In the book "Essential Mathematical Methods for Physicists" comes the following problem that I am trying to solve:
at first I could see that the first formula that is given as an answer is wrong since the hermite polynomial $ H_4 (x) $ does not match with the known since:
$\begin{align*} H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}(-1)^{2s}(2x)^{2s}\cfrac{(2n)!}{(2s)!(n-s)!}\\ H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[(-1)^{2}\right]^s\left[(2x)^{2}\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\ H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[1\right]^s\left[4x^2\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\ H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[4x^2\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\ H_{2(2)}(x)&=(-1)^{2}\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{(2\cdot 2)!}{(2s)!(2-s)!}\\ H_{4}(x)&=(1)\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{(4)!}{(2s)!(2-s)!}\\ H_{4}(x)&=\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{24}{(2s)!(2-s)!}\\ H_{4}(x)&=(4x^2)^{0}\cfrac{24}{(2\cdot 0)!(2-0)!}+(4x^2)^{1}\cfrac{24}{(2\cdot 1)!(2-1)!}+(4x^2)^{2}\cfrac{24}{(2\cdot 2)!(2-2)!}\\ H_{4}(x)&=(1)\cfrac{24}{(0)!(2)!}+(4x^2)\cfrac{24}{(2)!(1)!}+(16x^4)\cfrac{24}{(4)!(0)!}\\ H_{4}(x)&=(1)\cfrac{24}{(1)(2)}+(4x^2)\cfrac{24}{(2)(1)}+(16x^4)\cfrac{24}{(24)(1)}\\ H_{4}(x)&=12+(4x^2)(12)+(16x^4)\\ H_{4}(x)&=12+48x^2+16x^4\\ H_{4}(x)&=16x^4+48x^2+12 \\ \end{align*}$
But the real $H_4(x)$ is $H_4(x)=16x^4-48x^2+12$ so the $H_{2n}$ formula is wrong. But trying to find errors in the formula for $ H_ {2n + 1} $ I did not find any for the polynomials 1,3,5 so I think that this formula is correct.
If the generating function of the hermite polynomials is $g (x, t) = e^{-t^2 + 2tx} = \sum_{n = 0} ^ {\infty}H_n (x) \cfrac{t ^ n}{n!} $
How can the formula for $H_{2n + 1}$ be derived?
I tried using the formula $H_n(x)=\sum_{s=0}^{[n/2]}(-1)^s\cfrac{n!}{(n-2s)!s!}(2x)^{n-2s}$ ( which is just the equation 13.40 that mentions the problem) substituting in the value of 2n + 1 but couldn't deduce anything, as there were terms that I couldn't adjust to look like the one the book asks for an answer. Any help is really appreciated!
For $H_{2n}(x)$, the upper limit of their equation 13.40 is $\lceil (2n)/2\rceil=n$ and so the sum is $$H_{2n}(x)=\sum_{s=0}^n (-1)^s \frac{(2n)!}{(2n-2s)!s!}(2x)^{2n-2s}.$$ This is a descending sum, since it starts with highest power $(2x)^{2n}$ and terminates with $(2x)^0=1$. To make it ascending, we relabel $s\to n-s$ and so obtain
$$H_{2n}(x)=\sum_{s=0}^n (-1)^{n-s} \frac{(2n)!}{(2s)!(n-s)!}(2x)^{2s}.$$ This disagrees with the stated expansion in that it has $(-1)^s$ instead of $(-1)^{2s}$. Given that the latter would equal $1$ for all integers, though, I'm inclined to chalk that up to a typo. (By contrast, the case for odd $n$ does seem correct.)