# Connection between Hermite & Legendre polynomials

Prove that $$H_n(x)= 2^{n+1}e^{x^2}\int_x^\infty e^{-t^2}t^{n+1}P_n\left(\frac{x}t\right)dt,$$ where $H_n$ is Hermite polynomial & $P_n$ is Legendre polynomial

• What have you tried? What are your own thoughts? It is difficult to write an answer at the right level unless you first show what you know. May 3, 2013 at 15:14
• Try to prove that the integral satisfies Hermite equation. Sep 20, 2013 at 20:15

The Hermite polynomials $H_n(x)$ have the following explicit expression:

$$H_n(x)=\sum_{m=0}^{\lfloor n/2\rfloor}\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}\,x^{n-2m}\,.$$

On the other hand, the Legendre polynomials have the following explicit expression:

$$P_n(x)=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,,$$

which implies

$$t^{n+1}P_n(x/t)=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,t^{2k+1}\,.$$

For $k\geq0$, let $A_k=\int_x^\infty e^{-t^2}t^{2k+1}\,dt$. Integrating by parts we obtain

\begin{align*} A_k=&\,\int_x^\infty\frac{t^{2k}}{-2}e^{-t^2}(-2t)\,dt\\[2mm] =&\,\frac{t^{2k}e^{-t^2}}{-2}\biggl|_{t=x}^{t=\infty}-\int_x^\infty-ke^{-t^2}t^{2k-1}\,dt\\[2mm] =&\,\frac{x^{2k}e^{-x^2}}2+kA_{k-1}\,; \end{align*}

in particular $A_0=e^{-x^2}/2$, and so for each $k\geq0$ we have

$$\sum_{r=1}^k\frac{A_r-rA_{r-1}}{r!}=\sum_{r=1}^k\frac{x^{2r}e^{-x^2}}{2\cdot r!}$$

that is

$$\frac{A_k}{k!}-\frac{A_0}{0!}=\frac{e^{-x^2}}2\sum_{r=1}^k\frac{x^{2r}}{r!}\,.$$

Therefore we have, for $k\geq0$:

$$A_k=\frac{e^{-x^2}}2\sum_{r=0}^k\frac{k!}{r!}\,x^{2r}\,,$$

which in turn implies that

\begin{align*} 2^{n+1}e^{x^2}\int_x^\infty e^{-t^2}t^{n+1}P_n(x/t)\,dt=&\,2^{n+1}e^{x^2}\int_x^\infty e^{-t^2}\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}t^{2k+1}\,dt\\[2mm] =&\,2^{n+1}e^{x^2}\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\underbrace{\int_x^\infty e^{-t^2}t^{2k+1}\,dt}_{=A_k}\\[2mm] =&\,2e^{x^2}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,\frac{e^{-x^2}}2\sum_{r=0}^k\frac{k!}{r!}\,x^{2r}\\[2mm] =&\,\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,\frac{k!}{r!}\,x^{2r}\\[2mm] =&\,\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{r!(n-k)!(n-2k)!}\,x^{n-2(k-r)}\,. \end{align*}

Our objective is to show that

$$\sum_{m=0}^{\lfloor n/2\rfloor}\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}\,x^{n-2m}=\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{r!(n-k)!(n-2k)!}\,x^{n-2(k-r)}\,.$$

Note that when $k$ and $r$ ranges as in the right-hand side double sum above, then the quantity $k-r$ varies precisely on the set $\bigl\{0,1,\dots,\lfloor n/2\rfloor\bigr\}$. Given $m$ in this set, then the possible pairs $(k,r)$ with $k-r=m$ are precisely those satisfying $m\leq k\leq\lfloor n/2\rfloor$ and $r=k-m$, so the right-hand side sum above can be rewritten as

\begin{align*} &\,\sum_{m=0}^{\lfloor n/2\rfloor}\,\Biggl[\,\sum_{k=m}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{(k-m)!(n-k)!(n-2k)!}\Biggr]\,x^{n-2m}\\[2mm] =&\,\sum_{m=0}^{\lfloor n/2\rfloor}\,\Biggl[\,\sum_{k=m}^{\lfloor n/2\rfloor}(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\,m!\,\Biggr]\,x^{n-2m}\,, \end{align*}

so it remains to prove that for any $n,m$ with $0\leq m\leq\lfloor n/2\rfloor$ the following equality holds:

$$\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}=\sum_{k=m}^{\lfloor n/2\rfloor}(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\,m!\,,$$

or, equivalently,

$$(-1)^m2^{n-2m}\,\binom n{2m}\,\binom{2m}m=\sum_{k=0}^n(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\tag{\boldsymbol{\ast}}$$

(the summands with $k<m$ or $\lfloor n/2\rfloor<k\leq n$ are equal to $0$).

I already have a proof of this fact, but it is too long to write down, so I will include later. This proves the result.

Let $n$ and $m$ be as before, fixed. Let $f(z)=\sum_{j=0}^\infty(-1)^j\,\binom nj\,\binom jm\,z^j$ and $g(z)=\sum_{j=0}^\infty\binom{2j}n\,z^j$. Then the right-hand side of $\boldsymbol{(\ast)}$ is precisely the coefficient of $z^n$ in the power series expansion of $f(z)g(z)$. I use Mathematica to obtain the formulas for $f$ and $g$, and afterwards I $\$ construct $\ \$ try to construct a direct proof of the equalities (thanks Mathematica!). Now

\begin{align*} f(z)=&\,\sum_{j=m}^n(-1)^j\,\frac{n!j!}{j!(n-j)!m!(j-m)!}\,\frac{(n-m)!}{(n-m)!}\,z^j\\[2mm] =&\,\sum_{j=m}^n(-1)^j\,\binom{n-m}{j-m}\,\binom nm\,z^j\\[2mm] =&\,\sum_{j=0}^{n-m}(-1)^{j+m}\,\binom{n-m}j\,\binom nm\,z^{j+m}\\[2mm] =&\,\binom nm\,(-z)^m(1-z)^{n-m}\,. \end{align*}

On the other hand, denote by $\lceil x\rceil$ the least integer greater or equal than $x$. Then

$$g(z)=\sum_{\substack{j\geq0\\2j\geq n}}\binom{2j}n\,z^j=\sum_{j=\lceil n/2\rceil}^\infty\binom{2j}n\,z^j\,.$$

If $k=2j-n$ then $k\equiv n(\bmod\ 2)$, and $k\geq0$ iff$j\geq\lceil n/2\rceil$. Therefore the sum above can be rewritten as

\begin{align*} g(z)=&\,\sum_{\substack{k\geq0\\k\equiv n(\bmod\ 2)}}\binom{k+n}n\,z^{\,(k+n)/2}\\ =&\,\frac{(-1)^n}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[(-1)^k-(-1)^{n+1}\bigr]\,z^{\,(k+n)/2}\\ =&\,\frac{(-1)^nz^{n/2}}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[(-1)^k-(-1)^{n+1}\bigr]\,z^{\,k/2}\\ =&\,\frac{(-1)^nz^{n/2}}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[\bigl(-\sqrt z\bigr)^k-(-1)^{n+1}\bigr(\sqrt z\bigr)^k\bigr]\,.\\ \end{align*}

Now we use the fact that $(1-\alpha)^{-(n+1)}=\sum_{k=0}^\infty\binom{k+n}k\,\alpha^k=\sum_{k=0}^\infty\binom{k+n}n\,\alpha^k$ (for $|\alpha|$ small), obtaining

\begin{align*} g(z)=&\,\frac{(-1)^nz^{n/2}}2\Biggl[\frac1{\ \ \bigl[1-\bigl(-\sqrt z\bigr)\bigr]^{\,n+1}}-(-1)^{n+1}\frac1{\ \ \bigl[1-\sqrt z\,\bigr]^{\,n+1}}\Biggr]\\ =&\,\frac{(-1)^nz^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{n+1}}\,. \end{align*}

Therefore

\begin{align*} f(z)g(z)=&\,\binom nm\,(-z)^m(1-z)^{n-m}\,\frac{(-1)^nz^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{n+1}}\\ =&\,\binom nm\,z^m\,\frac{z^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{m+1}}\\ =&\,\binom nm\,\frac{z^m}{2(z-1)^{m+1}}\,z^{n/2}\ \sum_{k=0}^{n+1}\binom{n+1}k\,\bigl(\sqrt z\bigr)^{\,k}\bigl[(-1)^{n+1-k}-1\bigr]\\ =&\,\binom nm\,\frac{z^m}{(z-1)^{m+1}}\sum_{\substack{0\leq k\leq n+1\\k\equiv n(\bmod\ 2)}}-\binom{n+1}k\,z^{\,(k+n)/2}\\ =&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{\substack{0\leq k\leq n+1\\k\equiv n(\bmod\ 2)}}\binom{n+1}k\,z^{\,(k+n)/2}\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\,. \end{align*}

Taking $k=2j-n$, we see that $0\leq k\leq n+1$ iff $\frac n2\leq j\leq n+\frac12$, which implies

\begin{align*} f(z)g(z)=&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{j=\lceil n/2\rceil}^n\binom{n+1}{2j-n}\,z^j\,\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\\ =&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{j=0}^\infty\binom{n+1}{2j-n}\,z^j\,\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\\ \end{align*}

It remains to show that the coefficient of $z^{n-m}$ in $\Bigl[\sum_{j=0}^\infty\binom{n+1}{2j-n}\,z^j\,\Bigr]\sum_{r=0}^\infty\binom{r+m}m\,z^r$ is equal to $2^{n-2m}\binom{n-m}m$. Unfortunately, I was unable to prove this, but Mathematica says that it is true: taking $n=2\ell$ the software confirms that

$$2^{2(\ell-m)}\,\binom{2\ell-m}m-\sum_{j=\ell}^{2\ell-m}\binom{2\ell+1}{2(j-\ell)}\,\binom{2\ell-j}m=0\,,$$

and similarly it is confirmed that for $n=2\ell+1$ the equality

$$2^{2(\ell-m)+1}\,\binom{2\ell+1-m}m-\sum_{j=\ell+1}^{2\ell+1-m}\binom{2(\ell+1)}{2(j-\ell)-1}\,\binom{2\ell+1-j}m=0$$

holds.

• It is very frustrating to think hard about a question, only to be rewarded with an anonymous downvote. Sep 21, 2013 at 0:31
• Again, I don't have much of an idea about other interested parties' reactions, but I would be interested in more explanatory narrative to this answer. Or, if it is the point, an explicit claim that the relation is fundamentally combinatorial (which would surprise me, but I am willing to learn new things). Thus, as in my earlier comment/request-for-clarification, I am curious about where the formula manipulation come from, if one can say. Or, again, if there is a claim that the "true" argument is exactly "combinatorial", could this claim be made explicit? Just asking for context, in part... Sep 21, 2013 at 0:40
• ... e.g., the Hermite polynomials do have other definitions, in terms of eigenfunctions (up to Gaussians) for $-\Delta+x^2$, and such, which produces some of their properties "for general reasons", rather than by computational verification. That is, although I understand that others have other tastes, I am not a fan of "verification" of miraculous fact as proof (unless it is truly miraculous), but prefer, if possible, an a-priori reason that a formula or relation should hold. Or, at worst, to be told that there is no clear a-priori reason? ... all the worse that we have no recourse, ... Sep 21, 2013 at 0:44
• @paulgarrett I agree with you, believe me, I want to find a more systematic method of solution, but this is better than nothing. Be my guest to provide a more concise and solid answer. Sep 21, 2013 at 2:36
• Of course @paulgarrett is right in saying that a more intuitive explanation would be welcomed, but I too think it is unfair to award all this work with a downvote. While using the explicit formulae for the two special functions results in a fairly brutal calculation, one can at least follow the argument broadly and see that the author made every effort to be especially thorough. I hope that the original questioner accepts the answer and awards the bounty. Sep 21, 2013 at 7:25