Is $\frac{1}{(n) !}$ a good bound for the maximum possible error for the n^th degree Taylor polynomial about x = 0 approximating $\sin \left(\frac{x}{2}\right)$ on the interval [0, 1]?

Question

Quizplus · Accepted Answer

The maximum possible error for the nth degree Taylor polynomial approximation of a function is given by the remainder term of the Taylor series, which is not simply $\frac{1}{(n)!}$ but depends on the (n+1)th derivative of the function being approximated. For $\sin\left(\frac{x}{2}\right)$, the error bound involves the maximum value of the (n+1)th derivative of $\sin\left(\frac{x}{2}\right)$ on the interval [0, 1], scaled by $\frac{(x-0)^{n+1}}{(n+1)!}$, not just $\frac{1}{(n)!}$.

Is 1(n)!\frac{1}{(n) !}(n)!1​ A Good Bound for the Maximum Possible Error for the Error

Is $\frac{1}{(n) !}$ A Good Bound for the Maximum Possible Error for the Error