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)^2when 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 <br /> (1+x)(1-x)^2(1+x^2)^2<br /> 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}

 

[gabbagabbahey]

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}

Check your expansion of the B(1-x)^2(1+x^2)^2 term

 

[Nyasha]

Check your expansion of the B(1-x)^2(1+x^2)^2 term

Thanks l got G=0. I have now solved. I don't have access to Mathematica at the university library or home. How can you get it ?

 

[gabbagabbahey]

You can purchase a license off their website and download it to your computer, but it is fairly expensive (~$300 for a single home use license)...most University physics and math departments have several licenses for student use, so you should ask around your department before spending money you don't have to.

 

[Nyasha]

You can purchase a license off their website and download it to your computer, but it is fairly expensive (~$300 for a single home use license)...most University physics and math departments have several licenses for student use, so you should ask around your department before spending money you don't have to.

Thanks very much

 

[Dick]

There is a also a very capable free alternative to Mathematica called Maxima. http://maxima.sourceforge.net/ If you can get that installed then just type:
partfrac(x^2/(1-x^4)^2,x);