Higher-order derivatives follow this same pattern, so if \(f(x) =\sin x\) can be represented by a power series in \(x,\) then \[ \begin{eqnarray*} \sin x &=& f(0) +\frac{f^{\prime} (0) }{1!}x+\frac{ f^{\prime \prime} (0) }{2!}x^{2}+\frac{f^{\prime \prime \prime} (0) }{3!}x^{3}+\cdots\,=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}- \frac{x^{7}}{7!}+\cdots\\[4pt] &=& \sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !} \end{eqnarray*} \]
(b) To prove that the series actually converges to \(\sin x\) for all \(x\), we need to show that the remainders \(R_{2n+1}(x)\) and \(R_{2n}(x)\) approach zero. For \(R_{2n+1}(x)\) we have \[ R_{2n+1}(x)=(-1)^{n+1}\frac{f^{(2n+2)}(u)\,x^{2n+2}}{(2n+2)!} \]
where \(u\) is between \(0\) and \(x\). Since \(\vert f^{(2n+2)}(u)\vert = \vert \sin u \vert \leq 1\) for every number \(u\), then \[ 0\leq \vert R\,_{2n+1}(x)\vert =\frac{\left\vert f^{(2n+2)}(u)\right\vert }{(2n+2)!}\vert x\vert ^{2n+2}\leq \frac{ \vert x\vert ^{2n+2}}{(2n+2)!} \]
By the Ratio Test, the series \(\sum\limits_{k\,=\,0}^{\infty }\frac{|x|^{2k+2} }{( 2k+2) !}\) converges for all \(x\), so \[ \lim_{n\,\dot{\rightarrow}\infty }\frac{|x|^{2n+2}}{(2n+2)!}=0 \]
By the Squeeze Theorem, \(\lim\limits_{n\rightarrow \infty }\)\(\vert R_{2n+1}(x)\vert =0\) and therefore \(\lim\limits_{n\rightarrow \infty }\) \(R_{2n+1}(x)=0\) for all \(x.\)
A similar argument holds for the remainder \(\left\vert R_{2n}(x) \right\vert =\frac{\left\vert \cos u\right\vert x^{2n+1}}{(2n+1) !}.\)
We conclude that the series \(\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1) !}\) converges to \(\sin x\) for all \(x.\) That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{(2k+1)!} }} \]