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This question already has an answer here:

Why do the following 2 sequences give different answers?

  • n = 1.5

    Series[Gamma[0.5 - n - x], {x, 0, 2}]

  • Series[Gamma[-1 - x], {x, 0, 2}]

(..clearly the output from the second expression is correct since it has the 1/x pole..)


I want do similar Laurent series finding with products of Gamma functions like,

f[n_,x_] := (29/4)*((Gamma[
   0.5 - n - x] Gamma[(n + 1)/2 + x] Gamma[(n + 1)/2 + x])/(Gamma[
   1 - n/2] Gamma[1 - n/2] Gamma[1 + n + x]));

(at different values of "n")

I would think that doing, Series[f[1.5,x],{x,0,2}] should do. But is there something I need to be careful about given the above kind of discrepancy?

  • I eventually want to look at a sum of such "f" as above kind of functions and find the poles of the sum. Any cautions about that would be very helpful!
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marked as duplicate by rm -rf Jan 9 '14 at 23:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If you take the series Series[Gamma[-a - x], {x, 0, 2}] you get a series having Gamma[-a] now for a integer positive those functions are undefined. So, you can't just using one series to cover the whole behavior of this transcendental function. You have to have this in mind in the function you want to study. $\endgroup$ – Spawn1701D Apr 11 '13 at 6:10
  • $\begingroup$ @Spawn1701D This comment of yours is in reference to my "f"? What do you suggest I do? $\endgroup$ – user6818 Apr 11 '13 at 6:15
  • $\begingroup$ it extents to f because if, say, $0.5-n = -\kappa,\ \kappa\epsilon I$ the expansion will be different, because of the pole. $\endgroup$ – Spawn1701D Apr 11 '13 at 6:27
  • $\begingroup$ Why are you using inexact numbers? Try e.g. Series[1/2 - n - x, (* stuff *)]... $\endgroup$ – J. M. will be back soon Apr 11 '13 at 6:27
  • $\begingroup$ @J.M Changing 0.5 to 1/2 doesn't remove the discrepancy. What do I do? $\endgroup$ – user6818 Apr 11 '13 at 6:32