Solve the problem that involves partial fractions

[chwala]

Homework Statement

Given that; $$y=\frac {1+3x^2}{(1+x)^2(1-x)}$$

i. express $y$ in partial fractions.
ii. expand $y$ as a series in ascending powers of $x$, giving the first three terms.

Relevant Equations

partial fractions and binomial theorem

Let $$y=\frac {1+3x^2}{(1+x)^2(1-x)}= \frac {A}{1-x}+\frac {B}{1+x}+\frac {C}{(1+x)^2}$$
$$⇒1+3x^2=A(1+x)^2+B(1-x^2)+C(1-x)$$
$$⇒A-B=3$$
$$2A-C=0$$
$$A+B+C=1$$
On solving the simultaneous equations, we get $A=1$, $B=-2$ and $C=2$
therefore we shall have,
$$y=\frac {1}{1-x}+\frac {-2}{1+x}+\frac {2}{(1+x)^2}$$
$$y=2(1+x)^{-2} +(1-x)^{-1} -2(1+x)^{-1}$$
Now on using binomial theorem we shall get;
$$y=2(1-2x+3x^2+...)+(1+x+x^2+...)+ -2(1-x+x^2+...)$$
$$y=(6x^2-4x+2...)+(x^2+x+1+...)+(-2x^2+2x-2+...)$$
$$y=1-x+5x^2+...$$ Bingo, any other variation would be appreciated guys!

 

[DaveE]

What? I don't understand, do you have a question?
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686784/

 

[chwala]

What? I don't understand, do you have a question?
https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686784/

Am still posting..., i like sharing my approach in solving various math problems and then further explore other available ways of solving from esteemed members here...
I hope this is allowed in the forum...if not then do kindly advise me...

 

[DaveE]

Am still posting..., i like sharing my approach and further explore other available ways of solving from members...i hope this is allowed in the forum...if not then do advise me...

OK, looks better now.

 

[Mark44]

Here's a shorter version that uses a different technique.
$$\frac {1+3x^2}{(1+x)^2(1-x)} = \frac {A}{1-x}+\frac {B}{1+x}+\frac {C}{(1+x)^2}$$
$$⇒1+3x^2=A(1+x)^2+B(1-x^2)+C(1-x)$$
If x = 1, then 4 = 4A, so A =1
If x = -1, then 4 = 1(0) + B(0) + C(2), so C = 2
If x = 0, then 1 = 1(1) + B(1) + 2(1), so B = -2
Summarizing, A = 1, B = -2, C = 2

The technique you (@chwala) used equates the coefficients of the polynomials on either side of the equation. The technique I used relies on the fact that the equation $\frac {1+3x^2}{(1+x)^2(1-x)} = \frac {A}{1-x}+\frac {B}{1+x}+\frac {C}{(1+x)^2}$ is identically true: it must be true for all values of x. What I did was to pick values of x that made some of the terms on the right side vanish.