Review : Taylor Series — Question 2

Question 2 – Solution

Strategy. Use Taylor’s theorem for convergence and alternating tails for both upper and lower error bounds.

Step 1: Control all points simultaneously. Every derivative of $e^{-x}$ has absolute value $e^{-x}\le 1$ on $[0,1]$. Taylor’s theorem gives $$|e^{-x}-T_n(x)|\le\frac{x^{n+1}}{(n+1)!}\le\frac 1{(n+1)!}\longrightarrow 0.$$ This proves equality to the series and uniform convergence on the stated interval.

Step 2: Find a certified degree. Because $9!=362880<10^6<3628800=10!$, the smallest degree certified by this bound is $\boxed{n=9}$, using ten terms. Degree $8$ is not certified by the upper bound alone.

Step 3: Prove actual minimality. At $x=1$, the alternating tail has the sign of its first term, and its magnitude is strictly greater than the difference of its first two magnitudes: $$|e^{-1}-T_n(1)|>\frac 1{(n+1)!}-\frac 1{(n+2)!} =\frac{n+1}{(n+2)!}=:b_n.$$ Indeed, the remaining tail starts positive after factoring out its first sign and grouping consecutive pairs. Also $b_{n+1}/b_n=(n+2)/[(n+1)(n+3)]<1$ for $n\ge 0$. Hence for every $0\le n\le 8$, the error exceeds $b_8=9/10!>10^{-6}$. No smaller degree works on all of $[0,1]$.

Step 4: Enclose the value rationally. Odd partial sums lie below the limit and even ones above. Therefore $$\boxed{\frac{16687}{45360}=T_9(1)<e^{-1}<T_{10}(1) =\frac{16481}{44800}.}$$ The interval width is exactly $1/10!$; the upper endpoint comes from the positive first omitted term after $T_9$.