Express in partial fractions and expand the terms using binomial expansion up to $x^3$ [closed]

Question

$$ \frac{2}{(1-x)\left(1+x^{2}\right)} $$

This is then split into partial fractions

$$ \frac{A}{1-x}+\frac{B x+C}{1+x^{2}} $$

Computing this i had gotten

\begin{equation} 2=A\left(1+x^{2}\right)+(B x+C)(1-x) \end{equation} \begin{array}{l} 2=A+A x^{2}-B x-B x^{2}+C-C x \\ 2=(A-B) x^{2}+(B-C) x+(B+C) \end{array} A-B=0 b-c=0 A+c=2 B=1 A=1 C=1 As my final answer for i)

\end{array}$$ \frac{1}{1-x}+\frac{x+1}{1+x^{2}} $$

ii) "Expand $$ \frac{1}{1-x} $$ up to and including the term in \begin{equation} x^{3} \end{equation}

(1-x)^-1 =1+x+x^2+x^3.... as my final solution

iii), iv) and v) I am unsure where to go next, is it the same process for iii and iv again?

Edit:

iii) for my final answer I had gotten $$ \frac{1}{1+x^{2}}=\left(1+x^{2}\right)^{-1}=1-x^{2}+\left(x^{2}\right)^{2}-\left(x^{3}\right)^{3} $$

iv) $$\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^{2}} $$

simplfiyng

$$ \begin{array}{l} \left(1+x+x^{2}+x^{3}+\ldots\right)+(x+1)\left(1-x^{2}+. .\right) \\ =1+x+x^{2}+x^{3} . .+x+x^{3}+1-x^{2} \ldots \\ =2+2 x+0 x^{2}+0 x^{3} \end{array} $$

v) x<1

vi) $$ \begin{array}{l} (9+0.045)^{\frac{1}{2}} \\ {\left[\left(1+\frac{0.045}{9}\right]^{\frac{1}{2}}=3(1+0.005)^{\frac{1}{2}}\right.} \end{array} $$ So far this is my process

Answer 1

For iii), you can do $\frac{1}{1 + x^{2}}$ = $1 - x^{2} + x^{4}$...... This can be rewritten as $1 + 0x - x^{2} + 0x^{3} + x^{4}$......, though if this is a homework question, your teacher may or may not accept this.

For iv), you use the result in i) $$\frac{1}{1-x} + \frac{x}{1 + x^{2}} + \frac{1}{1+x^{2}}$$

The first term can be written as $1 + x + x^{2} + x^{3}...$

The second term can be written as $x*(1 - x^2 + x^4....)$ = $x - x^3 + x^5...$

The third term is the answer for iii), $1 - x^2 + x^4...$

Now we have to combine these terms. For sake of brevity I'll let you do it. You would take care of each term modulus 4 .

For v), remember that our expansion is equal to $$\frac{2}{(1 - x)(1+ x^2)}$$

So to find when our expansion converges, we can find when the above is finite, i.e. when it doesn't equal infinity. It equals infinity when the denominator equals 0 -- asuming this is limited to real numbers, that would be when x = 1. So the expansion converges for $x \neq 1$