# Series expansion of hypergeometric function with two variables

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? 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 $$g(x,y)$$)

• 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. – Jxx Sep 15 at 21:52
• Do you really need an expansion around x=0 and y=0? There is no limit at this point, causing problems. – Carl Woll Sep 15 at 22:19
• @CarlWoll How do you mean that? When I plot the function for some $s$ (those are integers by the way) it seems regular at $(0,0)$. – Jxx Sep 16 at 8:31
• 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}] – Carl Woll Sep 16 at 16:04
• @CarlWoll Yes true, one can see it also by setting either $x$ or $y$ to $0$. Let me think about it. – Jxx Sep 16 at 19:29