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MathGroup Archive 2006

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Re: Please help: How to use Mathematica to get Parametric solution for a transcendental equation?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69335] Re: Please help: How to use Mathematica to get Parametric solution for a transcendental equation?
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Thu, 7 Sep 2006 04:30:42 -0400 (EDT)
  • Organization: The University of Western Australia
  • References: <edjhha$lvs$1@smc.vnet.net>

In article <edjhha$lvs$1 at smc.vnet.net>, Leonxf <leonxf at gmail.com> 
wrote:

> The function is just one:
> 
> A*(1+G)+a*[(1+G)^t]=T, where A,a,t,T are all positive real numbers, and t is
> no larger than 1.

Consider the equivalent transcendental equation

  a g + b g^t == c

where a > 0, b > 0, c > 0, 0 < t <= 1, and you want to find g.

Using series reversion:

  http://mathworld.wolfram.com/SeriesReversion.html

one obtains a series expansion of the solution (about g == 1 to 2nd 
order):

  ser = InverseSeries[ Series[ a g + b g^t, {g, 1, 2}], c ]

                                2              2
      (-a - b + c)   (b t - b t ) (-a - b + c)                 3
  1 + ------------ + -------------------------- + O[-a - b + c]
        a + b t                        3
                            2 (a + b t)

One can choose a different expansion point and compute to higher order.

For, say, a -> 3, b -> 1, c -> 2.5 and t -> 0.7 one obtains

  Normal[ser]/. {a -> 3, b -> 1, c -> 2.5, t -> 0.7}

  0.5992586816180681

Using FindRoot, one obtains

 FindRoot[a g+b g^t == c /. {a -> 3, b -> 1, c -> 2.5, t -> 0.7}, {g, 1}]

 {g -> 0.600165573071457}

Cheers,
Paul

_______________________________________________________________________
Paul Abbott                                      Phone:  61 8 6488 2734
School of Physics, M013                            Fax: +61 8 6488 1014
The University of Western Australia         (CRICOS Provider No 00126G)    
AUSTRALIA                               http://physics.uwa.edu.au/~paul


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