# Partial Fractions

• Jul 13th 2008, 10:55 PM
kleyzam
Partial Fractions
Can someone help me with this question?

1. Using the result $x^3 + 1 = (x+1)(x^2 - x + 1)$, express $\frac{x^5 - 1}{x^2(x^3 + 1)}$ in partial fractions.
• Jul 13th 2008, 11:01 PM
Mathstud28
Quote:

Originally Posted by kleyzam
Can someone help me with this question?

1. Using the result $x^3 + 1 = (x+1)(x^2 - x + 1)$, express $\frac{x^5 - 1}{x^2(x^3 + 1)}$ in partial fractions.

You would want to polynomial long divide here, not PFD.
• Jul 13th 2008, 11:02 PM
kalagota
Quote:

Originally Posted by kleyzam
Can someone help me with this question?

1. Using the result $x^3 + 1 = (x+1)(x^2 - x + 1)$, express $\frac{x^5 - 1}{x^2(x^3 + 1)}$ in partial fractions.

$\frac{x^5 - 1}{x^2(x^3 + 1)}=\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{Dx + E}{x^2 - x + 1}$
• Jul 13th 2008, 11:04 PM
kalagota
Quote:

Originally Posted by Mathstud28
You would want to polynomial long divide here, not PFD.

i think, after that long division, s/he would still have a remainder in 2nd degree
• Jul 13th 2008, 11:14 PM
kleyzam
I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
$-x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1)$ $+ Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)$
• Jul 13th 2008, 11:15 PM
Mathstud28
Quote:

Originally Posted by kalagota
i think, after that long division, s/he would still have a remainder in 2nd degree

Yes, they will, but wow will it be much easier.
• Jul 13th 2008, 11:18 PM
kalagota
Quote:

Originally Posted by kleyzam
I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
$-x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1)$ $+ Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)$

completely solve your RHS.. then factor every powers of x.
$x^5:$
$x^4:$
$x^3:$
$x^2:$
$x:$
constant:
• Jul 14th 2008, 03:27 AM
ticbol
Quote:

Originally Posted by kleyzam
I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
$-x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1)$ $+ Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)$

In your LHS, the (-x^2 -1),
should that not be (x^5 -1)?

Then in your last term in the the RHS, the (Dx +E)(x^2)(x^2 -x +1),
should that not be (Dx +E)(x^2)(x +1)?
• Jul 14th 2008, 04:15 AM
kalagota
Quote:

Originally Posted by ticbol
In your LHS, the (-x^2 -1),
should that not be (x^5 -1)?

he already divided it..

and yes, the last term should be the one what ticbol noted.
• Jul 14th 2008, 04:25 AM
ticbol
Quote:

Originally Posted by kalagota
he already divided it..

He divided, or simplified, the original LHS, but he did not do that in the original RHS?

Then his final "equation or identity", even if the last term in the new RHS is corrected, is wrong.
Because if he multiplied the original RHS by (x^2)(x^3 +1), then he should have multiplied the original LHS by the same amount.

Clearly, (-x^2 -1) is not [(x^5 -1) / (x^2)(x^3 +1)] * [(x^2)(x^3 +1)].
• Jul 14th 2008, 04:29 AM
kalagota
Quote:

Originally Posted by ticbol
He divided, or simplified, the original LHS, but he did not do that in the original RHS?

Then his final "equation or identity", even if the last term in the new RHS is corrected, is wrong.
Because if he multiplied the original RHS by (x^2)(x^3 +1), then he should have multiplied the original LHS by the same amount.

Clearly, (-x^2 -1) is not [(x^5 -1) / (x^2)(x^3 +1)] * [(x^2)(x^3 +1)].

of course it is not equal to that..

what he meant was..

$\frac{x^5 -1}{(x^2)(x^3 +1)} = 1 + \frac{-x^2-1}{(x^2)(x^3 +1)}$

and $\frac{-x^2-1}{(x^2)(x^3 +1)}$ is the one he is solving now..
• Jul 14th 2008, 04:52 AM
ticbol
Quote:

Originally Posted by kalagota
of course it is not equal to that..

what he meant was..

$\frac{x^5 -1}{(x^2)(x^3 +1)} = 1 + \frac{-x^2-1}{(x^2)(x^3 +1)}$

and $\frac{-x^2-1}{(x^2)(x^3 +1)}$ is the one he is solving now..

I see.
Then what about the "1" in the new LHS? How would he incorporate that back into the final decomposition?
• Jul 14th 2008, 05:04 AM
kalagota
well, given the situation he can always replace it by

$\frac{x}{x}, \frac{x^2}{x^2}$, etc and add it to where it should be added.. :)
• Jul 14th 2008, 05:19 AM
ticbol
Quote:

Originally Posted by kalagota
well, given the situation he can always replace it by

$\frac{x}{x}, \frac{x^2}{x^2}$, etc and add it to where it should be added.. :)

You lost me there.

So the final decomposition might be:
x/x + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
or, (x^2 / x^2) + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
Etc....

And so the the final integral would be
= x +{INT[the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]dx} +C?

Umm...
• Jul 14th 2008, 05:27 AM
kalagota
Quote:

Originally Posted by ticbol
You lost me there.

So the final decomposition might be:
x/x + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
or, (x^2 / x^2) + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
Etc....

yes..

Quote:

Originally Posted by ticbol
And so the the final integral would be
= x +{INT[the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]dx} +C?

Umm...

i don't think he wants to integrate the whole thing.. but if so, yes, it would be like that..