Finding a Taylor Polynomial for \(f( x) =\sin x\)

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 \(y=\sin x\).