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Math Help - A Series ?

  1. #1
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    A Series ?

    For a specific algorithm, I figured that in the algorithm, for:

    n = 3, the answer is 1
    n = 4, the answer is 4
    n = 5, the answer is 10
    n = 6, the answer is 20
    n = 7, the answer is 35
    n = 8, the answer is 56
    n = 9, the answer is 84
    n = 10, the answer is 84
    n = 11, the answer is 120

    And so forth.

    I need the sum of all numbers in general, for n >= 3.

    I can't quite see what the general formula would be. It initially appeared to be a geometric series.

    So,

    1 + 4 + 10 + 20 + 35 + 56 + 84 + 120...

    It goes up in increments of: 3, 6, 10, 15, 21, 28, 36
    Where:
    3 to 6 --> 3
    6 to 10 --> 4
    10 to 15 --> 5
    15 to 21 --> 6
    x to y --> n + 1, where is n is the previous amount it went up by.

    Anyone see the generic formula/series?

    Thanks.
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  2. #2
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    Quote Originally Posted by AfterShock View Post
    For a specific algorithm, I figured that in the algorithm, for:

    n = 3, the answer is 1
    n = 4, the answer is 4
    n = 5, the answer is 10
    n = 6, the answer is 20
    n = 7, the answer is 35
    n = 8, the answer is 56
    n = 9, the answer is 84
    n = 10, the answer is 84
    n = 11, the answer is 120

    And so forth.

    I need the sum of all numbers in general, for n >= 3.

    I can't quite see what the general formula would be. It initially appeared to be a geometric series.

    So,

    1 + 4 + 10 + 20 + 35 + 56 + 84 + 120...

    It goes up in increments of: 3, 6, 10, 15, 21, 28, 36
    Where:
    3 to 6 --> 3
    6 to 10 --> 4
    10 to 15 --> 5
    15 to 21 --> 6
    x to y --> n + 1, where is n is the previous amount it went up by.

    Anyone see the generic formula/series?

    Thanks.
    Are you serious ( )?

    Here is a hint. It is a cubic sequence (not series).
    Thus,
    a_n=An^3+Bn^2+Cn+D

    You job is to find A,B,C,D.
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  3. #3
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    Quote Originally Posted by AfterShock View Post
    For a specific algorithm, I figured that in the algorithm, for:

    n = 3, the answer is 1
    n = 4, the answer is 4
    n = 5, the answer is 10
    n = 6, the answer is 20
    n = 7, the answer is 35
    n = 8, the answer is 56
    n = 9, the answer is 84
    n = 10, the answer is 84
    n = 11, the answer is 120

    And so forth.
    Assume the duplicate 84 is a mistake.

    Compute a diference table for this:

    Code:
    1   4  10  20  35  57  84  120
      3   6  10  15  21  28  36
        3  4   5   6   7   8
         1   1   1   1   1
    That the n-th differences are a constant is a characteristic of
    a n-th order polynomial, so here its the third differences are
    constant so the original sequence is a cubic in n.

    So fit a cubic to your original data to obtain the formula for the n-th
    term (note this is a bit of a risk as in general there is no guarantee
    that the data is produced this way, your really need to examine the
    underlying process to check this).

    RonL
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  4. #4
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    A result I discovered about sequencial differeneces would help you find the main coefficient of that polynomial.

    Since the number of terms until a trivial sequnce is 3 the polynomial is degree 3. Thus, the trivial sequcen (the last sequnce until a constant) must be 3!=6 but this is 1. Thus, you need it to be 1/6. So that when you multiply by the factorial of the degree it gives 1. That even further simplifies your problem.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    A result I discovered about sequencial differeneces would help you find the main coefficient of that polynomial.

    Since the number of terms until a trivial sequnce is 3 the polynomial is degree 3. Thus, the trivial sequcen (the last sequnce until a constant) must be 3!=6 but this is 1. Thus, you need it to be 1/6. So that when you multiply by the factorial of the degree it gives 1. That even further simplifies your problem.
    There is a procedure of Newton's that will give the polynomial from the
    difference table - I beleive its in Actons "Numerical methods that Work".

    RonL
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  6. #6
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    Yes, Perfect, I was serious . Even Math Majors do forget some of the simple stuff. I think I got it, though.

    C(n+3, 3) = (n + 1)(n + 2)(n + 3)/6
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  7. #7
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    Meh, not quite. Since n starts with 3!
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  8. #8
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    Quote Originally Posted by Tetris Princess
    Meh, not quite. Since n starts with 3!
    Not sure what you are trying to say.
    I mean that if you have a polynomial of degree 3 with no coefficient in front (except unity of course) the final difference is 3!
    (This holds in general).

    Thus, If you have Ax^3 then the final difference is A(3!)=1
    Thus, A=1/6

    Quote Originally Posted by CaptainBlank
    There is a procedure of Newton's that will give the polynomial from the
    difference table - I beleive its in Actons "Numerical methods that Work".
    Is not true that finding a result thyself if most satisfying?
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  9. #9
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    So, you're saying in general, 1/6 is constant, where A alternates to
    (n)(n+1)...(n+k)!

    But for the algorithm the n starts at 3. It can't be less than 3 or it's undefined. I will half to incorporate a sigma then, for n to start at 3 and to still get the same results. And,

    (1/6)*3! = 1
    (1/6)*4! = 4
    (1/6)*5! = 20

    (Thereby already skipping n = 5 [10])
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    Not sure what you are trying to say.
    I mean that if you have a polynomial of degree 3 with no coefficient in front (except unity of course) the final difference is 3!
    (This holds in general).

    Thus, If you have Ax^3 then the final difference is A(3!)=1
    Thus, A=1/6


    Is not true that finding a result thyself if most satisfying?
    Yes, but where do you start? (Note I did not recomend using it, just
    observed that it exists - I would not use it to do this problem -- but
    then I would not have found the op count formula by fitting to a table
    of specific cases anyway, I would have worked from the algorithm
    description and derived the formula directly (if at all possible)).

    RonL
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    Yes, but where do you start?
    Start what?
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  12. #12
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    Quote Originally Posted by ThePerfectHacker View Post
    Start what?
    In finding results for youself.

    Do we start from discovering that 1+1=2, and proceed to deduce the
    existance or rice pudding on our own?

    RonL
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  13. #13
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    Quote Originally Posted by CaptainBlank View Post
    In finding results for youself.

    Do we start from discovering that 1+1=2, and proceed to deduce the
    existance or rice pudding on our own?

    RonL
    I still do not understand you?
    ---
    Speaking more about sequencial differences is there a non-trivial answer to the problem,
    d(a_n)=a_n
    Where "d" is the derivative operator (i.e. subtraction).
    Note the Fibonacci sequence, almost works.

    It seems to be the fiboanny sequence plays the role of differencial equations for sequences (if there is such a thing) like the exponential function if differncial equaions.
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  14. #14
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    And the cubic sequence is? Since apparently my combinatorics formula doesn't work?
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  15. #15
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    Quote Originally Posted by AfterShock View Post
    And the cubic sequence is? Since apparently my combinatorics formula doesn't work?
    write the cubic as N(n)=a + b n + c n^2 + d n^3,

    then using the data you gave:

    n = 3, the answer is 1
    n = 4, the answer is 4
    n = 5, the answer is 10
    n = 6, the answer is 20

    we have a set of linear simultaneous equations:

    a+3b+9c+27d=1
    a+4b+16c+64d=4
    a+5b+25c+125d=10
    a+6b+36c+216d=20

    which can be solved for a, b, c, d.

    Now take these to QuickMath for solution and we get what is shown in the attachment

    RonL
    Attached Thumbnails Attached Thumbnails A Series ?-gash.jpg  
    Last edited by CaptainBlack; September 29th 2006 at 04:42 AM.
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