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Constructing Polynomials

  1. Aug 22, 2009 #1
    There's a question in Calculus by Spivak about polynomials and I was wondering about how to construct them to have specific roots or values at certain points. For example it says if
    [tex]x_{1}[/tex], ..., [tex]x_{n}[/tex] are distinct numbers, find a polynomial [tex]f_{i}[/tex] such that it's of degree n-1 which is 1 at [tex]x_{i}[/tex] and 0 at [tex]x_{j}[/tex] for [tex]j \neq i[/tex]. Now I know that for roots it's simply the product [tex]\prod (x-x_{j})[/tex] running from j=1 to n and [tex]j \neq i[/tex]. That evaluates the polynomial to 0 correctly. But I don't know how to add the additional condition of [tex]x_{i}[/tex] to make it evaluate to 1. I know I can't multiply another factor [tex](x-x_{k})[/tex] because that would increase the degree to n and it wouldn't do any good anyway. Is it a piecewise defined function? That seems a little too trivial. Any hints you could give me?
     
  2. jcsd
  3. Aug 22, 2009 #2
    hehe i remember doing this question a few months ago

    What value does the polynomial you have at the moment take at x_i?
     
  4. Aug 22, 2009 #3

    HallsofIvy

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    Multiply that product by some number, a: [tex]a\prod_{j\ne i} (x-x_{j})[/tex].

    Now you can choose a to make the value at [tex]x= x_i[/tex] 1. That is, you want [tex]a\prod (x_i-x_{j})= 1[/tex] so you must have [tex]a= \frac{1}{\prod (x_i- x_j)}[/tex].

    That gives [tex]\frac{\prod (x- x_j)}{\prod (x_i- x_j)}[/tex]

    In fact, by adding things of that type, you can get a polynomial of degree n-1 that takes on specified values at n points- the "Lagrange Interpolation Polynomial".
     
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