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Uniqueness of cubic interpolating polynomial

  1. Apr 1, 2005 #1
    This is a numerical analysis question, and I am trying to prove that the p(0), p'(0), p(1), p'(1) define a unique cubic polynomial, p. More precisely, given four real numbers, p00, p01, p10, p11, there is one and only one polynomial, p, of degree at most 3 such that p(0) = p00, p'(0) = p01, p(1) = p10, p'(1) = p11.

    I looked in a lot of numerical analysis textbooks, but the closest proof I can find proves the following: Given n+1 distinct points x0,..., xn and n + 1 values y0, ..., yn, there exists a unique polynomial p of degree of most n with the property that p(xj) = yj, for j = 0,...., n.

    For my problem I have no idea what I should do with p(0), p'(0), p(1), p'(1), the values and derivatives of p at 0 and 1. Thanks.
  2. jcsd
  3. Apr 1, 2005 #2


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    You know that a general cubic is P(x)= ax3+ bx2+ cx+ d.

    That is the cubic is determined by the 4 numbers, a, b, c, d.
    Of course, P'(x)= 3ax2+ 2bx+ c so

    P(0)= d= the given p(0)
    P'(0)= c= the given p'(0)
    P(1)= a+ b+ c + d= the given p(1)
    P'(1)= 3a+ 2b+ c= the given p'(1)

    That is four equations in four unknowns. As long at the equations are independent (the determinant of the coefficients on the left side is not 0), a, b, c, d are unique.
  4. Apr 1, 2005 #3
    Ok that makes sense. So I guess what you are using is the Fundamental Theorem of Linear Algebra -- determinant of the coefficients is not 0 iff the coefficients of the polynomial are unique? Thanks for your help.
  5. Apr 3, 2005 #4


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    An equivalent problem would be to show that the constraint:

    f(0) = 0
    f(1) = 0
    f'(0) = 0
    f'(1) = 0

    is satisfied only by the zero polynomial. (Note that if p and q are both solutions to your problem, (p-q) is a solution to this one)
  6. Apr 3, 2005 #5
    Ok thanks. :)
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