Given sin(x)=∑n=0∞(−1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}sin(x)=∑n=0∞(2n+1)!(−1)nx2n+1, what is the Maclaurin series for sin(x2)\sin(x^2)sin(x2)?
∑n=0∞(−1)nx4n+2(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n+1)!}∑n=0∞(2n+1)!(−1)nx4n+2
∑n=0∞(−1)nx4n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)nx4n+1
∑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
∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}∑n=0∞(2n)!(−1)nx2n