Question from MIT integration Bee 2023 final: Evaluate $\int^1_0 (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n})^2{\rm d}x$

Question

I am trying to evaluate the last question from MIT integration Bee 2023 Final. $$\int^1_0 \left (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n} \right )^2{\rm d}x$$ My approach is to divide $(0,1)$ into $1/2^n$ intervals and write the general term of the $y$-value. E.g. For $x \in (k/2^n, (k+1)/2^n)$, $$f(x)=\left (\sum^{n-1}_{k=0}\frac{\left\lfloor k/2^k\right\rfloor}{3^{n-k}}\right)^2$$ I know that the final integral is just summing up the areas of all the infinite rectangles but I can't solve it. Please help. Thank you. (The final answer of this question is $27/32$. Candidates were allowed to solve it within 4 minutes.)

Answer 1

Here is a slightly advanced solution: Define $X_1, X_2, \ldots$ on $[0, 1]$ by

$$ X_k (x) := [\text{$k$th digit in the binary expansion of $x$}] = \lfloor 2^k x\rfloor - 2 \lfloor2^{k-1}x\rfloor. $$

Then

\begin{align*} \sum_{n=0}^{\infty} \frac{\lfloor 2^n x \rfloor}{3^n} = \sum_{n=1}^{\infty} \frac{1}{3^n} \sum_{k=1}^{n} 2^{n-k}X_k = \sum_{k=1}^{\infty} \frac{X_k}{2^k} \sum_{n=k}^{\infty} \frac{2^n}{3^n} = \sum_{k=1}^{\infty} \frac{X_k}{3^{k-1}} . \end{align*}

Now by regarding $[0, 1]$ as a probability space with the probability measure $\mathrm{d}x$, we find that $X_1, X_2, \ldots$ are i.i.d. $\text{Bernoulli}(\frac{1}{2})$ variables. So,

\begin{align*} \int_{0}^{1} \left( \sum_{n=0}^{\infty} \frac{\lfloor 2^n x \rfloor}{3^n} \right)^2 \, \mathrm{d}x = \mathbf{E} \left[ \left( \sum_{k=1}^{\infty} \frac{X_k}{3^{k-1}} \right)^2 \right] = \sum_{j,k=1}^{\infty} \frac{1}{3^{j+k-2}} \mathbf{E}[X_j X_k]. \end{align*}

Using the independence, we get $\mathbf{E}[X_j X_k] = \frac{1}{4} + \frac{1}{4} \mathbf{1}_{\{j = k\}}$. Hence, the expectation reduces to

\begin{align*} \sum_{j,k=1}^{\infty} \frac{1}{3^{j+k-2}} \left( \frac{1}{4} + \frac{1}{4} \mathbf{1}_{\{j = k\}} \right) = \frac{1}{4} \left( \sum_{k=1}^{\infty} \frac{1}{3^{k-1}} \right)^2 + \frac{1}{4} \left( \sum_{k=1}^{\infty} \frac{1}{9^{k-1}} \right) = \boxed{\frac{27}{32}} \end{align*}

Answer 2

This answer uses the somewhat more advanced concept that, in Riemann integrals, any countable number of discontinuities (i.e., they have measure zero) can be ignored, such as in Thomae's function. First, define

$$f(x) = \sum_{n=0}^\infty\frac{\left\lfloor 2^nx\right\rfloor}{3^n}, \; \; g(y) = \int_{0}^{y}f^2(x)dx \tag{1}\label{eq1A}$$

Use the substitution $x = 2z \; \; \to \; \; dx = 2dz$ to get

$$g(1) = 2\int_{0}^{0.5}\left(\sum_{n=0}^\infty\frac{\left\lfloor 2^{n+1}z\right\rfloor}{3^n}\right)^2 dz \tag{2}\label{eq2A}$$

Next, for $0 \le z \le 0.5$, we have

$$\begin{equation}\begin{aligned} \sum_{n=0}^\infty\frac{\left\lfloor 2^{n+1}z\right\rfloor}{3^n} & = 3\sum_{n=0}^\infty\frac{\left\lfloor 2^{n+1}z\right\rfloor}{3^{n+1}} \\ & = 3\left(\sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z\right\rfloor}{3^{n}} -\frac{\lfloor2^{0}z\rfloor}{3^{0}}\right) \\ & = 3\left(\sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z\right\rfloor}{3^{n}}\right) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Substituting this into \eqref{eq2A} gives

$$g(1) = 18g(0.5) \; \; \to \; \; g(0.5) = \frac{g(1)}{18} \tag{4}\label{eq4A}$$

From \eqref{eq1A}, using the substitution $x = 1 - z \; \; \to \; \; dx = -dz$ in the second integral on the right of the first line below, and that $\lfloor m - a \rfloor = m - 1 - \lfloor a \rfloor$ for all integers $m$ and non-integers $a$, we get

$$\begin{equation}\begin{aligned} g(1) & = \int_{0}^{0.5}f^2(x)dx + \int_{0.5}^{1}f^2(x)dx \\ & = g(0.5) - \int_{0.5}^{0}\left(\sum_{n=0}^\infty\frac{\left\lfloor 2^n(1-z) \right\rfloor}{3^{n}}\right)^2 dz \\ & = g(0.5) + \int_{0}^{0.5}\left(\sum_{n=0}^\infty\frac{2^n - 1}{3^n} - \sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)^2 dz \\ & = g(0.5) + \int_{0}^{0.5}\left(\left(\frac{1}{1-\frac{2}{3}} - \frac{1}{1-\frac{1}{3}}\right) - \sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)^2 dz \\ & = g(0.5) + \int_{0}^{0.5}\left(\frac{3}{2} - \sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)^2 dz \\ & = g(0.5) + \int_{0}^{0.5}\left(\frac{9}{4} - 3\sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}} + \left(\sum_{n=0}^\infty\frac{\left\lfloor 2^{n+1}z\right\rfloor}{3^n}\right)^2\right)dz \\ & = g(0.5) + \frac{9}{8} - 3\int_{0}^{0.5}f(x)dx + g(0.5) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Next, using $x = 0.5 - z \; \; \to \; \; dx = -dz$, as well as $\lfloor 2^{0}(0.5 - z) \rfloor = \lfloor 2^{1}(0.5 - z) \rfloor = 0$ and $\lfloor 2^{0}z \rfloor = \lfloor 2^{1}z \rfloor = 0$ for $0 \lt z \le 0.25$, then

$$\begin{equation}\begin{aligned} \int_{0}^{0.5}f(x)dx & = \int_{0}^{0.25}f(x)dx + \int_{0.25}^{0.5}f(x)dx \\ & = \int_{0}^{0.25}f(x)dx - \int_{0.25}^{0}\sum_{n=0}^\infty\frac{\left\lfloor 2^n(0.5-z) \right\rfloor}{3^{n}}dz \\ & = \int_{0}^{0.25}f(x)dx + \int_{0}^{0.25}\left(\sum_{n=2}^\infty\frac{2^{n-1} - 1}{3^n} - \sum_{n=2}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)dz \\ & = \int_{0}^{0.25}f(x)dx + \int_{0}^{0.25}\left(\left(\frac{\frac{2}{9}}{1-\frac{2}{3}} - \frac{\frac{1}{9}}{1-\frac{1}{3}}\right) - \sum_{n=2}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)dz \\ & = \int_{0}^{0.25}f(x)dx + \int_{0}^{0.25}\left(\frac{1}{2} - \sum_{n=0}^\infty\frac{\left\lfloor 2^{n}z \right\rfloor}{3^{n}}\right)dz \\ & = \int_{0}^{0.25}f(x)dx + \frac{1}{8} - \int_{0}^{0.25}f(x)dx \\ & = \frac{1}{8} \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Substituting this into \eqref{eq5A}, and using \eqref{eq4A}, gives

$$\begin{equation}\begin{aligned} g(1) & = 2g(0.5) + \frac{9}{8} - 3\left(\frac{1}{8}\right) \\ g(1) & = \frac{g(1)}{9} + \frac{3}{4} \\ \frac{8g(1)}{9} & = \frac{3}{4} \\ g(1) & = \frac{27}{32} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Answer 3

$\newcommand{\d}{\,\mathrm{d}}$The first thing that jumps at me is that we should expand the squared series. Call this integral $I$. By expanding into odd and even terms, from $n\ge2$ (the corresponding $n=0,1$ summands is clearly zero by examining the floors) we have: $$\begin{align}I&=\sum_{n\ge2}3^{-n}\sum_{k=0}^n\int_0^1\lfloor2^kx\rfloor\lfloor2^{n-k}x\rfloor\d x\\&=\sum_{n\ge1}3^{-2n}\left(2\sum_{k=1}^{n-1}\int_0^1\lfloor2^kx\rfloor\lfloor2^{2n-k}\rfloor\d x+\int_0^1\lfloor2^nx\rfloor^2\d x\right)\\&+2\sum_{n\ge1}3^{-(2n+1)}\sum_{k=1}^n\int_0^1\lfloor2^kx\rfloor\lfloor2^{2n-k+1}x\rfloor\d x\end{align}$$

So we need to figure out how to evaluate $\int_0^1\lfloor2^ax\rfloor\lfloor2^bx\rfloor\d x$ where $a\le b$ are positive integers. To that end, we need to partition $(0,1)$ into fine subdivisions where each floor is constant: $$\begin{align}\int_0^1\lfloor2^ax\rfloor\lfloor2^bx\rfloor\d x&=\sum_{j=0}^{2^a-1}\sum_{i=0}^{2^{b-a}-1}\int_{(j2^{b-a}+i)\cdot2^{-b}}^{(j2^{b-a}+i+1)\cdot2^{-b}}\lfloor2^ax\rfloor\lfloor2^bx\rfloor\d x\\&=\sum_{j=0}^{2^a-1}\sum_{i=0}^{2^{b-a}-1}2^{-b}(j)(j2^{b-a}+i)\\&=\sum_{j=0}^{2^a-1}\sum_{i=0}^{2^{b-a}-1}j^22^{-a}+ij2^{-b}\\&=\sum_{j=0}^{2^a-1}(j^2\cdot2^{b-2a}+j\cdot(2^{b-2a-1}-2^{-a-1})\\&=\cdots\\&=\frac{1}{3}2^{b+a}-\frac{1}{4}(2^b+2^a)-\frac{1}{12}2^{b-a}+\frac{1}{4}\end{align}$$Using standard summation formulae.

We just insert $a,b$ into the above formula, then plug the result into the series expression for $I$. There are no longer any interesting details: it is just a tedious process of using the sum of a geometric series many many many many many times, and out pops the desired result.

This is a method. The high level of tedium associated to it makes me think this is not the fancy $4$-minute solution, but it is a solution that is reasonably easy to carry out reasonably quickly so long as you don't make mistakes with your algebra.

For example, I will demonstrate how to evaluate the subseries involving $\lfloor2^nx\rfloor^2$. We let $a=b=n$. $$\begin{align}\sum_{n\ge1}3^{-2n}\int_0^1\lfloor2^nx\rfloor^2\d x&=\sum_{n\ge1}3^{-2n}\frac{1}{3}2^{2n}-\sum_{n\ge1}3^{-2n}\frac{1}{4}2\cdot2^n-\sum_{n\ge1}3^{-2n}\left(\frac{1}{12}2^0+\frac{1}{4}\right)\\&=\frac{1}{3}\sum_{n\ge1}\left(\frac{4}{9}\right)^n-\frac{1}{2}\sum_{n\ge1}\left(\frac{2}{9}\right)^n+\frac{1}{6}\sum_{n\ge1}9^{-n}\\&=\frac{1}{3}\cdot\frac{4}{9}\cdot\frac{1}{1-\frac{4}{9}}-\frac{1}{2}\cdot\frac{2}{9}\cdot\frac{1}{1-\frac{2}{9}}+\frac{1}{6}\cdot\frac{1}{9}\cdot\frac{1}{1-\frac{1}{9}}\\&=\frac{4}{15}-\frac{1}{7}+\frac{1}{48}\end{align}$$It is probably sensible to leave these fractions expanded in case there are cancellations with the other series.

Answer 4

Here's an elementary solution:

Define

$$ I_b = \int_0^b \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx $$ The idea is to write $I_1$ in terms of $I_{1/2}$ in two different ways: one using the substitution $x \rightarrow 2x$, the other by splitting the integral range into two and writing both parts in terms of $I_{1/2}$. Then we solve the system of equations for $I_1$.

The substitution is straightforward, you just have to note that $\lfloor 2^1x\rfloor = 0$ for $0\le x<1/2$:

$$ \begin{align*} I_1 &= 2\int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^{n+1}x\rfloor}{3^n}\right)^2\,dx = 2\int_0^{1/2} \left(\sum_{n=\color{red}{2}}^\infty \frac{\lfloor 2^nx\rfloor}{3^{n-1}}\right)^2\,dx \\ &= 2\int_0^{1/2} \left(3\sum_{n=\color{red}{1}}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx = 18\,I_{1/2} \end{align*} $$

The other way to write $I_1$ is as follows:

$$ \begin{align*} I_1 &= \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx + \int_{1/2}^1 \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\right)^2\,dx \\ &= I_{1/2} + \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^n(x+1/2)\rfloor}{3^n}\right)^2\,dx \\ &= I_{1/2} + \int_0^{1/2} \left(\sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n} + \underbrace{\sum_{n=1}^\infty \frac{2^{n-1}}{3^n}}_{=1}\right)^2\,dx \\ &= 2\,I_{1/2} + 2\int_0^{1/2} \sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\,dx + \frac{1}{2} \end{align*} $$

Now the integral without the square is much easier to solve:

$$ \begin{align*} \int_0^{1/2} \sum_{n=1}^\infty \frac{\lfloor 2^nx\rfloor}{3^n}\,dx &= \sum_{n=1}^\infty \frac{1}{3^n} \int_0^{1/2} \lfloor 2^nx\rfloor\,dx \\ &= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n} \int_0^{2^{n-1}} \lfloor x\rfloor\,dx \\ &= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n} \sum_{k=1}^{2^{n-1}-1} k \\ &= \sum_{n=1}^\infty \frac{1}{3^n}\cdot\frac{1}{2^n}\cdot\frac{1}{2}2^{n-1}\left(2^{n-1}-1\right) \\ &= \frac{1}{8}\sum_{n=1}^\infty \left(\frac{2}{3}\right)^n - \frac{1}{4}\sum_{n=1}^\infty \left(\frac{1}{3}\right)^n \\ &= \frac{1}{8}\cdot 2 - \frac{1}{4}\cdot\frac{1}{2} = \frac{1}{8} \end{align*} $$

Recalling that $I_{1/2} = I_1/18$, substitute the values and solve for $I_1$:

$$ I_1 = 2\cdot\frac{1}{18}\,I_1 + 2\cdot\frac{1}{8} + \frac{1}{2} \\ \Rightarrow\frac{8}{9}\,I_1 = \frac{3}{4} \\ \Rightarrow I_1 = \frac{27}{32} $$