Question #116602

Find the taylor series for sin x at a=π

Expert's answer

f(x)=i=1f(i)(a)/i!(xa)if(x)=\sum\limits_{i=1}^{\infty}f^{(i)}(a)/i!*(x-a)^i

sin(π)=i=1f(2i1)(π)/(2i1)!(xπ)2i1sin(\pi)=\sum\limits_{i=1}^{\infty}f^{(2*i-1)}(\pi)/(2*i-1)!*(x-\pi)^{2*i-1}

The Taylor Series of sin(x) with center π : -(x-π)+1/6*(x-π)^3-1/120*(x-π)^5+1/5040*(x-π)^7+...




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