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Re: Inverse of Interpolating Function?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg120404] Re: Inverse of Interpolating Function?
*From*: Oliver Ruebenkoenig <ruebenko at wolfram.com>
*Date*: Thu, 21 Jul 2011 07:23:55 -0400 (EDT)
*References*: <201107210945.FAA06489@smc.vnet.net>
On Thu, 21 Jul 2011, Gary Wardall wrote:
> On Jul 20, 5:34=C2 am, gonzorascal <dj... at duke.edu> wrote:
>> Hello All,
>>
>> I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation. I am trying to find the period and pulse width of the oscillation. To do this I would like to have an inverse of my function y[t] (so I would have t[y]). I realize this function would be multivalued (that's what I want to find the period). =C2 I am not having success using Mathematica's InverseFunction[] or Reduce[] commands. Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)? Thank you.
>>
>> -GR
>
> gonzorascal,
>
> First make sure the function you wish to have an inverse function is a
> function that has an inverse function. A quick visual test is the
> Horizontal Test for Inverse Functions. If any horizontal line
> intersects the graph twice or more then the inverse is NOT an inverse
> function. Graph the function you wish to create an inverse function
> for. If it fails the horizontal test you may have ti restrict your
> function like is done in Trigonometry. If it passes try something like
> in my example.
>
>
> points = {{0, -1}, {1, 1}, {2, 3}, {3, 4}, {4, 6}, {5, 10}};=E2=80=A8ifun=
> Interpolation[points]
> Does the function pass the Horizontal Test?
>
> Plot[ifun[x], {x, 0, 5}]
>
> It does. Define:
>
> invifun[x_] := FindRoot[ifun[y] == x, {y, 0, 5}][[1]][[2]]
>
> Note that:
>
> ifun[invifun[4]]
>
> is 4.
>
> and
>
> invifun[ifun[3]]
>
> is 3.
>
> and look at the sketch.
>
> Plot[{invifun[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
> AspectRatio -> Automatic]
>
> We have an inverse function.
>
> I hope this will help.
>
> Good Luck
>
> Gary Wardall
>
>
You could also use NDSolve for that - that advantage would be that it will
generate an interpolation function
init = x /. FindRoot[ifun[x] == 0, {x, 0, 1}]
inf = t /.
First[NDSolve[{t'[a] == 1/(ifun'[t[a]]), t[0] == init},
t, {a, 0, 5}]]
Plot[{inf[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
AspectRatio -> Automatic]
Oliver
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