高斯超幾何函數（Gauss' Hypergeometric）

22 Jun, 2026

　　今天來分享暑假學到的數學，順便來練習LaTex輸入公式（我之前都是使用LibreOffice的語法來輸入公式，接著再用AI幫我轉換，但這次想來點不一樣的）。今天的這一個ODE的解法是來自Frobenius展開。那麼，就來開始上課吧。（順帶一提，這學期應數A+實在是很開心，本來還想說要完蛋了，畢竟段考的分數實在不可口）

Def. Analytic

We consider a function, f(x). If exists $ρ > 0$ such that $\forall x \in C^{\infty} (a - ρ, a + ρ)$ , then we say that f(x) is Analytic at x=a.

Thm. Analytic

If f(x) is analytic at x=a, f(x) can be expanded to be Taylor's Polynomial.

Frobenius Series Solution

We consider an ODE: $x^{2} y^{″} + x P (x) y^{'} + Q (x) y = 0$ .

If both P(x) and Q(x) are analytic at x=a and both $\frac{P (x)}{x}$ and $\frac{Q (x)}{x^{2}}$ are NOT analytic at x=a, then we say we have Regular Singular Point at x=a.（用Frobenius做規則奇點展開）
If both $\frac{P (x)}{x}$ and $\frac{Q (x)}{x^{2}}$ are analytic at x=a, then we say we have Regular Point at x=a.（用常點展開）
If both P(x) and Q(x) are analytic at x=a, then we say that we have Singular Point at x=a.（不可解析，因此暫不考慮）

Frobenius discovered that the series solutions of an ODE are usually like $\sum_{n = 0}^{\infty} a_n x^{n + r}$ where r is the Indicial Root（指標根）. Now, We consider two indicial roots, $r_1$ and $r_2$ , and we let $r_1 - r_2 \geq 0$ .

If $r_{1} - r_{2} \in ℤ$ , then $y_{1} = \sum_{n = 0}^{\infty} a_{n} x^{n + r_{1}}$ and $y_{2} = \sum_{n = 0}^{\infty} a_{n} x^{n + r_{2}}$
If $r_{1} - r_{2} = 0$ , then $y_{1} = \sum_{n = 0}^{\infty} a_{n} x^{n + r_{1}}$ and $y_{2} = y_{1} l n x + \sum_{n = 0}^{\infty} A_{n} x^{n + r_{2}}$
If $r_{1} - r_{2} \in ℕ$ , then $y_{1} = \sum_{n = 0}^{\infty} a_{n} x^{n + r_{1}}$ and $y_{2} = k y_{1} l n x + \sum_{n = 0}^{\infty} A_{n} x^{n + r_{2}}$ for some constant k.

Gauss' Hypergeometric ODE

Consider $x (1 - x) y^{″} + (c - (a + b + 1) x) y^{'} - a b y = 0$

Indicial Root

We rewrite the ODE first.（寫成Frobenius的標準形式） So, we have
$x^{2} y^{″} + x P (x) y^{'} + Q (x) y = x^{2} y^{″} + x (\frac{c - (a + b + 1) x}{1 - x}) y^{'} + \frac{- a b x}{1 - x} y = 0$ . We can note that both P(x) and Q(x) have a Regular Singular Point at x=0.
Solve Indicial Equation
$r (r - 1) + P (0) r + Q (0) = 0 ⟹ r_1 = 0 and r_2 = 1 - c$

General Solution for (y_1)

Now, we consider that $r_1 = 0 and r_2 = 1 - c where c \in ℝ \cap c > 0$

Set $y_{1} = \sum_{l = 0}^{\infty} a_{l} x^{l + r_{1}} = \sum_{l = 0}^{\infty} a_{l} x^{l}$ So, we obtain
$x (1 - x) \sum_{l = 2}^{\infty} a_{l} l (l - 1) x^{l - 2} + (c - (a + b + 1) x) \sum_{l = 1}^{\infty} a_{l} l x^{l - 1} - a b \sum_{l = 0}^{\infty} a_{l} x^{l} =$

\sum_{l = 2}^{\infty} a_{l} l (l - 1) x^{l - 1} - \sum_{l = 2}^{\infty} a_{l} l (l - 1) x^{l} + \sum_{l = 1}^{\infty} c a_{l} l x^{l - 1} - \sum_{l = 1}^{\infty} (a + b + 1) a_{l} l x^{l} - \sum_{l = 0}^{\infty} a b a_{l} x^{l} =

\sum_{l = 1}^{\infty} a_{l + 1} l (l + 1) x^{l} - \sum_{l = 2}^{\infty} a_{l} l (l - 1) x^{l} + \sum_{l = 0}^{\infty} c a_{l + 1} (l + 1) x^{l} - \sum_{l = 1}^{\infty} (a + b + 1) a_{l} l x^{l} - \sum_{l = 0}^{\infty} a b a_{l} x^{l} =

2 a_{2} x + \sum_{l = 2}^{\infty} a_{l + 1} l (l + 1) x^{l} - \sum_{l = 2}^{\infty} a_{l} l (l - 1) x^{l} + c a_{1} + c a_{2} x + \sum_{l = 2}^{\infty} c a_{l + 1} (l + 1) x^{l} - (a + b + 1) a_{1} x =

(- a b a_{0} + c a_{1}) + (- (a + b + 1 + a b) a_{1} + 2 (c + 1) a_{2}) x + \sum_{l = 2}^{\infty} (a_{l + 1} (l + 1) (l + c) - a_{l} (l + a) (l + b)) x^{l}

⟹ a_{l} + 1 = \prod_{m = 0}^{l} \frac{(a + m) (b + m)}{(c + m) (1 + m)} a_{0} = \frac{1}{(l + 1)!} \prod_{m = 0}^{l} \frac{(a + m) (b + m)}{(c + m)}

We take $a_0 = 1$ $y_1 = F (a, b, c; x) = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(a + m) (b + m)}{(c + m) (1 + m)}) x^{l}$

3.推廣

F (1, 1, 1; x) = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(1 + m) (1 + m)}{(1 + m) (1 + m)}) x^{l} = 1 + \sum_{l = 1}^{\infty} x^{l} = \frac{1}{1 - x}

for $| x | < 1$

F (1, b, b; x) = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(1 + m) (b + m)}{(b + m) (1 + m)}) x^{l} = 1 + \sum_{l = 1}^{\infty} x^{l} = \frac{1}{1 - x}

F (a, 1, a; x) = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(a + m) (1 + m)}{(a + m) (1 + m)}) x^{l} = 1 + \sum_{l = 1}^{\infty} x^{l} = \frac{1}{1 - x}

By Ratio Test, we have ${lim}_{l \to \infty} | \frac{a_{l + 1}}{a_{l}} | = {lim}_{l \to \infty} | \frac{x (a + l) (b + l)}{(c + l) (1 + l)} | = {lim}_{l \to \infty} | x | < 1$

5.Common Expansion $F (- n, b, b; - x) = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(- n + m) (b + m)}{(b + m) (1 + m)}) (- x)^{l} = 1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{n - m}{1 + m}) x^{l}$ $= 1 + \sum_{l = 1}^{\infty} \frac{n (n - 1) . . . (n - (l - 1)) (n - l) (n - (l + 1)) . . .2 * 1}{(1 * 2 * . . . * (l - 1) l) ((n - l) (n - (l + 1)) . . .2 * 1)} x^{l} ⟹ 1 + \sum_{l = 1}^{\infty} \frac{n!}{l! (n - l)!} x^{l} = (1 + x)^{n}$ $= 1 + \sum_{l = 1}^{\infty} \frac{n (n - 1) . . . (n - (l - 1)) (n - l) (n - (l + 1)) . . .2 \ * 1}{(1 \ * 2 \ * . . . \ * (l - 1) l) ((n - l) (n - (l + 1)) . . .2 \ * 1)} x^{l} ⟹ 1 + \sum_{l = 1}^{\infty} \frac{n!}{l! (n - l)!} x^{l} = (1 + x)^{n}$ （可以注意到的是，當 $l > n$ 時， $a_{n + 1}$ 會是0，因此級數收斂，所以 $F (- n, b, b; - x) = (1 + x)^{n}$ ）

$1 - n x F (1 - n, 1, 2; x) = 1 - n x (1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(1 - n + m) (1 + m)}{(2 + m) (1 + m)}) x^{l})$ $= 1 - n x (1 + \sum_{l = 1}^{\infty} (- 1)^{l} (\prod_{m = 0}^{l - 1} \frac{(n - 1) - m}{2 + m}) x^{l})$ $= 1 - n x (1 + \sum_{l = 1}^{\infty} (- 1)^{l} (\prod_{m = 0}^{l - 1} \frac{(n - 1) - m}{2 + m}) x^{l})$ $⟹ 1 - n x (1 + \sum_{l = 1}^{n} (- 1)^{l} \frac{(n - 1)!}{(l + 1)! (n - (l + 1))!} x^{l})$ $= 1 - n x + \sum_{l = 1}^{n} (- 1)^{l + 1} \frac{n!}{(l + 1)! (n - (l + 1))!} x^{l + 1} = 1 - n x + \sum_{l = 2}^{n} (- 1)^{l} (\binom{n}{l}) x^{l} = (1 - x)^{n}$ （可以注意到的是，當 $l > n$ 時， $a_{n + 1}$ 會是0，因此級數收斂，所以 $1 - n x F (1 - n, 1, 2; x) = (1 - x)^{n}$ ）

$x F (\frac{1}{2}, 1, \frac{3}{2}; - x^{2}) = x (1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(\frac{1}{2} + m) (1 + m)}{(\frac{3}{2} + m) (1 + m)}) (- x^{2})^{l})$ $= x (1 + \sum_{l = 1}^{\infty} \frac{\frac{1}{2} (\frac{1}{2} + 2) . . . . . . (\frac{1}{2} + (l - 1))}{(\frac{1}{2} + 1) (\frac{1}{2} + 2) . . . . . . (\frac{1}{2} + (l - 1)) (\frac{1}{2} + (l - 1) + 1)} (- x^{2})^{l})$ $= x (1 + \sum_{l = 1}^{\infty} \frac{\frac{1}{2}}{\frac{1}{2} + l} (- x^{2})^{l}) = x + \sum_{l = 1}^{\infty} \frac{(- 1)^{l} x^{2 l + 1}}{2 l + 1} = t a n^{- 1} x for | x | < 1$

$x F (\frac{1}{2}, \frac{1}{2}, \frac{3}{2}; x^{2}) = x (1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(\frac{1}{2} + m) (\frac{1}{2} + m)}{(\frac{3}{2} + m) (1 + m)}) (x^{2})^{l})$ $= x (1 + \sum_{l = 1}^{\infty} \frac{1}{2 l + 1} \frac{1}{2^{l}} \frac{1}{2^{l}} \frac{1 * (2 * 1) * 2 * (2 * 2) * 3 * (2 * 3) . . . . . . (2 l - 1) (2 * l)}{1 * 2 * . . . * l * (1 * 2 * . . . * l)} x^{2 l})$ $= x (1 + \sum_{l = 1}^{\infty} \frac{x^{2} l}{4^{l} (2 l + 1) (l!)^{2}}) = s i n^{- 1} x for | x | < 1$

$x F (1, 1, 2; - x) = x (1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(1 + m) (1 + m)}{(2 + m) (1 + m)}) (- x)^{l})$ $= x (1 + \sum_{l = 1}^{\infty} (\frac{1 * 2 * . . . . . . * l}{2 * 3 * . . . * l * (l + 1)}) (- x)^{l}) = x + \sum_{l = 1}^{\infty} (- 1)^{l} \frac{x^{2 l + 1}}{l + 1} = l n (1 + x)$

$2 x F (\frac{1}{2}, 1, \frac{3}{2}; x^{2}) = 2 x (1 + \sum_{l = 1}^{\infty} (\prod_{m = 0}^{l - 1} \frac{(\frac{1}{2} + m) (1 + m)}{(2 + m) (1 + m)}) (x^{2})^{l}) = 2 x + \sum_{l = 1}^{\infty} \frac{2 x^{2 l + 1}}{2 l + 1} = l n \frac{1 + x}{1 - x}$

General Solution for (y_2)

Now, we consider that $r_1 = 0 and r_2 = 1 - c where c \in ℝ \cap c < 0$

Set $y_2 = \sum_{l = 0}^{\infty} a_l x^{l + (1 - c)}$ . So, We have $x (1 - x) \sum_{l = 0}^{\infty} a_{l} (l + 1 - c) (l - c) x^{l + (1 - c) - 2} + (c - (a + b + 1) x) \sum_{l = 0}^{\infty} a_{l} (l + 1 - c) x^{l + (1 - c) - 1} - a b \sum_{l = 0}^{\infty} a_{l} x^{l + (1 - c)} =$ $\sum_{l = 0}^{\infty} a_{l} (l + 1 - c) x^{l - c} + \sum_{l = 0}^{\infty} a_{l} ((l + 1 - c) (l - a - b - c - 1) - a b) x^{l + (1 - c)}$ $= \sum_{l = 0}^{\infty} a_{l} (l + 1 - c) x^{l - c} + \sum_{l = 1}^{\infty} a_{l - 1} ((1 - c) (l - a - b - c - 2) - a b) x^{l - c}$ $= a_{0} (1 - c) c^{- c} + \sum_{l = 1}^{\infty} (a_{l} (l + 1 - c) + a_{l - 1} ((1 - c) (l - a - b - c - 2) - a b)) x^{l - c}$ $⟹ a_{0} = 0, a_{l} = \prod_{m = 1}^{l - 1} \frac{(a - c + m - 1) (b - c + m - 1)}{(1 + m) ((2 - c) + m)}$ $⟹ y_{2} = x^{1 - c} (1 + \sum_{l = 1}^{\infty} (\prod_{m = 1}^{l - 1} \frac{(a - c + m - 1) (b - c + m - 1)}{(1 + m) ((2 - c) + m)}) x^{l}) = x^{1 - c} F (a - c + 1, b - c + 1, 2 - c; x)$

#物理