Find the Taylor Polynomial $P_{7}( x)$ for $f( x) =\sin x$ at $0$.

Solution The derivatives of $f( x) =\sin x$ at $0$ are $$\begin{array}{rcl@{\qquad}rcl@{\qquad}rcl@{\qquad}rcl} f( x) &=&\sin x & f^\prime ( x) &=&\cos x & f^{\prime \prime} ( x) &=&-\!\sin x & f^{\prime \prime \prime} ( x) &=&-\!\cos x \\ f( 0) &=&0 & f^\prime ( 0) &=&1 & f^{\prime \prime} ( 0) &=&0 & f^{\prime \prime \prime} ( 0) &=&-1 \\ f^{( 4) }( x) &=&\sin x & f^{(5) }( x) &=&\cos x & f^{( 6)}( x) &=&-\!\sin x & f^{( 7) }( x) &=&-\!\cos x \\ f^{( 4) }( 0) &=&0 & f^{(5) }( 0) &=&1 & f^{( 6) }( 0) &=&0 & f^{( 7) }( 0) &=&-1 \end{array}$$

The Taylor Polynomial $P_{7}( x)$ for $\sin x$ at $0$ is $$P_{7}( x) =f( 0) +f^\prime ( 0) x+\dfrac{f^{\prime \prime} ( 0) }{2!}x^{2}+\dfrac{f^{\prime\prime \prime} ( 0) }{3!}x^{3}+\cdots +\dfrac{f^{( 7) }( 0) }{7!}x^{7}=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$$

Figure 18 shows the graph of $P_{7}$ superimposed on the graph of $y=\sin x$.