# Linear Spaces and Polynomial Functions

I am stuck with a Linear Algebra problem. I have attempted a solution but I do not know whether it's correct or not. The problem is as follows:

Let $C^1(\mathbb{R})$ be the vector space of all differentiable real-valued functions on the real line $\mathbb{R}$. Let $S\subseteq C(\mathbb{R})$ consist entirely of polynomial functions, and prove that the function $f\not\in \langle S\rangle$, where $f(x)=\sin{x}$

I have written that:

$$p(x)\in\langle S\rangle \mid p(x)=\sum a_n x^n \text{ and }a_n\in\mathbb{R}$$ and $$f(x)=\sum_0^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$

The only thing that comes to mind is if $\langle S\rangle$ permits finite linear combinations, this would disallow $f\in\langle S\rangle$. Is this assertion true? If not, if possible, please point me in the right direction.

NOTE: This is not a homework problem.

• A finite combination of polynomials will have at most finitely many zeros while $\sin(x)$ has infinitely many. – Gregory Grant Jun 16 '16 at 17:36
• Yes, linear span is defined to be linear combinations of finitely many elements of from the set - infinite linear combination is not allowed. But note that $\sin(x)$ is in the closure of the span, since now infinite sums are admissible through limiting procedure – user340297 Jun 16 '16 at 17:41

A finite combination of polynomials will have at most finitely many zeros while $\sin(x)$ has infinitely many.