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