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Re: Using FindRoot with free parameters

  • To: mathgroup at smc.vnet.net
  • Subject: [mg132572] Re: Using FindRoot with free parameters
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Sun, 13 Apr 2014 05:25:57 -0400 (EDT)
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  • References: <20140412091539.8767469FC@smc.vnet.net>

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,
> StevieP
>
>


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