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Is the mapping $$ f: \mathbb{R} \rightarrow [0,1], \ x \mapsto \sum_{n=1}^\infty \frac{\lfloor x^n \rfloor \mod 2}{2^n} $$ surjective?

  • If not, what is its image?

  • If yes, what can be said about images of intervals, besides the obvious $f([-1,1]) = \{0,\frac{2}{3},1\}$?

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    $\begingroup$ I guess yes. Given a sequence of required parities, you should be able to place $x$ "firmly" between two large integers of the correct parity, and then make quickly decreasing adjustments to get the correct parity of $\lfloor x^n\rfloor$ for every $n$. I recall seeing a long time ago the claim that there is an $x$ such that $\lfloor x^n\rfloor$ is prime for every $n$. I think it was proved this way. $\endgroup$ Oct 18, 2013 at 12:20
  • $\begingroup$ How do you get say $1/2$ or $1/4 + 1/32$? $x$ will not be integer, floor() must be odd very few times and even all the rest (or even very few times and odd all the rest). $\endgroup$
    – joro
    Oct 18, 2013 at 12:57

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I agree with Johan - it should be surjective. Here is a proof. I claim that $f([4,6])=[0,1]$. Let $[x,y]\subset [4,6]$ be such that $x^k=m$, and $y^k=m+1$. Then $y^{k+1}-x^{k+1}>x(y^k-x^k)=x\geq 4$. Which means $[x,y]$ contains inside itself two intervals $[x',y']$, and $[y',z']$ such that $(x')^{k+1}=m'$, $(y')^{k+1}=m'+1$, $(z')^{k+1}=m'+2$. Then one proceed recursively.

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  • $\begingroup$ Thank you! -- If I'm not overlooking some subtlety, this proves surjectivity. What remains is the part of the question asking for images of intervals. Here in particular short intervals lying a little above 1 or a little below -1 are of interest, as for long enough intervals of big enough numbers one can just use your method to show that the image is all of $[0,1]$. $\endgroup$
    – Stefan Kohl
    Oct 18, 2013 at 20:57
  • $\begingroup$ At least one can say that the image of any interval going beyond [-1,1] is uncountable by a similar argument. One will still have freedom for every $\ell$-th term of the sum where $\ell$ is sufficiently big. $\endgroup$
    – Victor
    Oct 18, 2013 at 21:44
  • $\begingroup$ Hm... Me on the contrary I started to believe that the image of [1,1+ϵ] is nowhere dense... $\endgroup$
    – Victor
    Oct 18, 2013 at 23:41

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