Inequality proof

  • Thread starter Townsend
  • Start date
  • #1
221
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:

Answers and Replies

  • #2
honestrosewater
Gold Member
2,105
5
Yeah, that's good. The more fancy meaningless symbols, the better. :biggrin:
 
  • #3
221
0
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
honestrosewater
Gold Member
2,105
5
Okay, I'll take your word for it. It would be nice if someone were around to explain it to me. :wink:
 
  • #5
221
0
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
honestrosewater
Gold Member
2,105
5
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
Johnny5
I know the solution, but I won't say to give other people a chance. It's not that hard.
 
  • #8
wolram
Gold Member
4,267
557
Its packman eating flies, so the ansewer must be MxBxHs
 

Related Threads on Inequality proof

  • Last Post
11
Replies
262
Views
27K
  • Last Post
10
Replies
227
Views
20K
  • Last Post
6
Replies
144
Views
16K
Replies
5
Views
5K
Replies
65
Views
13K
Replies
49
Views
5K
Top