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Math Help - sigma reccurence

  1. #1
    Junior Member
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    sigma reccurence

    f:N-->N
    f(0) = 1

    1+\sum_{j=0}^{n-1} jf(j), n>=1

    i.e.
    1 + ((0*f*0) + (1*f*1) + (2 * f *2) ....(n-1)*f*(n-1)) <-- I'm sure this isn't right

    then f(4) =

    1+ ((0*4*0) + (1*4*1) + (2 * 4 *2) + (3 * 4 *3))
    = 1 + 0 + 4 + 16 + 36
    = 55

    Have I worked this correctly?
    Thanks for your time.
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  2. #2
    Super Member
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    Thanks
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    No doesn't appear correct
    How about this recurrence based on your defintion
    f(n) = n.f(n-1)
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  3. #3
    Junior Member
    Joined
    Jun 2010
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    Am I correct in assuming f stands for function, f is not a numeric variable. only the n inside the function(n-1) is substituted for a number.

    f(n) = n.f(n-1) < -- ok, I see, that makes sense. The the final value goes up to n(n-1)

    so, how about

    1 + ( 1(f(n-1)) + (2(f(n-1)) ....(n(f(n-1)))

    then f(4) =

    1+ 1(1-1) + 2(2-1) + 3(3-1) + 4(4-1)
    = 1+ 0 + 2 + 6 + 12
    = 21
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