Partial Fractions: Success for A Level Maths
Partial Fractions – Introduction
When studying advanced mathematics at the A Level, one topic that often comes up is partial fractions. In this blog post, we will delve into the concept of partial fractions, discussing what they are and how they relate to A Level Maths.
Definition and Purpose:
Partial fractions are a method used to break down complex rational expressions into simpler fractions. This technique is particularly useful when dealing with integrals, differential equations, or algebraic expressions involving rational functions.
Partial Fractions - The Process
To express a rational function as a sum of partial fractions, we follow a systematic process:
Factorisation:
First, we factorise the denominator of the rational function. This step involves decomposing the denominator into irreducible factors, which cannot be factored further.
Setting Up Partial Fractions:
Once the denominator is factored, we set up the partial fractions. For each distinct factor in the denominator, we write a separate fraction with an unknown numerator over that factor.
Determining Unknown Numerators:
Next, we need to determine the unknown numerators in the partial fractions. To do this, we either use a suitable method, such as the method of undetermined coefficients, or equate coefficients of corresponding powers of the variable in the original expression.
Combining Fractions:
After finding the unknown numerators, we combine the fractions into a single expression. This will give us the original rational function expressed as a sum of partial fractions.
Application:
Partial fractions have various applications across different areas of mathematics:
Integration:
When integrating rational functions, expressing them as partial fractions allows us to simplify the integration process. By breaking down a complex rational expression into simpler fractions, we can integrate each fraction separately using well-known integration techniques.
Solve Differential Equations:
Partial fractions are also useful in solving linear ordinary differential equations. By converting the differential equation into its algebraic form using partial fractions, we can solve it more easily.
Simplify Algebraic Expressions:
Sometimes, partial fractions are used to simplify algebraic expressions involving rational functions. Breaking down a complex expression into partial fractions makes it easier to manipulate and solve for specific variables.
Partial Fractions - Example 1
Part b)
\begin{aligned} \frac{6 x-2}{(x-3)(x+1)} & \equiv \frac{A}{x-3}+\frac{B}{x+1} \\ & \equiv \frac{A(x+1)+B(x-3)}{(x-3)(x+1)} \\ 6 x-2 & \equiv A(x+1)+B(x-3) \\ & \equiv A x+A+B x-3 B \\ & \equiv(A+B) x+(A-3 B) \end{aligned}Equating coefficients
6=A+BEquate constants
-2=A-3 BSubtracting
\begin{aligned} & 8=4 B \\ & B=2 \end{aligned}Substituting to find A
6=A+2So A = 4
Partial Fractions – Example 2
So
\frac{A(x-1)(2 x+1)+B x(2 x+1)+C x(x-1)}{x(x-1)(2 x+1)}Which gives
\begin{aligned} \therefore 6 x^2+5 x-2 \equiv & A(x-1)(2 x+1) \\ & +B x(2 x+1)+C x(x-1) \end{aligned}x = 1:
\begin{aligned} 6+5-2 & =0+B \times 1 \times 3+0 \\ 9 & =3 B \\ B & =3 \end{aligned}x= 0
\begin{aligned} \mathrm{O}+\mathrm{O}-2 & =A \times(-1) \times 1+\mathrm{O}+\mathrm{O} \\ -2 & =-A \\ A & =2 \end{aligned}Let x=-\frac{1}{2}
\begin{aligned} \frac{6}{4}-\frac{5}{2}-2 & =0+0+C \times\left(-\frac{1}{2}\right) \times\left(-\frac{3}{2}\right) \\ -3 & =\frac{3}{4} C \\ C & =-4 \end{aligned}Which gives:
\frac{6 x^2+5 x-2}{x(x-1)(2 x+1)} \equiv \frac{2}{x}+\frac{3}{x-1}-\frac{4}{2 x+1}So A=2, B=3 \text { and } C=-4 \text {. }
Partial Fractions Example 3
Solution
\begin{aligned} & \frac{11 x^2+14 x+5}{(x+1)^2(2 x+1)} \equiv \frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(2 x+1)} \\ & \equiv \frac{A(x+1)(2 x+1)+B(2 x+1)+C(x+1)^2}{(x+1)^2(2 x+1)} \end{aligned}So \begin{gathered} 11 x^2+14 x+5 \\ \equiv A(x+1)(2 x+1)+B(2 x+1)+C(x+1)^2 \end{gathered}
Let x = – 1
\begin{aligned} 11-14+5 & =A \times 0+B \times-1+C \times 0 \\ 2 & =-1 B \\ B & =-2 \end{aligned}Let x=-\frac{1}{2}
This gives:
\begin{aligned} \frac{11}{4}-7+5 & =\mathrm{A} \times \mathrm{O}+\mathrm{B} \times \mathrm{O}+C \times \frac{1}{4} . \\ \frac{3}{4} & =\frac{1}{4} C \\ C & =3 \\ 11 & =2 A+C \\ 11 & =2 A+3 \\ 2 A & =8 \\ A & =4 \end{aligned}So:
\begin{aligned} & \frac{11 x^2+14 x+5}{(x+1)^2(2 x+1)} \\ & \equiv \frac{4}{(x+1)}-\frac{2}{(x+1)^2}+\frac{3}{(2 x+1)} \end{aligned}And A=4, B=-2 \text { and } C=3 \text {. }
Conclusion:
Partial fractions provide a powerful tool in simplifying and solving complex rational expressions in A Level Maths. By breaking down such expressions into simpler fractions, we can integrate, solve differential equations, and simplify algebraic expressions more effectively. Understanding and mastering the concept of partial fractions is essential for tackling advanced mathematical problems at the A Level.