- #1

God's Pen

- 12

- 0

for [tex]

i=1,2,...,(n+1)

[/tex] let [tex]P_{i}(X)=\frac{\prod_{1\leq j\leq n+1,j\neq i}(X-a_j)}{\prod_{1\leq j\leq n+1,j\neq i}(a_i-a_j)}[/tex]

prove that [tex]

(P_1,P_2,...P_{n+1})

[/tex] is basis of [tex]

\mathbb{R}_{n}[X]

[/tex].

i already have an answer but i don't understand some of it.

...

we have [tex]B=\{P_1, \cdots , P_{n+1} \} \subset \mathbb{R}_n[x][/tex] and [tex]\dim_{\mathbb{R}} \mathbb{R}[x]=n+1.[/tex] so, in order to show that [tex]B[/tex] is a basis for [tex]\mathbb{R}_n[x],[/tex] we only need

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

[tex]c_j \in \mathbb{R}.[/tex] put [tex]x=a_i[/tex] to get [tex]c_i=0. \ \Box[/tex]

...

what i don't understand,

why should we have [tex]P_j(X)[/tex] and not [tex]P_i(X)[/tex]

why we had to find [tex]P_j(a_i)[/tex] ?

and why we have [tex]\sum_{j=1}^{n+1}[/tex] and not [tex]\sum_{i=1}^{n+1}[/tex] ?

thank you so much.