Re: FindRoot repeatedly evaluating function

• To: mathgroup at smc.vnet.net
• Subject: [mg121084] Re: FindRoot repeatedly evaluating function
• From: Peter Pein <petsie at dordos.net>
• Date: Fri, 26 Aug 2011 05:25:43 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <j35afa\$oe3\$1@smc.vnet.net>

```Am 25.08.2011 13:08, schrieb Simon Pearce:
> Hi Mathgroup,
>
> When I use FindRoot[f[y],y] I find that the inner function f is evaluated 3 or 4 times at each value of y (or at least very similar values), even if y is far from the root. This has obvious implications to the speed of my code.
> Can anyone explain why this is the case, and tell me any way to stop it from repeatedly evaluating f? If I use f[a]:=f[a]=... then it uses the stored result, but I don't want to store thousands of such real valued expressions.
>
> The following simple code shows the essence of the problem, using Print to show where the function is evaluated and its value there.
>
> f[a_?NumericQ]:=Module[{sol},
>    sol=NDSolve[{x''[S]-x'[S]+x[S]==0,x[0]==1,x'[0]==a},x,{S,0,10}][[1]];
>    Print[{a,x[10]/.sol}]; x[10]/.sol ]
> FindRoot[f[y],{y,6}]
>
> Thanks,
> Simon Pearce
> Postdoctoral Researcher
> The Centre for Plant Integrative Biology
> School of Biosciences
> University of Nottingham
>
>
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Hi!

Sorry, I do not know why FindRoot behaves like this but the following is
fast enough IMHO:

In[1]:= a0 =a /. First[Solve[ x[10] == 0 /.
First[DSolve[
{ x''[S] - x'[S] + x[S] == 0, x[0] == 1, x'[0] == a},
x, S]],
a]] // FullSimplify

N[a0,20]

Out[1]= 1/2 (1 - Sqrt[3] Cot[5 Sqrt[3]])
Out[2]= 1.4029569306878377766

Regards,

Peter

```

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