answer: 4x 1/(3^5 * 5!) #11 - the taylor seri...
answer: 4x 1/(3^5 * 5!) #11 - the taylor series for a function f about x = 0 is given by sum from n = 1 to infinity of ((-1)^(n + 1) * x^n)/n! and converges to f for all numbers x. if the fourth - degree taylor polynomial for f about x = 0 is used to approximate f(1/3), what is the alternating series error bound?
Answer
# Answer:
$\frac{1}{3^{5}\cdot5!}$
# Explanation:
## Step1: Recall error - bound formula
For an alternating series $\sum_{n = 1}^{\infty}(- 1)^{n + 1}a_{n}$ (where $a_{n}\gt0$, $a_{n+1}\leq a_{n}$ and $\lim_{n\rightarrow\infty}a_{n}=0$), the error bound $E_N$ when using the sum of the first $N$ terms $S_N$ to approximate the sum $S$ of the series is $|E_N|\leq a_{N + 1}$.
## Step2: Identify relevant terms for Taylor series
The Taylor series of a function $f(x)$ about $x = 0$ is $\sum_{n=1}^{\infty}\frac{(-1)^{n + 1}x^{n}}{n!}$. When using the fourth - degree Taylor polynomial (sum of the first 4 terms) to approximate $f(\frac{1}{3})$, we consider the alternating - series error bound. Here, $x=\frac{1}{3}$ and since we used $N = 4$ terms, the error bound is given by the absolute value of the fifth - term of the series.
## Step3: Calculate the fifth - term
The general term of the series is $a_{n}=\frac{|x|^{n}}{n!}$. Substituting $x=\frac{1}{3}$ and $n = 5$ into the formula, we get $a_{5}=\frac{(\frac{1}{3})^{5}}{5!}=\frac{1}{3^{5}\cdot5!}$. So the alternating - series error bound is $\frac{1}{3^{5}\cdot5!}$.