Re: Re: y=f(t) vs t=f(y)
- To: mathgroup at smc.vnet.net
- Subject: [mg6327] Re: [mg6305] Re: [mg6267] y=f(t) vs t=f(y)
- From: Lou Talman <me at talmanl.mscd.edu>
- Date: Sat, 8 Mar 1997 23:26:50 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Wouter L. J. MEEUSSEN wrote: > > At 09:49 6-03-97 -0500, Larry Smith wrote: > > I would appreciate anyone helping me with using Mathematica to solve > > the following (geometrically, numerically, etc) > > > > I need to find an example of a function y=f(t) such that f'(0)=1 but t > > is not a function of y in any neighborhood of 0. I just arbitrarily > > picked f'(0)=1 you could pick something with value of 1. But the trick > > is that t is not a function of y in this neighborhood. Any > > suggestions? > > > > Larry > > larry.smith at clorox.com > > or > > lsmith at tcusa.net > > > > 601-939-8555 ext 255 > > > > > > > hm, what about: > > t[x_]:=Which[x<=0,-1+x,x>0,1+x] > > make a plot, Plot[t[x],{x,-2,2}], and you see that > the inverse function is: > > x[t_]:=Which[t<-1,(t+1),t<1,0,t>=1,(t-1)] > and > Plot[x[t],{t,-3,3}] > > and there you have it : the flat piece for x[t] between t=-1 and t=1 > causes the function x[t] to be independent on t in that area. > > Look at it again, and enjoy... > (whoever gave u this problem deserves a prize for didactics, it's a gem) > > wouter > > Dr. Wouter L. J. MEEUSSEN > eu000949 at pophost.eunet.be > w.meeussen.vdmcc at vandemoortele.be > Unfortunately, Meeussen's solution misses the point, which is that f'[0] is to be 1. This can't happen if f' is continuous at zero, for in that case the Inverse Function Theorem would guarantee the local invertibility of f--which we are to avoid. The secret is to force the derivative to have a discontinuity at the origin. Try f[t] = t + A t^2 Sin[1/t], if t != 0 f[0] = 0, where A is any convenient constant. Larger values for A make for more easily understood pictures. It is rather easy to see that f'[0] = 1 by examining the appropriate difference quotient. It's also easy to see that if A > 1, then f'[t] changes sign infinitely many times in any interval that contains t = 0. From the latter observation it follows that f cannot be monotonic on any such interval, hence cannot be invertible on any such interval. --Lou Talman