You are not logged in.
Last edited by Kurre (2009-05-18 20:41:16)
Offline
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.
Offline
sure! it's right
\sum_{k=1}^n \zeta_k^m =0
Offline
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.
Offline
Offline
sure! it's right
\sum_{k=1}^n \zeta_k^m =0
it does not hold for all m and n
Offline
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.
Offline
#15let k,n be positive integers, a a nonzero real, k<n+1 . Show that:
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)
Offline
Offline