1
$\begingroup$

One set of solutions to this is $f(x) = \frac{c}{x}$ for constant $c$. Are these the only solutions?

$\endgroup$
5
  • 1
    $\begingroup$ What are your thoughts? Did you try anything? $\endgroup$
    – Arnaldo
    Mar 31, 2017 at 12:38
  • 1
    $\begingroup$ Provide some context for your question and tell us about your previous efforts. Is $f(x)$ assumed continuous? $\endgroup$
    – mlc
    Mar 31, 2017 at 12:40
  • 1
    $\begingroup$ First of all, $c \neq 0$. Moreover, observe that $f(x)=1-x$ is also a solution. $\endgroup$ Mar 31, 2017 at 12:41
  • 2
    $\begingroup$ Why are you mute ? This site works with dialog... $\endgroup$
    – Jean Marie
    Mar 31, 2017 at 12:50
  • 1
    $\begingroup$ If $f$ is a solution and $g$ is any function, then $g(x) g(1-x)f(x)$ is another solution. $\endgroup$
    – Joel Cohen
    Mar 31, 2017 at 12:58

3 Answers 3

9
$\begingroup$

Let $f$ be a solution to the equation, and let $g(x):=(x+ \frac{1}{2})f(x + \frac{1}{2})$. Then $g$ is an even function (i.e. $ g(x)=g(-x)$). Conversly, let $h$ be any even function. Then by setting $f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$, we get $xf(x)=(1-x)f(1-x)$. Thus the set of all solutions to this equation is exactly the set of functions of the form: $$f(x):=\frac{h\left(x-\frac{1}{2}\right)}{x}$$ Where $h$ is an even function.


Some examples:

\begin{align*} f(x) &= \frac{c}{x} \\ \\ f(x) &= \frac{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}{x} = 1-x \\ \\ f(x) &=\frac{\cos\left(x-\frac{1}{2}\right)}{x} \\ \end{align*}

$\endgroup$
3
  • $\begingroup$ Thanks very much! That's very helpful! $\endgroup$
    – user431110
    Mar 31, 2017 at 13:44
  • $\begingroup$ @RichardP Please consider approving the answer. $\endgroup$ Mar 31, 2017 at 18:53
  • $\begingroup$ Thanks! I'm afraid I'm a bit new to the site! Your answer is extremely helpful to me. I'd tried the handful of tricks I know around functional equations, but nothing was helping. $\endgroup$
    – user431110
    Apr 1, 2017 at 6:34
3
$\begingroup$

No, since $f(x)=1-x$ is also a solution, as I wrote in the comments.

$\endgroup$
1
$\begingroup$

Among degree one polynomials, you can find solutions. Suppose $f(x)=a+bx$. Then you require

$$\frac{a+bx}{a+b(1-x)}=\frac{1-x}{x}$$ $$(a+bx)x=(1-x)(a+b-bx)$$ $$ax+bx^2=a+b -bx-(a+b)x+bx^2$$ $$0 = (a+b)(1-2x)$$ so that $b=-a$. You therefore have a family of functions $$f_a(x)=a\cdot(1-x)$$ which are solutions.

You can try the same procedure with other types of functions with arbitrary coefficients and see what you might get.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .