Question

In Exercises \(19-24,\) find the \(n\) th Taylor polynomial centered at \(c\). $$ f(x)=\frac{2}{x^{2}}, \quad n=4, \quad c=2 $$

Short answer

The 4th Taylor Polynomial is \(P_{4}(x) = (\frac{1}{2}) - (\frac{1}{2})(x-2) + \frac{(\frac{3}{4})(x-2)^{2}}{2} - \frac{(\frac{3}{4})(x-2)^{3}}{6} + \frac{(\frac{15}{8})(x-2)^{4}}{24}\)

Step-by-step solution

Find the first few derivatives

Begin by calculating the first few derivatives of \(f(x)\) until the required order \(n=4\). The first derivatives are: \n\(f'(x)=-\frac{4}{x^{3}}\), \n\(f''(x)=\frac{12}{x^{4}}\), \n\(f'''(x)=-\frac{48}{x^{5}}\), \n\(f''''(x)=\frac{240}{x^{6}}\).

Evaluate the derivatives at \(x=2\)

With these derivatives in hand, the next step is to evaluate each at \(x=c=2\). Doing so gives: \n\(f(2)=\frac{2}{2^{2}}=\frac{1}{2}\), \n\(f'(2)=-\frac{4}{2^{3}}=-\frac{1}{2}\), \n\(f''(2)=\frac{12}{2^{4}}=\frac{3}{4}\), \n\(f'''(2)=-\frac{48}{2^{5}}=-\frac{3}{4}\), \n\(f''''(2)=\frac{240}{2^{6}}=\frac{15}{8}\).

Use the Taylor Polynomial formula

Finally, use the Taylor Polynomial formula to generate the 4th degree polynomial.The Taylor Polynomial of degree 4 centered at \(x=c\) for a function \(f\) is given by: \n\(P_{4}(x) = f(c) + f'(c)(x-c) + \frac{f''(c)(x-c)^{2}}{2!} + \frac{f'''(c)(x-c)^{3}}{3!} + \frac{f''''(c)(x-c)^{4}}{4!}\).Plugging in the relevant values yields: \n\(P_{4}(x) = (\frac{1}{2}) - (\frac{1}{2})(x-2) + \frac{(\frac{3}{4})(x-2)^{2}}{2} - \frac{(\frac{3}{4})(x-2)^{3}}{6} + \frac{(\frac{15}{8})(x-2)^{4}}{24}\).Finally, simplifying these fractions and recomputing gives the final answer.