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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.

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    $\begingroup$ The series of $e^{3x}$ is $\sum_{n=0}^{\infty } \frac{(3x)^{n}}{n!}$ and not $\sum_{n=0}^{\infty } \frac{3x^{n}}{n!}$ $\endgroup$ Commented Aug 24, 2020 at 4:47

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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.

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