I'm having slight difficulties in understanding the Taylor polynomial for
$$g(x)=f(cx), \space c \in \mathbb{R}$$
The way I look at it is to start with:
$$T_{n, x_0}g(x)=\sum_{m=0}^n \frac{g^{(m)}(x_0)}{m!}(x-x_0)^m$$ $$=\sum_{m=0}^n \frac{f^{(m)}(c x_0)}{m!}(x-x_0)^m$$ $$=f(cx_0) + f'(cx_0)(x-x_0)+\frac{f''(cx_0)}{2}(x-x_0)^2+ \cdot\cdot\cdot$$
But I'm having trouble understanding, what can I do with the derivatives $f'(cx_0)$, $f''(cx_0)$ ... ?
Like can I move the constant out? Because what I'm trying to show is that
$$T_{n, x_0} g(x) = T_{n, cx_0} f(cx)$$
OR do I have to "map" the $x_0$ so that
$$=\sum_{m=0}^n \frac{f^{(m)}(c x_0)}{m!}(x-x_0)^m$$
is actually
$$=\sum_{m=0}^n \frac{f^{(m)}(c x_0)}{m!}(x-cx_0)^m$$
because the $x_0$ has been shifted by $c$?
Your Taylor series for $f$ is not correct. To see this, put $y = cx$, and write out the Taylor series of $f$ w.r.t. $y$ at the point $y_{0} = cx_{0}$. You get: $$ f(y) = \sum_{m \geq 0} {f^{(m)} \over m!}(y_{0}) \; (y - y_{0})^{m} = \sum_{m \geq 0} {f^{(m)} \over m!}(c x_{0}) \; (cx - cx_{0})^{m}. $$ If I understand your question correctly, this should clear it up.