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Re: FindRoot and parameters in NIntegrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg124511] Re: FindRoot and parameters in NIntegrate
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Sun, 22 Jan 2012 07:22:55 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201201211018.FAA10170@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

Here are two fixes:

FindRoot[{1 + a + b, Integrate[a + b x, {x, 0, 1}]}, {{a, 0}, {b, 0}}]

{a -> 1., b -> -2.}

Clear[f]
f[0., 0.] = 0.;
f[a_?NumericQ, b_?NumericQ] := NIntegrate[a + b x, {x, 0, 1}]
FindRoot[{1 + a + b, f[a, b]}, {{a, 0}, {b, 0}}]

{a -> 1., b -> -2.}

The first error message in your post results from FindRoot trying to  
evaluate NIntegrate[a + b x, {x, 0, 1}] before giving a and b numerical  
values. The second results from evaluating it with a == b == 0.

The solutions above avoid both problems.

Bobby

On Sat, 21 Jan 2012 04:18:00 -0600, gac <g.crlsn at gmail.com> wrote:

> I am using Mathematica 8 to solve for parameters in an integral and an  
> auxiliary equation.  A prototypical problem is
>
> FindRoot[{1 + a + b , NIntegrate[a + b x, {x, 0, 1}]}, {{a, 0}, {b, 0}}]
>
> Mathematica reports the correct answer for the parameters a and b:
>
> {a -> 1., b -> -2.}
>
> But it also reports two error messages:
>
> NIntegrate::inumr: The integrand a+b x has evaluated to non-numerical  
> values for all sampling points in the region with boundaries {{0,1}}. >>
>
> NIntegrate::izero: Integral and error estimates are 0 on all integration  
> subregions. Try increasing the value of the MinRecursion option. If  
> value of integral may be 0, specify a finite value for the AccuracyGoal  
> option. >>
>
> I would appreciate any advice or comments on this problem or the error  
> messages.
>
> Thanks.
>
> gac
>


-- 
DrMajorBob at yahoo.com



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