Question

Express `f(x) = (3x)/ [(1 +x)(1+2x^2)]` in partial fractions?

let `f(x) = (3x)/ [(1 +x)(1+2x^2)]`
(i) Express f(x) in partial fractions.
(ii) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in `x^3`

Answer 1

The partial fraction is `f(x)=(-1)/(1+x)+(2x+1)/(1+2x^2)`. The expansion is `=3(x-x^2-x^3)+o(x^3)`

Explanation:

Perform the decomposition into partial fractions

`(3x)/((1+x)(1+2x^2))=A/(1+x)+(Bx+C)/(1+2x^2)`

`=(A(1+2x^2)+(Bx+C)(1+x))/((1+x)(1+2x^2))`

The denominators are the same, compare the numerators

`3x=A(1+2x^2)+(Bx+C)(1+x)`

Let `x=0`, `=>`, `0=A+C`

Let `x=-1`, `=>`, `-3=3A`, `=>`, `A=-1`

`=>`, `C=-A=1`

Coefficients of `x^2`

`0=2A+B`, `=>`, `B=-2A=2`

Therefore,

`(3x)/((1+x)(1+2x^2))=(-1)/(1+x)+(2x+1)/(1+2x^2)`

The Taylor expansions are

`-1/(1+x)=-1+x-x^2+x^3 + o(x^3)`

`(2x+1)/(1+2x^2)=1+2x-2x^2-4x^3+o(x^3)`

You can also obtain theses expansions by performing a long division

Therefore,

`f(x)=-1/(1+x)+(2x+1)/(1+2x^2)=0+3x-3x^2-3x^3+o(x^3)`

Answer 2

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

Explanation:

Let

`{3x}/{(1+x)(1+2x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1+2x^2}`

`3x=(2A+B)x^2+(B+C)x+(A+C)`

Comparing the corresponding coefficients on both the sides, we get

`2A+B=0 \ ......(1)`

`B+C=3\ ............(2)`

`A+C=0\ .......(3)`

Subtracting (2) from (3), we get

`A-B=-3\ ......(4)`

Adding (1) & (4) we get

`2A+B+A-B=0-3`

`3A=-3`

`A=-1`

setting `A=-1` in (4), we get

`-1-B=-3`

`B=2`

Setting `A=-1` in (3), we get

`-1+C=0`

`C=1`

Now, setting the values of `A, B, C`, we get following partial fractions

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