Taylor series for sin(x) centered at pi

Question

I haven't been able to figure this problem out.

For the Taylor series I got: $$\sin{x}-0 = 0 - (x - \pi ) + 0+ \frac{1}{6} (x-\pi)^3 + 0 - \frac{1}{120} (x-\pi)^5 + o (x^5) $$

For the series in sigma form I made it: $$ (-1)^n \frac{(x-\pi)^n}{n!} $$

However, this is incorrect, because I assume the sign is incorrect, but I can't figure out how to make this form right.

Answer 1

You want to take the Taylor series of $\sin(x)$ at $x=\pi$. This is equivalent to taking the Taylor series of $\sin(x+\pi)$ at $x=0$.

$$\sin(x+\pi)=-\sin(x)=-\left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)=\sum_{n=0}^\infty\frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$$

$$\sin(x)=\sum_{n=0}^\infty\frac{(-1)^{n+1}(x-\pi)^{2n+1}}{(2n+1)!}$$