Review : Taylor Series — Question 4

Question 4 – Solution

Strategy. Convolve by degree, then bound the two truncation errors and the discarded product terms separately.

Step 1: Collect only the requested powers. Using the entire exponential and cosine series, $$[x^2]F=1-\tfrac 12=\tfrac 12,\quad [x^4]F=\tfrac 12-\tfrac 12+\tfrac 1{24}=\tfrac 1{24},$$ $$[x^6]F=\tfrac 16-\tfrac 14+\tfrac 1{24}-\tfrac 1{720}=-\tfrac{31}{720}.$$ Thus $\boxed{P_6(x)=1+x^2/2+x^4/24-31x^6/720}$.

Step 2: Recover a derivative and its parity. We have $F^{(6)}(0)=6!(-31/720)=-31$. Both factors are even and analytic everywhere, so all odd coefficients vanish. The next possible power after degree six is eight, proving $F-P_6=O(x^8)$ near $0$.

Step 3: Retain the discarded product explicitly. Direct multiplication gives $$AB-P_6=-\frac{23}{360}x^8+\frac 1{160}x^{10}-\frac 1{4320}x^{12}.$$ These terms belong to the product of polynomials, not to $P_6$.

Step 4: Certify the error on the whole interval. Write $E=e^{x^2}$ and $C=\cos x$. Then $EC-P_6=(E-A)C+A(C-B)+(AB-P_6)$. For $|x|\le 1/2$, $|C|\le 1$, $0<A\le e^{1/4}<4/3$ (compare the exponential series with the geometric series). Taylor’s theorem and the alternating cosine tail give $$|E-A|\le\frac 43\frac{|x|^8}{24},\qquad |C-B|\le\frac{|x|^8}{8!}.$$ Consequently, $$|F-P_6|\le\frac{4/3}{24\cdot 256}+\frac{4/3}{40320\cdot 256} +\frac{23}{360\cdot 256}+\frac 1{160\cdot 1024} +\frac 1{4320\cdot 4096}<0.000473<0.0005.$$ The bound is conservative but rigorous; no unbounded $O$-term is used as a numerical certificate.