Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

? asked in Science & MathematicsMathematics · 1 decade ago

Given n points and n functions, find a linear combination of n functions that fits all n points(details below)?

Fit f(x) = Σ[k=1, n] a_k*g_k(x) to n points, where

①g_1, ..., g_n are linearly independent real functions given.

②all points given are in the domain of all g_1, ..., g_n.

③a_1, ..., a_n are real numbers you pick.

Intuitively, I can set up n simultaneous equations in n unknowns. But is it always the case?

1 Answer

Relevance
  • 1 decade ago
    Favorite Answer

    I assume that the given functions are defined on some interval J, and are linearly

    independent as functions on J.

    In other words, if c_1 g_1(x) + ... + c_n g_n(x) = 0

    for *every* x in J, then all the coefficients c_k = 0.

    Also n points p_1, ..,p_n are given.

    It is quite possible that g_k(p_r) = 0 for all k & r.

    Trivial example with n = 1:

    Given function g(x) = x

    Given point p = 0.

    --------------

    What is needed is that the n functions g_k are linearly independent

    functions restricted to the given set S of n points {p_k : 1=<k =< n}

    Proof is trivial: the set F(S) of real functions on S is n dimensional,

    and the restrictions of the n g's to S are linearly independent, so are

    a basis of F(S). That is exactly what is required.

    ---

    Major example: the polynomials 1, x, x^2, ..., x^(n-1). There is a standard

    formula for finding a polynomial P of degree (n-1) with P(x_k) = a_k; it's

    a sum of terms a_k Q_k(x) where Q_k(x_k) = 1 and Q_k(x_j) = 0 otherwise.

Still have questions? Get your answers by asking now.