# Find reasonable upper-bound on error in approximating Taylor series $f(x)=\sin(5x)$

The question pertains to the upper-bound on the error of a Taylor polynomial. I understand the basics of Taylor polynomials and Lagrange error calculation, but how is the $|x-1| ≤ 0.1$ used here in the process of solving for error?

Find a reasonable upper-bound on the error in approximating $f(x)=\sin(5x)$ by its 3rd order Taylor polynomial $P_{3}(x)$ about $a=1$ valid for all values of $x$ such that $|x-1| ≤ 0.1$

$$|E_{n}(x)|≤\frac{M_{n+1}|x-a|^{n+1}}{(n+1)!}$$

$M_{n+1}=\max(|f^{(n+1)}(x)|,$0.9 ≤ |x| ≤ 1.1$)$

So far:

Maximum absolute value is $M=59.904$ from $f^{(n+1)}(0.9)$

Would the next step be to add this to Lagrange formula as such?

$$|E_{3}(x)|≤\frac{(59.904)|1.1-1|^{4}}{(4)!}$$

• It affects the interval where you compute $M_{n+1}$, as well as the $|x-a|^{n+1}$ term. – Ian Sep 3 '17 at 15:34
• So, interval would be [1, 1.1] since $a = 1$ and $|x-1| ≤ 0.1$? Is a similar rule followed in other cases? – Bgeo25 Sep 3 '17 at 15:43
• You have an absolute value so the interval goes both ways, giving $[0.9,1.1]$. It's not really specific to Taylor problems, it's just a basic property of absolute value. – Ian Sep 3 '17 at 15:44