Partial fraction decomposition

[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);