6
$\begingroup$

Find all functions $f$ such that: $$f(f(n))=f(f(n+2)+2)=n, f(0)=1$$

I used the following logic: if $f(k)=f(l)$ then $f(f(k))=f(f(l))$ so $k=l$.

And, for all numbers $m$ there exists a number $p=f(m)$ such that $f(p)=m$.

So, each number has an inverse through $f$ and this inverse must be unique.

Let $g$ be a function mapping a number to its inverse through $f$, i.e. $g(m)=p$ is the unique number such that $f(p)=m$.

Then we can apply $g(f(x))=x$.

Taking $g$ on both sides of the original equation, we get: $f(n)=f(n+2)+2$. By induction backwards and forwards, we see that $f(2k)=-2k+1$ for all $k$.

We also see $f(0)=1\implies f(f(0))=f(1)\implies f(1)=0$ so $f(2k+1)=-2k$ for all $k$ by induction.

So, $f(x)=1-x$.

Does this logic work? It seems that defining an inverse is a bit of a complicated way to solve, but it seems to come logically.

$\endgroup$
6
  • 1
    $\begingroup$ The part about the inverse can be skipped by simply noting that $f$ is injective. Then we can already conclude that $f(n) = f(n+2)+2$. $\endgroup$
    – player3236
    Commented Nov 10, 2020 at 11:05
  • 2
    $\begingroup$ minor complaint - in the future, please have at least one (non-math) word in your query's title, so that reviewers can right click to open the page. $\endgroup$ Commented Nov 10, 2020 at 11:07
  • $\begingroup$ @player3236 Ok, so is it in fact the only solution $(1-x)$? $\endgroup$
    – aman
    Commented Nov 10, 2020 at 11:13
  • 2
    $\begingroup$ You have shown $(1-x)$ is the only solution by induction pretty nicely. $\endgroup$
    – player3236
    Commented Nov 10, 2020 at 11:17
  • $\begingroup$ Looks good, see also Solving an Olympiad functional equation $f(f(n))=f(f(n+2)+2)=n$ and Proof verification: system of functional equation, well and many more: approach0.xyz/search/… :) $\endgroup$
    – Sil
    Commented Nov 11, 2020 at 20:41

0

You must log in to answer this question.

Browse other questions tagged .