# Tag Info

### Prove that $f(\mathbb R) = \mathbb R$. Partially solved.

[Note: There have been some important technical revisions since the first version of this answer, which was flawed.] I realize this is quite old and everyone has likely moved on, but since it got ...
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### Prove that $g(z)$ is identically zero

Denote by $D(R)$ the disk centered in zero with radius $R$. Let $M(R)$ be the maximum modulus of $g$ on the closed disk $D(R)$. Let $K$ be $M(10)$. Then for all $w\in D(81)$ we find a $z$ in $D(9)$ ...
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### Does there exist an $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x^3 + 1$?

This spells out an argument proposed by user tomasz. Each orbit of the action of $g(x)=x^3+1$ on $\mathbb R$ is a copy of $\mathbb N$, with $g$ acting on each orbit by $n\mapsto n+1$. Let $X$ be the ...
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### find all the polynomial f that satisfies $f(x^2)+f(x)f(x+1)=0$.

Zero-set You have correctly identified two transformations which send roots of $f$ to roots of $f$. Explicitly, setting $u(z)=z^2$ and $v(z)=(z-1)^2$, we have $u(S)\subset S$ and $v(S)\subset S$. Let'...
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### functional equation, continuous functions.

Disclaimer: im 99% sure there's an cleaner solution, but i have been at this for hours now, im just happy to be done. If someone knows an better way PLEASE do let me know Here we go! Claim 1: $f$ is ...
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1 vote

With the domain $(1,\infty)$ there are infinitely many solutions, all of which can be found as follows: take any function $g:[2,4) \to \Bbb R$ and define $f$ recursively as follows: $f(x)= \begin{... • 17.2k 1 vote ### Binary with digit 2 allowed Unless I made a mistake, my brute force implementation seems to indicate the answer to your first question above is$51$. Here's a list of solutions where$x=2\$ that perhaps provides a bit more ...
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