Problem 10 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=\cos 2 x$$

Step-by-Step Solution

Answer

Question: Find the first four nonzero terms of the Maclaurin series for the function $$f(x) = \cos(2x)$$, express the power series in summation notation, and determine its interval of convergence. Answer: The first four nonzero terms of the Maclaurin series for the function $$f(x) = \cos(2x)$$ are $$1 - 2x^2 + \frac{2}{3}x^4$$. The power series can be represented as $$\sum_{n=0}^{\infty}\frac{(-1)^n (2x)^{2n}}{(2n)!}$$. The interval of convergence is $$( -\infty, \infty )$$.

Step 1: Finding the first four nonzero terms of the Maclaurin series

To find the first four nonzero terms of the Maclaurin series, we use the formula: $$f(x) \approx f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots$$ where $$f^{(n)}(0)$$ is the nth derivative of the function evaluated at 0. First, we need to find the derivatives of $$f(x) = \cos(2x)$$. $$f'(x) = -2\sin(2x)$$ $$f''(x) = -4\cos(2x)$$ $$f'''(x) = 8\sin(2x)$$ $$f^{(4)}(x) = 16\cos(2x)$$ Now, we evaluate the derivatives at 0: $$f(0) = \cos(0) = 1$$ $$f'(0) = -2\sin(0) = 0$$ $$f''(0) = -4\cos(0) = -4$$ $$f'''(0) = 8\sin(0) = 0$$ $$f^{(4)}(0) = 16\cos(0) = 16$$ Now we substitute into the Maclaurin series formula up to the 4th derivative: $$f(x) \approx 1 + 0\cdot x - \frac{4x^2}{2!} + 0\cdot \frac{x^3}{3!} + \frac{16x^4}{4!}$$ $$f(x) \approx 1 - 2x^2 + \frac{2}{3}x^4$$ So, the expansion for the given function is: $$f(x) \approx 1 - 2x^2 + \frac{2}{3}x^4$$

Step 2: Writing the power series using summation notation

To express the series in summation notation, we'll use a representation for the coefficients of the x terms. We can rewrite the expansion as follows: $$f(x) \approx \sum_{n=0}^{\infty}\frac{(-1)^n (2x)^{2n}}{(2n)!}$$ This will represent the power series in a compact form.

Step 3: Determine the interval of convergence

To find the interval of convergence, we'll use the ratio test. For the series $$\sum a_n$$, the ratio test states: If $$\lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| = L$$, then: 1. If $$L < 1$$, the series converges. 2. If $$L > 1$$, the series diverges. 3. If $$L = 1$$, the test is inconclusive. So, let's apply the ratio test for our series: $$\lim_{n \to \infty} \left|\frac{(-1)^n(2x)^{2n}}{(2n)!} \frac{(2(n+1))!}{(-1)^{n+1} (2x)^{2(n+1)}}\right|$$ Simplify: $$\lim_{n \to \infty} \left|\frac{(2n+2)(2n+1)(2x)^2}{(2x)^2}\right|$$ $$\lim_{n \to \infty} |(2n+2)(2n+1)|$$ Since it's not dependent on x, the interval of convergence is $$-\infty < x < \infty$$, or in interval notation, $$( -\infty, \infty )$$. #Conclusion# We found that the first four nonzero terms of the Maclaurin series for $$f(x) = \cos(2x)$$ are $$1 - 2x^2 + \frac{2}{3}x^4$$, and the power series can be represented as $$\sum_{n=0}^{\infty}\frac{(-1)^n (2x)^{2n}}{(2n)!}$$. The interval of convergence is $$( -\infty, \infty )$$.