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Exponential function question

  1. Aug 30, 2011 #1
    why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) ) always seems to be when k=36?
     
  2. jcsd
  3. Aug 30, 2011 #2

    SteamKing

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    What is n?
     
  4. Aug 30, 2011 #3
    integers
    0<n<LIM
     
  5. Aug 31, 2011 #4
    ^ What's "LIM"? Do you mean n is any positive integer?

    I suppose the job to be done here is to find which value of k maximizes the expression in question. Let [itex]Q_n(k)[/itex] be that expression. The problem is to find [itex]k_0[/itex] such that, for each n, [itex]Q_n(k_0) \geq Q_n(k)[/itex] for all k -- and subsequently, to show that apparently [itex]k_0 = 36[/itex]. Is this what you're asking?
     
  6. Aug 31, 2011 #5
    Yes, you are correct. Thanks for stating it in a better way.
     
  7. Aug 31, 2011 #6
    Well, I feel real dumb... Nevermind this thread, I had an issue with the application I was using to compute the results. Sorry to all who spent any time on this.
     
  8. Aug 31, 2011 #7

    Interesting function none the less.

    t=(((n-k)*(k-1)/(2k)) + (k-1))/(n-1)

    n^(t)

    when k=1 then t=0 and n^(t)=k

    when k=n^(1/2) then t=0.5 and n^(t)=k

    when k=n then t=1 and n^(t)=k
     
  9. Sep 1, 2011 #8
    WolframAlpha shows some interesting results at well. The series expansion shows terms involving double factorial numbers.

    http://www.wolframalpha.com/widgets/view.jsp?id=daf29fc2857c2b71d6be58dcc6e7ef49 [Broken]

    http://www.wolframalpha.com/widgets/view.jsp?id=b8ba4c95900e275211a98d8bd1b0a53c [Broken]
     
    Last edited by a moderator: May 5, 2017
  10. Sep 1, 2011 #9
    Looking at its deviation from k is very interesting:

    k - n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )


    http://dl.dropbox.com/u/13155084/nt-k-4.png [Broken]
    http://dl.dropbox.com/u/13155084/nt-k-9.png [Broken]
    http://dl.dropbox.com/u/13155084/nt-k-16.png [Broken]
    http://dl.dropbox.com/u/13155084/nt-k-25.png [Broken]
    http://dl.dropbox.com/u/13155084/nt-k-36.png [Broken]
    http://dl.dropbox.com/u/13155084/nt-k-49.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  11. Sep 4, 2011 #10
    so basically the roots of the function:

    log(n,k) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

    are

    k=1
    k=n^(1/2)
    k=n

    Is this a correct statement?
     
    Last edited by a moderator: May 5, 2017
  12. Sep 5, 2011 #11
    more information

    "Another link into Eulers Generalized Pentagonal Numbers and the divisor function d(n):



    For our Divisor summatory function we have:

    D(n) = SUM(d(n)) :

    for k = 0 --> floor [sqrt n]
    SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)


    The notable difference in the equation from the published version is the:

    (n - k^2)/k (congruence of squares)

    which is derived from the

    z = (n - k^2)/2k + i n^(1/2)

    forming a parabolic coordinate system.


    The function (n - k^2)/2k forms a divisor symmetry centered on the square-root of n.

    Example:

    k = divisors of n {1,2,3,4,6,9,12,18,36}
    n = 36


    +17.5, +8.0, +4.5, +2.5, 00.0, -2.5, -4.5, -8.0, -17.5




    key results:
    sqrt(n) = 0
    Sum Terms = 0

    Offsetting by -((n-1)/2) = -17.5 and taking the absolute values gives us:


    0, 9.5, 13, 15, 17.5, 20, 22, 25.5, 35

    Key results:
    sqrt(n) = (n-1)/2;




    Another way to generate these terms is:

    ((n-k)*(k-1)/2k) + (k-1)

    The key ratio here being the (k-1)/2k function.

    reducing this ratio sequence we get:

    01/04, 01/03, 03/08, 02/05, 05/12, 03/07, 07/06, 04/09, 09/20, 05/11, 11/24, 06/13, 13/28, 07/15, 15/32, 08/17, 17/36

    or

    01 01 03 02 05 03 07 04 09 05 11 06 13 07 15 08 17
    04 03 08 05 12 07 16 09 20 11 24 13 28 15 32 17 36


    Showing a direct connection to Eulers Generalized Pentagonal Numbers and the divisor function d(n)



    **

    A026741 ( n if n odd, n/2 if n even. ) = xx, 00, 01, 01, 03, 02, 05, 03, 07, 04, 09, 05, 11, 06, 13, 07, 15, 08, 17
    A022998 ( If n is odd then n else 2*n. ) = 00, 01, 04, 03, 08, 05, 12, 07, 16, 09, 20, 11, 24, 13, 28, 15, 32, 17, 36

    A026741 = Partial sums give Generalized Pentagonal Numbers A001318 = 00, 01, 02, 05, 07, 12, 15, 22, 26, 35, 40, 51, 57
    A022998 = Partial sums give Generalized Octagonal Numbers A001082 = 00, 01, 05, 08, 16, 21, 33, 40, 56, 65, 85, 96, 120 "
     
    Last edited: Sep 5, 2011
  13. Sep 5, 2011 #12
  14. Sep 7, 2011 #13
    The divisor symmetry still shows up nicely. For example 36:

    (ln(x)/ln(36)) - (((36-x)*(x-1)/(2*x)) + (x-1))/(36-1))

    http://dl.dropbox.com/u/13155084/36.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  15. Sep 9, 2011 #14
    for the function

    f(n,k) = ( (ln(k)/ln(n)) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

    the

    local minimum = (((n-1)/2)-sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))
    local maximum = (((n-1)/2)+sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))

    ex: n=49

    49 minimum = (24-sqrt(576-49 log^2(7)))/(log(7))
    49 maximum = (24+sqrt(576-49 log^2(7)))/(log(7))


    and min * max = n
     
  16. Sep 14, 2011 #15
    the contour plot shows the divisor function very nicely:

    (n-k^2)/2k mod .5

    http://www.wolframalpha.com/input/?i=ContourPlot[Mod[(-k^2+++n)/(2+k),+0.5],+{k,+-2,+2},+{n,+-4,+4}]
     
  17. Sep 14, 2011 #16
    Last edited by a moderator: May 5, 2017
  18. Sep 19, 2011 #17
    A better way to plot this:

    e^(pi i ( ((n-k^2)/(2k)) / ((n-1)/2) ) ) for k=1 to 36, n=36

    http://dl.dropbox.com/u/13155084/unit%20circle.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
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