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Math Help - Deducing polynomials technique

  1. #1
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    Deducing polynomials technique

    Hi guys,

    I was hoping someone would be able to tell me the name of this technique (described below) I used so I can go away and do some reading about it and hopefully learn it in a more structured way. Also if there are any similar techniques for deducing polynomials from series (like expanding the series) could you tell me the names of them so I can do some reading about them too. Or any other techniques for finding the sum of squares. Thanks.

    The technique:

    So I was trying to deduce the sum of squares:

    \sum_{x=1}^{n}x^2=x^3/3+x^2/2+x/6

    and I used the technique where you go down the levels of the difference. i.e
    the first few terms of this series is:
    1,5,14,30,55,91,140
    difference of these terms is:
    4,9,16,25,36,49
    and the diff. of the diff.:
    5,7,9,11,13
    and the level 3 diff.:
    2,2,2,2
    so now we can see that since it's level 3 .
    \sum_{x=1}^{n}x^2=O(x^3)
    comparing this to x^3 series:
    series:
    1,8,27,64,125,216,343
    level 1:
    7,19,37,61,91,127
    level 2:
    12,18,24,30,36
    level 3:
    6,6,6,6
    comparing level 3's of x^3 and  \sum^n_{x=1}x^3 we get:
    \sum_{x=1}^{n}x^2=x^3/3+O(x^2)

    Now the level 2 difference of \sum_{x=1}^{n}x^2-x^3/3 is:
    1,1,1,1,1
    and level 2 difference for x^2 is:
    2,2,2,2,2
    so we now have:
    \sum_{x=1}^{n}x^2=x^3/3+x^2/2+O(x)

    Finally the level 1 difference of \sum_{x=1}^{n}x^2-x^3/3-x^2/2 is:
    1/6,1/6,1/6,1/6,1/6,1/6
    and for x is:
    1,1,1,1,1,1

    \sum_{x=1}^{n}x^2=x^3/3+x^2/2+x/6

    and I got the right result. Now what's this technique called?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    should be n^3/3 + n^2/2 + n/6
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  3. #3
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    Newton's finite difference method
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