Partial Fractions
Partial fractions are a technique used to decompose rational functions into simpler components. This method is particularly useful in calculus, especially for integration, and in solving certain types of differential equations, discussed further in topic 5.
Decomposition of Rational Functions
The process of partial fraction decomposition involves breaking down a complex rational function into a sum of simpler fractions. To perform partial fraction decomposition, the following conditions need to be met:
- The polynomial degree of the numerator is less than the degree of the denominator (e.g. \frac{x-4}{x^2+4x+3}
- The denominator can be factored into a product of linear terms (at most two distinct linear terms)
The condition that the numerator's degree must be less than the denominator's is crucial. If this is not the case, long division must be performed first to obtain a polynomial remainder and a rational function that meets this condition.
General Form of Partial Fractions
For a rational function with two distinct linear factors in the denominator, the general form of partial fraction decomposition is:
$$\frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Where $P(x)$ is a polynomial of degree less than 2, and $A$ and $B$ are constants to be determined.
Method for Finding Coefficients
To find the coefficients $A$ and $B$, we can use the following steps:
- Multiply both sides of the equation by the denominator of the original fraction.
- Expand and simplify the right-hand side.
- Equate coefficients of like terms on both sides.
- Solve the resulting system of linear equations.
Let's decompose $\frac{2x+1}{x^2+x-2}$ into partial fractions.
Step 1: Factor the denominator: $x^2+x-2 = (x-1)(x+2)$
Step 2: Set up the partial fraction form:
$$\frac{2x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$
Step 3: Multiply both sides by $(x-1)(x+2)$:
$2x+1 = A(x+2) + B(x-1)$
Step 4: Expand the right-hand side:
$2x+1 = Ax + 2A + Bx - B = (A+B)x + (2A-B)$
Step 5: Equate coefficients:
$2 = A+B$ $1 = 2A-B$
Step 6: Solve the system of equations:
Adding the equations: $3 = 3A$, so $A = 1$ Substituting back: $B = 2-A = 1$
Therefore, the partial fraction decomposition is:
$$\frac{2x+1}{x^2+x-2} = \frac{1}{x-1} + \frac{1}{x+2}$$
Students often forget to check if the degree of the numerator is less than the degree of the denominator before attempting partial fraction decomposition. If it isn't, long division must be performed first.
Higher-order partial fractions
The most straightforward form of higher-order partial fractions is:
$$\frac{P(x)}{(ax+b)(cx+d)(ex+f)...}$$
In this case, the decomposition is the same as for the case with two distinct linear factors.
$$\frac{P(x)}{(ax+b)(cx+d)(ex+f)...}=\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}+...$$
However, this will get very long very quickly, so in exams the denominator is mostly limited to cubics.
Repeated factors
For some polynomials, there will be a repeated factor, i.e.
$$\frac{P(x)}{(ax+b)^2(cx+d)}$$
In this case, the decomposition will be of the form:
$$\frac{P(x)}{(ax+b)^2(cx+d)} = \frac{A}{ax+b}+\frac{B}{(ax+b)^2}+\frac{C}{cx+d}$$
The term with the quadratic denominator $\frac{B}{(ax+b)^2}$ is needed to form the repeated factor when the fractions are added together. Otherwise, the denominator would simply be $(ax+b)(cx+d)$ without the square.
Students often forget that a repeated root requires a term in the decomposition with a higher-order denominator. When there is a factor $(ax+b)^n$ in the denominator, there needs to be terms in the decomposition with denominators $(ax+b), (ax+b)^2, ... (ax+b)^n$.
Irreducible factors
Sometimes, you will end up with an irreducible factor, for example:
$$\frac{x}{(x-3)(x^2+4)}$$
$(x^2+4)$ is not factorable, as its roots are imaginary. Therefore, the partial fraction decomposition would simply be:
$$\frac{x}{(x-3)(x^2+4)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+4}$$
