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