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MathGroup Archive 1999

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Re: FindRoot question (number of variables)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg15764] Re: [mg15743] FindRoot question (number of variables)
  • From: Daniel Lichtblau <danl>
  • Date: Sun, 7 Feb 1999 02:03:47 -0500 (EST)
  • References: <199902050842.DAA10047@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Doug Webb wrote:
> 
> Hello,
>   This is a simplified example of I problem I have run into with
> FindRoot. Does anybody have any idea why it's doing what it is? And if
> so, how I can make it do what I want, or some other work around? Thanks
> for any info. Here's the example:
> 
> define a function that takes one input, and outputs two results:
> 
> r[x_] := {x, x^2}
> 
> It does what's expected:
> 
> r[4]
> {4,16}
> 
> Now, find a solution using FindRoot:
> 
> FindRoot[ r[x] == {5, 25}, {x, 2}]
> 
> returns an error (rather than the obvious answer of x->5):
> 
> FindRoot::"frnum":
>     "Function {{-3., -21.}} is not a length 1 list of numbers at {x} =
> {2.}."
> 
> It seems that FindRoot REQUIRES you to solve for as many variables as
> your function returns. Even though when entered seperatly, r[5] == {5,
> 25} returns "true". Why? Any ideas? Or any ideas how to get around this
> "feature" of FindRoot? Thanks for any input.
> 
>       Doug
>       D_Webb at prodigy.net
>       Douglas.S.Webb at maf.nasa.gov


I do not know offhand of a way to have FindRoot work around this
limitation that #variables must equal #equations (possibly another
reader will have ideas about this). But notice you can simply resort to
FindMinimum instead.

In[5]:= FindMinimum[Evaluate[Apply[Plus, Thread[r[x] - {5,25}]^2]],
{x,2}]

                 -28
Out[5]= {3.187 10   , {x -> 5.}}

Daniel Lichtblau
Wolfram Research


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