# Is it possible that Taylor Remainder be a negative number?

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?

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

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.