use the error bound for taylor polynomials to give a good bound for the error for the nth degree taylor…

Explanation:

Step1: Recall Taylor - Remainder formula

The 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 $f(x)=\cos(x)$ centered at $a = 0$, we have $f^{(k)}(x)=\pm\cos(x)$ or $f^{(k)}(x)=\pm\sin(x)$.

Step2: Find an upper - bound for $|f^{(n + 1)}(c)|$

The derivatives of $y = \cos(x)$ are $f^{(1)}(x)=-\sin(x)$, $f^{(2)}(x)=-\cos(x)$, $f^{(3)}(x)=\sin(x)$, $f^{(4)}(x)=\cos(x)$ and the cycle repeats. So, $|f^{(k)}(x)|\leq1$ for all $x$ and all non - negative integers $k$. In particular, for $c$ between $0$ and $x\in[0,1.4]$, $|f^{(n + 1)}(c)|\leq1$.

Step3: Substitute into the error - bound formula

Since $a = 0$ and $|f^{(n + 1)}(c)|\leq1$, the error bound formula $|R_n(x)|=\left|\frac{f^{(n + 1)}(c)}{(n+1)!}(x - a)^{n + 1}\right|$ becomes $|R_n(x)|\leq\frac{|x|^{n + 1}}{(n+1)!}$.

Step4: Consider the interval $[0,1.4]$

On the interval $[0,1.4]$, $|x|\leq1.4$. So, $|R_n(x)|\leq\frac{(1.4)^{n + 1}}{(n+1)!}$ for $x\in[0,1.4]$.

Answer:

$\frac{(1.4)^{n + 1}}{(n+1)!}$