# Product of two generalized hypergeometric functions ${}_1F_2 \times {}_1F_2$

My question concerns the product of the 2 generalized hypergeometric functions $$_1F_2(a-1/2;a,2a-1;-4x)_1F_2(b-1/2;b,2b-1;-4x)$$ under the conditions that all the values are real and >zero, and a not equal to b. I was wondering if this product can be simplified to another generalized hypergeometric function. I tried the double-summation approach using Bailey's transform (Slater's 1966 book, p. 58-) and managed to reduce the inner, n-summation to a terminating, nearly-poised $$_4F_3(1-a-k,2-2a-k,b-1/2,-k;3/2-a-k,b,2b-1;1)$$ but could go no farther than that, assuming my lengthy derivation is correct. Any ideas? Note that the above function has a positive, +1 argument, and is not Saalschutzian, with which most of the recent research seems to be concerned. Slater has a long list of summation results in Appendix III of her book, but i would say all of them have very restrictive relations between the constants involved. Any help would be deeply appreciated

## 2 Answers

Probably a stupid question.

Why not to use $$\, _1F_2\left(a-\frac{1}{2};a,2 a-1;-4 x\right)=$$ $$-2 a x^{\frac{1}{2}-a} \Gamma (a)^2 J_a\left(2 \sqrt{x}\right) J_{a+1}\left(2 \sqrt{x}\right)+x^{1-a} \Gamma (a)^2 J_{a+1}\left(2 \sqrt{x}\right){}^2+x^{-a} \Gamma (a+1)^2 J_a\left(2 \sqrt{x}\right){}^2$$

or $$\, _1F_2\left(a-\frac{1}{2};a,2 a-1;-4 x\right) \, _1F_2\left(b-\frac{1}{2};b,2 b-1;-4 x\right)=$$ $$\Gamma (a)^2 \Gamma (b)^2 (a \, _0\tilde{F}_1(;a+1;-x)-x \, _0\tilde{F}_1(;a+2;-x)){}^2 (b \, _0\tilde{F}_1(;b+1;-x)-x \, _0\tilde{F}_1(;b+2;-x)){}^2$$

i know...i am not a very smart guy LOL!

i appreciate the quick answer, and i am familiar with the relation between $$_0F_1$$ and the Bessel function of the 1st kind. This means a lot to me.

but tragically, it's not what i am looking for. I am trying to convert the product of 2 hypergeometric functions (as stated in my question) to a single hypergeometric function - not to a polynomial of another hypergeometric function. The answer posted is even less compact than what i already have, and it is in powers of $$_0F_1$$. I know that $$(_0F_1)^2$$ produces a $$_2F _3$$, which can be simplified to a $$_1F _2$$ depending on the similarity of the parameters. This would bring us back to a polynomial in $$_1F _2$$ or Bessel functions, back to square one. The approach i am trying to follow begins by expressing each hypergeometric function in its infinite sum, resulting in a double sum, followed by Bailey's Transform. This reduces the double sum to a single sum of weighted, terminating $$_4F _3$$ sequence, as stated in the original question. I was wondering if there exists a simplification for the terminating $$_4F _3$$ sequence - because if there is, the sum of the weighted $$_4F _3$$ sequence reduces to a single, compact hypergeometric function - maybe i wasnt clear in the original post, apologies. As an example, Slater (p. 245) states that under certain conditions, a terminating $$_4F _3$$ reduces as follows: $$_4F _3 [a/2, 1/2+a/2,b+k,-k; b/2, b/2+1/2,1+a; 1]= (b-a)_k/(b)_k$$

but is not applicable to my case. Slater's book is almost 55 years old, and i am sure much progress has been made since the publication of her book. Mathematica and Maple were of no help either, assuming i used them correctly. Regardless, i still appreciate the time taken to answer the question - many thanks Claude.