# Finding Lagrange Error Bound

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!

• Hello. What does 'all values of $f_1$ mean? Commented May 17, 2021 at 13:43
• Should be fixed, thanks for pointing that out.
– joe
Commented May 17, 2021 at 13:46
• Then all you need is $\displaystyle |f(1) - P_3(1)| = \left\lvert \frac{1^4}{4!}f^{(4)}(\xi)\right\rvert$ for $\xi \in (0,1)$ and the right side is bounded by $\frac{4}{5} \cdot \frac{1}{4!}$. That gives the best estimate for $k$ given the available information. Commented May 17, 2021 at 13:50

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.