# Power series $x f''(x) + f'(x) + xf(x) = 0$

Find a power series with radius of convergence $R = \infty$ such that $$f(x) = \sum_{n=1}^{\infty} a_{n}x^{n}$$ satisfies $$x f''(x) + f'(x) + xf(x)= 0, \forall \mbox{ } x \in \mathbb R.$$

How should I go about solving this question? I have a gut feeling that it has to do with trig functions because I know that the terms in each power series have to cancel out in the resulting addition to satisfy the equation.

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Write $f'(x)$ and $f''(x)$ with a power series, rewrite terms in order to transform the LHS into a power series, then identity coefficients. – Davide Giraudo Apr 21 '12 at 20:54
If you just need one solution then $a_n=0$ for all $n$ should work... :-) – Henning Makholm Apr 21 '12 at 20:54

$$\begin{array}{lcr} f(x) &=& a_0 +& a_1x +& a_2x^2 +\cdots+& a_nx^n +\cdots \\[2ex] f'(x) &=& a_1 +& 2a_2x +& 3a_3x^2 +\cdots+& (n+1)a_{n+1}x^n +\cdots \\ xf''(x) &=& & 1\cdot2a_2 x +& 2\cdot3a_3x^2 +\cdots+& n(n+1)a_{n+1}x^n +\cdots \\ xf(x) &=& & a_0x +& a_1x^2 +\cdots+& a_{n-1}x^n +\cdots \\ \hline 0 &=& a_1 +& (4a_2+a_0)x +& (9a_3+a_1)x^2 +\cdots+& ((n+1)^2a_{n+1}+a_{n-1})x^n +\cdots \end{array}$$
In order for the bottom row to be $0$, each of the coefficients must be, so we have $a_1=0$ and $(n+1)^2a_{n+1}+a_{n-1}=0$ for $n\ge 1$. Setting $k=n+1$ the latter equation becomes $$a_k = -\frac{a_{k-2}}{k^2} \qquad\text{for }k\ge 2$$ So all of the odd coefficients must be zero, and the even coefficients all determine each other. Every solution must be a multiple of the one with $a_0=1$, where we have $$a_{2m} = \frac{(-1)^m}{2^2\cdot 4^2\cdot 6^2 \cdots (2m)^2} = \frac{(-1)^m}{m!^2 2^{2m}}$$ I trust you can verify the convergence radius of this one yourself.
(This solution is the Bessel function $J_0$).
I think you have a typo in the recursion formula for $a_k$. – David Mitra Apr 21 '12 at 21:34
@David: Thanks. Signs are annoying; one should always work in characteristic 2 where ${-}={+}$. – Henning Makholm Apr 21 '12 at 22:04
@HenningMakholm : No. It seemed in the same spirit as you "$+=-$" comment. – Michael Hardy Apr 22 '12 at 19:10