Partial fraction decomposition

1. Mar 20, 2009

Nyasha

1. The problem statement, all variables and given/known data
Find the partial fraction decomposition of :
$$\frac{x^2}{(1-x^4)^2}$$

3. 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 ?

2. Mar 20, 2009

jdougherty

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.

3. Mar 20, 2009

Dick

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.

Last edited: Mar 20, 2009
4. Mar 20, 2009

Nyasha

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

5. Mar 20, 2009

gabbagabbahey

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

6. Mar 20, 2009

Nyasha

$$(1-x^4)(1-x^4)=(1-x^2)^2 (1-x^2)^2$$

7. Mar 20, 2009

gabbagabbahey

No, I meant $1-x^4$ is a reducible quartic....factor that

8. Mar 20, 2009

Nyasha

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

9. Mar 20, 2009

gabbagabbahey

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....

10. Mar 21, 2009

Nyasha

"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 :

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

11. Mar 21, 2009

gabbagabbahey

Yes!

Careful:

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

12. Mar 21, 2009

Nyasha

Uhhmmm, which means

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

13. Mar 21, 2009

gabbagabbahey

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?

14. Mar 21, 2009

Nyasha

Yes that what l mean, is it correct ?

15. Mar 21, 2009

gabbagabbahey

Yes, now determine the constants....

16. Mar 21, 2009

Nyasha

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

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

Last edited: Mar 21, 2009
17. Mar 21, 2009

djeitnstine

Compare coefficients.

18. Mar 21, 2009

gabbagabbahey

That doesn't look right at all, be careful with your multiplication!

19. Mar 21, 2009

gabbagabbahey

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

20. Mar 21, 2009

Nyasha

Does this look correct:

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

Last edited: Mar 21, 2009