[Math] Estimate the maximum possible error in the approximation – Taylor Polynomials

calculusmultivariable-calculus

I am given the following example.

Estimate the maximum possible error in the approximation.

$cos(x) = T_{\displaystyle4,\frac{\displaystyle\pi}{\displaystyle2}}$ for $x \in \left[\frac{\displaystyle\pi}{\displaystyle2} -0.2, \frac{\displaystyle\pi}{\displaystyle2} + 0.2\right]$. $\\$

$R_{\displaystyle4,\frac{\pi}{2}}(x) \le \frac{\displaystyle|-\sin(z)|}{\displaystyle5!} |x-\frac{\pi}{2}|^5$ $\\$

$\Rightarrow R_{\displaystyle4,\frac{\displaystyle\pi}{\displaystyle2}}(x) \le \frac{\displaystyle1}{\displaystyle 5!}(0.2)^5 $

I don't understand how the example went from $|x-\frac{\pi}{2}|^5$ to $(0.2)^5$. Please specify the logical reasoning behind this and exactly how it is achieved. Thank you.

Best Answer

We have that

$$\dfrac{\pi}{2}-0.2\le x\le\dfrac{\pi}{2}+0.2$$

and that

$$ \cos(x)\approx 1-\dfrac{\left(x-\frac{\pi}{2}\right)^2}{4}+\dfrac{\left(x-\frac{\pi}{2}\right)^4}{24}$$

Thus we know that

$$ \dfrac{\pi}{2}-0.2\le x\le\dfrac{\pi}{2}+0.2$$

from which it follows that

$$ -0.2\le x-\dfrac{\pi}{2}\le 0.2$$

One should point out, however, that since this is an alternating series, a better estimation of the error would be to use the next term of the series to get $$ R<\left\vert\dfrac{\left(x-\frac{\pi}{2}\right)^6}{6!}\right\vert\le\dfrac{(0.2)^6}{720} $$