Let $f(x)$ and $g(x)$ be functions such that $f(x) = x\cdot g(x)$. Can somebody show why the Maclaurin polynomial of $f(x)$ is the same as the Maclaurin polynomial of $g(x)$, multiplied by $x$?
Here is what I got:
In the following, $T^f_{n,0}(x)$ and $T^g_{n,0}(x)$ denote the $n$-th Maclaurin polynomials for $f(x)$ and $g(x)$.
What's confusing me is that the last term of $T^f(x)$ is not the same as the last term of $x\cdot T^g(x)$.
As mentioned in a comment, since the $n$-th Maclaurin polynomial $T_{n,0}$ is degree $n$, the desired identity must be
$$T^f_{n,0}(x) = x\,T^g_{n-1,0}(x) \qquad\left(\;\text{or}\;\;T^f_{n+1,0}(x) = x\,T^g_{n,0}(x)\;\right)$$
in order for the degrees of the polynomials on each side to match.
You can compress the derivation this way, where I'll save space by not writing the "$(0)$" in "$f^k(0)$" and "$g^k(0)$": $$T^f_{n,0}(x) = \sum_{k=0}^nf^k\frac{x^k}{k!}=\sum_{k=1}^nkg^{k-1}\frac{x^k}{k!}=x\sum_{k=1}^ng^{k-1}\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^{n-1}g^k\frac{x^k}{k!} = x\;T^g_{n-1,0}(x)$$ (Note the shifting summation indices, which the reader should justify.) Looked at this way, the form of $T^g_{n-1,0}(x)$ appears pretty naturally.