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Timeline for Scale invariance distribution

Current License: CC BY-SA 4.0

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Dec 6, 2020 at 17:57 comment added Debbie I figured it out!
Dec 6, 2020 at 17:15 comment added Iosif Pinelis @user758469 : (i) For a real $x$, the standard notation $x\mod 1$ denotes the fractional part of $x$. (ii) The sequence $(\log_{10}(2^n)\mod1)=(n\log_{10}2\mod1)$ is equidistributed (not uniform!) on $[0,1)$ by the equidistribution theorem en.wikipedia.org/wiki/Equidistribution_theorem, because $\log_{10}2$ is an irrational number. (iii) Use $\{\cdot\}$ for sets and $(\cdot)$ for sequences. (iv) Dealing with the sequences in question, you cannot use probabilities, because such probabilities are undefined -- you need to use number-theoretic densities instead.
Dec 6, 2020 at 15:44 comment added Debbie $P(d) = Prob(d\le x < d+1) = Prob(\log_{10}(d) \le y < \log_{10}(d+1)) = \int_{\log_{10}(d)}^{\log_{10}(d+1)}1\,dy = \log_{10}(\frac{d+1}{d})$. Hence $P(d) = \log_{10}(\frac{d+1}{d})$
Dec 6, 2020 at 15:44 comment added Debbie Also is this related to my following observation? (if so, how?): Consider a sequence of powers of 2, and write each element in this sequence in scientific notation $x_m \times 10^{n_m}$. Then create a second sequence of first significant digits of $x_m$ (denote as $d_m$). Define $y_m = \log_{10}(x_m)$ and consider it in the interval $x \in [0,1)$. Is this $y_m$ same as the sequence of $\{\log_{10}(2^n) mod1\}$? If I assume its uniform, then the probability distribution follows simply:
Dec 6, 2020 at 15:44 comment added Debbie "Because 2 is not a rational power of 10, the sequence $\{y_{n}\} =\{\log_{10}(2^n) mod1\}$ is uniformly distributed on [0,1)" First of all what does $\log_{10}(2^n) mod1$ mean? Isn't anything mod 1 equal to 0? Second why is this uniform?
Dec 6, 2020 at 3:38 history edited Iosif Pinelis CC BY-SA 4.0
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Dec 6, 2020 at 3:31 history answered Iosif Pinelis CC BY-SA 4.0