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Lagrange polynomials

  1. Dec 22, 2009 #1
    hello everyone:smile:
    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.
     
  2. jcsd
  3. Dec 23, 2009 #2
    Hmmm..er....I dont know if I have understood your question properly ...but let me try to answer...


    Your question is..why have we used j instead of i in the summation? Well..as far as I can see...i and j are just dummy variables ..it does not matter whether you use i or j or k....

    You can very well use i if you want...but then you have to always remember that i denotes a general element....then..to prove the coefficients as zero, you cant put x=a_(i) as i is now reserved for the general element..you have to use some other letter in that case...


    Now, as for the question as to why we had to find P_(j)(a_i)...by putting x=a_(i), you are reducing all elements of the sum to zero except the term with index j=i. This term is
    c_i * P_(i)(a_i)=c_i.
     
  4. Dec 23, 2009 #3
    thank you.
    but if i and j are just dummy variables and it doesn't matter which one we put.
    how could we have [tex]P_i(a_i)=1 [/tex] and [tex]P_j(a_i)=0 [/tex] ??
    ???
     
  5. Dec 24, 2009 #4
    Welll...first thing you should realize is that even if you use i or j or k or a or b or c, the two expressions are fundamentally different...


    [tex]P_{i}(a_{i})[/tex] has the the index for [tex]P[/tex] and [tex]a[/tex] as the same..

    [tex]P_{j}(a_{i})[/tex] has the the index for [tex]P[/tex] and [tex]a[/tex] as different..of course, the assumption here being that, whatever i and j are, they are not equal..i.e if we put i=1, then j cannot be 1....


    See the definition of [tex]P_{i}(X)[/tex]......In the numerator, there is a product...the product contains all terms of the form [tex]X-a_{j}[/tex] such that j is not equal to i...

    Like, if i=3, then the product contains [tex](X-a_{1}),(X-a_{2}),(X-a_{4}),(X-a_{5}) etc[/tex] but not [tex](X-a_{3})[/tex]....


    Then, you can see why putting [tex]X=a_{3}[/tex] wont make the expression to zero..but [tex]X=a_{j}[/tex]..... such that j is not three......will reduce the whole term to zero...
     
  6. Dec 24, 2009 #5
    i think i'm getting somewhere,thank u so mush for your help.
    :smile:
     
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