Partial fraction decomposition

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

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
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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...

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.

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
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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.

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
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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

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
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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

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
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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. 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