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.
>
>
- References:
- Inverse Interpolation
- From: WetBlanket <wyvern864@gmail.com>
- Inverse Interpolation