Alternating Series Estimation Theorem

Chris*

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Jan 9, 2007
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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 ex\displaystyle e^x to calculate e0.2\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:
e0.2=n=0(0.2)nn!\displaystyle e^{-0.2}=\sum_{n=0}^{\infty}\frac{(-0.2)^n}{n!}
=10.2+12!(0.2)213!(0.2)3+14!(0.2)415!(0.2)5+16!(0.2)6...\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 16!(0.2)6=8.8×108\displaystyle \frac{1}{6!}(0.2)^6=8.\overline{8}\times 10^{-8}, so by the Alternating Series Estimation Theorem, e0.2=n=05(0.2)nn!=0.81873\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 15!(0.2)5\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. :)
 
4 terms will give it to within the desired accuracy.

e0.2=1+(.2)+(.2)22+(.2)36+(.2)424=1.22140\displaystyle e^{0.2}=1+(.2)+\frac{(.2)^{2}}{2}+\frac{(.2)^{3}}{6}+\frac{(.2)^{4}}{24}=1.22140

e.2=1.22140275816\displaystyle e^{.2}=1.22140275816

There, that is withinn 5 places. After the 0 it wonders off.

The e series converges rather quickly by comparison.
 
Ah well, it's actually e.2\displaystyle e^{-.2} but I know what you mean. Any idea on why the book's answer would choose the first five terms rather than the first four? I know it would be more accurate, but it threw me off a bit.

BTW, thanks for the help!
 
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