Exercise 1.4.11* (Polynomial basis)

Proof. We define $\left(\genfrac{}{}{0.0pt}{}{x}{k}\right)$ as the formal polynomial

(If $k=0$, $P_0(x)=\frac{1}{0!}\underset{i\in \varnothing }{\prod <!--\; FUNCTION\; APPLICATION\; -->}(x-i)=1$.)

Then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(P_k)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^{k-1}\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x-i)=k$, and the leading coefficient is $1∕k!$.

We must prove that $(P_0,P_1,\dots ,P_n)$ is a basis of the vector space ${ℝ}_n[x]$ of polynomials with real coefficients and degree at most $n$. Here $\dim <!--\; FUNCTION\; APPLICATION\; -->({ℝ}_n[x])=n+1$, and the family $(P_0,P_1,\dots ,P_n)$ contains $n+1$ polynomials, thus it is sufficient to prove that $P_0,\dots ,P_n$ are linearly independent, which is true since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(P_i)=i$, so

Explicitly, $P_0=1\ne 0$, so $P_0$ is linearly independent. Reasoning by induction, assume that $P_0,\dots ,P_k$ are linearly independent.

If ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^{n+1}{a}_i{P}_i=0$, assume for contradiction that $a_{n+1}\ne 0$. Then

This is a contradiction, since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(P_{n+1})=n+1$. Therefore $a_{n+1}=0$. Then ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n{a}_i{P}_i=0$, where $P_0,\dots ,P_n$ are linearly independent, thus $a_0=\cdots =a_n=0$, together with $a_{n+1}=0$. The induction is done.

So $(P_0,P_1,\dots ,P_n)$ is a basis of the vector space ${ℝ}_n[x]$. There exist (unique) real numbers $c_k$ such that

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