Proof that Polynomials Form a Basis I'm not even sure this is a true statement, but can someone prove that the polynomials for a basis for continuous functions? This seems to be motivation for Taylor series, and several of the eigenbasis in quantum mechanics. 
 A: The polynomials form a vector space, and that is certainly not a basis for anything. They are dense in the space of continuous functions on a compact interval with the uniform norm (Weierstraß's theorem), and in many other interesting function spaces as well. The eigenspaces in quantum mechanics is certainly a good example.
This density is not a good motivation for Taylor series, however. They are better motivated by a desire to get really good local approximations around a point; that they often converge to the function they are made from, seems almost coincidental.
When you talk about a basis, you are perhaps thinking of a basis for the polynomials. Any sequence $(p_n)_{n=0}^\infty$ of polynomials, where $p_n$ has exact degree $n$, forms a basis for the polynomials. In quantum mechanics, it is common to get such a basis by using Gram–Schmidt orthogonalization to the natural basis $1,x,x^2,\ldots$, which typically gives you a basis for some Hilbert space of functions.
I could go on (and on), but this topic is too big to handle in this format.
A: Relevant 
You can very-well approximate continuous functions using polynomials.
However, this polynomial is not necessarily a truncated Taylor Series.
A: They are not.  By definition, a subset $B$ of a vector space $V$ allows you to express every vector $f\in V$ as a linear combination of vectors in $B$ -- that is a weighted finite sum of vectors in $B$.  So $f=\exp$ cant't obviously not be exressed as a finite sum of polynomials.
