How to use Taylor's Theorem to obtain an upper bound for an error approximation

Question

$e \approx 1 + 1 + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} + \frac{1^5}{5!}$ must find upper bound for this but I don't see what I should be doing. The remainder/error is given by $\frac{f^{n+1}(z)(x-a)^{n+1}}{(n+1)!}$ but what is the function? I represented the approximation to be $\sum_{n=0}^5 \frac{1^n}{n!}$ this doesn't help as I cannot take the derivative of a factorial (can i?)

Answer 1

If I understand correctly you want to approximate $e^1$ by the value of the function $f(x)=e^x$ which has the value $f(0)=1$.

By the Taylor theorem you have that $f(1)=\sum_{n=0}^5 \frac{f^{(n)}(0)}{n!}(1-0)^n+R_5(f)$.

Evaluating $R_5(f)$ is by the Lagrange form of the reminder which is the value of the next factor in the sum $\frac{f^{(6)}(c)}{6!}(1-0)^6$ when $c\in(0,1)$ [lagrange mid value theorem for derivatives]