Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Need to figure out a formula that calculates -1,-1,+1,+1,-1,-1,+1,+1

  1. Jul 1, 2012 #1
    So I need a formula that calculates -1,-1,+1,+1,-1,-1,+1,+1,... for a value of n that is not a piece wise.

    So far I have come up with (-1)^[2n/3] but I don't like using greatest integer

    I also did:
    [cos(n × ∏)]!
    The notation is not setup correctly and anyone that knows how to do it proper let me know. It is supposed to function like this:

    cos(n∏)cos((n-1)∏)cos((n-2)∏)cos((n-3)∏)... until cos((n-n)∏)

    I have dug in algebraically but don't think there is a solution, anyone have any thoughts on this?
     
  2. jcsd
  3. Jul 1, 2012 #2

    jedishrfu

    Staff: Mentor

    -cos(n*pi/2) --> -1, 0, 1, 0 ...
    -sin(n*pi/2) --> 0, -1, 0, 1 ...
     
  4. Jul 1, 2012 #3
    The period for your sequence is 4, so it is represented as a discrete fourier series:
    [tex]
    x_n = \sum_{m = 0}^{3}{a_{m} \, \exp \left(\frac{2 \pi i \, m \, n}{4} \right)},
    [/tex]
    where the coefficients are found from:
    [tex]
    a_{m} = \frac{1}{4} \, \sum_{n = 0}^{3}{x_{n} \, \exp \left(-\frac{2 \pi i \, m \, n}{4} \right)}
    [/tex]

    Do the calculation by using [itex]x_0 = x_1 = -1, x_2 = x_3 = +1[/itex] for [itex]a_{0,1,2,3}[/itex], and express the complex exponentials via trigonometric functions through the Euler identity:
    [tex]
    e^{i \alpha} = \cos \alpha + i \sin \alpha
    [/tex]
     
  5. Jul 1, 2012 #4
    I am not familiar with this. What does it do?
    It looks complex.
     
  6. Jul 2, 2012 #5
    [tex]
    i^{(n)(n+1)}
    [/tex]
     
  7. Jul 2, 2012 #6

    Curious3141

    User Avatar
    Homework Helper

    Nice and simple.:approve:
     
  8. Jul 2, 2012 #7

    jedishrfu

    Staff: Mentor

    nice solution, need to mention for all n>0.
     
  9. Jul 2, 2012 #8
    That's awesome :)
    I played with [tex]i[/tex] but gave up on it (apparently too quickly!)
    Very simple solution
     
  10. Jul 2, 2012 #9
    You can usually get sin or cos to get you a periodic sequence nicely, especially if you don't want to use i, so here's one more way:[tex]\sqrt{2}\cos\left(\frac{\pi}{4} + \frac{n\pi}{2}\right)[/tex]
     
  11. Jul 2, 2012 #10

    jedishrfu

    Staff: Mentor

    or my solution: cos(n*pi/2 + pi) + sin(n*pi/2 + pi)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook