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Is there an infinite series composed possibly of periodic functions for a function $f(k)$ with the property that if, $ k\equiv b $ mod a, $f(k)=1$, and if not $f(k)=0$,

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I have no idea what you are asking. –  copper.hat Nov 28 '12 at 23:51
    
I reworded it, sorry –  Ethan Nov 28 '12 at 23:53
    
Do you mean "if $\;k\equiv b \bmod a$, then $f(k) = 1$, and if not, $f(k) = 0$", $\;a, b$ integers? –  amWhy Nov 28 '12 at 23:57
    
Hi Ethan, unfortunately I am still lost. The word Fourier disappeared, was that intentional? –  copper.hat Nov 29 '12 at 0:03
    
Yes I meant for k to be congruent to b mod a, and the word fourier series did disapeer, I don't belive I know enough about fourier series or much less harmonic analysis, to start incorperating them into my question. –  Ethan Nov 29 '12 at 0:09

2 Answers 2

up vote 0 down vote accepted

I'm interpreting the question as asking for a trigonometric function $f(x)$ such that for integers $k$, $f(k) = 1$ if $k \equiv b \mod a$ and $f(k) = 0$ otherwise, where $b$ is an integer and $a$ is a positive integer.
You could take $$f(x) = a^{-1} \sum_{j=0}^{a-1} e^{2 \pi i (x-b)/a}$$

If you insist on an infinite series rather than a finite sum, just add terms such as $c_n \sin(n \pi x)$ that are $0$ on the integers, with appropriate coefficients.

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Ive seen this formula before, but I sorta need it to be non finite, I don't know much about such series, so I wouldn't know how to construct the coefficients you speak of, I would appreciate it, if you could help me out with that. –  Ethan Nov 29 '12 at 0:12
    
The $c_n$ can be anything, as long as the series converges. Take $c_n = 2^{-n}$, for example, if you don't want them to be $0$. –  Robert Israel Nov 29 '12 at 0:25

I'm not sure what you mean by periodic function.

You could have $\sum \sin\left( \cfrac{1}{n^2}\right)$ and since you have $\sin\left( \cfrac{1}{n^2}\right) \sim_0 \cfrac{1}{n^2}$ and $\sum \cfrac{1}{n^2}$ converges, it would converge.

But if what you mean by periodic is that every $p$ numbers, you get the same value again then, if you have one non-zero value, $\sigma u_n$ will diverge because $(u_n)_n$ won't converge to $0$ (because $\exists \epsilon > 0, \forall n_0 \in \mathbb{N}, \exists n \ge n_0, u_n > \epsilon$, you would need to take $\epsilon$ smaller in norm than your non-zero value and then for $n$ you just pick the next index at which you get that non-zero value).


Apparently, you're in the second case.

$\exists \epsilon > 0, \forall n_0 \in \mathbb{N}, \exists n \ge n_0, u_n > \epsilon$

You can take $\epsilon = \cfrac{1}{2}$ and $n=b+ka$ for some $k\in\mathbb{N}$ and you can find one such $n\ge n_0, \forall n_0$.

So $(u_n)_n$ doesn't converge to 0 so $\sum u_n$ has no chance of converging.

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