# Problem of the Week #7 - May 14th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW. I'll just say it's no fun unless more people participate!

I'm going to keep the POTWs at this level of difficulty for a couple more weeks, hoping to get more people to bite!

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Problem: The Bessel function of order 1 is defined by

$J_1(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{n!(n+1)!2^{2n+1}}$

(a) Show that $J_1(x)$ satisfies the differential equation

$x^2J_1^{\prime\prime}(x)+xJ_1^{\prime}(x)+(x^2-1)J_1(x) = 0$

(b) Show that $J_0^{\prime}(x) = -J_1(x)$, where

$J_0(x) = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(n!)^22^{2n}}$

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• Sudharaka and Jameson

#### Chris L T521

##### Well-known member
Staff member
This week's problem was answered correctly by Sudharaka. Here's his solution.

a)

\begin{eqnarray}x^2J_1^{\prime\prime}(x)+xJ_1^{ \prime}(x)+(x^2-1)J_1(x) &=&x^2\sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n) x^{2n-1}}{n!(n+1)!2^{2n+1}}+x\sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)x^{2n}}{n!(n+1)!2^{2n+1}}+(x^2-1)\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{n!(n+1)!2^{2n+1}}

\\&=&

\sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n) x^{2n+1}}{n!(n+1)!2^{2n+1}}+\sum_{n=0}^{\infty} \frac{(-1)^n(2n+1)x^{2n+1}}{n!(n+1)!2^{2n+1}}+\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+3}}{n!(n+1)!2^{2n+1}}-\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{n!(n+1)!2^{2n+1}}

\\&=&

\sum_{n=0}^{\infty}\frac{(-1)^n 4n(n+1) x^{2n+1}}{n!(n+1)!2^{2n+1}}+\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{n!(n+1)!2^{2n+1}}

\\&=&

\sum_{n=1}^{\infty}\frac{(-1)^n x^{2n+1}}{(n-1)!n!2^{2n-1}}+\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{n!(n+1)!2^{2n+1}}

\\&=&

\sum_{n=0}^{\infty}\frac{(-1)^{n+1} x^{2n+3}}{n!(n+1)!2^{2n+1}}+\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{n!(n+1)!2^{2n+1}}

\\&=&

0

\end{eqnarray}

$\therefore x^2J_1^{\prime\prime}(x)+xJ_1^{\prime}(x)+(x^2-1)J_1(x) = 0$

b)\begin{eqnarray}J_0^{\prime}(x)&=&

\sum_{n=0}^{\infty}\frac{(-1)^n(2n)x^{2n-1}}{(n!)^22^{2n}}

\\&=&

\sum_{n=1}^{\infty}\frac{(-1)^n x^{2n-1}}{(n-1)!n!2^{2n-1}}

\\&=&

\sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n+1}}{n!(n+1)!2^{2n+1}}

\\&=&

-\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{n!(n+1)!2^{2n+1}}

\\&=&

-J_{1}(x)

\end{eqnarray}

$\therefore J_0^{\prime}(x)=-J_{1}(x)$

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