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Let $x_0,x_1,\cdots,x_n$ be $n+1$ nodes of interpolation to some $f$ and let $L(f)=\sum_{i=0}^n{f(x_i)l_i(x)}$ be a linear transformation. Show that $L$ have the property of $L(q)=q$ for all polynomial $q$ of grade at most $n$.

My process: By the Lagrange interpolation we have that $L(q)(x)=\sum_{i=0}^n{q(x_i)l_i(x)}=\sum_{i=0}^n{q(x_i)\prod_{i=0\\j\not=i}^n \left[\frac{x-x_j}{x_i-x_j}\right]}$, so if $x=x_k$ for some $k=0,1,\cdots,n$ we have two cases:

  1. If $k=i$ then $\prod_{i=0\\j\not=i}^n \left[\frac{x_k-x_j}{x_i-x_j}\right]=1$
  2. If $k\not=i$ then $\prod_{i=0\\j\not=i}^n \left[\frac{x_k-x_j}{x_i-x_j}\right]=0$

So $L(q)(x_k)=\sum_{i=0}^n{q(x_i)\prod_{i=0\\j\not=i}^n \left[\frac{x_k-x_j}{x_i-x_j}\right]}=q(x_k)$

But if $x\not=x_k$ for all $k=0,1,\cdots,n$ i don't know how to prove it, can you help me with some hint, please.

Lennis Mariana
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2 Answers2

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By Lagrange interpolation basics, the $l_i(x)$ are polynomials of degree $n$.
The linear combination of polynomials of degree $n$ produces a polynomial of degree from $0$ to $n$.

Given $n+1$ points there is a unique polynomial of degree $[0,n]$ that passes through them. For instance if you have $10$ points that are aligned , the interpolating polynomial will be the line of degree $1$ through them, or of degree $0$ if the line is horizontal.

The Lagrange interpolation produce a unique polynomial passing through the $n+1$ points. If these are actually corresponding $y_k=q(x_k)$ to a polynomial $q(x)$ of lower degree (over-interpolation), you get that polynomial: the coefficients of higher powers will be null.

To demonstrate this you shall connect Lagrange to Newton Interpolation by divided differences, where, if the points are generated by a lower degree polynomial then the differences of higher order are null and the Newton interpolation has the same degree as $q(x)$.

You can find some help to this regard in this related post, or this other

G Cab
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Question :
Let $\displaystyle l_i(x) = \prod_{\substack{0\leqslant j \leqslant n \\ j \neq i}}\frac{x-x_j}{x_i-x_j}$, $l_i$ is of degree $n$. Prove that $\displaystyle \forall f\in\mathbb{R}_n[x], f = \sum_{i=0}^{n}{f(x_i)l_i(x)}$ .

Linear Algebra method :
We claim that $(l_i)_{0\leqslant i \leqslant n}$ is a linearly independent vector family in $\mathbb{R}_n[x]$.
Proof: $\displaystyle \sum_{i=0}^n a_il_i = 0 \implies \forall x \in \mathbb{R}, \sum_{i=0}^n a_il_i(x) = 0$. Take $x=x_0,x_1,\cdots,x_n$ and we get $a_0 = a_1 = \cdots = a_n = 0$ by using the fact that $l_i(x_i) = 1 $ and $l_i(x_j) = 0 $ for $j \neq i$. Thus $(l_i)_{0\leqslant i \leqslant n}$ is linearly independent. Hence it forms a basis of $\mathbb{R}_n[x]$, noted as $\mathcal{B} = (l_i)_{0\leqslant i \leqslant n}$.

Now we can represent all polynomials of degree $n$ under the basis $\mathcal{B}$, that is
$f\in\mathbb{R}_n[x] \implies \exists!\{\lambda_0,\cdots,\lambda_n\}\in\mathbb{R^n}, f = \lambda_0l_0 + \cdots + \lambda_nl_n$.
Then $\forall x_i \in \{x_0,\cdots,x_n\}, f(x_i) = \lambda_0l_0(x_i) + \cdots + \lambda_il_i(x_i) + \cdots +\lambda_nl_n(x_i) = 0+\cdots+\lambda_i+\cdots+0 = \lambda_i$.

Thus the proposition is proved.

Alternatively, we can use the Lagrange theorem about the degree of polynomial. Put $P = f - \sum_{i=0}^{n}{f(x_i)l_i}$, then $\deg(P) \leqslant n$, whereas $f$ and $\sum_{i=0}^{n}{f(x_i)l_i(x)}$ both have $n+1$ distinct roots so as $P$, so from the Lagrange theorem we know that every coefficient of $P$ should be $0$, that is $f = \sum_{i=0}^{n}{f(x_i)l_i}$.

Pion Chan
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