I am having trouble using the Hermite generating function to determine $e^{t^2}\cos(2xt)$. I know the generating function is $e^{2tx-t^2}=\sum_{n=0}^\infty (-1)^n \frac{t^n}{n!}H_n(x)$ but can't seem to get anywhere. Can anyone help direct me on this?
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One may replace $t$ by $it$, getting $$ e^{2itx+t^2}=\sum_{n=0}^\infty (-1)^ni^n\frac{t^n}{n!}H_n(x) $$ then one may take the real part: $$ e^{t^2}\cos2xt=\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}H_{2n}(x) $$ where we have assumed that $t,x$ are real numbers.
Olivier Oloa
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Where does the cosine come from? I can see how removing the imaginary would give the right side but not how you got the left. – Aksel'sRose Jul 26 '16 at 15:44
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From De Moivre's formula: $e^{i\alpha}=\cos(\alpha)+i \sin(\alpha)$. Do you see it? Here: $e^{2itx}=\cos(2tx)+i \sin(2tx)$ – Olivier Oloa Jul 26 '16 at 15:45
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ah okay that makes sense. Thank you! – Aksel'sRose Jul 26 '16 at 15:54
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@Aksel'sRose You are welcome. – Olivier Oloa Jul 26 '16 at 15:56