# Partial fraction decomposition

## Homework Statement

Find the partial fraction decomposition of :
$$\frac{x^2}{(1-x^4)^2}$$

## The Attempt at a Solution

$$\frac{x^2}{(1-x^4)^2}=\frac{A}{(1-x^4)}+\frac {B}{(1-x^4)^2}$$

$$=A(1-x^4)+B$$

when x=1

$$1=A(1-1^4)+B$$

Hence B=1 and A=0

$$\frac{x^2}{(1-x^4)^2}=\frac{0}{(1-x^4)}+ \frac{1}{(1-x^4)^2}$$

How come according to the answers at the back of the book l am wrong. Where have l done a mistake ?

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When x = 1, you have
$1 = A(1-1^{4})+B = A(0)+B = B$
So B = 1 works, but what values of A satisfy that equation? What values don't? If there are multiple values A can have, then you need to determine which is the correct one.

Dick
Homework Helper
Well, 0/(1-x^4)^2+1/(1-x^4)^2 clearly does not equal x^2/(1-x^4)^2, now does it? You should probably factor the denominator completely and then look up what partial fractions look like if you have a quadratic in the denominator.

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Well, 0/(1-x^4)+1/(1-x^4)^2 clearly does not equal x^2/(1-x^4)^2, now does it? You should probably factor the denominator completely and then look up what partial fractions look like if you have a quadratic in the denominator.

Isn't the denominator already in a factored form ? Should l expand it ?

gabbagabbahey
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Isn't the denominator already in a factored form ? Should l expand it ?
The denominator is a reducible quartic...I'll give you a hint to factoring it: $(1-x^a)(1+x^a)=$____?

The denominator is a reducible quartic...I'll give you a hint to factoring it: $(1-x^a)(1+x^a)=$____?
$$(1-x^4)(1-x^4)=(1-x^2)^2 (1-x^2)^2$$

gabbagabbahey
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No, I meant $1-x^4$ is a reducible quartic....factor that No, I meant $1-x^4$ is a reducible quartic....factor that I am getting very confused. Show me an example which has nothing to do with this question maybe l will understand what you are trying to say

gabbagabbahey
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I'm not sure how much simpler I can make it....do you really not know how to factor $1-x^4$? It is something you should have been taught in highschool....

I'm not sure how much simpler I can make it....do you really not know how to factor $1-x^4$? It is something you should have been taught in highschool....

"Reducible quartic" send me a little bit off track $$1-x^4=(1-x^2)(1+x^2)=(1+x)(1-x)(1+x^2)$$

Will this mean l will be left with :

$$\frac{x^2}{(1-x^4)^2}=\frac{A}{(1+x)}+\frac {B}{(1-x)}+\frac{Cx+D}{1+x^2}$$

gabbagabbahey
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"Reducible quartic" send me a little bit off track $$1-x^4=(1-x^2)(1+x^2)=(1+x)(1-x)(1+x^2)$$
Yes! Will this mean l will be left with :

$$\frac{x^2}{(1-x^4)^2}=\frac{A}{(1+x)}+\frac {B}{(1-x)}+\frac{Cx+D}{1+x^2}$$
Careful:

$$\frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}$$

Yes! Careful:

$$\frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}$$

Uhhmmm, which means

$$\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{1-x}+\frac{D}{1-x}+\frac{Ex+F}{1+x^2)}^2$$

gabbagabbahey
Homework Helper
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Uhhmmm, which means

$$\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{1-x}+\frac{D}{1-x}+\frac{Ex+F}{1+x^2)}^2$$
Your LaTeX is a little a sloppy; surely you mean:

$$\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1+x)^2}+\frac{C}{(1-x)}+\frac{D}{(1-x)^2}+\frac{Ex+F}{(1+x^2)}+\frac{Gx+H}{(1+x^2)^2}$$

right?

Your LaTeX is a little a sloppy; surely you mean:

$$\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{(1+x)}+\frac{B}{(1+x)^2}+\frac{C}{(1-x)}+\frac{D}{(1-x)^2}+\frac{Ex+F}{(1+x^2)}+\frac{Gx+H}{(1+x^2)^2}$$

right?

Yes that what l mean, is it correct ?

gabbagabbahey
Homework Helper
Gold Member
Yes, now determine the constants....

Yes, now determine the constants....
$$x^2=A(1-x)^2(1+x)+B(1-x)^2+C(1-x)(1+x)^2+D(1+x^2)^2+(Ex+F)(1-x)^2+(Gx+H)(1-x)^2$$

when x=1

$$1= A(1-1)^2(1+1)+ B(1-1)^2+C(1-1)(1+1)^2+D(1+1)^2+(Ex+F)(1-1)^2+(Gx+H)(1-1)^2$$

$$4D=1$$

$$D=\frac{1}{4}$$

when x=-1

$$1=A(1+1)^2(1-1)+B(1+1)^2+C(1+1)(1-1)^2+\frac{1}{4}(1+1)^2+(Ex+F)(1+1)^2+(Gx+H)(1--1)^2$$

when x=o

0=A+B+C+\frac{1}{4}+(Ex+F)+ (Gx+H)

It got so many unknowns and only three equations, how do l get around this last hurdle

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djeitnstine
Gold Member
Compare coefficients.

gabbagabbahey
Homework Helper
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$$x^2=A(1-x)^2(1+x)+B(1-x)^2+C(1-x)(1+x)^2+D(1+x^2)^2+(Ex+F)(1-x)^2+(Gx+H)(1-x)^2$$
That doesn't look right at all, be careful with your multiplication!

gabbagabbahey
Homework Helper
Gold Member
Compare coefficients.
There is an easier way than solving the system of equation this method results in.

Try plugging in x=i and x=-i for starters

That doesn't look right at all, be careful with your multiplication!
Does this look correct:

$$A(1+x)(1-x)^2(1+x^2)^2+B(1-x)^2(1+x^2)^2+C(1-x)(1+x^2)^2(1+x)^2+D(1+x^2)^2(1+x)^2+(Ex+F)(1+x)^2(1+x^2)(1-x)^2+(Gx+H)(1-x)^2(1+x)^2$$

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gabbagabbahey
Homework Helper
Gold Member
That's better! Now start plugging in points....I recommend using 1,-1,0,i,-i,2,-2 and 3

That's better! Now start plugging in points....I recommend using 1,-1,0,i,-i,2,-2 and 3
Gabbagabbahey thanks for your help but l can't think l can continue for today it is already 4:00am. First thing when l wake up tomorrow l will try to solve it using these points and then show you the solution when l get it. Again thanks very much for your help

gabbagabbahey
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You're welcome! ....get a good night's sleep!:zzz:

You're welcome! ....get a good night's sleep!:zzz:
When x=1

$$D=\frac{1}{16}$$

when x=-1

$$B=\frac{1}{16}$$

when x=0

$$0=A+\frac{1}{16}+C+\frac{1}{16}+F+H$$

The calculus text book says l should solve for the coefficients by method of equating them. So my question is should l expand this $$(1+x)(1-x)^2(1+x^2)^2$$ and then equate the coefficients ? I was wondering if there is an easier way to equate the coefficients without going through all this algebra

gabbagabbahey
Homework Helper
Gold Member
When x=1

$$D=\frac{1}{16}$$

when x=-1

$$B=\frac{1}{16}$$

when x=0

$$0=A+\frac{1}{16}+C+\frac{1}{16}+F+H$$

The calculus text book says l should solve for the coefficients by method of equating them. So my question is should l expand this $$(1+x)(1-x)^2(1+x^2)^2$$ and then equate the coefficients ? I was wondering if there is an easier way to equate the coefficients without going through all this algebra