So I was given the following prompt:
"Let $f$ be a function that has derivatives of all orders for all real numbers, and let $P_3(x)$ be the third-degree Taylor polynomial for $f$ about $x=0$. The Taylor series for $f$ about $x=0$ converges at $x=1$, and $|f^{(n)}(x)|\leq\frac{n}{n+1}$, for $1\leq n\leq4$ and all values of $x$. What is the smallest value of $k$ for which the Lagrange error bound guarantees that $|f(1)-P_3(1)|\leq k$?"
I guess I'm confused about what the application of the formula used to find the Lagrange bound would look like in a situation like this. I understand that given an equation I would have to apply it to the formula of $|S-S_k|\leq b_{k+1}$ where the error is bounded by $b_{k+1}$, but I'm confused over how I'd apply that knowledge in this context. Any help would be appreciated!
Best Answer
The Lagrange error bound of a Taylor polynomial gives the worst-case scenario error of the Taylor approximation on some interval. It levarages the fact that a Taylor-approximation of order n has an error of order n+1. More preciesely $$(\forall x_0 \in I), (\forall n \in \mathbb{Z_+}), (\exists \xi \in I), \text{ s.t. } f(x) - T_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-x_0)^{n+1}$$ This error term can be bounded by above as follows: $$\bigg| \frac{f^{(n+1)}(\xi)}{(n+1)!} \bigg| \le \max_{t \in I} \bigg|\frac{f^{(n+1)}(t)}{(n+1)!} \bigg| = M_n$$ And this is the Lagrange error bound of the n-th order Taylor polynomial.