can be written as a product of factors having the form
x^2 n− 2 xncos nθ + 1 = 2 n−^1
n∏− 1
r=0
(x^2 − 2 xcos(θ+
2 rπ
n ) + 1) (12.126)
This is accomplished by considering the function x^2 n− 2 xncos nθ +1 and then dividing
it by xnand defining
un=xn−2 cos nθ +x−n (12.127)
and then verifying that un can be written as
un=(xn−^1 +x−n+1)(x−2 cos θ+x−^1 )
+ 2 cos θ(xn−^1 −2 cos[(n−1)θ] + x−n+1)−(xn−^2 −2 cos[(n−2)θ] + x−(n−2))
(12.128)
or in terms of the undefinition
un= (xn−^1 +x−n+1)u 1 + 2un− 1 cos θ−un− 2 (12.129)
Observe that un is divisible by u 1 if both un− 1 and un− 2 are also divisible by u 1. To
show this is true verify that
u 2 =x^2 −2 cos 2θ+x−^2 = (x−2 cosθ+x−^1 )(x+ 2 cos θ+x−^1 )
and consequently u 2 is divisible by u 1. Using equation (12.129) write
u 3 = (x^2 +x−^2 )u 1 + 2 u 2 cos θ−u 1
to show u 3 is divisible by u 1. Continuing in this fashion u 4 , u 5 , u 6 ,... , u n− 2 , u n− 1 , are all
divisible by u 1. This demonstrates that x^2 − 2 xcos θ+1 is a factor of x^2 n− 2 xncos nθ +1.
Since θis an arbitrary angle, replace θby θ+ 2 rπ/n ,ran integer constant, to show
x^2 − 2 xcos
(
θ+^2 rπ
n
)
+ 1 is a factor of x^2 n− 2 xncos[n
(
θ+^2 rπ
n
)
] + 1
for r= 0, 1 , 2 ,.. ., n − 1.
Using the trigonometric identity
cos nθ = cos[n(θ+^2 rπ
n
)]
for ran integer, one can say that the factors of
x^2 n− 2 xncos nθ + 1
