       Re: Re: Showing that a hypergeometric expression is 0?

• To: mathgroup at smc.vnet.net
• Subject: [mg73480] Re: [mg73435] Re: Showing that a hypergeometric expression is 0?
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Mon, 19 Feb 2007 01:29:15 -0500 (EST)

```On 16 Feb 2007, at 07:03, David W.Cantrell wrote:

> Paul Abbott <paul at physics.uwa.edu.au> wrote:
>>> [Message also posted to: comp.soft-sys.math.mathematica]
>>>
>>> How can I use Mathematica to show that
>>>
>>> n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, -1] +
>>> Hypergeometric2F1Regularized[k, -n, k - n, -1] +
>>> k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, -1] -
>>> k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, -1]
>>>
>>> is identically 0?
>>>
>>> I presume that I could do that by hand, perhaps without much
>>> trouble.
>>> But I also have far, far messier expressions involving
>>> hypergeometric
>>> functions which I need to show to be identically 0, and I really
>>> don't
>>> want to have to do so by hand!
>>
>> One approach is to use FunctionExpand to convert
>> Hypergeometric2F1Regularized ->
>> Hypergeometric2F1 and then apply
>>
>>    http://functions.wolfram.com/07.24.17.0039.01
>>
>> in the form of a generic rule:
>>
>>    Hypergeometric2F1[b_, (m_.) - n, c_, -1] :>
>>     (n! Hypergeometric2F1[-n, 1 - b, 1, 2])/((-1)^n Pochhammer[1 -
>> b, n])
>>     /;
>>        c == b + m - n
>>
>> If you FunctionExpand and then FullSimplify the result you obtain
>>
>>   (Gamma[1 - k] Gamma[1 + n] (k (1 + k - n) Hypergeometric2F1[1 -
>> k, -n,
>>    1, 2] - (1 + k - 2 n) (k - n) Hypergeometric2F1[-k, -n, 1, 2])
>> Sin[(k
>>    - n) Pi])/
>>     ((-1)^n (k - n) (1 + k - n) Pi)
>>
>> Simplification of this result for Element[{k,n}, Integers] yields 0.
>> Observe that
>>
>>   Simplify[Gamma[1 - k] Sin[(k - n) Pi], Element[{k,n}, Integers]]
>>
>>   0
>>
>> Another approach might be to show that (z+1) is a factor of
>>
>> n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, z] +
>> Hypergeometric2F1Regularized[k, -n, k - n, z] +
>> k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, z] -
>> k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, z]
>>
>> for arbitrary z.
>
> Many, many thanks for your response, Paul!
>
> BTW, if anyone has other ways of showing that the expression is 0,
> I'd love
> to see them.
>
> David Cantrell
>

Here is an informal argument which, while of course not a proof,
indicates that the result ought to be true and perhaps a formal proof =

could be constructed along these lines. We define

f[x_] := n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, x] +
Hypergeometric2F1Regularized[k, -n, k - n, x] +
k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, x] -
k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, x]

and simply compute

Table[FullSimplify[Limit[Normal[f[x] + O[x]^p], x -> -1],
(n | k) =E2=88=88 Integers], {p, 1, 5}]

{(k*n)/Gamma[k - n + 2], (k*(k + 1)*(n - 1)*n)/
Gamma[k - n + 3], (k*(k + 1)*(k + 2)*(n - 2)*(n - 1)*
n)/(2*Gamma[k - n + 4]),
(k*(k + 1)*(k + 2)*(k + 3)*(n - 3)*(n - 2)*(n - 1)*n)/
(6*Gamma[k - n + 5]), (k*(k + 1)*(k + 2)*(k + 3)*
(k + 4)*(n - 4)*(n - 3)*(n - 2)*(n - 1)*n)/
(24*Gamma[k - n + 6])}

The answers strongly suggest that if p>n the  Taylor series truncated
at p  is actually 0 and hence so is the full Taylor series. (This is
because the numerators contain products of the form  (n-p+1) ... (n -
4)*(n - 3)*(n - 2)*(n - 1). While this is, of course, not a proof it
suggest a possible proof, by induction on p.

Andrzej Kozlowski

```

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