1

I've tried many different things myself, but I can't find any exact function that seems to follow this path.

First, I tried thinking that $f(1)$ would most likely also be 1, so then $f(1)f(1/2)=1$, which narrows it down to a function where $f(1/2)=1$.

After this, I was thinking it might be parabolic, because if $f(1)=f(1/2)$ then the vertex would be along the line $x=\dfrac{3}{4}$, which leads to an entire infinite set of functions that allow it.

This is where I started feeling confused, because I wasn't sure how to find the coefficients of such a parabolic function, or how to prove it was a solution to all real inputs.

Edit: solution below

if f(1)f(1/2)=1, then both f(1) and f(1/2) must be 1, but f(1/2) must exist so that f(1/2)f(0)=1/2, but f(0) has to equal zero.

Jack Phillips
  • 84
  • 2
    Please edit your question to show us what you've tried and where it failed. – Robert Shore Feb 09 '23 at 21:46
  • 1
    I wouldn't expect this to be something you recognize, though anything is possible. Rather I think you are meant to construct suitable functions. Or at least to try to. Start by noticing what happens if $x=0$. – lulu Feb 09 '23 at 21:50
  • Thanks for the idea @lulu. This makes me think it would be sinusoidal at first, with crests between 1/2 and 1, and a trough at 0. – Jack Phillips Feb 09 '23 at 22:17
  • 1
    This doesn't exactly make sense though, because if $f(1)f(1/2)=1$, then both $f(1/2)$ and $f(1/2)$ must be 1, but $f(1/2)$ must exist so that $f(1/2)f(0)=1/2$, but $f(0)$ has to equal zero. This makes me think that there is no such function that exists, or there is a flaw in my logic. – Jack Phillips Feb 09 '23 at 22:24
  • 1
    There are no functions. You are right. – Umesh Shankar Feb 09 '23 at 22:25
  • 1
    There you go! Note that you don't actually know that $f(0)=0$. You could have $f(-\frac 12)=0$. But that also leads to a contradiction. – lulu Feb 09 '23 at 22:27
  • 1
    Note: the site prefers not to have unanswered questions. If you are so inclined, you could post your solution below. That way, users here could comment on it. – lulu Feb 09 '23 at 22:28
  • But what if f(1) and f(1/2) exist so that they are reciprocals of each other? – Jack Phillips Feb 09 '23 at 22:35
  • Why would $f(1)f(\frac12)=1$ imply that $f(1)=f(\frac12)=1$? There are lots of pairs of real numbers that multiply to $1$. – Greg Martin Feb 09 '23 at 23:42
  • That was my question, and now I'm wondering if my solution isn't valid. – Jack Phillips Feb 09 '23 at 23:44

1 Answers1

5

Evaluating the given identity at $x=\frac12$ and $x=-\frac12$ yields $f(\frac12)f(0) = \frac12$ and $f(-\frac12)f(-1) = -\frac12$; multiplying these together yield $$ \bigl( f(0)f(-\tfrac12) \bigr) f(0)f(-1) = -\tfrac14 \ne 0. $$ However, the given identity at $x=0$ gives $f(0)f(-\frac12) = 0$, which contradicts the above calculation. Therefore there are no such functions.

Greg Martin
  • 78,820