# Taylor series: Assuming the x-value for which there is maximum magnitude of 6th derivative?

Problem

Suppose you know that $$f^n(4)=\frac{(-1)^nn!}{3^n(n+1)}$$ and the Taylor series of $$f$$ centered at $$4$$ converges to $$f(x)$$ for all $$x$$ in the interval of convergence. Show that the fifth-degree Taylor polynomial approximates $$f(5)$$ with error less than $$0.0002$$.

I can prove the statement but in order to do so I have to assume that the maximum value of $$|f^6(x)|$$ on $$3\leq x\leq 5$$ is at $$x=4$$. Is there something I am missing where I can conclude the maximum value is at $$x=4$$ based on the information the problem provides?

You can estimate it directly as follows:

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(4)}{n!}(x-4)^n = \sum_{n=0}^{\infty}\frac{(-1)^n}{3^n(n+1)}(x-4)^n$$

has a radius of convergence of $$3$$.

So,

$$f(5) = \sum_{n=0}^{\infty}\frac{(-1)^n}{3^n(n+1)}$$

The searched for difference to the corresponding value of the $$5$$th- degree Taylor polynomial is

$$\left|f(5) - \sum_{n=0}^{\color{blue}{5}}\frac{(-1)^n}{3^n(n+1)}\right| = \left|\sum_{n=\color{blue}{6}}^{\infty}\frac{(-1)^n}{3^n(n+1)}\right|$$ $$= \frac{1}{3^6}\left|\sum^{\infty}_{n=6}\frac{(-1)^n}{3^{n-\color{blue}{6}}(n+1)}\right| = \frac{1}{3^6}\left|\sum^{\infty}_{n=\color{blue}{0}}\frac{(-1)^n}{3^n(n+\color{blue}{7})}\right| \leq \ldots$$

Now note that the last series is an alternating one ($$S = \sum_{n=0}^{\infty}(-1)^na_n$$) with decreasing $$a_n \geq 0$$. So, you can always estimate $$S \leq S_n + a_{n+1}$$ where $$S_n =\sum_{k=0}^{n}(-1)^ka_k$$:

Hence, $$\ldots \leq \frac{1}{3^6}\left( \underbrace{\left(\frac{1}{7} - \frac{1}{3\cdot 8}\right)}_{=S_2} + \underbrace{\frac{1}{3^2\cdot 9}}_{= a_3}\right) \boxed{< 0.0002}$$

• How did you get from $$\left|\sum_{n=\color{blue}{6}}^{\infty}\frac{(-1)^n}{3^n(n+1)}\right|$$ to $$\frac{1}{3^6}\left|\sum^{\infty}_{n=6}\frac{(-1)^n}{3^n(n+1)}\right|$$
– user532874
Nov 11, 2019 at 18:38
• The largest brackets are supposed to be absolute value signs right?
– user532874
Nov 11, 2019 at 19:16
• I do not understand your question. The brackets in the last line are brackets and are not supposed to indicate any absolute value. The expression within the brackets is positive. For the estimation I use at the end you may refer to en.wikipedia.org/wiki/Alternating_series_test Nov 11, 2019 at 19:46