91Ó°ÊÓ

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: $$f(x) \approx 1 - 2x^2 + \frac{2}{3}x^4$$

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.

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 $$( -\infty, \infty )$$.