This worksheet is designed to give you one or two questions covering every important idea we’ve covered this term, mostly with a bit of explanation or revision notes before the question(s).

Every question is taken from the textbook so, if stuck, you can check with Solution Bank. (You can of course email me, too).

The worksheet will prepare you for the test, which you’re also asked to do outside class time, between now and 1 November. The test will cover exactly the same ground.

**Year 12 Further Maths homework to Wednesday 1 November**

Do ten questions from this question sheet:

Some of the questions you’ve done for your 18 October homework, and some you’ve done for your 20 October test. Please select some new ones.

Q.21 looks like there’s a typo in it, and they mean to describe the equation as:

x^{4} + ax^{3} + bx^{2} + cx + d = 0

I think Q.17 may have a typo in it, too. The square root involves a complicated surd. With the calculator method you can only get an approximate square root: you can’t see that the decimal you get on the calculator is in the way you can see that 1.414213562 is √2. On the other hand, maybe they only want an approximate answer in decimals: they don’t say: give the answer as an exact surd. I’d skip Q.17.

Other than that: choose questions nearer the end of the question sheet if you’re confident and want to push yourself, choose questions nearer the start if you’re feeling less confident.

I’ll post worked answers for all the questions here soon.

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**Geometry with complex numbers**

What are the coordinates of the reflection of the point (1,0) in the line ?

A Freddo and fame for every well-reasoned attempt at a solution.

Answers to Mr Thomas by Friday 3 November.

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1. Prove that if a, b, c is a “Pythagorean triple” of whole numbers, then abc must be divisible by 2.

2. In fact there are many numbers bigger than 2 which can be proved to divide exactly into abc for all “Pythagorean triples”. Find some more “Pythagorean triples”, and make a good guess what that number is.

3. Prove your guess.

No attempts this time, which I put down to UCAS frenzy and the fact that I’m not yet able to post copies of the prize puzzle around the Sixth Form Building.

Solution: If a,b,c is a Pythagorean triple, then 60 | abc

Call a,b,c a *primitive* Pythagorean triple (PPT) if it is a Pythagorean triple and a,b,c have no common factors. (So 3,4,5 is a PPT. 6,8,10 is a Pythagorean triple, but not a PPT.)

If 2|abc for every PPT, then 2|abc for every Pythagorean triple, and if 60|abc for every PPT, then 60|abc for every Pythagorean triple. So we can just work with PPTs.

**Claim 1**: If a, b, c is a PPT, then one of a and b is even, the other is odd, and c is odd.

**Proof**: They can’t all be even, or 2 would be a common factor, and it wouldn’t be a PPT. If two of them were even, then the third would be. So at most one can be even.

They can’t all be odd, because Odd^{2} + Odd^{2} = Even.

So exactly one is even.

Odd^{2} + Odd^{2} is of form 4N + 2 for some N, so can’t = Even^{2}.

So c is odd.

So just one of a and b must be even. ∎

From now on, assume that b is the even one.

**Claim 2**: In every PPT, b is not just even but divisible by 4.

**Proof**: c and a are odd, so they are of form 2s+1 and 2t+1 for some s and t.

Therefore b^{2} = (4s^{2} + 4s + 1) – (4t^{2} + 4t + 1) = 4(s-t)(s+t+1)

If s and t are both odd or both even, then (s-t) is even

If s and t are opposite parities, then (s+t+1) is even

Either way, b^{2} is divisible by 8.

But if b were not divisible by 4, i.e. of form (4r+2) for some r, then

b^{2} = 16r^{2} + 16r + 4, which cannot be divisible by 8.

So b must be divisible by 4 ∎

**Claim 3**: abc is divisible by 60 for any PPT (and therefore for any Pythagorean triple)

**Proof**: We know b is divisible by 4 from claim 2. We will prove that one of a, b is divisible by 3; and that one of a, b, c is divisible by 5. Since 3×4×5 = 60, that is enough.

c^{2} = a^{2} + b^{2} in clock arithmetic with 3 hours [1]

But the only square numbers in clock arithmetic with 3 hours are 0 and 1

So the only way equation [1] can be satisfied is if it is of form 1 = 0 + 1

And so either a = in clock arithmetic with 3 hours or b = 0 in clock arithmetic with 3 hours

c^{2} = a^{2} + b^{2} in clock arithmetic with 5 hours [2]

But the only square numbers in clock arithmetic with 5 hours are 0 and 1 and 4

So the only way equation [2] can be satisfied is if it is of form 0 = 1 + 4 or 4 = 0 + 4 or 1 = 0 + 1

And so one of a, b, or c = 0 in clock arithmetic with 5 hours, in other words one of them is divisible by 5 ∎

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Example: find w so that

**Method 1: calculator**

Convert to r ∠ θ form: w^{2} ≈ 34 cis 2.06

Mentally, square-root the modulus and halve the argument

w ≈ √34 cis 1.03

Convert back to a+bi form: w ≈ 3+5i (round your answer back to a neat value, to correct the calculator error introduced when halving a decimal approximation of the argument)

Check that you’ve rounded back to an exactly correct answer:

(3+5i)^{2} = -16 + 30i

Write your answer: w = ± (3+5i)

(See below for a tweak to this method)

**Method 2: Equate modulus and real part**

Let square root = w = a + bi

Then

Solve those simultaneous equations for a^{2} and b^{2}

a^{2}=9 and b^{2}=25

Only problem now is to work out the valid signs for a and b. . In this case a and b must have the same sign (both negative or both positive) because . (If then a and b would have to have opposite signs, one positive, one negative).

So w = ± (3+5i)

**Method 3: Equate real and imaginary parts** (the method in the old Edexcel textbook)

Let square root = w = a + bi

Then

Substitute from the second equation into the first

Multiply up to get a quadratic equation in

Solve: a^{2} = 9, so a = ± 3, and b = ± 5

So w = ± (3+5i)

**Method 1 (calculator) tweaked**

You can minimise the approximation problem in the calculator method by using the calculator rather than your head to halve the argument. As in these images (poor quality, sorry):

This will usually give you the exact answer in the exact form you want it. The likely exception is if you’re finding something like √(−1 + 2√2 i), where the answer is (1 + √2 i).

The calculator will give you (1 + 1.414213562 i), and it won’t convert that to (1 + √2 i) even if you press the S-D key.

Workaround: if you get an answer for real or imaginary part of the square root which isn’t a whole number, square your answer to see if it’s an exact square root of a whole number (in A level, it pretty much always will be).

It’s worth memorising the facts that

√2 ≈ 1.414

√3 ≈ 1.732

√5 ≈ 2.236

Often come in useful, like it comes in useful to be able to remember your way home from school, or to a friend’s house, without having to look at Google Maps.

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Part A: Abass – redo Q.10 and 11 from p.23

Aniqa – redo Q.6, 7, 8 from p.23

Callum – complete p.23 Q.13

David – redo p.26 Q.5 and Q.7

Enoch – finish p.26 from booklet

Helen – no Part A

Howard – finish p.26 from booklet, Q.5-6-7. Find √(4+4i) by using the Classwiz calculator.

Jeffrey – redo Q.8 and 9 from page 23 of booklet.

Lara – p.23 Q.1, 2, 3, 8; p.26 Q.1, 4, 5.

Obi – redo p.23 Q.10 and p.26 Q.1

Mya – Find quadratic equations with roots:

a. 5 ± 7i

b. 7 ± 5i

Reece – Redo p.23 Q.10 and 11

No Part C this week.

Part B: complex numbers workbook p.28. Click here for an answer sheet. Please mark your own work before handing it in.

Click here for worked examples of how to solve quartic equations (equations with x^{4}

No Part A this week (but Mohaned, Brenda, and Genie need to do a lot of catch-up)

Part B: Complete Ex.3E Q.1, 2 and 6

Write out the Edexcel proof of De Moivre’s Theorem (p.28 and 29 of the textbook) as concisely and briefly as you can (while still being satisfactory to an Edexcel examiner).

Complete Activity 6 in the Möbius transformations booklet

Part C: Ex.6F Q.2

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Jeffrey Sylvester won the prize by more or less completing all parts of the puzzle except the last two.

*n!* is pronounced “*n* factorial” and means

For example,

• *n!* is the number of possible ways of arranging *n* different items. For example, if there are 30 students in a class and 30 chairs, then there are 30! possible seating plans. Work out 30! on your calculator.

30! = 265252859812191058636308480000000

Big, eh?

• List these formulas in order of how fast they increase when *n* is big and gets bigger and bigger:

If *n* is of order 10^{k}, for big *k*, then:

1000n is of order 10^{k+3}

n^{1000} is of order 10^{3000}

1000^{n} is of order 10^{3k}

n! is of order (k−0.43)10^{k} + k/2 (see Stirling’s approximation.

So those are the formulas in order of size, and n! is the biggest by a big margin.

• Triangular numbers (e.g. ) are to addition what *n!* is to multiplication. Work out a quick test you can do on a calculator to check whether a given number *n* is a triangular number without having to calculate all the triangular numbers up to that size. Use your test to find out which of 2278 and 2277 is a triangular number (one of them is).

If T is a triangular number, then

n(n+1) = 2T for some whole number n

The quadratic formula tells us that for this quadratic to have whole-number roots, 8T+1 must be a perfect square; and if 8T+1 is a perfect square, then the quadratic has whole-number roots, and T is a triangular number.

You can check on your calculator that 8×2278 + 1 is a perfect square (it’s 135^{2}).

So 2278 is a triangular number and 2277 isn't.

• 1 = 1! = T_{1} is both a factorial and a triangular number. Using your quick test, find two other numbers which are both factorials and triangular numbers. (They’re not very big). Nobody yet knows whether there are other numbers, apart from those three, which are both factorials and triangular numbers. A check on a computer has shown there are none less than 10^{77337}.

6 = 3! = 3rd triangular number, and

120 = 5! = 15th triangular number

are the two numbers other than 1 which are known to be both factorials and triangular numbers.

• Primes get thinner on the ground as numbers get bigger, but not very, very thin: it has been proved that for every number *n* > 1, there is always a prime *p* in the range *n < p < 2n*. Use that fact to prove that *n!* can never be a square number if *n*>1

Suppose n is even, so n=2m for some whole number m. Then there is a prime p bigger than m and smaller than 2m.

Since p is smaller than 2m, it must be appear in the multiplication sum to work out (2m)!, so it must be a prime factor of (2m)!

For (2m)! to be a square number, every one of its prime factors must appear in its factorisation an even number of times (i.e. 2, for example, must appear in the factorisation as 2^{2} or 2^{4} or 2^{6} or something, not as 2 or 2^{3} or such).

So for (2m)! to be a square number, p must appear in its factorisation 2 times, or 4 times, or 6 times, or some even number of times.

Since p>m, 2p>2m, so in fact p can appear in the prime factorisation of (2m)! only once. And 1 is an odd number.

So (2m)! can’t be a square number.

If n is odd, it is (2m+1) for some whole number m. There is a prime p between m and 2m. 2p > 2m, and since 2p is even, also 2p > 2m+1, and the same argument applies as for n even.

• isn’t a square number, but you can get a square number by removing one term from the multiplication. Which one? Generalise to prove that for any *n* >2, you can get a square number from by removing one term from the multiplication, and say which one.

Sorry, I mis-stated the problem, putting 2n where I should have put 4n.

= 4n

× (4n-1) × (4n-1)

× (4n-2) × (4n-2) × (4n-2)

…

× 3 × 3 × 3 ….. × 3

× 2 × 2 × 2 ….. × 2 × 2

In that multiplication, every odd number appears an even number of times, and every even number, an odd number of times.

So, if we remove one appearance in the multiplication of every even number, then every number in the multiplication will appear an even number of times, and we have a square number.

If we remove (2n)! from the multiplication, then we remove one appearance each of 2n, 2n-1, 2n-2,… 3, 2, so we remove one appearance in the multiplication of the even numbers 4n, 4n-2, 4n-4, …., 6, 4, only we put back in a factor 2 × 2 × 2 …. × 2 = 2^{2n}. But 2^{2n} is a square number, so that’s all right.

For example,

are both square numbers

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Click here to download pdf notes

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**Maths prize 29 September 2017: factorials**

*n!* is pronounced “*n* factorial” and means

For example,

• *n!* is the number of possible ways of arranging *n* different items. For example, if there are 30 students in a class and 30 chairs, then there are 30! possible seating plans. Work out 30! on your calculator.

• List these formulas in order of how fast they increase when *n* is big and gets bigger and bigger:

• Triangular numbers (e.g. ) are to addition what *n!* is to multiplication. Work out a quick test you can do on a calculator to check whether a given number *n* is a triangular number without having to calculate all the triangular numbers up to that size. Use your test to find out which of 2278 and 2277 is a triangular number (one of them is).

• 1 = 1! = T_{1} is both a factorial and a triangular number. Using your quick test, find two other numbers which are both factorials and triangular numbers. (They’re not very big). Nobody yet knows whether there are other numbers, apart from those three, which are both factorials and triangular numbers. A check on a computer has shown there are none less than 10^{77337}.

• Primes get thinner on the ground as numbers get bigger, but not very, very thin: it has been proved that for every number *n* > 1, there is always a prime *p* in the range *n < p < 2n*. Use that fact to prove that *n!* can never be a square number if *n*>1

• isn’t a square number, but you can get a square number by removing one term from the multiplication. Which one? Likewise, isn’t a square number, but you can get a square number by removing one of the terms: which? Generalise to prove that for any *n* >2 you can get a square number from by removing one term from the multiplication, and say which one.

A Freddo (or vegetarian chocolate) and fame for everyone who can make a good attempt for at least three parts of this problem. Answers to Mr Thomas by Friday 29 September.

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Tegan: calculate the Maclaurin series for y = cos^{2} x, up to the term in x^{4}, by multiplying these two series (which are the same as each other)

Umut: using the method of differences, find

Michelle: calculate the Maclaurin series for y = cos^{2} x, up to the term in x^{4}, in two ways. First, by multiplying these two series (which are the same as each other)

Then, by differentiating y = cos^{2} x to find

and so

Bolaji: By the quickest method you can find, calculate Maclaurin series up to the term in x^{4} for y = cos^{2}x and y = sin^{2}x.

Calculate cos(π/3) from the first few terms of the Maclaurin series:

and compare your answer to what you get as cos(π/3) from your calculator if set to radians rather than degrees.

Brenda: By the quickest method you can find, calculate Maclaurin series up to the term in x^{3} for:

y = e^{3x}

and y = ln(1+2x)

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