Question

How to find the coefficient of any term in the power series expansion?

find the coefficient of `x^4` in the power series expansion (`x^2`-1)/(`x^2`+1)(`x^2`+2)

Answer 1

`(x^2-1)/((x^2+1)(x^2+2)) =-1/2+5/4x^2-13/8x^4+...`

Explanation:

Perform partial fraction decomposition:

`(x^2-1)/((x^2+1)(x^2+2)) =A/(x^2+1) + B/(x^2+2)`

`(x^2-1)/((x^2+1)(x^2+2)) =(A(x^2+2) +B(x^2+1))/((x^2+1)(x^2+2))`

`x^2-1 = (A+B)x^2 +(2A+B)`

`{(A+B=1),(2A+B=-1):}`

`{(A=-2),(B=3):}`

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

Consider now the sum of a geometric series:

`sum_(n=0)^oo q^n = 1/(1-q)` for `absq<1`

we have:

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

Let `q= -x^2/2` so we can see that:

`3/(x^2+2) = 3/2 sum_(n=0)^oo (-x^2/2)^n`

`3/(x^2+2) = 3/2 sum_(n=0)^oo (-1)^n x^(2n)/2^n`

SImilarly:

`2/(x^2+1) = 2sum_(n=0)^oo (-1)^n x^(2n)`

Then:

`(x^2-1)/((x^2+1)(x^2+2)) = 3/2 sum_(n=0)^oo (-1)^n x^(2n)/2^n - 2sum_(n=0)^oo (-1)^n x^(2n)`

and grouping the terms of the same index:

`(x^2-1)/((x^2+1)(x^2+2)) = sum_(n=0)^oo (-1)^n (3/2^(n+1)-2)x^(2n)`

`(x^2-1)/((x^2+1)(x^2+2)) =-1/2+5/4x^2-13/8x^4+...`