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Continuous Function sum_1^infty[ sin(n^2 x)/n ]

  1. Dec 6, 2006 #1
    I was wondering if someone could lend assistance in answering if the function

    f(a) = sum_{n=1}^{infinity} sin(n^2 a) / n

    is continuous at a=0?

    So this becomes showing |f(x)-f(0)| = |f(x)| < epsilon if there is some |x - 0 | = |x| < delta(epsilon).

    I was looking at a different problem which I know to be continuous where the denominator is n^2 instead of n using two different sums,

    sum_{n=1}^{n=m} |sin(n^2 x)/n^2| < sum_{n=1}^{n=m} |n^2 x / n^2|

    = m|x|

    for 0 < x < 1 (effectively limiting the size of the delta chosen) and m found by the integral value of n at which |n^2 x| < 1 is no longer true.

    And another summation beyond the value m

    sum_{m+1}^{infinity} |sin(n^2 x) /n^2| = sum_{m+1}^{infinity} |1 / n^2|

    and using the fact that the last sum is less than the integral of 1/n^2 from m to infinity which equals 1 / m.

    Long story short,

    |sum_{n=1}^{infinity} sin(n^2 x) / n^2| < m*|x| + 1/m

    Choosing m to be the floor of 1/x^{1/2} and since 0<x<1 means that

    |sum_{n=1}^{infinity} sin(n^2 x) / n^2| <|x|^{1/2} + |x|^{1/2}

    So the abs. value of this function must be less than 2|x|^{1/2}

    If we choose a delta(epsilon) = epsilon^2 / 4, it will guarentee f is less than epsilon. However, if the denominator of f is n instead of n^2 this tactic becomes useless as second summation of 1/n is divergent, and using either combination of logic results in a divergent sum. The very fact that 1/n is divergent smacks of this function lacking continuity.

    So any hints, or suggestions on approaching this? I don't really even have a clue as to whether or not it is continuous... If anyone has approached this problem before and could just say "continuous" or "not continuous" that might help. :)

    In any case thanks for reading and (maybe) replying!

    -Nelson
     
  2. jcsd
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