The solution to this is ugly, right?

[Alec Mihailovs]

From: "Alec Mihailovs" <alec at mihailovs.com>
Sent: Friday, May 30, 2008 11:49 PM


>>> Solve:
>>>    (u+c)^(-1/2) + (u-c)^(-1/2) = 1 for u.
>
> The solution can be expressed as
>
> u = 1+3* 3F2(-1/2, 1/4, 3/4; 1/3, 2/3; -4 c^2/27)
>
> where 3F2 is a hypergeometric function.

Just 2 more comments.

1. That can be written as

u = 4 + 3 sum_{n=1}^oo (-1)^(n+1) (4n n)/(2n-1)/64^n  c^(2n)

with (4n n) being a binomial coefficient. The beginning of the series looks 
like

u = 4 + (3/16)c^2 - (7/1024)c^4 + (33/65536)c^6 - (195/4194304)c^8 + ...

and I didn't find the numerators in the OEIS.

2. Starting from the initial equation, r + (1-r) = 1 with 0 < r <= 1/2, we 
get

c = (1/r^2 - 1/(1-r)^2) / 2 and u = (1/r^2 + 1/(1-r)^2) / 2.

Substituting that in the formula for the solution, we get an identity

3F2(-1/2, 1/4, 3/4; 1/3, 2/3; -(1-2r)^2 / (27 r^4 (1-r)^4)) =

(1 - 2r + 4r^3 - 2r^4) / (6 r^2 (1-r)^2)  for 0 < r < 1.

In particular, for r = 1/3 that gives

3F2(-1/2, 1/4, 3/4; 1/3, 2/3; -27/16) = 37/24

Alec Mihailovs