Show that the integration rule is precise if it is precise for the basis of polynomials

Question

Let an integration rule:$$\sum_{i=1}^N w_i f(x_i)$$ Show that if the integration rule is precise for $\{p_0(x)=1, p_1(x)=x\ldots,p_n(x)=x_n\}$ then it is precise for every polynomial with $\deg (p) \le n$

My try:
Let some $p(x)$ with $\deg (p) \le n$. Then,
$$\int_a^b p(x)\ dx = \int_a^b \sum_{i=1}^n c_i x^i = \sum_{i=1}^n c_i \int_a^b x^i = \sum_{i=1}^n c_i \sum_{j=1}^N w_j p_i(x_j) = \sum_{j=1}^N w_j \sum_{i=1}^n c_i p_i(x_j) = \sum_{j=1}^N w_j p(x_j)$$

So yes, the integration rule is precise for this arbitrary $p$. Now, basically I've added the $N$ annotation (they didn't wrote it in the original question) and I wanted to know if there's a relation between $N$ (the number of weights) to $n$ (the size of the polynomial vector space)

TL;DR
What is the number of weights/points in this case (which I annotated as $N$)?

Thanks!

Answer 1

Let $P_n$ the real vector space of all polynomial of degree $\le n$.

Let $B$ its canonical basis : $B=(e_0:t\mapsto 1,e_1:t\mapsto t,\cdots,e_n:t\mapsto t^n)$

Let $a,b$ two real numbers.

If the linear maps $\Phi:f\mapsto\sum_{i=1}^Nw_if(x_i)$ and $f\mapsto\int_a^bf(t)\,dt$ coincide of $B$, then they are equal.

Proof

For any $f\in P_n$, there exist unique $(a_0,\cdots,a_n)\in\mathbb{R}^{n+1}$ such that $f=\sum_{k=0}^na_ke_k$; hence by linearity :

$$\int_a^bf(t)\,dt=\sum_{k=0}^na_k\int_a^be_k(t)\,dt=\sum_{k=0}^na_k\Phi(e_k)=\Phi\left(\sum_{k=0}^na_ke_k\right)=\Phi(f)$$

What connection between $n$ and $N$ ?

There is no need to exist any connection between those integers if we only want to guaranty the previous point.

The Prop. 1 below asserts that taking $N=n+1$ is ok.

Of course, what we wish is an integration method that possibly works for a larger class of polynomials and which would also be a starting point for a general integration process for arbitrary functions (regular enough ones, though).

Prop. 1 Given $n\in\mathbb{N}$ and given pairwise distinct real numbers $t_0,\cdots,t_n$ (not necessary in $[a,b]$), there exist a unique $(\lambda_0,\cdots,\lambda_n)\in\mathbb{R}^{n+1}$ such that :

$$\forall f\in P_n,\int_a^bf(t)\,dt=\sum_{k=0}^n\lambda_kf(t_k)\tag{*}$$

Proof - The family of linear maps $\varphi_k:P_n\to\mathbb{R},f\mapsto f(t_k)$ is a basis of the dual space $P_n^\star$. Hence the linear map $P_n\to\mathbb{R},f\mapsto\int_a^bf(t)\,dt$ can be written, in a unique way, as a linear combination of $\varphi_0,\cdots,\varphi_n)$.

Prop. 2 Given $n\in\mathbb{N}$, there is exists a unique choice for the numbers $t_0,\cdots,t_n$ such that the formula (*) extends in fact to all $f\in P_{2n+1}$. Those numbers are the roots of the (monic) polynomial which spans the orthogonal in $P_{n+1}$ of $P_n$, with respect to the inner product defined by :

$$\left=\int_a^bP(t)Q(t)\,dt$$

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The next step is the so-called Gauss quadrature method.

Roughly speaking, the interval $[a,b]$ is divided into $q$ equal parts and an approximation of the integral of $f$ on each part is obtained with Prop. 2. Summing those approximations yields an approximate value of $\int_a^bf(t)\,dt$. To learn the details, you should start with this Wikipedia article.