# Piecewise function with integer conditions

I'm trying to define (and plot) a piecewise function with the following structure:

$$f(x) = \begin{cases} g(x) & nT \leq x \leq \frac{2n+1}{2}T\\ -g(x) & \frac{2n+1}{2}T \leq x \leq (n+1)T \end{cases}$$

where n is an integer, n >= 0 and T is a constant.

The idea is that it cycles between positive and negative after a full period is complete.

My attempt was:

f[x_] := Piecewise[{
{g[x], (n T <= x <= ((2 n + 1)/2) T) && Element[n, Integers] && (n >= 0)},
{-g[x], (((2 n + 1)/2) T <= x <= (n + 1) T) && Element[n, Integers] && (n >= 0)}}]


But when I try a test value for x in f[x] it shows n in the conditions for the output, so it seems to not get past the conditions. I believe plotting fails for this reason too.

I have never tried to use Integers in Mathematica.

• Have you specified a value for T ? – Sektor Jul 15 '15 at 15:23
• Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Dr. belisarius Jul 15 '15 at 15:44
• Sektor: T is specified. Patrick Stevens: Okay, thank you. Using Floor[x/T] instead of n worked. f[x_] := Piecewise[{{g[x], Floor[x/T] T <= x <= ((2 Floor[x/T] + 1)/2) T}, {-g[x], ((2 Floor[x/T] + 1)/2) T <= x <= (Floor[x/T] + 1) T}}] I'm not exactly sure how I would simplify the second condition. belisarius: Thank you, I can see that your version works. – Canada709 Jul 15 '15 at 15:51
• It won't let me accept an answer either, I think because I was not properly signed in originally when I posted the question. Apparently I can comment now though – Canada709 Jul 15 '15 at 15:53

Your problem is that you haven't told Piecewise that n can be arbitrary, so Mathematica thinks it's constant. You'd be better off defining n as Floor[x/T] for the first comparison, rather than asking Mathematica to find an n such that the definition holds (or prove that n doesn't exist).

It looks like you could simplify the second condition to True, by the way: Piecewise checks the pieces in order, so there's no need to give a complicated final condition.

A way following your line of thought:

T = Pi;
g[x_] := Sin@x
f[x_?NumericQ] := Piecewise[
{{g[x] , Quiet@Resolve@Exists[n, n ∈ Integers && (n + 1/2)T <= x <= (n + 1)T]}},
-g[x]]
Plot[f[x], {x, 0, 10}] I would do it this way instead:

Plot[-g[x] SquareWave[x/T], {x, 0, 10}] And from there you can get an equivalent Piecewise form:

PiecewiseExpand[-g[x] SquareWave[x/T]]


$$\begin{cases} -g(x) & 0\leq (\frac{x}{T} \bmod 1)<\frac{1}{2} \\ g(x) & \text{True} \end{cases}$$