I am trying to find an algebraic way to solve for the intersection of a step function and another function (in this case linear but any polynomial function in general). The example is given as follows :
Line : $y = 2x+100 $ and
Step : $y = -200 \lfloor {-x/150} \rfloor +400$
I tried solving this by setting them equal but I arrive at a logical issue in which
$2x-300 = -200 \lfloor {-x/150} \rfloor $
I have no interval to assign the floor function to as it depends completely on the neighbouring value of $x$. I don't know of any other manipulation to solve this without either plotting the functions or guessing values of the step function which may intersect the line as well. Any ideas?
Best Answer
First of all, it is a bit cleaner to write the step function with ceiling rather than floor: $y = 200 \lceil x/150\rceil + 400$.
As for solving the problem: rather than guessing values, we can start with the inequality $$x/150 \le \lceil x/150\rceil < x/150 + 1.$$ (Rounding up to the next integer always increases the value we get, but by no more than $1$.) Therefore if $2x-300 = 200\lceil x/150\rceil$, we have $$200(x/150) \le 2x-300 < 200(x/150+1)$$ which simplifies to $450 \le x < 750$. This means that $3 \le x/150 < 5$, and $\lceil x/150\rceil$ is $3$, $4$, or $5$.
We get an intersection point for each case:
In all cases, we should check that we did not get an extraneous solution: $\lceil x/150 \rceil$ matches what it is assumed to be in the case. All three checks are satisfied, so $x=450$, $x = 550$, and $x = 650$ give us the three intersection points.
If there were more cases than this, I would want to "automate" checking them. Write $x = 150a - b$ where $0 \le b < 150$, so that $\lceil x/150\rceil = a$, and solve $$2(150a-b) - 300 = 200a$$ for $b$ in terms of $a$. We get $b = 50a - 150$; the condition that $0 \le b < 150$ tells us $3 \le a < 6$ (again), and the values $a=3,4,5$ give us $b=0,50,100$, for $x = 3\cdot150-0$, $x=4\cdot150-50$, and $x=5\cdot150-100$: the same solutions we found earlier.