Chapter 15: Problem 9
Derive the indicial equation for the Riemann DE.
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 15: Problem 9
Derive the indicial equation for the Riemann DE.
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeDerive the series expansion of the Bessel function of the first kind from that of the confluent hypergeometric series and the expansion of the exponential. Check your answer by obtaining the same result by substituting the power series directly in the Bessel DE.
Consider the function \(v(z) \equiv z^{r}(1-z)^{s} F\left(\alpha_{1}, \beta_{1} ; \gamma_{1} ; 1 / z\right)\) and assume that it is a solution of HGDE. Find a relation among \(r, s, \alpha_{1}, \beta_{1}\), and \(\gamma_{1}\) such that \(v(z)\) is written in terms of three parameters rather than five. In particular, show that one possibility is $$ v(z)=z^{\alpha-\gamma}(1-z)^{\gamma-\alpha-\beta} F(\gamma-\alpha, 1-\alpha ; 1+\beta-\alpha ; 1 / z) . $$ Find all such possibilities.
The linear combination $$ \begin{aligned} \Psi(\alpha, \gamma ; z) \equiv & \frac{\Gamma(1-\gamma)}{\Gamma(\alpha-\gamma+1)} \Phi(\alpha, \gamma ; z) \\ &+\frac{\Gamma(\gamma-1)}{\Gamma(\alpha)} z^{1-\gamma} \Phi(\alpha-\gamma+1,2-\gamma ; z) \end{aligned} $$ is also a solution of the CHGDE. Show that the Hermite polynomials can be Written as $$ H_{n}\left(\frac{z}{\sqrt{2}}\right)=2^{n} \Psi\left(-\frac{n}{2}, \frac{1}{2} ; \frac{z^{2}}{2}\right) $$.
Spherical Bessel functions are defined by $$ f_{i}(z) \equiv \sqrt{\frac{\pi}{2}}\left(\frac{Z_{l+1 / 2}(z)}{\sqrt{z}}\right) . $$ Let \(f_{l}(z)\) denote a spherical Bessel function "of some kind." By direct differentiation and substitution in the Bessel equation, show that $$ \begin{array}{l} \text { (a) } \frac{d}{d z}\left[z^{l+1} f_{l}(z)\right]=z^{l+1} f_{l-1}(z), \\\ \text { (b) } \frac{d}{d z}\left[z^{-l} f_{l}(z)\right]=-z^{-l} f_{l+1}(z) . \end{array} $$ (c) Combine the results of parts (a) and (b) to derive the recursion relations $$ f_{l-1}(z)+f_{l+1}(z)=\frac{2 l+1}{z} f_{l}(z) $$ $$ l f_{l-1}(z)-(l+1) f_{l+1}(z)=(2 l+1) \frac{d f_{l}}{d z} $$
By differentiating the hypergeometric series, show that \(\frac{d^{n}}{d z^{n}} F(\alpha, \beta ; \gamma ; z)=\frac{\Gamma(\alpha+n) \Gamma(\beta+n) \Gamma(\gamma)}{\Gamma(\alpha) \Gamma(\beta) \Gamma(\gamma+n)} F(\alpha+n, \beta+n ; \gamma+n ; z) .\)
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