Answer
$|R_4|\leq2.025\times10^{-5}$
Work Step by Step
$cos(0.3)\approx1-\frac{(0.3)^2}{2!}+\frac{(0.3)^4}{4!}$
$f(x)=cosx$
$x=0.3$
$a=0$ (function is centered at 0)
$n=4$
To find the error of the function, use Lagrange Error Bound. The formula for Lagrange Error Bound is:
$|R_n|\leq\frac{max|f^{n+1}c||x-a|^{n+1}}{(n+1)!}$
$|R_4|\leq\frac{max|f^{4+1}c||0.3-0|^{4+1}}{(4+1)!}$
$|R_4|\leq\frac{max|f^{5}c||0.3|^{5}}{5!}$
To find the $max|f^{5}c|$:
$f(x)=cosx$
$f'(x)=-sinx$
$f''(x)=-cosx$
$f'''(x)=sinx$
$f^4(x)=cosx$
$f^5(x)=-sinx$
$max|f^{5}c|=max|-sinc|=max|sinc|$
The maximum of sin(c) will always be $1$ because the upper limit for the range of sinx is always $1$. Therefore, the $max|sinc|=1$:
$|R_4|\leq\frac{1||0.3|^{5}}{5!}$
$|R_4|\leq2.025\times10^{-5}$
