1. ## Lagrange polynomials

hello
for $i=1,2,...,(n+1)$
let
$P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}$,
prove that $(P_1,P_2,...P_{n+1})$ is a basis of $\mathbb{R}_{n}[X]$.
thanks.

$P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}" title="P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}" />$

2. Originally Posted by Raoh
hello
for $i=1,2,...,(n+1)$
let
$P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}$,
prove that $(P_1,P_2,...P_{n+1})$ is a basis of $\mathbb{R}_{n}[X]$.
thanks.

$P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}" title="P_{i}(X)=\frac{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(X-a_{j})}_{\prod_{1\leq&space;j\leq&space;n+1,j\neq&space;i}(a_{i}-a_{j})}" />$
obviously we also need to have $a_i \neq a_j,$ for $i \neq j.$ anyway, again, we have $B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x]$ and $\dim_{\mathbb{R}} \mathbb{R}[x]=n+1.$ so, in order to show that $B$ is a basis for $\mathbb{R}_n[x],$ we only need

to show that $B$ is linearly independent. first see that for any $i$ we have $P_i(a_i)=1$ and $P_j(a_i)=0,$ for $i \neq j.$ now, to show that $B$ is linearly independent, suppose that $\sum_{j=1}^{n+1}c_j P_j(x)=0,$ for some

$c_j \in \mathbb{R}.$ put $x=a_i$ to get $c_i=0. \ \Box$

3. Originally Posted by NonCommAlg
obviously we also need to have $a_i \neq a_j,$ for $i \neq j.$ anyway, again, we have $B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x]$ and $\dim_{\mathbb{R}} \mathbb{R}[x]=n+1.$ so, in order to show that $B$ is a basis for $\mathbb{R}_n[x],$ we only need

first see that for any $i$ we have $P_i(a_i)=1$ and $P_j(a_i)=0,$ for $i \neq j.$ now, to show that $B$ is linearly independent, suppose that $\sum_{j=1}^{n+1}c_j P_j(x)=0,$ for some $c_j \in \mathbb{R}.$ put $x=a_i$ to get $c_i=0. \ \Box$
if i put $x=a_i$ we'll have $P_j(a_i)=0$, no ?
thank you.

4. Originally Posted by Raoh
if i put $x=a_i$ we'll have $P_j(a_i)=0$, no ?
thank you.
yes, for a fixed $i,$ we have $P_i(a_i)=1$ and $P_j(a_i)=0,$ for all $j \neq i.$ by the way, a few words in my previous post were missing. i just edited my post. so you should take a look at it again.

5. Originally Posted by NonCommAlg
yes, for a fixed $i,$ we have $P_i(a_i)=1$ and $P_j(a_i)=0,$ for all $j \neq i.$ by the way, a few words in my previous post were missing. i just edited my post. so you should take a look at it again.
what if i suppose $\sum_{i=1}^{n+1}c_i P_i(X)=0$ for some $c_i$ and put $X=a_i$ to get $c_i=0$ is it okay ?
thanks.

6. Originally Posted by Raoh
what if i suppose $\sum_{i=1}^{n+1}c_i P_i(X)=0$ for some $c_i$ and put $X=a_i$ to get $c_i=0$ is it okay ?
thanks.
yes, just for the index of your sum you should use $j$ instead of $i$ because when you put $x=a_i$ you're assuming that $i$ is fixed. (do as i did in my solution!)

7. Originally Posted by NonCommAlg
yes, just for the index of your sum you should use $j$ instead of $i$ because when you put $x=a_i$ you're assuming that $i$ is fixed. (do as i did in my solution!)
of course i'm going to do as you did in your solution but at least let me know how did you do it.
my question,what do you mean by a fixed $i$,i don't understand your last post,would you please explain it to me.