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I am hoping to evaluate a simpler expression for the following: $${}_3F_2(-n,1,1; a, (a+3)/2; -1)$$ Here $n,a \in \mathbb{N}$ and $a$ is odd. I am also interested in the asymptotics in $n \in \mathbb{N}$ for fixed $a$. There are numerous transformations for hypergeometric series with unit argument, but I was not able to find any with negative unity as the argument. Considering that two of the parameters in the numerator are unity, I was hoping that a nice expression would be nearby.

I would appreciate if someone can help with references or ideas.

1 Answers 1

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With $f_n$ denoting the hypergeometric function, it can be checked to satisfy the inhomogeneous rank 2 recurrence equation: $$ (n+a+1)\left(n+\frac{a+5}{2}\right)f_{n+2} - 3 (n+2)\left(n+\frac{a+3}{2}\right) f_{n+1} + 2 (n+1)(n+2) f_n = \frac{a^2-1}{2} $$

In[52]:= Table[-(1/2) (a^2 - 1) + (n + a + 1) (n + (a + 3)/2 + 1) y[       n + 2] - 3 (n + 2) (n + (a + 3)/2) y[n + 1] +      2 (n + 1) (n + 2) y[n] /.     y -> Function[k,       HypergeometricPFQ[{-k, 1, 1}, {a, (a + 3)/2}, -1]], {n, 0,     6}] // Together  Out[52]= {0, 0, 0, 0, 0, 0, 0} 

From this recurrence equation, using techniques of Birkhoff and Trjitzinsky, "Analytic theory of singular difference equations", Acta Math., 60 (1932), pp. 1–89, one gets: $$ f_n = 2^n n^{(1-3a)/2} \left(c_1 + \mathcal{O}\left(n^{-1}\right) \right) + \frac{1}{n} \left(c_2 + \mathcal{O}\left(n^{-1}\right) \right) + \frac{a^2-1}{2} \frac{\log(n)}{n} \left(1 + \mathcal{O}\left(n^{-1}\right) \right) $$

Also note, that for special case of $a=1$, closed-form is easy to find: $$ f_n = {}_3F_2\left(-n,1,1; 1,2; -1\right) = {}_2F_1\left(-n,1; 2; -1\right) = \frac{2^{n+1}-1}{n+1} $$

One could also try use the following integral representation for $f_n$, valid for $a>1$: $$ f_n = \frac{a+1}{2} \int_0^1 \left(1-t\right)^{(a-1)/2} \cdot {}_2F_1\left(-n, 1; a; -t\right) \mathrm{d} t $$

In[79]:= Table[  Simplify[HypergeometricPFQ[{-n, 1, 1}, {a, (a + 3)/2}, -1] ==     Assuming[     a > 1, ((a + 1)/2)*      Integrate[(1 - t)^((a - 1)/2)*        HypergeometricPFQ[{-n, 1}, {a}, -t], {t, 0, 1}]]], {n, 1, 5}]  Out[79]= {True, True, True, True, True} 
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    Wow! Thanks @Sasha! I did not know any such techniques were available. Could you suggest a book for the techniques you have employed from Birkoff and Trjitzinsky?2012-09-19
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    @Ankur I do not know of such book, which is not to say one does not exist. You could also check Birkoff's [article](http://www.jstor.org/stable/1988577), freely available from JSTOR.2012-09-19
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    Could you tell which theorem from Birkoff and Trjitzinsky have you employed here? I went through the paper but unable to find a result that would suit my problem. Also, I think we are in the same town - if that's ok I could you even meet you in person to get your help.2012-09-21
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    @Ankur Well, I might have given you not quite the right article reference. See Bikhoff's article, "Formal theory of irregular singular difference equations", Acta Mathematica ( [SpringerLink](http://www.springerlink.com/content/f18753v776587770/)). We are indeed in the same town. Small world.2012-09-21
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    Unfortunately, in this paper too I was not able to find a theorem from which your solution is derived. I am mystified by how you were able to derive the solution in such a specific form. Could you edit the answer with a few more details or point me to a reference where something similar has been solved?2012-09-21
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    I tried substituting Eq (6) from Birkhoff's paper to determine the coefficients in the homogeneous version of the recursion. But that does not lead me to any term like $2^n n^{(1-3a)/2}$. Could you clarify how you get this term? It would be of great help if you could provide some details or an outline of your calculations. Thanks!2012-09-24
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    @Ankur I will try to provide more details tonight. Essentially this is a method of indeterminate coefficients, like you tried. Substitute into eq. 6 the anzats $f_n = n^{c_1 n} c_2^n n^{c_3}$. Then consider asymptotic behavior of $f_{n+1}/f_n$. Substitute back into the recurrence equation, and match coefficients. I will also try to provide an alternative (calculus-based) derivation based on the integral representation mentioned in the post, where I would use $$ {}_2 F_1(-n,1,a,-t) = (a-1) \int_0^1 (1-u)^{a-1} (1+t u)^n \mathrm{d} u $$ valid for $a>1$ and $n \geqslant 0$.2012-09-24
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    I tried the method of undetermined coefficients and was able to find two linearly independent solutions for the *homogeneous* version of the recursion: $2^nn^{(1-3a)/2}(1+ O(1/n))$ and $\frac{1}{n}(1+ O(1/n))$. But I was not able to obtain the particular solution that you have found. Could you please add a few details to your answer?2012-09-30
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    @Ankur Since the inhomogeneous recurrence equation reads $g_n = \frac{a^2-1}{2}$, it can be homogenized, as $g_{n+1}-g_n = 0$. This implies order 3 recurrence homogeneous recurrence equation for $f_n$. You can apply similar technique to find the asymptotic behavior of $f_n$. Substitution back into the inhomogeneous system determines relation between two of three free coefficients. I hope this is clear enough.2012-10-02
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    @Sasha What is a good reference for recurrence relations for 4F3 along the lines of the one at the beginning of your answer? Thanks.2015-06-16
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    The paper referenced, Birkhoff and Trjitzinsky, "Analytic theory of singular difference equations", is at https://projecteuclid.org/download/pdf_1/euclid.acta/14858879902018-05-16