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#26 2009-05-18 20:40:28

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Kurre's Exercises

Last edited by Kurre (2009-05-18 20:41:16)

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#27 2009-05-22 07:58:04

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Kurre's Exercises

Hi Kurre;

Will you please post your solution to #7. I have been using summation by parts, abels transformation, exponential substitution to get a geometric sum and all the trig identities I know.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#28 2009-05-22 10:39:45

kean
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Registered: 2009-05-17
Posts: 8

Re: Kurre's Exercises

sure! it's right

\sum_{k=1}^n \zeta_k^m =0

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#29 2009-05-22 15:27:34

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Kurre's Exercises

Hi Bo Li;

Welcome to the forum.

You forgot to enclose your latex between {math}{/math} with [ replacing { and ] replacing } so it looks like this:

Anyway, how does this prove # 14


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#30 2009-05-23 00:04:13

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Kurre's Exercises

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#31 2009-05-23 00:31:08

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Kurre's Exercises

kean wrote:

sure! it's right

\sum_{k=1}^n \zeta_k^m =0

it does not hold for all m and n

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#32 2009-05-23 00:37:58

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Kurre's Exercises

Thanks Kurre for providing the answer to #7.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#33 2009-05-24 21:40:00

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Kurre's Exercises

#15let k,n be positive integers, a a nonzero real, k<n+1 . Show that:


both with real analysis and by using residue calculus

edit: i did a mistake so i dont know if its possible to do this using residues, but that does not mean it must be impossible

Last edited by Kurre (2009-05-25 06:43:09)

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#34 2009-05-27 21:12:01

Kurre
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Registered: 2006-07-18
Posts: 280

Re: Kurre's Exercises

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