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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)|,$0.9 ≤ |x| ≤ 1.1$)$

[Edit]

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?

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

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  • $\begingroup$ It affects the interval where you compute $M_{n+1}$, as well as the $|x-a|^{n+1}$ term. $\endgroup$ – Ian Sep 3 '17 at 15:34
  • $\begingroup$ So, interval would be [1, 1.1] since $a = 1$ and $|x-1| ≤ 0.1$? Is a similar rule followed in other cases? $\endgroup$ – Bgeo25 Sep 3 '17 at 15:43
  • $\begingroup$ 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. $\endgroup$ – Ian Sep 3 '17 at 15:44

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