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Q. 64

Expert-verifiedFound in: Page 681

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Exercise 64-68 concern with the bessel function.

What is the interval for convergence for ${J}_{0}\left(x\right)?$

The series converges for every x .

We have to find out the interval for convergence for ${J}_{p}\left(x\right)$

We denote the the given function as

${J}_{p}\left(x\right)=\sum _{k=0}^{\mathrm{\infty}}\u200a\frac{(-1{)}^{k}}{k!(k+p)!{2}^{2k+p}}{x}^{2k+p}$$\begin{array}{r}{J}_{0}\left(x\right)=\sum _{k=0}^{\mathrm{\infty}}\u200a\frac{(-1{)}^{k}}{k!(k+0)!{2}^{2k+0}}{x}^{2k+0}\\ =\sum _{k=0}^{\mathrm{\infty}}\u200a\frac{(-1{)}^{k}}{(k!{)}^{2}{2}^{2k}}{x}^{2k}\end{array}$

For finding the convergence of the function we will do the ratio test for absolute convergence we will assume ${b}_{k}=\frac{(-1{)}^{k}}{(k!{)}^{2}{2}^{2k}}{x}^{2k}$ therefore the next term will be ${b}_{k+1}=\frac{(-1{)}^{k+1}}{\left[\right(k+1)!{]}^{2}{2}^{2(k+1)}}{x}^{2(k+1)}$ implies that

role="math" $\begin{array}{r}\underset{k\to \mathrm{\infty}}{lim}\u200a\left|\frac{{b}_{k+1}}{{b}_{k}}\right|=\underset{k\to \mathrm{\infty}}{lim}\u200a\left|\frac{\left[\right(k+1)!{]}^{2}{2}^{2(k+1)}}{}{x}^{2(k+1)}\right|\frac{(-1{)}^{k}}{(k!{)}^{2}{2}^{2k}{x}^{2k}}\mid \\ =\underset{k\to \mathrm{\infty}}{lim}\u200a\left|-\frac{1}{(k+1{)}^{2}4}{x}^{2}\right|\end{array}$

Now we will be evaluating the limit $k\to \infty $

so, $\underset{k\to \mathrm{\infty}}{lim}\u200a\left|{x}^{2}\right|\frac{-1}{(k+1{)}^{2}}=0$ zero is the value no matter what is the value of x .

Hence by the ratio test it is clear that this series absolutely converges for every value of x

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