Can You Prove This Complex Sequence Inequality?

In summary, the given inequality shows that for every complex sequence c_1, c_2, ..., c_n and for every integer 1 \leq H < N, the sum of the squares of the absolute values of the first N terms of the sequence is less than or equal to the sum of the squares of the absolute values of the terms multiplied by a factor of 4N divided by H+1, plus the sum of the absolute values of the terms multiplied by a factor of the sum of the first N terms of the sequence. This bound is important and can be further explained by a knowledgeable mathematician.
  • #1
Townsend
232
0
Show that for each complex sequence [tex]c_1, c_2, ..., c_n[/tex] and for each integer [tex]1 \leq H < N[/tex] one has the inequality

[tex]
| \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)|)
[/tex]

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... :tongue2:
 
Last edited:
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  • #2
Yeah, that's good. The more fancy meaningless symbols, the better. :biggrin:
 
  • #3
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:
 
  • #4
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me. :wink:
 
  • #5
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?
 
  • #6
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.
 
  • #7
I know the solution, but I won't say to give other people a chance. It's not that hard.
 
  • #8
Its packman eating flies, so the ansewer must be MxBxHs
 

1. What is a complex sequence inequality?

A complex sequence inequality is an inequality that involves a sequence of complex numbers, which are numbers that include a real and imaginary component. It is used to compare the values of different complex numbers in a sequence.

2. How do you prove a complex sequence inequality?

To prove a complex sequence inequality, you must show that the inequality holds true for all values in the sequence. This can be done through mathematical induction, where you first show that the inequality is true for the first few values in the sequence, and then assume it is true for a general term in the sequence to prove that it holds for all values.

3. What are some strategies for proving complex sequence inequalities?

Some strategies for proving complex sequence inequalities include using properties of complex numbers, such as their magnitudes and arguments, and applying mathematical identities and inequalities, such as the triangle inequality and Cauchy-Schwarz inequality.

4. Can you give an example of proving a complex sequence inequality?

Sure, for example, we can prove the inequality |z| < |z+1| for all complex numbers z in the sequence {z_n} = {1, 2, 3, ...}. First, we show that the inequality holds true for the first value in the sequence, z_1 = 1, since |1| = 1 < |1+1| = |2|. Then, we assume that the inequality holds for a general term in the sequence, z_k, and prove that it holds for the next term, z_k+1. Using the triangle inequality, we have |z_k+1| = |z_k + 1| ≤ |z_k| + |1| < |z_k| + |z_k| = 2|z_k|. By our assumption, we know that |z_k| < |z_k+1|, so we can substitute and get |z_k+1| < 2|z_k| < 2|z_k+1|, which is true for all values in the sequence. Therefore, the inequality holds for all values in the sequence, and it is proven.

5. Why are complex sequence inequalities important in mathematics?

Complex sequence inequalities are important in mathematics because they allow us to compare and analyze the behavior of complex numbers in a sequence. They are also used in many areas of mathematics, such as calculus, number theory, and analysis, to prove theorems and solve problems involving complex numbers. Additionally, understanding complex sequence inequalities can help improve problem-solving skills and critical thinking in mathematics.

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