combinatorial formula for Hermite polynomials

Question

From wikipedia, the Hermite polynomial has a combinatorial description as $$ H_r (x)= r!\sum^{[r/2]}_{s=0}\frac{(-1)^s}{s! (r-2s)!}(2x)^{r-2s} $$ I am trying to express the following equation in terms of the Hermite polynomials $$ f_r (x)= r!\sum^{[r/2]}_{s=0}\frac{(v)^s}{s! (r-2s)!}(x)^{r-2s} $$ but am unable to figure out how to do this, as it is slightly different.

Answer 1

From $$ H_r (x)= r!\sum^{[r/2]}_{s=0}\frac{(-1)^s}{s! (r-2s)!}(2x)^{r-2s} $$ we have \begin{align*} H_r (ix/(2\sqrt{v})) &= r!\sum^{[r/2]}_{s=0}\frac{(-1)^s}{s! (r-2s)!}(ix/\sqrt{v})^{r-2s}\\ & = r!\sum^{[r/2]}_{s=0}\frac{(-1)^s (\sqrt{v}/i)^{2s - r} }{s! (r-2s)!}x^{r-2s}\\ & = r!\sum^{[r/2]}_{s=0}\frac{(-1)^s (-i\sqrt{v})^{2s - r} }{s! (r-2s)!}x^{r-2s}\\ & = r!\sum^{[r/2]}_{s=0}\frac{(-1)^s (-i)^{2s} \sqrt{v}^{2s}(-i\sqrt{v})^{-r} }{s! (r-2s)!}x^{r-2s}\\ & = r!\sum^{[r/2]}_{s=0}\frac{v^{s}(-i\sqrt{v})^{-r} }{s! (r-2s)!}x^{r-2s}\\ &= (-i\sqrt{v})^{-r} f_r(x) \end{align*} if I haven't made any mistakes.