So I have a homework question asking me to find the absolute error of the sine function. I’m told the real equation is

$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots $$

And to approximate using

$$ \sin(x) \approx x $$

For values $ x = 0.1 , 0.5 , 1.0 $

So I know I just want to find the absolute value of the difference if the two numbers, but I need to know the actual value of $\sin(x)$ to determine these values and I can’t do that because my calculator is most likely making some kind of rounding error and trying to solve analytically returns an infinite sum. I know the question is supposed to be simple, but hey, I’m simple myself. The question just feels very catch 22.

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    $\begingroup$ What's "absolute error" mean? Are you simply wanting to find an upper bound for the error? If so, note that this is an alternating series, who's terms are decreasing in magnitude for the given values of $x$. $\endgroup$ – Simply Beautiful Art Sep 3 '19 at 23:56
  • $\begingroup$ @SimplyBeautifulArt I’m not entirely sure to be honest, the question just asks for forward/backward error using absolute error (instead of relative error) - which I believe requires me to use an exact value for the actual value of sin(x), which I think is impossible. So I’m probably misunderstanding something. I believe if the question is asking me for big O I could say it is just x, but the way I’m reading it my teacher is asking for something else. $\endgroup$ – saucin' Sep 4 '19 at 0:04

Supposing the intention is to give the actual error, you can simply do $|\sin(x)-x|$ into some calculator. Supposing what they wanted was actually an upper bound to this absolute error, it suffices to notice this is an alternating series with terms decreasing in magnitude, and hence we have the upper bound of $|\sin(x)-x|<\frac{|x|^3}{3!}$ for all $|x|\le1$.


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