Re: Using FindRoot with free parameters

• To: mathgroup at smc.vnet.net
• Subject: [mg132590] Re: Using FindRoot with free parameters
• From: steviep <ssplotkin at gmail.com>
• Date: Mon, 14 Apr 2014 23:00:59 -0400 (EDT)
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```>
> f[n_, x_] = Exp[-3 x^2 + 2] BesselI[n, 4 x];
>
>
> xf[alpha_?NumericQ, n_?NumericQ] :=
>  Module[{x}, x /.
>    FindRoot[f[n, x] - alpha, {x, 2}][[1]]]
>
>
> xf[3, 0]
>
>
> 1.11426
>
>
> xf[alpha, n] /. {alpha -> 3, n -> 0}
>
>
> 1.11426
>
>
>
> Bob Hanlon
>
>
>
>
> On Sat, Apr 12, 2014 at 5:15 AM, steviep2
> <ssplotkin at gmail.com> wrote:
>
> > Hi,
> > I want to define a function of 2 parameters that uses FindRoot. I.e. I
> > have a known but complicated function f[n_,x_] = "complicated function of
> > (n,x)". I want to find the value of x where f[n,x] == alpha, and I want to
> > call this a function xf[alpha_,n_].
> >
> > So my attempts (this is non-working code) looks something like this:
> >
> > xf[alpha_, n_] =  Function[{x}, x /. FindRoot[f[n,x] - alpha, {x, 2}]]
> >
> > or
> > xf[alpha_, n_] =  Function[x /. FindRoot[f[n, x] - alpha, {x, 2}]][alpha,
> > n]
> >
> > It seems this is a pretty simple question-- basically using FindRoot but
> > holding off on substituting in the parameters until later. Is there a
> > simple solution?

Thanks,
I had actually come up this solution in the meantime:

xf[alpha_, n_] = Function[{alpha, n}, x /. FindRoot[f[n, x] - alpha, {x, 2}]][alpha,n]

which gives the same answer. It also gives a bunch of warnings, as does Module when "=" is used instead of ":=".

Is one method preferred over another? Thanks in any event.
-S

```

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