Re: Non-linear ecuation... -Reply

[Michael Hanke]

Hi,

The explanations below were not very satisfactory for me. So I played a little
bit with the example. If I did not make any mistake, I got the following
solutions starting with the initial guess identical to one:

l2 =

    0.1696
    0.8485
    1.5093
    0.3211
    5.8123
    8.6608
    0.1994
   25.0683
    0.7635  

They seem to be a solution. I was using a straighforward damped Newton method
and a rathet sophisticated affine invariant Newton method. They needed 8 and 7
Newton steps, respectively, for solving this system.

To be honest, an implementation of a trust-region method and the back-tracking
line search along NR did not converge.

Hope that helps,

Michael


Gerrit Visser wrote on Fre, 24 Mär 2000:
> I've played around with the equations, and I can see why octave can't
> solve it. Some of the variables have a very large influence on the
> equations. At a guess, some of the gradients must be in the 10^9 region
> or higher. To see what I mean, set up a "cost" function which has the
> 
> >>> Dirk Laurie <address@hidden> 23/March/2000 05:59pm >>>
> Cederik skryf:
> > 
> > I have the next set of nine non-linear ecuations:
> > 
> > ln(a)-ln(f)+i+4g+(19720/8314)=0
> > ln(b)-ln(f)+2g+h-(192420/8314)=0
> > ln(c)-ln(f)+h+i-(200240/8314)=0
> > ln(d)-ln(f)+2h+i-(395790/8314)=0
> > ln(e)-ln(f)+2g=0
> > a+c+d-2=0
> > 4a+2b+2e-14=0
> > b+c+2d-3=0
> > a+b+c+d+e-f=0
> > 
> > Where:
> > x[1]=a
> > x[2]=b
> > x[3]=c
> > x[4]=d
> > x[5]=e
> > x[6]=f
> > x[7]=g
> > x[8]=h
> > x[9]=i
> > 
> > Octave can't solve it... (non-convergent), but in fact i know that set of
> > ecuations has solution. Because is a Book example. 
> > Octave can solve it if i put the initial x's [a,b,c,d,e,f,g,h,i] as the
> > exactly solutions. But that way isn't usefull for me... 
> > Any sugestions?
> > 
> 
> 
> 
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|  Michael Hanke                Royal Institute of Technology   |
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