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Re: Another MPFIT question [message #62828 is a reply to message #62827]

Sun, 12 October 2008 01:29

MichaelT
Messages: 52
Registered: May 2006

Member

On Oct 12, 4:02 am, Vince Hradil <vincehra...@gmail.com> wrote:
> How's this
> re-parameterize our problem to -
> p[0]=c1, p[1]=c1+c2, p[2]=c1+c2+c3 and p[3]=c1+c2+c3+c4
> then limit all these to [0,1]

Many thanks Vince! That gave me the right idea!

I had to implement it a tiny little bit differently, though. It turned
out that in your suggested case c1...4 would not have been limited to
[0, 1] but [-1, 1].
So I had to do it like this:

p[0]=1-c1, p[1]=1-c1-c2, p[2]=1-c1-c2-c3 and p[3]=1-c1-c2-c3-c4

function myfunc, x, p
c1 = 1 - p[0]
c2 = p[0]-p[1]
c3 = p[1]-p[2]
c4 = p[2]-p[3]
return, somefunctionof(c1,c2,c3,c4)
end

Have a good Sunday!
Michael

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[Message index]

		Re: Another MPFIT question By: MichaelT on Sun, 12 October 2008 03:56
		Re: Another MPFIT question By: MichaelT on Sun, 12 October 2008 03:42
		Re: Another MPFIT question By: MichaelT on Sun, 12 October 2008 01:29
		Re: Another MPFIT question By: Vince Hradil on Sat, 11 October 2008 19:02

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