Is this Hermite polynomial identity known?
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In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$ are Hermite polynomials and
$r=sqrt{x^2+y^2+z^2}$. Is this identity known?
mp.mathematical-physics orthogonal-polynomials
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
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up vote
4
down vote
favorite
In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$ are Hermite polynomials and
$r=sqrt{x^2+y^2+z^2}$. Is this identity known?
mp.mathematical-physics orthogonal-polynomials
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$ are Hermite polynomials and
$r=sqrt{x^2+y^2+z^2}$. Is this identity known?
mp.mathematical-physics orthogonal-polynomials
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
In some physics related problem, I found out the curious identity
$$sumlimits_{n_1+n_2+n_3=n}frac{n!}{n_1!,n_2!,n_3!},H_{2n_1}(x),H_{2n_2}(y),H_{2n_3}(z)=frac{H_{2n+1}(r)}{2r},$$
where $H_n(x)=(-1)^ne^{x^2}frac{d^n}{dx^n}e^{-x^2}$ are Hermite polynomials and
$r=sqrt{x^2+y^2+z^2}$. Is this identity known?
mp.mathematical-physics orthogonal-polynomials
mp.mathematical-physics orthogonal-polynomials
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
asked Nov 29 at 5:31
Zurab Silagadze
10.7k2569
10.7k2569
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
12
down vote
accepted
If we define the generating functions $F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$ and $G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
If we define the generating functions $F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$ and $G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
add a comment |
up vote
12
down vote
accepted
If we define the generating functions $F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$ and $G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
add a comment |
up vote
12
down vote
accepted
up vote
12
down vote
accepted
If we define the generating functions $F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$ and $G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
If we define the generating functions $F(x,t)=sum_{n=0}^{infty}H_{2n}(x)frac{t^n}{n!}$ and $G(x,t)=sum_{n=0}^{infty}H_{2n+1}(x)frac{t^n}{n!}$ then your identity is equivalent to
$$F(x,t)F(y,t)F(z,t)=frac{Gleft(sqrt{x^2+y^2+z^2},tright)}{2sqrt{x^2+y^2+z^2}}.$$
This is in turn an immediate corollary to the fact that we have
$$F(x,t)=frac{1}{(1+4t)^{1/2}}expleft(frac{4tx^2}{1+4t}right)$$
$$G(x,t)=frac{2x}{(1+4t)^{3/2}}expleft(frac{4tx^2}{1+4t}right)$$
and these have appeared in the literature in various contexts. Here is a paper that has a physics style proof, and here is one that derives it from the exponential generating function of $H_n(x)H_n(y)$.
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
answered Nov 29 at 7:03
Gjergji Zaimi
61.3k4160305
61.3k4160305
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
add a comment |
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
1
1
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
Thanks for the references! I proved the identity not by using this new generating function (which was unknown for me) but by using $H_n(x)=e^{-frac{1}{4},frac{d}{dx}},(2x)^n$, analogous formula for scalar Hermite polynomials (with $frac{d}{dx}$ replaced by Laplacian) and relations to the Laguerre polynomials. Of course this specific generating function makes the proof almost trivial.
– Zurab Silagadze
Nov 29 at 7:48
1
1
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
There is a typo in the above comment: the correct formula is $H_n(x)=e^{-frac{1}{4},frac{d^2}{dx^2}},(2x)^n$.
– Zurab Silagadze
Nov 29 at 8:24
add a comment |
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