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Math Help - How to define a quadratic series algebraically?

  1. #1
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    How to define a quadratic series algebraically?

    If I have a series of numbers such as 1, 4, 26, 137, how do I work out an equation that will define the function that produces it?
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  2. #2
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    Quote Originally Posted by ribbie View Post
    If I have a series of numbers such as 1, 4, 26, 137, how do I work out an equation that will define the function that produces it?
    this sequence of numbers you listed is not generated by a "quadratic".

    my calculator gets a cubic equation to fit rather closely ...

    y = \frac{35}{3} x^3 - \frac{121}{2} x^2 + \frac{617}{6} x - 53
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  3. #3
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    Quote Originally Posted by ribbie View Post
    If I have a series of numbers such as 1, 4, 26, 137, how do I work out an equation that will define the function that produces it?
    Code:
     1    4     26      137
       3    22     111
         19    89
            70
    Watch the left-hand numbers 1,3,19,70:

    1 + 3(n-1) + 19(n-1)(n-2) / 2! + 70(n-1)(n-2)(n-3) / 3! : see the pattern?

    Simplifies to:
    (70n^3 - 363n^2 + 617n - 324) / 6

    You can now get next term (407) and see what it "gives":
    Code:
     1    4     26      137       407
       3    22     111      270
         19    89      159
            70     70
                0
    OK?
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  4. #4
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    Amazing! Thanks to both of you!
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  5. #5
    Member rtblue's Avatar
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    I do agree that this function is more suited to a cubic, but if you wanted to derive a quadratic equation from a set of data, you could use Wilmer's method, or you could simply substitute points into the standard form of a quadratic, and solve for a, b, and c.
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