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MathGroup Archive 2007

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Re: Newbie question on FindRoot and NIntergrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80461] Re: Newbie question on FindRoot and NIntergrate
  • From: Chris Chiasson <chris.chiasson at gmail.com>
  • Date: Thu, 23 Aug 2007 01:02:23 -0400 (EDT)
  • References: <fagu3e$937$1@smc.vnet.net>

On Aug 22, 4:02 am, "Hoa Bui" <hoabu... at gmail.com> wrote:
> Dear all,
>
> Please help me with this problem:
>
> I have a series of points:
> In[9]:=points
> Out[9]={{0.0001,0.359381},{0.0002866,0.403984},{0.000821394,0.454122},{0.00235411,0.510482},{0.00674688,0.573838},{0.0193365,0.645056}}
>
> Define one of my function as the trapezoidal area created by these
> points up to some x:
> linNx = Interpolation[points, InterpolationOrder -> 1];
> ff[x_] := Integrate[linNx[s], {s, 0.0001, x}]
>
> Define the second function as a power law:
> gg[x_] := 0.9 x^(1/0.9)
>
> My third function is the root of the equation:
> hh[x_] := FindRoot[ff[y] == gg[x], {y, 0.0001, 0.019}];
>
> I can evaluate h at specific values of x, e.g. h[0.001] ({y ->
> 0.00107654}), h[0.01] ({y -> 0.0101302}), etc...
>
> Now say I want to use hh[x] in an integral,
> NIntergrate[hh[x],{x, 0.0001, 0.019}]
> obviously it doesn't work, and I have tried using Solve instead of
> FindRoot but it also did not output a numerical value for the integral
> because the inverse function is not in closed form or something..
>
> Is there a way for me to compute the integral of hh[x] ?
>
> Thank you all so much,
> Hoa Bui

I forgot to mention that the technique I used in sol2 is called
homotopy continuation. (Dr.) Carl Woll showed me how to do it (a while
back).

--
http://chris.chiasson.name/



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