I am presently struggling in finishing this problem:
Determine the Taylor series representation of the function $f(x)=e^{3x}$. Then, determine the approximate value of $e^{1.2}$ using the series with a relative approximate error of less than 0.1%.
Finding the Taylor series representation is easy. It is equal to $\sum_{n=0}^{\infty } \frac{3x^{n}}{n!}$.
What should I do with the second part of the problem? If the formula $f(x)=T_{n}(x)+R_{n}(x)$ is not applicable for this case, then I couldn't find a formula elsewhere that relates the formula for relative error and the approximate value. I suspect that Taylor Polynomial should be applied. If it is to be applied, then how can I obtain the degree based on the relative approximate error?
What I actually thought is to use the relative error formula directly since I already know the values for the relative error (0.001) and the true value ($e^{1.2}$). However, I know that that is the wrong approach because there are given values in the problem.
Any help would be appreciated. Thanks.
It follows from the Taylor formula with Lagrange remainder that $$ 0 \le e^x - \sum\limits_{n = 0}^N {\frac{{x^n }}{{n!}}} \le \frac{{x^{N + 1} }}{{(N + 1)!}}e^x $$ for all $x\geq 0$ and $N\geq 0$. Thus $$ 0 \le 1 - e^{ - x} \sum\limits_{n = 0}^N {\frac{{x^n }}{{n!}}} \le \frac{{x^{N + 1} }}{{(N + 1)!}} $$ for all $x\geq 0$ and $N\geq 0$. Thus, you want to find an $N$, such that $$ \frac{{1.2^{N + 1} }}{{(N + 1)!}} < \frac{1}{{1000}}. $$ It is found that any $N\geq 6$ is fine.