Review : Power Series — Question 4

Question 4 – Solution

Strategy. An exact remainder supplies a sharp interval bound and exposes what fails when points approach the boundary as the degree changes.

Step 1: Retain the exact geometric remainder. Multiplication gives $(1-x)S_N=1-x^{N+1}$. Therefore $$\boxed{f(x)-S_N(x)=\frac{x^{N+1}}{1-x}.}$$ For $0<x<1$ it is positive. For $-1<x<0$, its sign is $(-1)^{N+1}$; at $x=0$ it is zero. The denominator is positive throughout the interval.

Step 2: Find the least degree on the compact interval. For $|x|\le 1/2$, $$|f-S_N|\le\frac{(1/2)^{N+1}}{1-1/2}=2^{-N}.$$ Equality holds at $x=1/2$, so this is the exact maximum, not merely a convenient estimate. Since $2^{-9}>10^{-3}$ and $2^{-10}<10^{-3}$, $\boxed{N=10}$ is least. This retains $\boxed{11\text{ terms}}$, including the constant.

Step 3: Select a moving point with fixed relative error. The relative error simplifies to $\boxed{|x|^{N+1}}$. Taking $x_N=2^{-1/(N+1)}$ gives a point in $(0,1)$ with relative error exactly $1/2$ for every $N\ge 0$. These points approach $1$ as $N$ increases.

Step 4: Distinguish fixed-point from uniform accuracy. At each fixed $|x|<1$, both errors tend to zero. Yet for every fixed $N$, $x^{N+1}/(1-x)\to+\infty$ as $x\to 1^-$, so the supremum of the absolute error on $(-1,1)$ is infinite. The supremum of relative error is $1$. Neither error converges uniformly to zero on that whole interval. The plot shows slower approximation near the right boundary; on any fixed $[-r,r]$, $r<1$, the bound $r^{N+1}/(1-r)\to 0$ does give uniform convergence.