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chwala

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- 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!

$$⇒1+3x^2=A(1+x)^2+B(1-x^2)+C(1-x)$$

$$⇒A-B=3$$

$$2A-C=0$$

$$A+B+C=1$$

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!

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