Prove that the error of the Taylor expansion of 
satisfies the following inequality.
![\[ \sin x = \sum_{k=1}^n \frac{(-1)^{k-1} x^{2k-1}}{(2k-1)!} + E_{2n}(x), \qquad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d3599a838bc1249137b704071bbf6aa6_l3.png "Rendered by QuickLaTeX.com")
Proof. Since the derivatives of are always 
, or we know that for 
we have 
. (In other words, the 
st derivative is bounded above by 1 and below by 
.) Therefore, we can apply Theorem 7.7 (p. 280 of Apostol) to estimate the error in Taylor’s formula at 
with 
and 
. For 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-075f0ab3ad3445588eca5bbec695f624_l3.png "Rendered by QuickLaTeX.com")
Next, (from the second part of Theorem 7.7) if 
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9feb7ad0b365c754fef38beb9097eff9_l3.png "Rendered by QuickLaTeX.com")
the last inequality is false cause |E|>0 and x^{2n+1}<0
For the x<0 part, shouldn't 'x' be '-x'? The definition uses |x|.