Basic doubt on Taylor's polynomial I have a doubt about a general situation in where I am asked to calculate $f(x)$ with a certain precision. How can I compute the number of terms of the Taylor polynomial needed for that?
For example if I wanted to calculate $\dfrac{1}{\sqrt e}$ from the function $e^x$ with a precision of $.01$ how can I know how many terms I need? I think it has something to do with Taylor's remainder.
Thanks for your help!
 A: Taylor's Theorem states that if a function is of class $C^{n+1}$ ($n+1$ times continuously differentiable) then the difference between the function and its Taylor Polynomial at some point $x$ will be $$\text{Error}=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$ where $a$ is the center of the Taylor expansion and $c$ is some number $c\in(a,x)$. ($n$ is the degree of the approximation, or $1$ less than the number of terms you have.)
So if you can bound the derivative in that expression, then you can bound the error. In your case, the function is $e^x$. We have $$\frac{d^{n+1}}{dx^{n+1}}e^x=e^x$$ so the derivative, on the interval $(a,b)$, is at most $e^{\max\{a,b\}}$ because $e^x$ is strictly increasing everywhere. In your case $b=-1/2$ so we get $$|\text{Error}|\le \frac{e^0}{(n+1)!}|-1/2-0|^{n+1}=\frac{1}{2^{n+1}(n+1)!}$$ This is an excellent bound. To get $0.01$ precision just take $n$ large enough so that $$\frac{1}{2^{n+1}(n+1)!}\le 0.01$$ (I believe $n=3$ will work)
With $n=3$, we have $$e^{-1/2}\approx 1-{1\over 2}+\frac{1}{2!\cdot 2^2}-\frac{1}{3!\cdot 2^3}=0.604166...$$ The true answer is $0.6065306597126334...$, and the error is about $0.002<0.01$.
A: For each $n$ you can usually tell whether $n$ terms are enough by computing an upper bound of the remainder. Recall that
$f(x)=\Sigma_{k=0}^nf^{(k)}(x_0)(x-x_0)^k/k! + f^{(n+1)}(x_*)(x-x_0)^{n+1}/(n+1)!$,
where $x_*\in[x_0,x]$.
If you can produce a bound on $f^{(n+1)}(x_*),x_*\in[x_0,x]$ (which is easy enough for convex functions such as $e^x$), then you can pick $n$ where the remainder is below your desired precision threshold.
