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Consider two curves $y=f(x)$ and $y=h(x)$, both are one-to-one (invertible). Define $g(x)= f^{-1}(h(x))$ (inverse of $h(x)$). Assuming the scheme $X_{n+1}= g(X_n)$ converges to a fixed point $x^*$, show that at $x^*$ the curves $f$ and $h$ intersect.

I have tried graphing various functions to see how I can prove this but to get started I don't really even get what to do here... I think this is fixed point iteration but I am not good a proving proofs so I need some help getting started.

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At the fixed point $x^*=g(x^*)$ but this just means $x^*=f^{-1}(h(x^*))$ that is also $f(x^*)=h(x^*)$, that is the curves intersect at $x^*$. –  Jon Jan 12 '12 at 21:34

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I will post a full answer after having seen that comments are better as a suggestion. The idea is to use the iteration technique to find the intersection between two curves $y=f(x)$ and $y=h(x)$. This is achieved by defining

$$g(x)=f^{-1}(h(x))$$

assuming an inverse exists for at least one of these curves. Now, we iterate the equation

$$x_{n+1}=g(x_n)$$

until a fixed point $x^*$ is reached, provided the criteria for such a point to exist are fulfilled. At the fixed point one has by definition

$$x^*=g(x^*).$$

Now we turn back to our definition of $g(x)$ and we have

$$x^*=f^{-1}(h(x^*)).$$

Inverting $f$

$$f(x^*)=g(x^*)$$

and so $x^*$ is the intersection point for the two curves.

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