Taylor approximation is different from the actual value

Question

The problem is to use the $n$th Taylor polynomial to approximate the value of $F(x)=x^{2}\cos{x}$ with $n=3,x_{0}=-2,x=-1.09$.

I found that the $3$rd Taylor polynomial with remainder term is \begin{align*} x^{2}\cos{x} =& 4\cos(2)+[-4\cos(2)+4\sin(2)][x+2]+[-\cos(2)-4\sin(2)][x+2]^{2}+ \\ & [2\cos(2)+ \dfrac{1}{3}\sin(2)][x+2]^{3} + [\dfrac{(x+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))] \end{align*}

approximating $x=-1.09$ using the Taylor polynomial,

\begin{align*} (-1.09)^{2}\cos(-1.09) =& 4\cos(2)+[-4\cos(2)+4\sin(2)][(-1.09)+2]+[-\cos(2)-4\sin(2)][(-1.09)+2]^{2}+ \\ & [2\cos(2)+ \dfrac{1}{3}\sin(2)][(-1.09)+2]^{3} + [\dfrac{((-1.09)+2)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))] \\ =& 0.0938985 + [\dfrac{((0.91)^{4}}{24}][-8\cos(\xi(x))+16\sin(\xi(x))] \end{align*}

However, $(-1.09)^{2}\cos(-1.09)=0.549479$ which is a lot of decimals different from Taylor approximation of $0.0938985$.

Is there something that I miss?

Answer 1

This just means that teh third order is not sufficient in particular because $-1.09$ is very far away from $-2$.

To make the story short, what you have is $$x^2 \cos(x)=\sum_{n=0}^\infty \frac{\left(-n^2+n+4\right) \cos \left(2-\frac{\pi n}{2}\right)+4 n \sin \left(2-\frac{\pi n}{2}\right)}{n!}(x+2)^n$$ So, if you truncate to $O((x+2)^{p+1})$ you will have the following results $$\left( \begin{array}{cc} 2 & 0.4926841712 \\ 3 & 0.0938985194 \\ 4 & 0.6047229336 \\ 5 & 0.5723469909 \\ 6 & 0.5466012534 \\ 7 & 0.5489493571 \\ 8 & 0.5495411050 \\ 9 & 0.5494858556 \\ 10 & 0.5494781119 \\ 11 & 0.5494788050 \\ 12 & 0.5494788702 \\ 13 & 0.5494788647 \\ 14 & 0.5494788644 \end{array} \right)$$

Answer 2

Expand just the cosine series around $x=-2$ and you will have: $$x^2cos x = x^2\left(cos(2) + sin(2)[x+2] - \frac{cos(2)[x+2]^2}{2} - \frac{sin(2)[x+2]^3}{6}\right )$$ $$\\$$ (Expanding the $x^2$ in a Taylor polynomial and multiplying everything by brute force also works, but the Taylor series for $x^2$ will just reduce back to $x^2$ anyway, so it is not necessary)

$$\\$$ Evaluated at $x=-1.09$:

$$ (-1.09)^2cos(-1.09) \approx 0.549479$$ $$\\$$ $$(-1.09)^2\left(cos(2) + sin(2)[0.91] - \frac{cos(2)[0.91]^2}{2} - \frac{sin(2)[0.91]^3}{6}\right )\approx 0.557713$$