Taylor series of a piecewise function

Question

Can the Taylor series of a piecewise function be determined? Assume this example: $$f(x)=\begin{cases}\sin x&\mbox{ }x\ge -\frac{\pi}{2}\\ -1 &\mbox{ }x\lt -\frac{\pi}{2},\end{cases}$$ then $f(x)$ is differentiable everywhere. The Taylor series should not be piecewise: $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$ How can I get the coefficients? Expanding at one point, for example $a=-\frac{\pi}{2}$ doesn't work – in that case, it yields pure sine.

Answer 1

Taylor's series of $f$ at a point $a$ needs that $f$ is infinitly differentiable at $a$. There is no hope for this at point $-\frac{\pi}{2}$, you've chosen the (only) bad one.