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I want to represent the function:

\begin{equation} f(x)=e^{-a(x-b)^{2}} \end{equation}

where, $0<a<1$, $x\in\mathbb{R}$, and $b\in\mathbb{R}$.

As a power series for an integral I am working on. First, I re-wrote $f(x)$ using the power series definition of the exponential to get,

\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}(x-b)^{2n} \end{equation}

Using binomial expansion on the $(x-b)^{2n}$ yields

\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}\sum_{k=0}^{2n}\binom{2n}{k}(-b)^{2n-k}\,x^{k} \end{equation}

I want to then simplify this expression by switching the order of summation. Here is what I did:

\begin{equation} \begin{aligned} f(x) &= \sum_{k=0}^{\infty}x^{k} \sum_{n=k/2}^{\infty}\binom{2n}{k}\frac{(-a)^{n}}{n!}(-b)^{2n-k}\\ &=\sum_{k=0}^{\infty}\frac{(-\sqrt{-a}x)^{k}}{\frac{k}{2}!}\,{_{1}}F_{1}\left[\frac{k+1}{2},\frac{1}{2},-ab^{2}\right] \end{aligned} \end{equation}

where, ${_{1}}F_{1}(a,b,z)$ is the confluent hypergeometric function of the 1st kind.

At this point the form of $f(x)$ makes the solution to my integral much simpler. However, I am not sure if switching the order of summation like this is allowed under these circumstances. Looks like the sum has an imaginary component that one would not expect. Any thoughts?

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  • $\begingroup$ You have turned a function which for $a > 0$ is very well behaved for large $x$ into an alternating sum, each term of which diverges for large $x$. Also, I believe the transformation to confluent hypergeometric function that you use, while formally correct, applies only for negative $a$ if you want convergence. So I would be very uncomfortable, especially since I assume your integral will cover regions of large $x$. $\endgroup$ – Mark Fischler Jul 22 '16 at 0:16
  • $\begingroup$ I am integrating over $\mathbb{R}^{+}$ and $a$ is positive. I know something isn't right but not sure what went wrong. I do believe everything up to the binomial expansion is good. Should be a way to convert to single sum. $\endgroup$ – Aaron Hendrickson Jul 22 '16 at 0:36
  • $\begingroup$ @AaronHendrickson: what is your ultimate goal in creating this expansion? Are you simply trying to integrate the function over $\mathbb{R}^+$? $\endgroup$ – robjohn Jul 22 '16 at 1:08
  • $\begingroup$ @robjohn I am trying to evaluate a difficult (at least for me) integral see link below. Expanding the exponential allows me to evaluate the integral while maintaining my limits of integration. I am still stuck with a power series at the end but hoping it converges quickly enough to be reasonably approximated with a partial sum. ( math.stackexchange.com/questions/1643790/…). $\endgroup$ – Aaron Hendrickson Jul 22 '16 at 1:17
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The rearrangement appears to be okay. Assuming that $b$ is positive, if we start by switching variables to $z=-x$, then we're looking at $$ f(z) = e^{-a(b+z)^2} $$ which is entire, so the series expansion $$ f(z) = \sum_{n=0}^\infty \frac{(-a)^n}{n!}(b+z)^{2n} $$ converges absolutely for all $z$. In particular for positive $z$, binomial expansion on $(b+z)^{2n}$ will simply break each term of the absolutely convergent series into finitely many terms that all have the same sign, so the broken-up sequence is still absolutely convergent and can be rearranged freely.

When, by rearranging, you get everything down to a power series in $z$, you know that this power series will converge correctly for every positive $z$, which means that it converges (absolutely) for every $z$ where $|z|$ is smaller than a positive number -- that is, for every $z$. And by the identity theorem, then, it converges towards the right thing everywhere in the complex plane.

Finally, changing the sign of some of the coefficients to get a series in $x$ instead of $z$ is of course not going to change the convergence properties.

I lack the prerequisites to verify your final ${}_1F_1$ form, though.

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    $\begingroup$ While these series do converge absolutely, they are not very useful numerically since the terms grow quite large before finally converging when $b+z$ is large. If a unilateral series can be obtained, such as is done in this answer, the numerics become better. $\endgroup$ – robjohn Jul 22 '16 at 1:23
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    $\begingroup$ @robjohn: I don't know what the OP is going to use the final series for, but my point here is that the rearranged series must be the Maclaurin expansion of $f$, so it ought to be useful for small $z$ even if $b+z$ is large. $\endgroup$ – Henning Makholm Jul 22 '16 at 1:29
  • $\begingroup$ @robjohn I took a look at that answer you linked in your comment. Not sure how you got the recursion formula for the coefficients. I recognize that for my problem $\Omega (x)=e^{-a(x-b)^{2}}$, $\Omega' (x)=-2a(x-b)\Omega (x)$. $\endgroup$ – Aaron Hendrickson Jul 22 '16 at 18:10
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    $\begingroup$ @Aaron: Once you have $\Omega'(x)=-2a(x-b)\Omega(x)$ (and you know $\Omega$ is entire) you can expand your differential equation as $$ c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \cdots \\ = 2abc_0 + (2abc_1-2ac_0)x + (2abc_2-2ac_1)x^2 + (2abc_3-2ac_2)x^3 + \cdots $$ and the two sides can only be equal for all $x$ if the coefficients match for each power of $x$, so we get $$ (n+1)c_{n+1} = 2abc_n - 2ac_{n-1} $$ which produces a recurrence for the $c_n$s. $\endgroup$ – Henning Makholm Jul 22 '16 at 19:21
  • $\begingroup$ @Henning Makholm Thank you for your help. I used the method you explained to come up with a solution here. I am just about done but had one last question if your able to help. Thanks. $\endgroup$ – Aaron Hendrickson Jul 24 '16 at 21:51

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