- #26

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Did l do something wrong for when x=0 ? I would also like to know if l must expand [tex]Check your math again....

(1+x)(1-x)^2(1+x^2)^2

[/tex] before equating the coefficients ?

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- #26

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Did l do something wrong for when x=0 ? I would also like to know if l must expand [tex]Check your math again....

(1+x)(1-x)^2(1+x^2)^2

[/tex] before equating the coefficients ?

- #27

gabbagabbahey

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

- #28

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When x=-1I get B=D=1/8...

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

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

[tex]B=\frac{1}{(1--1)^2(1+(-1)^2)^2)}=\frac{1}{16}[/tex]

I can't find where l made the mistake ?

- #29

gabbagabbahey

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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...

- #30

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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.

- #31

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After plugging in all the points l got :

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...

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

[tex]E=F=0[/tex]

[tex]G=\frac{-1}{8}[/tex]

[tex]H=\frac{1}{8}[/tex]

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

[tex]\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}[/tex]

- #32

gabbagabbahey

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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.After plugging in all the points l got :

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

[tex]E=F=0[/tex]

[tex]G=\frac{-1}{8}[/tex]

[tex]H=\frac{1}{8}[/tex]

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

- #33

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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" ?

- #34

gabbagabbahey

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

- #35

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

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

[tex]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[/tex]

[tex]+(Ex+F)(1+x)^2 (1+x^2)(1-x)^2+(Gx+H) (1-x)^2(1+x)^2[/tex]

[tex]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)+[/tex][tex]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)[/tex][tex]+(Ex+F)(x^8-3x^4+x^2+1)+(Gx+H)(x^4-2x^2+1)[/tex]

[tex]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[/tex]

[tex]\therefore G=\frac{-1}{16}[/tex]

- #36

gabbagabbahey

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Check your expansion of the [tex]B(1-x)^2(1+x^2)^2[/tex] termFor H l got [tex]\frac{-1}{4}[/tex] but for some reason l can't find G=0

[tex]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[/tex]

[tex]+(Ex+F)(1+x)^2 (1+x^2)(1-x)^2+(Gx+H) (1-x)^2(1+x)^2[/tex]

[tex]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)+[/tex][tex]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)[/tex][tex]+(Ex+F)(x^8-3x^4+x^2+1)+(Gx+H)(x^4-2x^2+1)[/tex]

[tex]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[/tex]

[tex]\therefore G=\frac{-1}{16}[/tex]

- #37

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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 ?Check your expansion of the [tex]B(1-x)^2(1+x^2)^2[/tex] term

- #38

gabbagabbahey

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- #39

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Thanks very much

- #40

Dick

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