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

Find the Taylor series for \( f(x) \) centered at the given value of \( a \). [Assume \( f(x) = \sin(x) \), \( a = \pi \)]

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

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

\[ R = \infty \]

Solution

To find the Taylor series for \( f(x) = \sin(x) \) centered at \( a = \pi \), we follow these steps: \[ f(x) = \sin(x) \] First, express \( x \) as \( \pi + h \), where \( h = x - \pi \): \[ h = x - \pi \] Substitute \( x = \pi + h \) into \( f(x) \): \[ f(\pi + h) = \sin(\pi + h) \] Using the sine addition formula: \[ \sin(\pi + h) = -\sin(h) \] Thus, \[ f(\pi + h) = -\sin(h) \] Now, recall the Taylor series expansion for \( \sin(h) \) around \( h = 0 \): \[ \sin(h) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} h^{2n+1} \] Therefore, \[ -\sin(h) = -\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} h^{2n+1} \] Substitute back \( h = x - \pi \): \[ f(x) = -\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x - \pi)^{2n+1} \] This is the Taylor series for \( \sin(x) \) centered at \( \pi \). The radius of convergence \( R \) for the Taylor series of \( \sin(x) \) is: \[ R = \infty \]