I get $f(f(a)) = a^2 + 1 = f(f(-a))$, and so $f(a)^2 + 1 = f(a^2 + 1) = f(-a)^2 + 1$, so $f(a) = f(-a)$ or $f(a) = -f(-a)$, but then I donot know what to do next. Thanks for any help.
$\begingroup$
$\endgroup$
7
-
1$\begingroup$ If one could read Chinese or use a translator, there's this same discussion on Zhihu: zhihu.com/question/340104755 $\endgroup$user632577– user6325772019-08-14 06:09:17 +00:00Commented Aug 14, 2019 at 6:09
-
$\begingroup$ Above link shows how to construct a solution of the given functional equation. In fact there is a construction in English also. But I don't see uniqueness there. Is it clear from that post that $f(1)$ is uniquely determined? $\endgroup$Kavi Rama Murthy– Kavi Rama Murthy2019-08-14 06:23:07 +00:00Commented Aug 14, 2019 at 6:23
-
1$\begingroup$ @KaviRamaMurthy Ah no, the second highest voted answer in the same thread showed that if we drop analyticity of $f$ then it's anywhere in $(1,5)\setminus\{2\}$ $\endgroup$user632577– user6325772019-08-14 06:26:12 +00:00Commented Aug 14, 2019 at 6:26
-
$\begingroup$ OP should provide some context. If $f(1)$ cannot be determined uniquely by the given equation I don't think this question is appropriate. $\endgroup$Kavi Rama Murthy– Kavi Rama Murthy2019-08-14 06:38:31 +00:00Commented Aug 14, 2019 at 6:38
-
$\begingroup$ See jstor.org/stable/2321556?seq=1#page_scan_tab_contents. $\endgroup$Feng– Feng2019-08-14 08:15:00 +00:00Commented Aug 14, 2019 at 8:15
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
A piece of information is still required
$f(f(x)) = 1+x^2$
Let's say $y = f(x)$ is the function, then $y = g(x)$ is the inverse function
$f(x) = g(1+x^2)$, $f(1) = g(2)$
$f(f(x)) = 1+x^2 = f(f(-x))$ $f(f(x)) = f(f(-x))$ Therefore $f(x) = f(-x)$ $f(x)$ is a even function
$f(f(x)) = 1+x^2$, say $x → g(x)$
$f(x) = 1+g(x)^2$, $f(1) = 1+g(1)^2$
$1+g(x)^2 = g(1+x^2)$
Say $x → f(x)$ $1+x^2 = g(1+f(x)^2)$, $f(1+x^2) = 1+f(x)^2$
$f(x) = $
$1+f(x)^2 = f(1+x^2)$
I can't find $f(1)$ unless I knew $f(n)$
$f(1+x^2) = 1+f(x)^2$, Remember $f(x) = 1+g(x)^2$
$f(1+x^2) = 1+(g(x)^2+1)^2$, $f(1+x^2) = g(x)^4+2*g(x)^2+2$
Recall $f(x) = g(1+x^2)$, say $1+x^2→x$ $g(x) = f(\sqrt(x-1))$
$f(1+x^2) = f(\sqrt(x-1))^4+2*f(\sqrt(x-1))^2+2$
