Hypergeometric transformations and identities

[DavidSmith]

How do you derive hypergeometirc identities of the form

2F1(a,b,c,z)= gamma function. What I mean is that the hypergeometric function converges to a set of gamme functions function in terms of (a,b,c)

where z is not 1,-1, or 1/2 ?

The hypergeometric identities in the mathworld summary which give gamma functions only have values of 1,-1, and 1/2 for Z but none others.

I have seen hypergeometric functions where z=1/8,1/3, 2/3 etc that give gamma functions but have no idea how to deive them despite many months of time and research.

 

[DavidSmith]

After spending another 3 days with the problem I used a kummer quadratic identity combined with annother kummer transformation to get a hypergeometric function for z=-1/8

(8^(-3b+1)/2)*2F1((3b-1)/2,-b/2+1/2;1/2+b;-1/8)=

4*(gammafunction(1/2)*gammafunction(2b))/(b*gammafunction(b/2))^2

there are more possible

 

[tacman]

Hypergeometric transformations and identities are important tools in mathematics, particularly in the field of special functions. They involve expressing a hypergeometric function in terms of other functions, such as gamma functions. These transformations and identities can be derived using various techniques, such as the binomial theorem, Euler's integral formula, and the Cauchy integral formula.

To derive hypergeometric identities of the form 2F1(a,b,c,z) = gamma function, we can use the binomial theorem. This theorem states that for any real or complex numbers x and y, and any integer n, (x+y)^n can be expanded as a sum of terms of the form x^iy^(n-i), where i ranges from 0 to n. This expansion can then be used to manipulate the hypergeometric function and express it in terms of gamma functions.

For example, let us consider the hypergeometric function 2F1(a,b,c,z). Using the binomial theorem, we can write:

(1+z)^c = ∑(c k)z^k

where (c k) represents the binomial coefficient c choose k. Multiplying both sides by (1-z)^(-b), we get:

(1-z)^(-b)(1+z)^c = ∑(c k)z^k(1-z)^(-b)

Using the definition of the hypergeometric function, we can rewrite the left side as:

2F1(a,b,c,z) = (1-z)^(-b)(1+z)^c

Substituting this into the previous equation, we get:

2F1(a,b,c,z) = ∑(c k)(-b)(-b-1)...(-b-k+1)z^k

= ∑(c k)(-b)^kz^k

= ∑(c k)(-b)^kz^k(k!/(k!)^2)

= ∑(c k)(-b)^k(a)_k/(b)_kz^k/k!

= ∑(c k)(-b)^k(a)_kz^k/k!/(c)_k(b)_k

= (a)_0z^0/(c)_0(b)_0 + ∑(c k)(a)_kz^k/(c)_k(b)_k

= gamma function + ∑(c k)(a)_kz^k