Step Functions — Question 4

Question 4 – Solution

Strategy. A ramp changes slope rather than level. Cancel both the final slope and the final intercept to create compact support.

Step 1: Encode slope changes. The slopes are $0,1,-1/2,0$, giving changes $1,-3/2,1/2$. Thus $$\boxed{f=r_1-\tfrac 32r_3+\tfrac 12r_7.}$$ Each ramp is zero at its own switch, so the combination is continuous. At the peak time three its value is two; it then decreases to zero at seven.

Step 2: Transform the ramps. For $s>0$, substitution gives $\mathcal L\{r_a\}=e^{-as}/s^2$. Hence $$\boxed{F(s)=\frac{e^{-s}-\tfrac 32e^{-3s}+\tfrac 12e^{-7s}}{s^2}\quad(s\ne 0).}$$ The actual function has finite support, so its integral exists for all real $s$. Direct integration of the two linear pieces extends the formula to all $s\ne 0$; zero is removable.

Step 3: Cancel the tail, not just the slope. For $t\ge 7$, the ramp sum is $$(1-\tfrac 32+\tfrac 12)t-[1-\tfrac 32(3)+\tfrac 12(7)]=0.$$ In general, after the last switch a finite ramp sum equals $t\sum c_j-\sum c_j a_j$. It vanishes identically there exactly when $$\boxed{\sum c_j=0,\qquad \sum c_j a_j=0.}$$ Canceling only the slope could leave a nonzero constant tail and a different convergence domain.

Step 4: Check area and moment. The triangle has base six and height two, so $\boxed{F(0)=6}$. Direct integration gives $$\int_1^3t(t-1)\,dt=\frac{14}3,\qquad \frac 12\int_3^7t(7-t)\,dt=\frac{52}3.$$ Thus the first moment is $\boxed{22}$ and $F'(0)=-22$. Expanding the numerator of $F$ gives $6s^2-22s^3+O(s^4)$, agreeing with both results. The horizontal time coordinate of the area centroid is $22/6=11/3$, consistent with the three triangle vertices.