7
$\begingroup$

The series $$\sum_{k=0}^\infty \frac{\exp(c k \beta)}{(k!)^\beta} $$ has come up when I'm trying to apply the methodology in this paper (http://www.ism.ac.jp/~eguchi/pdf/Robustify_MLE.pdf) to Poisson regression.

When $\beta = 1,$ it is $e^{e^c}$ and when $\beta = 2,$ it is $I_0(2e^c),$ the modified Bessel function of the first kind, but have no idea what happens at other values, or how to approach the question.

Does this series have either a closed form expression in terms of special functions, or an asymptotic expansion that is numerically useful for $\beta \in [1, 5]$ and $|c| < R$?

$\endgroup$
  • $\begingroup$ This sum was discussed in the book "Advanced Mathematical Methods for Scientists and Engineers" by Carl M. Bender and Steven A Orszag in section 6.7 Example 4. The result is (with $x=exp(c \beta)$) $(2 \pi)^{(1-\beta)/2} \beta^{-1/2} x^{(1-\beta)/(2 \beta)} exp(\beta x^{1/\beta})$. $\endgroup$ – Johannes Trost Nov 29 '14 at 16:15
  • $\begingroup$ The "result" in my above comment means the asymptotic expansion of the sum for large $x$. $\endgroup$ – Johannes Trost Dec 1 '14 at 12:04
  • $\begingroup$ ... and of course I would have to have given away my copy of Bender and Orszag :). If you put that as an answer, I'll accept it. $\endgroup$ – AatG Dec 1 '14 at 18:54
  • $\begingroup$ Buy one from Amazon ? Google ? However, I put my comment as an answer. Maybe you can accept it anyways. $\endgroup$ – Johannes Trost Dec 3 '14 at 8:55
5
$\begingroup$

I copied this from my comments above:

This sum was discussed in the book "Advanced Mathematical Methods for Scientists and Engineers" by Carl M. Bender and Steven A Orszag in section 6.7 Example 4. The asymptotic result is (with $x=\exp(c \beta)$) $(2 \pi)^{(1-\beta)/2} \beta^{-1/2} x^{(1-\beta)/(2\beta)}\exp(\beta x^{1/\beta})$.

Edit: Sketch of a proof following Bender's & Orszag's lines: Let us look at a somewhat simpler sum $$ \sum_{k=0}^{\infty} \frac{x^{k}}{k!^{\beta}}. $$ By examining the ratio of two consecutive terms of this sum we find that the terms are increasing until $k=k_{max}:=\lfloor x^{1/\beta}\rfloor$ and then decreasing. Using Stirling's formula and relaxing the requirement of $k$ being integer, we can expand $k!$ around $k_{0}$ by inserting $k=x^{1/\beta}+t$ $$ k! \sim x^{k/\beta} \exp(-x^{1/\beta}+t^{2}x^{-1/\beta}/2) \sqrt{2 \pi} x^{1/(2 \beta)} $$ for large $x$ and small $t$, i.e. $t^{3}\ll x^{2/\beta}$, which let us neglect higher terms in $t$. We find that the terms are sharply peaked for $k$ around $k_{0}$. Approximate the sum over $k$ by an integral over $t$, with the integration range extended to the whole real axis. (We can do this since the additional contributions to the integral are negligible.) Evaluating the integral gives the result.

Edit: Getting higher orders for any real $\beta$ is straight forward but tedious with the above (Laplace-) method. However, going after chapter 5.11 in Luke's "The Special Functions and their Approximation", Vol. 1, one can give a rather compact formula for any order, when restricting $\beta$ to positive integers. Actually, it is the asymptotic expansion of the hypergeometric function $_{0}F_{\beta-1}(;1,...,1;\exp(c \beta))$, which is the same as the OP's sum. Write $x=\exp(c \beta)$. Then $$ _{0}F_{\beta-1}(;1,...,1;x)= (2 \pi)^{(1-\beta)/2} \beta^{-1/2} \exp(\beta x^{1/\beta}) x^{\frac{1}{2 \beta}-\frac{1}{2}} \sum_{k=0}^{\infty} c_{k} \beta^{-k} x^{-k/\beta} $$ where $c_{k}$ is recursively defined as $$ c_{k}=\frac{1}{\beta k} \sum_{s=1}^{\beta-1} c_{k-s} \sum_{r=0}^{s}\frac{(-1)^{s-r}}{r!(s-r)!} (r+k+\frac{1}{2}-\frac{\beta}{2})^{\beta} $$ and $c_{0}=1$, $c_{k}=0$ for $k<0$.

I checked the formula with Mathematica for several values of $\beta$ and orders and it works fine.

$\endgroup$
5
$\begingroup$

If $\beta$ is an integer $\ge 3$, it can be written as a hypergeometric $$ \mbox{$_0$F$_{\beta-1}$}(\ ;\, 1,\ldots,1;,\exp(c\beta))$$ where there are $\beta-1$ ones.

$\endgroup$
  • 1
    $\begingroup$ I think what you are saying is that this is how the hypergeometric is defined, but this does not really shed a huge amount of light on the matter :) $\endgroup$ – Igor Rivin Oct 31 '14 at 15:10
  • 1
    $\begingroup$ I agree, it does not by itself shed a lot of light on the matter, but it does provide a name that can be looked up. For example, you can ask Maple's FunctionAdvisor about hypergeom([],[1,1,1],t). In this case, you'll find a differential equation ${\frac {{\rm d}^{4}}{{\rm d}{t}^{4}}}f \left( t \right) =-6\,{\frac {{ \frac {{\rm d}^{3}}{{\rm d}{t}^{3}}}f \left( t \right) }{t}}-7\,{ \frac {{\frac {{\rm d}^{2}}{{\rm d}{t}^{2}}}f \left( t \right) }{{t}^{ 2}}}-{\frac {{\frac {\rm d}{{\rm d}t}}f \left( t \right) }{{t}^{3}}}+{ \frac {f \left( t \right) }{{t}^{3}}} $ $\endgroup$ – Robert Israel Oct 31 '14 at 16:31

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.