1
$\begingroup$

In mathematics, an inverse function is a function that "reverses" another function: if the function $f$ applied to an input $x$ gives a result of $y$, then applying its inverse function $g$ to $y$ gives the result $x$, and vice versa. i.e., $f(x) = y$ if and only if $g(y) = x$ , i'm interesting to check the relationship between multiplicative inverse function and the inverse compositional function ,Then my question here is related to the solution of the below functional equation

Question:
When does $ \displaystyle f^{-1}=\frac{1}{f}$ with $f$ a function mapping $\mathbb{R}^{*}$ to $\mathbb{R}$ and $f(x)\neq 0$ ?

Note: $f^{-1}$ is the compostional inverse of $f$ and $\displaystyle\frac{1}{f}$ is the multiplicative inverse of $f$

Edit: I have edited the question to define $f$ according to the answer below by Robert Israel.

$\endgroup$
4

3 Answers 3

10
$\begingroup$

You can't have $f^{-1}(x) = 1/f(x)$ with both sides defined for all $x \in \mathbb R$. Namely, if $t = f^{-1}(0)$ then $f^{-1}(t) = 1/f(t) = 1/0$ is undefined.

EDIT: With the correction that $f$ maps $\mathbb R^* = \mathbb R \backslash \{0\}$ to itself, here is one class of solutions. Take any $f_0$ that maps $(0,1]$ one-to-one onto $(-\infty,-1]$ with $f_0(1) = -1$. Then let $$ f(x) = \cases{f_0(x) & if $x \in (0,1]$\cr 1/f_0(1/x) & if $x \in (1,\infty)$\cr f_0^{-1}(1/x) & if $x \in (-1,0)$\cr 1/f_0^{-1}(x) & if $x \in (-\infty,-1]$\cr} $$

$\endgroup$
1
  • 1
    $\begingroup$ For example, $f_0(x)=-1+\tan\left(\frac{(x-1)\pi}2\right)$. $\endgroup$ Jan 26, 2017 at 2:46
7
$\begingroup$

You may like to look into the following article.

Robert Anschuetz II and H. Sherwood, When Is a Function's Inverse Equal to Its Reciprocal? The College Mathematics Journal Vol. 27, No. 5 (Nov., 1996), pp. 388-393.

$\endgroup$
4
  • 6
    $\begingroup$ @zeraouliarafik If only there were some sort of scientific hub where many behind-a-paywall papers were available. (It has this paper!) $\endgroup$
    – dvitek
    Jan 26, 2017 at 0:53
  • $\begingroup$ Thanks for this orientation , really i have checked bookzz.org but i don't find it available $\endgroup$ Jan 26, 2017 at 0:55
  • $\begingroup$ This is freely and legitimately available on Jstor. $\endgroup$
    – user44143
    Jan 26, 2017 at 4:53
  • 1
    $\begingroup$ @MattF. only if you sign up, which means it's not really freely available. But I get your point. $\endgroup$
    – David Roberts
    Jan 26, 2017 at 5:36
6
$\begingroup$

You can download a copy of my University of Central Florida thesis that covers this topic in even greater detail than the Journal article below:

http://stars.library.ucf.edu/rtd/3139/

Robert Anschuetz

$\endgroup$
1
  • $\begingroup$ Thanks , your paper is already montioned and cited above by the answer of T. Amdeberhan $\endgroup$ Mar 3, 2017 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.