Answer
$|R_5|\leq0.003775391428$
Work Step by Step
$e\approx1+1+\frac{1^2}{2!}+\frac{1^3}{3!}+\frac{1^4}{4!}+\frac{1^5}{5!}$
$f(x)=e^x$
$x=1$
$a=0$ (the function is centered at a=0)
$n=5$
To find the error of the function, use Lagrange Error Bound. The formula for Lagrange Error Bound is:
$|R_n|\leq\frac{max|f^{n+1}c||x-a|^{n+1}}{(n+1)!}$
$|R_5|\leq\frac{max|f^{5+1}c||1-0|^{5+1}}{(5+1)!}$
$|R_5|\leq\frac{max|f^{6}c||1|^{6}}{6!}$
To find the $max|f^6c|$:
$f(x)=e^x$
$f'(x)=e^x$
$f''(x)=e^x$
$f'''(x)=e^x$
$f^4(x)=e^x$
$f^5(x)=e^x$
$f^6(x)=e^x$
Therefore:
$f^6(1)=e^1$.
$|R_5|\leq\frac{e|1|^{6}}{6!}$
$|R_5|\leq0.003775391428$
