Evaluate $\int_0^1(\sum_{n=0}^\infty \frac{[2^n x]}{3^n})^2$ where [.] represent the greatest integer fn.

Question

I tried to squeeze it by applying the inequality of GIF fn but ended up with a range $(\frac 34,3)$. Then I tried to squeeze it with the following inequality: $[2^nx] \ge 2^nx$ and, as $x\in[0,1]$, $2^nx \ge 3^n$. This gives $\frac{[2^nx]}{3^n} \ge 1$, and then integrating the result still ended up in the range of $(\frac 34,1)$.

The answer given is $\frac{27}{32}$.

Any method used would be apprciated.

Answer 1

Let $d_j(x)$ be the $j$th binary digit of $x$, so that the binary expansion of $x$ is $0.d_1(x)d_2(x)d_3(x)\dots$. Note that $$ [2^nx] = [d_1(x)\dots d_n(x).d_{n+1}(x)\dots] = d_1(x)\dots d_n(x) = \sum_{k=1}^n 2^{n-k}d_k(x). $$ (Here $d_1(x)\dots d_n(x)$ represents an $n$-bit number written in binary.) Then \begin{align*} \sum_{n=0}^\infty \frac{[2^nx]}{3^n} = \sum_{n=0}^\infty \frac1{3^n} \sum_{k=1}^n 2^{n-k}d_k(x) &= \sum_{k=1}^\infty 2^{-k}d_k(x) \sum_{n=k}^\infty \biggl( \frac23 \biggr)^n \\ &= \sum_{k=1}^\infty 2^{-k}d_k(x) \frac{(2/3)^k}{1-2/3} = \sum_{k=1}^\infty \frac{d_k(x)}{3^{k-1}}. \end{align*} Therefore \begin{align*} \int_0^1 \biggl( \sum_{n=1}^\infty \frac{[2^nx]}{3^n} \biggr)^2 \,dx &= \int_0^1 \biggl( \sum_{k=1}^\infty \frac{d_k(x)}{3^{k-1}} \biggr)^2 \,dx = \int_0^1 \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{d_j(x)d_k(x)}{3^{j+k-2}} \,dx. \end{align*} Note that $$ \int_0^1 d_j(x)d_j(x)\,dx = \frac12 \quad\text{and}\quad \int_0^1 d_j(x)d_k(x)\,dx = \frac14 \text{ for } j\ne k, $$ since "half the real numbers in $[0,1]$ have $1$ as their $j$th bit" and "a quarter of those real numbers have $1$ as their $j$th and $k$th bits". We conclude that \begin{align*} \int_0^1 \sum_{j=1}^\infty \sum_{k=1}^\infty \frac{d_j(x)d_k(x)}{3^{j+k-2}} \,dx &= \sum_{j=1}^\infty \sum_{k=1}^\infty \frac1{3^{j+k-2}} \int_0^1 d_j(x)d_k(x) \,dx \\ &= \sum_{j=1}^\infty \frac1{3^{j+j-2}} \frac12 + \sum_{j=1}^\infty \sum_{\substack{k=1 \\ k\ne j}}^\infty \frac1{3^{j+k-2}} \frac14 \\ &= \sum_{j=1}^\infty \frac1{3^{j+j-2}} \frac14 + \sum_{j=1}^\infty \sum_{k=1}^\infty \frac1{3^{j+k-2}} \frac14 \\ &= \frac94 \sum_{j=1}^\infty \frac1{9^j} + \frac94 \biggl( \sum_{j=1}^\infty \frac1{3^j} \biggr)^2 \\ &= \frac94 \frac{1/9}{1-1/9} + \frac94 \biggl( \frac{1/3}{1-1/3} \biggr)^2 = \frac{27}{32}. \end{align*}