xn-1/(x2n-2xncosnq+1)
By Arun Iyer on Monday, November 19, 2001
- 06:46 pm:
Could anyone please help me solve this problem...
The question is....
Express [xn-1/(x2n-2xncosnq+1)] as the sum of n partial
fractions with denominators quadratics in x...
love arun
By Kerwin Hui on Monday, November 19,
2001 - 08:23 pm:
Note that your denominator factorises over the complexes
to give xn-ei nq)(xn-e-i nq). Now it should be clear what
to do.
Kerwin
By Arun Iyer on Tuesday, November 20, 2001
- 07:02 pm:
Kerwin,
actually I did the factorization, though the actual problematic
part was the fact that they wanted n terms with denominators
quadratic in x....
i could not proceed any further....though i tried many different
ways...
love arun
By Kerwin Hui on Wednesday, November 21,
2001 - 02:29 am:
What you do is to factorise the two factors into products
of linear factors
|
Õ
| (x-ei(q+2mp/n))(x-e-i(q+2mp/n))
|
and group the appropriate conjugates together to give you a quadratic in x
with real coefficients, i.e. (x2-2xcos(q+2mp/n)+1).
Kerwin
By Arun Iyer on Thursday, November 22,
2001 - 06:50 pm:
yup!!i got ya....Kerwin!!
is the answer....
|
[1/nsinnq] |
n-1
å
r=0
|
[sin(q+2rp/n)]/[x2-2xcos(q +2rp/n)+1] |
|
Thanks for all the help!!
love arun
By Arun Iyer on Friday, November 23, 2001
- 06:20 pm:
Kerwin,in proving that result i used a result you mentioned
that is,
| x2n-2xncosnq+1= |
n-1
Õ
r=0
|
(x2-2xcos(q+2mp/n)+1)
|
I have been trying to prove this above result myself but no quite
successful...any help??
love arun
By Kerwin Hui on Saturday, November 24,
2001 - 11:38 am:
First, factorise (xn-ei nq) using result about
nth root of unity. Now multiply each factor with the appropriate factor of
(xn-e-i nq), and remember that 2cosy=ei y+e-i y.
Kerwin
By Arun Iyer on Saturday, November 24,
2001 - 05:56 pm:
Kerwin, I got it....
Thanks again for all the help..
love arun