Step Functions — Question 9

Question 9 – Solution

Strategy. Encode each upward increment and the final reset. A pointwise error bound becomes an integral certificate valid for a whole range of parameters.

Step 1: Include the final shutoff. There are increments $1/n$ at $k/n$, $1\le k\le n-1$, followed by a drop $(n-1)/n$ at one. Hence $$f_n(t)=\frac 1n\sum_{k=1}^{n-1}u_{k/n}(t)-\frac{n-1}{n}u_1(t),$$ $$\boxed{F_n(s)=\frac{\sum_{k=1}^{n-1}e^{-ks/n}-(n-1)e^{-s}}{ns}\quad(s\ne 0).}$$ Transforming steps gives the formula for $s>0$; finite-interval integration gives it for all $s\ne 0$.

Step 2: Evaluate area and the target. Each stair has width $1/n$, so $$\boxed{F_n(0)=\frac 1{n^2}\sum_{k=0}^{n-1}k=\frac{n-1}{2n}.}$$ Integration by parts on $[0,1]$ gives $$\boxed{R(s)=\frac{1-(1+s)e^{-s}}{s^2}\ (s\ne 0),\qquad R(0)=\tfrac 12.}$$ Both original functions have finite support, so both transform domains are the whole real line. Their apparent singularities at zero are removable.

Step 3: Convert pointwise error to weighted error. On each half-open stair, $0\le r-f_n<1/n$, and both functions vanish from one onward. For $s>0$, $$\boxed{0\le R(s)-F_n(s)\le\frac 1n\int_0^1e^{-st}\,dt =\frac{1-e^{-s}}{ns}.}$$ At zero the actual error is $1/(2n)$, while the continuous limiting bound is $1/n$. The bound is sufficient, not an equality.

Step 4: Certify a simultaneous tolerance. The function $b(s)=\int_0^1e^{-st}\,dt$ is decreasing for $s>0$, since its derivative is $-\int_0^1te^{-st}\,dt<0$. Its largest value on $s\ge 1$ is $1-e^{-1}$. Thus the certificate requires $$n\ge 100(1-e^{-1})\approx 63.2121,\qquad\boxed{n=64}.$$ For $n=63$ this bound exceeds $.01$ at $s=1$; that does not prove the actual error exceeds $.01$. The graph shows $n=4$ only to display the staircase geometry; it is not the certified design.