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Math Help - Fourier analysis; trigonometric sum.

  1. #1
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    Fourier analysis; trigonometric sum.

    Sn = Sum from k=1 to n of Ck*cos (kt), where Ck = 1/2^k.

    Prove Sn >= .7 for 0 <= t <= .1.

    This question is killing me.

    Here is what I've got so far:

    I estimated cos kt > .95 for 0 <= kt <= .3 which hits the first three terms... now I need to find a lower bound for all the terms, and I think (from previous examples) I'm going to have to use the difference form of the triangle inequality...? Any guidance would be greatly appreciated. Thanks.
    Last edited by cgiulz; September 5th 2009 at 01:15 PM.
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  2. #2
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    Hey again,

    So, I have a solution, but something is seriously off. If anyone gets a chance please correct any logical errors you may come across.

    Thank you.

    Ok, so I estimated cos u > .995 for 0<= t <= .1.
    Then, I set up the difference form of the triangle inequality:
    |1/2*cos (t) + 1/4*cos(2t)+..+1/2^k*cos(kt)|

    >= |1/2||cos(t)| - |1/4||cos(2t)|-..-|1/2^k||cos(kt)|.

    >=1/2*.995 - 1/4*.995-...-1/2^k*.995 (for 0 <= t <= .1)

    >=.7 ????--> (this is supposed to be my conclusion, but clearly this step is not true, or am I not seeing something).

    Thanks again.
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