1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Floor function experimenting

  1. Feb 10, 2013 #1
    Results from exploring the floor function and presented for general interest. Be good.

    When [itex] k = 1, 2, 3, ... [/itex] :

    (1) [itex] \left\lfloor \frac{1}{n}(k-1) + 1 \right\rfloor = \underbrace{1, ..., 1}_{n}, \underbrace{2, ..., 2}_{n}, \underbrace{3, ..., 3}_{n}, ... [/itex]

    (2) [itex] \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor = 1, \underbrace{2, 2}_{}, \underbrace{3, 3, 3}_{}, \underbrace{4, 4, 4, 4}_{} ... [/itex]
    (3) [itex] \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor - \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor = 1, \underbrace{1, 0}_{}, \underbrace{1, 0, 0}_{}, \underbrace{1, 0, 0, 0}_{}, ... [/itex]

    (4) [itex] \left\lfloor \sqrt{k-1} + \frac{3}{2} \right\rfloor = 1, \underbrace{2, 2}_{}, \underbrace{3, 3, 3, 3}_{}, \underbrace{4, 4, 4, 4, 4, 4}_{}, ... [/itex]
    (5) [itex] \left\lfloor \sqrt{k-1} + 1 \right\rfloor = 1, \underbrace{2, 2, 2}_{}, \underbrace{3, 3, 3, 3, 3}_{}, \underbrace{4, 4, 4, 4, 4, 4, 4}_{}, ... [/itex]

    (6) [itex] k - \left\lfloor k-1 \right\rfloor ^2 = 1, \underbrace{1, 2, 3}_{}, \underbrace{1, 2, 3, 4, 5}_{}, \underbrace{1, 2, 3, 4, 5, 6, 7}_{}, ... [/itex]

    (7) [itex] k - n \left\lfloor \frac{1}{n} (k-1) \right\rfloor = \underbrace{1, 2, 3, ..., n}_{}, \underbrace{1, 2, 3, ..., n}_{}, \underbrace{1, 2, 3, ..., n}_{}, ... [/itex]

    (8) [itex] k - \frac{1}{2} \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor = 1, \underbrace{1, 2}_{}, \underbrace{1, 2, 3}_{}, \underbrace{1, 2, 3, 4}_{}, ... [/itex]
    (9) [itex] 1 - k + \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor + \frac{1}{2} \left\lfloor \sqrt{2k} + \frac{1}{2} \right\rfloor \left\lfloor \sqrt{2k} - \frac{1}{2} \right\rfloor= 1, \underbrace{2, 1}_{}, \underbrace{3, 2, 1}_{}, \underbrace{4, 3, 2, 1}_{}, ... [/itex]

    In particular when presenting the series product [itex](\sum_{j=1}^n a_j)(\sum_{j=1}^n b_j)[/itex] as the sum of the terms of the nxn matrix [itex] \begin{bmatrix} a_1 b_1 & a_1 b_2 & ... & a_1 b_n \\ a_2 b_1 & a_2 b_2 & ... & ... \\ ... & ... & ... & ... \\ a_n b_1 & ... & ... & a_n b_n \end{bmatrix} [/itex] it can be seen that the horizontal a's follow (1) while the respective b's accord to (7) above.

    Judging from this we get the result [itex] (\sum_{j=1}^n a_j)(\sum_{j=1}^n b_j) = (\sum_{j=1}^{n^2} a_{\left\lfloor \frac{1}{n}(j-1) + 1 \right\rfloor} b_{j - n \left\lfloor \frac{1}{n} (j-1) \right\rfloor}) [/itex].
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Floor function experimenting
Loading...