Hideous Partial Differential decomposition

[chengeto]

Find the partial fraction decomposition of \(\displaystyle \frac{x^2}{(1-x^4)^2}\)

Attempt to solution:

\(\displaystyle \frac{x^2}{(1-x^4)^2}=\frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}\)

\(\displaystyle \frac{x^2}{(1+x)^2(1-x)^2(1+x^2)^2}=\frac{A}{1+x}+\frac{B}{(1+x)^2}+\frac{C}{1-x}+\frac{D}{(1-x^2)}+\frac {Ex+F}{1+x^2}+\frac {Gx+H}{(1+x^2)^2}\)


\(\displaystyle x^2=A(1+x)(1-x)^2(1+x^2)^2+B(1-x)^2(1+x^2)^2+C(1-x)(1+x^2)^2(1+x)^2+D(1+x^2)^2(1+x)^2\)
\(\displaystyle +(Ex+F)(1+x)^2 (1+x^2)(1-x)^2+(Gx+H) (1-x)^2(1+x)^2\)

when x=-1

\(\displaystyle B(1-x)^2(1+x^2)^2=x^2\)

\(\displaystyle B=\frac{1}{(1--1)^2(1+(-1)^2)^2)}=\frac{1}{16}\)

when x=1


\(\displaystyle D=\frac{1}{16}\)

\(\displaystyle x^2=A(x^7-x^6+x^5-x^4-x^3+x^2-x+1)+B(x^6-x^5+3x^4-4x^3+3x^2-2x+1)+\)\(\displaystyle C(-x^7-x^6-x^5-x^4+x^3+x^2+x+1)+ D(x^6+2x^5+3x^4+4x^3+3x^2+2x+1)\)
\(\displaystyle +(Ex+F)(x^8-3x^4+x^2+1)+(Gx+H)(x^4-2x^2+1)\)\(\displaystyle (Gx+H)(x^4-2x^2+1)\)


Equating coefficients



\(\displaystyle x^9\rightarrow 0=E\)

\(\displaystyle x^8\rightarrow 0=F\)

\(\displaystyle x^7\rightarrow 0=A-C\)

\(\displaystyle \therefore A=C\)

\(\displaystyle x^6\rightarrow 0=-A+\frac{1}{16}-C+\frac{1}{16} \rightarrow \frac{-1}{8}=-2A\)

\(\displaystyle \therefore A=C=\frac{1}{16}\)

\(\displaystyle x^5\rightarrow 0=A-B-C+\frac{1}{8}+G \rightarrow 0=\frac{1}{16}-\frac{1}{16}-\frac{1}{16}+\frac{1}{8}+G\)

\(\displaystyle \therefore G=\frac{-1}{8}\)

\(\displaystyle x^4\rightarrow \frac{-3}{16}=-A+3B-C+H \rightarrow \frac{-3}{16}=-2A+3B\)

\(\displaystyle \therefore H=\frac{1}{8}\)


Guys how come the answers at the back of the textbook are saying l am wrong. Did l use the wrong approach to solve this question ? Is there an easier one ?

 

[royhaas]

Perhaps you could try the decomposition of \(\displaystyle \frac{x}{1-x^4}\) and then square it.

 

[chengeto]

Perhaps you could try the decomposition of \(\displaystyle \frac{x}{1-x^4}\) and then square it.

Thanks for the help, l have actually been able to solve it.

 

[BigGlenntheHeavy]

Or easier yet if you have a trusty TI-89 graphic calculator; just plugged in the nasty equation and hit expand (F2-3), and walla, you have it.

 

[chengeto]

Or easier yet if you have a trusty TI-89 graphic calculator; just plugged in the nasty equation and hit expand (F2-3), and walla, you have it.

Good idea but I will be cheating on myself because we are not allowed to use a calculator in the final exam. So l might have to learn to solve this questions using the brute force method

 

[galactus]

I tend to agree with pka on this matter. Some topics are best left to the calculators and have become rather redundant in that they still make

students do them by hand. PFD's are one of those things. Teach mathematics and not these mundane, archaic details.

I plugged this into my V200 and it easily gives me:

\(\displaystyle \frac{-1}{4(x^{2}+1)^{2}}+\frac{1}{16(x-1)^{2}}+\frac{1}{16(x+1)^{2}}+\frac{1}{16(x+1)}-\frac{1}{16(x-1)}\)

 

[chengeto]

I tend to agree with pka on this matter. Some topics are best left to the calculators and have become rather redundant in that they still make

students do them by hand. PFD's are one of those things. Teach mathematics and not these mundane, archaic details.

I plugged this into my V200 and it easily gives me:

\(\displaystyle \frac{-1}{4(x^{2}+1)^{2}}+\frac{1}{16(x-1)^{2}}+\frac{1}{16(x+1)^{2}}+\frac{1}{16(x+1)}-\frac{1}{16(x-1)}\)

When l plugged this into a friend s TI-89 l got. It looks a bit different from what you got.

\(\displaystyle \frac{x^2}{(1-x^4)^2}=\frac{\frac{1}{16}}{1+x}+\frac{\frac{1}{16}}{(1+x)^2}+\frac{\frac{1}{16}}{1-x}+\frac{\frac{1}{16}}{(1-x)^2}-\frac{\frac{1}{4}}{(1+x^2)^2}\)

 

[galactus]

That's the same thing.

 

[chengeto]

That's the same thing.

Okay thanks