How would I write the following polynomial in terms of the Hermite polynomials, $H_n(z)$?
\begin{equation}
P_n(z) = \sum_{k=0}^{[n/2]} \frac{n!a^k}{k!(n-2k)!}(2a z)^{(n-2k)}
\end{equation}
I have noticed $H_n(iaz)$ appears close, but the $a^k$ factor is puzzling.
This expression is obtained from a series expansion of the Taylor series of
\begin{equation}
\frac{d^n}{dx^n}e^{-ax^2}
\end{equation}
Is there a set of polynomials that can be found from
\begin{equation}
\frac{d^n}{dx^n}e^{-ax^m}
\end{equation}
where $m=2,3,\dots$?
Best Answer
You want $a^{k+n-2k}=a^{n-k}$ so what if you take in the argument $(-a)^{n/2}$ and multiply by that factor, you will get $$(-a)^{n/2}H_n((-a)^{1/2}\cdot z)=(-a)^{n/2}\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{k!(n-2k)!}(-1)^k(2\sqrt{-a}\cdot z)^{n-2k}=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{k!(n-2k)!}(-1)^k(2z)^{n-2k}(-a)^{\frac{2n-2k}{2}}$$ $$=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{k!(n-2k)!}(-1)^k(2z)^{n-2k}(-a)^{n-2k+k}=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{k!(n-2k)!}(-1)^k(-2az)^{n-2k}(-a)^k=P_n(-z).$$ So $P_n(z)=(-a)^{n/2}H_n(-(-a)^{n/2}z).$