Can You Prove This Complex Sequence Inequality?

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    Inequality Proof
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The discussion revolves around proving the inequality involving a complex sequence and its components. Participants express interest in the inequality |∑_{n=1}^N c_n|^2 ≤ (4N/(H+1))(∑_{n=1}^N |c_n|^2 + ∑_{h=1}^H |ρ_N(h)|). There is a playful tone, with some participants joking about the complexity of the symbols and the need for a knowledgeable mathematician to explain the proof. One participant mentions having a solution but prefers to let others attempt it first. There is a suggestion that complex numbers can be treated as ordered pairs of real numbers, which could simplify understanding for those unfamiliar with complex analysis. Overall, the thread highlights a mix of mathematical inquiry, humor, and a collaborative spirit in tackling the problem.
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Show that for each complex sequence c_1, c_2, ..., c_n and for each integer 1 \leq H < N one has the inequality

<br /> | \sum_{n=1}^N c_n|^2 \leq \frac{4N}{H+1} ( \sum_{n=1}^N |c_n|^2 + \sum_{h=1}^H | \rho_N(h)|)<br />

Any one...matt grime perhaps? :wink:

note: if anyone actually wants to work this out let me know and I will fill in the missing parts...but don't ask me to do it... :-p
 
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Yeah, that's good. The more fancy meaningless symbols, the better. :biggrin:
 
honestrosewater said:
Yeah, that's good. The more fancy meaningless symbols, the better. :biggrin:

There is an important part missing but it is a true bound...not just meaningless... :smile:
 
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me. :wink:
 
honestrosewater said:
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me. :wink:

Yeah...it would take a really smart...creative...mathematician to do so...who would be able to do that I wonder?
 
Townsend said:
Yeah...it would take a really smart...creative...
... and patient. I've never even worked with complex numbers before. You can just treat them as ordered pairs of real numbers, right? I think I'd like that approach.
 
I know the solution, but I won't say to give other people a chance. It's not that hard.
 
Its packman eating flies, so the ansewer must be MxBxHs
 

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