6
$\begingroup$

Working on a problem in spectral theory, I need to study the asymptotics of a confluent hypergeometric function (here $(a)_0=1$ and $(a)_s=a(a+1)\cdots(a+s-1)$ denote the Pochhammer symbol) $$ \mathbf{M}(a,b,z)=\sum_{s=0}^{+\infty}\frac{(a)_s}{\Gamma(b+s)s!}z^s,\quad\text{as}\ z\to+\infty. $$ In my case $-1<a<0$ and $b=1$, and I'm only interested in real $z$.

I had a look in Abramowitz–Stegun (13.5.1, where $M(a,b,z)=\Gamma(b)\mathbf{M}(a,b,z)$), and found that, as $z\to+\infty$, we have the expansion $$ \mathbf{M}(a,1,z)\sim \frac{e^{i\pi a}}{\Gamma(1-a)}z^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s} + \frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}. $$ What is worrying me is the factor $e^{i\pi a}$ in the first term. It is complex (in fact, non-real), and everything else in the expansion is real for positive $z$. Also, from the definition of $\mathbf{M}$ we see that it should be real for positive $z$. I have also had a look in 13.7.2, where the same expansion is given. It is also the same in the book Asymptotics and special functions by Frank Olver, and I get the same from Mathematica. Thus, I believe that the expansion above is correct.

In the asymptotic expansion above the term with the $e^{i\pi a}$ factor is small in comparison with the second one. In fact, some sources hint that it can be neglected (compare 13.7.1). As it happens, I want to keep that term, even if it is small. Thus, I think I can state my questions as follows:

  1. Why is the real-valued function having complex terms in its asymptotic expansion?
  2. I'm only considering real $z$. Will the expansion of $\mathbf{M}(a,1,z)$ above still be valid if I replace $e^{i\pi a}$ with its real part, $\cos(\pi a)$?
  3. Could it be that the imaginary part somehow cancels? (I don't see how it could.)
$\endgroup$
2
$\begingroup$

When z is real and tends to $\infty$ (that is, $z>0$) the second part of this double series is dominant, that is, is exponentially bigger that the first part and then, in asymptotic sense, you can avoid it to obtain

$$\mathbf{M}(a,1,z)\sim \frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}$$

so, the complex terms in its asymptotic expansion dissapear. Note that when $z$ is real and negative, that is, $z\to-\infty$, is the first part that is dominant and taking account that $z^{-a}=e^{i\,\pi\,a}(-z)^{-a}$, the imaginary part also dissapear

$$\mathbf{M}(a,1,z)\sim \frac{1}{\Gamma(1-a)}(-z)^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s}$$

where here $-z$ is positive.

$\endgroup$
  • $\begingroup$ Thank you for your answer. What you write is clear to me, but I cannot throw away the smaller part. I should perhaps have been more precise in my question. I know that for $-1<a<0$ there is a unique zero of $a\mapsto \mathbf{M}(a,1,z)$ as $z$ is large (positive). I want to find the asymptotics of this zero as $z\to+\infty$. The only term that can help me is $1/\Gamma(a)$, and as $a\to 0^-$ this term tends to zero. However, to get the correct asymptotics ($a\sim -z e^{-z}$) I need the first term from the non-dominate part of the expansion. $\endgroup$ – mickep Apr 4 '16 at 12:09
  • $\begingroup$ Yes, maybe you must be more explicit in your question and explain better this last purpose. $\endgroup$ – ppooppii Apr 4 '16 at 12:48

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.