Partial fractions in calculus

Question

Why are multiplicity expanded in the manner they are expanded? For example, $$ \frac{x^2}{(x-2)^2 \cdot (x-9)}=\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-9)}$$

Why is the $(x-2)^2$ expanded with two terms $A$ and $B$? And also if it had been another higher power $n$, we would expand in the same manner in which the powers decrement? Why do we do this?

Answer 1

Given the following $$ \frac{x^2}{(x−2)^2} $$

We can express it as a sum of two parts $$ \frac{A}{x-2} + \frac{B}{(x-2)^2}$$

or just one part $$ \frac{Dx + E}{(x-2)^2}$$

These different parts, are, in fact, equal $$\frac{A}{x-2} + \frac{B}{(x-2)^2} = \frac{Dx + E}{(x-2)^2}$$

We can multiply $\frac{A}{x-2}$ by $\frac{x-2}{x-2}$, yielding $$\frac{Ax-2A}{(x-2)^2} + \frac{B}{(x-2)^2} = \frac{Dx + E}{(x-2)^2}$$

Combining fractions and like terms yields $$\frac{Ax + B -2A}{(x-2)^2} = \frac{Dx + E}{(x-2)^2}$$

So as you can see, $A = D$ and $B -2A = E$

Why different techniques?

Usually we want the numerators to be just constants, rather than linear equations, because it makes solving for the unknowns much easier.

Take, for example, the following $$ \frac{x^2}{(x−2)^3(x+3)^2(x-5)} $$

We can express it as a sum of five parts $$ \frac{x^2}{(x−2)^3(x+3)^2(x-5)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{x+3} + \frac{E}{(x+3)^2} + \frac{F}{x-5}$$

Compare it with expressing it as three fractions rather than five $$\frac{x^2}{(x−2)^3(x+3)^2(x-5)} = \frac{Gx^2+Hx+I}{(x-2)^3} + \frac{Jx + K}{(x+3)^2} + \frac{L}{x-5}$$

The first one is much easier to work with when trying to find the variables.

Using numerators with more constants $$x^2 = A(x-2)^2(x+3)^2(x-5) + B(x-2)(x+3)^2(x-5) + C(x+3)^2(x-5)+D(x-2)^3(x+3)(x-5) + E(x-2)^3(x-5) + F(x-2)^3(x+3)^2$$

Using numerators with less constants $$x^2 = (Gx^2+Hx+I)(x+3)^2(x-5) + (Jx + K)(x-2)^3(x-5) + L(x-2)^3(x+3)^2$$

Although the first one looks much longer, it is much easier to plug in values. When you plug in $2, -3,$ and $5$, nearly all the terms vanish except for a few. Performing algebraic simplification is much easier.

Ex. For $x=2$ $$4 = C(4+3)^2(4-5)$$ $$4 = -49C$$

Finding C would require much more effort and foiling if $(x-2)$, $(x+3)$, and $(x-5)$ were not split up strategically in powers. And yes, this pattern repeats as $n$ increases. This pattern works for quadratics, and higher order polynomials for, at least, $n > 1$.

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

Answer 2

First, note that the order (degree) of the polynomial in the numerator is less than that of the denominator. Otherwise one must do polynomial long division.

Second, in decomposing a fraction into a sum of simpler fractions, each fraction must have a polynomial of one less order (degree) in the numerator than that of denominator. For example: $$\begin{align}1) \ \ \frac{x^2}{(x-2)^2 \cdot (x-9)}&=\frac{\overbrace{Ax+B}^{1st}}{\underbrace{(x-2)^2}_{2nd}}+\frac{\overbrace{C}^{0th}}{\underbrace{x-9}_{1st}};\\ 2) \ \ \frac{x}{(x-2)^3 \cdot (x-9)^2}&=\frac{\overbrace{Ax^2+Bx+C}^{2nd}}{\underbrace{(x-2)^3}_{3rd}}+\frac{\overbrace{Dx+E}^{1st}}{\underbrace{(x-9)^2}_{2nd}}.\end{align}$$ Third, the fractions can be further decomposed to get constant numbers at the numerators: $$1)\ \ \frac{Ax+B}{(x-2)^2}=\frac{A(x-2)+2A+B}{(x-2)^2}=\frac{A}{x-2}+\frac{2A+B}{(x-2)^2};\\ 2) \ \ \frac{Ax^2+Bx+C}{(x-2)^3}=\frac{A(x-2)^2+(4A+B)(x-2)+4A+2B+C}{(x-2)^3}=\\ \frac{A}{x-2}+\frac{4A+B}{(x-2)^2}+\frac{4A+2B+C}{(x-2)^3}.$$ Reference: 1) Partial fraction decomposition. 2) Polynomial long division..