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I am trying to prove an identity involving Hermite polynomials using other identities from Wikipedia, but I can't find the way. I have checked the identity in Mathematica for many values of $n$ and it holds for all values of $n$ I have tried. The identity is

$$ \frac{1}{n!} \Big( \text{He}_n(x) \Big)^2 = \sum_{k=0}^n {n\choose k} \frac{1}{k!} \, \text{He}_{2k} (x) \, ,$$

where $\text{He}_n(x)$ is the probabilists' Hermite polynomial

$$\text{He}_n(x) = (-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}} \, .$$

Any ideas?

MBolin
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1 Answers1

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This proof uses operator methods. From the wiki page we know $$(1)\quad \operatorname{He}_n(x) =\exp{\big(-\frac{1}{2}\frac{d^2}{dx^2}\big)} \ x^n $$ and also from Mehler's formula $$(2) \quad \frac{\big(\operatorname{He}_n(x)\big)^2}{n!} = [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)}$$ where $[u^n]$ is the "coefficient of" operator. We need one more lemma. (Someone else has done this before me, but I don't have a reference.) $$ (3) \quad \exp{\big(a \frac{d^2}{dx^2}\big)} \exp{\big(b\ x^2\big)} = \frac{1}{\sqrt{1-4\ a\ b}}\exp{\big(\frac{b\ x^2}{1-4a\ b}\big)} $$ Begin proof of lemma: Use well-known Gaussian formula $$ \exp{(a\ t^2)} = \frac{1}{2\sqrt{a\pi}} \int_{-\infty}^\infty du \exp{(t\ u)} \exp{\big(\frac{-u^2}{4a}\big)}$$ Substitute $t=\frac{d}{dx}.$ Use operational form of Taylor series $\exp{(u d/dx)}f(x) = f(x+u).$ Then the left-hand side of (3) becomes $$ \frac{1}{2\sqrt{a\pi}} \int_{-\infty}^\infty du \exp{\big(\frac{-u^2}{4a}\big)} \exp{(b(x+u)^2)} $$ Use the penultimate equation again, and algebra. End proof of lemma:

By eqs. (1) & (2) the OP's formula is equivalent to showing $$\exp{\big(-\frac{1}{2}\frac{d^2}{dx^2}\big)} \sum_{k=0}^n \binom{n}{k} \frac{x^{2k}}{k!} = [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)} $$ which is equivalent to showing $$ L_n(-x^2) = \exp{\big(\frac{1}{2}\frac{d^2}{dx^2}\big)} [u^n] \frac{1}{\sqrt{(1-u^2)}} \exp{\big(\frac{u}{1+u} x^2\big)} $$ where I've used the well-known polynomial expression for Laguerre polynomials. Use eq (3) on the right-hand side of the previous equation. With algebra it is seen that this is equivalent to $$ L_n(-x^2) = [u^n] \frac{1}{1-u} \exp{\big(\frac{u\ x^2}{1-u} \big) } $$ This is a well-known formula; see Gradshteyn & Ryzhik 8.975.1

K.defaoite
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user321120
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  • Do you have a reference for Mehler’s formula? – MBolin Jul 06 '20 at 09:52
  • @MBolin Mehler's formula is on the wiki page you linked to, although it is expressed in the physicist's notation for Hermite polys. I think it's generalization is also called the Hardy-Hill formula. – user321120 Jul 06 '20 at 14:38
  • Also I think you miss a factor of $(1-4ab)^{-1/2}$ on the RHS of the Gaussian equation (3), but I think it doesn't affect the next equations. – MBolin Jul 06 '20 at 15:51
  • In the wiki page I linked I think it's written somewhat differently, so I don't know what are the "coefficients of operators" $[u^n]$ you mention. If you could explain how to get one formula from the other, defining these coefficients, that would be great. – MBolin Jul 06 '20 at 16:41
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    @MBolin Thanks for the catch on the typo; I corrected it in the answer. The 'coefficient of' operator is really quite simple. If $c(u) = \sum c_n u^n,$ then $[u^n] \ c(u) = c_n.$ There is a paper by D. Knuth, "Bracket notation for the 'coefficient of' operator" in the arXiv (arXiv:math/9402216v1) that you might find interesting. Basically to get eq. (2), I used the Mehler formula in the wiki, but set both x and y to $x/\sqrt{2}.$ Then I use the map from the physicists' notation (in which the formula is stated) to the probabilicists'. The $[u^n]$ extracts that square of ther Hermite polys. – user321120 Jul 06 '20 at 16:49
  • OK. Please take a look at my new question as well, which is a generalization of the above: https://math.stackexchange.com/questions/3748422/new-hermite-polynomial-identity-vol-ii – MBolin Jul 07 '20 at 10:26