0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica 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
  • 1
    $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.