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.

June 21st, 2014, 07:46 AM  #22  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 Math Focus: Mainly analysis and algebra  Quote:
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! 
June 21st, 2014, 07:53 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 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.

June 21st, 2014, 08:00 AM  #25 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 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 $(n1)$th prime, $\omega_1=1$. 
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 $i$.

June 21st, 2014, 09:01 AM  #27 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  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: 6,555 Thanks: 2148 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. 
June 21st, 2014, 09:29 AM  #29  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,555 Thanks: 2148 Math Focus: Mainly analysis and algebra  Quote:
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.


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