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I came across the following hypergeometric function recently: $$ _2F_1(1-n,p-2n+1;p-n+1;x) $$ where $p > 0$ is a non-integer constant, $n$ some large positive integer, and $x > 0$ a small constant.

I'd like to know its growth order. I am expecting it to decay like $(1-x)^{2n}$ or so but I have no idea how to prove it. Invoking the identity $$_2F_1(a,b;c;x) = (1-x)^{c-a-b} {}_2F_1(c-a,c-b;c;x)$$ I get $$ _2F_1(1-n,p-2n+1;p-n+1;x) = (1-x)^{2n-1} {}_2F_1(p,n;p-n+1;x) $$ so I am expecting that when $n$ is sufficiently large, $_2F_1(p,n;p-n+1;x)$ grows very slowly (with respect to $n$), slower than $n^{p-\epsilon}$ for some small $\epsilon$ that may depend on $x$ and $p$. I tried numerical values in Mathematica, and it seems that $_2F_1(p,n;p-n+1;x)$ is positive, slowly increasing and upper bounded by $1$ when $n$ is sufficiently large. In fact, proving it is upper bounded by some constant (could depend on $p$ and $x$) would be already good for me.

I tried to expand it into series and compared the term of the same index for $n$ and $n+1$, but this did not work.

If it is well-known, can anyone point me to some reference?

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  • $\begingroup$ It's not explicitly stated in your question what kind of growth are considering, forcing the readers to guess. Do you mean growth in $x$, growth in $n$, or something else? Also, are considering the growth wrt one variable, with all others fixed, or do you need some uniform bound with respect to some domain of the remaining variables? $\endgroup$ – Igor Khavkine Mar 2 '16 at 21:33
  • $\begingroup$ @IgorKhavkine sorry, I've updated the question. Growth is w.r.t. $n$ and the constant bound can depend on $p$ and $x$. $\endgroup$ – user58955 Mar 2 '16 at 21:49
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    $\begingroup$ You may want to take a look at DLMF§15.12, in particular at formula 15.12.7. Though, there the large parameter may have the sign opposite to yours. Still, at the bottom of that page, there are also some hints about how to combine their formulas and some transformations to obtain other asymptotic forms. $\endgroup$ – Igor Khavkine Mar 2 '16 at 22:10
  • $\begingroup$ @IgorKhavkine Thanks for the reference. Just curious, is there a simple way to say, for instance, for all large $n$, $_2F_1(p,n;p-n+1;x)$ have the same sign? $\endgroup$ – user58955 Mar 3 '16 at 1:28
  • $\begingroup$ All I can say that it would be simple for me. $\endgroup$ – Igor Khavkine Mar 3 '16 at 5:33

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