I have a function $g(x,y)$ that contains a product of hypergeometric functions, both involving the variables $x$ and $y$. I try to do a series expansion in the two variables as recommended in this answer, however one of the two hypergeometrics does not get expanded. What can I do to avoid that?

enter image description here

The code:

g = (8 (1 + 2 s) (t x - I t y)^(2 + s) (t x + I t y) Hypergeometric2F1[1/2, -1 - s, -(1/2) - s, (t x + I t y)/(t x - I t y)] Hypergeometric2F1[1/2, 3 + s,7/2 + s, (t x - I t y) (t x + I t y)])/(1 + s);
Normal[Series[g, {t, 0, 6}]] /. t -> 1

(set $t$ to $1$ in order to retrieve the original function $g(x,y)$)

  • 1
    $\begingroup$ It's quite clear what happens by the way: the $t$ gets canceled in the argument of the hypergeometric function, maybe revealing a weak point of the answer linked above. $\endgroup$ – Jxx Sep 15 at 21:52
  • $\begingroup$ Do you really need an expansion around x=0 and y=0? There is no limit at this point, causing problems. $\endgroup$ – Carl Woll Sep 15 at 22:19
  • $\begingroup$ @CarlWoll How do you mean that? When I plot the function for some $s$ (those are integers by the way) it seems regular at $(0,0)$. $\endgroup$ – Jxx Sep 16 at 8:31
  • $\begingroup$ For example, the following limit is indeterminate Limit[Hypergeometric2F1[ 1/2, -1 - s, -(1/2) - s, (t x + I t y)/(t x - I t y)], {x, y} -> {0, 0}] $\endgroup$ – Carl Woll Sep 16 at 16:04
  • $\begingroup$ @CarlWoll Yes true, one can see it also by setting either $x$ or $y$ to $0$. Let me think about it. $\endgroup$ – Jxx Sep 16 at 19:29

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