Thread: Hypergeometric function

Hello everyone, this is my first post and might be trivial.

After some calculations, i come up with the following equality:

2F1(1,1/n,1+1/n,a c)==2F1(1,1/n,1+1/n,b c) (1)

where, 2F1 is Gauss' hypergeometric function, n,c real positive quantities.

Obviously, the equality holds when a==b (2). My problem is that i want to show that (2) is not only sufficient but also necessary, i.e. (1) holds only when (2) holds.

Any ideas?
TIA

Try to prove it isn't necessary and derive a contradiction.

Originally Posted by Randomas

Hello everyone, this is my first post and might be trivial.

After some calculations, i come up with the following equality:

2F1(1,1/n,1+1/n,a c)==2F1(1,1/n,1+1/n,b c) (1)

where, 2F1 is Gauss' hypergeometric function, n,c real positive quantities.

Obviously, the equality holds when a==b (2). My problem is that i want to show that (2) is not only sufficient but also necessary, i.e. (1) holds only when (2) holds.

Any ideas?
TIA

Some members may be able to help you, but since hypergeometric functions are not super common outside of special function theory, some background would be appreciated.

Ok, we define Gauss' hypergeometric function by an infinite sum as follows:

= =

Where (.)n is Pochhammer's symbol. Changing the notation in my example:

2F1(1,1/m,1+1/m,x*d)==2F1(1,1/m,1+1/m,y*d) (1)

where m,d positive real quantities. The equality holds when x==y (2)

I want to show that (2) is a necessary condition.

Assuming that x<>y, (1) can be written in the form:

(frac1)*((x-y)d)+(frac2)*((x^2-y^2)*d^2)+(frac3)*...==0 (3)

It is sufficient to show that (3) cannot hold... Obviously, we can show this if each term is positive (negative), thus the LHS is positive (negative) as an infinite sum of positive (negative) values.

I think that this is true if we assume (additionally) that |x|<=1,|y|<=1. Am I right? Is there an elegant way to present this?

TIA