Find the Maclaurin series for f(x)=sin(x)f(x) = \sin(x)f(x)=sin(x) and identify the radius of convergence.
∑n=0∞(−1)nx2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)nx2n+1 with radius of convergence R=∞R = \inftyR=∞
∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}∑n=0∞(2n)!(−1)nx2n with radius of convergence R=∞R = \inftyR=∞
∑n=0∞x2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!x2n+1 with radius of convergence R=1R = 1R=1
∑n=1∞(−1)n+1x2n−1(2n−1)!\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n-1)!}∑n=1∞(2n−1)!(−1)n+1x2n−1 with radius of convergence R=1R = 1R=1