Partial fraction decomposition

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Homework Statement


Find the partial fraction decomposition of :
[tex]\frac{x^2}{(1-x^4)^2}[/tex]



The Attempt at a Solution



[tex]\frac{x^2}{(1-x^4)^2}=\frac{A}{(1-x^4)}+\frac {B}{(1-x^4)^2}[/tex]

[tex]=A(1-x^4)+B[/tex]

when x=1

[tex]1=A(1-1^4)+B[/tex]

Hence B=1 and A=0

[tex]\frac{x^2}{(1-x^4)^2}=\frac{0}{(1-x^4)}+ \frac{1}{(1-x^4)^2}[/tex]

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

Answers and Replies

  • #2
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When x = 1, you have
[itex]1 = A(1-1^{4})+B = A(0)+B = B[/itex]
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
Dick
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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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  • #4
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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 ?
 
  • #5
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: [itex](1-x^a)(1+x^a)=[/itex]____?
 
  • #6
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The denominator is a reducible quartic...I'll give you a hint to factoring it: [itex](1-x^a)(1+x^a)=[/itex]____?
[tex](1-x^4)(1-x^4)=(1-x^2)^2 (1-x^2)^2[/tex]
 
  • #7
gabbagabbahey
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No, I meant [itex]1-x^4[/itex] is a reducible quartic....factor that:smile:
 
  • #8
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No, I meant [itex]1-x^4[/itex] is a reducible quartic....factor that:smile:
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
gabbagabbahey
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I'm not sure how much simpler I can make it....do you really not know how to factor [itex]1-x^4[/itex]? It is something you should have been taught in highschool....
 
  • #10
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I'm not sure how much simpler I can make it....do you really not know how to factor [itex]1-x^4[/itex]? It is something you should have been taught in highschool....

"Reducible quartic" send me a little bit off track:smile:


[tex]1-x^4=(1-x^2)(1+x^2)=(1+x)(1-x)(1+x^2)[/tex]


Will this mean l will be left with :

[tex]\frac{x^2}{(1-x^4)^2}=\frac{A}{(1+x)}+\frac {B}{(1-x)}+\frac{Cx+D}{1+x^2}[/tex]
 
  • #11
gabbagabbahey
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"Reducible quartic" send me a little bit off track:smile:


[tex]1-x^4=(1-x^2)(1+x^2)=(1+x)(1-x)(1+x^2)[/tex]
Yes!:smile:


Will this mean l will be left with :

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

[tex]\frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}[/tex]
 
  • #12
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Yes!:smile:




Careful:

[tex]\frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}[/tex]

Uhhmmm, which means

[tex]\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[/tex]
 
  • #13
gabbagabbahey
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Uhhmmm, which means

[tex]\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[/tex]
Your LaTeX is a little a sloppy; surely you mean:

[tex]\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}[/tex]

right?
 
  • #14
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Your LaTeX is a little a sloppy; surely you mean:

[tex]\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}[/tex]

right?

Yes that what l mean, is it correct ?
 
  • #15
gabbagabbahey
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Yes, now determine the constants....
 
  • #16
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Yes, now determine the constants....
[tex]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[/tex]

when x=1

[tex]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[/tex]

[tex]4D=1[/tex]

[tex]D=\frac{1}{4}[/tex]

when x=-1

[tex]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[/tex]


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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  • #17
djeitnstine
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Compare coefficients.
 
  • #18
gabbagabbahey
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[tex]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[/tex]
That doesn't look right at all, be careful with your multiplication!
 
  • #19
gabbagabbahey
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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
 
  • #20
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That doesn't look right at all, be careful with your multiplication!
Does this look correct:

[tex]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[/tex]
 
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  • #21
gabbagabbahey
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That's better!:smile:

Now start plugging in points....I recommend using 1,-1,0,i,-i,2,-2 and 3
 
  • #22
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That's better!:smile:

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
 
  • #23
gabbagabbahey
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You're welcome!:smile:....get a good night's sleep!:zzz:
 
  • #24
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You're welcome!:smile:....get a good night's sleep!:zzz:
When x=1

[tex]D=\frac{1}{16}[/tex]

when x=-1

[tex]B=\frac{1}{16}[/tex]

when x=0

[tex]0=A+\frac{1}{16}+C+\frac{1}{16}+F+H[/tex]


The calculus text book says l should solve for the coefficients by method of equating them. So my question is should l expand this [tex](1+x)(1-x)^2(1+x^2)^2[/tex] and then equate the coefficients ? I was wondering if there is an easier way to equate the coefficients without going through all this algebra
 
  • #25
gabbagabbahey
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When x=1

[tex]D=\frac{1}{16}[/tex]

when x=-1

[tex]B=\frac{1}{16}[/tex]

when x=0

[tex]0=A+\frac{1}{16}+C+\frac{1}{16}+F+H[/tex]


The calculus text book says l should solve for the coefficients by method of equating them. So my question is should l expand this [tex](1+x)(1-x)^2(1+x^2)^2[/tex] 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....
 

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