# Regarding approximation theorem

1. Dec 2, 2005

### 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

2. Dec 2, 2005

### bigplanet401

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:
\begin{align*} 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)} \, .\\ \end{align*}
(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
$$P(f) = P(Sn) + \sum_{|n| > N} |c_n|^2 \, .$$
Take it from there.

Last edited: Dec 2, 2005
3. Dec 3, 2005

### standardflop

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

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