A Maclaurin series is a power series

which provides an approximation for y near x=0.

It is a way of “coding” functions into a standard form, i.e. power series.

It doesn’t work for all functions, but it does work for a lot of them.

The coefficient can be calculated from

For many functions the approximation can be made as good as you like, or even as good as you like for x as big as you like, by calculating enough terms of the power series.

Maclaurin series can be used

- to find approximate solutions to differential equations which can’t be solved otherwise
- to calculate values of functions like for different values of x
- to make it easier to differentiate functions
- to make it possible to integrate functions which can’t be integrated otherwise.

Click here for pdf on Maclaurin series

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A train goes round a circular track at a varying speed, but in a constant pattern (i.e. its speed at each point on the track is the same every time it goes round). Prove there must be at least one pair of diametrically-opposite points on the track at both of which the train has exactly the same speed.

This is the simplest case of the Borsuk-Ulam theorem, proved by the Polish mathematician Karol Borsuk in the 1930s. The more general case says that on a sphere (e.g. the Earth) there must be at least one pair of diametrically-opposite points which have identical temperature and barometric pressure, that for any three continuous functions defined on a hypersphere (4-dimensional “sphere”) there must be at least one pair of diametrically-opposite points which have identical values of those three functions, and so on.

The general case requires ideas from algebraic topology, a branch of maths often taught as a third-year option in uni maths courses.

Borsuk developed many of his ideas at the Scottish Café in Lviv (a city then in Poland, now in Ukraine), a meeting place for mathematicians where they kept a list of mathematical problems (“the Scottish book”: bit.ly/sc-bk) and prizes were given for solving them. The café is shown below.

A Freddo (or vegan chocolate) for every proof (for the simple case), or good attempt. Answers to Mr Adebayo, Mr Osborn, or Mr Thomas by 29 June.

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**Most helpful to other students**

Conner

Wei Kong

Alex

**Best questions and objections in class**

Jetmir

**Best diagrams**

Taija

**Hardest working**

Tobi

**Neatest work**

Victor

Nii

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Statics and work-energy questions as pdf and as doc.

Worked answers to statics questions

Worked answers to work-energy questions

**Ideas coming from this work**

Click here for summary as pdf or here as odt

**Plan of campaign**

**First**: draw your diagram (or use the diagram in the question paper). Mark in all the forces on the rod. Use a ruler. Draw long lines of action for the forces. Mark the angles (“α” and “not α”). Put all the information from the question into the diagram. Make the diagram big so you can put all the info in, and make corrections without making a mess.

**Second**: write your three equations

Resolve horizontally

Resolve vertically

Take moments about a chosen point

and simplify them as much as you can.

**Third**: do algebra to find your unknowns from your three equations.

**Choosing the point to take moments round**

In Gold 4 Question 7, for example, you could choose A or C. Both will give you the same information. A is better because moments round A gives you an equation with fewer unknowns in it.

100 (2 cos α) = 3 R_{2}

That’s good. Plug in cos α = (2√2)/3, which you can get from doing a little right-angled triangle with α and sin α = 1/3, and then:

R_{2} = (400 √2)/9

Straight away, just from that one equation.

Some students say: take moments around C instead, because then you get μ (which is what you want to find) in the moments equation. Sounds sensible, but isn’t sensible. You can’t get more information by taking moments round a different point. You can only get the same information in a simpler or more complicated form. Simpler is better.

You have three equations, resolving vertically, resolving horizontally, moments: μ will appear in at least one one of them. It’s easier for you if it appears in only one than if it appears in two or three.

The three simultaneous equations

μR_{1} = R_{2}

R_{1} = 10

R_{2} = 5

are simpler to solve than three equations every one of which contains all three of μ, R_{1}, and R_{2}. Aren’t they?

Take moments around the point which will give you fewest unknowns (usually: only one unknown) in your moments equation.

**What is an unknown?**

Sometimes in these equations you’re given the weight or mass of the rod, sometimes you’re just told weight = W or mass = m.

Sometimes you’re given the length of the rod, say 4 metres, sometimes you’re just told length = 4a or something like that.

Almost always, W or m or a are not really unknowns in your working. They are given quantities which just happen to be represented by letters rather than numbers. Your real unknowns will be things like the reactions and frictional forces.

**Which directions to resolve forces?**

In some mechanics questions with inclined planes, it is best to resolve along the plane and perpendicular to the plane.

In M2 statics? No. Best to resolve *vertically* and *horizontally*.

You’re pretty sure to have a number of vertical and horizontal forces on the rod (weight, normal reaction at ground, friction at ground, sometimes normal reaction and friction at a wall). There’ll be more forces that are vertical and horizontal than ones which are along the rod and perpendicular to the rod. So resolving *vertically* and *horizontally* is better.

**How to find moments when rod is resting on a drum or peg**

No surprise that when Edexcel collects together its hardest past-paper questions on statics, a number of them are about rods resting on pegs or drums. In those cases you have a force which is *not* vertical or horizontal: the normal reaction from the peg or drum on the rod. If the peg or drum is not smooth, i.e. there is friction from the peg or drum acting on the rod, you have two forces which are not vertical or horizontal.

Resolve those forces into horizontal and vertical components so that you can resolve all forces on the rod horizontally and vertically.

To calculate the moment of the force R_{2} around A in the diagram below, you can add the moment of the vertical component R_{2} cos α and the moment of the horizontal component R_{2} sin α

That gives you an anticlockwise moment of: 3 cos^{2}α + 3 R_{2} sin^{2}α

Since cos^{2}α + sin^{2}α = 1

moment = 3 R_{2}

You can get that more directly by forgetting for the time being about resolving vertically and horizontally. Just calculate directly the vertical distance from A to the line of action of R_{2}. That distance = 3 (along the rod). So moment = 3 R_{2}

**Using your calculator**

The general rule in pure maths is not to use your calculator until you have to, and to leave an expression like (400 √2)/9 as just (400 √2)/9 .

In mechanics, however, everything is approximate. You’re told that the rod is 4 metres long, but on a warm day it might be 4.001 metres. So there’s nothing lost by writing

R_{2} = 62.9 rather than R_{2} = (400 √2)/9

Nothing lost. Something gained if you find R_{2} = 62.9 easier to manage and to understand than R_{2} = (400 √2)/9

So in Gold 4 Q.7 you’d have

Moments around A: … R_{2} = 62.9

Resolving horizontally:… R_{2} = 3 μ R_{1}

Resolving vertically:… 3 R_{1} + 2.828 R _{2} = 300

Plug R_{2} = 62.9 into the third equation to get R_{1} , and then plug both R_{1} and R_{2} into the second equation to get μ.

Don’t give results in mechanics to more than 2 or 3 sig figs, unless specially asked. The exam papers tell you to take g = 9.8, which is correct only to two sig figs anyway. Writing μ = 0.5142594772 is worse, not better, than writing μ = 0.51, because the longer way of writing suggests that you have some accuracy to the tenth decimal place. Which you don’t.

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Here are seven results about parabolas, of the sort often proved by laborious algebra for A level exams, proved from a single diagram by geometry. Results 2 and 5 are also proved, but are important as steps to prove the other results rather than for themselves.

For once, by the way, it’s not just Edexcel to blame for the turn from elegant geometry to brute-force algebra. By chance I found demands for algebraic proofs of a couple of these results in a Cambridge Tripos paper of 1840.

For simplicity, the diagram omits the actual parabola itself. P and Q are points on a parabola. F is the focus. DE is the directrix. PT and QT are the tangents at P and Q.

By the focus-directrix definition of a parabola, FP=DP and FQ=EP.

1. TP bisects ∠ FPD (and similarly TQ bisects ∠ FQD)

Proof: The line bisecting the angle is the line with points equally distant from F and D. By definition of a parabola, the tangent at P, i.e. the line most closely approximating the parabola, will also have its points equally distant from F and D. Therefore the tangent and the line bisecting the angle are identical. (Click here for a more formal argument).

2. PDTF is a kite (and so is QETF)

Proof: since TP bisects ∠ FPD, and FP=DP, TP is the perpendicular bisector of FD.

3. Every ray travelling parallel to the axis of a parabolic mirror is reflected by that mirror to the focus F

Proof: The broken line PN is the normal at P. What we need to prove is that ∠ XPN = ∠ FPN.

∠ FPN + ∠ FPT = π/2 (because normal ⊥ tangent)

∠ XPN + ∠ FPN + 2 × ∠ FPT = π (straight line)

Therefore ∠ XPN = ∠ FPN.

4. TF bisects ∠ PFQ

Proof: Since DT = TF (kite) and ET = TF (kite), △ DTE is isoceles.

∠ TDE = ∠ TED = γ, say

∠ TFP = ∠ TDP = π/2 − γ

Similarly ∠ TFQ = π/2 − γ, therefore ∠ TFP = ∠ TFQ

5. The two kites are similar

Proof: ∠ DTE = π − 2 γ

Adding up all the angles round the point T,

∠ DTE + 2 × ∠ DTP + 2 × ∠ ETQ = 2π

Halving, π/2 − γ + ∠ DTP + ∠ ETQ = π

But from triangle TDP, π/2 − γ + ∠ DTP + ∠ DPT = π

Therefore ∠ DPT = ∠ ETQ

By a similar argument ∠ EQT = ∠ DTP

Therefore the angles of the two kites are identical, therefore the kites are similar.

6. PF.QF = TF^{2}

Proof: By similarity PF/TF = TF/QF (the ratio of the longer side to the shorter side is the same in both kites).

7. T is halfway between DP and EQ (or y_{T} = ½ (y_{P} + y_{Q}))

Proof: Triangle DTE is isosceles.

8. The foot of the line from the focus to a tangent lies on the tangent at the vertex (or: R and S are on the tangent at the vertex)

Proof: R is halfway along the line FD because PDTF is a kite, and so it is also halfway between F and DE horizontally, i.e. on the line parallel to the directrix going through the vertex, which is the tangent at the vertex. This also shows y_{R} = ½ (y_{F} + y_{P}).

9. T lies on the directrix if and only if PT ⊥ QT

Proof: That happens if and only if γ = 0 and ∠ DTE = π, in which case adding all the angles round T tells us:

π + 2 × ∠ DTP + 2 × ∠ ETQ = 2π

i.e. ∠ DTP + ∠ ETQ = π/2

∠ FTP + ∠ FTQ = π/2; but ∠ PTQ = ∠ FTP + ∠ FTQ.

If P and Q are both on the same side of the axis of the parabola, the diagram is different in that the kites are arrow-shape rather than ordinary kite-shape, but the argument is the same.

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If you’re sitting A levels in England, you probably are worried when you add up all the slips and fumbles you know you’ve made, and reckon you’ve lost ten marks.

It’s cold comfort. But most maths exams aren’t like that.

The Cambridge Mathematical Tripos never publishes detailed marks, but it is known that in one year in the 1860s the total possible marks were 17,000.

The best student got 7634, i.e. he dropped about 10,000 marks. The second best got 4123. All those getting about 1500 (i.e. dropping about 15,500 marks) were awarded first-class degrees.

The same source told me about what happened in 1880. Women had not been allowed to sit the exam. Charlotte Angas Scott got permission to sit it, and got the eighth highest mark. She was not allowed to get a degree, so she wasn’t counted in the ranking, and a male student was formally ranked eighth highest. Students protested.

“The man read out the names and when he came to ‘eighth’, before he could say the name, all the undergraduates called out ‘Scott of Girton’, and cheered tremendously, shouting her name over and over again with tremendous cheers and waving of hats”.

Scott was allowed to become a lecturer at Cambridge. She did postgraduate research there with Britain’s most eminent mathematician, Arthur Cayley, who personally supported equality for women; but could get a degree recognising that research, done in Cambridge, only from the University of London.

She moved to the USA and become a professor of mathematics there. One of her results was, appropriately, a proof of the “fundamental theorem” of Max Noether, father of Emmy Noether. Emmy would be the foremost woman mathematician of the 20th century, and, forced out of Germany by the Nazis, would move to work at the same college that Scott had worked at.

Cambridge continued to refuse to give women degrees until 1948.

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Please mark every past paper before handing it in. You can hand it in to either Mr Osborn or Mr Thomas, and Mr Thomas will check it and provide feedback and help.

**For Wednesday 24 May**

● FP3 practice paper A, FP3 practice paper B, in the main FP3 past-papers booklet

● FP2 June 2013 “withdrawn” paper and mark scheme

● FP2 “extra” paper 4 – https://mathsmartinthomas.wordpress.com/2017/04/14/extra-fp2-practice-papers/

**Lessons Friday 26 May**

● Vectors practice sheet (see here for answers).

● Issues from past papers

● Start on FP3 sample paper

**For Monday 5 June**

● FP3 sample paper, in the booklet

● FP3 “extra” paper A, from Booklet of extra FP3 practice papers (mark schemes for the questions which make up these papers are in your main FP3 past-papers booklet https://mathsmartinthomas.files.wordpress.com/2016/02/fp3_frost.pdf)

● FP2 “extra” papers 5 and 6 – https://mathsmartinthomas.wordpress.com/2017/04/14/extra-fp2-practice-papers/

(FP2 exam is Wednesday 7 June)

**For Wednesday 14 June**

● FP3 “extra” papers B, C, D, E from Booklet of extra FP3 practice papers (mark schemes for the questions which make up these papers are in your main FP3 past-papers booklet https://mathsmartinthomas.files.wordpress.com/2016/02/fp3_frost.pdf)

**For Wednesday 21 June**

● FP3 “extra” papers F, G, H, I from Booklet of extra FP3 practice papers (mark schemes for the questions which make up these papers are in your main FP3 past-papers booklet)

(FP3 exam is Monday 26 June)

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With thanks to Daisy Thomas.

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You use the Intersecting Chords Theorem.

If the triangle is a different shape, then its top vertex will be outside the circle. But the proof can be adapted. You just need the Intersecting Secants Theorem in place of the Intersecting Chords Theorem.

The proof is due to Sidney H. Kung (in the Mathematical Magazine of the Mathematical Association of America, December 1990).

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