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
> >
> >
> >
>
> 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

```

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