Notes · Further MathematicsUK · A-Levels

Further algebra and functions

This topic gathers three powerful algebraic techniques: the relationships between the roots of a polynomial and its coefficients, the summation of finite series using standard results and the method of differences, and the Maclaurin series expansion of a function. Together they let you extract information about roots without solving, sum unfamiliar series exactly, and approximate functions by polynomials with a stated range of validity.

5 sections~13 min reading time3 competenciesLevel Standard 1 · Advanced 4

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Exam profile

AO1 · Apply standard series results and root-coefficient relationships accuratelyAO2 · Derive Maclaurin series and telescoping sums, justifying each stepAO3 · Combine techniques to sum unfamiliar series and to approximate functions

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basic level

AS Further Mathematics covers the roots of quadratics and cubics, the standard summation results and simple series.

higher level

The full A-Level adds quartics, transforming the roots of a polynomial, the method of differences, and Maclaurin series with their ranges of validity.

Depth

Text

Contents · 5 sections

Further algebra and functions

Roots of polynomials and their coefficients#

AdvancedAQA 7367 §3.4DfE Further Mathematics

A cubic and its three real roots

Fig. 1The cubic

(x-1)(x-2)(x-3)

has roots

1, 2, 3

whose sum

6

equals

-b/a

from the coefficients.

The roots of a polynomial are linked to its coefficients by simple symmetric relationships, so a great deal can be found out about the roots without ever solving the equation. For a quadratic

ax^2 + bx + c = 0

with roots

\alpha, \beta

, comparing with

a(x-\alpha)(x-\beta)

gives

\alpha + \beta = -\dfrac{b}{a}

and

\alpha\beta = \dfrac{c}{a}

. These are the first instances of the general pattern.

For a cubic

ax^3 + bx^2 + cx + d = 0

with roots

\alpha, \beta, \gamma

, the relationships are

\sum\alpha = -\dfrac{b}{a}

\sum\alpha\beta = \dfrac{c}{a}

and

\alpha\beta\gamma = -\dfrac{d}{a}

; and for a quartic there is one further relation, with the product of all four roots equal to

+\dfrac{e}{a}

. The signs alternate, matching the coefficients of the expanded product

a\prod(x - \alpha_i)

. These are known as the relationships between roots and coefficients (Vieta's formulae).

The examinable skill is to compute symmetric functions of the roots from these building blocks. For example

\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta

, and

\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\sum\beta\gamma}{\alpha\beta\gamma}

. Any expression that is symmetric in the roots can be rewritten in terms of the elementary symmetric functions

\sum\alpha

\sum\alpha\beta

and

\alpha\beta\gamma

, and hence in terms of the coefficients.

This gives a route to results that would be impossible by direct solution — a cubic with irrational or complex roots still has a rational sum of squares of its roots, computed cleanly from the coefficients. The technique also underpins the next section, where a polynomial whose roots are a transformation of the original roots is constructed by working with these symmetric functions or by direct substitution.

\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}

Cubic: roots and coefficients

For $ax^3+bx^2+cx+d=0$ ; the signs alternate.

\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)

A symmetric function

Rewrite any symmetric function in terms of the elementary ones.

Worked example

A symmetric function of the roots

The roots of $x^3 + 2x^2 - 5x + 1 = 0$ are $\alpha, \beta, \gamma$ . Find $\alpha^2 + \beta^2 + \gamma^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ .

Read off the symmetric functions
Here $a=1$ , so $\sum\alpha = -2$ , $\sum\alpha\beta = -5$ and $\alpha\beta\gamma = -1$ .
Sum of squares
$\alpha^2+\beta^2+\gamma^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = (-2)^2 - 2(-5) = 4 + 10 = 14$ .
$\alpha^2+\beta^2+\gamma^2 = 4 + 10 = 14$
Sum of reciprocals
$\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma} = \dfrac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \dfrac{-5}{-1} = 5$ .

Result: $\alpha^2+\beta^2+\gamma^2 = 14$ and $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma} = 5$ .

Typical mistakes

Getting the signs of the root-coefficient relationships wrong; remember the alternating $-b/a, +c/a, -d/a$ pattern.
Trying to evaluate a symmetric function term by term instead of rewriting it in terms of the elementary symmetric functions.

Active revision

The cubic $2x^3 - 3x^2 + 4x - 1 = 0$ has roots $\alpha, \beta, \gamma$ . Find $\alpha + \beta + \gamma$ , $\alpha\beta + \beta\gamma + \gamma\alpha$ and $\alpha^2 + \beta^2 + \gamma^2$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Transforming the roots of a polynomial#

AdvancedAQA 7367 §3.4DfE Further Mathematics

Frequently a problem asks for a polynomial whose roots are a simple transformation of the roots of a given polynomial — for instance

2\alpha, 2\beta, 2\gamma

, or

\alpha + 1

, or

\dfrac{1}{\alpha}

, or

\alpha^2

. There are two standard methods, and the substitution method is usually the quicker and less error-prone.

In the substitution method, if the new roots are

y = g(x)

where

x

is a root of the original polynomial, you invert to get

x = g^{-1}(y)

and substitute into the original equation. For roots doubled,

y = 2x

x = y/2

, and replacing every

x

y/2

(then clearing fractions) gives the new polynomial. For roots increased by

1

y = x + 1

x = y - 1

; for reciprocal roots,

y = 1/x

x = 1/y

, and clearing the denominator produces a polynomial whose coefficients are those of the original in reverse order.

The alternative is to build the new polynomial from its own symmetric functions, computed from the old ones. If the new roots are

\alpha + 1

, etc., then their sum is

\sum(\alpha+1) = \sum\alpha + 3

(for a cubic), their pairwise-product sum is

\sum(\alpha+1)(\beta+1)

, and so on — each expanded and evaluated using the known elementary symmetric functions. This is more laborious but is a good check and is sometimes demanded explicitly.

A recurring special case is the reciprocal transformation: the equation whose roots are the reciprocals of the roots of

ax^n + \dots + k = 0

is obtained by reversing the coefficients,

kx^n + \dots + a = 0

. Recognising such structural facts saves time. Whichever method is used, always check the transformed polynomial by verifying that its constant and leading coefficients are consistent with the transformed product and sum of the roots.

\text{New roots } y = g(x): \ \text{substitute } x = g^{-1}(y) \text{ into the original equation}

The substitution method

Invert the transformation and replace the variable, then clear fractions.

Worked example

Doubling the roots

The cubic $x^3 - 2x^2 + 3x - 5 = 0$ has roots $\alpha, \beta, \gamma$ . Find a cubic with roots $2\alpha, 2\beta, 2\gamma$ .

Set up the substitution
Let $y = 2x$ , so $x = \dfrac{y}{2}$ . Substitute into the original equation.
Substitute
$\left(\dfrac{y}{2}\right)^3 - 2\left(\dfrac{y}{2}\right)^2 + 3\left(\dfrac{y}{2}\right) - 5 = 0$ , i.e. $\dfrac{y^3}{8} - \dfrac{y^2}{2} + \dfrac{3y}{2} - 5 = 0$ .
Clear fractions
Multiply by $8$ : $y^3 - 4y^2 + 12y - 40 = 0$ .
$y^3 - 4y^2 + 12y - 40 = 0$
Check
New sum of roots $= 4 = 2\times 2 = 2\sum\alpha$ ; new product $= 40 = 2^3\times 5 = 8\,\alpha\beta\gamma$ . Both consistent.

Result: $y^3 - 4y^2 + 12y - 40 = 0$ .

Typical mistakes

Substituting $y = g(x)$ directly instead of $x = g^{-1}(y)$ — you must replace the original variable.
Forgetting to clear fractions after a reciprocal or scaling substitution, leaving a non-polynomial expression.

Active revision

The equation $x^3 - 4x^2 + x + 6 = 0$ has roots $\alpha, \beta, \gamma$ . Find a cubic equation with integer coefficients whose roots are $2\alpha, 2\beta, 2\gamma$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Summation of series: standard results#

StandardAQA 7367 §3.4DfE Further Mathematics

The three standard results, proved by induction in the Proof topic, are the workhorses of series summation:

\sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1)

\sum_{r=1}^{n} r^2 = \tfrac{1}{6}n(n+1)(2n+1)

and

\sum_{r=1}^{n} r^3 = \tfrac{1}{4}n^2(n+1)^2

. Committing these to memory (or being able to reproduce them quickly) is essential, since almost every summation question reduces to them.

Summation is linear:

\sum (a\,f(r) + b\,g(r)) = a\sum f(r) + b\sum g(r)

, and

\sum_{r=1}^{n} c = cn

for a constant

c

. So any polynomial in

r

can be summed by splitting it into the standard pieces. For instance

\sum_{r=1}^{n} (r^2 + 3r) = \sum r^2 + 3\sum r

, each evaluated by its standard result and then combined and factorised.

A frequent extra step is summing between limits other than

1

and

n

. To evaluate

\sum_{r=m}^{n} f(r)

, use

\sum_{r=m}^{n} = \sum_{r=1}^{n} - \sum_{r=1}^{m-1}

: compute the sum up to

n

and subtract the sum up to

m-1

. This avoids re-deriving anything and handles, for example, the sum of the squares from

11

30

in two standard evaluations.

The final answer is expected in a fully factorised form, and reaching it cleanly is a skill in itself: after adding the standard results over a common denominator, take out the largest common factor (typically

n

(n+1)

\tfrac{1}{6}

) before expanding any remaining bracket. A tidy factorised answer both earns the presentation marks and makes the result easy to check at small

n

\sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1), \quad \sum_{r=1}^{n} r^2 = \tfrac{1}{6}n(n+1)(2n+1), \quad \sum_{r=1}^{n} r^3 = \tfrac{1}{4}n^2(n+1)^2

The three standard results

Proved by induction; used to sum any polynomial in $r$ .

\sum_{r=m}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{m-1} f(r)

Changing the lower limit

Subtract the sum up to $m-1$ from the sum up to $n$ .

Worked example

Summing a polynomial in r

Show that $\displaystyle\sum_{r=1}^{n} r(r+2) = \tfrac{1}{6}n(n+1)(2n+7)$ .

Expand and split
$\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r) = \sum r^2 + 2\sum r$ .
Insert the standard results
$= \tfrac{1}{6}n(n+1)(2n+1) + 2\cdot\tfrac{1}{2}n(n+1) = \tfrac{1}{6}n(n+1)(2n+1) + n(n+1)$ .
Take out the common factor
$= \tfrac{1}{6}n(n+1)\big[(2n+1) + 6\big] = \tfrac{1}{6}n(n+1)(2n+7)$ .
$\sum_{r=1}^{n} r(r+2) = \tfrac{1}{6}n(n+1)(2n+7)$
Check
At $n=1$ : LHS $=1\cdot 3 = 3$ ; RHS $=\tfrac{1}{6}(1)(2)(9) = 3$ . Agrees.

Result: $\displaystyle\sum_{r=1}^{n} r(r+2) = \tfrac{1}{6}n(n+1)(2n+7)$ .

Typical mistakes

Misremembering a standard result (a wrong $\tfrac{1}{6}$ or $2n+1$ ) — check it against $n=1$ or $n=2$ .
Forgetting that $\sum_{r=1}^{n} c = cn$ , treating a constant term as if it summed to $c$ .

Active revision

Find $\displaystyle\sum_{r=1}^{n} (2r-1)^2$ in fully factorised form, and hence evaluate $\sum_{r=1}^{20}(2r-1)^2$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

The method of differences#

AdvancedAQA 7367 §3.4DfE Further Mathematics

The method of differences sums a series whose general term can be written as a difference of consecutive values of some function: if

f(r) = g(r) - g(r+1)

, then the sum telescopes. Writing out the terms, almost everything cancels:

\sum_{r=1}^{n} \big(g(r) - g(r+1)\big) = g(1) - g(n+1)

, because each

-g(r+1)

is cancelled by the

+g(r+1)

of the next term.

The central skill is to express the summand in this difference form, and for the rational summands that appear at A-Level the tool is partial fractions. For example

\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}

, which is exactly

g(r) - g(r+1)

with

g(r) = \dfrac{1}{r}

; summing telescopes to

1 - \dfrac{1}{n+1}

. Recognising that a partial-fraction split produces a telescoping difference is the insight the method rests on.

Care is needed when the difference spans more than one step, for instance

f(r) = g(r) - g(r+2)

, where the cancellation leaves the first two and the last two terms:

\sum_{r=1}^{n} \big(g(r)-g(r+2)\big) = g(1) + g(2) - g(n+1) - g(n+2)

. Writing out the first few and last few terms explicitly, and lining up the cancellation, prevents the common error of dropping or keeping the wrong boundary terms.

The method of differences also delivers the sum to infinity when the series converges: letting

n \to \infty

, any

g(n+1)

that tends to

0

simply drops out, so

\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)} = 1

. This connects finite summation to the idea of a convergent infinite series and previews the limiting arguments used in Further Calculus. It is also a reminder always to state the boundary terms that survive, since they carry the whole value of the sum.

\sum_{r=1}^{n} \big(g(r) - g(r+1)\big) = g(1) - g(n+1)

Telescoping sum

Consecutive terms cancel, leaving only the first and last.

Worked example

A telescoping sum

Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$ and deduce the sum to infinity.

Partial fractions
$\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ , which is $g(r) - g(r+1)$ with $g(r) = \dfrac{1}{r}$ .
Write out and cancel
$\left(\tfrac{1}{1}-\tfrac{1}{2}\right) + \left(\tfrac{1}{2}-\tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n}-\tfrac{1}{n+1}\right)$ ; all the interior terms cancel.
Surviving terms
$\sum_{r=1}^{n} \dfrac{1}{r(r+1)} = 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$ .
$\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}$
Sum to infinity
As $n\to\infty$ , $\dfrac{1}{n+1}\to 0$ , so $\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)} = 1$ .

Result: $\sum_{r=1}^{n} \dfrac{1}{r(r+1)} = \dfrac{n}{n+1}$ ; the sum to infinity is $1$ .

Typical mistakes

Cancelling terms carelessly and keeping or dropping the wrong boundary terms, especially when the difference spans two or more steps.
Forgetting to write the surviving terms as the final answer, or mis-stating the sum to infinity by not checking that the tail tends to zero.

Active revision

Using partial fractions and the method of differences, find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$ and hence state $\displaystyle\sum_{r=1}^{\infty} \frac{1}{r(r+1)}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Maclaurin series and approximation#

AdvancedAQA 7367 §3.4DfE Further Mathematics

Maclaurin approximations to e^x

Fig. 2The degree-1, 2 and 3 Maclaurin polynomials hug

\mathrm{e}^x

near

x=0

and each captures more of the curve.

A Maclaurin series expresses a function as an infinite polynomial in

x

, built from the function's derivatives at

0

f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \cdots

. Provided the function and all its derivatives exist at

0

and the series converges, this matches the function near

x = 0

, and truncating after a few terms gives a polynomial approximation whose accuracy is best close to the origin and degrades further away.

Four standard expansions must be known:

\mathrm{e}^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots

(valid for all

x

);

\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots

and

\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots

(valid for all

x

); and

\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots

(valid only for

-1 < x \leq 1

). The binomial series

(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \cdots

(valid for

|x| < 1

when

n

is not a non-negative integer) completes the set.

Series for related functions are usually obtained not by differentiating repeatedly but by combining or substituting into the standard results. For example the series for

\mathrm{e}^{2x}

comes from replacing

x

2x

in the

\mathrm{e}^x

series, and the series for

\mathrm{e}^x\sin x

from multiplying two known series and collecting terms up to the required power. This is faster and less error-prone than computing high-order derivatives.

The range of validity is not an afterthought: a Maclaurin series equals its function only where the series converges. For

\mathrm{e}^x

\sin x

and

\cos x

this is everywhere, but for

\ln(1+x)

and the binomial series it is a bounded interval, and using the series outside that interval gives nonsense. The figure of

\ln(1+x)

against its cubic approximation shows the polynomial hugging the curve near

0

and peeling away beyond

x = 1

, a vivid reminder to quote the interval of validity whenever a non-entire series is used.

f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Maclaurin series

The coefficients are the derivatives of $f$ at $0$ divided by factorials.

\mathrm{e}^x = \sum_{r=0}^{\infty} \frac{x^r}{r!}, \qquad \ln(1+x) = \sum_{r=1}^{\infty} \frac{(-1)^{r+1}x^r}{r} \ \ (-1 < x \leq 1)

Two standard series

$\mathrm{e}^x$ converges for all $x$ ; the logarithm series only on $(-1,1]$ .

Range of validity of the ln(1+x) series

Fig. 3The cubic approximation matches

\ln(1+x)

for

|x|<1

but diverges beyond

x=1

, the edge of the interval of validity.

Worked example

A series by substitution

Find the Maclaurin series of $\ln(1+2x)$ up to the term in $x^3$ , state the range of validity and estimate $\ln 1.2$ .

Substitute into the standard series
Replace $x$ by $2x$ in $\ln(1+x) = x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \cdots$ : $\ln(1+2x) = 2x - \tfrac{(2x)^2}{2} + \tfrac{(2x)^3}{3} - \cdots$ .
Simplify
$= 2x - 2x^2 + \tfrac{8}{3}x^3 - \cdots$ .
$\ln(1+2x) = 2x - 2x^2 + \tfrac{8}{3}x^3 - \cdots$
Range of validity
Valid when $-1 < 2x \leq 1$ , i.e. $-\tfrac12 < x \leq \tfrac12$ .
Estimate ln 1.2
$\ln 1.2 = \ln(1 + 2(0.1))$ , so put $x = 0.1$ : $2(0.1) - 2(0.1)^2 + \tfrac{8}{3}(0.1)^3 = 0.2 - 0.02 + 0.002667 \approx 0.18267$ (true value $0.18232$ ).

Result: $\ln(1+2x) = 2x - 2x^2 + \tfrac{8}{3}x^3 - \cdots$ for $-\tfrac12 < x \leq \tfrac12$ ; $\ln 1.2 \approx 0.1827$ .

Typical mistakes

Using the $\ln(1+x)$ or binomial series outside its interval of convergence, where it no longer represents the function.
Dropping factorials in the denominators, or making a sign slip in the alternating $\sin/\cos/\ln$ series.

Active revision

Find the Maclaurin series of $\ln(1+2x)$ up to and including the term in $x^3$ , state its range of validity, and use it to estimate $\ln 1.2$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Further algebra and functions

Key points

A symmetric function of the roots

Key points

Doubling the roots

Key points

Summing a polynomial in r

Key points

A telescoping sum

Key points

A series by substitution

Sources