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Regarding approximation theorem

  1. Dec 2, 2005 #1
    The effect of a 2pi periodic function f is defined as
    [tex] P(f) = 1/(2\pi) \int_{-\pi}^\pi |f(t)|^2 \ dt[/tex]
    and Parsevals Theorem tells us that
    [tex] P(f) = \sum_{n=\infty}^\infty |c_n|^2 [/tex]. Now, it seems rather intuituve that the effect of the N'te partial sum is
    [tex] P(Sn) = \sum_{n=-N}^N |c_n|^2 [/tex] But what is the in-between math argument? And furthermore, how can i proove that the inequality [itex] P(Sn)/P(f) \geq \delta [/itex] is satisfied only if
    [tex] \sum_{|n|>N} |c_n|^2 \leq (1-\delta)P(f) [/tex]

  2. jcsd
  3. Dec 2, 2005 #2
    First Part:
    You can define "partial sum" (almost) any way you want here. If you have a series
    \sum_{n = A}^B \text{ (something)} \, ,
    any of the following are partial sums:
    f(S_1) &= \sum_{n = A^\prime}^B \text{ (something)} \, , \quad (A^\prime > A) \, ;\\
    f(S_2) &= \sum_{n = A}^{B^\prime} \text{ (something)} \, , \quad (B^\prime < B) \, ;\\
    f(S_\text{crazy}) &= \sum_{n = -1249742}^{197494614} \text{ (something)} \, .\\
    (The limits in the last expression are both between A and B.)
    You have it right. All you've done is written down a nice, symmetric partial sum, which is the one that will solve the...
    Second Part:
    P(f) = P(Sn) + \sum_{|n| > N} |c_n|^2 \, .
    Take it from there.
    Last edited: Dec 2, 2005
  4. Dec 3, 2005 #3
    Sorry, but im not sure i understand what you're trying to say. Shouldn one use the definition of effect to show that
    [tex] P(Sn) = \sum_{n=-N}^N |c_n|^2 [/tex]

    Yes. Ok. I see, that greatest value of [tex] \sum_{|n| > N} |c_n|^2 [/tex] is bound to be no higher than P(f), and only when [itex] \delta = 0 [/itex]. And ofcourse when [itex] \delta = 1 [/itex], [tex] \sum_{|n| > N} |c_n|^2 \leq 0[/tex]. But i cant see why all the in-between values of delta must satisfy
    [tex] \sum_{|n| > N} |c_n|^2 \leq (1-\delta ) P(f)[/tex]
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