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Summation of a sequence by parts.

  1. Dec 6, 2009 #1
    I hope can someone clarify this for me.

    I have a sequence f(of n) which is like this:

    fn(x) = [tex] 0-- if--x<\frac{1}{n+1}[/tex]
    is = [tex] sin^2(x/pi)--if--\frac{1}{n+1}<=x<=\frac{ 1}{n}[/tex]
    is = [tex]0--if--\frac{1}{n}<x[/tex]

    (the - are for spaces because I don't know how to do it. Nothing is negative)

    Then there is a summation (sigma) of f(of n). I don't understand how can there be such a summation for sequence by parts or how it would look like. Can anyone explain please?
  2. jcsd
  3. Dec 6, 2009 #2


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    Science Advisor

    Given any positive x, there exist N such that x< 1/(N+1) and M such that x> 1/M. That is, the sum is a finite sum, between N and M, for every x.

    For example, if x= 1/3, then, Since 1/3< 1/(0+1), f0(1/3)= 0. Since 1/3< 1/(1+1), f1(1/3)= 0. Since 1/(2+1)= 1/3< 1/2, f2(1/3)= [itex]sin^2(1/(3\pi)[/itex]. Since 1/(3+1)< 1/3= 1/3, f3(1/3)= [itex]sin^2(1/(3\pi))[/itex]. Since 1/3< 1/4, f4(1/3)= 0 and fn(1/3)= 0 for all larger n.

    The sum is [itex]f(x)= 2sin^2(1/(3\pi))[/itex].

    In fact, it looks to me like, if x= 1/n for some integer n, [itex]f(x)= 2sin^2(1/(\pi x))[/itex] while if x is any other number, [itex]f(x)= sin^2(1/(\pi x))[/itex] since if x= 1/n, we have both x= 1/n and x= 1/((n-1)+1) and so sum two terms, while if x is not, we have only 1/(n+1)< x< 1/n for a single n.
  4. Dec 7, 2009 #3
    Hmmm, I think I get it. Thanks.
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