Regarding approximation theorem

[standardflop]

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

Thanks

 

[bigplanet401]

First Part:
You can define "partial sum" (almost) any way you want here. If you have a series
<br /> \sum_{n = A}^B \text{ (something)} \, ,<br />
any of the following are partial sums:
<br /> \begin{align*}<br /> f(S_1) &amp;= \sum_{n = A^\prime}^B \text{ (something)} \, , \quad (A^\prime &gt; A) \, ;\\ <br /> f(S_2) &amp;= \sum_{n = A}^{B^\prime} \text{ (something)} \, , \quad (B^\prime &lt; B) \, ;\\<br /> f(S_\text{crazy}) &amp;= \sum_{n = -1249742}^{197494614} \text{ (something)} \, .\\<br /> \end{align*}<br />
(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:
Notice
<br /> P(f) = P(Sn) + \sum_{|n| &gt; N} |c_n|^2 \, .<br />
Take it from there.

 

[standardflop]

First Part:
You can define "partial sum" (almost) any way you want here. If you have a series
<br /> \sum_{n = A}^B \text{ (something)} \, ,<br />
any of the following are partial sums:
<br /> \begin{align*}<br /> f(S_1) &amp;= \sum_{n = A^\prime}^B \text{ (something)} \, , \quad (A^\prime &gt; A) \, ;\\ <br /> f(S_2) &amp;= \sum_{n = A}^{B^\prime} \text{ (something)} \, , \quad (B^\prime &lt; B) \, ;\\<br /> f(S_\text{crazy}) &amp;= \sum_{n = -1249742}^{197494614} \text{ (something)} \, .\\<br /> \end{align*}<br />
(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...

Sorry, but I am not sure i understand what you're trying to say. Shouldn one use the definition of effect to show that
P(Sn) = \sum_{n=-N}^N |c_n|^2

Second Part:
Notice
<br /> P(f) = P(Sn) + \sum_{|n| &gt; N} |c_n|^2 \, .<br />
Take it from there.

Yes. Ok. I see, that greatest value of \sum_{|n| &gt; N} |c_n|^2 is bound to be no higher than P(f), and only when \delta = 0. And ofcourse when \delta = 1, \sum_{|n| &gt; N} |c_n|^2 \leq 0. But i can't see why all the in-between values of delta must satisfy
\sum_{|n| &gt; N} |c_n|^2 \leq (1-\delta ) P(f)