Find the composition of a multivariate function with itself

[docnet]

Homework Statement

psb

Relevant Equations

psb

Solution attempt:

$$F(F(x,y))=(z,w)$$

is the map given by

$$x=z$$
$$y=w$$

 

[mitochan]

I observe for x=-1,y=0, u=0/0,v=0/0 not defined so this point should be excluded.

 

[fresh_42]

Homework Statement:: psb
Relevant Equations:: psb

View attachment 278552Solution attempt:

$$F(F(x,y))=(z,w)$$

is the map given by

$$x=z$$
$$y=w$$

No. The map is given if you express ##z=z(u,v)=z(u(x,y),v(x,y))=z(x,y)##. You have to substitute the ##u,v## with their definitions in terms of ##x,y##.

 

[mitochan]

You have to substitute the with their definitions in terms of .

I assume after the substitution he got this result
##F^2=E##
##F=F^{-1}##

 

[fresh_42]

I assume after the substitution he got this result
##F^2=E##
##F=F^{-1}##

Maybe, but I guess there is no way but calculation to be sure. It looks a bit like a local Lie group.

 

[mitochan]

I should appreciate it if you would tell me what kind of transformation this ##F = F^{-1}## is. Named? Usage?

 

[fresh_42]

It is called reflection, or involution.

 

[mitochan]

Thanks. And for this special idempotent transformation, is it used in Physics ?

 

[fresh_42]

Thanks. And for this special idempotent transformation, is it used in Physics ?

I corrected it. Idempotent functions are ##F^2=F##. We have theoretically ##F^3=F## and any element with ##F^n=F## qualifies to be idempotent, but Wiki said idempotent functions are those with ##n=2##.

So let's call it reflection or involution. These terms already suggest their usage. Any reflection satisfies ##F^2=1:## reflected twice is the original. This is a geometric property but as such also used in physics whenever geometry plays a role (crystallography, particle physics, mechanics). Other examples for involutions are matrix transposition, complex conjugation or the Legendre transformation.

 

[docnet]

As Fresh_42 was explaining, ##F## is a "Möbius transformation" at its core, an involution about a circle in an offset coordinate system. That's how I got my answer by thinking ##F=F^{-1}##. Although my answer is probably wrong, because ##F## is not defined everywhere on the ##uv## plane to start with. I wonder if the composite map would be different if I computed everything out?

This problem was assigned last semester in differential geometry by the person who invented the Ricci flow and was partially credited for solving the Poincaré conjecture. This was probably too difficult to solve for most undergraduate students in differential geometry.

 

[PeroK]

We can construct this transformation as follows. First, we have something simpler: $$u = \frac{2x}{x^2 + y^2}, \ \ v = \frac{2y}{x^2 + y^2}$$ Leads to the same inverse transformation: $$x = \frac{2u}{u^2 + v^2}, \ \ y =\frac{2v}{u^2 + v^2}$$ Now, if we let ##x' = x +1## and ##u' = u +1##, and we let:
$$u' = \frac{2x'}{x'^2 + y^2}, \ \ v = \frac{2y}{x'^2 + y^2}$$ With the same inverse transformation: $$x' = \frac{2u'}{u'^2 + v^2}, \ \ y =\frac{v}{u'^2 + v^2}$$ Then express that in terms of ##x, u## we have:
$$u +1 = \frac{2(x+1)}{(x+1)^2 + y^2}, \ \ v = \frac{2y}{(x+1)^2 + y^2}$$ With, again, the same inverse transformation. Finally, we re-express ##u##:
$$u = \frac{1 - x^2 - y^2}{(x+1)^2 + y^2}$$ And we see that it's the same transformation as above.

 

[mitochan]

Thanks @PeroK

Although my answer is probably wrong, because is not defined everywhere on the plane to start with. I wonder if the composite map would be different if I computed everything out?

The transformation for (x,y)=(-1,0) ## u(-1,0)=-1, v(-1,0)=0##, otherwise as above mentioned would be defined for all xy, uv plane.