Further Pure — Algebra and series: A-level Further Maths OCR revision notes

4.05

Roots of polynomials

Roots and coefficients 4.05

Key results

Quadratic $a x^{2} + b x + c = 0$ with roots $α, β$ : $α + β = - \frac{b}{a}$ , $α β = \frac{c}{a}$ .
Cubic $a x^{3} + b x^{2} + c x + d = 0$ with roots $α, β, γ$ : $\sum α = - \frac{b}{a}$ , $\sum α β = \frac{c}{a}$ , $α β γ = - \frac{d}{a}$ .
Quartic $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ : $\sum α = - \frac{b}{a}$ , $\sum α β = \frac{c}{a}$ , $\sum α β γ = - \frac{d}{a}$ , $α β γ δ = \frac{e}{a}$ .

Notes

These come from expanding $a (x - α) (x - β) (x - γ)$ and comparing coefficients with the polynomial; the signs alternate, starting with minus.
$\sum α β$ means the sum over every pair of different roots: three terms for a cubic, six for a quartic.
The results hold for complex roots too. In a real cubic with roots $p \pm q i$ and $r$ , the sum of the roots is $2 p + r$ , which is real, as it must be.

Symmetric functions of the roots 4.05

Key results

$α^{2} + β^{2} + γ^{2} = (\sum α)^{2} - 2 \sum α β$ .
$\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{\sum α β}{α β γ}$ .
For a quadratic: $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)$ .

Method

Write the required expression in terms of $\sum α$ , $\sum α β$ and $α β γ$ only, by expanding a power of $\sum α$ and subtracting the unwanted terms; then substitute.

In practiceThe roots of $2 x^{3} - 3 x^{2} + 4 x - 1 = 0$ are $α, β, γ$ . Find $\sum α^{2}$ and $\sum \frac{1}{α}$ .

$\sum α = \frac{3}{2}, \sum α β = \frac{4}{2} = 2, α β γ = \frac{1}{2}$ $- \frac{b}{a}$ , $\frac{c}{a}$ , $- \frac{d}{a}$
$\sum α^{2} = (\sum α)^{2} - 2 \sum α β = \frac{9}{4} - 4 = - \frac{7}{4}$ negative, so the roots are not all real
$\sum \frac{1}{α} = \frac{\sum α β}{α β γ} = \frac{2}{1/2} = 4$

Notes

A symmetric function is unchanged by swapping any two roots. Every symmetric polynomial in the roots can be written in terms of the basic sums above.
If $α^{2} + β^{2} + γ^{2}$ comes out negative, the roots cannot all be real, since squares of reals are non-negative. This is a quick way to show an equation has non-real roots.

Equations with transformed roots 4.05

Method

To find an equation whose roots are $g (α), g (β), g (γ)$ : let $w = g (x)$ , rearrange to make $x$ the subject, and substitute into the original equation.
Roots $α + k$ : substitute $x = w - k$ . Roots $k α$ : substitute $x = \frac{w}{k}$ , then clear fractions. Roots $\frac{1}{α}$ : substitute $x = \frac{1}{w}$ , then multiply through by $w^{n}$ .
Roots $α^{2}$ : substitute $x = w$ , collect the terms containing $w$ on one side, and square both sides.

In practiceThe roots of $x^{3} - 2 x^{2} + 5 x - 3 = 0$ are $α, β, γ$ . Find a cubic with roots $2 α + 1$ , $2 β + 1$ , $2 γ + 1$ .

$w = 2 x + 1 ⟹ x = \frac{w - 1}{2}$ make $x$ the subject
$(\frac{w - 1}{2})^{3} - 2 (\frac{w - 1}{2})^{2} + 5 \cdot \frac{w - 1}{2} - 3 = 0$ substitute into the original equation
$(w - 1)^{3} - 4 (w - 1)^{2} + 20 (w - 1) - 24 = 0$ multiply through by 8
$w^{3} - 7 w^{2} + 31 w - 49 = 0$ check: the new roots sum to $2 \sum α + 3 = 7$

Notes

Substitution is usually faster and safer than finding the new sums of roots one by one, especially for a cubic or quartic.
The shift $x = w - 1$ is a translation of the graph one unit right, which is why every root moves up by 1.

Worked examples

Worked example

The roots of:

2 x^{3} - 3 x^{2} + 4 x - 5 = 0

are $α, β, γ$ .

Find $α^{2} + β^{2} + γ^{2}$ , and explain what it shows about the roots.

Show worked solution

\sum α = \frac{3}{2}

and:

\sum α β = \frac{4}{2} = 2

So:

\sum α^{2} = (\frac{3}{2})^{2} - 2 (2) = \frac{9}{4} - 4 = - \frac{7}{4}

A sum of squares of real numbers cannot be negative, so the roots are not all real: the cubic has one real root and two non-real conjugate roots.

Worked example

The roots of:

x^{3} + 2 x^{2} - 3 x + 1 = 0

are $α, β, γ$ .

Find a cubic equation with integer coefficients whose roots are $2 α, 2 β, 2 γ$ .

Show worked solution

Let $w = 2 x$ , so:

x = \frac{w}{2}

Then:

\frac{w ^{3}}{8} + 2 \cdot \frac{w ^{2}}{4} - 3 \cdot \frac{w}{2} + 1 = 0

Multiplying by 8:

w^{3} + 4 w^{2} - 12 w + 8 = 0

4.06

The method of differences

The method of differences 4.06

Key results

If $u_{r} = f (r) - f (r + 1)$ , then $r = 1 \sum n u_{r} = f (1) - f (n + 1)$ .
$r = 1 \sum n \frac{1}{r ( r + 1 )} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ .

Method

Express the general term as a difference, often by partial fractions.
Write out the first three rows and the last two, one under another.
Cross out the terms that cancel, and add what survives at the start and the end.

In practiceFind $r = 1 \sum n \frac{2}{( r + 1 ) ( r + 2 )}$ .

$\frac{2}{( r + 1 ) ( r + 2 )} = \frac{2}{r + 1} - \frac{2}{r + 2}$ partial fractions
$r = 1 : \frac{2}{2} - \frac{2}{3}, r = 2 : \frac{2}{3} - \frac{2}{4}, \dots, r = n : \frac{2}{n + 1} - \frac{2}{n + 2}$ write out the rows
$r = 1 \sum n \frac{2}{( r + 1 ) ( r + 2 )} = \frac{2}{2} - \frac{2}{n + 2} = \frac{n}{n + 2}$ the first and last survive; check $n = 1$ : $\frac{1}{3}$

Notes

Writing out rows is not optional working: it is how you see which terms survive, and it is what earns the method marks.
If a sum to infinity is asked for, let $n \to \infty$ in the result: here $r = 1 \sum \infty \frac{1}{r ( r + 1 )} = 1$ .

Differences with a gap 4.06

Key results

If $u_{r} = f (r) - f (r + 2)$ , two terms survive at each end: $r = 1 \sum n u_{r} = f (1) + f (2) - f (n + 1) - f (n + 2)$ .

Method

Example: $\frac{1}{r ( r + 2 )} = \frac{1}{2} (\frac{1}{r} - \frac{1}{r + 2})$ . Each $\frac{1}{r + 2}$ cancels with the $\frac{1}{r}$ two rows later, so $\frac{1}{1}$ and $\frac{1}{2}$ survive at the start and $\frac{1}{n + 1}$ , $\frac{1}{n + 2}$ at the end.

In practiceFind $r = 1 \sum n \frac{1}{r ( r + 2 )}$ .

$\frac{1}{r ( r + 2 )} = \frac{1}{2} (\frac{1}{r} - \frac{1}{r + 2})$ a gap of 2
$survivors: \frac{1}{1}, \frac{1}{2} (start); - \frac{1}{n + 1}, - \frac{1}{n + 2} (end)$ two terms at each end
$\sum = \frac{1}{2} (\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{3}{4} - \frac{2 n + 3}{2 ( n + 1 ) ( n + 2 )}$ check $n = 1$ : $\frac{3}{4} - \frac{5}{12} = \frac{1}{3}$

Notes

Count the gap from the partial fractions: a difference $f (r) - f (r + k)$ leaves $k$ terms at each end.
Partial fractions with three terms, such as $\frac{1}{r} - \frac{2}{r + 1} + \frac{1}{r + 2}$ , also telescope; list enough rows to see the pattern before cancelling.

Worked example

Show that:

r = 1 \sum n \frac{1}{r ( r + 2 )} = \frac{3}{4} - \frac{2 n + 3}{2 ( n + 1 ) ( n + 2 )}

Show worked solution

\frac{1}{r ( r + 2 )} = \frac{1}{2} (\frac{1}{r} - \frac{1}{r + 2})

Listing rows, every $\frac{1}{r + 2}$ cancels with a $\frac{1}{r}$ two rows below, leaving:

\frac{1}{2} (1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{3}{4} - \frac{1}{2} \cdot \frac{( n + 2 ) + ( n + 1 )}{( n + 1 ) ( n + 2 )} = \frac{3}{4} - \frac{2 n + 3}{2 ( n + 1 ) ( n + 2 )}

4.06

Standard sums

The standard sums 4.06

Key results

$r = 1 \sum n 1 = n$ , $r = 1 \sum n r = \frac{1}{2} n (n + 1)$ .
$r = 1 \sum n r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ .
$r = 1 \sum n r^{3} = \frac{1}{4} n^{2} (n + 1)^{2}$ .

Notes

Sums are linear: $\sum (a u_{r} + b v_{r}) = a \sum u_{r} + b \sum v_{r}$ . But $\sum u_{r} v_{r} \neq = \sum u_{r} \sum v_{r}$ ; expand the product first.
The picture on the sheet is a proof of the first result: two copies of $1 + 2 + \dots + n$ fill an $n \times (n + 1)$ rectangle.
The formula for $\sum r^{3}$ is the square of the formula for $\sum r$ , a coincidence worth remembering.
Deriving $\sum r^{2}$ by differences: $(r + 1)^{3} - r^{3} = 3 r^{2} + 3 r + 1$ . Summing from $r = 1$ to $n$ , the left side telescopes to $(n + 1)^{3} - 1$ , so $3 \sum r^{2} + \frac{3}{2} n (n + 1) + n = (n + 1)^{3} - 1$ . Rearranging, $3 \sum r^{2} = n^{3} + \frac{3}{2} n^{2} + \frac{1}{2} n = \frac{1}{2} n (n + 1) (2 n + 1)$ , so $\sum r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ .

Using the standard sums 4.06

Method

Expand the general term into powers of $r$ , split the sum, substitute the standard results, and factorise. Take out common factors such as $n (n + 1)$ early; it keeps the algebra small.
For a sum that does not start at 1: $r = k \sum n u_{r} = r = 1 \sum n u_{r} - r = 1 \sum k - 1 u_{r}$ .
For a sum to $2 n$ , replace $n$ by $2 n$ throughout the standard formula.

In practiceShow that $r = 1 \sum n r (r + 1) = \frac{1}{3} n (n + 1) (n + 2)$ .

$\sum r (r + 1) = \sum r^{2} + \sum r$ expand, then split the sum
$= \frac{1}{6} n (n + 1) (2 n + 1) + \frac{1}{2} n (n + 1)$ standard results
$= \frac{1}{6} n (n + 1) [(2 n + 1) + 3]$ take out $\frac{1}{6} n (n + 1)$ early
$= \frac{1}{6} n (n + 1) (2 n + 4) = \frac{1}{3} n (n + 1) (n + 2)$ check $n = 2$ : $2 + 6 = 8 = \frac{1}{3} \cdot 2 \cdot 3 \cdot 4$

Notes

Subtract up to $k - 1$ , not $k$ : the term $r = k$ belongs to the sum.
Check any final formula with $n = 1$ and $n = 2$ by adding the terms directly.

Worked example

Show that:

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

and hence evaluate:

r = 11 \sum 20 r (r + 1)

Show worked solution

\sum r (r + 1) = \sum r^{2} + \sum r = \frac{1}{6} n (n + 1) (2 n + 1) + \frac{1}{2} n (n + 1) = \frac{1}{6} n (n + 1) (2 n + 1 + 3) = \frac{1}{3} n (n + 1) (n + 2)

Then:

r = 11 \sum 20 = \frac{1}{3} (20) (21) (22) - \frac{1}{3} (10) (11) (12) = 3080 - 440 = 2640

4.01

Proof by induction

The structure of a proof by induction 4.01

Method

Basis: show the statement is true for the first value, usually $n = 1$ .
Assumption: assume it is true for $n = k$ , for some positive integer $k$ .
Inductive step: using the assumption, show it is true for $n = k + 1$ . Write down the target ( $n = k + 1$ version) first, so you know what you are aiming for.
Conclusion: "The statement is true for $n = 1$ , and if it is true for $n = k$ then it is true for $n = k + 1$ . So, by induction, it is true for all positive integers $n$ ."

In practiceProve that $r = 1 \sum n (2 r - 1) = n^{2}$ for every positive integer $n$ .

$n = 1 : 2 (1) - 1 = 1 = 1^{2}$ basis
$Assume r = 1 \sum k (2 r - 1) = k^{2}$ assumption, for some positive integer $k$
$r = 1 \sum k + 1 (2 r - 1) = k^{2} + (2 (k + 1) - 1) = k^{2} + 2 k + 1 = (k + 1)^{2}$ inductive step: add the next term
Conclusion: true for $n = 1$ , and if true for $n = k$ then true for $n = k + 1$ ; so, by induction, true for all positive integers $n$ .

Notes

Both halves are needed. A correct inductive step with no basis proves nothing; a basis with no step proves only one case.
The concluding sentence carries a mark of its own. It must state both the basis and the implication.

Induction for series, divisibility and matrices 4.01

Method

Series: $r = 1 \sum k + 1 u_{r} = r = 1 \sum k u_{r} + u_{k + 1}$ . Replace the first sum by the assumed formula, add the next term, and simplify towards the formula with $n = k + 1$ .
Divisibility: let $f (n)$ be the expression. Show that $f (k + 1) - f (k)$ , or $f (k + 1) - m f (k)$ for a suitable $m$ , is a multiple of the divisor. Then $f (k + 1)$ is a sum of multiples of it.
Matrices: $M^{k + 1} = M^{k} M$ . Replace $M^{k}$ by the assumed form, multiply out, and show each entry has the $n = k + 1$ form.

In practiceProve that $f (n) = 5^{n} + 3$ is divisible by 4 for every positive integer $n$ .

$f (1) = 5 + 3 = 8 = 4 \times 2$ basis
$f (k + 1) - f (k) = 5^{k + 1} - 5^{k} = 4 \cdot 5^{k}$ the difference is a multiple of 4
$f (k + 1) = f (k) + 4 \cdot 5^{k}$ if $4 ∣ f (k)$ , both terms are multiples of 4
So $4 ∣ f (k)$ implies $4 ∣ f (k + 1)$ . With the basis, by induction $4 ∣ 5^{n} + 3$ for all positive integers $n$ .

Notes

In the divisibility step, choose $m$ to cancel the fastest-growing term: for $7^{n} - 3^{n}$ , use $f (k + 1) - 3 f (k) = 7^{k + 1} - 3 \cdot 7^{k} = 4 \cdot 7^{k}$ , which is visibly divisible by 4.
Induction can also prove results about recurrence relations and derivatives, for example a formula for $\frac{d ^{n}}{d x ^{n}} (x e^{x})$ .

Worked examples

Worked example

Prove by induction that $7^{n} - 3^{n}$ is divisible by 4 for every positive integer $n$ .

Show worked solution

Let:

f (n) = 7^{n} - 3^{n}

Basis: $f (1) = 4$ , divisible by 4.

Assume $f (k)$ is divisible by 4.

Then:

f (k + 1) - 3 f (k) = 7^{k + 1} - 3^{k + 1} - 3 \cdot 7^{k} + 3^{k + 1} = 7^{k} (7 - 3) = 4 \cdot 7^{k}

So:

f (k + 1) = 3 f (k) + 4 \cdot 7^{k}

a sum of multiples of 4.

True for $n = 1$ , and true for $n = k$ implies true for $n = k + 1$ , so by induction $7^{n} - 3^{n}$ is divisible by 4 for all positive integers $n$ .

Worked example

Prove by induction that:

(31 - 4 - 1)^{n} = (1 + 2 n n - 4 n 1 - 2 n)

for all positive integers $n$ .

Show worked solution

Basis: $n = 1$ gives:

(31 - 4 - 1)

Assume true for $n = k$ .

Then:

M^{k + 1} = (1 + 2 k k - 4 k 1 - 2 k) (31 - 4 - 1) = (3 + 6 k - 4 k 3 k + 1 - 2 k - 4 - 8 k + 4 k - 4 k - 1 + 2 k) = (3 + 2 k 1 + k - 4 - 4 k - 1 - 2 k)

which is:

(1 + 2 (k + 1) k + 1 - 4 (k + 1) 1 - 2 (k + 1))

True for $n = 1$ , and true for $k$ implies true for $k + 1$ , so true for all positive integers $n$ by induction.