Partial fraction decomposition

[geoffrey159]

Homework Statement

What is the partial fraction decomposition in $\mathbb{R}[X]$ of $F = \frac{1}{X^{2n} - 1 }$, $n\ge 1$.

Homework Equations

The Attempt at a Solution

Is this correct ?

$F = \frac{1}{2n}(\frac{1}{X-1} - \frac{1}{X+1} + 2 \sum_{k = 1}^{n-1} \frac{ \cos (\frac{k\pi}{n})X - 1 }{ X^2 - 2\cos (\frac{k\pi}{n}) X +1 })$

I have done the decomposition in $\mathbb{C}[X]$ and grouped the terms with the form
$\frac{a}{X - w} + \frac{b} {X - \bar w }$, with $w \in \mathbb{C}-\mathbb{R}$ , in order to get the terms that are in the sum.
The polynomials at the denominator are irreducible in $\mathbb{R}[X]$, and the numerators have degree strictly less than the degree of the denominators. It looks correct to me but I have a doubt

 

[epenguin]

Unless you know different, to me the question just reads you have to factorise (X^2N - 1) - you know factorise (X² - 1), (X⁴ - 1), etc into ornery real factors as far as your can go and then use those to get real partial fractions.

 

[geoffrey159]

Hello,

What I've learned is that you can break any polynomials with real coefficients into a product of monomials and/or polynomials of degree two with $\triangle < 0$. However, I couldn't find it like that on this exemple.

If you write $D = X^{2n} - 1$, then it has $2n$ complex zeros which are the $w_k = e^{\frac{ik\pi}{n}}$. Among these zeros, two are real : $w_0 = 1$ and $w_n = -1$. The others $2(n-1)$ zeros are complex conjugates, so
$D = (X-1)(X+1)\prod_{k = 1} ^ {n-1} (X-w_k)(X-\bar w_k)$

The partial fraction decomposition will be

In $\mathbb{C}[X]$:
$F = \frac{a}{X-1} + \frac{b}{X+1} + \sum_{k = 1}^{n-1} \frac{\alpha_k}{X-w_k} + \frac{\beta_k }{X-\bar w_k }$

I find
$a = 1/2n$, $b = - 1 / 2n$, $\alpha_k = w_k / 2n$, $\beta_k = \bar w_k/2n$

Also,

$F = \frac{a}{X-1} + \frac{b}{X+1} + \sum_{k = 1}^{n-1} \frac{(\alpha_k+\beta_k) X - \alpha_k\bar w_k - \beta_k w_k }{ X^2 - 2\cos (\frac{k\pi}{n}) X +1 }$

The polynomial $X^2 - 2\cos (\frac{k\pi}{n}) X +1$ is irreducible in $\mathbb{R}[X]$ ($\triangle$ < 0), $\alpha_k+\beta_k = \frac{1}{n}\cos (\frac{k\pi}{n})$ is real, as well as $\alpha_k\bar w_k + \beta_k w_k = \frac{1}{n}$. So luckily, I get the decomposition in $\mathbb{R}[X]$ from tha one in $\mathbb{C}[X]$.

The thing is that I tend to do a lot (really a lot) of mistakes, and I would like to know if you agree.

 

[epenguin]

I see what you mean now and it appears to me along the right lines but have not time to think about it till later today hopefully
You could I think get the a and b without factorising the rest - probably more convenient.

 

[epenguin]

Have you arrived at any further conclusions? I'm sorry I haven't been able to help any more in the last days. On the one hand I have found it quite difficult and things going fuzzy, things going out of focus. And on the other I was following a line of thought of my own of just factorising the polynomials algebraically. In reality your approach makes more sense, however this actually led me to one of the most famous of all theorems, but I will say more if you come back (see my sig). Also a different method is a check.

Normally if anyone asked if a partial fraction result is right I would just say well recombine it adding the fractions into one and see if it gives you the starting fraction. But I have been finding it difficult to see how to do that in your general case, though I'm maybe starting to see the light now.

Another thing you can always do is calculate a particular case. That can find out an error, though if it doesn't it may not be conclusive. Maybe this shows a mistake of a sign in yours though I wouldn't go to the stake for it.
......
Edit since yesterday:
For (x⁴ - 1) your formula seems to work.
Since
$\frac{1}{x^2 - 1} = ½ [\frac{1}{x - 1} - \frac{1}{x + 1} ]$

$\frac{1}{x^4 - 1} = ½ [ \frac{1}{x^2 - 1} - \frac{1}{x^2 + 1} ]$

The first RHS term is just the separate two terms in your form (it is optional to not keep these separate but to combine them in a single formula with a Σ up to n)..

The sum Σ has only one term, this is n=2, k=1. Cos(π/2) = 0 so his formula gives a

$\ \frac{-1}{x^2 + 1}$

in accord with what I was getting algebraically

I am not getting this accord for the next case. Maybe I have made a mistake. Anyone else is welcome to take this over.

......
Anyway, doing it for (x⁶ - 1): this factorises algebraically to

(x² - 1)(x² + x + 1)(x² - x + 1).

In partial fractions
$\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = ½ [ \frac{x + 1}{x^2 + x + 1} - \frac{x - 1}{x^2 - x + 1} ]$

which can also be written

$½ [ \frac{x\ cos\frac{π}{6} + 1}{x^2 + 2x\ cos\frac{π}{6} + 1} - \frac{x\ cos\frac{π}{6} - 1}{x^2 - 2x\ cos\frac{π}{6} + 1} ]$

Again that can be written

$½ [ \frac{x\ cos\frac{π}{6} + 1}{x^2 + 2x\ cos\frac{π}{6} + 1} + \frac{x\ cos\frac{2π}{6} + 1}{x^2 + 2x\ cos\frac{π}{6} + 1} ]$

which appears close to but as far as I can see ATM not quite the same as yours.

 

[geoffrey159]

Anyway, doing it for (x⁶ - 1): this factorises algebraically to

(x² - 1)(x² + x + 1)(x² - x + 1).

Hi,

Unless I'm wrong, $\cos (\frac{\pi}{3}) = 1/2,\ \cos (\frac{2\pi}{3}) = -1/2$, so
$x^2 - x + 1 = x^2 - 2x\cos (\frac{\pi}{3}) + 1$
$x^2 + x + 1 = x^2 - 2x\cos (\frac{2\pi}{3}) + 1$

Following what you said, the factorisation of $x^6 - 1$ is
$x^6-1 = (x-1)(x+1)(x^2 - 2x\cos (\frac{\pi}{3}) + 1)(x^2 - 2x\cos (\frac{2\pi}{3}) + 1)$

All the factors are irreducible so the partial fraction decomposition should have all these factors at the denominator. How did you get the pi/6 ?

 

[epenguin]

Glad you came back.
Yes I think the 6 should be 3 in the formulae but I think this was only a mistake in final writing out so if you change this 6 to 3 the rest still stands. (I.e. right or wrong, makes sense :D).
I have to take a break now but will change that when I can come back.
Overlapping posts - I made an addition to my last post5.

 

[epenguin]

OK I have rewritten that section. My numerator doesn't quite agree with your formula but maybe the result is suggestive?

...doing it for (x⁶ - 1): this factorises algebraically to

(x² - 1)(x² - x + 1)(x² + x + 1).

In partial fractions
$\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = ½ [ - \frac{x - 1}{x^2 - x + 1} + \frac{x + 1}{x^2 + x + 1} ]$

which can also be written

$- ½ [ \frac{2 x\ cos\frac{π}{3} - 1}{x^2 - 2x\ cos\frac{π}{3} + 1} - \frac{2x\ cos\frac{π}{3} + 1}{x^2 + 2x\ cos\frac{π}{3} + 1} ]$

Again that can be written

$- ½ [ \frac{2x\ cos\frac{π}{3} - 1}{x^2 - 2x\ cos\frac{π}{3} + 1} + \frac{2x\ cos\frac{2π}{3} - 1}{x^2 - 2x\ cos\frac{2π}{3} + 1} ]$

which still appears close to but as far as I can see ATM not quite the same as yours.

 

[Ray Vickson]

OK I have rewritten that section. My numerator doesn't quite agree with your formula but maybe the result is suggestive?

...doing it for (x⁶ - 1): this factorises algebraically to

(x² - 1)(x² + x + 1)(x² - x + 1).

In partial fractions
$\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = ½ [ \frac{x + 1}{x^2 + x + 1} - \frac{x - 1}{x^2 - x + 1} ]$

which can also be written

$½ [ \frac{2x\ cos\frac{π}{3} + 1}{x^2 + 2x\ cos\frac{π}{3} + 1} - \frac{2 x\ cos\frac{π}{3} - 1}{x^2 - 2x\ cos\frac{π}{3} + 1} ]$

Again that can be written

$½ [ \frac{2x\ cos\frac{π}{3} + 1}{x^2 + 2x\ cos\frac{π}{3} + 1} + \frac{2x\ cos\frac{2π}{3} + 1}{x^2 + 2x\ cos\frac{π}{3} + 1} ]$

which still appears close to but as far as I can see ATM not quite the same as yours.

I'm no sure I have followed all the details you and the OP have written, so I will do it my way. The $2n$ roots of 1 are +1, -1 and
$$r_{\pm k} = \exp{ \left(\pm \frac{i \, k} {2 \pi} \right) }, \: k = 1, 2, \ldots, n-1$$
So, when we factor $x^{2n} - 1$ we can multiply together the factors $(x - r_k)$ and $(x - r_{-k})$ to get a quadratic factor with real coefficients.

 

[geoffrey159]

@epenguin
Hello, I don't understand all your calculations, but simply :

$\frac{x+1}{x^2+x+1} = - \frac{2\cos (\frac{2\pi}{3})x - 1}{x^2-2\cos (\frac{2\pi}{3})x+1}$
$\frac{x-1}{x^2-x+1} = \frac{2\cos (\frac{\pi}{3})x - 1}{x^2-2\cos (\frac{\pi}{3})x+1}$

We get the same thing, don't we ?

 

[geoffrey159]

I'm no sure I have followed all the details you and the OP have written, so I will do it my way. The $2n$ roots of 1 are +1, -1 and
$$r_{\pm k} = \exp{ \left(\pm \frac{i \, k} {2 \pi} \right) }, \: k = 1, 2, \ldots, n-1$$
So, when we factor $x^{2n} - 1$ we can multiply together the factors $(x - r_k)$ and $(x - r_{-k})$ to get a quadratic factor with real coefficients.

Yes it is what I did, but there is a mistake in $r_{\pm k}$ I think. It should be equal to 1 at the power 2n.

 

[Ray Vickson]

Yes it is what I did, but there is a mistake in $r_{\pm k}$ I think. It should be equal to 1 at the power 2n.

Yes: $r_{\pm k} = \exp(\pm i \pi k/n))$. So,
$$(x - r_k)(x - r_{-k}) = x^2 - 2 \cos(\pi k/n) x +1$$

 

[epenguin]

Sorry you couldn't follow because there were still some misprints, and expressions were not in best order to make my point and I now summarise what I'm doing in #8:
the first line is the factorised expression arranged in more suitable order
the second line is ordinary partial fractions
the third line is a trigonometrical form
the fourth line is applying cos(π/3) = - cos(2π/3)

the aim being trying to hammer the expressions to make them most closely possible resemble the terms in your Σ. The denominators do but the numerators not quite.