Question

In Exercises \(19-24,\) find the \(n\) th Taylor polynomial centered at \(c\). $$ f(x)=\frac{1}{x}, \quad n=4, \quad c=1 $$

Short answer

The 4th Taylor polynomial centered at \(c = 1\) for the function \(f(x) = \frac{1}{x}\) is \(P_4(x) = 1 - (x - 1) + \frac{(x - 1)^2}{2} - \frac{(x - 1)^3}{6} + \frac{(x - 1)^4}{24}\).

Step-by-step solution

Find the derivatives

First, it is necessary to determine the first 4 derivatives of \(f(x)\) at \(x = 1\). The derivatives are calculated as follows: \[f'(x) = -\frac{1}{x^2}\] \[f''(x) = \frac{2}{x^3}\] \[f'''(x) = -\frac{6}{x^4}\] \[f^{(4)}(x) = \frac{24}{x^5}\] Moreover, since \(x = 1\), all derivatives at \(x = 1\) will have the value as their coefficient. Therefore, \(f'(1) = -1, f''(1) = 2, f'''(1) = -6,\) and \(f^{(4)}(1) = 24.\)

Substitute these values into the Taylor series formula

The derivatives above are then substituted into the Taylor series formula: \[P_4(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4\]

Simplify the polynomial

After substituting \(f(1) = 1, f'(1) = -1, f''(1) = 2, f'''(1) = -6, f^{(4)}(1) = 24\), we simplify to get the final polynomial: \[P_4(x) = 1 - (x - 1) + \frac{(x - 1)^2}{2} - \frac{(x - 1)^3}{6} + \frac{(x - 1)^4}{24}\]