Solved: TEAM PROJECT. Hypergeometric Equation, Series, and

Chapter 5, Problem 5.1.75

(choose chapter or problem)

Hypergeometric Equation, Series, and Function. Gauss's hypergeometric ODE \({ }^{5}\) is

(15)          x (1 - x) y “ +[c -(a + b + 1) x] y ‘ - aby = 0.

Here, a, b, c are constants. This ODE is of the form \(p_{2} y^{\prime \prime}+p_{1} y^{\prime}+p_{0} y=0\), where \(p_{2}, p_{1}, p_{0}\) are polynomials of degree 2, 1, 0, respectively. These polynomials are written so that the series solution takes a most practical form. namely,

\(y_{1}(x)=1+\frac{a b}{1 ! c} x+\frac{a(a+1) b(b+1)}{2 ! c(c+1)} x^{2}\)

(16)

\(+\frac{a(a+1)(a+2) b(b+1)(b+2)}{3 ! c(c+1)(c+2)} x^{3}+\cdots\)

This series is called the hypergeometric series. Its sum \(y_{1}(x)\) is called the hypergeometric function and is denoted by F(a, b, c ; x). Here, \(c \neq 0,-1,-2, \cdots\). By choosing specific values of a, b, c we can obtain an incredibly large number of special functions as solutions of (15) [see the small sample of elementary functions in part (c)]. This accounts for the importance of (15)

(a) Hypergeometric series and function. Show that the indicial equation of (15) has the roots \(r_{1}=0\) and \(r_{2}=1-c\). Show that for \(r_{1}=0\) the Frobenius method gives (16). Motivate the name for (16) by showing that

\(F(1,1,1 ; x)=F(1, b, b ; x)=F(a, 1, a ; x)=\frac{1}{1-x}\)

(b) Convergence. For what a or b will ( 16) reduce to a polynomial? Show that for any other a, b, c \((c \neq 0,-1,-2, \cdots)\) the series (16) converges when |x| < 1.

(c) Special cases. Show that

\((1+x)^{n}=F(-n, b, b ;-x)\)

\((1-x)^{n}=1-n x F(1-n, 1,2: x)\)

\(arctan x=x F\left(\frac{1}{2}, 1 , \frac{3}{2} ;-x^{2}\right)\)

\(\arcsin x=x F\left(\frac{1}{2}, \frac{1}{2}, \frac{3}{2} ; x^{2}\right)\)

\(\ln (1+x)=x F(1,1,2 ;-x),\)

\(\ln \frac{1+x}{1-x}=2 x F\left(\frac{1}{2}, 1, \frac{3}{2} ; x^{2}\right).\)

Find more such relations from the literature on special functions.

(d) Second solution. Show that for \(r_{2}=1-c\) the Frobenius method yields the following solution (where \(c \neq 2.3 .4 . \cdots)\) :

\(v_{2}(x)=x^{1-c}\left(1+\frac{(a-c+1)(b-c+1)}{1 !(-c+2)} x\right.\)

(17)

\(+\frac{(a-c+1)(a-c+2)(b-c+1)(b-c+2)}{2 !(-c+2)(-c+3)} x^{2}\)

\(+\cdots).\)

Show that

\(y_{2}(x)=x^{1-c} F(a-c+1, b-c+1,2-c ; x)\).

(e) On the generality of the hypergeometric equation. Show that

(18)      \(\left(t^{2}+A t+B\right) \ddot{y}+(C t+D) \dot{y}+K y=0\)

with \(\dot{y}=d y / d t\), etc., constant A, B, C, D, K, and \(t^{2}+A t+B=\left(t-t_{1}\right)\left(t-t_{2}\right), t_{1} \neq t_{2}\), can be reduced to the hypergeometric equation with independent variable

\(x=\frac{t-t_{1}}{t_{2}-t_{1}}\)

and parameters related by \(C t_{1}+D=-c\left(t_{2}-t_{1}\right)\), C = a + b + 1, K = ab. From this you see that (15) is a "normalized form" of the more general (18) and that various cases of (I8) can thus be solved in terms of hypergeometric functions.