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## #26 2009-05-19 18:40:28

Kurre
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### Re: Kurre's Exercises

Last edited by Kurre (2009-05-19 18:41:16)

## #27 2009-05-23 05:58:04

bobbym

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### 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.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #28 2009-05-23 08:39:45

kean
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### Re: Kurre's Exercises

sure！ it's right

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

## #29 2009-05-23 13:27:34

bobbym

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### 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.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

Kurre
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## #31 2009-05-23 22:31:08

Kurre
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### 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

## #32 2009-05-23 22:37:58

bobbym

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### 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.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #33 2009-05-25 19:40:00

Kurre
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### 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 (2009-05-26 04:43:09)

Kurre
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