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Chapter 6: Problem 13

A solution of Bessel's equation, \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-n^{2}\right) y=0\), can be found using the guess \(y(x)=\sum_{j=0}^{\infty} a_{j} x^{j+n} .\) One obtains the recurrence relation \(a_{j}=\frac{-1}{j(2 n+j)} a_{j-2} .\) Show that for \(a_{0}=\left(n ! 2^{n}\right)^{-1}\), we get the Bessel function of the first kind of order \(n\) from the even values \(j=2 k\) : $$ J_{n}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(n+k) !}\left(\frac{x}{2}\right)^{n+2 k} $$

Short Answer

Expert verified
The Bessel function of the first kind of order n using the given guess and recurrence relation is \( J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(n+k) !}\left(\frac{x}{2}\right)^{n+2 k} \)

Step by step solution

01

Identify the Initial Guess, Recurrence Relation, and Coefficient Value

Firstly, note that the initial guess for the solution of the Bessel equation is \( y(x)=\sum_{j=0}^{\infty} a_{j} x^{j+n} \). Then, the values of the coefficients \( a_j \) in the series are calculated from the recurrence relation \( a_{j}=\frac{-1}{j(2 n+j)} a_{j-2} \), with the given value for \( a_0 \) as \((n !2^{n})^{-1}\).
02

Substitute Even Indices

Now, instead of using all indices j, let's use even indices \( j=2k \), since the recurrence relation is two-step. Substituting \( j=2k \) into the series gives \( y(x)=\sum_{k=0}^{\infty} a_{2k} x^{2k+n} \).
03

Use the Recurrence Relation

For the recurrence relation, substitute j by 2k to get \( a_{2k} = \frac{-1}{(2k)(2n+2k)} a_{2k-2} \). This indicates that each even-indexed coefficient can be written in terms of the previous even-indexed coefficient.
04

Compute the Coefficients a_{2k}

Use the recursion relation iteratively to find the coefficients \( a_{2k} \). Begin at \( a_{0} \) and use the recursion formula to get the terms \( a_2, a_4, a_6, . . . \). This gives \( a_{2k} = \frac{(-1)^k}{2^{2k} (k!)^2} a_0 \). After substituting the given \( a_0 = (n! 2^n)^{-1} \), we obtain \( a_{2k} = \frac{(-1)^k}{k!(n+k) !} \left(\frac{1}{2}\right)^{n+2 k} \).
05

Substitute a_{2k} into the Series

Substitute \( a_{2k} \) in our summation to give the Bessel function of the first kind of order n: \( J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(n+k) !}\left(\frac{x}{2}\right)^{n+2 k} \)

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Most popular questions from this chapter

Use integration by parts to show \(\Gamma(x+1)=x \Gamma(x)\)

Prove Green's identity \(\int_{a}^{b}(u \mathcal{L} v-v \mathcal{L} u) d x=\left.\left[p\left(u v^{\prime}-v u^{\prime}\right)\right]\right|_{a} ^{b}\) for the eral Sturm-Liouville operator \(\mathcal{L}\).

The coefficients \(C_{k}^{p}\) in the binomial expansion for \((1+x)^{p}\) are given by $$ C_{k}^{p}=\frac{p(p-1) \cdots(p-k+1)}{k !} $$ a. Write \(C_{k}^{p}\) in terms of Gamma functions. b. For \(p=1 / 2\), use the properties of Gamma functions to write \(C_{k}^{1 / 2}\) in terms of factorials. c. Confirm your answer in part b. by deriving the Maclaurin series expansion of \((1+x)^{1 / 2}\)

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

Find the eigenvalues and eigenfunctions of the given Sturm-Liouville lems: a. \(y^{\prime \prime}+\lambda y=0, y^{\prime}(0)=0=y^{\prime}(\pi)\) b. \(\left(x y^{\prime}\right)^{\prime}+\frac{\lambda}{x} y=0, y(1)=y\left(e^{2}\right)=0\).
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