Use a suitable Maclaurin Series to estimate to five decimal place accuracy.
A) 0.00915
B) 0.00905
C) 0.00920
D) 0.00910
E) 0.00900

Question

Use a suitable Maclaurin Series to estimate  to five decimal place accuracy.
A) 0.00915 
B) 0.00905 
C) 0.00920 
D) 0.00910 
E) 0.00900

Quizplus · Accepted Answer

The Maclaurin series of $e^x$ is $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+...$

Substituting $x=-0.02$ (since $11ead0bc_8b64_f653_99a0_69b094302617_TB6988_11=-0.02$ in hexadecimal notation), we get:
$$e^{-0.02} \approx 1+(-0.02)+\frac{1}{2}(-0.02)^2+\frac{1}{6}(-0.02)^3+...$$

Using the formula for the error in the Maclaurin series estimation, we can bound the error as:
$$|R_n(-0.02)| \leq \frac{e^{c}0.02^{n+1}}{(n+1)!}$$
where $|f^{(n+1)}(x)| \leq e^c$ for some $c$ between 0 and $-0.02$.

Taking $n=3$ (since this gives us a remainder of five decimal place accuracy), we get:
$$|R_3(-0.02)| \leq \frac{e^c0.02^{4}}{4!}< 5 	imes 10^{-6}$$

Solving this inequality for $e^c$, we get $e^c< 5.3$, which means that $c< \ln(5.3) \approx 1.667$.

Thus, we can estimate $e^{-0.02}$ to five decimal place accuracy as:
$$e^{-0.02} \approx 1+(-0.02)+\frac{1}{2}(-0.02)^2+\frac{1}{6}(-0.02)^3 \approx 0.00910$$

Therefore, the answer is D.

Use a Suitable Maclaurin Series to Estimate to Five