Find $\,\,e^{-1/4}\,$ to within $0.0002$ I am using Taylor's Inequality to solve this problem but this formula (page 607) is getting on my nerves and the textbook does not do a good job of explaining how to do this problem and I've gone to khanacademy and patrickjmt and they do not help me.
But I will try to interpret the problem anyways.  I know that M is the max value for the nth term I want but added to another.  The remainder must be less than or equal to $.0002$, so $R_n(x) \leq .0002$.  I'm guessing that the max value of M is $1$ but I'm not sure.  I do not know how to solve for what n is then.  I do know the Taylor series expansion for $e^x$ though but I do not know what term I should stop at.  Please help?  I'm extremely frustrated.
 A: It is easiest if we take advantage of the fact that the series for $e^{-1/4}$ is an Alternating Series. But instead we will use the harder to work with, and generally less useful, inequality of the text.
Suppose that the last term we use in the approximation is $\dfrac{x^n}{n!}$, where $x=-\dfrac{1}{4}$.
Then the error we make is $\dfrac{f^{(n+1)}(z)}{(n+1)!}x^{n+1}$, where all we know about $z$ is that it is between $-\frac{1}{4}$ and $0$.
Now any derivative of $e^t$, including the $(n+1)$-th, is $e^t$. So we want to estimate $e^z$. Since $z$ is between $-\frac{1}{4}$ and $0$, we can say for sure that $e^{-1/4}\lt e^z\lt e^0=1$.
Thus for sure our error has absolute value less than $\dfrac{1}{(n+1)!}\left(\dfrac{1}{4}\right)^{n+1}$.
Make this less than $0.0002$. Because of the $(n+1)!$, the inequality is unpleasant to attack algebraically. So we just fool around. Try $n=3$. It works! Try to be cheaper, using $n=2$: the inequality is not satisfied, so $n=2$ may not be good enough.
A: We have 
$$ e^{-1/4}=\sum_{n=0}^\infty\frac{(-1)^n}{4^nn!}.$$
This is an alternating series with summands decreasing in absolute value. Therefore, the error will we less than the first summand you drop. That is, if you compute the approximation
$$ e^{-1/4}\approx \sum_{n=0}^N\frac{(-1)^n}{4^nn!},$$
then the error will be $<\frac{1}{4^{N+1}(N+1)!}$. Note that already $N=3$ leads to $\frac{1}{4^{N+1}(N+1)!}\approx 0.00016$. Therefore the approximation
$$ e^{-1/4}\approx 1-\frac1{4}+\frac1{16\cdot 2}-\frac1{64\cdot 6}=\frac{299}{384}$$
is good enough for your purpose.
A: First note that the series is $$e^x=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\dots$$
When $x$ is negative the series is a convergent alternating series and [once the terms are decreasing in magnitude] the error is at most the magnitude of the last term you use (also of the first term you don't use).
$0.0002=\frac 1{5000}$
Note that the term in $x^3$ has denominator $64\times 6$ which is a little under 400. The term in $x^4$ has denominator a bit under $16 \times 400 =6400 \gt 5000$
So you can safely terminate at the $x^3$ term.
