Hello, my friend and I were working on a textbook problem and came upon a confusing solution.
The problem said,
Use the Maclaurin series for \(\displaystyle e^x\) to calculate \(\displaystyle e^{0.2}\) correct to five decimal places.
The solution manual ended up using the Alternating Series Estimation Theorem to approximate the answer. Which, up til now, I thought I understood.
Here's the solution:
\(\displaystyle e^{-0.2}=\sum_{n=0}^{\infty}\frac{(-0.2)^n}{n!}\)
\(\displaystyle =1-0.2+\frac{1}{2!}(0.2)^2-\frac{1}{3!}(0.2)^3+\frac{1}{4!}(0.2)^4-\frac{1}{5!}(0.2)^5+\frac{1}{6!}(0.2)^6-...\)
But \(\displaystyle \frac{1}{6!}(0.2)^6=8.\overline{8}\times 10^{-8}\), so by the Alternating Series Estimation Theorem, \(\displaystyle e^{-0.2}=\sum_{n=0}^{5}\frac{(-0.2)^n}{n!}=0.81873\)
First of all, when they say correct to five decimal places do they mean Some number<0.00001? In class, when my teacher asked us to solved a problem correct to four decimal places, they wrote the answer as less than 0.0001.
If this is the case, wouldn't \(\displaystyle \frac{1}{5!}(0.2)^5\) be the first term that gives the desired approximation?
For some of the other answers in the book, it seems like when they ask for a number correct to 5 decimal places, the answer just has to have 5 zeroes after the decimal point...
Any help would be appreciated... This place has been so helpful to me in the past! Thanks in advance.
The problem said,
Use the Maclaurin series for \(\displaystyle e^x\) to calculate \(\displaystyle e^{0.2}\) correct to five decimal places.
The solution manual ended up using the Alternating Series Estimation Theorem to approximate the answer. Which, up til now, I thought I understood.
Here's the solution:
\(\displaystyle e^{-0.2}=\sum_{n=0}^{\infty}\frac{(-0.2)^n}{n!}\)
\(\displaystyle =1-0.2+\frac{1}{2!}(0.2)^2-\frac{1}{3!}(0.2)^3+\frac{1}{4!}(0.2)^4-\frac{1}{5!}(0.2)^5+\frac{1}{6!}(0.2)^6-...\)
But \(\displaystyle \frac{1}{6!}(0.2)^6=8.\overline{8}\times 10^{-8}\), so by the Alternating Series Estimation Theorem, \(\displaystyle e^{-0.2}=\sum_{n=0}^{5}\frac{(-0.2)^n}{n!}=0.81873\)
First of all, when they say correct to five decimal places do they mean Some number<0.00001? In class, when my teacher asked us to solved a problem correct to four decimal places, they wrote the answer as less than 0.0001.
If this is the case, wouldn't \(\displaystyle \frac{1}{5!}(0.2)^5\) be the first term that gives the desired approximation?
For some of the other answers in the book, it seems like when they ask for a number correct to 5 decimal places, the answer just has to have 5 zeroes after the decimal point...
Any help would be appreciated... This place has been so helpful to me in the past! Thanks in advance.