Question

Find the Maclaurin series for the function. (See Example \(7 .)\) $$ h(x)=x \cos x $$

Short answer

The Maclaurin series for the function \(h(x) = x \cos x\) is: \(h(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\).

Step-by-step solution

Finding the Derivatives at x=0

First, compute the derivatives of \(x\cos(x)\) at \(x = 0\). To ease the process, apply the product rule. The product rule states that if we have two functions, let's say \(u(x)\) and \(v(x)\), differentiated with respect to \(x\), then the derivative is \(u'(x)v(x) + u(x)v'(x)\).\n\nSo, applying the product rule, we get: \(h'(x) = \cos(x) - x\sin(x)\). Evaluating this derivative at \(x = 0\), we get \(h'(0) = 1\).\n\nProceed similarly to get the other derivatives: \(h''(x) = -2\sin(x), h''(0) = 0\), \(h'''(x) = -2\cos(x), h'''(0) = -2\), \(h''''(x) = 4\sin(x), h''''(0) = 0\), etc.

Representing the Maclaurin Series

The Maclaurin series representation is given by: \(f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \dotsb\)\n\nSubstitute the derivatives of \(h(x)\) at 0 obtained in the first step: \(h(x) = 0 + 1x + \frac{0x^2}{2!} + \frac{-2x^3}{3!} + \frac{0x^4}{4!} + \dotsb\) This simplifies to: \(h(x) = x - \frac{x^3}{3!}\)

Generalizing the Representation

Notice that only odd degrees of \(x\) have non-zero terms in this series. Therefore, we can express this function more succinctly by summing over the odd terms: The general Maclaurin series representation of this function is given by: \(h(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dotsb, or h(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)