Re: Solving a numerical integration
- To: mathgroup at smc.vnet.net
- Subject: [mg5476] Re: [mg5440] Solving a numerical integration
- From: fransm at win.tue.nl (Frans Martens)
- Date: Wed, 11 Dec 1996 03:15:57 -0500
- Sender: owner-wri-mathgroup at wolfram.com
Pere Llosas wrote:
> I would like to solve an equation of this kind
>
> NIntegrate[f[n],{n,0,x}]==0 (f[x_]=Sqrt[1+x+x^2...)
> where x is the searched value, and f cannot be integrated
> analytically.
>
> NSolve[NIntegrate[f[n],{n,0,x}]==0,x], tries to evaluate
> NIntegrate[f[n],{n,0,x}] before assignin a numerical value to x and
> returns an error.
>
> How could this calculation be done without having to write a
> program that searches the root?
The equation NIntegrate[f[n],{n,0,x}]==0 has root x = 0 and the
NSolve function tries to compute the inverse of the function x |->
NIntegrate[f[n],{n,0,x}] .
The function FindRoot is more suitable. Here is an example with a
second root in the neigbourhood of x = 4.8 .
In[25]:=
Clear[int,f]
int[x_]:=NIntegrate[f[t],{t,0,x},
AccuracyGoal -> 6];
f[x_]:=Sqrt[1+x+x^2]-3;
In[28]:=
FindRoot[int[x]==0,{x,4.8},
Jacobian -> f[x]]
Out[28]=
{x -> 4.66231}
>>>>> The whole above without messages <<<<<<<<<
There are two precautions:
1)The option AccuracyGoal in NIntegrate is set to 6 because the
integral int[x] equals zero for the root x.
2)FindRoot uses the method of Newton-Raphson and setting the option
Jacobian prevents the symbolic computation of int'[x] . Note that
int'[x] equals f[x].
You must have a global idea of the roots of the original equation.
Frans Martens
Eindhoven
The Netherlands