Re: InverseFunction: how to manage?

• To: mathgroup at smc.vnet.net
• Subject: [mg124303] Re: InverseFunction: how to manage?
• From: Peter Breitfeld <phbrf at t-online.de>
• Date: Mon, 16 Jan 2012 17:10:52 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <jesv8m\$cp0\$1@smc.vnet.net>

```I think, you have to use InverseFunction[t] not InverseFunction[f[f]]
eg. like this:

In[33]:= t[f_] :=
2 (EllipticF[f/2 - \[Pi]/4, 2] + EllipticF[\[Pi]/4, 2])

In[34]:= angle[time_] = InverseFunction[t][time]

During evaluation of In[34]:=
InverseFunction::ifun: Inverse functions are being used.
Values may be lost for multivalued inverses. >>

Out[34]= 1/2 (\[Pi] -
4 JacobiAmplitude[1/2 (-time + 2 EllipticF[\[Pi]/4, 2]), 2])

In[35]:= angle[1]

Out[35]= 1/2 (\[Pi] -
4 JacobiAmplitude[1/2 (-1 + 2 EllipticF[\[Pi]/4, 2]), 2])

In[39]:= angle[.25] // Chop

Out[39]= 0.0156249

"Dr. Wolfgang Hintze" wrote:

> I apologize for asking this very elementary question but how do I
> manage InverseFunction?
>
> Here is an example
>
> When I solve the equation of motion for a pendulum
>
>     f''[t] == Cos[f[t]], f[0]== 0, f'[0] == 0
>
> I get (with paper and pencil) the time t as a function of the angle f
> thus
>
>     t[f_] = Integrate[1/Sqrt[Sin[u]], {u, 0, f}]
>     Out[22]=
>     2*(EllipticF[f/2 - Pi/4, 2] + EllipticF[Pi/4, 2])
>
> Now I want the the angle as a function of time (f[t]) like this
>
>     "f[t_] = InverseFunction[t[f]]"
>
> But this does not work. I also tried to define t as a pure function
>
> t = 2*(EllipticF[#1/2 - Pi/4, 2] + EllipticF[Pi/4, 2]) &
>
> but again, I have not seen a way to invert this, and for instance carry
> out Plot[f,{t,0,2 Pi}].
>
> Thanks in advance for any hints.
>
> Best regards,
> Wolfgang
>
>
>
>

--
_________________________________________________________________
Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de

```

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