Question: Find the Maclaurin series for \( f(x) \) using the definition of a Maclaurin series. (Assume that…

Find the Maclaurin series for \( f(x) \) using the definition of a Maclaurin series. (Assume that \( f \) has a power series expansion. Do not show that \( R \)-infinity.)

\[ f(x) = \sin\left(\frac{\pi x}{5}\right) \]

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

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

\[ R = \square \]

Solution

Identify the Maclaurin series expansion for \( f(x) = \sin\left(\frac{\pi x}{5}\right) \) and determine its radius of convergence. Recall the Maclaurin series for \( \sin(x) \): \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] Substitute \( x \) with \( \frac{\pi x}{5} \): \[ \sin\left(\frac{\pi x}{5}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n \left(\frac{\pi x}{5}\right)^{2n+1}}{(2n+1)!} \] Simplify the expression: \[ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n \pi^{2n+1}}{5^{2n+1} (2n+1)!} x^{2n+1} \] To determine the radius of convergence \( R \), apply the ratio test. However, since the Maclaurin series for \( \sin(x) \) converges for all real numbers and the substitution involves scaling by \( \frac{\pi}{5} \), the radius of convergence remains infinite. \[ R = \infty \]