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Prove an inequality for the error of the Taylor polynomial of cos x

Prove that the error in the Taylor expansion of \cos x satisfies the following inequality.

    \[ \sum_{k=0}^n \frac{(-1)^k x^{2k}}{(2k)!} + E_{2n+1} (x), \qquad |E_{2n+1} (x)| \leq \frac{|x|^{2n+2}}{(2n+2)!}. \]


Proof. As in the previous exercise we know all of the derivatives of f(x) = \cos x are bounded by \pm 1 (since all of the derivatives are -\sin x, -\cos x, \sin x, or \cos x). Therefore we can apply Theorem 7.7 (page 280) with a = 0, m = -1 and M = 1. For x > 0 we have

    \begin{alignat*}{3}  & m \frac{(x-a)^{n+1}}{(n+1)!} &\leq E_n (x) &\leq M \frac{(x-a)^{n+1}}{(n+1)!} \\[9pt]  \implies & - \frac{x^{n+1}}{(n+1)!} &\leq E_n (x) &\leq \frac{x^{n+1}}{(n+1)!} \\[9pt]  \implies & - \frac{x^{2n+2}}{(2n+2)!} &\leq E_{2n+1} (x) &\leq \frac{x^{2n+2}}{(2n+2)!} \\[9pt]  \implies & \phantom{-} | E_{2n+1} (x) | &\leq \frac{x^{2n+2}}{(2n+2)!}. \end{alignat*}

For x < 0 we have

    \begin{alignat*}{3}  & m \frac{(a-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1}E_n (x) &\leq M \frac{(a-x)^{n+1}}{(n+1)!} \\[9pt]  \implies & - \frac{(-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1} E_n (x) &\leq \frac{(-x)^{n+1}}{(n+1)!} \\[9pt]  \implies & - \frac{(-x)^{2n+2}}{(2n+2)!} &\leq (-1)^{2n+2} E_{2n+1} (x) &\leq \frac{(-x)^{2n+2}}{(2n+2)!} \\[9pt]  \implies & - \frac{x^{2n+2}}{(2n+2)!} &\leq E_{2n+1} (x) &\leq \frac{x^{2n+2}}{(2n+2)!} \\[9pt]  \implies & \phantom{-} | E_{2n+1} (x) | &\leq \frac{x^{2n+2}}{(2n+2)!}. \qquad \blacksquare \end{alignat*}

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