How to get sums like these in the form of the Cauchy Product?

Question

I know that the question isn't very well-worded, please feel free to change it to something better.

I have this sum:

$$\sum_{k=0}^{\infty} \sum_{l=0}^{k} a_lx^ll!a_{k-l}x^{k-l+1}(k-l)!\frac{1}{(k+1)!}$$

This is the result of convoluting a polynomial of the form $\sum_{n=0}^{\infty} a_n x^n$ with itself, so we'd have $\int_{0}^{x} (\sum_{n=0}^{\infty} a_n t^n)(\sum_{n=0}^{\infty} a_n (x-t)^n)dt$, which can be rewritten as $\sum_{k=0}^{\infty}\sum_{l=0}^{k}a_la_{k-l}\int_{0}^{x}t^l{(x-t)}^{k-l}dt$ which evaluates to the sum above.

I would like to find a way to put in a form such that it can be written as a product of two sums, that is $$\sum_{j=0}^{\infty}\sum_{k=0}^{j} b_{k}c_{j-k}$$

Most of it is able to be written in this form, where $b_k=a_kx^kk!$ and $c_{j-k}=a_{j-k}x^{j-k+1}(j-k)!$. However, I still have that $\frac{1}{(j+1)!}.$ What I want to know is if there's any way I could algebraically manipulate that so that it can completely be in the form I want.

Thank You.

Answer 1

Not sure if I correctly understood but this can be done as follows.

Start writing the series obtained convoluting as $$ \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \sum_{\ell=0}^{k} \left( a_{\ell}\ a_{k-\ell}\ \ell!\ (k-\ell)! \right) x^{k+1} $$

Note that this is a power series with coefficients $A_k = \dfrac{1}{(k+1)!} \sum_\limits{\ell=0}^{k} \left( a_{\ell}\ a_{k-\ell}\ \ell!\ (k-\ell)! \right)$.

We want to write $\sum_\limits{k=0}^{\infty} A_k\ x^{k+1} = \sum_\limits{k=0}^{\infty} \sum_\limits{\ell=0}^{k} b_l\ c_{k-l}\ x^k = \left( \sum_\limits{m=0}^{\infty} b_m\ x^m \right) \left( \sum_\limits{n=0}^{\infty} c_n\ x^n \right) $, which still also a power series with coefficients $P_k = \sum_\limits{\ell=0}^{k} b_l\ c_{k-l}$.

Since a power series uniquely determines a function, it follows that the coeffiecients of both must be equal. That is, $P_0 = 0$ and $ P_{k+1} = A_k $. Also, because $P_0 = b_0 c_0$ we have that $b_0 = 0$ or $c_0 = 0$. Without loss of generality suppose that $b_0 \neq 0$; then $c_0 = 0$.

We gonna write $c_n$ as a function of $a_k$ and $b_m$.

For $k = 0$ $$ A_0 = a_{0}^2 = P_1 = b_0 c_1 + b_1 c_0 = b_0 c_1 \implies c_1 = \frac{a_0^2}{b_0} $$

For $k = 1$ $$ A_1 = \frac{1}{2} \left( a_{0} a_{1} + a_{1} a_{0} \right) = a_{0} a_{1} = P_2 = b_0 c_2 + b_1 c_1 + b_2 c_0 = b_0 c_2 + b_1 c_1\\ \implies c_2 = \frac{1}{b_0} \left( a_{0} a_{1} - b_1 c_1 \right) $$

For $k = 2$ $$ A_2 = \frac{1}{3} \left( 4 a_0 a_2 + a_1^2 \right) = P_3 = b_0 c_3 + b_1 c_2 + b_2 c_1\\ \implies c_3 = \frac{1}{b_0} \left( \frac{1}{3} \left( 4 a_0 a_2 + a_1^2 \right) - b_1 c_2 - b_2 c_1 \right) $$

And so on... Hence $$ c_{n+1} = \frac{1}{b_0} \left( A_n - \sum_{\ell=1}^{k-1} b_{\ell}\ c_{k-\ell} \right) $$

This means that $b_n$ is a sequence that we are free to determine as long as we choose $c_n$ as above. Thus we may choose a sequence which make its respective power series to have infinite radius of convergence to avoid any problem, say $b_m = \dfrac{1}{m!}$; so $\sum_\limits{m=0}^{\infty} b_i\ x^i = e^x$. We are done.

Answer 2

Put plainly, you want to express the Integral Convolution of a given polynomial with itself by means of the product of two polynomials to be determined.

Keeping general, that is unspecified, the coefficients $a_k$ of the auto-convolved polynomial, that would mean to be able to express $$ {{l!\left( {k - l} \right)!} \over {\left( {k + 1} \right)!}} = {1 \over {\left( {k + 1} \right)\left( \matrix{ k \cr l \cr} \right)}} = f(l)g(k - l) $$ which is not possible, since the binomial is not splittable in such a multiplicative way.
In fact, that would imply $$ \eqalign{ & \left( \matrix{ k \cr l \cr} \right) = f(l)\,g(k - l)\quad \Rightarrow \quad \left( \matrix{ k \cr l \cr} \right) = \left( \matrix{ k \cr k - l \cr} \right) = f(l)\,f(k - l)\quad \Rightarrow \cr & \Rightarrow \quad \left( {n + m} \right)! = F(n)\,F(m) \cr} $$ which cannot be for every $n$ and $m$.

That means that it is not assured that the polynomial resulting from the convolution be in general factorizable , except from having the constant term null and thus allowing to separate a $x$.

For example $$ \int_0^x {\left( {1 + t} \right)\left( {1 + x - t} \right)dt} = {{x^{\,3} } \over 6} + x^{\,2} + x = x\left( {{{x^{\,2} } \over 6} + x + 1} \right) $$