Relationship between Taylor and Weierstrass theorem

I'm not a mathematician, but these two theorems sound related to me.

Taylor's theorem. Every k-times differentiable function can be approximated in a neighborhood around a given point by a k-th order polynomial to an arbitrary degree.

Weierstrass theorem. Every continuous function defined on a closed interval $[a, b]$ can be approximated to an arbitrary degree by a polynomial function.

(The statements are probably not precise.) I always wondered what is the underlying relationship between those two? Does one imply the other, or are they each special cases of some more general result?

Morally, these two theorems shouldn't say much about each other. If either were to imply the other, it seems less absurd that Weierstrass implies Taylor: if all we know is Taylor then we only have very weak data about a generic continuous function (if $f$ were to be absolutely continuous then we could perhaps use its antiderivative to some end, but generally speaking, no). But this direction also seems unlikely, because Weierstrass gives no control at all on the degree of the polynomial.
As rych points out, the two kinds of approximations suggested by these two theorems are different. Weierstrass' approximation is uniform which means that no point in the approximation can be further than a specified tolerance from the original. But Taylor's approximation is a much more subtle condition: no point $p(x)$ in the approximation can be further than $\varepsilon (x-a)^k$ from $f(x)$. This is a stronger condition: it means, for instance, that $p(a)=0$, but this need not be true at any point of a Weierstrass approximation.
• I agree that the $\varepsilon(x-a)^k$ property is strong, but the $p(a)=0$ property in particular doesn't seem so strong to me. If I fixed a point $a$, I could approximate to $\varepsilon/2$ using Weierstrass, then shift the polynomial to get $p(a) = 0$ and the result still approximates to $\varepsilon$. Oct 30 '20 at 2:51