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Thread: Finding whether a polynomial exists

  1. #1
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    Finding whether a polynomial exists

    Given a set of values how can one determine whether any polynomial (say p) of degree atmost 3 and having real coefficients exists such that p(i) = x$_i$ for all i ?
    E.g;
    For the set of values:
    0,1,2,3,4
    there exists such a polynomial but for values:0,1,2,4,5 there isn't.
    Is it related to some theorem?
    Thanks.
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  2. #2
    Super Member TheChaz's Avatar
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    Your question could use some clarification. Why is the degree at most 3? Was that your choice?
    What are the values 0,1,2,3,4,5...??
    Given n points, you can create a polynomial of degree n-1 ...
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  3. #3
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    Given any n (x,y) pairs there exist a polynomial of degree at most n-1 that passes through those points. It might happen that some of the n equations that are given by those values are not independent. Specifically, in order that there be a polynomial of degree 3, exactly 4 of the n equations must be independent, the other n- 4 equations linear combinations of those 4.
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  4. #4
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    Hello, pranay!

    Given a set of values how can one determine whether any polynomial.
    say p(x) of degree at most 3 and having real coefficients exists
    such that p(i) = x_i for all \,i ?

    E.g;
    For the set of values: \{0,1,2,3,4\}, there exists such a polynomial

    but for values: \{0,1,2,4,5\}, there isn't.

    Is it related to some theorem?

    This is a rehash of what HallsofIvy and TheChas said . . .


    Given 2 points, there is always a line that passes through both points.
    . . That is, there is a first-degree polynomial for them.

    Given 3 points, there is always a parabola that passes through all the points.
    . . That is, there is a second-degree polynomial for them.
    There may or may not be a line (first-degree polynomial) for the 3 points.

    Given 4 points, there ia always a cubic that passes through all the points.
    . . That is, there is a third-degree polynomial for them.
    There may or may not be a line or parabola for the 4 points.

    Given 5 points, there is always a quartic that passes through all the points.
    . . That is, there is a fourth-degree polynomial for them.
    There may or may not be a line or parabola or cubic for the 5 points.

    And the pattern should be clear . . .

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