Hello friends.

I was just wondering if there is an easy proof for the following:

I may have found a horrid proof using the rational root theorem, but I have not checked the details.Quote:

For ever ,

Any input is appreciated!

Printable View

- May 8th 2010, 08:25 PMDrexel28Sum of nth powers
Hello friends.

I was just wondering if there is an easy proof for the following:

Quote:

For ever ,

Any input is appreciated! - May 8th 2010, 08:57 PMroninpro
You can probably try to show that the set is linearly independent over the field . The result will follow immediately.

It seems to be fairly straightforward if you know some field theory. (But I haven't actually tried it, so I could be wrong.) - May 8th 2010, 09:01 PMDrexel28
- May 8th 2010, 11:50 PMBacterius
Wouldn't it be possible to somehow get rid of the roots by taking the inverses of the exponents while keeping the sum equivalent ? Dealing with powers seems simpler somehow.

This is just an idea, I guess it would be possible by messing around with logs and using their irrational properties. (Wondering) - May 8th 2010, 11:57 PMDrexel28
- May 9th 2010, 12:05 AMBacterius
I was dreading it :D

I meant, expressing your statement using a sum of powers instead of a sum of roots. Some changes would be required, but for instance :

- or even better -

Seems easier than dealing with somehow.

But this is not gonna be "elementary" anyway :(