Differentiating a Bessel function power series

Question

For a function defined as $$\begin{align} y(x) &:= \sqrt{\frac{\pi x}{2}} J_{1/2}(x) \\ &= \sqrt{\frac{\pi x}{2}} \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \Gamma (k + 3/2)} \left( \frac{x}{2} \right)^{1/2 + 2k} \\ &= \sqrt{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \Gamma (k + 3/2)} \left( \frac{x}{2} \right)^{1 + 2k} \end{align}$$

I need to show that $y'' = -y$ and I have the relations $\Gamma (k+1) = k \Gamma (k)$ and $\Gamma (1/2) = \sqrt{\pi}$. I'm not getting very far with this... Differentiating twice we get:

$$\begin{align} y''(x) &= \sqrt{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k 2k (1+2k)}{k! \Gamma (k + 3/2)} \left( \frac{x^{2k-1}}{2^{1+2k}} \right) \end{align}$$

I've tried using the gamma function relations I've been given in different ways but I can't seem to get each series in a form where it's clear to see $y'' = - y$. Any help would be appreciated.

Answer 1

You need to account for the term of

$$y''(x)=\sqrt{\pi} \sum\limits_{k=0}^{\infty} \frac{(-1)^k\, 2k\, (1+2k)}{k!\, \Gamma(k+3/2)} \left(\frac{x^{2k-1}}{2^{1+2k}}\right)\tag{1}$$

evaluates to zero at $k=0$ and when evaluated at $k+1$ is equivalent to the negative of the term of

$$y(x)=\sqrt{\pi} \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{k!\, \Gamma(k+3/2)} \left(\frac{x}{2}\right)^{1+2k}\tag{2}$$

evaluated at $k$.

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In other words

$$y''(x)=\sqrt{\pi} \sum\limits_{k=1}^{\infty} \frac{(-1)^k\, 2k\, (1+2k)}{k!\, \Gamma(k+3/2)} \left(\frac{x^{2k-1}}{2^{1+2k}}\right)$$ $$=\sqrt{\pi} \sum\limits_{k=0}^{\infty} \frac{(-1)^{k+1}\, 2 (k+1)\, (1+2 (k+1))}{(k+1)!\, \Gamma\left(k+1+3/2\right)} \frac{x^{2 (k+1)-1}}{2^{2 (k+1)+1}}\tag{3}$$

and I'll leave it to you to show formula (3) is the negative of formula (2) when simplified using the identity $\Gamma \left(k+\frac{5}{2}\right)=\left(k+\frac{3}{2}\right)\, \Gamma\left(k+\frac{3}{2}\right)$.