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How do I write this in summation notation?

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex](\sin x) (\cos x)^{n - 1} + (\sin x) (\cos x)^{\frac{n - 1}{2}} + (\sin x) (\cos x)^{\frac{\frac{n - 1}{2}}{2}} + (\sin x) (\cos x)^{\frac{\frac{\frac{n - 1}{2}}{2}}{2}} ...[/itex]

    n is an odd number and the series ends when (n-1)/2^k = 1, and the last term ends up being sin(x) only (no cos factor). For example, if n was 5, then 4 / 2^2 would be the last term in the series, so it looks like cos^1(x)sin(x) would be the last term, but the last term is actually sin(x) only. So if n was 5, the series would look like:

    [itex](\sin x) (\cos x)^4 + (\sin x) (\cos x)^{2} + (\sin x) [/itex]


    2. Relevant equations
    N/A

    3. The attempt at a solution
    I got [itex]\sin(x)\sum_{k=0}^{m}\cos^{2^{m-k}}(x)[/itex] for [itex]n=2^m+1[/itex] but the last term in this equation is sin(x)cos(x), not sin(x).
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    vela

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    You can't write it the way you want to because you have an inconsistency. You say that series ends when the exponent of cosine is 1, but you also say that the series ends when the exponent of cosine is 0.
     
  4. Feb 12, 2013 #3
    Also, do you want this to be for just odd n as you state at the beginning? Or [itex] n = 2^m + 1[/itex] as you state later? If the former case, generic odd n will almost never result in [itex] \frac{n-1}{2^k} = 1[/itex], so what is your termination condition in that case?
     
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