Welcome to Further Algebra and Functions
Welcome to one of the most powerful and elegant chapters in CCEA A2 1 Pure Mathematics! In standard A Level Mathematics, algebra often involves finding individual numerical solutions or graphing simple curves. In Further Mathematics, we take a giant leap forward: we begin to explore the underlying structure and symmetries of equations, sum up thousands of algebraic terms in seconds, and rewrite complex curves as friendly polynomials using infinite series.
Don't worry if these ideas sound abstract right now. We will break every concept down step-by-step with clear recipes, memory aids, and common pitfalls to watch out for. Let's dive in!
1. Roots of Polynomial Equations
When solving an equation such as a quadratic, cubic, or quartic, we usually try to find the roots. But did you know that you can know almost everything about the relationships between roots without solving the equation at all? This relies on the relationships between the coefficients and the roots discovered by the French mathematician François Viète.
The General Principle: The Sign Alternation Rule
For any polynomial equation written in standard descending order:
\(a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \dots + a_1 x + a_0 = 0\)
If we divide every term by the leading coefficient \(a_n\), the coefficients relate directly to the sums and products of the roots, with alternating signs: negative, positive, negative, positive...
1.1 Quadratic Equations (\(n = 2\))
For the equation \(a x^2 + b x + c = 0\) with roots \(\alpha\) and \(\beta\):
Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
Product of roots: \(\alpha \beta = \frac{c}{a}\)
1.2 Cubic Equations (\(n = 3\))
For the cubic equation \(a x^3 + b x^2 + c x + d = 0\) with roots \(\alpha\), \(\beta\), and \(\gamma\):
Sum taken one at a time: \(\sum \alpha = \alpha + \beta + \gamma = -\frac{b}{a}\)
Sum of products taken two at a time: \(\sum \alpha \beta = \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}\)
Product taken three at a time: \(\alpha \beta \gamma = -\frac{d}{a}\)
1.3 Quartic Equations (\(n = 4\))
For the quartic equation \(a x^4 + b x^3 + c x^2 + d x + e = 0\) with roots \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\):
Sum taken one at a time: \(\sum \alpha = \alpha + \beta + \gamma + \delta = -\frac{b}{a}\)
Sum of products taken two at a time: \(\sum \alpha \beta = \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{c}{a}\)
Sum of products taken three at a time: \(\sum \alpha \beta \gamma = \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = -\frac{d}{a}\)
Product of all four roots: \(\alpha \beta \gamma \delta = \frac{e}{a}\)
Memory Aid for Signs: Always write out the signs above the coefficients from left to right starting after the leading term: \((-)\), \((+)\), \((-)\), \((+)\).
Symmetric Expressions: Key Algebraic Identities
In exam questions, you are often asked to evaluate expressions involving combinations of roots. Because these expressions are symmetric (swapping any two roots leaves the expression unchanged), they can always be rewritten entirely in terms of the standard sums \(\sum \alpha\), \(\sum \alpha \beta\), and \(\alpha \beta \gamma\).
Here are the essential identities you must master:
1. Sum of squares: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)\)
In shorthand notation: \(\sum \alpha^2 = (\sum \alpha)^2 - 2 \sum \alpha \beta\)
2. Sum of reciprocals: \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \gamma \alpha + \alpha \beta}{\alpha \beta \gamma} = \frac{\sum \alpha \beta}{\alpha \beta \gamma}\)
3. Sum of cubes: \(\alpha^3 + \beta^3 + \gamma^3\)
Since each root satisfies the original cubic \(a x^3 + b x^2 + c x + d = 0\), substitute each root and add the equations together:
\(a \sum \alpha^3 + b \sum \alpha^2 + c \sum \alpha + 3d = 0\)
(Notice that the constant term \(d\) appears 3 times, once for each root, giving \(3d\).)
Forming New Equations with Transformed Roots
A classic CCEA exam question gives you a polynomial with roots \(\alpha, \beta, \gamma\) and asks you to find a new polynomial whose roots are transformed (for example, roots that are \(2\alpha, 2\beta, 2\gamma\) or \(\alpha + 3, \beta + 3, \gamma + 3\) or \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)).
The Direct Substitution Method:
Let \(w\) represent the new root in terms of the old root \(x\). Rearrange this relationship to make \(x\) the subject, then substitute this expression for \(x\) into the original equation.
Worked Example:
The cubic equation \(x^3 - 3x^2 + 2x - 5 = 0\) has roots \(\alpha, \beta, \gamma\). Find an equation with integer coefficients whose roots are \(2\alpha + 1, 2\beta + 1, 2\gamma + 1\).
Step 1: Set \(w = 2x + 1\).
Step 2: Rearrange for \(x\): \(x = \frac{w - 1}{2}\).
Step 3: Substitute into the original cubic:
\(\left(\frac{w - 1}{2}\right)^3 - 3\left(\frac{w - 1}{2}\right)^2 + 2\left(\frac{w - 1}{2}\right) - 5 = 0\)
Step 4: Multiply the entire equation by \(8\) to clear fractions:
\((w - 1)^3 - 6(w - 1)^2 + 8(w - 1) - 40 = 0\)
Step 5: Expand and collect like terms:
\((w^3 - 3w^2 + 3w - 1) - 6(w^2 - 2w + 1) + 8(w - 1) - 40 = 0\)
\(w^3 - 9w^2 + 23w - 55 = 0\)
Common Mistake to Avoid: When dealing with transformed roots like \(\frac{1}{\alpha}\), students often try to compute all the new sums manually. While this works, substituting \(x = \frac{1}{w}\) is ten times faster and far less prone to arithmetic slips!
Key Takeaway for Roots of Polynomials: Always write down the values of \(\sum \alpha\), \(\sum \alpha \beta\), etc., before doing anything else. For transforming equations, the substitution \(w = g(x) \implies x = g^{-1}(w)\) is your most powerful tool.
2. Summation of Finite Series
In standard maths, you learned how to sum Arithmetic and Geometric progressions. In Further Mathematics, we can find the exact sum of polynomials in \(r\) (such as \(\sum r^2\) or \(\sum r(r+1)\)) and rational functions using standard formulae and the Method of Differences.
2.1 Standard Summation Formulae
You need to be completely confident using these standard results:
1. Constant sum: \(\sum_{r=1}^{n} 1 = n\)
2. Sum of the first \(n\) natural numbers: \(\sum_{r=1}^{n} r = \frac{1}{2}n(n + 1)\)
3. Sum of squares: \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\)
4. Sum of cubes: \(\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2 = \left(\sum_{r=1}^{n} r\right)^2\)
Did you know? The sum of the first \(n\) cubes is always equal to the square of the sum of the first \(n\) integers! For example: \(1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36\), which is exactly \((1 + 2 + 3)^2 = 6^2 = 36\).
2.2 Using Linearity and Factorisation
Because the summation operator \(\sum\) is linear, you can split expressions term-by-term:
\(\sum_{r=1}^{n} (a r^2 + b r + c) = a\sum_{r=1}^{n} r^2 + b\sum_{r=1}^{n} r + c\sum_{r=1}^{n} 1\)
Top Exam Strategy — Do NOT fully expand polynomials early!
When simplifying your answer, look for common algebraic factors like \(n\), \((n + 1)\), or numerical fractions like \(\frac{1}{6}\) immediately. Expanding everything into a messy cubic or quartic polynomial and then trying to factorise it is a recipe for errors.
Worked Example:
Show that \(\sum_{r=1}^{n} r(r + 2) = \frac{1}{6}n(n + 1)(2n + 7)\).
Step 1: Expand the bracket inside the sum:
\(\sum_{r=1}^{n} (r^2 + 2r) = \sum_{r=1}^{n} r^2 + 2\sum_{r=1}^{n} r\)
Step 2: Substitute the standard formulae:
\(= \frac{1}{6}n(n + 1)(2n + 1) + 2 \cdot \left(\frac{1}{2}n(n + 1)\right)\)
\(= \frac{1}{6}n(n + 1)(2n + 1) + n(n + 1)\)
Step 3: Take out the common factor \(\frac{1}{6}n(n + 1)\):
\(= \frac{1}{6}n(n + 1)\left[ (2n + 1) + 6 \right]\)
\(= \frac{1}{6}n(n + 1)(2n + 7)\) (Fully factorised and neat!)
2.3 Sums Between Limits (\(r = k\) to \(n\))
If a question asks for the sum starting from a value other than \(r = 1\), use the subtraction rule:
\(\sum_{r=k}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{k-1} f(r)\)
Common Mistake to Avoid: To find the sum from \(r = 11\) to \(n = 30\), subtract the sum up to \(r = 10\), not \(r = 11\). You want to keep the 11th term in your final answer!
2.4 The Method of Differences (Telescoping Series)
When terms can be rewritten as the difference between two consecutive terms of another sequence, say \(f(r) - f(r+1)\), almost all terms cancel out in a cascading chain (like a collapsible telescope).
How to apply the Method of Differences:
1. Use partial fractions to split the algebraic fraction into the form \(\frac{A}{r + a} - \frac{B}{r + b}\).
2. Write out the first 2 or 3 terms explicitly.
3. Write out the last 2 or 3 terms explicitly.
4. Clearly show which terms cancel.
5. Combine the surviving terms into a single simplified expression.
Worked Example:
Find \(\sum_{r=1}^{n} \frac{1}{r(r + 1)}\).
Step 1: Split using partial fractions: \(\frac{1}{r(r + 1)} = \frac{1}{r} - \frac{1}{r + 1}\).
Step 2: List the terms systematically:
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:
Every \(-\frac{1}{k}\) cancels with the subsequent \(+\frac{1}{k}\).
Surviving terms: \(1 - \frac{1}{n + 1} = \frac{(n + 1) - 1}{n + 1} = \frac{n}{n + 1}\).
Key Takeaway for Series: Look at the structure first. If it is polynomial in \(r\), use standard results and factorise common terms early. If it is a rational function, think partial fractions and the method of differences.
3. Maclaurin and Taylor Series
In calculus, polynomial functions like \(y = ax^2 + bx + c\) are simple to differentiate, integrate, and evaluate. Functions like \(\sin x\), \(e^x\), or \(\ln(1+x)\) are much harder to compute by hand. Maclaurin and Taylor series allow us to approximate virtually any smooth curve as an infinite polynomial!
3.1 Maclaurin Series
The Maclaurin expansion represents a function \(f(x)\) around the point \(x = 0\):
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots + \frac{f^{(r)}(0)}{r!}x^r + \dots\)
where \(f^{(r)}(0)\) denotes the \(r\)-th derivative of \(f(x)\) evaluated at \(x = 0\), and \(r! = r \times (r-1) \times \dots \times 1\).
Standard Maclaurin Series (Provided in CCEA Formulae Sheet)
You should be intimately familiar with the following series and their ranges of validity:
\(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^r}{r!} + \dots \quad \text{for all } x\)
\(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots + \frac{(-1)^r x^{2r+1}}{(2r+1)!} + \dots \quad \text{for all } x\)
\(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots + \frac{(-1)^r x^{2r}}{(2r)!} + \dots \quad \text{for all } x\)
\(\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots + \frac{(-1)^{r+1} x^r}{r} + \dots \quad \text{for } -1 < x \le 1\)
\((1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \quad \text{for } |x| < 1\)
Notice the symmetry! \(\sin x\) is an odd function, so it contains only odd powers of \(x\). \(\cos x\) is an even function, so it contains only even powers of \(x\). Both alternate signs (\(+, -, +, -\dots\)).
Building New Series: Two Methods
Method A: Direct Differentiation
Used when dealing with a brand new or unfamiliar function (e.g., \(f(x) = \sec x\) or \(f(x) = \ln(\cos x)\)).
Find \(f(0), f'(0), f''(0), f'''(0)\), then substitute into the general Maclaurin formula.
Method B: Substitution & Algebraic Combination
Used when the function is built from standard components (e.g., \(e^{-2x}\), \(\cos(3x)\), or \(e^x \sin x\)). This is much quicker!
Worked Example:
Find the Maclaurin series for \(f(x) = e^x \cos x\) up to and including the term in \(x^3\).
Step 1: Write out the standard expansions up to the \(x^3\) power:
\(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots\)
\(\cos x = 1 - \frac{x^2}{2} + \dots\) (Note: the next term is \(x^4\), so we stop here)
Step 2: Multiply the polynomials together, ignoring powers greater than \(x^3\):
\(e^x \cos x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) \left(1 - \frac{x^2}{2}\right)\)
\(= 1\left(1 - \frac{x^2}{2}\right) + x\left(1 - \frac{x^2}{2}\right) + \frac{x^2}{2}(1) + \frac{x^3}{6}(1) + \dots\)
\(= 1 - \frac{x^2}{2} + x - \frac{x^3}{2} + \frac{x^2}{2} + \frac{x^3}{6} + \dots\)
Step 3: Collect like powers of \(x\):
Constant term: \(1\)
\(x\) term: \(x\)
\(x^2\) term: \(-\frac{1}{2}x^2 + \frac{1}{2}x^2 = 0\)
\(x^3\) term: \(-\frac{1}{2}x^3 + \frac{1}{6}x^3 = -\frac{1}{3}x^3\)
Result: \(e^x \cos x = 1 + x - \frac{1}{3}x^3 + \dots\)
3.2 Taylor Series
What if we want to approximate a function near a point other than \(x = 0\), say near \(x = a\)? That is where the Taylor series comes in.
The Taylor series of \(f(x)\) expanded about the point \(x = a\) in ascending powers of \((x - a)\) is:
\(f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \dots + \frac{f^{(r)}(a)}{r!}(x - a)^r + \dots\)
Alternatively, letting \(x = a + h\) (so that \(x - a = h\)), the series in powers of \(h\) is:
\(f(a + h) = f(a) + f'(a)h + \frac{f''(a)}{2!}h^2 + \frac{f'''(a)}{3!}h^3 + \dots\)
Note: The Maclaurin series is simply a special case of the Taylor series where \(a = 0\).
Key Takeaway for Maclaurin & Taylor Series: Always check if you can use known standard series first to save time. When expanding about \(x = a\), evaluate all derivatives strictly at the center point \(x = a\), not at \(x = 0\).
Quick Reference Summary Box
1. Roots of Polynomials:
Sign pattern: \((-), (+), (-), (+)\dots\)
\(\sum \alpha = -\frac{b}{a}\), \(\sum \alpha\beta = \frac{c}{a}\), \(\alpha\beta\gamma = -\frac{d}{a}\)
Transformations: Substitute \(x = g^{-1}(w)\) into the equation.
2. Summation of Series:
\(\sum r = \frac{1}{2}n(n+1)\)
\(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\)
\(\sum r^3 = \frac{1}{4}n^2(n+1)^2\)
Factorise common terms early; use partial fractions for telescoping cancellations.
3. Maclaurin & Taylor:
Maclaurin (about \(0\)): \(f(x) = \sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} x^r\)
Taylor (about \(a\)): \(f(x) = \sum_{r=0}^{\infty} \frac{f^{(r)}(a)}{r!} (x - a)^r\)
Remember: \(0! = 1\), and always mind the range of validity for series like \(\ln(1+x)\).