Step Functions — Question 8

Question 8 – Solution

Strategy. Moment constraints form a small linear system. The step differences then expose cancellation in the transform numerator.

Step 1: Solve the two moment equations. The unit-width intervals give $$1+B+C=0,\qquad \tfrac 12+\tfrac 32B+\tfrac 52C=0.$$ Subtracting the first equation from twice the second yields $2B+4C=0$. Combining this with $B+C=-1$ gives $\boxed{B=-2,\ C=1}$. The coefficient determinant is $5/2-3/2=1\ne 0$, so the solution is unique.

Step 2: Encode changes and factor the transform. The levels $1,-2,1,0$ have jumps $-3,+3,-1$. Hence $$f(t)=1-3u_1(t)+3u_2(t)-u_3(t),$$ $$\boxed{F(s)=\frac{1-3e^{-s}+3e^{-2s}-e^{-3s}}s =\frac{(1-e^{-s})^3}s\quad(s\ne 0).}$$ This follows first by transforming steps for $s>0$, or directly on the three finite intervals for any $s\ne 0$.

Step 3: Interpret the removable zero. Compact support gives convergence for every real $s$. Since $1-e^{-s}=s+O(s^2)$, $$\boxed{F(s)=s^2+O(s^3),\quad F(0)=F'(0)=0,\quad F''(0)=2.}$$ Directly, the second moment is $$\int_0^3t^2f(t)\,dt=\tfrac 13-2(\tfrac 73)+\tfrac{19}3=\boxed 2,$$ consistent with differentiating the finite-support transform twice.

Step 4: Separate two constraints from identity. The pulse is one throughout its first interval, so it is not zero. For $s>0$, both $1-e^{-s}$ and $s$ are positive, giving $\boxed{F(s)>0}$. The zero area and first moment eliminate only the constant and linear terms near $s=0$. They do not eliminate all weighted integrals. The second moment already witnesses the remaining information. The graph makes the signed cancellation visible without replacing it by a pointwise sign claim.