Re: cos(PI/N)
- From: JohnCreighton_@xxxxxxxxxxx
- Date: 27 Dec 2005 22:04:12 -0800
JohnCreighton_@xxxxxxxxxxx wrote:
> > 2*cos(pi/n)
> > =exp(i*pi/n)+exp(-i*pi/n)
> > =exp(i*pi)^(1/n)+exp(-i*pi)^(1/n)
> > = (-1)^(1/n) + (-1)^(1/n)
>
> Hmmm.....Thinking about what this means the first root in the sum
> should be the root with the smallest angle greater then zero. I think
> this would be called the principle root and the root chosen for the
> second term in the sum should be the complex conjugate of the root
> chosen in the first term in the sum. So the answer should be written
> as:
> 2*cos(pi/n)=principle nth root of (-1) + conjugate of the principle nth
> root of (-1)
>
> So basically we are saying if we have a complex vector in polar form
> exp(-i*pi/n), then the cosine is the real part of this. Well, Duh
> (slaps self on forehead). Although the result is obvious it is useful
> for computer algebra systems which wish to reduce the number of
> functions that the use to represent a mathematical expression. The
> advantage of this is it reduces the function space simplification
> routines need to search though. It however does not seem useful as a
> method of computing the cosine.
>
> Perhaps the original poster had in mind and answer based on trig
> identities of cosines with known radical solutions.
>
After some more thought if we want to obtain an expression that is easy
to compute we can recognize that the problem reduces to finding roots
of negative one. If we want these roots in rectangular form we can
write:
(x+i*y)^n=-1
Before we solve this we recognize that all the roots lie in a circle.
If n is odd then
(x=-1, y=0 is a solution and if n is even {x=0, y=1} {x=0,y=-1} are two
solutions solution). Notice also that:
x=-1, y=0 is the solution for n=1 and {x=0, y=1} {x=0,y=-1} are the
solutions for n=2. After the equation is expanded and rearranged so
p(x)=0 these solutions can be divided out. Remember we are in search of
the solution with the smallest angle. From the algebra done so far it
appears that the solution for the smallest angles can be found by
dividing the solution out for the largest angel, and then doing the
same for the remaining polynomial until we are left with a polynomial
that can be factored using the quadratic formula. I'll give this some
more thought.
.
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