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xn-1/(x2n-2xncosn q+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- einq)(xn-e-inq). 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 P(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]Sn-1 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)=Pr=0n-1(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-einq) using result about nth root of unity. Now multiply each factor with the appropriate factor of (xn-e-inq), and remember that 2 cos y=eiy+e-iy.

Kerwin

By Arun Iyer on Saturday, November 24, 2001 - 05:56 pm:

Kerwin, I got it....

Thanks again for all the help..
love arun