$x^{n - 1} / (x^{2 n} - 2 x^{n} \cos n θ + 1)$

By Arun Iyer on Monday, November 19, 2001 - 06:46 pm:

Could anyone please help me solve this problem...
The question is....

Express

[x^{n - 1} / (x^{2 n} - 2 x^{n} \cos n θ + 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

x^{n} - e^{i n θ}) (x^{n} - e^{- i n θ})

. 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

Π (x - e^{i (θ + 2 m π / n)}) (x - e^{- i (θ + 2 m π / n)})

and group the appropriate conjugates together to give you a quadratic in

x

with real coefficients, i.e.

(x^{2} - 2 x \cos (θ + 2 m π / n) + 1)

Kerwin

By Arun Iyer on Thursday, November 22, 2001 - 06:50 pm:

yup!!i got ya....Kerwin!!

is the answer....

[1 / n \sin n θ] \sum_{r = 0}^{n - 1} [\sin (θ + 2 r π / n)] / [x^{2} - 2 x \cos (θ + 2 r π / 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,

x^{2 n} - 2 x^{n} \cos n θ + 1 = Π_{r = 0}^{n - 1} (x^{2} - 2 x \cos (θ + 2 m π / 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

(x^{n} - e^{i n θ})

using result about

n

th root of unity. Now multiply each factor with the appropriate factor of

(x^{n} - e^{- i n θ})

, and remember that

2 \cos y = e^{i 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