Re: Help finding x of hypergeometric 2F1[a,b,c,x] ?
- To: mathgroup at smc.vnet.net
- Subject: [mg71927] Re: Help finding x of hypergeometric 2F1[a,b,c,x] ?
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Mon, 4 Dec 2006 06:39:02 -0500 (EST)
- Organization: The University of Western Australia
- References: <ekmet6$8mj$1@smc.vnet.net>
In article <ekmet6$8mj$1 at smc.vnet.net>, titus_piezas at yahoo.com wrote:
> Hello all,
>
> The above function is given by Mathematica as
> Hypergeometric2F1[a,b,c,d]. Define the ff,
>
> h1 = 2F1[a,b,1,1-x]
> h2 = 2F1[a,b,1,x]
>
> Problem: Given a,b where 0<(a,b)<1, and a+b=1, find the unique real
> number x with 0<x<1 such that,
>
> h1/h2 = sqrt[n]
>
> for arbitrary rational n>0.
I would define h1 and h2 as follows:
h1[a_][x_] = Hypergeometric2F1[a, 1 - a, 1, 1 - x]
h2[a_][x_] = Hypergeometric2F1[a, 1 - a, 1, x]
that is, as functions of one parameter a, and one variable x.
> For certain (a,b) namely, (1/2, 1/2), (1/3, 2/3), (1/4, 3/4) and (1/6,
> 5/6), closed-form solutions are known for x. For example, the first
> reduces to finding the "elliptic modulus k" and x can easily be given
> as x = ModularLambda[Sqrt[-n]]. Surprisingly though, it seems little
> is known for other (a,b).
Why surprisingly? New special functions, such as ModularLambda, are only
introduced if they are frequently encountered.
>
> Question: Is there Mathematica code to numerically evaluate x to
> arbitrary precision for any (a,b) and n?
Yes -- and it is called FindRoot. For example, with a -> 1/2
FindRoot[h1[1/2][x]/h2[1/2][x] == Sqrt[3/4], {x, 0.1},
WorkingPrecision -> 50]
This gives the same answer as
N[ModularLambda[Sqrt[-(3/4)]], 50]
> The fact that x is 0<x<1 greatly helps.
Not sure why this "greatly helps". It is just a condition for the
problem.
Cheers,
Paul
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