Proving $\frac{1}{e}\cosh(2x)=\sum_{n=0,1,2,...}^\infty\frac{1}{(2n)!}H_{2n}(x)$, where $H_{2n}(x)$ is the even order Hermite polynomial

Question

How can we prove the following relation $$\frac{1}{e}\cosh(2x)=\sum_{n=0, 1, 2, ...}^\infty\frac{1}{(2n)!}H_{2n}(x)$$ where the $H_{2n}(x)$ is the even order Hermite polynomial?

I didn't get any clue by doing $$\frac{1}{e}\left(\frac{e^{2x}+e^{-2x}}{2}\right)=\sum_{n=0, 1, 2, ...}^\infty\frac{1}{(2n)!}H_{2n}(x)$$Thanks beforehand.

Answer 1

Here's a hint: $\cosh(2x)$ and $\sinh(2x)$ satisfy the ODE $y''(x)=4 y(x)$. Use the recurrence relation $H_{m}'(x)=2mH_{m-1}(x)$ and differentiate the series. If the series satisfies the same ODE, it is a linear multiple of $\sinh(2x)$ or $\cosh(2x)$. It is easy to show that it is indeed proportional to $\cosh(2x)$ using a simple parity argument. Then, all that is left is to show this proportionality constant is $1/e$. If you're still struggling after this hint, I'll add more. But give it a try first.

Answer 2

Answering my own question.

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We can proceed by putting $t=1$ in the known equation $$e^{2tx-t^2}=\sum_{n=1}^\infty\frac{H_n(x).t^n}{n!}$$ we get $$e^{2x-1}=\sum_{n=1}^\infty\frac{H_n(x)}{n!}$$ and $$e^{2x-1}+e^{-2x-1}=\sum_{n=1}^\infty\frac{H_n(x)}{n!}+\sum_{n=1}^\infty\frac{H_n(x)}{n!}(-1)^n$$ $$=2\sum_{n=1}^\infty\frac{H_{2n}(x)}{(2n)!}$$as odd terms will be cancelled, $$\implies \frac{1}{2}\left(\frac{e^{2x}+e^{-2x}}{e}\right)=\sum_{n=1}^\infty\frac{H_{2n}(x)}{(2n)!}$$ So ultimately we can get, $$\frac{1}{e}\cosh(2x)\:=\sum_{n=0, 1, 2, ...}^\infty\frac{1}{(2n)!}H_{2n}(x)$$So we can prove it in this way also.