General maths terminology

⇒ ‘implies that’. For example:

2x = 8

⇒ x = 4

⇔ ‘equivalent’. For example:

2x + 10 = 4 ⇔ x + 5 = 2

Proof

Algebraic Proof (proof by deduction)

… If n can be any whole number, then 2n will always be an …?… number because it will give multiples of …?..

… If n can be any whole number, then an expression for an odd number can be 2n + …?…

… Consecutive whole numbers can be represented by n , n+…?.. , n+…?.. etc.

… Consecutive EVEN numbers can be written…?

… Consecutive ODD numbers can be written …?

… Most proof type questions involve three steps:

Expand

Simplify

Factorise, using a common factor that matches the question.

For example:

“Prove algebraically that the difference between the squares of any two consecutive even numbers is always a multiple of 4”:

(2n+2)² – (2n)²

(2n+2)(2n+2) – 4n²

4n² + 4n + 4n + 4 – 4n²

8n + 4

4(2n + 1) which is always a multiple of 4.

… But what if you had to prove that an expression is NOT a multiple of (for example) 4? Let’s say you simplified the algebra down to:

8n – 10

Now factorise using a common factor which matches the question. But first we have to make an adjustment so that the factorisation can work:

8n – 8 – 2

4(2n – 2) – 2

Here 4(2n – 2) is always a multiple of 4, but because we then subtract 2, the result can never be a multiple of 4.

… Prove that n(n+1)(n+2) is always divisible by 6

One of the factors must be a multiple of 3 and another factor must be a multiple of 2 (i.e. even). Therefore, the product must be a multiple of both 2 and 3 and hence is divisible by the LCM of 2 and 3 which is 6.

… Proving that a quadratic such as n² +3n -1 is odd for all integer values of n. The trick here is to sub-in an even number 2n in place of ‘n’ and deduce that the answer is always odd.

Then sub-in an odd number (2n+1) and deduce that the answer is always odd.

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Inequality proofs

… To show that x² + 6x + 18 > 2 – 0.5x for all values of x

We can rearrange this to make:

x² + 6.5x + 16 > 0

If the equation

x² + 6.5x + 16 = 0 has no real solutions (roots), then we know it doesn’t cut through the x axis and in this case will be floating above it as a ‘happy’ quadratic, i.e. always > 0.

This means that the original inequality must be true as well.

–

… Show that for a three digit integer, if the sum of the digits is divisible by nine, then the number itself is divisible by nine

Our three digit number could be “c,b,a”, so we could write our number as:

1a + 10b + 100c

(because the a is the units column, the b is the tens column etc.)

This can now be written as

a + b + 9b + c + 99c

= a + b + c + 9b + 99c

So if the sum of the digits = 9n, then a + b + c = 9n, and our number is now

= 9n + 9b + 99c

= 9(n + b + c)

Which is divisible by 9 :).

Disproving a conjecture by counter example

…. You may also be asked to prove other types of mathematical statements. For example, “if a < b < c < d then is a/b < c/d always true?”

In this case, try to find one single example where the statement is NOT true, then you have disproved it.

Proof by exhaustion

Proof by exhaustion depends on there being a small number of results so that it is manageable to find all possibilities. For example: https://youtu.be/XJIgql-vO-8

Proof by contradiction

To prove something by contradiction, we assume that what we want to prove is NOT true, and then show that the consequences of this are not possible – and so the original statement must be true!

Example 1: Prove by contradiction to show that there exist no integers a and b for which:

25a + 15b = 1

Start by assuming the opposite viewpoint, that the integers a and b DO exist for which 25a + 15b = 1

Now by factorising:

5(5a + 3b) = 1

So

5a + 3b = 1/5

5a + 3b is ALSO an integer because a and b are integers, but 1/5 is NOT an integer! So we have a contradiction in our opposite viewpoint and the original statement is therefore true :).

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Example 2: proving that √3 is irrational.

Here we take the opposite view, that √3 is rational in which case it can be written as a fraction:

√3 = a/b

where a/b is a fraction in its lowest form and so a and b have no common factor other than 1 (they are called ‘co-prime’).

3 = a²/b²

a² = 3b²

This means that a² is some multiple of 3… and so ‘a’ must also be a multiple of 3

(this is actually another theory which says that if a prime number p divides into a², then p also divides into a. For example, 3 divides into 36 and also divides into √36 = 6)

We can now say that

(3c)² = 3b²

9c² = 3b²

3c² = b²

Which means that b² is also some multiple of 3… and so ‘b’ must also be a multiple of 3.

But now we have both a and b that are multiples of 3, which means that they are NOT co-prime (i.e. they have a common factor other than 1) after all and our initial assumption was wrong! Hence by contradiction √3 is irrational. (Phew!)

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Example 3 (a classic example): proving that √2 is irrational.

Here we take the opposite view, that √2 is rational in which case it can be written as a fraction:

√2 = a/b

where a/b is a fraction in its lowest form and so a and b have no common factor other than 1 (they are called ‘co-prime’).

2 = a²/b²

a² = 2b²

This means that a² is some multiple of 2… and so is even. This means that a is also even, which we could call 2c

We can now say that

(2c)² = 2b²

4c² = 2b²

2c² = b²

Which means that b² is even, and so b must also be even.

But now we have both a and b that are multiples of 2, which means that they are NOT co-prime (i.e. they have a common factor other than 1) after all and our initial assumption was wrong! Hence by contradiction √2 is irrational. (Phew!)

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Example 4: Use proof by contradiction to show that there is no greatest positive rational number.

So here we take the opposite view: that there IS a greatest positive rational number. Let’s call this greatest positive rational number c.

But 2c is larger than c, so we have a contradiction in our opposite viewpoint, which therefore cannot be true.

This means that the original statement must be true.

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Example 5: Use proof by contradication to show that there are no positive integer solutions to x² – y² = 1

So we take the opposite view: that there ARE positive integer solutions to this equation.

(x – y)(x + y) = 1

The only possible way to make 1 by multiplying two positive integers is 1 x 1 = 1

So

x – y = 1

x + y = 1

Solving using simultaneous equations:

x = 1

y = 0

But y = 0 is NOT a positive integer, therefore our opposite view is false. So there are NOT any positive integer solutions to the equation.

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Example 6: Use proof by contradiction to show that if a is a rational number, and b is an irrational number, then a – b is irrational.

So we take the opposite view: we assume that a – b is rational.

So…

a – b = c/d where c and d are integers with no common factor other than 1 (co-prime)

a is also rational, so we can say that a = e/f where e and f are also co-prime.

So now we have:

e/f – b = c/d

b = e/f – c/d

b = (ed – cf)/fd

But this is a rational number as the numerator and denominator are integer values. So this contradicts the assumption that a – b is rational, therefore a – b must be irrational.

Ta da!

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Prove by contradiction that there are infinitely many prime numbers.

Assume that the statement is not true in that there are a finite number of primes (n of them). Write them as

p1, p2, p3, … pn

where p1 = 2

and pn is the largest prime number.

Multiply all of these primes together, and call it N:

N = p1 x p2 x p3 x … x pn

So N is a multiple of every prime number in the list.

Now,

N+1 = p1 x p2 x p3 x … x pn + 1

According to our assumption, N+1 should not be a prime number, since there are no prime numbers other than p1 to pn

However, if we divide N+1 by any of the prime numbers p1 to pn, this leaves a remainder of 1. Since there are no integers that divide 1 other than 1 itself, N+1 must be a prime number, or be divisible by another prime greater than pn.

This contradicts our original assumption.

Hence, there are infinitely many primes, and the statement is true.

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Proving the sum of a series

We also looked at how to prove, for example, that the sum of the first 200 integers = 20100

To do this, first write out the first few terms and the last few terms in the series. Then do the same in reverse underneath and add the two series together:

1 + 2 +…. + 199 + 200

200 + 199 + … + 2 + 1

201 + 201 + … + 201 + 201 = 2(sum of the first 200 integers)

So we can see that the sum of the first 200 integers = ( 201 × 200 ) / 2 = 20100

You can use this method to find the sum of many types of arith metic series.

Surds, Powers and solving equations

… Multiplying mixed surds and numbers. Surds can be combined with surds, and numbers with numbers.

… Simplifying surds using the ‘spot the square factor’ method. If you can spot the highest square factor, that helps to speed up the process, otherwise you can always simplify in two or more steps.

… Rationalising the denominator – simple fractions and more complicated fractions where the denominator is a mixture of a surd and a number.

… It is generally useful to try and replace surds with their power equivalent, so

root(5) can be written as 5^(1/2)

… Simplifying powers in and outside brackets, e.g. (8^1/3)^2 and 27^(2/3)

… Multiplying out ‘Double surd’ brackets.

… To solve power equations, it can be useful to convert bases to the same number. For example:

8^x – 2^x = 0

(2^3)^x – 2^x = 0

(2^x)^3 – 2^x = 0

We can then use a substitution y = 2^x to solve this:

y^3 – y = 0

y(y^2 – 1) = 0 … so either y = 0, 1 or -1

… Any number to the power of zero equals…?

… A negative power means? So 2^(-1) is…?

… Take care with equations where you need to square both sides to eliminate a square root. For example:

x – 3 = root(x + 9)

Squaring both sides…

(x – 3)(x – 3) = x + 9

… When solving two term quadratics equations such as:

x^2 – 7x = 0

Always try to factorise first. If you divide by x, you will lose one of the solutions! So:

x(x – 7) = 0

x = 0 or 7

Differentiation, gradients and stationary points

… The “rate of change of y” with how x changes is called “dy/dx” and this gives us the gradient of the curve at any particular x value.

… If y = f(x) then dy/dx can be written as f'(x).

The second differential d²y/dx² can be written as f”(x) … And so on!

… To differentiate a function with respect to x (and so find the gradient dy/dx, i.e. the rate of change of y), multiply by the power of x, and reduce the power by one. For example:

y = 3x² – 4x + 5

dy/dx = 6x – 4

(Note that 4x could be written 4x^1

and

5 could be written as 5x^0, where ^ means ‘to the power of’)

… Finding the differential of reciprocal functions, e.g. if y = 2/x

It is best to rewrite these so that the x is a numerator with a negative power, so:

y = 2x^(-2)

Then it is much easier to differentiate:

dy/dx = -4x^(-3) = -4/x³

… A function which is ‘increasing’ at a certain x value will have dy/dx > 0, which means that the gradient will be positive.

… A function which is ‘decreasing’ at a certain x value will have dy/dx < 0, which means that the gradient will be negative.

… If a function is “increasing on the interval [-3 , 1]”, this means that the curve has a positive gradient for -3 < x < 1.

… The minimum or maximum of a function’s curve (the stationary points) will have a gradient of zero, so you can find these points by setting dy/dx = 0.

Once you have found the x value of the maximum or minimum, substitute it back into the original function to find the y coordinate.

… Once you know the x values of the curve’s stationary points, substitute these x values into the second differential d²y/dx² to identify whether it is a local minimum or a maximum (the ‘nature’ of the stationary points):

If d²y/dx² > 0 , then the gradient is increasing and you have a local minimum – called ‘convex’ or ‘concave upwards’.

If d²y/dx² < 0 , then the gradient is decreasing and you have a local maximum – called ‘concave’ or ‘concave downwards’.

(Alternatively, you can put in values of x either side of the stationary points to see if the gradient f'(x) is positive or negative.)

… If d²y/dx² = 0 , then the gradient is neither increasing nor decreasing – it has stopped changing!… This COULD be a ‘point of inflection’… but could also be a minimum or a maximum.

How could we find out which?

A point P on a curve of y = f(x) is a point of inflection if f(x) is continuous there and the curve changes from concave to convex (or vice versa).

So that means that f”(x) will change sign on either side of P.

For the case of y = x^4 (or any other even power):

f'(x) = 4x³

f”(x) = 12x²

Since f”(x) ≥ 0 for all x values (because of the even power), the function never changes its concavity… Therefore f(x) has no inflection point.

We could also find the value of f”(x) either side of the stationary point to show that the sign doesn’t change.

Using Completing the Square for quadratic functions:

… The giveaway clue is when the question mentions “…can be written in the form a(x + b)² + c “, which is the classic completing the square format.

… You may be asked to compete the square for expressions where the ‘a’ value on the x² term is greater than 1. For example:

2x² – 8x + 3

2 [ x² – 4 ] + 3

2 [ (x + 2)² – 4 ] + 3

2 (x + 2)² – 8 + 3

2 (x + 2)² – 5

Completing the square can be used:

… To find the coordinates of the minimum or maximum vertex.

For example, for y = 2 (x + 2)² – 5 , the minimum value of this function must be when x = -2 (to make the squared part disappear, leaving just y = -5).

So the minimum turning point of the this curve is at (-2 , -5)

… To solve quadratic equations to find the coordinates where the curve crosses the x axis.

… To rewrite a function so you can identify the transformations that have been applied.

For example, the function y = 2x² – 8x + 3 can be rewritten as:

y = 2(x – 2)² – 5

If we compare with an ‘original’ function y = x², then we can see the transformations would be:

- A translation by vector (2,0)
- A stretch parallel to the y-axis, scale factor 2
- A translation by vector (0,-5)

For x-direction transformations (‘in the brackets’):

Translations first

Stretches and reflections

For y-direction transformations (‘outside the brackets’):

Stretches and reflections first

Translations last

… Completing the square can also be used to tidy up a ‘messy’ circle equation into the usual circle equation format so you can find the centre and the radius.

Quadratic Functions and the Discriminant (b² – 4ac)

… Using the Discriminant to discover if a function has distinct, equal or no real roots (solutions).

… Note that ‘root’ means the same thing as ‘solution’ (the values of x which make the equation ‘work’)

… The values of the Discriminant when a quadratic f(x) = 0 has two distinct real solutions (>0), one repeated solution (=0), or no real solutions (<0).

… Using the quadratic formula to solve quadratic equations (you may need to leave your answer in a simplified surd form).

… If a straight line is a tangent to a curve, then the roots of the simultaneous equation will be such that the Discriminant = 0 (as there is only one place – one solution – where the line touches the curve).

… If a straight line cuts through a curve function at two points, then the Discriminant > 0 as there will be two distinct solutions for the intersection points.

… If there is no intersection between two plotted functions, then the Discriminant < 0 , so no real solution ‘roots’

… By finding the gradient function, dy/dx and setting = 0, we can also use the Discriminant to test if there are 2 distinct stationary points, 1 single stationary point, or no stationary points.

… Solving Quadratic inequalities – it is essential to draw the graph of the curve first, having figured out where it crosses the x axis (solving the quadratic = 0).

Remember that the quadratic function (in x) gives you an ‘output’ y. From your graph you can see where the curve (the y value) is greater than > or less than < 0 (above or below the x axis) and so find the corresponding set of values for x which makes the inequality true.

However, please note:

- If there is ONE set of values of x which satisfies the inequality (e.g. a single drawn line on a number line), then you could write (for example) “x > -2 AND x < 3”. This can also be written as “-2 < x < 3”
- If there are TWO distinct sets of values for x , then you could write (for example) “x < -2 OR x > 3”.

Note that this CANNOT be written as “-2 > x > 3” because this statement implies an ‘AND’ and so is impossible!

… Note the ‘trick’ of multiplying both sides by x² for inequalities such as:

5 > 4/x

We can’t multiply by just x because x could be either a positive or a negative (which would involve us switching the inequality sign around).

Graph sketching

… Graph shapes that you need to remember are:

y = kx² (Quadratic)

y = kx³ (Cubic)

y = k/x (Reciprocal)

y = k^x (Exponential)

y = √x (square root)

y = sin(x)

y = cos(x)

… First identify the general shape of the function: e.g. Quadratic (‘Happy’ or ‘Sad’), cubic, reciprocal.

…The Table Function on your calculator is a useful tool to give you a basic shape of the function.

… If you can factorise a quadratic or cubic, this will enable you to find the roots where the equation = 0 (i.e. setting y = 0 is where the curve intercepts the x-axis).

… Put in x = 0 and see what you get (where the graph intercepts the y-axis). Note that you might get an ‘infinity’ situation if you are dealing with a reciprocal…

… See if you can identify a value of x for which the function tends to infinity (for example a value of x which makes a denominator zero). This will give you an asymptote line in x or y.

Put in an ‘x’ value just greater than, or just less than the asymptotic value – what happens to y? Does it get positive and huge? Negative and huge?

… What happens as x tends to positive infinity? Negative infinity?

Sometimes you will be able to cross out insignificant constants, e.g. if y = x/(x-5) then as x tends to positive infinity, then y tends to x/x = 1 because the ‘5’ becomes insignificant when added to infinity!

When sketching graphs, remember to show the coordinates of any points where the graph crosses the axes.

… For exponential functions such as y = e^(-at).

As t → ∞, then e^(-at) → 0, because of the negative power (reciprocal).

As t → 0, then e^(-at) → 1, because e^0 = 1

… If you can find dy/dx easily, see if you can solve dy/dx = 0 to find the stationary points.

Geometric Transformations using functions, e.g. If y = f(x) such as y = sin(x), then…

… If you change the x ingredients by modifying ‘inside the brackets’ e.g. f(x + 2), e.g. sin(x+2), then you are moving the curve in the x direction but everything is back-to-front… You’re in the ‘weird and wacky world of the brackets’… So this is actually a TRANSLATION of -2 in the x direction.

… If y = f(x), and you make y = f(2x), e.g. y = sin(2x), which is a change inside the brackets, then this would be a STRETCH of scale factor 1/2 in the x direction.

… If you are ‘outside’ the brackets e.g. f(x) + 2 , e.g. y = sin(x) + 2, then you are moving things in the Y direction – either a stretch, translation or a reflection. Outside the brackets is ‘normal’, so this would be a TRANSLATION of +2 in the y direction.

… If y = f(x), then y = 2f(x), e.g. y = 2sin(x), this would be a STRETCH of scale factor 2 in the y direction. Note that any points on the x-axis would not move.

… If y = f(x), and you make y = f(-x), e.g. y = sin(-x), then this is a REFLECTION in the y axis (you are still moving the curve in the x direction, as you’re ‘in the brackets’!)

… If y = f(x), then y = -f(x), e.g. y = -sin(x), would be a REFLECTION in the x-axis.

… Completing the square and factorising can be useful to identify what transformations have been applied to a function.

For example:

if the original function was y = sin(5x), describe the transformation given by

y = sin(5x + 10):

y = sin(5(x + 2))

So we started with y = f(x) and now we have y = f(x+2)

This is a translation of -2 in the x-direction.

Another example:

If the original function was y = 2x², describe the transformation given by

y = 2x² – 4x + 2

y = 2[x² – 2x] + 2

y = 2[(x-1)² – 1] + 2

y = 2(x-1)² – 2 + 2

y = 2(x-1)²

So we started with y = f(x) and now we have y = f(x-1)

This is a translation of +1 in the x-direction.

… Plotting the graphs of sin(A), cos(A) and tan(A) and using function transformations to create new curves. It can be useful to work ‘inside out’ when sketching a function transformation. For example:

if y = cos(x)

then sketch: y = 2cos(0.5x – 30).

- start by drawing the original cos curve,
- then stretch by SF x2 in x direction,
- then translate by +30 in x direction,
- then stretch SF x2 in y direction.

Geometric, Arithmetic and Recurrence relationships – sequences and series

Arithmetic

… a (or U1) is the first term, d is the common difference and l is the last term in the sequence.

… Formulae for the nth term of an arithmetic sequence

Un = a + (n – 1)d

… Sum of the first n terms for an arithmetic series,

Sn = (1/2)n(2a + (n-1)d)

or

Sn = n/2(a + l)

(the first formula is usually the more useful)

… Proof of Sn for an arithmetic series:

Write out the first two terms + … + last two terms, giving yourself plenty of room between each term.

Write out the same sequence, but backwards underneath the first (matching the columns)

Add the two equations.

1st term 2nd term (n-1)th term nth term

Sn = a + a+d + … + a+(n-2)d + a+(n-1)d

Sn = a+(n-1)d + a+(n-2)d + … + a+d + a

(+) —————————————————————–

2Sn = 2a+(n-1)d + 2a+(n-1)d + …+ 2a+(n-1)d + 2a+(n-1)d

So…

2Sn = n[2a+(n-1)d ]

Sn = (n/2)(2a+(n-1)d)

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Geometric

… Nth term for a geometric series, Un = ar^(n-1)

… Sum of the first n terms for a geometric series, and = a(1-r^n)/(1-r)

… Sum to infinity for a geometric series when the magnitude of r < 1, S = a/(1-r)

… When using logs (to find the term number), take care when dividing or multiplying both sides of an inequality by a log… if the log is negative you may need to flip the inequality sign around!

… Proof of Sn for a geometric series

Write out the first two terms + … + last two terms, giving yourself plenty of room between each term.

Write out the same sequence, but multiplied by r, underneath the first (matching the columns)

Subtract the two equations, noting that diagonal terms cancel out:

1st term 2nd term (n-1)th term nth term

Sn = a + ar + … + ar^(n-2) + ar^(n-1)

rSn = ar + ar^2 + … + ar^(n-1) + ar^n

(-) —————————————————————–

Sn – rSn = a – ar^n

Sn(1-r) = a(1-r^n)

Sn = a(1-r^n)/(1-r)

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… Sigma notation, for example:

Σ from k=1 (under the Σ) to 25 (above the Σ) of the sequence 2k – 1

This means add up all the terms from the 1st term (k=1) to the 25th term (k=25):

1 + 3 + 5 + … + 47 + 49

We can see that this is an arithmetic series summation with a=1, L=49, d=2 and n=25 and so we can use the appropriate arithmetic summation Sn formula.

… Sigma notation for summations which don’t start at the first term!… For example:

Σ (k=6 to 25) of the sequence 3k + 3

So we have:

21 + 24 + 27 + … + 75 + 78

We can see that this is a an arithmetic series summation with a=21, L=78, d=3

and… most importantly…

n=25-5=20 (this is because we need to subtract the first 5 terms as they are not required in the summation of terms 6 to 25).

So we can then go ahead an use the formula for an arithmetic series, Sn = …

… In the same way, you can also split up summations into two parts using sigma notation:

Σ(k=6 to 25) of 3k+3 = Σ(k=1 to 25) of 3k+3 – Σ(k=1 to 5) of 3k+3

… You may be asked to sum a sequence which is not arithmetic or geometric, for example:

Σ from k=3 to 6 of the sequence 1/k

To do this, carry out the summation manually or with your calculator sigma function:

= 1/3 + 1/4 + 1/5 + 1/6

… Sometimes you may need to work out how many terms are in the series. It can be useful to draw yourself a sequence term ‘list’, for example:

20 → 1st

21 → 2nd … so we are subtracting 19 to get from the term value to the term number

…etc

100 → = 100-19 = 81st term

… Note that you may need to split sigma summations into separate series, for example:

Σ(k=1 to 9) of [ (0.5^k) + 3k – 1]

can be split into

Σ(k=1 to 9) of [0.5^k] + Σ(k=1 to 9) of [3k – 1]

Because the first series is geometric, and the second is arithmetic.

Sigma Sign Strategy

… Remember this strategy for summation questions involving the sigma sign:

- Can you split up the sigma sign into separate sigma signs (adding terms)?
- Write down the first 4 terms for each series
- Identify whether each series is arithmetic, geometric or just a simple pattern
- Use an appropriate formula to work out each summation (or work out the summation from patterns).

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Recurrence relations

… Recurrence relations depend of the previous terms. U(n+1) is often the ‘next term’ in the series, e.g. U(n+1) = 2Un -3

… Recurrence relations can describe arithmetic, geometric and other sequences such as ‘alternating’ type.

… Sequences can be oscillating, for example:

U(n+1) = -Un

will result in 1, -1, 1, -1…

… To find the sum (sigma) of the first n terms of a recurrence relationship, first write out the beginning terms and see if you can spot a pattern. If it is actually a geometric or arithmetic sequence, use the corresponding formula to find Sn.

If it is an alternating type sequence, e.g. -0.5, 1, -0.5, 1, … Then find out how many of each term you have to work out the summation series ‘by hand’.

…. The nth term of a recurrence relation, Un, may tend to some limit L as n→∞ . This means that each successive term gets closer and closer to L.

It is possible to find L by substituting Un = L and U(n+1) = L into the recurrence formula. This is because infinitely far down the sequence, the previous term Un and the next term U(n+1) will both infinitely close to the value of L.

… It is often possible to check recurrence relations which tend to some value L as n→∞ using your calculator. Start with the first term in the formula and then replace that with ANS. Repeatedly pressing the = key will calculate the next term in the sequence.

… The ‘order’, k, of a recurrence relation tells us the period of any repeating pattern such that U[n+k] = U[n]

So for the periodic sequence

1, -1, 1, -1…

U1 = U3, so k = 2

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General Sequences and Series Notes

… Questions on series will often involve creating and solving two or three simultaneous equations.

… Keep expressions factorised where possible.

… When you factorise an expression for the nth term, look out for coefficients, e.g. the terms of the series Un = 4n(n+3) will all be multiples of 4

… The Sum to infinity for a geometric series is only possible when the magnitude of the common ratio, |r| < 1.

… Geometric series with a common ratio < 1: as n tends to infinity, Un will converge to a constant value L. This is because each successive term in is smaller by the same ratio. After thousands of terms, the next terms will be so small as to have negligible effect on the total sum – hence why the series converges.

Use the iterative series formula given to find this value. For example: if you are given that U(n+1) = 0.8U(n) +2 then when the number of terms n tends to infinity, then:

U(n+1) = U(n) = L

So L = 0.8L + 2 and L = 2/0.2 = 10

… The Geometric series formula for the nth term Un = ar^(n-1) can be confusing when looking at yearly compound interest (say 4%) payments on a starting amount of capital £A.

Start by writing that “year 0” = starting capital £A with no interest paid yet. This is your n=1 term of a geometric series.

Then “year 1” = (1.04)(A) which is your n=2 term of the series.

So Un = ar^(n-1) will work here using r = 1.04 and a = £A. If you are asked to calculate the balance after the 10th year, this will be when n = 11.

# Coordinate geometry

… It is usually a good idea to start by rewriting any equations that you are given in the form y = mx + c. It is them much easier to sketch them onto a graph and see what is going on.

… A large, neatly drawn diagram is well worth doing – it is surprising how much easier it is to spot the method in coordinate geometry questions!

… As soon as you have worked out a coordinate, do add it to your sketch to build up the ‘big picture’

… Whenever you read the word ‘intersection’ you will usually need to solve a simultaneous equation.

… ‘Vertex’ means corner

… A midpoint between two points can be found by taking the average of each coordinate value.

… The gradient of a line between 2 known points can be found by using the equation m = (y2 – y1)/(x2 – x1).

… If you know the coordinates of a point on a straight line (x1, y1) and the gradient (m), then the equation of the line can be found by using y – y1 = m(x – x1).

… The perpendicular bisector of a chord will always be a d…?… of the circle.

… Using completing the square to ‘organise’ a jumbled circle equation into its general form, so you can find the circle’s centre coordinates and its radius.

… Use Pythagoras to find the distance between two points (draw a right angled triangle with the hypotenuse joining the two points).

… The gradient of a line that is perpendicular to another line (gradient m) = -1/m

… By calculating the gradients of the lines which define a 4-sided shape, you can work out if the shape is a rectangle/square (perpendicular gradients) or a parallelogram/rhombus.

… Sometimes you may need to combine equations (e.g. from gradients) to form and solve a quadratic equation to find an unknown x or y value of a point.

… If a triangle has one of its sides along an axis and you know all of its vertex coordinates, then it is straight forward to use Area = (b × h)/2 .

… The general equation of a circle is (x – a)² + (y – b)² = r²

where

(a , b) are the centre coordinates of the circle

r is the radius

Try finding the centre and radius of the circle given by:

x² + 10x + y² – 6y = – 30

… Note that you may be asked to find the range of values of some constant k within the equation of a circle. For example:

(x – 3)² + (y + 2)² = 4+k

Because 4+k = r²

r = √(4+k)

and so 4+k > 0 because we cannot take the square root of a negative number, and if r=0 then we wouldn’t have a circle!

… To see if a straight line and a circle or a curve intersect, substitute for y to form a quadratic equation in x.

Then use the discriminant with one of the following conditions:

- The line and the curve intersect at two points (so there will be two ‘roots’ or ‘x’ solutions), or
- The line is a tangent to the curve (so there will only be one intersection point, i.e. one ‘root’), or
- The line doesn’t intersect the curve at all (so there will be no roots, i.e. no possible x values from the quadratic).

… For circle questions, a careful sketch is often essential. Seeing the geometry of the situation will then help massively :).

… When solving intersections (by substituting one equation into the other), be careful that you don’t lose information by dividing both sides by an ‘x’ factor. For example:

2x(3x – 1) = 2x

If you divide both sides by 2x, we will lose one of the solutions:

3x – 1 = 1

3x – 2 = 0

x = 2/3

We should have kept the 2x:

2x(3x – 1) – 2x = 0

6x^2 – 2x – 2x = 0

6x^2 – 4x = 0

3x^2 – 2x = 0

x(3x – 2) = 0

So the solutions are actually x = 0 or x = 2/3