The solution to this is ugly, right?

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

On Sat, May 31, 2008 at 12:49 AM, Alec Mihailovs <alec at> wrote:
> From: "Mitch Harris" <maharri at>
>> On Fri, May 30, 2008 at 1:40 PM, Jack Brennen <jb at> 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


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
comments that were sent to the mailing list but not submitted as formal
comments through the web site.

I've discovered a handful of places where you did not receive proper
credit.  These will be fixed at the next big update, tomorrow I hope.
[Authorship of the b-file for A116436 is one example.]

I'm sure I owe similar apologies to other people, so I will
copy this to the mailing list.

And, by the way, Thanks for all your contributions, as always!

- Do send important new sequences and comments, especially

- Don't send less important sequences or minor comments

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- Submissions that seem arbitrary will be silently deleted

I am starting to change the way contributions are processed.
There will be major changes to the OEIS this year.

But there is a large backlog that I must get caught up with first.
Then there will be a lot of new programs to be written.
So please be patient.

Thank you!        Neil Sloane

PS Extensions of existing sequences are welcomed, also b-files, also
corrections of bad errors.

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