Periodic Functions & Orthogonal Functions — Question 3

Question 3 – Solution

Strategy. Separate periodicity, assigned endpoint values, one-sided limits and integral properties.

Step 1: Locate the joins and establish minimality. The representation gives $s(x+2)=s(x)$. Its discontinuities occur precisely at odd integers. Any period must preserve this set, so a positive period is a positive even integer; two is therefore fundamental. At $c=2m+1$, $$\boxed{s(c)=-1,\qquad s(c-)=1,\qquad s(c+)=-1.}$$ Each increasing segment has its left endpoint included and its right endpoint excluded. The graph also includes the assigned value $s(3)=-1$.

Step 2: Repair oddness, not continuity. Away from odd integers, $s(-x)=-s(x)$. At an odd integer both $s(x)$ and $s(-x)$ equal $-1$, so pointwise oddness fails. Assigning zero at every odd integer produces an odd function and preserves two-periodicity. No assigned join value can reconcile the unequal one-sided limits, so the jumps remain.

Step 3: Integrate over any full period. The integral on $[-1,1]$ is $\int_{-1}^1x\,dx=0$. Reduce $a$ modulo two to $b\in[-1,1)$. Splitting at one and shifting the remaining piece by two gives $$\int_a^{a+2}s=\int_b^1s+\int_1^{b+2}s =\int_b^1s+\int_{-1}^b s=0.$$ Changing finitely many join values in any bounded interval leaves these Riemann integrals unchanged. Integral cancellation does not require pointwise oddness.

Step 4: Check a continuous periodic join. For $p$, both endpoint limits of $x(1-x)$ are zero, matching the assigned value at each integer. Thus $p$ is continuous everywhere. The left derivative at an integer is $-1$, while the right derivative is $1$, so it is not differentiable there. Value matching alone does not give derivative matching.