sum_{m = 0}^{n - 1} (- 1)^{m} \frac{(a)_{n - m} (a)_{m} (b)_{n - m} (b)_{m}}{(c)_{n - m} (c)_{m}} (\begin{array}{c} n \\ m \end{array}) γ_{m}} \\ γ_{n + 1} = (- 1)^{n} \frac{(c)_{n + 1} (n + 1)!}{(a)_{n + 1} (b)_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a)_{n + 1 - k} (a)_{k} (b)_{n + 1 - k} (b)_{k}}{(c)_{n + 1 - k} (c)_{k} (n + 1 - k)! k!} γ_{k} (n \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$

Gaussの超幾何関数 $f (x) =_{2} F_{1} (a, b; c; z) = \sum_{n = 0}^{\infty} \frac{(a)_{n} (b)_{n}}{(c)_{n} n!} z^{n}$ の逆数 $g (x)$ は下記の様に書ける。
$\begin{array}{r} {\begin{cases} g (z) = 1 - \sum_{n = 1}^{\infty} \frac{z^{n}}{n!} {\sum_{m = 0}^{n - 1} (- 1)^{m} \frac{(a)_{n - m} (a)_{m} (b)_{n - m} (b)_{m}}{(c)_{n - m} (c)_{m}} (\begin{array}{c} n \\ m \end{array}) γ_{m}} \\ γ_{n + 1} = (- 1)^{n} \frac{(c)_{n + 1} (n + 1)!}{(a)_{n + 1} (b)_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a)_{n + 1 - k} (a)_{k} (b)_{n + 1 - k} (b)_{k}}{(c)_{n + 1 - k} (c)_{k} (n + 1 - k)! k!} γ_{k} (n \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$

超幾何関数 $_{N + 1} F_{N} (a_{1}, a_{2}, . . ., a_{N + 1}; b_{1}, b_{2}, . . ., b_{N}; z)$ の逆数 $g (x) = \sum_{n = 0}^{\infty} b_{n} x^{n}$ がどうなるか予測してください。

これも同じ様に計算すればいいです。
$\begin{array}{r} {\begin{cases} b_{n} = - \frac{1}{n!} \sum_{k = 0}^{n - 1} (- 1)^{k} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k}} (\begin{array}{c} n \\ k \end{array}) γ_{k} (n \in N) \\ b_{0} = 1 \\ γ_{n + 1} = (- 1)^{n} \frac{(b_{1})_{n} (b_{2})_{n} \dots (b_{N})_{n}}{(a_{1})_{n} (a_{2})_{n} \dots (a_{N})_{n}} \sum_{k = 0}^{n} (- 1)^{k} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a_{1})_{n + 1 - k} (a_{2})_{n + 1 - k} \dots (a_{N + 1})_{n + 1 - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n + 1 - k} (b_{2})_{n + 1 - k} \dots (b_{N})_{n + 1 - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} (n - k)! k!} γ_{k} (n \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$
実際 $b_{0} = 1$
$\begin{array}{rcl} b_{n} & = & - b_{0} \sum_{k = 0}^{n - 1} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (n - k)!} (- 1)^{k} \frac{(a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} k!} γ_{k} \\ = & - \sum_{k = 0}^{n} (- 1)^{k} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} (n - k)! k!} γ_{k} + (- 1)^{n} \frac{(a_{1})_{n} (a_{2})_{n} \dots (a_{N + 1})_{n}}{(b_{1})_{n} (b_{2})_{n} \dots (b_{N})_{k} n!} γ_{n} \end{array}$
なので以下の式を得る。
$γ_{n + 1} = (- 1)^{n} \frac{(b_{1})_{n + 1} (b_{2})_{n + 1} \dots (b_{N})_{n + 1} (n + 1)!}{(a_{1})_{n + 1} (a_{2})_{n + 1} \dots (a_{N})_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a_{1})_{n + 1 - k} (a_{2})_{n + 1 - k} \dots (a_{N + 1})_{n + 1 - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n + 1 - k} (b_{2})_{n + 1 - k} \dots (b_{N})_{n + 1 - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} (n + 1 - k)! k!} γ_{k} (n \in N_{0})$
ゆえに
$\begin{array}{rcl} b_{n} & = & - \sum_{k = 0}^{n - 1} (- 1)^{k} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} (n - k)! k!} γ_{k} \\ = & - \frac{1}{n!} \sum_{k = 0}^{n - 1} (- 1)^{k} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k}} (\begin{array}{c} n \\ k \end{array}) γ_{k} \end{array}$

超幾何関数 $f (x) =_{N + 1} F_{N} (a_{1}, a_{2}, . . ., a_{N + 1}; b_{1}, b_{2}, . . ., b_{N}; z) = \sum_{n = 0}^{\infty} \frac{(a_{1})_{n} (a_{2})_{n} \dots (a_{N + 1})_{n}}{(b_{1})_{n} (b_{2})_{n} \dots (b_{N})_{n} n!} z^{n}$ の逆数 $g (x)$ は下記の様に書ける。
$\begin{array}{r} {\begin{cases} g (z) = 1 - \sum_{n = 1}^{\infty} \frac{z^{n}}{n!} \sum_{k = 0}^{n - 1} (- 1)^{k} \frac{(a_{1})_{n - k} (a_{2})_{n - k} \dots (a_{N + 1})_{n - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n - k} (b_{2})_{n - k} \dots (b_{N})_{n - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k}} (\begin{array}{c} n \\ k \end{array}) γ_{k} \\ γ_{n + 1} = (- 1)^{n} \frac{(b_{1})_{n} (b_{2})_{n} \dots (b_{N})_{n}}{(a_{1})_{n} (a_{2})_{n} \dots (a_{N})_{n}} \sum_{k = 0}^{n} (- 1)^{k} \sum_{k = 0}^{n} (- 1)^{k} \frac{(a_{1})_{n + 1 - k} (a_{2})_{n + 1 - k} \dots (a_{N + 1})_{n + 1 - k} (a_{1})_{k} (a_{2})_{k} \dots (a_{N + 1})_{k}}{(b_{1})_{n + 1 - k} (b_{2})_{n + 1 - k} \dots (b_{N})_{n + 1 - k} (b_{1})_{k} (b_{2})_{k} \dots (b_{N})_{k} (n + 1 - k)! k!} γ_{k} (n \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$

$ζ (2) =_{3} F_{2} (1, 1, 1 : 2, 2; 1) = \frac{π^{2}}{6}$ を用いると下記の様に書ける。
$\begin{array}{r} {\begin{cases} \frac{6}{π^{2}} = 1 - \sum_{n = 1}^{\infty} \sum_{k = 0}^{n - 1} \frac{γ_{k}}{(n + 1 - k)^{2} (k + 1)^{2}} \\ γ_{n + 1} = (- 1)^{n} (n + 2)^{2} \sum_{k = 0}^{n} (- 1)^{k} \frac{γ_{k}}{(n + 1 - k)^{2} k^{2}} (n \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$

級数 $S (x) = 1 + \sum_{n = 1}^{\infty} a_{n} x^{n}$ に関してその逆数 $T (x) = \sum_{n = 0}^{\infty} b_{n} x^{n}$ がどのようになるか予測してください。

$b_{0} = 1$ を用います。またある数列 ${γ_{n}}_{n \in N_{0}} \subset C$ を用いて次の様に書けたとしましょう。
$b_{n} = (- 1)^{n} a_{n} γ_{n}$
この場合は、
$\begin{array}{rcl} b_{n + 1} & = & - b_{0} \sum_{k = 0}^{n} a_{n + 1 - k} (- 1)^{k} b_{k} \\ = & - \sum_{k = 0}^{n + 1} (- 1)^{k} a_{n + 1 - k} a_{k} γ_{k} + (- 1)^{n + 1} a_{n + 1} γ_{n + 1} \end{array}$
より以下の条件が得られる。
$γ_{n + 1} = (- 1)^{n} \frac{1}{a_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} a_{n + 1 - k} a_{k} γ_{k} (k \in N_{0})$
まとめると次の様です。
$\begin{array}{r} {\begin{cases} b_{n} = - \sum_{k = 0}^{n - 1} (- 1)^{k} a_{n - k} a_{k} γ_{k} \\ b_{0} = 1 \\ γ_{n + 1} = (- 1)^{n} \frac{1}{a_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} a_{n + 1 - k} a_{k} γ_{k} (k \in N_{0}) \\ γ_{0} = 1 \\ T (z) = 1 - \sum_{n = 1}^{\infty} z^{n} \sum_{k = 0}^{n - 1} (- 1)^{k} a_{n - k} a_{k} γ_{k} \end{cases} \end{array}$

級数 $S (x) = 1 + \sum_{n = 1}^{\infty} a_{n} x^{n}$ に関してその逆数 $T (x)$ は下記の様に書ける。
$\begin{array}{r} {\begin{cases} T (z) = 1 - \sum_{n = 1}^{\infty} z^{n} \sum_{k = 0}^{n - 1} (- 1)^{k} a_{n - k} a_{k} γ_{k} \\ γ_{n + 1} = (- 1)^{n} \frac{1}{a_{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} a_{n + 1 - k} a_{k} γ_{k} (k \in N_{0}) \\ γ_{0} = 1 \end{cases} \end{array}$