Question #27cc4

Mar 18, 2015

Let me start from a very simple illustrative example.

Say, we want to decompose $\frac{x + 1}{{x}^{2}}$ into partial fractions.
The correct procedure calls for a representation
$\frac{x + 1}{{x}^{2}} = \frac{A}{x} + \frac{B}{{x}^{2}}$,

which results in
$x + 1 = A \cdot x + B$

$\frac{x + 1}{{x}^{2}} = \frac{1}{x} + \frac{1}{{x}^{2}}$.

Without $\frac{A}{x}$ this would not be solvable since the equation would look like
$\frac{x + 1}{{x}^{2}} = \frac{B}{{x}^{2}}$,

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:
$\frac{{x}^{2} + x + 1}{{x}^{3}} = \frac{A}{x} + \frac{B}{{x}^{2}} + \frac{C}{{x}^{3}}$,

that results in

${x}^{2} + x + 1 = A \cdot {x}^{2} + B \cdot x + C$,
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
$\frac{{x}^{2} + 1}{{x}^{3}} = \frac{1}{x} + \frac{1}{{x}^{3}}$.

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.