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Solve v = f(x) for x

  1. Jun 10, 2007 #1
    Hi

    I have a summation formula that can calculate the compound sum of terms

    (a + c0) + (a + c1) + (a + c2) ... + (a + cm) to any level. Or put another way, it can sum the terms, sum the sums of the terms sum the sums of the sums of the terms etc.

    Given
    a = element of reals
    c = element of reals
    l = element of naturals
    m = element of naturals

    where
    a = 1
    c = 1
    l = 1
    m = 4

    10 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]

    where
    a = 1
    c = 2
    l = 2
    m = 4

    30 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]

    etc

    What I'm wondering is, if given a compound sum and an unknown m, is it possible to do some sorcery and solve for m?

    For example, with the simple case of summing naturals

    v = [tex] \frac{n(n + 1)}{2}[/tex]
    n = [tex]\frac{(sqrt(8v + 1) - 1)}{2}[/tex]

    Can the same sort of thing be done with the above compound sum formula?
     
    Last edited: Jun 10, 2007
  2. jcsd
  3. Jun 10, 2007 #2
    Omitting all the extraneous stuff from above, it becomes

    [tex]v = \frac{(m + l)!(al + a + cm)}{m!(l + 1)!}
    [/tex]

    Rearranging terms and expanding factorials I got this

    [tex]v = \frac {(al + a + cm)(m + 1)(m + 2)(m + 3) ... \times (m + l)}{1 \times 2 \times 3 ... \times (l + 1)}[/tex]

    But what I don't know is if there is any way to rearrange the above to isolate the m term. Anyone know if there is a general way to do that?
     
    Last edited: Jun 10, 2007
  4. Jun 10, 2007 #3

    Office_Shredder

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    Your notation is incredibly confusing... why do you need four variables to define an arithmetic sequence? At worst you should only need three
     
  5. Jun 10, 2007 #4
    Sorry. Highest level math I took in school was Algebra 2 (twenty years ago) so I'm kind of making it up as I go along.

    It's to allow for compound summation. for example

    a = real = starting point
    c = real = constant difference
    m = natural = zero based term index
    l = natural = zero based summation level

    Example 1:
    a = 1, c = 1, m = 5, l = 4

    Sum table:
    0 1 2 3 4 5 m
    ----------------------------------------
    1 1 1 1 1 1 c
    ----------------------------------------
    1 2 3 4 5 6 l = 0
    1 3 6 10 15 21 l = 1
    1 4 10 20 35 56 l = 2
    1 5 15 35 70 126 l = 3
    1 6 21 56 126 252 l = 4

    Plugged into formula
    [tex] 252 = \frac{(5 + 4)!(1 \times 4 + 1 + 1 * 5)}{5!(4 + 1)!}[/tex]

    Example 2:
    a = 1, c = 5, m = 5, l = 4

    Sum table:
    0 1 2 3 4 5 m
    ----------------------------------------
    0 5 5 5 5 5 c
    ----------------------------------------
    1 6 11 16 21 26 l = 0
    1 7 18 34 55 81 l = 1
    1 8 26 60 115 196 l = 2
    1 9 35 95 210 406 l = 3
    1 10 45 140 350 756 l = 4

    Plugged into formula
    [tex] 756 = \frac{(5 + 4)!(1 \times 4 + 1 + 5 * 5)}{5!(4 + 1)!}[/tex]

    What the formula does isn't as much of a concern as how to isolate "m"
     
    Last edited: Jun 10, 2007
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