I need to approximate $e^{0.1}$ using Taylor polynomials. My error margin is $5\cdot 10^{-5}$. I also need to determine of what degree should the polynomials be in order to be within the margin.
I'm very new to Taylor series, and I don't understand how should I approach the problem.
What I know is that:
$e^{0.1}=\sum_{n=0}^\infty \frac{(0.1)^n}{n!}$. If I let $T_n(x)$ be the partial sum of the series, my Taylor polynomial is
$T_n(0.1)=\sum_{i=0}^n \frac{(0.1)^i}{i!}=1+0.1+0.01*\frac{1}{2!}+...$
I understand that, as $n\to\infty$, the more precise my approximation is. What I don't understand is how can I determine at what point my approximation will be within the error margin? How can I find the $k$ such that $e^{0.1}-T_n(0.1)<5*10^{-5}$ for $n\geq k?$
Solution 1: You may want to use one of the explicit formulas for the remainder in Taylor polynomial (see https://en.wikipedia.org/wiki/Taylor%27s_theorem), e.g. Lagrange form:
$$R_k(x)=\frac{f^{(k+1)}(\xi)}{(k+1)!}(x-a)^{k+1}$$
for some $\xi$ between $x$ and $a$. In your case, $a=0, x=0.1, f^{(n)}(\xi)=e^\xi$ for every $n$, and so the error term is:
$$R_k(0.1)=\frac{e^\xi}{(k+1)!}0.1^{k+1}$$
Also $e^\xi<e^{0.1}$ as $0<\xi<0.1$ so you only need to try to solve:
$$\frac{e^{0.1}}{(k+1)!}0.1^{k+1}<5\cdot 10^{-5}$$
which you can check directly for some small values of $k$. I will leave it to you to finish the calculation.
Solution 2: You may want to approximate the remainder by bounding it "from above" by a geometric progression:
$$\begin{array}{rl}\frac{x^{k+1}}{(k+1)!}+\frac{x^{k+2}}{(k+2)!}+\frac{x^{k+3}}{(k+3)!}+\ldots&=\frac{x^{k+1}}{(k+1)!}\left[1+\frac{x}{k+2}+\frac{x^2}{(k+2)(k+3)}+\ldots\right]\\&<\frac{x^{k+1}}{(k+1)!}\left[1+\frac{x}{k+2}+\frac{x^2}{(k+2)^2}+\ldots\right]\\&=\frac{x^{k+1}}{(k+1)!}\frac{1}{1-\frac{x}{k+2}}\end{array}$$
so it is enough to make sure you've found $k$ such that:
$$\frac{0.1^{k+1}}{(k+1)!}\frac{1}{1-\frac{0.1}{k+2}}<5\cdot 10^{-5}$$
which is again checked easily with a small $k$.