Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I'm tried to find a certain expansion of the Hypergeometric1F1 function using Mathematica:

FullSimplify[Series[Hypergeometric1F1[-n, -2 n, 2 I k x], {x, Infinity, 1}, 
Assumptions -> {n \[Element] Integers, x \[Element] Reals, k \[Element] Reals, n >= 0}]]

were here $n>=0$ and integer and $k,x>0=0$ and I'm trying to obtain the leading order terms in the expansion as $x\to\infty$. You can see the result Mathematica gives by copy and pasting the above, but I think it must not be taking my assumptions into account as it gives factors like $\Gamma[-2 n]/\Gamma[-n]$ which are not well defined for non-negative integer $n$.

How can I obtain this expansion using Mathematica in a way were it will take into account the assumptions about the parameters correctly?

share|improve this question
    
You might find the output from this to be suggestive: Assuming[k \[Element] Reals, Limit[k^(-#) (2 # - 1)!! x^# Abs[ Hypergeometric1F1[-#, -2 #, 2 I k /x]], x -> 0] & /@ Range[0, 9]]. As far as that Gamma ratio goes, the output of 2 (-1)^n Gamma[2 n + 1] / Gamma[n + 1] Gamma[-2 n]/Gamma[-n] // FullSimplify suggests one way to simplify it. –  whuber Mar 21 '13 at 19:33
    
@whuber thanks. I was looking for $x\to\infty$ limit rather than $x\to 0$. Could you give me some more info about what your first line is doing here? as for your second line are you suggesting that this ratio is infact quite well defined? –  fpghost Mar 21 '13 at 20:07
1  
Take a closer look: I have replaced your $x$ by $1/x$, so as my $x\to 0$, you learn about the asymptotic behavior at $\infty$. And yes, the ratio is well-defined if it is viewed as a removable singularity of $\Gamma(-2z)/\Gamma(-z)$. Perhaps the most obvious demonstration is in terms of the infinite product expansion of $1/\Gamma$, which makes it immediate that every pole of $\Gamma(-2z)$ (for integral $z$) is canceled by a pole of $\Gamma(-z)$; the ratios of their residues is where the double factorial comes in. –  whuber Mar 21 '13 at 20:12
    
So why does Evaluate[Gamma[-2]/Gamma[-1]] in Mathematica give complex infinity? How could I make Mathematica spit out a complex number instead? –  fpghost Mar 21 '13 at 22:10
    
It gives infinity for the same reason that, say, evaluating Infinity/Infinity gives an indeterminate result: neither Gamma[-2] nor Gamma[-1] is finite. You can, however, make sense of such a ratio as a "removable singularity" of some function, such as the limit as $z\to 1$ of $\Gamma(-2z)/\Gamma(-z)$. However, there is no unique way to do this. For a example, compute that limit and compare it to the limit as $z\to 1$ of $\Gamma(2 z)/\Gamma(-z^2)$: you will find they are different, even though both expressions have equally valid claims to be called "Gamma[-2]/Gamma[-1]". –  whuber Mar 22 '13 at 0:27
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.