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The function $g$ is strictly positive. Let the function $f$ be defined as

$$f(x) = \int_0^x g(u) du$$

Is there a way to express $f^{-1}(x)$ in terms of $g$?

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    $\begingroup$ In general $f$ need not be invertible, i.e., $f^{-1}$ need not exist. Do you mean to suppose, e.g., that $g$ is strictly positive (in which case $f$ is strictly increasing and hence invertible)? $\endgroup$ – Travis Willse Oct 6 '15 at 22:22
  • $\begingroup$ Yes, $g$ can be assumed to be strictly positive. $\endgroup$ – Halbort Oct 6 '15 at 22:23
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If $g$ takes on both negative and positive values, or is zero on some interval, then $f$ is not invertible, as mentioned in comments.

Assume $g$ is strictly positive (or strictly negative), hence $f^{-1}$ exists and is differentiable by inverse function theorem.

Then $f(f^{-1}(x))=x$, so by differentiating, we get that $f'(f^{-1}(x))(f^{-1})'(x) =1$, i.e, $g(f^{-1}(x))(f^{-1})'(x) = 1$.

Thus we see that $f^{-1}$ satisfies the differential equation $$y' = \frac{1}{g(y)}$$

For some functions $g$ this can be solved exactly (for example, $g(y) = e^y$ or $g(y) = y^2+1$), while for others it cannot be solved exactly (for example $g(y) = e^{y^2}$). Hence, you can get this differential formula but no explicit solution in general.

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  • $\begingroup$ To be clear, if $g$ is strictly negative, $f$ is strictly decreasing and so is again invertible. $\endgroup$ – Travis Willse Oct 6 '15 at 22:47
  • $\begingroup$ @Travis: Added, thanks. $\endgroup$ – shalop Oct 6 '15 at 23:13
  • $\begingroup$ This is a concise and complete solution! Well done. +1 $\endgroup$ – Mark Viola Oct 6 '15 at 23:27
  • $\begingroup$ This is exactly what I was looking for. Thank you: +1 $\endgroup$ – Halbort Oct 7 '15 at 0:13

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