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Sequences problem help

  1. Mar 16, 2010 #1

    Mentallic

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    I want to express

    [tex]...-2\pi+\theta,\theta, 2\pi-\theta, 2\pi+\theta, 4\pi-\theta...[/tex]

    in terms of a variable integer k.

    e.g. ...-x,0,x, 2x, 3x... = kx, k E Z

    So I was thinking expressing it as so: [tex]2k\pi \pm \theta[/tex]
    but I believe it can be expressed in another way to avoid the [itex]\pm[/itex], using (-1)k somehow.

    How can this be done?
     
  2. jcsd
  3. Mar 16, 2010 #2

    Filip Larsen

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    Re: Sequences

    The example sequence you gave seems to indicate you want "two plusses", "a minus", "a plus", and "a minus". Is this a mistake? If you want the sequence to alternate between plus and minus for theta and still use every integer k for [itex]2\pi[/itex] you may be able to use something like
    [itex]\pi(k+1)k + (-1)^k \theta[/tex]
     
  4. Mar 16, 2010 #3

    uart

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    Re: Sequences

    Yeah I'm not sure about your example, but I think this will do what you want.

    [tex]\lfloor k/2 \rfloor 2\pi + (-1)^k \theta [/tex]
     
  5. Mar 16, 2010 #4

    Mentallic

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    Re: Sequences

    My example was just trying to illustrate what I meant, I didn't mean for it to be similar to my problem :smile:
    I think you might have a typo: [tex]\pi(k+1)k[/tex] ?

    Anyway, I solved it.

    [tex]...-\theta,\theta,2\pi-\theta,2\pi+\theta,4\pi-\theta...=\frac{\pi}{2}\left(2k+1\right)+(-1)^k\left(\theta-\frac{\pi}{2}\right), k \in Z[/tex]

    edit: I was wrong :biggrin:
     
    Last edited: Mar 16, 2010
  6. Mar 16, 2010 #5

    Mentallic

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    Re: Sequences

    uart, what does that symbol you used mean?
     
  7. Mar 16, 2010 #6

    Filip Larsen

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    Re: Sequences

    Try insert integers in [itex](k+1)k[/itex] and see what you get. It corresponds to the integer truncation uart used in his post.

    For [itex]k=0[/itex] you then get [itex]\pi - \theta[/itex]. Is that included in your sequence?
     
  8. Mar 16, 2010 #7

    uart

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    Re: Sequences

    It's called "floor(x)" and it means the "largest integer less than or equal to x"
     
  9. Mar 16, 2010 #8

    Mentallic

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    Re: Sequences

    But k(k+1) grows at a quadratic rate, not linear.
    For k=0,1,2... k(k+1)=0,2,6,12,20,30...
    which grows too rapidly.


    Yes sorry, I fixed up the error.


    Aha, like how computers truncate decimal expansions rather than rounding off by default :smile:
     
  10. Mar 16, 2010 #9

    Filip Larsen

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    Re: Sequences

    Oh my, you are quite right. What I really was trying to do was to give the same sequence as uart also gave, but some part of my brain must have been a bit too creative with trying to get rid of the integer truncation while another part must have been sleeping instead of catching the nonsense. I do apologise.
     
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