Post History
#2: Post edited
by
Richard
·
2024-08-11T21:45:20Z (3 months ago)
- 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.
- 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)
#1: Initial revision
by
Richard
·
2024-08-11T21:41:51Z (3 months ago)
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.