Recent content by bobn

its easier to solve this logically, the term is distance between two consecutive roots of unity, so, for n = 3 its equi triangle, 4 its square, so as n tends to infinity it becomes a circle, and sum of distance between all points will be its perimeter on unit circle. so the limit tends to 2 pi.

seems so at first glance, let me think for any contradictions..

but as its modulus, common factor will 1; absolute value of n roots of equity will be 1. ??

f(x) =sum | e^k - e^k/e| where e = e^2pi i/n n roots of unity. f(x) =sum |e^k|/|e| *[|e-1|] = n*[|e-1|] = 2n sin pi/n ; which diverges...

did lil. bit manipulation but it seems weird now.

is it 2??

tried in this way, I am getting 0, but that shouldn't be...

thx for reply tiny-tim

haa, I am stupid, 1/√n > 1/n implies, (1- 1/√n) < (1 - 1/n).

I tried integration then applying limit as n tends to infinity, for k = 1, it becomes a circle, but as k increases, points decrease hence it should be wrong.

Lt (1 – 1/√2)* (1 – 1/√3)…… (1 – 1/√n+1) n->∞ There must some simple calculation to find this, but I cannot get this, limit exists since each tern is less than 1, and limit is more than 0. so limit exists between (1 0).

Office thanks for your reply. More over, h(x) will be of degree 6 since, a and b are of degree 3,2 respectively with gcd 1. so, a+b will be of degree 6.

Say f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0. [Given] There exists , h(x) a polynomial with in the form of Ci*Xi where Cis are integer coefficients and h(a+b) = 0. since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., for each a^i*b^i = a^j*(a^3+a+1)*b^k*(...

I think this should do. f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0. h(a+b) = ci*xi. This should be zero since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., a^i*b^i = a^j*(a^3+a+1)*b^k*( b^2+b-3) = 0. Similarly k(ab) = 0

yea, all ai s should be integers, but not necessarily roots right.