20 Hyper Geometric
if \(a\) and/or \(b\) are negative integers, then it will become a polynomial of degree \(n\).
General Form¶
something
\[ (1-x^2) \]
Standard Form¶
\[ x(1-x) y'' + \Big[c - (a+b+1)x \Big]y' - (ab)y = 0 \]
where \(a, b, c\) are real constants
\(x=0, x=1\) are the regular singular points of equation
By Frobenius Series method, at regular singular points, we get 2 initial roots
- \(m=0\)
- \(m=1-c\)
| \(m=0\) | \(m=1-c\) | |
|---|---|---|
| Solution \(y\) | \(1 + \frac{a \cdot b}{1 \cdot c}x + \frac{a(a\textcolor{hotpink}{+1}) \cdot b(b\textcolor{hotpink}{+1})}{1(1\textcolor{hotpink}{+1}) \cdot c(c\textcolor{hotpink}{+1})} x^2 + \\ \frac{a(a\textcolor{hotpink}{+1})(a\textcolor{orange}{+2}) \cdot b(b\textcolor{hotpink}{+1})(b\textcolor{orange}{+2})}{1(1\textcolor{hotpink}{+1})(1\textcolor{orange}{+2}) \cdot c(c\textcolor{hotpink}{+1})(c\textcolor{orange}{+2})} x^3 + \dots\) | \(x^{\textcolor{hotpink}{1-c}} \times F(a + \textcolor{hotpink}{1-c}, b+ \textcolor{hotpink}{1-c}, 2-c, x)\) |
| \(F(a, b, c , x) = F(b, a, c, x)\) Commutative |
| Constant | Outcome |
|---|---|
| \(a \le 0\) or \(b \le 0\) | Series breaks off into finite polynomial |
| \(c\le 0\) | Solution doesn’t exist |