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Suppose I have the function $_2F_1\left(a,b;c;x^2\right)$ with $a=\frac{3}{4}+\frac{k}{4}$, $b=\frac{3}{4}-\frac{k}{4}$ and $c=\frac{1}{2}$. I want to know the behaviour about $x=1.\,$ I go to DLMF equation 15.10.21 and choose $$ w_1\left(x^2\right) = {\frac {\Gamma \left( c \right) \Gamma \left( c-a-b \right)}{ \Gamma \left( c-a \right) \Gamma \left( c-b \right) }} \, w_3\left(x^2\right) +{\frac {\Gamma \left( c \right) \Gamma \left( a+b-c \right)}{\Gamma \left( a \right) \Gamma \left( b \right) }} \, w_4\left(x^2\right). $$ Since $w_4$ is singular at $x=1$ (and it should be finite) I expected this typical constraint that $a$ or $b$ must be some $n\leq 0 \in \mathbb{Z}$ for the second term to vanish. Now the first term has $\Gamma(-1)$. Is that a problem or can I absorb this into a constant? However if I do so, then the original function $w_1$ is not really defined. Does this mean the solution is not valid unless $k=3$?

PS: Actually if $k=4n+3$ $(n\geq 0)$ then the second term vanishes and in the first term the poles cancel?

Somos
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Diger
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2 Answers2

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The first thing to notice is that the equation is symmetric in $\, a,b \,$ which comes from changing $\, k \,$ to $\, -k. \,$ Thus, without loss of generality suppose $\, k\ge 0. \,$ Let us define some notation: $\, f_3 := (\Gamma(c)\Gamma(c-a-b))/(\Gamma(c-a)\Gamma(c-b)), \quad f_4 := (\Gamma(c)\Gamma(a+b-c))/(\Gamma(a)\Gamma(b)) $ which are multiplied by $\, w_3(z) \,$ and $\, w_4(z) \,$ respectively. The equation is $\, w_1(z) = w_3(z) f_3 + w_4(z) f_4. \,$ Note that we will also add $\,\epsilon\,$ to $\,a\,$ to probe sensitivity to parameters. We are interested in the values of the functions at $\, z=1. \,$ For all integer $\, k, \,$ we have $\, w_3(1) = 1. \,$ We have the cases:

  • If $\, k \,$ is even, $\, w_1 \,$ has a pole at $\,z=1,\,$ but the factor $\, f_3 \,$ has an $\, \epsilon \,$ pole, and $\, w_4 \,$ has an $\, \epsilon \,$ pole, but its factor $\, f_4 \,$ is finite.

  • If $\, k=1 \,$ then both $\, w_1 \,$ and $\, w_4 \,$ have a $\, z=1 \,$ pole while $\, f_3 = 0. \,$

  • If $\, k>1 \,$ and $\, k=4n+1 \,$ then $\, w_1 \,$ has a pole at $\,z=1,\,$ while $\, f_3 \,$ has an $\, \epsilon \,$ pole, and $\, w_4 \,$ has an $\, \epsilon \,$ pole while $\, f_4 \,$ is finite.

  • If $\, k = 4n+3, \,$ then $\, w_1, w_3, f_3 \,$ are all finite, while $\, f_4 = 0 \,$ but $\, w_4 \,$ has both a $\, z=1 \,$ pole and an $\, \epsilon \,$ pole. Note that $\, w_1(z) \,$ is a polynomial in $\, z \,$ with $\, w_1(1) = 1. \,$

Somos
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  • I'm not quite on par with you here: If $k$ is even the second term ($w_4$) is finite, but the first term is actually infinite! If $k=1$ then I get $1/4$ and $1$ for the first and second term respectively. When $k=4n+1$ the first term is infinite and the second is finite. When $k=4n+3$ the first term is finite and the second is 0. – Diger Jun 18 '18 at 01:04
  • Yes, I seem to have confused $\Gamma$ with $1/\Gamma.$ – Somos Jun 18 '18 at 01:28
  • So what do these (automatic) infinities mean, when e.g. $k$ is even? Is there no expansion about $x=1$ in this case? – Diger Jun 18 '18 at 01:32
  • "If $k=4n+3$, then $f_1$ is infinite but is very sensitive to the values of $a,b$ and $f_2=0$."

    Again you probably mean "then $f_1$ is finite, but very sensitive..." ?

    What I did here is because this term is not clearly defined I replaced $a$ and $b$ by $a+\epsilon$ and $b+\epsilon$ and finally let $\epsilon \rightarrow 0$.

     – Diger Jun 18 '18 at 11:52
  • But it is true: I could have also replaced it by $a+\epsilon$ and $b+2\epsilon$ and get a different answer. However: How do I decide if I have nothing but the differential equation I obtained this solution from. There is no "hint" on the precise value... – Diger Jun 18 '18 at 11:58
  • Actually the last is also true for $k=1$. I obtain different values on the chosen $\epsilon$. – Diger Jun 18 '18 at 12:05
  • "... and $w_4$ has an $\epsilon$ pole..." What $\epsilon$ pole are you talking about? $w_4$ has a $z$ pole... Where is the epsilon pole for $w_4$ ? I just see $(1-z)^{c-a-b} \sim (1-z)^{-1+{\cal O}(\epsilon)}$ about $z=1$. – Diger Jun 18 '18 at 13:02
  • I note that $\epsilon$ is added to $a$ to probe sensitivity to parameters. – Somos Jun 18 '18 at 13:20
  • I see you are right. If I set $a=\frac{3+k}{4}+\epsilon$ and $b=\frac{3-k}{4}+\epsilon$ then $w_4$ happens to be $$(1-z)^{-1-2\epsilon} , _2F_1\left(\frac{-1-k}{4} - \epsilon, \frac{-1+k}{4} - \epsilon ; -2\epsilon ; 1-z \right)$$ – Diger Jun 18 '18 at 13:30
  • But then even in the case $k=4n+3$ the second term does not vanish, because of the additional $\epsilon$ pole due to $w_4$. – Diger Jun 18 '18 at 13:38
  • I guess I see it's because of the logarithmic singularity as Maxim pointed out. – Diger Jun 18 '18 at 13:53
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Let $$F(x) = {_2F_1}\left( \frac {3+k} 4, \frac {3-k} 4; \frac 1 2; x \right), \quad k \geq 0.$$ If $(3-k)/4$ is an integer, $F(x)$ becomes a polynomial and we have $$F(x) = F(1) + O(|1-x|) = \frac {(-1)^{(k-3)/4} \sqrt \pi \,\Gamma \left( \frac {k+5} 4 \right)} {\Gamma \left( \frac {k-1} 4 \right)} + O(|1-x|).$$ Otherwise this is the logarithmic case, the formulas for which can be found here. The leading term is $$F(x) = \frac {\sqrt \pi} {\Gamma\left( \frac {3+k} 4 \right) \Gamma\left( \frac {3-k} 4 \right) (1-x)} + O(|\!\ln(1-x)|).$$

Maxim
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