# The solution to this is ugly, right?

Mitch Harris maharri at gmail.com
Mon Jun 2 05:01:39 CEST 2008

```On Sat, May 31, 2008 at 12:49 AM, Alec Mihailovs <alec at mihailovs.com> wrote:
> From: "Mitch Harris" <maharri at gmail.com>
>> On Fri, May 30, 2008 at 1:40 PM, Jack Brennen <jb at brennen.net> wrote:
>>>
>>> Solve:
>>>
>>>   (u+c)^(-1/2) + (u-c)^(-1/2) = 1 for u.
>>>
>>> The solution for u is still a quartic depending on c, but
>>> it will at least appear a bit less ugly.
>>
>> namely:
>> -4 c^2 + c^4 +
>> 4 c^2 u +
>> (8 - 2 c^2) u^2 +
>> -4 u^3 +
>> u^4
>> ==0
>
> I think, it is wrong. I get
> u^4 - 4 u^3 - 2c^2 u^2 + 4c^2 u + c^4 + 4c^2 = 0

Yes. I must have made a mistake in my manipulations. On redoing, I get
the same as yours.

> 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.

That is something I don't know how to do.
- that is one solution out of 4, right? any particular one?
- how do you derive that? Is it  straightforward to read directly
from the coefficients in the equation?
- any properties of the solution that are more apparent from the
hypergeometric?

Mitch

Dear Franklin,   I'm trying to get caught up with "Comments",
of which there are nearly 2000 in the current stack (I'm planning to

I'm writing to apologize to you for not making full use
of the comments you have sent in during the past year, in particular
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I've discovered a handful of places where you did not receive proper
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[Authorship of the b-file for A116436 is one example.]

I'm sure I owe similar apologies to other people, so I will
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And, by the way, Thanks for all your contributions, as always!

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