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I am trying to write a shifted exponential function as a power series. I am aware of the power series definition of the exponential function, i.e. $e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}$. However, for this application, this definition is not as useful for reasons I will not go into detail about.

That said, I want to come up with a power series representation for the following shifted exponential function

\begin{equation} y = e^{-a(gx-b)^{2}} \end{equation}

The following is what I have done...

Taking the derivative of $y$ w.r.t. $x$ allows for the following differential equation to be derived.

\begin{equation} y'+2ag(gx-b)y = 0 \end{equation}

Setting this up as a power series gives

\begin{equation} \begin{aligned} 0 &=\sum_{n=1}^{\infty} n \,c_{n}(gx-b)^{n-1}+2ag\sum_{n=0}^{\infty}c_{n}(gx-b)^{n+1}\\ &=\sum_{n=0}^{\infty} (n+1) \,c_{n+1}(gx-b)^{n}+2ag\sum_{n=1}^{\infty}c_{n-1}(gx-b)^{n}\\ &=c_{1}+\sum_{n=1}^{\infty}\left((n+1) \,c_{n+1}+2ag\,c_{n-1}\right)(gx-b)^{n} \end{aligned} \end{equation}

Setting the coefficients equal to zero gives a solution for $c_{1}$ and a recurrence relation

\begin{equation} \begin{aligned} c_{1}=0 & \quad\quad\text{for} \ n=0\\ c_{n+1} =-\frac{2ag}{(n+1)}c_{n-1} & \quad\quad\text{for} \ n=1,2,3,\dots \end{aligned} \end{equation}

If I start plugging values for $n$ into the recurrence equation I get

\begin{equation} \begin{aligned} n=1\to\quad &c_{2}=-\frac{2ag}{(2)}c_{0} &&\\ n=2\to\quad &c_{3}=-\frac{2ag}{(3)}c_{1} &&=0\\ n=3\to\quad &c_{4}=-\frac{2ag}{(4)}c_{2} &&=\frac{(2ag)^{2}}{(4)(2)}c_{0}\\ n=4\to\quad &c_{5}=-\frac{2ag}{(5)}c_{3} &&=0\\ n=5\to\quad &c_{6}=-\frac{2ag}{(6)}c_{4} &&=-\frac{(2ag)^{3}}{(6)(4)(2)}c_{0}\\ \end{aligned} \end{equation}

This recurrence relation can be solved to be

\begin{equation} c_{2k}=\frac{(-2ag)^{k}}{(2k)!!}c_{0} \quad \text{for} \ k=1,2,3,\dots \end{equation}

which gives a power series solution for the shifted exponential

\begin{equation} e^{-a(gx-b)^{2}} = c_{0}\sum_{k=1}^{\infty}\frac{(-2ag)^{k}}{(2k)!!}(gx-b)^{k} \end{equation}

Here is my question: How do I solve for $c_{0}$? Sorry if this is a simple question but I couldn't figure it out.

Aaron Hendrickson
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  • Um, $c_0$ is the constant term in your series, so it has to be equal to the function value at $x=0$. Or am I missing something? – hmakholm left over Monica Jul 24 '16 at 21:54
  • It does seem like I'm missing something: Why are you developing your function as a power series in $(gx-b)$ rather than a series in $x$? You can get a series in $(gx-b)$ much easier simply by plugging $-a(gx-b)^2$ into the known series for the exponential function. – hmakholm left over Monica Jul 24 '16 at 21:58
  • I ended up using binomial theorem to expand the $(gx-b)^{k}$ then switch the order of summation and sum over $n$ giving $-2(1-e^{-ab})c_{0}\sum_{n=0}^{\infty}\frac{(n+1)}{(2n+2)!!}(2agx)^{n}$. So $c_{0}$ is simply $e^{-ab^{2}}$? – Aaron Hendrickson Jul 24 '16 at 22:02
  • @Henning Makholm I tried solving for a series in $x$ directly. The recurrence formula was really difficult to solve. By solving in $(gx-b)$ I was able to get a recurrence relation I could solve and then did the expansion/reordering to get the final series in $x$. – Aaron Hendrickson Jul 24 '16 at 22:30
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    You made an error in differentiating, there's a 'g' missing. – Burrrrb Jul 24 '16 at 23:25
  • @Asemismaiel I fixed the error although there are still issues with my math somewhere. Thanks for the help. – Aaron Hendrickson Jul 24 '16 at 23:52

2 Answers2

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If your solution is a power series, then you can solve for $c_0$ with the initial value of your ODE. Typically, $c_0 = y(0)$ if your series is built about $x=0$. So if that is the case then $c_0=y(0)=e^{-ab^2}$

Z-Math
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Plug in $x=\frac{b+\frac{1}{2a}}{g}$ in the two sides of the last equation you get: $$e^{-\frac{1}{4a}} =c_0 \cdot\sum_{k=1}^{\infty}\frac{1}{k!!}$$ Mathematica gives the value $\frac{1}{2} \left[-2 + 2 \sqrt{e} + \sqrt{2 e \pi} \cdot erf(\frac{1}{\sqrt{2}}) \right]$ for the series on the RHS, which gives a very weird value for $c_0$.

Burrrrb
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