Find the Taylor series of f(x)=sin(x)f(x) = \sin(x)f(x)=sin(x) at a=πa = \pia=π.
∑n=0∞(−1)n(x−π)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi)^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n(x−π)2n+1
∑n=0∞(x−π)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(x-\pi)^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(x−π)2n+1
∑n=0∞(−1)n+1(x−π)2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^{n+1} (x-\pi)^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n+1(x−π)2n+1
∑n=0∞(−1)n(x−π)2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi)^{2n}}{(2n)!}∑n=0∞(2n)!(−1)n(x−π)2n