Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Consider a series: $$\sum_{t=0}^\infty \frac{8^{-11-2t}(22+4t)!}{t!(11+t)!(11+2t)!(32+t)}$$

Sum[(8^(-11 - 2 t) (22 + 4 t)!)/(t! (11 + t)! (11 + 2 t)!) 1/(32 + t),{t,0,Infinity}]

I enter it into Mathematica, and get the following: $$\frac{26880307030942914706481517908268600094665992\sqrt2}{2617883526123366341980485070559163273300330975\pi}\tag1$$

26880307030942914706481517908268600094665992 Sqrt[2]/(2617883526123366341980485070559163273300330975 Pi)

So far so good. Now I change $32$ to q and get the new answer: $$\frac{88179{\;}_3 F_2\left(\frac{23}4,\frac{25}4,q;12,q+1;1\right)}{1073741824q}\tag2$$

88179 HypergeometricPFQ[{23/4, 25/4, q}, {12, 1 + q}, 1]/(1073741824 q)

OK, not bad. Let's now do %/.q->32. But... it remains in almost unchanged form of $$\frac{88179{\;}_3 F_2\left(\frac{23}4,\frac{25}4,32;12,33;1\right)}{34359738368}\tag3$$ I now try doing Simplify, FullSimplify, but it doesn't change. I tried checking $(1)$ and $(2)$ for equality with ==, and still got unevaluated expression. Using Reduce over it gave me True with strange problem of

Reduce::ztest1: Unable to decide whether numeric quantity /* difference of LHS and RHS multiplied by denominator */ is equal to zero. Assuming it is. >>

Still, computing difference between ${}_3 F_2\left(\frac{23}4,\frac{25}4,32;12,33;1\right)$ and $\frac{923600316834709429658044469155615723122343517467181056\sqrt2}{230842351450032320669497193036836458276349885044525\pi}$ with N[...,30] gives me almost zero ($10^{-74}$), so this function indeed must be simplifiable.

So, how do I convince Mathematica to simplify such hypergeometric function expressions?

share|improve this question
    
Just an observation, FullSimplify[(sum2 /. q -> 32) - sum1] gives 0 without warnings. –  b.gatessucks Dec 7 '13 at 9:15
    
Oh, indeed, using FullSimplify instead of Reduce gives True without warnings. Still, it doesn't reveal $(1)$ without need to put it into expression. –  Ruslan Dec 7 '13 at 9:17
    
@Nasser added some code –  Ruslan Dec 7 '13 at 9:56
add comment

1 Answer

up vote 4 down vote accepted
expr2 = Sum[(8^(-11 - 2 t) (22 + 4 t)!)/(t! (11 + t)! (11 +2 t)!) 1/(q + t), {t, 0, Infinity}]
% /. q -> 32

Mathematica graphics

FunctionExpand[%]

Mathematica graphics

share|improve this answer
    
Wow, didn't know of FunctionExpand. This is great. Thanks. –  Ruslan Dec 7 '13 at 10:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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