Welcome to Further Algebra and Functions (CCEA AS 1 Pure Mathematics)
Welcome to one of the most powerful and rewarding chapters in your CCEA AS Further Mathematics (Subject Code: 2330) journey! In standard A Level mathematics, algebra gives us the tools to solve equations and model curves. In Further Mathematics, we take these algebraic tools much further.
In this chapter, you will master four core algebraic techniques:
• Roots of Polynomial Equations: Discovering hidden links between coefficients and roots without actually solving the equations.
• Summation of Series: Adding up thousands of terms using neat algebraic patterns and standard formulae.
• Partial Fractions & Rational Expressions: Breaking down complicated algebraic fractions into simpler, manageable building blocks.
• Maclaurin Series: Turning complex curves and functions into infinite polynomials that are easy to evaluate.
Don't worry if some of these topics sound intimidating right now! We will break down every single concept into small, easy-to-digest steps with clear examples and examiner tips.
---1. Roots of Polynomial Equations (Vieta’s Relations)
When solving a quadratic equation \(ax^2 + bx + c = 0\), we usually find the roots using factorisation or the quadratic formula. But what if we do not need to know the exact individual roots, or what if finding them is too messy? Vieta’s Relations give us a direct shortcut connecting the roots to the coefficients \(a\), \(b\), and \(c\).
The Fundamental Relations for Quadratics
For any quadratic equation \(ax^2 + bx + c = 0\) (where \(a \neq 0\)) with roots \(\alpha\) and \(\beta\):
• Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
• Product of roots: \(\alpha\beta = \frac{c}{a}\)
Memory Trick: Notice the alternating signs! The sum starts with a minus sign \(\left(-\frac{b}{a}\right)\), while the product has a plus sign \(\left(+\frac{c}{a}\right)\).
Symmetric Functions of Roots
A symmetric expression is an expression that does not change if you swap \(\alpha\) and \(\beta\). To evaluate these, we rewrite them purely in terms of \((\alpha + \beta)\) and \((\alpha\beta)\):
1. Sum of Squares:
Expanding \((\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2\), rearranging gives:
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
2. Sum of Cubes:
Expanding \((\alpha + \beta)^3 = \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta)\), rearranging gives:
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
3. Sum of Reciprocals:
Combining into a single fraction:
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)
4. Difference Squared:
\((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
Forming New Equations with Transformed Roots
You will often be asked to find a new quadratic equation whose roots are related to \(\alpha\) and \(\beta\) (for example, roots \(2\alpha, 2\beta\), or \(\alpha + 2, \beta + 2\), or \(\frac{1}{\alpha}, \frac{1}{\beta}\)).
There are two reliable methods to do this:
Method A: Sum and Product Method
1. Calculate the New Sum of the transformed roots.
2. Calculate the New Product of the transformed roots.
3. Write the new equation as \(w^2 - (\text{New Sum})w + (\text{New Product}) = 0\).
Method B: Direct Substitution Method
Let \(w\) represent the new root in terms of \(x\) (e.g., \(w = g(x)\)). Rearrange this to find \(x = g^{-1}(w)\), and substitute this expression for \(x\) directly into the original equation \(ax^2 + bx + c = 0\). Simplify and expand!
Step-by-Step Worked Example
The equation \(2x^2 - 5x + 1 = 0\) has roots \(\alpha\) and \(\beta\). Find a quadratic equation with integer coefficients whose roots are \(\alpha + 3\) and \(\beta + 3\).
Step 1: Identify original values from \(2x^2 - 5x + 1 = 0\):
\(a = 2\), \(b = -5\), \(c = 1\)
\(\alpha + \beta = -\left(-\frac{5}{2}\right) = \frac{5}{2}\)
\(\alpha\beta = \frac{1}{2}\)
Step 2: Using Method A (Sum and Product):
\(\text{New Sum} = (\alpha + 3) + (\beta + 3) = (\alpha + \beta) + 6 = \frac{5}{2} + 6 = \frac{17}{2}\)
\(\text{New Product} = (\alpha + 3)(\beta + 3) = \alpha\beta + 3(\alpha + \beta) + 9 = \frac{1}{2} + 3\left(\frac{5}{2}\right) + 9 = \frac{1}{2} + \frac{15}{2} + 9 = 8 + 9 = 17\)
Step 3: Form the new equation in \(w\):
\(w^2 - \left(\frac{17}{2}\right)w + 17 = 0\)
Multiply through by \(2\) to obtain integer coefficients:
\(2w^2 - 17w + 34 = 0\)
Alternative check via Method B: Let \(w = x + 3 \implies x = w - 3\).
Substitute into original: \(2(w - 3)^2 - 5(w - 3) + 1 = 0\)
\(2(w^2 - 6w + 9) - 5w + 15 + 1 = 0\)
\(2w^2 - 12w + 18 - 5w + 16 = 0 \implies 2w^2 - 17w + 34 = 0\). Both methods yield the exact same answer!
Key Takeaways & Pitfalls: Roots of Polynomials
• Watch the negative sign: A very common error in CCEA exams is writing \(\alpha + \beta = \frac{b}{a}\). Always remember the negative sign: \(-\frac{b}{a}\).
• Integer coefficients: If the question asks for integer coefficients, always clear fractions at the end.
2. Summation of Series and Standard Sum Formulae
Sigma notation \(\sum\) represents adding together a sequence of terms. Instead of manually adding hundreds of terms, we use standard algebraic summation formulae.
Standard Summation Formulae
You need to know and be able to apply these standard sum identities:
• Sum of Constants:
\(\sum_{r=1}^n 1 = n\)
(Adding the number 1 to itself \(n\) times equals \(n\). If summing a constant \(c\), \(\sum_{r=1}^n c = cn\).)
• Sum of First \(n\) Integers:
\(\sum_{r=1}^n r = \frac{1}{2}n(n+1)\)
• Sum of First \(n\) Squares:
\(\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)\)
• Sum of First \(n\) Cubes:
\(\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2 = \left(\sum_{r=1}^n r\right)^2\)
Linearity Property of Summation
Just like expanding brackets, you can split sums over addition/subtraction and pull out constant multipliers:
\(\sum_{r=1}^n (a r^2 + b r + c) = a\sum_{r=1}^n r^2 + b\sum_{r=1}^n r + c\sum_{r=1}^n 1\)
Evaluating Sums Over Sub-intervals
What if the sum does not start at \(r = 1\)? If you need to sum from \(r = k\) to \(n\), calculate the sum up to \(n\) and subtract the terms that come before \(k\):
\(\sum_{r=k}^n f(r) = \sum_{r=1}^n f(r) - \sum_{r=1}^{k-1} f(r)\)
Crucial Examiner Warning: A classic trap is subtracting up to \(k\). If the sum starts at \(r = 10\), the 10th term is part of the sum! Therefore, you must subtract the sum up to \(k - 1 = 9\).
The Method of Differences (Telescoping Series)
When summing rational expressions, you can often split the general term \(f(r)\) into partial fractions of the form \(g(r) - g(r+1)\) or \(g(r) - g(r+2)\). When you write out the sum term-by-term, almost all intermediate terms cancel out like collapsing dominoes.
Step-by-Step Worked Example (Method of Differences)
Show that \(\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}\).
Step 1: Express the term in partial fractions:
\(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\)
Step 2: Write out the first few terms and the last few terms of the series:
For \(r = 1\): \(\left(1 - \frac{1}{2}\right)\)
For \(r = 2\): \(\left(\frac{1}{2} - \frac{1}{3}\right)\)
For \(r = 3\): \(\left(\frac{1}{3} - \frac{1}{4}\right)\)
\(\dots\)
For \(r = n-1\): \(\left(\frac{1}{n-1} - \frac{1}{n}\right)\)
For \(r = n\): \(\left(\frac{1}{n} - \frac{1}{n+1}\right)\)
Step 3: Cancel the intermediate terms:
Notice that \(-\frac{1}{2}\) cancels with \(+\frac{1}{2}\), \(-\frac{1}{3}\) cancels with \(+\frac{1}{3}\), and so on up to \(-\frac{1}{n}\) and \(+\frac{1}{n}\).
Step 4: Collect the surviving terms (the first and the last):
\(\text{Sum} = 1 - \frac{1}{n+1} = \frac{n+1 - 1}{n+1} = \frac{n}{n+1}\)
Key Takeaways & Pitfalls: Summation
• Constant term error: \(\sum_{r=1}^n 5 = 5n\), not \(5\).
• Factorising completely: When summing polynomial terms, always look to take out common factors such as \(\frac{1}{6}n(n+1)\) early rather than expanding out giant cubic expressions.
3. Partial Fractions and Rational Expressions
Partial fractions allow us to reverse the process of adding algebraic fractions. They are essential for summation using the method of differences and for advanced integration.
Proper vs Improper Algebraic Fractions
Before splitting into partial fractions, you must compare the degree (highest power of \(x\)) of the numerator \(P(x)\) and the denominator \(Q(x)\):
• Proper Fraction: \(\text{Degree of Numerator} < \text{Degree of Denominator}\). You can proceed straight to partial fractions.
• Improper Fraction: \(\text{Degree of Numerator} \ge \text{Degree of Denominator}\). You must perform polynomial algebraic division first to obtain a polynomial quotient plus a proper remainder fraction.
The Three Denominator Cases
Once your rational expression is in proper form, factorise the denominator completely. The form of the partial fractions depends on the nature of the factors:
Case 1: Distinct Linear Factors
\(\frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}\)
Case 2: Repeated Linear Factors
\(\frac{P(x)}{(ax+b)^2(cx+d)} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{cx+d}\)
(Notice the repeated factor requires two terms: one to the power of 1, and one squared.)
Case 3: Irreducible Quadratic Factors
An irreducible quadratic factor cannot be factorised into real linear terms (e.g., \(x^2 + 1\) or \(ax^2 + bx + c\) where \(b^2 - 4ac < 0\)). Its numerator must be linear \((Ax + B)\):
\(\frac{P(x)}{(ax^2+bx+c)(dx+e)} = \frac{Ax+B}{ax^2+bx+c} + \frac{C}{dx+e}\)
Step-by-Step Worked Example (Repeated Linear Factor)
Express \(\frac{4x^2 + 5x + 3}{(x+1)^2(x+2)}\) in partial fractions.
Step 1: Set up the partial fraction template:
\(\frac{4x^2 + 5x + 3}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}\)
Step 2: Multiply through by the common denominator \((x+1)^2(x+2)\):
\(4x^2 + 5x + 3 = A(x+1)(x+2) + B(x+2) + C(x+1)^2\)
Step 3: Substitute strategic values of \(x\) to find constants:
• Let \(x = -1\):
\(4(-1)^2 + 5(-1) + 3 = A(0) + B(-1+2) + C(0)\)
\(4 - 5 + 3 = B(1) \implies B = 2\)
• Let \(x = -2\):
\(4(-2)^2 + 5(-2) + 3 = A(0) + B(0) + C(-2+1)^2\)
\(16 - 10 + 3 = C(1) \implies C = 9\)
• Equate coefficients of \(x^2\) to find \(A\):
\(4 = A + C \implies 4 = A + 9 \implies A = -5\)
Step 4: Write down the final decomposition:
\(\frac{4x^2 + 5x + 3}{(x+1)^2(x+2)} = -\frac{5}{x+1} + \frac{2}{(x+1)^2} + \frac{9}{x+2}\)
Key Takeaways & Pitfalls: Partial Fractions
• Check for improper fractions first: If numerator degree \(\ge\) denominator degree, failure to divide first is an immediate exam trap.
• Irreducible quadratics need \(Ax + B\): Never put just a single constant \(A\) over an irreducible quadratic denominator.
4. Maclaurin Series (Power Series Expansions)
A Maclaurin series allows us to approximate complicated functions (such as trigonometric, exponential, and logarithmic functions) as polynomials in ascending powers of \(x\). This makes calculus, limits, and approximations vastly simpler.
The Maclaurin Expansion Formula
For a function \(f(x)\) that can be differentiated repeatedly at \(x = 0\):
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots + \frac{f^{(n)}(0)}{n!}x^n + \dots\)
Note on Factorials: Recall that \(0! = 1\), \(1! = 1\), \(2! = 2 \times 1 = 2\), \(3! = 3 \times 2 \times 1 = 6\), and \(4! = 24\).
Standard Procedure for Finding a Maclaurin Series
Step 1: Write down \(f(x)\) and compute its successive derivatives: \(f'(x)\), \(f''(x)\), \(f'''(x)\), etc., up to the power required by the question.
Step 2: Evaluate the original function and every derivative at \(x = 0\) to get numerical values: \(f(0)\), \(f'(0)\), \(f''(0)\), \(f'''(0)\).
Step 3: Substitute these values into the Maclaurin expansion formula, not forgetting the factorial denominators \(2!\), \(3!\), etc.
Step 4: Simplify the coefficients.
Step-by-Step Worked Example
Find the Maclaurin series expansion of \(f(x) = \ln(1 + 2x)\) up to and including the term in \(x^3\).
Step 1 & 2: Differentiate and evaluate at \(x = 0\)
• \(f(x) = \ln(1 + 2x) \implies f(0) = \ln(1) = 0\)
• \(f'(x) = 2(1 + 2x)^{-1} \implies f'(0) = 2(1)^{-1} = 2\)
• \(f''(x) = 2(-1)(2)(1 + 2x)^{-2} = -4(1 + 2x)^{-2} \implies f''(0) = -4(1)^{-2} = -4\)
• \(f'''(x) = -4(-2)(2)(1 + 2x)^{-3} = 16(1 + 2x)^{-3} \implies f'''(0) = 16(1)^{-3} = 16\)
Step 3: Substitute into the Maclaurin formula
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)
\(f(x) = 0 + 2x + \frac{-4}{2!}x^2 + \frac{16}{3!}x^3 + \dots\)
Step 4: Simplify
\(f(x) = 2x - \frac{4}{2}x^2 + \frac{16}{6}x^3 + \dots\)
\(f(x) = 2x - 2x^2 + \frac{8}{3}x^3 + \dots\)
Key Takeaways & Pitfalls: Maclaurin Series
• Don't forget the factorials: A very common error is writing \(\frac{f'''(0)}{3}x^3\) instead of \(\frac{f'''(0)}{3!}x^3\) (which has a denominator of \(6\)).
• Chain Rule accuracy: Take extra care when using the chain rule on composite functions (like \(2(1+2x)^{-1}\) above where the inner derivative produces extra factors of \(2\)).
Quick Revision Checklist
Before sitting your AS 1 examination, make sure you can confidently:
• State and use Vieta’s formulas \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\) for quadratics.
• Express symmetric combinations (\(\alpha^2+\beta^2\), \(\alpha^3+\beta^3\), \(\frac{1}{\alpha}+\frac{1}{\beta}\)) in terms of \(\alpha+\beta\) and \(\alpha\beta\).
• Form new equations using both substitution and sum/product methods.
• Apply standard series formulae for \(\sum 1\), \(\sum r\), \(\sum r^2\), and \(\sum r^3\).
• Sum series over sub-intervals using \(\sum_{r=k}^n f(r) = \sum_{r=1}^n f(r) - \sum_{r=1}^{k-1} f(r)\).
• Apply the method of differences by canceling telescoping terms.
• Decompose distinct linear, repeated linear, and irreducible quadratic rational expressions into partial fractions (and perform polynomial division for improper fractions).
• Derive Maclaurin series expansions up to a required power of \(x\) by finding successive derivatives evaluated at \(x=0\).