$f(x) = |x − n|$ 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.
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- Sketch the graph of
$y=\lceil x \rceil ^ 2.$ - Find an expression for the sum
$1^2+2^2+\ldots+n^2.$ - Evaluate
$\int_{-2}^{2} (|x|-1)\ dx.$
- Sketch the graph of
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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?
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… specifically, the number and height of the increasing and decreasing staircases pattern.
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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?
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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$m=4:$