## Linear independence of polynomials

One of the exercises this week asked for a proof of linear independence for the set

$\{x^i\}_{i\in {\bf N}}$

inside the polynomials $R[x]$ with real coefficients. However, note that the polynomials here are regarded as *functions* from $R$ to $R$. Thus, it amounts to showing that if

$c_0+c_1x+\cdots c_nx^n=0$

as a function, then all $c_i$ have to be zero. This does require proof. One quick way to do this is to note that all polynomial functions are differentiable. And if

$f(x)=c_0+c_1x+\cdots c_nx^n$

is the zero function, then so are all its derivatives. In particular,

$f^{(i)}(0)=0$

for all $i$. But $f^{(i)}(0)=i!c_i.$ Thus, $c_i=0$ for all $i$.

One possible reason for confusion is that there is another ‘formal’ definition of $R[x]$ by simply identifying a polynomial with its sequence of coefficients. That is, you can think of an element of $R[x]$ as a function $f:N \rightarrow R$ that has *finite support* in that $f(i)=0$ for all but finitely many $i$. With this definition, the polynomial $x^i$ becomes identified with the function $e_i$ that sends $i$ to 1 and everything else to zero. If you take this approach, the linear independence also becomes formal. But in this problem, you are defining $R[x]$ as a function in its variable. This of course is the natural definition you’ve been familiar with at least since secondary school.

Here are two questions:

1. If you think of two polynomials $f$ and $g$ as functions from $N$ to $R$ with finite support, what is a nice way to write the product $fg$?

2. What is the advantage of this formal definition?