# Product of Power Series of Different Powers

I am trying to find the product $M$ of two power series of the form

$$M=\left(\sum_{n=0}^{\infty}a_{n}\, x^{2n}\right) \left(\sum_{n=0}^{\infty}b_{n}\, x^{n}\right)$$

where, $a_{n}=\frac{(-ag^{2})^{n}}{n!}$, and $b_{n}=\frac{(2ag)^{n}}{n!}$.

The product of the two series could be found with the standard formula (discrete convolution) if both series contained powers of $x^{n}$. I have tried to find a way to calculate the product but am not making progress. One potential issues is that $a_{n}$ is alternating and would become imaginary if the square root is taken. How can I calculate this product?

P.S- I suspect the final answer will be an infinite sum over confluent hypergeometric functions.

I am working on an integral of the form

$$\int_{0}^{\infty} x\, e^{-a(gx-b)^{2}}\, e^{-\mu x}\, {_{1}}F_{1}[-\alpha,-\beta,\lambda x] \ dx$$

If I keep my limits of integration and write the exponential as a power series I can solve the integral. There is no way I can find to solve the integral if I substitute $u=x-b$. I tried tackling this by writing the exponential in quesiton as:

\begin{aligned} e^{-a(gx-b)^{2}} &= \sum_{n=0}^{\infty}\frac{(-a)^{n}(gx-b)^{2n}}{n!}\\ &= \sum_{n=0}^{\infty}\frac{(-a)^{n}}{n!}\sum_{k=0}^{2n}\binom{2n}{k}(-b)^{2n-k}(gx)^{k} \end{aligned}

Switching the order of summation allows for a solution as a single sum:

$$e^{-a(gx-b)^{2}} =\sum_{k=0}^{\infty}\, \frac{(-a)^{k/2}(-g)^{k}}{\frac{k}{2}!}\,{_{1}}F_{1}\left(\frac{k+1}{2};\frac{1}{2},-ab^{2}\right)\, x^{k}$$

This sum has imaginary terms for odd $k$ and is not particularly useful for my purposes.

• Define $c_n$ by $c_{2n+1}=0$ and $c_{2n}=a_n$. Then $M=(\sum c_n x^n)(\sum b_n x^n)$. – Kelenner Jul 26 '16 at 14:22
• In this case, two exponential functions? – André Nicolas Jul 26 '16 at 14:29
• @Andre Nicolas Yes this is the product of two exponential functions. I need the final series to be of the form $\sum c_{n}x^{n}$ because I am integrating the $x^{n}$. – Aaron Hendrickson Jul 26 '16 at 14:31
• If you use that $M^{\prime}(x)=(-2ag^2x+2ag)M(x)$, you get that $$(m+1)d_{m+1}=-2ag^2d_{m-1}+2agd_m$$ But I do not know if this is useful for you. – Kelenner Jul 26 '16 at 14:50
• The $a_{k/2}$ are here when $k$ is even. Rewrite completely the formula to see that $$d_n=\sum a_l b_{n-2l}$$ where we sum over the $l$ such that $0\leq 2l\leq n$ – Kelenner Jul 26 '16 at 14:56

We can use the standard formula with a slight variation:

We obtain \begin{align*} \left(\sum_{k=0}^\infty a_kx^{2k}\right)\left(\sum_{l=0}^\infty b_lx^l\right) &=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}a_kb_l\right)x^n\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_kb_{n-2k}\right)x^n\tag{2} \end{align*}

Comment:

• In (1) the condition for the inner sum is $2k+l=n$ to respect the even powers $x^{2k}$ and all powers $x^l$.

• In (2) we use the floor function to set the upper limit of the inner sum and use $l=n-2k$.

For anyone who is interested: Using the answer posted by @Markus Scheuer, I found a series representation for the exponential function in the additional information part of the question I posted.

\begin{aligned} e^{-a(gx-b)^{2}} &= e^{-ag^{2}x^{2}}e^{2abgx}e^{-ab^{2}}\\ &=\left(\sum_{n=0}^{\infty}\frac{(-ag^{2})^{n}}{n!}x^{2n}\right) \left(\sum_{n=0}^{\infty}\frac{(2abg^{2})^{n}}{n!}x^{n}\right)e^{-ab^{2}}\\ &=e^{-ab^{2}} \sum_{n=0}^{\infty}\left(\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{(-ag^{2})^{k}}{k!}\frac{(2abg)^{n-2k}}{(n-2k)!}\right)\, x^{n}\\ &=e^{-ab^{2}} \sum_{n=0}^{\infty}U\left(-\frac{n}{2},\frac{1}{2},ab^{2}\right)\, \frac{(2\sqrt{a}g)^{n}}{n!}\, x^{n} \end{aligned}

where, $U(a,b,z)$ is the confluent hypergeometric function of the second kind (a.k.a. Kummer U function).