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June 21st, 2014, 07:40 AM   #21
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Well, it isn't valid. No condition is imposed upon $f$ -- neither differentiability nor continuity. Indeed, you can see that $f$ is not continuous.
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June 21st, 2014, 07:46 AM   #22
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Yes. For example, $1$ and $\pi$ are $\Bbb Q$-linearly independent.
So $A$-linearly independent means there are no $a_i \in A$ such that $\sum a_i \omega_i = 0$.

Meaning that $f(y) = c$ fits the easy question, but only if the $\omega_i$ can be found. I'm a bit stuck after $1$ and $\mathbb{i}$.
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June 21st, 2014, 07:50 AM   #23
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Indeed. I'll spill the beans, seeing that you're almost nearly done : There is no triplet $\{\omega_1, \omega_2, \omega_3\}$ in $\Bbb C$ pairwise $\Bbb R$-linearly dependent, as $\Bbb R$ is spanned over $\Bbb C$ by $\{1, i\}$.

Now try the harder question!
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June 21st, 2014, 07:53 AM   #24
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Yes, I was going to say that $\{1, \mathbb{i} \}$ being a basis for $\mathbb{C}$ means that any basis has exactly two members and thus any set of more than two meembers is linearly dependent.
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June 21st, 2014, 08:00 AM   #25
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So $f(x) = c, \omega_i = \sqrt{i} \; 1 \le I \le 3$ works.

We can have any number of $\omega$ where $\omega_n$ is the square root of the $(n-1)$th prime, $\omega_1=1$.
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June 21st, 2014, 09:00 AM   #26
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No, it doesn't. $\sqrt{i} = \frac1{\sqrt{2}} + i\frac1{\sqrt{2}}$ is $\Bbb R$-linearly dependent of $i$.
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June 21st, 2014, 09:01 AM   #27
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Yes, I was going to say that $\{1, \mathbb{i} \}$ being a basis for $\mathbb{C}$ means that any basis has exactly two members and thus any set of more than two meembers is linearly dependent.
And doesn't that mean there is no such triply periodic function $f$?
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Last edited by mathbalarka; June 21st, 2014 at 09:04 AM.
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June 21st, 2014, 09:19 AM   #28
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Well, as I say, $f(x) = c$ works for any three numbers $\omega_i$. It's not so much the function that can't exists as the linearly independent $\omega_i$.

I'm not sure that any more interesting fuunction would exist, even for the second question.
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June 21st, 2014, 09:29 AM   #29
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No, it doesn't. $\sqrt{i} = \frac1{\sqrt{2}} + i\frac1{\sqrt{2}}$ is $\Bbb R$-linearly dependent of $i$.
That $i$ was an index, not $\sqrt{-1}$. That post was looking at the second part of your question.

If you prefer
$$f(x) = c \qquad \omega_k = \sqrt{k} \qquad 1 \le k \le 3$$

I'm becoming certain that a constant function is the only type that can have multiple, linearly independent periods.
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June 21st, 2014, 09:39 AM   #30
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Oh, yeah, I meant a nonconstant example, sorry.
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