If I may modify my previous post, the situation is probably a little more complex than that.

Instead of saying "Right, this is what you do," I should have said "You are within your rights to do that." Yours may not be the only possible definition of the reflection in a curve. The well known "reflection" in a circle is inversion.

Two points P and Q are inversions of each other in a circle with center O and radius r iff the three points O, P, Q are colinear and OP·OQ = r².

my problem now is how tofind the equation of the new line. i realise this is probably a bit too much to go into easily, but do you know where i can find it out?

thanks again
ian

What I asked was, "What is it you see in your mind's eye when you talk about those reflections?" The reason I asked was because there may be several kinds of reflections - that's all.

>my problem now is how to find
>the equation of the new
>line. i realise this
>is probably a bit too
>much to go into easily,
>but do you know where
>i can find it out?
>

It's pure analytic geometry and/or some Calculus. The most essential part of you query is being able to find the point on a given curve nearest to a given point.

Say, you are given a point (x₀, y₀) whose reflection in y = x² you intend to find. You'll have to minimize the sqaure of the distance:

d(x) = (x - x₀)² + (x² - y₀)²

Equate the derivative to 0 and see if you can solve the equation. If you can, assume x₁ is a solution. Then the reflection point is given by

(x, y) = 2(x₁, y₁) - (x₀, y₀)

consider the set of axis "y" against "f(x)".

the line "y=f(x)" is a straight line through the origin.

the line "y=x" is the inverse of f(x). i.e. "y=(f^-1)(f(x))".

reflect this in the 'straight line' "y=f(x)" and you get
"y=f(f(x))", i think

this means that y=x reflected in y=f(x) gives y=ff(x)!

i realise that there's a problem with,for example f(x)=x^2 but i know that this comes from the squaring. it can be solved by reflecting y=mod(x)
im trying to do the same for y=g(x) reflected in y=f(x), but im not sure. at the moment i think its y=fg(-1)f(x).

sorry to bother you again, but thanx for your help. none of my tutors bother to help me.

thanx!

The function f(x) must be monotone for that to make sense.
>
>the line "y=f(x)" is a straight
>line through the origin.

Yes. It's the first quadrant's diagonal.
>
>the line "y=x" is the inverse
>of f(x). i.e. "y=(f^-1)(f(x))".

Yes.

>reflect this in the 'straight line'
>"y=f(x)" and you get
>"y=f(f(x))", i think

Yes. That's true.

>this means that y=x reflected in
>y=f(x) gives y=ff(x)!

That's right, but why the exclamation mark?

>i realise that there's a problem
>with,for example f(x)=x^2 but i
>know that this comes from
>the squaring. it can
>be solved by reflecting y=mod(x)

That I do not know.

>sorry to bother you again, but
>thanx for your help.
>none of my tutors bother
>to help me.

So, what are they for?