To bound the Error of the approximation $\sin(x)\approx x$ for $-\frac{\pi}4\le x\le \frac{\pi}{4}$ I used Taylor Remainder formula and I get $R_2(1)=\frac{-\sqrt{2}}{12}$. I want to make sure that is it possible that this value be a negative number?

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    $\begingroup$ It's possible that the remainder has a negative sign, however we say the error is the absolute value of the remainder. $\endgroup$ – Ian Dec 27 '20 at 19:23
  • $\begingroup$ @Ian Thank you I got it. $\endgroup$ – Soheil Dec 27 '20 at 19:24

Yes it's possible.

Let's take a Taylor development around $0$ :

$$f(x) = a_0 + a_1 x + a_2 x^2 + ...{}{}{}$$

You can have a function $f$ with whatever coefficient you want.

Then if you take one such, for example, all $a_k<0$, then when you evaluate with a $x>0$, you always get a negative remainder.

A simple example is to take a polynomial with only negative coefficients : then its Taylor development is the polynomial (if you're at a rank high enough).


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