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

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,102

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.**

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**kean****Member**- Registered: 2009-05-17
- Posts: 8

sure！ it's right

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,102

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.**

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

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,102

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.**

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

**#15**let 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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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

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