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Math Help - Polynomial

  1. #1
    Junior Member Hardwork's Avatar
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    Polynomial

    Question: If x_{1}, ... , x_{n} are distinct numbers, find a polynomial f_{i} of degree n-1 which is 1 at x_{i} and 0 at x_{j} for j \ne i. Hint: the product of all (x-x_{i}) for j \ne i is zero at x_{j} if j \ne i. .

    Confusion: The part of the question that I don't understand is the 'which is 1 at x_{i} and 0 at x_{j} for j \ne i.' What are x_{i} and x_{j}? Does it mean that for any two numbers we choose from the sequence x_{1}, ..., x_{n}, we get f_{i} as 0, for one, and 1, for the other? And how would you go about finding this polynomial?
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  2. #2
    MHF Contributor chisigma's Avatar
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    The polynomial is...

    \displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})}

    Kind regards

    \chi \sigma
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  3. #3
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by chisigma View Post
    The polynomial is...

    \displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})}

    Kind regards

    \chi \sigma
    Can you do a bit of explanation, please, if you don't mind? For example, it's not apparent to me that it's of degree n-1.
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  4. #4
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    Quote Originally Posted by Hardwork View Post
    The part of the question that I don't understand is the 'which is 1 at x_{i} and 0 at x_{j} for j \ne i.' What are x_{i} and x_{j}?
    The problem asks to find, for every i=1,\dots,n, a polynomial f_i with the given property, which is a total of n polynomials.

    For example, it's not apparent to me that it's of degree n - 1.
    In the denominator, you have a product of numbers. In the numerator, you have a product of n-1 expressions of the form x-c for various constants c.

    This is similar to the interpolation polynomial in the Lagrange form (the link uses notation \ell_j instead of f_j).
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