Question

In Exercises 11-14, prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=\cos x $$

Short answer

The Maclaurin series for \(f(x) = \cos x\) is: \[ f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \] It converges for all \(x\) due to the Alternating Series Test.

Step-by-step solution

Write down the Maclaurin Series

The Maclaurin series for a function \(f(x)\) is given by: \[ f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} + \ldots \]

Calculate the derivatives

For the function \(f(x) = \cos x\), calculate the first few derivatives at \(0\): \(f(0) = \cos 0 = 1\), \(f'(0) = -\sin 0 = 0\), \(f''(0) = -\cos 0 = -1\), \(f'''(0) = \sin 0 = 0\). Note that the pattern repeats every four terms.

Construct the Maclaurin series

Substitute these values into the Maclaurin series. This yields: \[ f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \] This is the Maclaurin series for \(f(x) = \cos x\)

Prove convergence

The terms of the series become smaller and smaller as \(x\) increases, and alternate in sign. Therefore, this series converges for all \(x\), according to the Alternating Series Test