Partial fraction decomposition

[Nyasha]

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 ?

 

[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.

 

[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.

 

[Nyasha]

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]

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: $(1-x^a)(1+x^a)=$____?

 

[Nyasha]

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]

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

 

[Nyasha]

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]

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 high school...

 

[Nyasha]

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 high school...

"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]

"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:

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

 

[Nyasha]

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]

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?

 

[Nyasha]

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]

Yes, now determine the constants...

 

[Nyasha]

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

 

[djeitnstine]

Compare coefficients.

 

[gabbagabbahey]

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

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

 

[gabbagabbahey]

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

 

[Nyasha]

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

Does this look correct:

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

 

[gabbagabbahey]

That's better!

Now start plugging in points...I recommend using 1,-1,0,i,-i,2,-2 and 3

 

[Nyasha]

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]

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

 

[Nyasha]

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 textbook 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]

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 textbook 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

Check your math again...

 

[Nyasha]

Check your math again...

Did l do something wrong for when x=0 ? I would also like to know if l must expand $$(1+x)(1-x)^2(1+x^2)^2$$ before equating the coefficients ?

 

[gabbagabbahey]

I get B=D=1/8...

 

[Nyasha]

I get B=D=1/8...

When x=-1

$$B(1-x)^2(1+x^2)^2=x^2$$


$$B(1--1)^2(1+(-1)^2)^2=(-1)^2$$


$$B=\frac{1}{(1--1)^2(1+(-1)^2)^2)}=\frac{1}{16}$$


I can't find where l made the mistake ?

 

[gabbagabbahey]

You didn't make a mistake...i did... 2^4=16 not 8

Anyways... you can either expand everything out and compare coefficients, or continue plugging in points...x=i and x=-i make finding G and H very easy...

 

[Nyasha]

You didn't make a mistake...i did... 2^4=16 not 8

Anyways... you can either expand everything out and compare coefficients, or continue plugging in points...x=i and x=-i make finding G and H very easy...

Man, thanks very much for your help. I really appreciate the time you spend with me on this question.

 

[Nyasha]

You didn't make a mistake...i did... 2^4=16 not 8

Anyways... you can either expand everything out and compare coefficients, or continue plugging in points...x=i and x=-i make finding G and H very easy...

After plugging in all the points l got :

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

$$E=F=0$$


$$G=\frac{-1}{8}$$

$$H=\frac{1}{8}$$


Which means l end up with ( I am not 100% sure if this is correct) :


$$\frac{x^2}{(1-x^4)^2}=\frac {x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{\frac{1}{16}}{(1+x)}+\frac{\frac{1}{16}}{(1+x)^2}+\frac{\frac{1}{16}}{(1-x)}+\frac{\frac{1}{16}}{(1-x)^2}+\frac{0}{(1+x^2)}+\frac{\frac{-x}{8}+\frac{1}{8}}{(1+x^2)^2}$$

 

[gabbagabbahey]

After plugging in all the points l got :

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

$$E=F=0$$


$$G=\frac{-1}{8}$$

$$H=\frac{1}{8}$$


Which means l end up with ( I am not 100% sure if this is correct) :

It isn't; you should have G=0 and H=-1/4. Incidentally, Mathematica has a built in function called 'Apart' which will decompose any fraction...it is a useful tool for checking your answers.

 

[Nyasha]

It isn't; you should have G=0 and H=-1/4. Incidentally, Mathematica has a built in function called 'Apart' which will decompose any fraction...it is a useful tool for checking your answers.

I couldn't find the built in function apart on the web. However l double checked my work and saw that H=-1/4 and G=-1/16. Do you have a link to this built in function called "apart" ?

 

[gabbagabbahey]

Here is the mathematica documentation on 'Apart'. Do you have access to Mathematica at your university/home?

You should get G=0 and H=-1/4...if you show me your calcs for that I will point out your errors

 

[Nyasha]

Here is the mathematica documentation on 'Apart'. Do you have access to Mathematica at your university/home?

You should get G=0 and H=-1/4...if you show me your calcs for that I will point out your errors

For H l got $$\frac{-1}{4}$$ but for some reason l can't find G=0

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


$$x^2=A(x^7-x^6+x^5-x^4-x^3+x^2-x+1)+B(x^6-x^5+3x^4-4x^3+3x^2-2x+1)+$$$$C(-x^7-x^6-x^5-x^4+x^3+x^2+x+1)+ D(x^6+2x^5+3x^4+4x^3+3x^2+2x+1)$$$$+(Ex+F)(x^8-3x^4+x^2+1)+(Gx+H)(x^4-2x^2+1)$$

$$x^5\rightarrow 0=A-B-C+2D-3E+G \rightarrow 0=\frac{1}{16}-\frac{1}{16}-\frac{1}{16}+\frac{1}{8}+G$$

$$\therefore G=\frac{-1}{16}$$