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Wide mathjax copied from [here](https://math.codidact.com/posts/292225) to test with:

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 The first case $\langle \frac{\partial f}{\partial z} , g \rangle$. Change of variable
	\begin{align*}
		u   =r^2 \\ 
		 \qquad du  =2r dr 
	\end{align*}
	\begin{align*}
	(k+1)\int_{0}^{\infty} e^{-r^2}r^{2k+1}dr=\int_{0}^{\infty} e^{-u}u^{k +1/2}\frac{1}{2\sqrt{u}}du = \frac{1}{2}\int_{0}^{\infty}e^{-u}u^{k}du = (k+1)\frac{\Gamma\left((k+1)/1\right)}{2}=(k+1)\frac{\Gamma (k+1)}{2}=\frac{(k+1)!}{2}
	\end{align*}
The second case $\langle f, zg \rangle $. The same change of variables
	\begin{align*}
	(k+1)\int_{0}^{\infty} e^{-r^2}r^{2k+3}dr=\int_{0}^{\infty} e^{-u}u^{k +3/2}\frac{1}{2\sqrt{u}}du = \frac{1}{2}\int_{0}^{\infty}e^{-u}u^{k+1}du = \frac{\Gamma\left((k+2)/1\right)}{2}=\frac{\Gamma (k+2)}{2}=\frac{(k+1)!}{2}
	\end{align*}
Hence, the inner products are equals. 
(In the last equality i used the gauss integral and its relation with Gamma function. Look at in Wikipedia)