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Langrange interpolation polynomial and Euclidian division

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ##x_1,...,x_n## be distinct real numbers, and ## P = \prod_{i=1}^n(X-x_i)##.
    If for ##i=1...n ##, ##L_i = \frac{\prod_{j \neq i}^n(X-x_j)}{\prod_{j\neq i}(x_i-x_j)}##, show that for any polynomial A (single variable and real coefs), the rest of the euclidian division of A by P is ##\sum_{i=1}^n A(x_i)L_i ##

    2. Relevant equations

    ## Q = \sum_{i=1}^n A(x_i)L_i ##

    3. The attempt at a solution

    Hello, could you tell me if my proof is correct please? It seems correct to me but I have a doubt.

    Let ##(B,R)## the quotient and the rest of the euclidian division of A by P. We have that ##\text{deg}(R)<\text{deg}(P) = n ##, and ##A = BP+R##.

    Since ## Q(x_i) = A(x_i) = R(x_i)##, then for all ##i = 1...n ##, ##(R-Q)(x_i) = 0 ##, so ##X-x_i## divides ##R-Q##
    But for ##i\neq j##, ##X-x_i## and ##X-x_j ## are relatively prime and divide ##R-Q##, so ## P | R-Q##.
    However, ##\text{deg}(R-Q) \le \max(\text{deg}(R),\text{deg(Q)}) \le n-1 < n =\text{deg}(P) ##, so ## R-Q = 0 ##
     
    Last edited: May 15, 2015
  2. jcsd
  3. May 16, 2015 #2

    wabbit

    User Avatar
    Gold Member

    Seems perfectly fine to me.
    You could formulate it in a marginally more direct way perhaps, e.g. saying A-Q is zero at each x_i, hence it divides P, etc. - but this is equivalent to what you say.
     
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