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 show that \( R_n(x) \to 0 \).]

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

\[ f(x) = \sum_{n=0}^{\infty} \left( \frac{(-1)^n}{(2n+1)!} (x - \pi)^{2n+1} \right) \]

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

\[ R = \infty \]

Solution

Step 1: Identify the function and the point of expansion. \[ f(x) = \sin(x) \] \[ a = \pi \] Step 2: Compute the derivatives of \( f(x) \) at \( x = \pi \). \[ f(x) = \sin(x) \] \[ f’(x) = \cos(x) \] \[ f”(x) = -\sin(x) \] \[ f'''(x) = -\cos(x) \] \[ f^{(4)}(x) = \sin(x) \] Continuing this pattern, the derivatives at \( x = \pi \) are: \[ f(\pi) = \sin(\pi) = 0 \] \[ f’(\pi) = \cos(\pi) = -1 \] \[ f”(\pi) = -\sin(\pi) = 0 \] \[ f'''(\pi) = -\cos(\pi) = 1 \] \[ f^{(4)}(\pi) = \sin(\pi) = 0 \] And so on. Step 3: Write the Taylor series expansion of \( f(x) \) around \( a = \pi \). \[ f(x) = f(\pi) + f’(\pi)(x - \pi) + \frac{f”(\pi)}{2!}(x - \pi)^2 + \frac{f'''(\pi)}{3!}(x - \pi)^3 + \cdots \] Substituting the computed derivatives: \[ \sin(x) = 0 - 1(x - \pi) + 0 + \frac{1}{3!}(x - \pi)^3 + \cdots \] This simplifies to: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}(x - \pi)^{2n+1} \] Step 4: Determine the radius of convergence \( R \). Since \( \sin(x) \) is an entire function (analytic everywhere in the complex plane), the radius of convergence is infinite. \[ R = \infty \]