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Let $S(x,t)=e^{2xt-t^2}=\sum^\infty_{n=0}\frac{H_n(x)}{n!}t^n$ If we now differentiate each term with respect to $x$ we find: \begin{align*} \frac{\partial S}{\partial x}=2tS&=\sum^\infty_{n=0}2\frac{H_n(x)}{n!}t^{n+1} \\&=\sum^\infty_{n=1}2\frac{H_{n-1}(x)}{(n-1)!}t^{n} \\&=\sum^\infty_{n=0}2\frac{H_n'(x)}{n!}t^{n} \end{align*}

I cant see how the last step is done, turn the $H_{n-1}$ into a $H_n'(x)$? Any help would be great thankyou

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$H_n'(x)=2nH_{n-1}(x)$ this is the result I want to get to but I cant see where the derivative springs up from. –  JamesT Feb 14 '13 at 10:57
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up vote 2 down vote accepted

You have $S(x,t)= e^{xt-t^2}$. Thus it is easy to check that $\partial_x S= 2 t S$.

Now use the definition of $H_n(x)$, $$ \tag{1} S(x,t) = \sum_{n=0}^\infty \frac{H_n(x)}{n!} t^n.$$

Taking the derivative of (1) with respect to $x$, we have (exchanging the summation and differentiation) $$ \tag{2} \partial_x S(x,t) = \sum_{n=0}^\infty \frac{H_n'(x)}{n!} t^n.$$

Similarly, multiplying (1) by $2t$ yields $$\tag{3} 2 t S(x,t) = \sum_{n=0}^\infty \frac{2 H_n(x)}{n!} t^{n+1} = \sum_{n=1}^\infty \frac{2 H_{n-1}(x)}{(n-1)!} t^{n} .$$

Now, we know that (2) should be equal to (3); $$ \sum_{n=0}^\infty \frac{H_n'(x)}{n!} t^n = \sum_{n=1}^\infty \frac{2 H_{n-1}(x)}{(n-1)!} t^{n}$$ for all $t$ which can only be true of each coefficient of $t^n$ matches; $$ \frac{H_0'(x)}{0!} = 0 \Rightarrow H_0'(x)=0 $$ and $$ \frac{H_n'(x)}{n!} = \frac{2 H_{n-1}(x)}{(n-1)!} \Rightarrow H_n'(x) =2 n H_{n-1}(x)$$ for $n\geq 1$.

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That is excellent, Thank you very much :) –  JamesT Feb 14 '13 at 11:18
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