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I have two different functions $y_1=f_1(x)$ and $y_2=f_2(x)$, both invertible but quite complex. I am able to find their inverse functions numerically, i.e. $f^{-1}_1(x)$ and $f^{-1}_2(x)$, by solving for $x_1$ and $x_2$ two equations $$ f_1(y_1) - x_1 = 0, \\ f_2(y_2) - x_2 = 0, $$ for specific realizations of $y_1$ and $y_2$. Since $f_1 \neq f_2$, I get two different solutions for both equations $x_1 \neq x_2$. The problem is that I would like to solve equation on those inverse functions. The goal is to find a single point $x_0$ for $$ f^{-1}_1(x_0) - f^{-1}_2(x_0) =0 $$ Is this possible? If I plug in the above inverse equations, assuming that $x_1=x_2$, into the main equation I get $f_1(y_1)=f_2(y_2)$ which is a contradiction. Is there a method to solve this problem, even numerically? If not possible, maybe finding $\min |f^{-1}_1(x) - f^{-1}_2(x)|$ would be an option?

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If I well understand (which is not sure), you have two given functions $f_1(x)$ and $f_2(x)$

The inverse function of $y=f_1(x)$ is $x=f_1^{-1}(y)$

The inverse function of $y=f_2(x)$ is $x=f_2^{-1}(y)$

You want to find a root $y_0$ of the equation $f_1^{-1}(y)=f_2^{-1}(y)$

So, you first have to $$\text{solve}\quad f_1(x)=f_2(x)\quad \text{for } x\quad \text{which leads to}\quad x_0$$ Then, knowning $x_0$, compute : $$y_0=f_1(x_0)\quad \text{or}\quad =f_2(x_0)$$

$y_0$ is a root of $f_1^{-1}(y)=f_2^{-1}(y)$

Note that, in order to avoid confusion, I changed the symbol $x_0$ in the wording of the question to the symbol $y_0$ but anyways this doesn't matter : any symbol used doesn't change the solution.

For intuitive understanding only, draw $y=f_1(x)$ and $y=f_2(x)$ on a graph. On the same graph draw the symetric curves relatively to the line $y=x$. Those two last curves represent the inverse functions. The intersection represents the root of the inverse equation. This point is the symetrical point representing the root of the direct equation. That is why we solve first the direct equation $f_1(x)=f_2(x)$ : The intersection point is $(x_0\:,\:y_0)$ while it is $(y_0\:,\:x_0)$ on the inverse curves.

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This is not always possible.

Take $f_1(x)=x$ and $f_2(x)=x+1$.

Then $f_1^{-1}(y)=y$ and $f_2^{-1}(y)=y-1$.

Therefore for any $y_0$:

$$f_1^{-1}(y_0)-f_2^{-1}(y_0)=y_0-y_0+1=1\neq 0.$$

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  • $\begingroup$ Do you know how to derive conditions which have to be satisfied for the above to work? $\endgroup$ – wlq Feb 9 '16 at 17:21

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