Taylor Polynomial with E

Question

Estimate $e^{0.1}$ to $6$ decimal places using a Taylor polynomial about $0$. Use error bounding to prove that your estimate is accurate to at least 6 decimal places.

I think I will have to use the formula for Error bounding in some way. However, I am not sure how to compute enough derivatives to form the entire polynomial, and I am also not sure how to make sure that the estimate is accurate to 6 places. Does anyone have ideas?

Answer 1

Use the Lagrange form of the remainder: for all positive integers $n$, $$ e^x = \sum_{k=0}^n \frac{x^k}{k!} + R_n(x) $$ where $$R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} $$ for some $\xi$ between $0$ and $x$ (here $f(x) = e^x$). Obviously, $f^{(n)}(x) = e^x$ for all $x$, which is strictly greater than one on $(0,\infty)$, so this error is maximized at $\xi=\frac1{10}$. So we have the inequality $$ \frac{\left(e^{\frac1{10}}\right)}{(n+1)!}e^{\frac1{10}} < 0.0000005. $$ Clearly the left-hand side is monotone decreasing in $n$, and the inequality is satisfied for $n=9$ but not for $n=8$. So take $n=9$, for an error of $$ 2.783427*10^{-7}. $$

Answer 2

Starting from @Math1000's answer, suppose that you want $k$ decimal places, you need to solve $$\frac{\left(e^{\frac1{10}}\right)}{(n+1)!}e^{\frac1{10}} = 10^{-k}\implies (n+1)!=e^{\frac1{5}}10^k$$ For the time being, let $m=n+1$ and we need the inverse factorial of $m$.

Have a look at @rogjohn's answer in this post and notice that the approximation of $$m!=a \implies m \sim \exp\left(1+W\left(\frac{\log \left(\frac{a^2}{2 \pi }\right)}{2 e}\right) \right)-\frac 12$$ Applied to your case, this would give $$n \sim \exp\left(1+W\left(\frac{10 k \log (10)+2-5 \log (2 \pi )}{10 e}\right)\right)-\frac 32$$ and you will use $\lceil n \rceil$.

Below is given a table for a few values of $k$ $$\left( \begin{array}{cccc} k & n & \lceil n \rceil & \frac{\left(e^{\frac1{10}}\right)}{(\lceil n \rceil+1)!}e^{\frac1{10}}\\ 1 & 2.52768 & 3 & 5.08918 \times 10^{-2}\\ 2 & 4.00593 & 5 & 1.69639 \times 10^{-3}\\ 3 & 5.27585 & 6 & 2.42342 \times 10^{-4}\\ 4 & 6.43054 & 7 & 3.02927 \times 10^{-5}\\ 5 & 7.50863 & 8 & 3.36586\times 10^{-6} \\ 6 & 8.53069 & 9 & 3.36586\times 10^{-7} \\ 7 & 9.5093 & 10 & 3.05987\times 10^{-8} \\ 8 & 10.4528 & 11 & 2.54989\times 10^{-9} \\ 9 & 11.3672 & 12 & 1.96146\times 10^{-10} \\ 10 & 12.2567 & 13 & 1.40106\times 10^{-11} \\ 11 & 13.1248 & 14 & 9.34027\times 10^{13} \\ 12 & 13.9741 & 14 & 9.34027\times 10^{13} \end{array} \right)$$