Review : Taylor Series — Question 5

Question 5 – Solution

Strategy. Integrate a uniformly convergent series, then use the integrated alternating tail as an error certificate.

Step 1: Justify the interchange. On $[0,1/2]$, $|(-t^2)^k/k!|\le(1/4)^k/k!$, whose sum is finite. The Weierstrass test gives uniform absolute convergence. Integrating yields $$I=\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{2k+1}k!(2k+1)}.$$ The positive term magnitudes decrease strictly to zero.

Step 2: Choose a certified truncation. The first omitted magnitude is $b_m=1/[2^{2m+3}(m+1)!(2m+3)]$. It decreases with $m$. For $m=3$, $b_3=1/110592>10^{-6}$, whereas $$b_4=\frac 1{2703360}<3.70\times 10^{-7}<10^{-6}.$$ Thus $\boxed{m=4}$ is the smallest index certified by this bound, using five terms.

Step 3: Supply an exact enclosure. The even partial sum is an upper bound and the next odd sum a lower bound: $$\boxed{S_4-\frac 1{2703360}=S_5<I<S_4 =\frac 12-\frac 1{24}+\frac 1{320}-\frac 1{5376}+\frac 1{110592}.}$$ Numerically, $S_4\approx 0.4612813637$. Reporting $I\approx 0.46128136$ introduces less than $4\times 10^{-9}$ additional rounding error, so its total absolute error is less than $3.74\times 10^{-7}$.

Step 4: Interpret the signed area. Set $p_8(t)=1-t^2+t^4/2-t^6/6+t^8/24$. The alternating tail is strictly negative for $t>0$ in this interval, so $p_8(t)>e^{-t^2}$. Hence $S_4-I$ is the positive area between the curves. The scaled gap below makes the very small difference visible; it vanishes at $t=0$.