Question

Question #27cc4

Answer 1

Let me start from a very simple illustrative example.

Say, we want to decompose `(x+1)/(x^2)` into partial fractions.
The correct procedure calls for a representation
`(x+1)/(x^2)=A/x+B/(x^2)`,

which results in
`x+1=A*x+B`

and the answer
`(x+1)/(x^2)=1/x+1/(x^2)`.

Without `A/x` this would not be solvable since the equation would look like
`(x+1)/(x^2)=B/(x^2)`,

that cannot be solved for `B` to make it an identity.

The idea behind having members of the decomposition with every power of a denominator from 1 to maximum is to be able to accommodate any polynomial in the numerator that can contain a variable `x` in any power from 1 to maximum. If we restrict our decomposition only to powers occurring in the denominator, we will only be able to match nominators that are multiple of that power.

Similar example:
`(x^2+x+1)/(x^3)=A/x+B/(x^2)+C/(x^3)`,

that results in

`x^2+x+1=A*x^2+B*x+C`,
which can be solved as `A=1, B=1, C=1`.

Without all three components we would not be able to accommodate any polynomial in the numerator. But, if the numerator does not contain a certain power of `x`, we could have omitted a particular member of the decomposition like in
`(x^2+1)/(x^3)=1/x+1/(x^3)`.

But it's a very peculiar case and, when talking about general rule, it's advisable to retain members with all possible powers of the members of the denominator, from 1 to maximum.