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

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

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


Proof. Since the derivatives of \sin x are always \cos x, -\sin x, -\cos x, or \sin x we know that for f(x) = \sin x we have |f^{(n+1)}(x)| \leq 1. (In other words, the n+1st derivative is bounded above by 1 and below by -1.) Therefore, we can apply Theorem 7.7 (p. 280 of Apostol) to estimate the error in Taylor’s formula at a= 0 with m = -1 and M = 1. For x > 0 this gives us

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

Next, (from the second part of Theorem 7.7) if 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+1}}{(2n+1)!} &\leq (-1)^{2n+1} E_{2n}(x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt]  \implies & - \frac{(-x)^{2n+1}}{(2n+1)!} &\leq -E_{2n} (x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt]  \implies & \phantom{-} |E_{2n} (x)| &\leq \frac{x^{2n+1}}{(2n+1)!}. \qquad \blacksquare \end{alignat*}

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