Proving that $zJ''(z)+J'(z)+zJ(z)=0$

Question

In my complex analysis book there is the following problem that I'm having some trouble solving:

Consider the function: $$J(z)=\sum_{n\geq0} \frac{(-1)^n}{(n!)^2} \left(\frac{z}{2}\right)^{2n}$$ 1 - Determine the radius of convergence of this power series.

2 - Show that: $$zJ''(z)+J'(z)+zJ(z)=0$$

I already solved number one and got that the radius of convergence is $+\infty$, but I'm having some trouble with the second one

---

My approach: Lets rewrite:

$$J(z)=\sum_{n\geq0} \frac{(-1)^n}{(n!)^2} \left(\frac{z}{2}\right)^{2n}=\sum_{n\geq0} \frac{(-1)^n}{2^{2n}(n!)^2} z^{2n}$$

We can define a sequence $\alpha_n$ to be:

• $1$, if $n = 0$
• $0$, if $n$ is odd
• $\frac{(-1)^{\frac{n}{2}}}{\left(\left(\frac{n}{2}\right)!\right)^2 2^{n}}$, if $n$ is even

Then we end up with:

$$J(z)=\sum_{n \geq 0}\alpha_n z^n$$

So now we know that:

• $$zJ(z)=\sum_{n \geq 0}\alpha_n z^{n + 1}$$
• $$J'(z)=\sum_{n \geq 1}n\alpha_n z^{n - 1}=\sum_{n \geq 2}n\alpha_n z^{n - 1}$$
• $$zJ''(z)=\sum_{n \geq 2}n(n - 1)\alpha_n z^{n - 1}$$

So if we sum all these 3 we end up with:

$$ zJ''(z)+J'(z)+zJ(z)= $$ $$\sum_{n \geq 2}n(n - 1)\alpha_n z^{n - 1} + \sum_{n \geq 2}n\alpha_n z^{n - 1} + \sum_{n \geq 0}\alpha_n z^{n + 1}=$$ $$=\sum_{n \geq 2} n^2\alpha_n z^{n - 1} + \sum_{n \geq 0}\alpha_n z^{n + 1}=$$ $$=\sum_{n \geq 1} (n + 1)^2\alpha_{n + 1} z^{n} + \sum_{n \geq 1}\alpha_{n - 1} z^{n}=$$ $$=\sum_{n \geq 1} \left[(n + 1)^2\alpha_{n + 1} + \alpha_{n - 1}\right]z^{n}$$

But I don't know how to proceed now. How can I prove that this is equal to $0$?

Answer 1

Introducing the $\alpha_n$ is making a lot of extra work. Just compute directly:

Note that $zJ(z) = \sum_{n \ge 0} 2 {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$

$J'(z) = \sum_{n \ge 0} (n+1) {- 1 \over (n+1)^2} {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$,

$zJ''(z) = \sum_{n \ge 0} 2(n+1)(n+{1 \over 2}) {- 1 \over (n+1)^2} {(-1)^n \over (n!)^2} ({z \over 2})^{2n+1}$.

Hence the coefficient of $({z \over 2})^{2n+1}$ in $zJ(z)+J'(z)+zJ''(z)$ is ${(-1)^n \over (n!)^2}(2 -{1 \over n+1} - {2 (n+{1 \over 2}) \over n+1} ) = 0$.

Answer 2

You are almost done. If $n\geq 1$, then $$ (2n)^2 \alpha _{2n} + \alpha _{2n - 2} = (2n)^2 \frac{{( - 1)^n }}{{n!^2 2^{2n} }} + \frac{{( - 1)^{n - 1} }}{{(n - 1)!^2 2^{2n - 2} }} \\ = \frac{{( - 1)^n }}{{(n - 1)!^2 2^{2n - 2} }} + \frac{{( - 1)^{n - 1} }}{{(n - 1)!^2 2^{2n - 2} }} = 0 $$ and $$ (2n + 1)^2 \alpha _{2n + 1} + \alpha _{2n - 1} = (2n + 1)^2 \cdot 0 + 0 = 0. $$

Answer 3

You've shown $[z^{2n}]J=\frac{(-1/4)^n}{n!^2}$; that's a good start. Since $J$ is even, $zJ^{\prime\prime}+J^\prime+zJ$ is odd. Its $z^{2n+1}$ coefficient is$$\begin{align}[z^{2n}]J^{\prime\prime}+[z^{2n+1}]J^\prime+[z^{2n}]J&=(2n+1)(2n+2)[z^{2n+2}]J^{\prime\prime}+(2n+2)[z^{2n+2}]J+[z^{2n}]J\\&=(2n+2)^2[z^{2n+2}]J+[z^{2n}]J,\end{align}$$which you can verify is $0$.