1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.