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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

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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. –  Mårten W May 3 '13 at 15:14
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Try to prove that the integral satisfies Hermite equation. –  Felix Marin Sep 20 '13 at 20:15

1 Answer 1

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.

ADDENDUM

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.

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While I don't know whether or not this addresses the question in a context that suits the questioner, it is dismayingly opaque to me. E.g., please, what is the reason, if any, that there is such a relation? Is it truly just computational and formulaic? ... so that any small error is fatal? –  paul garrett Sep 21 '13 at 0:26
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It is very frustrating to think hard about a question, only to be rewarded with an anonymous downvote. –  Matemáticos Chibchas Sep 21 '13 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... –  paul garrett Sep 21 '13 at 0:40
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@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. –  Matemáticos Chibchas Sep 21 '13 at 2:36
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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. –  Bennett Gardiner Sep 21 '13 at 7:25

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