3
$\begingroup$

Let $n>2$ be odd, and let $x\in [0,1]$. I would like to calculate the Taylor expansion of $$ x^{2-n} \, _2F_1\left(-\frac{n}{2}-1,-n;2-\frac{n}{2};x^2\right) $$ at $x=1$ leaving $n$ non specified. However, I get inconsistent results with Series. For instance, consider the minimal example in which the value of $\textrm{Hypergeometric2F1}[- 5/2, -3, -5, 1]$ is calculated

Hypergeometric2F1[-1 - n/2, -n, -n - 2, x] /. n -> 3 /. x -> 1

which gives $\frac{1}{32}$, while

Series[  Hypergeometric2F1[-1 - n/2, -n, -n - 2, (1 - z)], {z, 0, 
   0}] /. n -> 3

gives $\frac{3 \sqrt{z}}{16}$ which for $z=0$ vanishes.

Do you have any suggestion on how to correct this behavior and on how to solve my original problem?

$\endgroup$
  • $\begingroup$ That's very strange. 2F1 with a negative integer in the first group is a polynomial. Your example is (32 - 48 x + 18 x^2 - x^3)/32, or (1 + 15 z + 15 z^2 + z^3)/32. $\endgroup$ – The Vee Jul 6 '16 at 9:31
  • $\begingroup$ @The Vee, yes, in general, hypergeometric functions whose numerator parameters are nonpositive integers degenerate to a polynomial, since some of the associated Pochhammer symbols in the series expansion become zero. $\endgroup$ – J. M. will be back soon Jul 6 '16 at 10:45
  • $\begingroup$ @J.M. Sorry for a misleading wording. I meant the behaviour of Series was strange because the reduction to a polynomial is so obvious. $\endgroup$ – The Vee Jul 6 '16 at 10:53
  • $\begingroup$ Anyway: Sum[(n (n^2 - 4) (-1)^k Binomial[n, k] x^(2 k - n + 2))/(n (n^2 - 4) + 4 k^2 (3 n - 2 k) - k (6 n^2 - 8)), {k, 0, n}], is prolly the series you want, but I derived that without using Mathematica. $\endgroup$ – J. M. will be back soon Jul 6 '16 at 11:02
  • $\begingroup$ @J.M., Thanks. It looks like a Laurent expansion at x=0. I was looking for the expansion at $x=1$. $\endgroup$ – Ettore Minguzzi Jul 6 '16 at 11:36
3
$\begingroup$

I realized I did not answer what the OP was asking in my earlier answer. That was the reason for the misbehaviour, here's the solution to the problem:

The solution can be obtained in a piecewise format: the coefficient at $(x-1)^l$ is $$\alpha_l = \begin{cases} a_l + b_l & l ≤ n+2\ \mbox{and even}, \\ a_l + c_l & l ≤ n+2\ \mbox{and odd}, \\ a_l & \mbox{otherwise}, \end{cases}$$ where $$\begin{aligned} a_l &= (-1)^l {n+l-3 \choose l} {_3F_2}\left(-1-\frac n2, -n, -\frac{n-3}2;\ -\frac{n-3+l}2, -\frac{n-4+l}2;\ 1\right), \\ b_l &= \frac{\left(-1-\frac n2\right)_{\tilde l} (-n)_{\tilde l} (l+1)}{(2-n)_{\tilde l} \tilde l!} {_3F_2}\left(-1-\frac n2+\tilde l, -n+\tilde l, -\frac{n-3}2+\tilde l;\ 1+\tilde l, \frac32;\ 1\right), \\ c_l &= \frac{\left(-1-\frac n2\right)_{\tilde l} (-n)_{\tilde l}}{(2-n)_{\tilde l} \tilde l!} {_3F_2}\left(-1-\frac n2+\tilde l, -n+\tilde l, -\frac{n-3}2+\tilde l;\ 1+\tilde l, \frac12;\ 1\right), \\ \tilde l &= \lfloor(l+n-1)/2\rfloor. \end{aligned}$$ Note that the formulas for $b_l$ and $c_l$ differ in one of the denominator arguments and in a prefactor. All the $_3F_2$'s involved have a finite number of terms as both $-n$ and $-n+\tilde l$ are guaranteed to be nonpositive integers ($\tilde l \le n$).

This was obtained by an explicit expansion of the hypergeometric function $$x^{2-n} {_2F_1}\left(-1-\frac n2, -n;\ 2-\frac n2;\ x^2\right)$$ in $x = 1+z$, using the generalized binomial theorem on $(1+z)^{2-n+2m}$, and extracting the coefficient for $z^l$ by hand. I then separated the cases where the Pochhammer symbol resulting from the latter step was zero, where it was a product of negative numbers only and positive numbers only. (Equivalently, the cases of positive and negative powers of $(1+z)$.) Mathematica then helped simplify the resulting sum over $m$. Unfortunately this is one of the cases where I found doing most of the work easier than trying to convince MMA of using all the assumptions correctly and at a proper time. Even trying to simplify the above piecewise expression brings in new trouble (namely introducing Indeterminates from cases of 0*ComplexIninifty).

I can't guarantee there's no typo in the above. Here's MMA code that should do the same:

With[{n = 7}, Table[
  (-1)^l*Binomial[n + l - 3, l]*
    HypergeometricPFQ[
     {-1 - n/2, -(n - 3)/2, -n},
     {-(n - 3 + l)/2, -(n - 4 + l)/2},
     1]
   + If[l > n + 2, 0,
    Pochhammer[-1 - n/2, ll]*Pochhammer[-n, ll]*
      If[EvenQ[l], l + 1, 1]/Pochhammer[2 - n/2, ll]/ll!*
      HypergeometricPFQ[
       {-1 - n/2 + ll, -n + ll, -(n - 3)/2 + ll},
       {ll + 1, If[EvenQ[l], 3/2, 1/2]},
       1] /. ll -> Floor[(l + n - 1)/2]
    ], {l, 0, 20}]]
CoefficientList[Series[
  With[{n = 7}, (1 + x)^(2 - n)*
    Hypergeometric2F1[-1 - n/2, -n, 2 - n/2, (x + 1)^2]],
  {x, 0, 20}], x]
% == %%

{8192, 16384, 15360, 7168, 1792, 0, 0, 0, 0, 0, 56, -168, 350, -616, 981, -1461, 2073, -2835, 3766, -4886, 6216}

{8192, 16384, 15360, 7168, 1792, 0, 0, 0, 0, 0, 56, -168, 350, -616, 981, -1461, 2073, -2835, 3766, -4886, 6216}

True

$\endgroup$
  • $\begingroup$ Nicely done. It probably should be noted that the hypergeometric functions that show up in the coefficients are in fact polynomial cases as well. $\endgroup$ – J. M. will be back soon Jul 7 '16 at 17:39
2
$\begingroup$

It's a problem of genericity. The output of the command Series[ Hypergeometric2F1[-1 - n/2, -n, -n - 2, (1 - z)], {z, 0, 0}] (prior to substituting for $n$) is valid for almost all real values of $n$ but fails for those which are integer. The reason is that the hypergeometric function changes behaviour in these:

hg = Hypergeometric2F1[-1 - n/2, -n, -n - 2, (1 - z)];
Plot[Evaluate@Table[hg, {n, 2.94, 3.06, 0.02}], {z, 0, 50}]

enter image description here

I'm looking for a reference explaining this behaviour, I'll put it here when I have one.

Edit: Not entirely sure but I think it's using this reduction which has its validity restricted to $c \not\in \mathbb{Z}$.

$\endgroup$
0
$\begingroup$

Somehow this question morphs from asking about Hypergeometric2F1[-1 - n/2, -n, 2-n/2, z] to Hypergeometric2F1[-1 - n/2, -n, -n-2, z]? I'd expect the former is correct and, if so, you can use one of the functional identities (easily turned into a replacement rule) to obtain the series you are after:

x^(2-n) Hypergeometric2F1[-n/2-1,-n,2-n/2,x^2]/.
Hypergeometric2F1[a_,b_,c_,z_]:> Gamma[c-a-b] Gamma[c]*
Hypergeometric2F1[a,b,a+b+1-c,1-z]/(Gamma[c-a] Gamma[c-b])+
Gamma[a+b-c] Gamma[c] (1-z)^(c-a-b)*
Hypergeometric2F1[c-a,c-b,c+1-a-b,1-z]/(Gamma[a] Gamma[b])//FullSimplify

followed by

Series[%, {x, 1, 3}]

The leading term in the expansion is $\frac{2^{n+1} \Gamma \left(2-\frac{n}{2}\right) \Gamma \left(\frac{n+3}{2}\right)}{\sqrt{\pi }}$. Amusingly, the value of the alternative hypergeometric at $x=1$ is the reciprocal of this expression!

BTW, the correct expansion coefficients for $n=7$ are as follows:

{8192, 16384, 15360, 7168, 1792, 0, 0, 0, 0, 0, 112, -336, 700, 
 -1232, 1962, -2922, 4146, -5670, 7532, -9772, 12432}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.