       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' 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,

where A is any convenient constant.  Larger values for A make for more easily
understood pictures.

It is rather easy to see that f' = 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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