calculus - calculator allowed\n1. the fourth - degree maclaurin polynomial for cos x is given by 1…

Explanation:

Step1: Recall Lagrange error - bound formula

The Lagrange error - bound for the $n$th - degree Taylor polynomial $P_n(x)$ of a function $f(x)$ centered at $a$ is given by $|R_n(x)|=\left|\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^{n + 1}\right|$, where $c$ is some number between $a$ and $x$. For a Maclaurin polynomial ($a = 0$), it becomes $|R_n(x)|=\left|\frac{f^{(n + 1)}(c)}{(n+1)!}x^{n + 1}\right|$. The Maclaurin series for $y=\cos x$ is $\sum_{k = 0}^{\infty}\frac{(-1)^k}{(2k)!}x^{2k}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$. The fourth - degree Maclaurin polynomial for $\cos x$ is $P_4(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}$, so $n = 4$.

Step2: Find the $(n + 1)$th derivative of $y=\cos x$

The derivatives of $y = \cos x$ follow a pattern: $y'=-\sin x$, $y''=-\cos x$, $y'''=\sin x$, $y^{(4)}=\cos x$, $y^{(5)}=-\sin x$. So, $f^{(n + 1)}(x)=f^{(5)}(x)=-\sin x$. Then, $|f^{(5)}(c)|=|-\sin c|\leq1$ for all $c$.

Step3: Identify $x$ value

We want to approximate $\cos\frac{\pi}{6}$, so $x=\frac{\pi}{6}$.

Step4: Calculate the error bound

Using the Lagrange error - bound formula $|R_4(x)|=\left|\frac{f^{(5)}(c)}{(5)!}x^{5}\right|$. Since $|f^{(5)}(c)|\leq1$, we have $|R_4\left(\frac{\pi}{6}\right)|\leq\frac{1}{5!}\left(\frac{\pi}{6}\right)^{5}$. We know that $5!=5\times4\times3\times2\times1 = 120$ and $\left(\frac{\pi}{6}\right)^{5}=\frac{\pi^5}{7776}$. So, $|R_4\left(\frac{\pi}{6}\right)|\leq\frac{\pi^5}{120\times7776}\approx\frac{306.019}{933120}\approx3.28\times 10^{-4}$.

Answer:

$|R_4\left(\frac{\pi}{6}\right)|\leq\frac{\pi^5}{120\times7776}\approx3.28\times 10^{-4}$