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Discrete mathematics and its application 2.4 problem 26

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Find a formula for when m [tex]\sum[/tex] k=0 the flooring function of[k1/3 ] ,m is a positive integer.

    2. Relevant equations

    n[tex]\prod[/tex] j=m aj



    3. The attempt at a solution

    the flooring function of[k1/3] = K

    the summation of K is [tex]\frac{m(m+1)}{2}[/tex]

    There's a table of useful summation formulas in the Discrete Mathematics and Its Application sixth edition textbook pg.157 and that's where I got the summation formula for K. Just plug in m in the formula for the summation of K.
     
  2. jcsd
  3. Feb 15, 2010 #2
    omit the second part

    2. Relevant equations

    nLaTeX Code: \\prod j=m aj
    ....not part of the problem
     
  4. Feb 15, 2010 #3
    omit the second part

    2. Relevant equations

    nLaTeX Code: \\prod j=m aj
    ....not part of the problem
     
  5. Feb 15, 2010 #4

    Mark44

    Staff: Mentor

    It's not clear to me what you're asking. When [tex]m\sum_{k = 0}^? floor(k^{1/3})[/tex] does what?
     
  6. Feb 15, 2010 #5
    I have to find the summation formula for floor(K1/3):; and m is the on top of the summation symbol.
     
  7. Feb 15, 2010 #6

    Mark44

    Staff: Mentor

    Start by expanding the summation:
    floor(1) + floor(21/3) + floor(31/3) + ... + floor(m1/3).

    Use enough terms so that you can find out how many of the terms will be equal to 1, to 2, to 3, and so on. That's what I would start with.
     
  8. Feb 15, 2010 #7
    I'd be willing to offer some help but I can't quite read you equations! Could you try and format them a bit better?
     
  9. Feb 15, 2010 #8

    Mark44

    Staff: Mentor

    GogoDancer12 wants to find a closed for expression for
    [tex]\sum_{k = 0}^m \lfloor k^{1/3}\rfloor[/tex]

    The [itex]\lfloor[/itex] and [itex]\rfloor[/itex] symbols are for the "floor" function, the greatest integer less than or equal to the specified argument.
     
  10. Feb 15, 2010 #9
  11. Feb 15, 2010 #10

    Mark44

    Staff: Mentor

    Again, try expanding as I suggested in post #6. You're going to get a bunch of terms that are 1, a bunch that are 2, and so forth. See if you can figure a way to count how many 1s, 2s, and so forth.
     
  12. Feb 15, 2010 #11
    after expanding the summation I got this :

    1+1+1+1+1+1+m
     
  13. Feb 15, 2010 #12

    Mark44

    Staff: Mentor

    And that works if m is, say, 64?
     
  14. Feb 15, 2010 #13
    after expanding the summation some more I got this:
    1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+...+m

    and I'm still lost.
     
    Last edited: Feb 15, 2010
  15. Feb 15, 2010 #14

    Mark44

    Staff: Mentor

    How many 1s are there? How many 2s? 3s? Can you find any pattern? At what value of k do the 1s turn to 2s? Do the 2s turn to 3s? Are you sure the last number will be m?
     
  16. Feb 16, 2010 #15
    Here's a tip: look at the inverse function of [itex]x^{1/3}[/itex], which is [itex]x^{3}[/itex]. Then you'll see this that:

    [tex]\left\lfloor k^{1/3}\right\rfloor=1 \Rightarrow k \in \left[1,2^{3}\left[=\left[1,8\left[[/tex]

    [tex]\left\lfloor k^{1/3}\right\rfloor=2 \Rightarrow k \in \left[2^{3},3^{3}\left[=\left[8,27\left[[/tex]

    and so on...

    From this you should be able to count the number of 1,2,etc.
     
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