Why is $e^x$ having a different series expansion in terms of the Hermite polynomials?

Question

I had asked a question yesterday relating to "Evaluation of the first three terms of $e^x$ in terms of the Hermite polynomials".

Yes, my problem for that question was solved and I was successfully able to evaluate the integrals.

But what struck me was that the final expression was:

$$f(x)=e^{\frac{1}{4}}\left[\frac{1}{8}H_2(x)+\frac{1}{2}H_1(x)+H_0(x)\right]+\frac{1}{2\sqrt{\pi}}H_1(x)$$

which is not the same as the Taylor expansion of $e^x$ if we replace $H_n(x)$ by the tabulated values. But the working out and the formulae are correct as evident in the the hyperlinked question.

Is what I have done correct. But then why do we have that $e^{\frac{1}{4}}$ in wfront?

Someone please help.

Answer 1

The two expansions need not to be the same. You can imagine that the function $f(x)$ is a vector, if $u_n(x)$ is a basis then

$$ f(x) = \sum_n \langle u | f\rangle u_n(x) = \sum_n a_n u_n(x) $$

where $\langle \cdot | \cdot\rangle$ denotes the inner product. Depending on the selection of the basis $u_n(x)$, the coefficients of the expansion $a_n$ are going to be different, and so is going to be the value of the RHS sum, should it be truncated at a given number $n$

Answer 2

I will use the statistician's Hermite polynomials, the answer is still the same.

The coefficients in both a Taylor expansion $$ f(x) = \sum_{n=0}^\infty a_n x^n $$ and the Hermite expansion $$ f(x) = \sum_{n=0}^\infty d_n H_n(x) $$ are both related to the $n^\textrm{th}$ derivative of the function. It is well known that for the Taylor expansion, the coefficients are proportional to the $n^\textrm{th}$ derivative evaluated at the expansion point (let's take zero) $$ a_n = \frac{1}{n!} \frac{\partial^nf(0)}{\partial x^n} $$ whereas the coefficient in the Hermite expansion is the weighted integral of the $n^\textrm{th}$ derivative over the entire real line $$ d_n = \frac{1}{n!} \int_{-\infty}^\infty \frac{\partial^nf(x)}{\partial x^n}\exp\left(-\frac{x^2}{2}\right)\frac{dx}{\sqrt{2\pi}} $$ For $e^x$ all the derivatives are $e^x$ and the integral over the real line is $e^{\frac{1}{2}}$. Therefore the Hermite expansion is $$ e^x = e^{\frac{1}{2}}\sum_{n=0}^\infty \frac{H_n(x)}{n!} $$ To see how this can be related back to the Taylor expansion, we instead calculate the Hermite expansion of $e^{x\sqrt{t}}$ and find $$ e^{x\sqrt{t}} = e^{\frac{1}{2}t}\sum_{n=0}^\infty \frac{\sqrt{t}^nH_n(x)}{n!} $$ Making the substitution $y=x\sqrt{t}$ leads to $$ e^{y} = e^{\frac{1}{2}t}\sum_{n=0}^\infty \frac{\sqrt{t}^nH_n\left(\frac{y}{\sqrt{t}}\right)}{n!} $$ The quantity $\sqrt{t}^nH_n\left(\frac{y}{\sqrt{t}}\right)$ is known as the Hermite martingale, but note it is still monic, ie the coefficient of the highest order power ($n$) is $1$, all other powers are multiplied by a power of $t$.

Taking the $t\rightarrow 0$ limit of the RHS leads to the Taylor series, since $$ \lim_{t\rightarrow 0} \sqrt{t}^nH_n\left(\frac{y}{\sqrt{t}}\right) = y^n. $$