Question: Find the Taylor series for \( f(x) \) centered at the given value of \( a \). [Assume that \(…

Find the Taylor series for \( f(x) \) centered at the given value of \( a \). [Assume that \( f \) has a power series expansion. Do not use the Taylor series.]

\[ f(x) = \sin(x), \quad a = \pi \]

\[ f(x) = \sum_{n=0}^{\infty} \left( \square \right) \]

Find the associated radius of convergence, \( R \).

\[ R = \square \]

Solution

To find the Taylor series for \( f(x) = \sin(x) \) centered at \( a = \pi \) and determine its radius of convergence, we follow these steps: First, we compute the derivatives of \( f(x) \) at \( x = \pi \). \[ f(x) = \sin(x) \] \[ f(\pi) = \sin(\pi) = 0 \] \[ f’(x) = \cos(x) \] \[ f’(\pi) = \cos(\pi) = -1 \] \[ f”(x) = -\sin(x) \] \[ f”(\pi) = -\sin(\pi) = 0 \] \[ f'''(x) = -\cos(x) \] \[ f'''(\pi) = -\cos(\pi) = 1 \] \[ f''''(x) = \sin(x) \] \[ f''''(\pi) = \sin(\pi) = 0 \] Next, we observe the pattern in the derivatives: \[ f^{(n)}(\pi) = \begin{cases} 0 & \text{if } n \text{ is even} \\ (-1)^{\frac{n-1}{2}} & \text{if } n \text{ is odd} \end{cases} \] Now, the Taylor series expansion of \( f(x) \) around \( a = \pi \) is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\pi)}{n!} (x - \pi)^n \] Substituting the derivatives: \[ f(x) = \sum_{n=0}^{\infty} \begin{cases} 0 \cdot \frac{(x - \pi)^n}{n!} & \text{if } n \text{ is even} \\ \frac{(-1)^{\frac{n-1}{2}}}{n!} (x - \pi)^n & \text{if } n \text{ is odd} \end{cases} \] Simplifying, we include only the non-zero terms: \[ f(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} (x - \pi)^{2k+1} \] To find the radius of convergence \( R \), we observe that the Taylor series for sine has an infinite radius of convergence, regardless of the center point. \[ R = \infty \] Final Answers: Taylor series for \( f(x) = \sin(x) \) centered at \( a = \pi \): \[ f(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} (x - \pi)^{2k+1} \] Radius of convergence: \[ R = \infty \]