# Finding the approximate value given Taylor series representation and relative approximate error

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.

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

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.