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I might be missing some background knowledge on this subject, but nevertheless I am interested. In some cases like this, the answers talk about finding the taylor series for $\cos(x)$ and then substituting the first two terms of the series in instead of $\cos(x)$ i.e in the link they solve for $x$ in

$$\frac{x^2}{2} \leq \frac{1}{2}\cdot 10^{-8}$$

Rather than $$1-\cos(x) \leq \frac{1}{2}\cdot 10^{-8}$$

So my question is - why is this allowed? And why not include the 3rd term from the Taylor series as well?

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    $\begingroup$ Because the "tail" of the series (all terms after the first two) will not sum to more that the acceptable error. $\endgroup$ – user247327 Sep 17 at 16:34
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This is allowed because the Taylor series for $\cos x$ is an alternating series, and they use Leibniz' theorem for alternating series: the error is bounded by the first missing term (in absolute value) and has the same sign.

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    $\begingroup$ As long as the terms are decreasing in absolute value. $\endgroup$ – marty cohen Sep 17 at 18:01
  • $\begingroup$ @martycohen: Yes, but in the case of $\cos x$, they're ultimately decreasing, and I suppose $|x|<1$ in practise. $\endgroup$ – Bernard Sep 17 at 19:52
  • $\begingroup$ The Lagrange remainder gives a cleaner argument in this particular case: $|\cos(x)-1|=|\cos(x)-1-0x|=\cos(\xi) \frac{x^2}{2}$ for some $\xi$ between $0$ and $x$. In particular, the $|x|<1$ assumption is not technically necessary to get $|\cos(x)-1| \leq x^2/2$, though of course without this assumption the bound gets rather bad. $\endgroup$ – Ian Sep 21 at 15:55

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