I have the following equation:$x^2=\cos x$ and calculating the Taylor series of $3rd$ degree around $0$ I've got: $x\approx \pm\sqrt{\frac{2}{3}}$
However, now I need to prove that if x is a solution, then $\frac{\sqrt{5}}{3}\le|x|\le\frac{\sqrt{7}}{3}$
Bounding the remainder of the Taylor expansion we have: $|R|\le \frac{|x^3|}{6}$ ,then I don't know what to do, what should I do next?.
Any help will be appreciated.
Best Answer
The function $$f(x):=\cos x-x^2=1-{3\over2}x^2+{1\over24}x^4-\ldots$$ is monotonically decreasing for $x>0$, since $f'(x)=-\sin x -2x<0$ for positive $x$. Therefore it has at most one positive zero $\xi$. When $|x|<1$ the subsequent terms of the Taylor series are decreasing in absolute value; therefore we know that $$1-{3\over2}x^2<f(x)<1-{3\over2}x^2+{1\over24}x^4\qquad(0<x<1)\ .$$ It follows that $$f\left({\sqrt{5}\over3}\right)>1-{3\over2}\cdot{5\over9}={1\over6}>0$$ and $$f\left({\sqrt{7}\over3}\right)<1-{3\over2}\cdot{7\over9}+{49\over24\cdot 81}<-{1\over6}+{54\over24\cdot 81}=-{5\over36}<0\ .$$ Therefore $\xi$ lies between the claimed bounds.