I want the expression for the following:
Sum[(1/Factorial[(l + p)])*(x y)^l*LaguerreL[p, l, x^2]*
LaguerreL[p, l, y^2], {l, 0, Infinity}]
When I write this in Mathematica, it just gives back the same thing written in mathematical form, not the solution.
If I specify the value of p, then I get closed form expression, but I am not getting for a general p.
It can be expressed by The Srivastava and Daoust multivariable hypergeometric function.
I don’t know if this triple sum can be simplified.
$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \sum _{u=0}^{\infty } \frac{\left((1)_u
(-p)_m (-p)n (1+p)u\right) \left(\left(a x^2\right)^m (c x y)^u \left(b y^2\right)^n\right)}{\Gamma (1+p) \left((1){m+u} (1){n+u}\right) (m! n! u!)}$
The result for any positive integer n is the sum of two terms: (1) an Exp[x y] term multiplied by a polynomial in x and y and (2) a polynomial in x and `y’. The first term is
Exp[x y] LaguerreL[n, (x - y)^2]/n!
but I haven’t figured out a general form for the second polynomial (assuming there is a closed form for that polynomial). So far for that second polynomial the sum of the x and y exponents is always even and the coefficient for x^i y^j is the same as the coefficient for x^j y^i. That’s all I’ve determined. Maybe someone else can see the pattern.
Another solution:
Going from a single sum to a triple sum will probably not help me. I would need a closed form solution because this will then go inside an integral.
How did you get this form?
Do you have any reason to think there is a closed form?
Questions like this should always indicate why it is reasonable to expect a solution.
Most integrals, series don’t have one.
Why do you want closed-form expression ?
From here we have:
L_p^l\left(a x^2\right)=\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a x^2\right)^k}{k!}
Then we have:
.
$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\\sum _{l=0}^{\infty } \frac{(c x y)^l \left(\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a
x^2\right)^k}{k!}\right) \sum _{j=0}^p \frac{(-1)^j \binom{p+l}{p-j} \left(b y^2\right)^j}{j!}}{(l+p)!}=\\sum _{k=0}^p \sum _{j=0}^p \left(\sum _{l=0}^{\infty } \frac{(-1)^{j+k} \left(a x^2\right)^k (c x
y)^l \left(b y^2\right)^j \binom{l+p}{-j+p} \binom{l+p}{-k+p}}{j! k! (l+p)!}\right)=\\sum _{k=0}^p \sum _{j=0}^p \frac{(-1)^{j+k} \left(a x^2\right)^k \left(b y^2\right)^j \binom{p}{-j+p} \binom{p}{-k+p}
, _2F_2(1,1+p;1+j,1+k;c x y)}{j! k! p!}$
I believe there probably will be a closed form. I saw the Hardy-Hille formula on Wikipedia:Laguerre polynomials - Wikipedia which is of the similar form, just with an extra factorial in the numerator. Further, when specifying the p values from 0 to 10, Mathematcia does give closed form expressions for those particular p values.
I need closed form expression because this is a small part of my project.
