How can one find the inverse of $$ f(x) = \mathrm{a} \left(1 + \frac{\mathrm{c}}{(1+x^\mathrm{b})^{-\frac{1}{\mathrm{b}}} - \mathrm{c}}\right) \cdot (1+x^{-\mathrm{b}})^\frac{1}{\mathrm{b}} $$ with $a, b, c$ positive constants?

I have not any experience with inverting such functions, and I cannot find any textbooks that contain such inversion techniques. Any relevant suggestions would be appreciated very much.

The reason I want this inversion is to find $\frac{\mathrm{d}x}{\mathrm{d}a}$ for this equation: $$ a \left(1 + \frac{\mathrm{c}}{(1+x^\mathrm{b})^{-\frac{1}{\mathrm{b}}} - \mathrm{c}}\right) \cdot (1+x^{-\mathrm{b}})^\frac{1}{\mathrm{b}} = 1$$

  • $\begingroup$ have you proved that an inverse function exists? $\endgroup$ – Dr. Sonnhard Graubner May 7 '15 at 19:24
  • $\begingroup$ No, but I have taken it from a paper. In this particular paper, $a$ is a variable. He takes $\frac{\mathrm{d}x}{\mathrm{d}a}$ for the equation: $a \left(1 + \frac{c}{(1+x^b)^{-\frac{1}{b}} - c}\right) \cdot (1+x^{-b})^\frac{1}{b} = 1$ so I suppose he knows somehow that it exists. $\endgroup$ – Konstantinos May 7 '15 at 19:26
  • $\begingroup$ Wow @epimorphic nice trick! Math is beautiful! :) math.stackexchange.com/questions/292590/… I will edit my question so as with copy/paste you can get the accepted answer. $\endgroup$ – Konstantinos May 7 '15 at 23:24
  • $\begingroup$ First, I am completely covered by your comment. Second, the link is this: goo.gl/UrPZZ3 I modified the original equation for simplicity. What I am referring to is actually equation (10), p. 18 of the paper. $a$ refers to the whole expression on the RHS. The LHS is a product of equations found on pages 5-6 (Eq. 1, 2, and τ(x)) . For intuition, the slides at goo.gl/7VJFmC might help. Paper is published now (couldn't find a link). $c$ refers to $\rho$ which must be $\rho \in (0,1)$. $\endgroup$ – Konstantinos May 8 '15 at 12:18
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    $\begingroup$ A simpler expression for $f(x)$ is $\;\displaystyle \frac a{(1+x^{-b})^{-\large{\frac 1b}}-c\,x}$. $\endgroup$ – Raymond Manzoni May 10 '15 at 12:22

Looking at the source paper (Michaillat and Saez, "Aggregate Demand, Idle Time, and Unemployment" (2014)) that you linked, I think the function that you'd want to invert is actually $$ g(x) = \left(1 + \frac{c}{(1+x^b)^{-1/b} - c}\right)^{\epsilon - 1} (1 + x^{-b})^{-1/b} $$ (where I guess you set $\epsilon = 2$ for simplicity) so that $g(x) = a$.

Anyways, I suspect that neither $f$ nor $g$ are invertible using elementary operations because you have $x$ trapped inside multiple levels of not-so-well-related expressions. In Mathematica, the related Solve and InverseFunction functions quickly give up, while Reduce had to be manually stopped because it wasn't returning anything within a reasonable amount of time.

But notice that the authors of the paper didn't actually bother to find a concrete expression for the inverse (redactions and emphasis mine):

Since [$g$ is a strictly increasing function of $x$], equation $(10)$ implicitly defines $x$ as an increasing function of [$a$]. (p. 18)

The idea here is simply that a strictly increasing function always has a strictly increasing inverse (with the domain and the codomain restricted appropriately). We may not have a way to compute the inverse directly, but it exists.

To calculate $\frac{dx}{da}$ at some $a$, find/estimate the $x$ such that $g(x) = a$ and apply the well-known formula $$ [g^{-1}]'(a) = \frac{1}{g'(x)} $$ for the derivative of inverse functions.


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