Show that the digit $d$($0<d<10$) occurs as a first digit in power of 2 with frequency $log_{10} \frac{d+1}{d}$.

Question

Show that the digit $d$($0<d<10$) occurs as a first digit in power of $2 $ with frequency $log_{10} \frac{d+1}{d}$.

I think that this can be proved using Birkhoff Ergodic theorem. Any help?

Answer 1

Although one answer has been posted, I wanted to put some light on the way you have asked the question. The flavour would be the same as Mr alex.jordan.

Now observe if $d$ be the first digit of a number $n$ then $\exists r \in \Bbb N$ s.t $d10^r \leq 2^n < (d+1)10^{r}$.

Taking $\log_{10}$ both side we get $\log_{10}d + r \leq \log_{10}n< \log_{10}(d+1)+r $ Now when we take $(\mod 1)$ we'll get

$d10^r \leq n\log_{10}2 < (d+1)10^{r}$ iff $n\log_{10}2(\mod 1) \in [\log_{10}d, \log_{10}(d+1)]$

Now consider the map $T:[0,1) \to [0,1)$ s.t $x \mapsto x+ \alpha$ where $\alpha=\log_{10}2$ is irrational.

Then you check that the map $T$ is ergodic.

Consider $I=[\log_{10}d, \log_{10}(d+1)] \subseteq [0,1)$ and $f=\chi_I$

Now, Birkhoff ergodic theorem along with the equivalent conditions of ergodicity states that $\lim_{n \to \infty} \frac 1n |\{0 \leq j \leq n-1: T^j(x) \in I\}|=\lim_{n \to \infty} \frac 1n \sum_{j=0}^{n-1}f(T^j(x))=\lim_{n \to \infty} \frac 1n \sum_{j=0}^{n-1}\chi_I(T^j(x))\to \mu(I)$ for almost every $x \in [0,1)$.

Now the notable thing is that here for irrational rotation the fact happens for all $x \in [0,1)$ in particular putting $x=0$ we get

$\lim_{n \to \infty} \frac 1n |\{0 \leq j \leq n-1: T^j(x) \in I\}|=\lim_{n \to \infty} \frac 1n |\{0 \leq j \leq n-1:$ the first digit of $2^i$ is $d\}|=\mu(I)=\log_{10}\frac{d+1}d$

Hence proved.

Answer 2

I don't know that theorem by name, but maybe it is like this argument. Consider the sequence $\left(2^n\right)_{n\in\mathbb{N}}$.

To capture its first digit, consider the sequence $\left(\left\{\log_{10}(2^n)\right\}\right)_{n\in\mathbb{N}}$ where $\{x\}$ is the fractional part of $x$. These numbers determine the leading digit, since the integer part of $\log_{10}(2^n)$ will just mean mutliplication by some whole number power of $10$ when you exponentiate back base $10$.

Now this sequence is the same as $\left(\left\{n\cdot\log_{10}(2)\right\}\right)_{n\in\mathbb{N}}$ so it's really just whole number multiples of some irrational number reduced "mod" 1 to within $[0,1)$. As such, it will be dense in $[0,1)$, uniformly distributed.

So you just need to cut up $[0,1)$ into subintervals where exponentiating base $10$ changes the leading digit. The partition is $$[0\mid\log_{10}(2)\mid\log_{10}(3)\mid\log_{10}(4)\mid\log_{10}(5)\mid\log_{10}(6)\mid\log_{10}(7)\mid\log_{10}(8)\mid\log_{10}(9)\mid1)$$ where the end numbers are of course $\log_{10}(1)$ and $\log_{10}(10)$. The length of each subinterval is in the form $\log_{10}(d+1)-\log_{10}(d)$, giving you your result.