Let $f(x) = |x − n|$ and let $g_m (x) = f(f(\ldots f (\lfloor x \rfloor) \ldots))$ where $f$ is applied $m$ times. Find $\int_0^{mn} g_m(x)\ dx$ as a function of $n$ and $m,$ where $n$ and $m$ are positive integers.
1. Sketch the graph of $y=\lceil x \rceil ^ 2.$
2. Find an expression for the sum $1^2+2^2+\ldots+n^2.$
3. Evaluate $\int_{-2}^{2} (|x|-1)\ dx.$

Have you tried sketching $g_m(x)$ for simple values of $n$ and $m?$

Why not start with $f(\lfloor x \rfloor)$ and then successively add more applications of $f?$

Could you quantify any emerging patterns?

… specifically, the number and height of the increasing and decreasing staircases pattern.

Consider the bounds of the integral, $0$ to $mn.$ How does this relate to the widths of the staircase pattern?

You may have different expressions for the two cases, based on the parity of $m.$

Have you tried finding the area of each individual increasing and decreasing staircase section?

A sketch of the graph is extremely useful for this question; a sample plot is given below.

First we note that $\lfloor x\rfloor$ is an increasing staircase that has the first flat piece at $y=0.$ Then applying $f$ once subtracts $n$ and flips the negative part up, which gives us a decreasing staircase from $y=n$ down to $y=1,$ then an increasing staircase from $y=0$ to infinity. A second application of $f$ again subtracts and flips, but this time we get an increasing staircase from $0$ to $(n-1),$ then a decreasing staircase from $n$ to $1,$ then an increasing staircase to infinity. A total of $m$ applications of $f$ will form $\frac{m}{2}$ bumps, each of which is $2n$ units wide. Hence the integral upper bound $mn$ tells us to find the area under the bumps only.

Let’s label the value of the integral $I$ and for simplicity, split up the problem into two cases: where $m$ is odd and where it is even. For $m=2k,$ we have $k$ increasing step patterns from $0$ to $(n-1)$ and $k$ decreasing step patterns from $n$ to $1.$ Since the steps are of unit width, we sum these arithmetic progressions: $I=k\frac{n(n-1)}{2}+k\frac{n(n+1)}{2}=kn^2=\frac{mn^2}{2}.$

For $m=2k+1,$ we have $(k+1)$ decreasing step patterns from $n$ to $1$ and $k$ increasing step patterns from $0$ to $(n-1).$ Hence our expression for the total area becomes $I=(k+1)\frac{n(n+1)}{2}+k\frac{n(n-1)}{2}=kn^2+\frac{n(n+1)}{2}=\frac{mn^2+n}{2}.$

Plot of $g_m(x)$ for $n=3$ and $m=4:$