Question

find this

Answer 1

\[\sum_{k=1}^{2014}\frac{1}{(1 + \omega_k )^2}\]where \(1,\omega_1 , \omega_2, \cdots, \omega _{2014}\) are the 2015th roots of unity. So I did this. \[x^{2015} - 1 = 0 \Rightarrow x = 1, \omega_1, \cdots , \omega_{2014}\]\[(x+1)^{2015}-1 = 0 \Rightarrow x= 1+1, 1 + \omega_1 , \cdots , 1 + \omega_{2014}\]\[\left(\frac{1}{\sqrt{x}} + 1\right)^{2015}-1=0\Rightarrow x = \frac{1}{4}, \frac{1}{(1+\omega_1)^2}, \cdots , \frac{1}{(1+\omega_{2014})^2}\]

Answer 2

Now the only problem I'm having is expressing \(\left(\frac{1}{\sqrt{x}}+1 \right)^{2015} - 1\) as a degree-2015 polynomial. After that, I'm all set.

Answer 3

Oh hmm not sure but slight confirmation, sum is supposed to start at 0 to match what you're saying is that right?

Answer 4

nah.

Answer 5

I mean what I posted is the original problem, but it hardly makes any difference.

Answer 6

Oh well you closed it so you must have solved it?

Answer 7

Good timing gotta go anyways

Answer 8

no... I haven't

Answer 9

bye