高斯超幾何函數（Gauss' Hypergeometric）

　　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 \(\exists \rho > 0\) such that \(\forall x \in C^\infty(a - \rho, a + \rho)\), 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: \(x^2 y'' + xP(x)y' + Q(X)y =0 \). 

• 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 \( r_1 - r_2 \notin\mathbb{Z}\), 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 \(y_2 =y_1 lnx + \sum_{n=0}^{\infty}A_n x^{n+r_2}\) 
• If \( r_1 - r_2 \in \mathbb{N}\), then \(y_1 = \sum_{n=0}^{\infty} a_n x^{n+r_1}\) and \(y_2 =ky_1 lnx + \sum_{n=0}^{\infty}A_n x^{n+r_2}\) for some constant k. 

Gauss' Hypergeometric ODE 

    Consider \(x(1-x)y'' + (c-(a+b+1)x)y'-aby=0\) 

Indicial Root 

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

General Solution for \(y_1\) 

    Now, we consider that \(r_1 = 0 \text{ and } r_2 = 1-c\text{ where }c\in\mathbb{R} \cap c>0\) 

1. Set \(y_1=\sum_{l=0}^{\infty}a_l x^{l+r_1}=\sum_{l=0}^{\infty}a_l x^l\) So, we obtain
   \(x(1-x)\sum_{l=2}^{\infty}a_l l(l-1) x^{l-2}+(c-(a+b+1)x)\sum_{l=1}^{\infty}a_l l x^{l-1}-ab\sum_{l=0}^{\infty}a_l x^l \)
   \( =\sum_{l=2}^{\infty}a_l l(l-1) x^{l-1}-\sum_{l=2}^{\infty}a_l l(l-1) x^{l}+\sum_{l=1}^{\infty}ca_l l x^{l-1}\)
   \(-\sum_{l=1}^{\infty}(a+b+1)a_l l x^{l}-\sum_{l=0}^{\infty}aba_l x^l\)
   \( =\sum_{l=1}^{\infty}a_{l+1} l(l+1) x^{l}-\sum_{l=2}^{\infty}a_l l(l-1) x^{l}+\sum_{l=0}^{\infty}ca_{l+1} (l+1) x^{l}\)
   \(-\sum_{l=1}^{\infty}(a+b+1)a_l l x^{l}-\sum_{l=0}^{\infty}aba_l x^l\)
   \( =2a_2 x+ \sum_{l=2}^{\infty}a_{l+1} l(l+1) x^{l}-\sum_{l=2}^{\infty}a_l l(l-1) x^{l}\)
   \(+ca_1+ca_2 x+\sum_{l=2}^{\infty}ca_{l+1} (l+1) x^{l}-(a+b+1)a_1 x\)
   \(-\sum_{l=2}^{\infty}(a+b+1)a_l l x^{l}-aba_0-aba_1 x - \sum_{l=0}^{\infty}aba_l x^l\)
   \(=(-aba_0+ca_1)+(-(a+b+1+ab)a_1+2(c+1)a_2)x\)
   \(+\sum_{l=2}^{\infty}(a_{l+1} (l+1)(l+c)-a_l (l+a)(l+b))x^l\)
   \(\implies a_l+1 = \prod_{m=0}^{l}\frac{(a+m)(b+m)}{(c+m)(1+m)}a_0=\frac{1}{(l+1)!}{\prod_{m=0}^{l}\frac{(a+m)(b+m)}{(c+m)}}\) 
2. 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. \(F(1,1,1;x)=1+ \sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(1+m)(1+m)}{(1+m)(1+m)})x^l=1+\sum_{l=1}^{\infty}x^l=\frac{1}{1-x} \text{ for } |x| <1\)
   \(F(1,b,b;x)=1+ \sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(1+m)(b+m)}{(b+m)(1+m)})x^l=1+\sum_{l=1}^{\infty}x^l=\frac{1}{1-x} \text{ for } |x| <1\)
   \(F(a,1,a;x)=1+ \sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(a+m)(1+m)}{(a+m)(1+m)})x^l=1+\sum_{l=1}^{\infty}x^l=\frac{1}{1-x} \text{ for } |x| <1\) 
4. By Ratio Test, we have 
   \( \lim_{l \to \infty} |\frac{a_{l+1}}{a_l}|=\lim_{l \to \infty}|\frac{x(a+l)(b+l)}{(c+l)(1+l)}|=\lim_{l \to \infty}|x| <1\) if F(a,b,c;x) is absolutely convergent for some x. In conclusion, F(a,b,c;x) is absolutely convergent if \(a,b,c\in\mathbb{R} \cap c>0\) 
5. Common Expansion（常見展開式）
   \(F(-n,b,b;-x)=1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(-n+m)(b+m)}{(b+m)(1+m)})(-x)^l=1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{n-m}{1+m})x^l\)
   \(=1+\sum_{l=1}^{\infty}\frac{n(n-1)...(n-(l-1))(n-l)(n-(l+1))...2*1}{(1*2*...*(l-1)l)((n-l)(n-(l+1))...2*1)}x^l \implies 1+\sum_{l=1}^{\infty}\frac{n!}{l!(n-l)!}x^l=(1+x)^n\)（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\(F(-n,b,b;-x)=(1+x)^n\)）
   
   \(1-nxF(1-n,1,2;x)=1-nx(1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(1-n+m)(1+m)}{(2+m)(1+m)})x^l)\)
   \(=1-nx(1+\sum_{l=1}^{\infty}(-1)^l (\prod_{m=0}^{l-1} \frac{(n-1)-m}{2+m})x^l)\)
   \(=1-nx(1+\sum_{l=1}^{\infty}(-1)^l \frac{((n-1)(n-2)...(n-l))((n-(l+1))(n-(l+2))...2*1)}{(2*3*...(l+1))((n-(l+1))(n-(l+2))...2*1)})\)
   \(\implies 1-nx(1+\sum_{l=1}^{n}(-1)^l \frac{(n-1)!}{(l+1)!(n-(l+1))!}x^l)\)
   \(=1-nx+\sum_{l=1}^{n}(-1)^{l+1}\frac{n!}{(l+1)!(n-(l+1))!}x^{l+1}=1-nx+\sum_{l=2}^{n}(-1)^l \binom{n}{l}x^l=(1-x)^n\)（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\(1-nxF(1-n,1,2;x)=(1-x)^n\)）
   
   \(xF(\frac{1}{2},1,\frac{3}{2};-x^2)=x(1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(\frac{1}{2} + m)(1+m)}{(\frac{3}{2} + m)(1+m)})(-x^2)^l)\)\(=x(1+\sum_{l=1}^{\infty}\frac{\frac{1}{2}(\frac{1}{2}+2)......(\frac{1}{2}+(l-1))}{(\frac{1}{2}+1)(\frac{1}{2}+2)......(\frac{1}{2}+(l-1))(\frac{1}{2}+(l-1)+1)}(-x^2)^l)\)
   \(=x(1+\sum_{l=1}^{\infty}\frac{\frac{1}{2}}{\frac{1}{2}+l}(-x^2)^l)=x+\sum_{l=1}^{\infty}\frac{(-1)^l x^{2l+1}}{2l+1}={tan^{-1}} x \text{ for } |x|<1\)
   
   \(xF(\frac{1}{2},\frac{1}{2},\frac{3}{2};x^2)=x(1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(\frac{1}{2} + m)(\frac{1}{2}+m)}{(\frac{3}{2} + m)(1+m)})(x^2)^l)\)
   \(=x(1+\sum_{l=1}^{\infty}\frac{1}{2l+1}\frac{1}{2^l}\frac{1}{2^l}\frac{1*(2*1)*2*(2*2)*3*(2*3)......(2l-1)(2*l)}{1*2*...*l*(1*2*...*l)}x^{2l})\)
   \(=x(1+\sum_{l=1}^{\infty}\frac{x^2l}{4^l (2l+1)(l!)^2})={sin^{-1}} x \text{ for } |x|<1\)
   
   \(xF(1,1,2;-x)=x(1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(1+m)(1+m)}{(2+m)(1+m)})(-x)^l)\)
   \(=x(1+\sum_{l=1}^{\infty}(\frac{1*2*......*l}{2*3*...*l*(l+1)})(-x)^l)=x+\sum_{l=1}^{\infty}(-1)^l \frac{x^{2l+1}}{l+1}=ln(1+x)\)
   
   \(2xF(\frac{1}{2},1,\frac{3}{2};x^2)=2x(1+\sum_{l=1}^{\infty}(\prod_{m=0}^{l-1}\frac{(\frac{1}{2} +m)(1+m)}{(2+m)(1+m)})(x^2)^l)=2x+\sum_{l=1}^{\infty}\frac{2x^{2l+1}}{2l+1}\)
   \(=ln\frac{1+x}{1-x}\) 

General Solution for \(y_2\) 

    Now, we consider that \(r_1 = 0 \text{ and } r_2 = 1-c\text{ where }c\in\mathbb{R} \cap c<0\) 

    Set \(y_2=\sum_{l=0}^{\infty}a_l x^{l+(1-c)}\). So, We have
\( x(1-x)\sum_{l=0}^{\infty}a_l (l+1-c)(l-c) x^{l+(1-c)-2}\)
\(+(c-(a+b+1)x)\sum_{l=0}^{\infty}a_l (l+1-c)x^{l+(1-c)-1}-ab\sum_{l=0}^{\infty}a_l x^{l+(1-c)} \)
\( =\sum_{l=0}^{\infty}a_l (l+1-c) x^{l-c}+\sum_{l=0}^{\infty}a_{l} ((l+1-c)(l-a-b-c-1)-ab)x^{l+(1-c)} \)
\( =\sum_{l=0}^{\infty}a_l (l+1-c) x^{l-c}+\sum_{l=1}^{\infty}a_{l-1} ((1-c)(l-a-b-c-2)-ab)x^{l-c} \)
\(=a_0 (1-c)c^{-c} +\sum_{l=1}^{\infty}(a_l (l+1-c) +a_{l-1} ((1-c)(l-a-b-c-2)-ab))x^{l-c} \)
\(\implies a_0=0, a_l=\prod_{m=1}^{l-1}\frac{(a-c+m-1)(b-c+m-1)}{(1+m)((2-c)+m)} \)
\(\implies y_2=x^{1-c} (1+\sum_{l=1}^{\infty}(\prod_{m=1}^{l-1}\frac{(a-c+m-1)(b-c+m-1)}{ (1+m)((2-c)+m)})x^l)\)
\(=x^{1-c}F(a-c+1,b-c+1,2-c;x)\)