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 June 21st, 2014, 07:40 AM #21 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Well, it isn't valid. No condition is imposed upon $f$ -- neither differentiability nor continuity. Indeed, you can see that $f$ is not continuous. Thanks from eddybob123
June 21st, 2014, 07:46 AM   #22
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Quote:
 Originally Posted by mathbalarka 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 $\mathbb{i}$.

 June 21st, 2014, 07:50 AM #23 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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! Thanks from eddybob123
 June 21st, 2014, 07:53 AM #24 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra 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. Thanks from eddybob123
 June 21st, 2014, 08:00 AM #25 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra 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, $\omega_1=1$. Thanks from eddybob123
 June 21st, 2014, 09:00 AM #26 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory No, it doesn't. $\sqrt{i} = \frac1{\sqrt{2}} + i\frac1{\sqrt{2}}$ is $\Bbb R$-linearly dependent of $i$. Thanks from eddybob123
June 21st, 2014, 09:01 AM   #27
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Quote:
 Originally Posted by v8archie 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 $f$?

Last edited by mathbalarka; June 21st, 2014 at 09:04 AM.

 June 21st, 2014, 09:19 AM #28 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,394 Thanks: 2101 Math Focus: Mainly analysis and algebra 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. Thanks from eddybob123
June 21st, 2014, 09:29 AM   #29
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Quote:
 Originally Posted by mathbalarka 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.

 June 21st, 2014, 09:39 AM #30 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Oh, yeah, I meant a nonconstant example, sorry. Thanks from eddybob123

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