# Thread: sum

1. ## sum

$f(x)=\frac{x^2-x-1}{(x+2)(x-3)}$

Obtain an expression of f(x) in ascending power of 1/x up to the term in 1/x^3 . Determine the set of values of x which this expansion is valid .

My work .

$f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

$=(x+2)^{-1}-(x-3)^{-1}$

$=\frac{1}{x}(1-\frac{3}{x})^{-1}-\frac{1}{x}(1+\frac{2}{x})^{-1}$

When i continue simplifying , i got $\frac{5}{x^2}+\frac{5}{x^3}$

Am i correct ?

The other thing is to find the set of values of x , is it
$

|-\frac{3}{x}|<1
$

and $|\frac{2}{x}|<1$ ??

2. Originally Posted by thereddevils
f(x)=\frac{x^2-x-1}{(x+2)(x-3)}

Obtain an expression of f(x) in ascending power of 1/x up to the term in 1/x^3 . Determine the set of values of x which this expansion is valid .

My work .

$f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

$=(x+2)^{-1}-(x-3)^{-1}$

$=\frac{1}{x}(1-\frac{3}{x})^{-1}-\frac{1}{x}(1+\frac{2}{x})^{-1}$

When i continue simplifying , i got $\frac{5}{x^2}+\frac{5}{x^3}$

Am i correct ?

The other thing is to find the set of values of x , is it
$

|-\frac{3}{x}|<1
$

and $|\frac{2}{x}|<1$ ??

your function

$f(x)=\frac{x^2-x-1}{(x+2)(x-3)}$

or

$f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

if your question at the first form I think the solution will be

$f(x)=\frac{x^2-x-1}{(x+2)(x-3)}$

$f(x)=\frac{1-\dfrac{1}{x}-\dfrac{1}{x^2}}{\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{3}{x}\right)}$

the denominator should not equal zero in each fraction

so $x\ne 0$ and

$1+\frac{2}{x}\ne 0 \Rightarrow x\ne -2$ and

$1-\frac{3}{x} \ne 0 \Rightarrow x\ne 3$

so the values of x all real numbers expect {-2,0,3}

if

$f(x)=\frac{1}{x-3} - \frac{1}{x+2}$

$f(x)=\frac{5}{(x-3)(x+2)}$

$f(x)=\dfrac{\dfrac{5}{x^2}}{\left(1-\dfrac{3}{x}\right)\left(1+\dfrac{2}{x}\right)}$

and the values of x is the same as the first form

3. Originally Posted by Amer
your function

$f(x)=\frac{x^2-x-1}{(x+2)(x-3)}$

or

$f(x)=\frac{1}{x-3}-\frac{1}{x+2}$

if your question at the first form I think the solution will be

$f(x)=\frac{x^2-x-1}{(x+2)(x-3)}$

$f(x)=\frac{1-\dfrac{1}{x}-\dfrac{1}{x^2}}{\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{3}{x}\right)}$

the denominator should not equal zero in each fraction

so $x\ne 0$ and

$1+\frac{2}{x}\ne 0 \Rightarrow x\ne -2$ and

$1-\frac{3}{x} \ne 0 \Rightarrow x\ne 3$

so the values of x all real numbers expect {-2,0,3}

if

$f(x)=\frac{1}{x-3} - \frac{1}{x+2}$

$f(x)=\frac{5}{(x-3)(x+2)}$

$f(x)=\dfrac{\dfrac{5}{x^2}}{\left(1-\dfrac{3}{x}\right)\left(1+\dfrac{2}{x}\right)}$

and the values of x is the same as the first form

Well , that doesn't answer my question . However , the answer to this question is $1+\frac{5}{x^2}+\frac{5}{x^3}$ , i just cant figure out where is my '1' .

And also the ranges of x which makes this expansion valid ??