91Ó°ÊÓ

We will need the derivatives of the function \(f(x) = \sin x\) evaluated at \(x = 0\) in order to create the Maclaurin series. Find the first few derivatives: \(f(x) = \sin x\) \(f'(x) = \cos x\) \(f''(x) = -\sin x\) \(f'''(x) = -\cos x\) Evaluate them at \(x = 0\): \(f(0) = \sin 0 = 0\) \(f'(0) = \cos 0 = 1\) \(f''(0) = -\sin 0 = 0\) \(f'''(0) = -\cos 0 = -1\)

The Taylor series formula for a function \(f(x)\) centered at \(a\) is: \[T_n(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\] Since we have a Maclaurin series (centered at \(a=0\)), the formula becomes: \[T_n(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\] Using the derivatives evaluated at \(x=0\) (from Step 1), we can write the first few terms of the Maclaurin series: \(T_n(x) = 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} -1 \cdot \frac{x^3}{3!} + ... \) This simplifies to: \(T_n(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...)

Now we will find the interval of convergence using the ratio test for the series: \[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\] The ratio test states that if: \(\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\) converges, and the result is less than 1, then the series converges. Let's apply the ratio test: \(\lim_{n \to \infty} \left|\frac{(-1)^{n+1} x^{2(n+1)+1}}{(2(n+1)+1)!}\cdot \frac{(2n+1)!}{(-1)^n x^{2n+1}}\right|\) Simplify and cancel out terms: \(\lim_{n \to \infty} \left|\frac{x^2}{(2n+3)(2n+2)(2n+1)}\right|\) Since the only variable term is \(x^2\), the result is less than 1 for all x values, which means the series converges for all x. Therefore, the interval of convergence is: \((- \infty, \infty)\) So, the Maclaurin series for the function \(f(x) = \sin x\) is: \[T_n(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\] And the interval of convergence is \((- \infty, \infty)\).