- #1
sarrah1
- 66
- 0
Hi All
remember the summation
$\sum_{k=0}^{m}{m \choose k}{a}^{k} \sum_{j=0}^{n} {n \choose j} \frac{{b}^{n-j}{c}^{j}}{(k+j)!}$ ....... (1)
I showed that the special case , that is when $a=b=c=1$ becomes
$\sum_{j=0}^{m+n}{m+n \choose j}\frac{1}{j!}$ ..........(2)
to prove it, I considered
$\sum_{k=0}^{m}{m \choose k}\sum_{j=0}^{n} {n \choose j} \frac{1}{(k+j)!}{x}^{k+j}$ ...(3)
and substitute at the end $x=1$.
Using Laplace transform it becomes
$ s \left(\sum_{k=0}^{m}{m \choose k}\frac{1}{{s}^{k+1}}\right)\left(\sum_{j=0}^{n}{m \choose j}\frac{1}{{s}^{k+1}}\right)$
the inside of the 2 brackets are Laplace transform of Laguerre polynomials so I did a convolution to obtain inverse Laplace
$\frac{d}{dx}\int_{0}^{x} \,{L}_{m}(t) {L}_{n}(x-t) dt $ ...... (4)
which gives an associate Laguerre polynomial and after some algebraic operations operations I arrived at (2). Note that Laguerre in (2) is divergent $ n\implies\infty$ I have in my case $m=n$. But (1) is convergent for say $a=b=c=1/2$. So I need to redo the proof in the presence of the constants, unless from (1) using Vandermonde convolution for combinations or else I get (2) directly without this hectic process. The idea is that I need to find the condition on the constants $a,b,c$ for (1) to become convergent at infinity to zero. I can do it if someone helps in finding (4) when the argument in ${L}_{m}$ is $at$ and in ${L}_{n}$ is $b(x-t)$ , i.e.
$\int_{0}^{x} \,{L}_{m}(at) {L}_{n}(b(x-t)) dt $ ...
the book by Oldham, Myland and Spanier gives it for $a=b=1$ whereas Gradshtein and Ryzhik book also gives it for $a=b$, I need the case when $ a\ne b$
grateful
Sarrah
remember the summation
$\sum_{k=0}^{m}{m \choose k}{a}^{k} \sum_{j=0}^{n} {n \choose j} \frac{{b}^{n-j}{c}^{j}}{(k+j)!}$ ....... (1)
I showed that the special case , that is when $a=b=c=1$ becomes
$\sum_{j=0}^{m+n}{m+n \choose j}\frac{1}{j!}$ ..........(2)
to prove it, I considered
$\sum_{k=0}^{m}{m \choose k}\sum_{j=0}^{n} {n \choose j} \frac{1}{(k+j)!}{x}^{k+j}$ ...(3)
and substitute at the end $x=1$.
Using Laplace transform it becomes
$ s \left(\sum_{k=0}^{m}{m \choose k}\frac{1}{{s}^{k+1}}\right)\left(\sum_{j=0}^{n}{m \choose j}\frac{1}{{s}^{k+1}}\right)$
the inside of the 2 brackets are Laplace transform of Laguerre polynomials so I did a convolution to obtain inverse Laplace
$\frac{d}{dx}\int_{0}^{x} \,{L}_{m}(t) {L}_{n}(x-t) dt $ ...... (4)
which gives an associate Laguerre polynomial and after some algebraic operations operations I arrived at (2). Note that Laguerre in (2) is divergent $ n\implies\infty$ I have in my case $m=n$. But (1) is convergent for say $a=b=c=1/2$. So I need to redo the proof in the presence of the constants, unless from (1) using Vandermonde convolution for combinations or else I get (2) directly without this hectic process. The idea is that I need to find the condition on the constants $a,b,c$ for (1) to become convergent at infinity to zero. I can do it if someone helps in finding (4) when the argument in ${L}_{m}$ is $at$ and in ${L}_{n}$ is $b(x-t)$ , i.e.
$\int_{0}^{x} \,{L}_{m}(at) {L}_{n}(b(x-t)) dt $ ...
the book by Oldham, Myland and Spanier gives it for $a=b=1$ whereas Gradshtein and Ryzhik book also gives it for $a=b$, I need the case when $ a\ne b$
grateful
Sarrah