Re: Inverse Interpolation

• To: mathgroup at smc.vnet.net
• Subject: [mg120962] Re: Inverse Interpolation
• From: Heike Gramberg <heike.gramberg at gmail.com>
• Date: Fri, 19 Aug 2011 06:33:50 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201108180722.DAA11854@smc.vnet.net>

```I'm not entirely sure what your code is supposed to do but maybe this helps

list = {};
F = Cos[x];

Plot[F, {x, -1, 1}, PlotPoints -> 1000, PlotRange -> All,
AxesLabel -> {"x", "F"}, EvaluationMonitor :> AppendTo[list, {F, x}]]

Heike.

On 18 Aug 2011, at 09:22, WetBlanket wrote:

> Previously, the following code could be used to use the Plot functions
> algorithm to select point at which to evaluate a function to assist in
> obtaining a good numerical interpolation.  In Version 8 this code does
> not seem to work ( at least for me).  Can someone assist me by showing
> how this task is best accomplished in Version 8.  I use the Cos
> function in this example for simplicity.  Clearly, numerical
> interpolation is not needed to obtain an inverse for the Cos.
>
> list={};
> F = Cos[x];
>
> Plot[ (  ss=F;  AppendTo[list, {ss,x}]; ss), {x,-1,1}, PlotPoints-
>> 1000,
> PlotRange->All, AxesLabel->{"x","F"}]
>
> Thanks for the help.
>
>

```

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