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Re: calculation

Saeed Esmaily Rashid wrote:
> Hello!
> My name is Saeed and i'm studying physics. I have a problem which ihope
> someone can help me with it. I have an equation
> .001422409738*R^(73/80)*(R^(1/2)-1)^(-13/40) == .04
> i'm using the Solve function in Mathematica 3.0 to solve it for R, but
> it calculats endlessly and takes very long time. the question is that
> is there any way to optimize this equation or using another function in
> Mathematica 3.0 to make it faster to calculate?
> Regards

Your equation looks transcendental, meaning not solveable by algebra. 
It also appears to be ill conditioned numerically.  To see this, try:

Plot[.001422409738*R^(73/80)*(R^(1/2)-1)^(-13/40) ,{R,1,3}]

You will see that your equation has a vertical asymptote at R==1.  To
understand the effect this has, set 


Two test values are:


The difference in the argument is very small and the derivatives are
huge in the region of interest.  There are very few numerical root
finders that can handle this kind of equation without modification. 
The first thing you want to do is make sure that any numerical root
finder never tries values less than one since that makes the function
complex.  To do this, set
f[R_]=Abs[.001422409738*R^(73/80)*(R^(1/2)-1)^(-13/40)].  The next thing
is to use the correct routine.  Solve uses algebra until it gives up
and then calls NSolve which uses Gauss factoring.  FindRoot uses
Newtons Method if you give it one guess (and will attempt to calculate
symbolic derivates of your function) or secant method if you give it
two guesses.  With a big "no mans land" of possible arguments for R
less than one, secant method is the best in this case if you give it a
very good initial guess as in:

    Abs[.001422409738*R^(73/80)*(R^(1/2)-1)^(-13/40)] ==

which returns:


Remove the _nospam_ in the return address to respond.

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