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

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 . With this definition, the polynomial x^i becomes identified with the function e_i that sends 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.