> From: Oleg Kobchenko <[EMAIL PROTECTED]>
>
> > From: John Randall
> >
> > Raul Miller wrote:
> > > In other words, I am trying to find the inverse of a moderately
> > > complex set of functions describing curves.
> >
> > My understanding of the example is this. You have a function f and
> > its graph G={(x,f(x))}. You rotate this through an angle t to get a
> > graph G'. Now (where possible) for each u in R, you look at the
> > vertical line through (u,0) and see where it intersects G'. Choose
> > the maximum of these values, say v. Then v=g(u) gives a new function.
> >
> > If f is a polynomial, then solving the intersection problem for a
> > given u also involves polynomials. In general this will be hard.
> >
> > Are you looking for the implicit function theorem? This gives a local
> > inverse on the graph of an implicit function whenever the tangent line
> > is not horizontal.
>
> In parametric form, some function f:R->R, parameter t, angle a.
>
> G(t) = (x(t),y(t))
>
> x(t) = t
> y(t) = f(t)
>
> R[a] is rotation by a
>
> G' = R[a] mp G
>
> G'(t) = (x'(t),y'(t))
>
> x'(t) = cos(a) * t - sin(a)*f(t)
> y'(t) = sin(a) * t + cos(a)*f(t)
>
> Now we need a way to find explicit f'(x).
> Which as was discussed earlier, piece-wisely defined
> using a max (or min). So that
>
> G'max = {x'(t),max{y'(t)}} = {x,f'(x)}, x in range of x'(t)
>
> Something along these lines?
Isn't then
f'(x) = y'(x'^:_1(x))
Does it mean finding inverse x'(t), ie t = t'(x) = x'^:_1(x)?
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