How can one prove $e^n$ and $\ln(n)$, modulo 1, are dense in $[0,1]$, for $n=2,3,4...$?
Prove $e^n$ and $\ln(n)$, mod 1, for $n=2,3,4...$ is dense in $[0,1]$
2 Answers
They are not uniformly distributed, which would mean e.g.
$$\lim_{n\to\infty} \frac{|\{k
But they are dense in $[0,1]$ and that is the property you are looking for (as reflected by the edit of the question).
For the logarithm: Let $\epsilon>0$ be given. Find $N$ such that $\frac1N<\epsilon$.
Then $0<\ln(n+1)-\ln n<\frac1n<\epsilon$ for all $n>N$ (because the derivative of $\ln$ is the reciprocal). Therefore the numbers $\ln n\bmod1$ with $N For the exponential this is a bit more difficult.
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0I know that this is an old answer, but I don't understand the last paragraph. How does it prove that it is dense in $[0,1]$? – 2014-04-19
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0@math.n00b When computing modulo $1$, we may consier $[0,1]$ wrapped to a circle by identifying $0$ and $1$. On the tour from $N$ to $\approx eN+1$, the logarithm grows by $1$, i.e. we walk around the circle once, and we always walk forward in steps $<\epsilon$. Therefore we cannot "jump" over any interval of length$\epsilon$ without hitting it. - Revisiting this asnwer, however, i have still litttle more to say about the exponential thantahat it is "a bit more difficult", hm. – 2014-04-19
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0@math.n00b. For the log's case, only consider $\{log(2^n)\}$,then you will get the density property. – 2016-06-30
I do not know for $e^n$, but for $\log(n)$ modulo 1, it is not uniformly distributed, according to http://en.wikipedia.org/wiki/Equidistributed_sequence .