1
DCM ATHEMATICA 2015
E DITORS Zihan Zhou Zhengyuan Zhu
Junxiao Shen S UPERVISOR Dr Purchase
C OVER D ESIGNED B Y Zhengyuan Zhu & Zihan Zhou
A RTICLE C ONTRIBUTORS
Alex Boiardi
Henry Bradley
Alexander Cartwright
Andrew Hong
Keesje ’t Hooft
David Jaffe
Leslie Leung
Zeb Micic
Mr. Ottewill
Jan Rybojad
Junxiao Shen
Xiaofeng Xu
Andy Zhang
Zhengyuan Zhu
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E DITORS ’ N OTES
It is the time to look forward to Founder’s Day this year, whose theme is Persistence and Endurance. 2015 is the 100 th anniversary of Shackleton’s Great Imperial Trans Antarctic Expedition. Not surprisingly, this year’s DC Mathematica is tightly linked to the great spirits of a large number of mathematicians who devoted their whole lives to discover and explore mathematical principles behind simple questions. A large group of boys in the College have developed great enthusiasm to extend their vision beyond the syllabuses and they have made impressive progress. Not limited by their age, they eagerly show their discoveries by writing essays or short articles. During the editing process, we have received a considerable amount of articles and we have included some interesting mathematical puzzles for you to play with. Due to the limited length of this magazine, we are sorry that we cannot include all scripts we have received. Also thank you to Dr. Purchase for organising the editing again this year.
T HE E DITING T EAM . 9 TH J UNE 2015
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C ONTENTS Fooled by Randomness.............................................................1
By Zhengyuan Zhu (Y12)
Game Theory .............................................................................5
By Henry Bradley (Y13)
Latin in Mathematics...............................................................10
By Zeb Micic (Y8)
Markov Chain in Shackleton’s Voyage..................................12
By Andy Zhang (Y12)
Dr. Purchase’s Multiplication Problem.................................16
By Alex Boiardi (Y9)
Persistence and Endurance ....................................................18
By Junxiao Shen (Y12)
Coding in Mathematics ...........................................................22
By Andrew Hong (Y8)
Reflections on Reflections......................................................23
By Mr. Ottewill
Shackleton, Sextant and the…quadratic regression ...........25
By Jan Rybojad (Y12)
The Golden Ratio....................................................................33
By Alexander Cartwright (Y8)
Are Fourier Series of great importance?..............................34
By Xiaofeng Xu (Y13)
The Pancake Problem .............................................................41
By Keesje 't Hooft (Y13)
Triangle and Square .................................................................46
By David Jaffe (Y9)
What is the God's Number of the Rubik's Cube?..............47
By Leslie Leung (Y13)
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Fooled by Randomness
----Zhengyuan Zhu (Y12)
“Humans are often unaware of the existence of randomness. They tend to explain random outcomes as non-random. ” 1
People don’t understand how things are just random and try to make sense out of them. They would be making models out of stock data and look at how the pattern in the data is and forecast the future stock price, which exactly analysts and investors are doing in the investment banks. Is there really a sort of pattern and tendency behind the security price or just randomness? In this article, we would like to use some A-Level mathematics to make a sense of it. In order to understand how security prices work in the market, we need to understand the meaning of the term of the Efficient Market Hypothesis (EMH). It is a theory that states markets efficiently incorporate all public information. Security prices accurately reflect available information and respond rapidly to new information as soon as it becomes available 2 . The idea is that, if the market is really efficient, then any change from day to day has to be due only to news. However, news is essentially unpredictable. Thus, security prices have to do a random walk through time, which means any future movement in them is always unpredictable and changes are purely random. The random walk process can be described as follows: E(ε) = 0, for i ≠ s Using A-level statistics, we know that the mean of is zero and they are uncorrelated with any other previous values, meaning that is random. To test whether the stock market is really efficient, we are using the method called “auto-regression”, which is combined with PMCC (S1) and testing for zero correlation (S3). Autoregressive processes are used by investors to investigate the pattern in data. Auto-regressive models take into account past movements and future values are estimated based on the past values. E(ε i ε s ) = 0, Y t = Y t−1 + ε t ⟹ Y t − Y t−1 = ε t
1 Taleb, N. N. (2008) Fooled by Randomness: The Hidden Role of Chance in Life and in the Markets. 2nd edn. New York: Random House Publishing Group 2 Efficient-market hypothesis (2015) in Wikipedia. Available at: http://en.wikipedia.org/wiki/Efficient- market_hypothesis (Accessed: 12 April 2015)
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We investigate Shanghai Stock Exchange Index data (SSE) as an example (monthly closing price from 1991-2015). Let Shanghai Stock Exchange Index be 𝑖 , 𝑖 = 1, 2, 3 … . To find the differences (Y) of the original data (X), we take the following form.
𝑖 𝑖−1
) − ln( 𝑖−1 )
) = ln( 𝑖
𝑖
= ln (
Where X i
is approximately measured in percentage change. One observation is
lost due to differencing.
Figure 1. Shanghai Stock Exchange Index
for i = 2, 3 …, let 𝑖
=
In order to calculate the relationship between X i
and X i-1
𝑖+1 , and then the following formula (S1) can be used to calculate the sample correlation coefficient between X and Y:
𝑆
=
𝑆
√𝑆
− ̅ ) 2 𝑆
− ̅ ) 2
ℎ 𝑆
= ∑( 𝑖
− ̅ )( 𝑖
− ̅ ) , 𝑆
= ∑( 𝑖
= ∑( 𝑖
Using the Excel we get that the correlation coefficient = −0.0440 .
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Figure 2. Scatter Plot between X and Y
To be more convenient, we use the following notation to calculate the k th order of auto-correlation coefficient 3 .
∑ ( 𝑖 𝑖=𝑘+1
− ̅ )( 𝑖−𝑘
− ̅ )
𝑘
=
= 1, 2, 3 …
∑ ( 𝑖 𝑖=1
− ̅ ) 2
Therefore the above obtained coefficient is r 1 autocorrelation coefficient. Similarly we obtain r 2
which is the first order of
=0.0805, r 3
=0.0002 and r 4
=-
0.073 by letting 𝑖
= 𝑖+2
, 𝑖
= 𝑖+3
, 𝑖
= 𝑖+4
respectively.
To test whether sample correlation coefficient is significantly different from zero, we conduct the following test. The standard error ( SEr ) of the autocorrelation coefficient (r k ) is given:
2 𝑘−1
1 + 2 ∑ 𝑘 𝑖=1
𝑆𝐸( ) = √
The null and alternative hypothesis are shown below to test whether the r k
order auto-
correlation coefficient is significantly different from zero.
𝐻 0 : ≠ 0 The t statistic can be calculated (t statistic can be found in S4) − 𝑘 𝑆( ) For the up to 4 th order of auto-correlation, we obtain : 𝑘 = 0 . 𝐻 1 = 𝑘
1 290
−0.0440 0.0579
1
= −0.0440,
𝑆(1) = √
= 0.0587, =
= −0.749
3 Hanke, J. E., Reitsch, A. G. and Wichern, D. (2001) Business forecasting. United States: Prentice Hall
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0.0805 0.0588 0.0002 0.0592
2
= 0.0805,
𝑆(2) = 0.0588, =
= 1.369
3
= 0.0002,
𝑆(3) = 0.0592, =
= 0.0034
−0.0733 0.0592
4
= −0.0733,
𝑆(4) = 0.0592, =
= 1.238
Critical values of t statistic are given in the following table:
Two-tailed test
Significant level
10% 1.65
5%
1%
Critical value 2.59 Since the largest t-value obtained is 1.369, which is in the acceptance region ( −1.65 < 1.369 < 1.65) , there is no enough evidence to reject the null hypothesis at any significant level, indicating the Chinese stock market is indeed efficient. To conclude, the test we conducted for the Shanghai Stock Exchange security prices clearly shows that the market is working efficiently. However, if markets are really efficient, the serious question raised as to which roles these professional analysts in investment banks can play. Securities markets are flooded with thousands of intelligent, well-paid, and well-educated analysts. So are they just claiming free lunches but doing nothing in the bank? Of course not. One of the evidences is that some fund managers do consistently outperform the market over a very long period. There is mounting evidence that stock returns are predictable by using various complicated mathematical models 4 . Those that accept the EMH generally believe that the primary role of a portfolio manager consists of analyzing and investing appropriately based on an investor's tax considerations and risk profile. Optimal portfolios will vary according to factors such as age, tax bracket, risk aversion, and employment. The role of the portfolio manager in an efficient market is to tailor a portfolio to those needs, rather than to beat the market. 5 “Human beings overestimate causality, e.g. , they see elephants in the clouds instead of understanding that they are in fact randomly shaped clouds that appear to our eyes as elephants (or something else). They tend to view the world as more explainable than it really is. So they look for explanations even when there are none.” (Fooled by Randomness). 1.97
4 Pesaran, M. H. (2001) Market Efficiency and Stock Market Predictability 5 The Efficient Market Hypothesis (no date) Available at: http://www.investorhome.com/emh.htm (Accessed: 12 April 2015)
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Game Theory
----Henry Bradley (Y13)
Whether it be in the actions of enormous multinational companies or simply the everyday choices we make, game theory plays an enormous role in determining outcomes of our decisions. Game theory, is simply defined as the study of strategic reasoning and decision making. In this article I will present 3 problems, which on the face of it look simple, but actually are surprisingly complex due to game theory. The first problem illustrates how game theory can cause individuals not to cooperate even if it would appear it is in both their interests, it is known as the prisoners’ dilemma: There are 2 prisoners (A and B) in separate cell after committing a crime together. The prisoners cannot communicate with each other at all, but they both have a choice, they can either stay silent, or betray their fellow prisoner. They both have to make this choice, without knowledge of what the other prisoner has chosen, but knowing that the other prisoner also has to make the choice. Here is the deal offered to both prisoners
If prisoner A and B both stay silent, they each get 1 year in prison.
If prisoner A stays silent, but prisoner B betrays A, then A will get 10
years in prison, and B will be set free (and vice versa).
If both prisoner A and B betray each other, then both will serve 5 years
in prison. Clearly, it would seem that it would be in both their interests to both stay silent,
and simply serve the 1 year, however this is unlikely to be the case due to game theory incentives to betray. What is likely to happen is that both prisoners betray. The reason this outcome is likely is that both prisoners have an incentive to betray. For example, imagine you are prisoner B, if prisoner A has stayed silent, and you betray, you will be in prison for 1 year less than if you had stayed silent. If prisoner A has betrayed, if you betray you will be in prison for 5 years
Figure 1 – Table of outcomes
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(as opposed to 10 years if you stay silent). As a result, both prisoners will benefit more from betraying, regardless of what the other prisoner chose. As a result, the outcome if the prisoners are both logical will be something which is not in either prisoner’s interests – they both betray and get 5 years in prison. The next problem is not strictly game theory, but demonstrates well how a counterintuitive choice may be the right one, It is known as the Monty Hall problem. 6 You are on a game show, and there are 3 doors (A, B and C), you know that behind one of the doors is a brand new Ferrari, and there are goats behind the other 2. You of course do not know which door (A, B or C) has a Ferrari behind it, and which doors have goats behind them. Your aim is to win the Ferrari (I hope). You pick a door, say A for example. The game show host, who does know what is behind each door, opens another door (say B for the sake of argument) to reveal one of the two goats. There is now only 1 goat and the Ferrari left in either door A or C. You are then given a choice, do you want to open door A, and receive what is behind it, or do you want to switch to door C? Try to decide which door would be best to pick, before looking at the solution. Solution: On the face of it, it seems like it shouldn’t matter, there are 2 doors left, 1 has a goat, and 1 has a Ferrari. Therefore whichever door you choose gives you a 50% probability of winning the car, right? Wrong! In fact, you will have a higher probability of winning the Ferrari if you switch to door C, this is why:
Initially, on the first pick clearly you have a 2/3 chance of picking a
goat. Let’s imagine have picked a goat.
If you do pick one of the goats, and another goat is revealed, then if
you switch to door C, you will have 100% chance of picking the Ferrari (given that you picked a goat first [a 2/3 chance]).
This is because there is 1 goat behind the door the host opened, and
there is 1 goat behind the door you initially picked, therefore behind the other door must be the Ferrari.
On the other hand, lets imagine you picked the Ferrari initially (1/3
chance), if you switch, after 1 goat is revealed, you will have 100% chance of getting a goat (given that you picked the Ferrari first [a 1/3 chance]).
6 BBC (2013) Monty Hall problem: The probability puzzle that makes your head melt. Available at: http://www.bbc.co.uk/news/magazine-24045598 (Accessed: 13 May 2015)
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As a result if you initially picked a goat, and you switch, you will
definitely win the Ferrari, but if you initially picked the Ferrari and you switched, you will definitely receive a goat, but because the chance of initially picking a goat is higher (2/3 as opposed to 1/3), you will overall have a 2/3 chance of winning a Ferrari if you switch to door C, (and only a 1/3 chance if you stick with door A). If, despite my fantastic explanation you are still confused, don’t worry, one of the most influential and brilliant mathematicians of the 20 th Century, Paul Erdős, did not accept that you would have a higher chance of winning if you switched door until he was shown a computer simulation of thousands of outcomes of the game show. 7 The third and final problem is known as the Truel problem. This is one of the most fascinating game theory problems, as it again illustrates how counterintuitive decisions can lead to the best outcome. There are 3 people participating in a 3 way duel, a truel! Each person ( A, B and C) has a gun, and takes it in turns to aim and fire 1 shot of their gun, first A will shoot, then B will shoot, then C, and then A again and B and the cycle continues until there is only 1 person left alive. Each participant will die if they get hit by a single shot. However, person A is a bad shot, and only has a 1/3 (33%) chance of hitting one of the other participants if he aims at them, person B is a better shot and has a 2/3 (67%)chance, and person C is a great shot and has a 3/3 (100%) chance of hitting. Each participant knows the other participants chances of an accurate shot. If one of the participants gets killed, then of course they cannot themselves shoot. For example, if A goes first and shoots C and kills him, then it will be B’s turn, then if B misses it will be A’s turn, and so forth until only 1 participant is left alive. Imagine you are person A and you are going first, assuming that the other 2 participants are logical, should you aim at B? Should you aim at C? What should you do to maximise your chances of winning this truel? Try to come up with the answer before looking at the solution below.
Solution: What you probably thought, is that A should aim at C, because he is the more likely of the other participants to shoot accurately, and therefore he should be eliminated. It would be illogical to aim at B, because if you happened to kill B, then C would have a 100% chance of killing you, as it is their turn
7 Arbesman, S. and Pearson, G. (2014) Monty Hall, Erdos, and Our Limited Minds. Available at: http://www.wired.com/2014/11/monty-hall-erdos-limited-minds/ (Accessed: 13 May 2015)
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and they are always accurate. However, the way for A to maximise the chances of winning this Truel is to not aim at B or C in the first turn, but rather to intentionally miss both of them, by shooting the gun straight into the air! To show why this is the case, we will have to analyze several possible outcomes. Firstly, If A aims at B : If A hits B (a 33% chance) then A will definitely lose, because C will kill A right away. If A misses B, then B will aim at C (because if B doesn’t kill C, C will kill B next turn as B is more dangerous
Image 1
than A). If B hits C, then A and B will shoot at each other in turns until one of them is killed. If B misses C, then C will kill B, and then A will have 1 chance of shooting C, if A misses C will shoot A and the truel will be over. As a result the probability of A losing if he aims at B
= P(A hits B)+ P(A misses C)[P(B misses C)xP(A misses C)]+[P(B hits C)xP(B will win duel when A goes first)].
P(B will win duel when A goes first)]
= 1 − [(1/3) + (2/3)(1/3)2 + (2/3)2(1/3)2. . . . ] = 1 − (1/3)/(1 − 2/9) = 4/7
Therefore the probability of A losing if he aims at B = 1/3 + 2/3[(1/3 x 2/3) + (2/3 x 4/7)] = 139/189 . So if A aims at B, he will only have a 40/189 (26%) chance of winning. If A aims at C: If A hits C, then B and A will have a duel until one dies (with B going first), if A misses C, then B will aim at C, and if B hits C then A and B will have a duel until one dies (with A going first). If B misses C then C will kill B and A can aim at C, if A misses C then C will also kill A. Therefore, the probability of A losing if he aims at C first is: 8 =[P(A hits C)xP(B wins duel when B goes first)]+P(A misses C)[P(B hits C)xP(B wins duel when A goes first)+P(B misses C) x P(A misses C)]
P(B wins duel when B goes first)= (2/3)+(1/3)(2/3) 2 +(1/3) 2 (2/3) 3 ....=(2/3)/(1- 2/9)=6/7.
8 http://www.readthehook.com/79248/strange-true-aim-high-avoid-first-shot-truel
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Therefore, the probability of A losing if he aims at C first is: (1/3)(6/7)+2/3[(2/3)(4/7)+(1/3)(2/3)] =130/189. So if A aims at C he will only have a 49/189 (31%) chance of winning.
If A deliberately misses: If B hits C, then A and B will duel to the death, with A going first. If B misses C, then C will kill B and A will have aim at C, if A misses C will kill A. Therefore the probability of A losing if he deliberately misses is:
P(B hits C) x P(B wins duel when A goes first) + P(B misses C) x P(A misses C)
= (2/3)(4/7) + (1/3)(2/3) = 38/63
So if A deliberately misses, he will have a 25/63 (40%) chance of winning, the best chance by some margin! To put it simply, A deliberately missing guarantees that he will have the first shot in a 2 way duel, either with B or C (because B or C will aim at each other). If aims at B or C then there is always the possibility that he will be a in a 2 way duel, but not shooting first, which is clearly a disadvantage.
Puzzle No.1
How can I get the answer 24 by only using the numbers 8,8,3,3. You
can use add, subtract, multiply, divide, and parentheses.
Bonus rules: also allowed are logarithms, factorials and roots
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Latin in Mathematics
----Zeb Micic (Y8)
There are many instances in Mathematics where phrases of letters are derived from Latin. In this (brief) article I will go over some of the most common examples.
Perhaps, the most known and the most used is ‘e.g.’, this is derived from ‘exempli gratia’, which means ‘for example’
Another common one is: ‘i.e.’ (id est), literally ‘that is’.
Take note of, ‘n.b.’ - ‘note well’. This is a way of saying ‘take note’ (no pun intended!)
I have included some more, here, in tabular format’:
Latin
English
‘as infinitum’
‘to infinity’, a process or operation that can be carried out endlessly.
‘a fortiori’
‘with stronger reason’ (e.g. if every multiple of four is even, then a fortiori every multiple of four isA even).
‘ipso facto’
‘by that very fact’
‘per impossibile’
(this one is self explanatory!)
‘quod erat demonstrandum’ (QED)
‘that which was to have been proved’
‘quod erat faciendum’ (QEF)
‘that which was to have been shown’
For the sake of completeness I have included some other common Latin terms: ‘ad hoc’ (for immediate purpose), ‘alma mater’ (old university/college), ‘A.D’ or ‘anno Domini’ (in the year of our Lord), ‘bona fide’ (in good faith), ‘de facto’ (in actual fact), ‘erratum’ (error), ‘et cetera’ (and so forth), ‘ibid’ (in the same place) and the old favorite, ‘P.S’ (‘post scriptum’).
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This short paper has only touched on a few of the most common ones, I have not included the dozens, in not hundreds, of the letters from the Latin alphabet used in mathematics (e.g. ‘x’ used to denote an unknown variable). There are also many other Greek letters used in Mathematics, but that is a different story!
I would like to thank Mr. M. J. Emson for drifting ‘off topic’ ever so slightly, to introduce us to this!
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Markov Chain in Shackleton’s Voyage
----- Andy Zhang (Y12)
Ernest Shackleton was at South Georgia, preparing for his last journey toward the Antarctica. He attempts to estimate the number of days it might take to get to the Antarctica, by calculating the distribution of different directions of wind (south, north, east or west) every day. With some help from a few statisticians and mathematicians, the directions of wind follows some underlying rules:
After a day of S wind, the probability of getting S on the next day is 0.9,
N is 0.075, EW is 0.025
After a day of N wind, the probability of getting N on the next day is
0.8, S is 0.15, EW is 0.05
After a day of EW wind, the probability of getting EW on the next day
is 0.5, N is 0.25, S is 0.25
Most importantly, the conditional probability of the future and past
states are independent, given the present state. This means that the states before 1 n states are not relevant to this conditional probability.
Here is an illustrative graph,showing the conditional probability
between each state.
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The question is: After n days, what is the distribution of wind directions among these days? This continuing process is called a Markov chain, discovered by Andrey Markov, a Russian mathematician during Shackleton’s time.
Here is the formal definition:
n X XXX ...... 3 2 1
,
,
A sequence of random variables:
has Markov property
the future and past states are conditionally independent, if given the present state.
Pr(
X
x
X
x
,....,
X
x
Pr( ) X
x
X
x
)
n
1
1
1
n
n
n
1
n
n
Interestingly, this possibility distribution can be calculated by a matrix algorithm process called the transition matrix.
.
.
.
50 250 250 050 80 150 025 0 075 0 90 . .
.
.
Pr
.
.
Each row represents present states and each column represents the next states. For example, on row 1 column 1, it represents the probability of going from S wind to S wind.
th
n )1 ( state, starting
The transition matrix above shows the distribution for
th n state:
Pr x x
from
n
n
1
th n state is North
th
n )3 ( state if the
What about the distribution of the
wind?
X x
( Pr 1 n x
Pr Pr)
n
n
3
2
2
2
X
x
Pr
Pr )Pr (
n
n
1
3 Pr
X
n
13
3
5.0 25.0 25.0 05.0 8.0 15.0 025 .0 075 .0 9.0
010
3
16125 .0 37125 .0 4675 .0 07425 .0 56825 .0 3575 .0 04675 .0 17875 .0 7745 .0
010
07425 .0 56825 .0 3575 .0
You simply multiply the position matrix
010 by 3 Pr .
3 Pr , I will demonstrate the matrix
To be more illustrative with the working of
2 Pr :
algorithm in
2
5.0 25.0 25.0 05.0 8.0 15.0 025 .0 075 .0 9.0
For the sum of first row first column:
25.0 025 .0 15.0 075 .09.09.0 . This
th
n )2 ( state, given that on the
means the possibility of having a south wind on
th n state is a south wind.
2 Pr will be the possibility of having a certain
Following the same algorithm,
th n state is a certain
th
n )2 ( state, given that on the
direction of wind on
direction of wind. Thus, if you plug in the position matrix, the possibility
th mn )
(
distribution of each direction of wind on
state will be displayed.
Now our general formula can be concluded:
th mn )
(
On the
day of the voyage, the distribution of the directions of wind
should be
m Pr
Position matrix ×
Shackleton is not really sure about how many days exactly they will reach Antarctica, thus he wants to work out the probability distribution in a very long term, i.e. when total number of days tend to infinity.
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Interestingly, when I keep multiplying the matrix itself many times, the outcome converges to the result below. This implies that over an infinity number of runs, the probability of S wind is 0.625, N wind is 0.3125, and EW wind is 0.0625.
0.625 0.3125 0.0625 0.625 0.3125 0.0625 0.625 0.3125 0.0625
𝑃 =
lim 𝑁→∞
15
Dr. Purchase’s Multiplication Problem
----Alex Boiardi (Y9)
I’ll be explaining how to solve this problem. First, as there are two three digit numbers. THESE numbers must be 0.
Then I noticed that the second column had: 2, 9 and X to add up to 6, as 9+2=11, the numbers in the second column must add up to 16; so 16-11=5 giving me this. Then due to this, I realised that in the last column, it had an equation of X+1(from carried number) which=5---so the answer is 4. Then I realised that I had a number of 45200 which made me think, what numbers multiply to 2? 1x2 so I put a 1 HERE and not a 2 as 5/2 doesn’t give a full number, and since the answer is 1, this means that the top multiple must be 452. This is derived from × 2 . So I thought, what number times by 5 gives a reminder that also ends with a 0 which adds to make 9. I placed 1 in the box and noticed that it wouldn’t work as 1 × 5 doesn’t give a remainder to create 9; so I placed a 2 and then noticed that 2 × 5 = 10 which then means that 2 × 4 = 8 but when the carrying 1 is added, produces a 9-a perfect fit. Since there’s a 2, 2 × 2 = 4 making this.
16
The last number needed to finish the problem easily is THIS. We need a number when timed by 4 produces 22 but since 22/4 isn’t a whole number, THIS can’t times 4 to make 22.
The answer can’t be 4 as when timed by 4 produces 16 which means that we would need a carried number of 6 from the previous multiplication which is impossible as 4 × 5 = 20 leaving a carried number of 2. The answer can’t be 6 as 6 × 4 = 24 which is larger than 22. So the only option left is 5- see how it works. 5 × 5 = 25 leaving a carrying number of 2; 5 × 4 = 20 but when added with the carried number 2 leaves us with 22- just the number needed.
So now that we have discovered the number, we carry it out like a normal multiplication. So 5 × 2 = 10 leaving 0 with the 1 as a carrying number. 5 × 5 = 25 but when added with the carrying number equals 26 leaving a 6 in the box and a carrying number of 2 which is added to THIS. Since we have all these numbers, we just add them together leaving this.
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Persistence and Endurance
----Junxiao Shen (Y12)
Introduction
One hundred years ago, a ship floats on the sea, lonely and helpless. Aman with a grey coat stood on the deck, thinking: this trip is going to take a long time. When he first stepped onto the ship, he knew this was going to be a long and hard trip. This man is Ernest Shackleton. 50 years later, another adventure happens in Britain. However this is a trip which floats on paper and ink – known as the proof to Fermat’s last theorem. This theorem was first proposed by Fermat in the form of a note scribbled in the margin of his copy of the ancient Greek text arithmetic by Diophantus. First of all, Fermat’s last theorem states that no three positive integers a , b , c can satisfy the equation + = No one could prove this until Andrew Wiles found the solution in 1994. Great success, regardless of whether mathematics or in exploration, all have enormous similarities. Ernest Shackleton was a polar explorer who led three British expeditions to the Antarctic. It is Andrew Wiles who was the first person to provide a complete proof for Fermat’s Last Theorem. Although they are experts in a completely different areas, they both developed their interests in their childhood. For an instance, Andrew Wiles began to attempt to prove Fermat’s last theorem using textbook methods at the age of ten. After they both grew up, they soon started out on the careers they would like to be. Shackleton became shipmaster and an explorer while Wiles earned a PhD from Cambridge Mathematics department in 1980. However, there will always be difficulty and frustration throughout the way to success. Shackleton’s ship, Endurance, became trapped in pack ice and was slowly crushed, which was a great frustration for Shackleton. Wiles also experienced the difficulty when he need to solve one single conjecture for a whole year or more. However, they had never gave it up and to the contrary, they became stronger and more experienced because of those difficulties. “Where there is a will there is a way” –quoted from a Chinese ancient saying. A strong will seems to be a common characteristic for every great people ever since, including Wiles and Shackleton. Whatever the situation is, they will always think out a way to make a bright future, to help them out of this impasse. What’s more, what they have learned from their own experience and only work by themselves would not be enough ,how they reach success or get rid of difficulties is not only be helped by others or working in a team but also learn from the base built by the great previous pioneer. Wiles might not have proved Fermat last theorem if there was no other great mathematicians
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provided other modern mathematics. It is worth mentioning that the Taniyama- Shimura conjecture is a vital point in the development of the proof of Fermat’s last theorem. Taniyama-Shimura conjecture states that every elliptic curve over the rational numbers is uniformed by a modular form. So to prove Fermat’s last theorem, wiles had to prove the Taniyama-Shimura conjecture. Last but not least, the biggest similarity between the proof for Fermat’s last theorem and Shackleton’s is duration. Wiles spent 7 years to prove the theorem and Shackleton spent 18 years to explore the South Pole. It is really hard to imagine that such a simple theorem can cost over 400 hundred years to generate a book long solution. Here I will demonstrate the proof for n equal to 3.
Proof
Note: the mathematical argument below follows closely to the source 9 . However, the explanations are all in my own words.
First of all, we assume that the three non-zero integers x , y and z which are
pairwise coprime （ greatest common divisor is 1 ） and not all positive are the
3 + 3 + 3 = 0
solution to the equation
Let us assume that x , y and z are all odds and 3 , 3 and 3 are clearly all odds, hence the sum of them is odd. But this is obviously not true as their sum should be zero. First of all, x , y and z cannot be all even because they are coprime. Therefore, there is at least one even and one odd number. Furthermore, if the third number is even, 3 + 3 + 3 is not zero again. Hence we come to the conclusion that there are two odd numbers and one even number in x , y and z.
Hence Z may be assumed to be even to prevent loss of generality.
If = , substitute this into the original equation, we get 2 3 = 3 , which implies that x is even, so this is a contradiction.
Since x doesn’t equal y and they are both odd, their sum and difference are all even.
We express + 2 = 2 , − = 2 here u and v are coprime and one of them is even and the other is odd.
9 Proof of Fermat’s Last Theorem for specific exponents (2015) in Wikipedia. Available at: http://en.wikipedia.org/wiki/Proof_of_Fermat’s_Last_Theorem_for_specific_exponents (Accessed: 1 May 2015)
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Therefore, = + and = – , substitute these two into the original equation, we get – 3 = ( + ) 3 + ( – ) 3 = 2 ( 2 + 3 2 ) Since 2 + 3 2 is odd and z 3 is even, u is even. Then we know that the greatest common divisor of 2 u and u 2 + 3 v 2 is either 1 (case A) or 3 (case B).
Proof for case A
2 = 3 2 + 3 2 = 3
There is a lemma states that if s is odd and if it satisfies an equation 3 = 2 + 3 2 then it can be written as s = 2 + 3 2 where e and f are two
coprime integers. Therefore, we get
= ( 2 − 9 2 ) = 3 ( 2 − 2 ) Because u is even and v is odd then e is even and f is odd. Substitute u into the equation for r 3 we get: 3 = 2 = 2 ( – 3 ) ( + 3 ) The factors of r 3 are coprime because 3 cannot divide e as a result, the three factors must individually equal cubes of smaller integers as follows
−2 = 3 − 3 = 3 + 3 = 3
Add the three equations up we get a solution 3 + 3 + 3 = 0 . Therefore, by the argument of infinite descent (the original equation can generate infinitely many of the same equations which makes the original equation unsolvable, the condition for this argument is that the numbers evolved cannot all be negative.), the original solution x , y , z is impossible.
Proof for case B
Because the greatest common divisor of 2 u and is 3, 3 divides u . U can be expressed as 3 w where w is a smaller integer. Substitute u =3 w into the equation for – z 3
− 3 = 6 (9 2 + 3 2 ) = 18 (3 2 + 2 )
Since v and w are coprime, then 18 w and 3 2 + 2 are coprime and hence they are each the cube of smaller integers, r and s
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18 = 3 3 2 + 2 = 3
Because of the lemma mentioned above, = 2 + 3 2 then we get
= ( 2 − 9 2 ) = 3 ( 2 − 2 )
Therefore, the expression for 18 w becomes
r 3 = 18 w = 54 f ( e 2 − f 2 ) = 54 f ( e + f ) ( e − f ) = 3 3 ×2 f ( e + f ) ( e − f ).
Use the similar method above, because the three factors of r 3 are all coprime, they can be expressed as
−2 = 3 + = 3 − = 3 Which also yields a solution 3 + 3 + 3 = 0 . Finally, be the argument of infinite descent, the original solution x , y , z is impossible.
Then, the proof for Fermat’s last theorem when n =3 is done.
Conclusion
As a concluded above, Persistence and endurance are the most common points between Shackleton’s expedition and Wiles’s proof of Fermat’s last theorem. As we can see, being persistence with something you like and never giving up will eventually lead to a day when you will reach the highest mountain in your heart. The process of fighting the difficulties is the process you beat and surpass yourself again and again. Last but not least, to all the readers, Rome was not built in one day, one need to be unremitting with one’s dream so that one can realize it.
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Coding in Mathematics
----Andrew Hong (Y8)
When you think of spies and secret agents, you might think of lots of things such as nifty gadgets, foreign travel, dangerous missiles and more. You probably wouldn't think of mathematics. But you should. Cracking codes and unravelling the true meaning of secret messages involves loads of maths, from simple addition, to data handling and logical thinking. In fact, some of the most famous code breakers in history have been mathematicians who have been able to use quite simple maths to uncovered plots, identify traitors and influence battles.
Example:
This is the letter sent by Mary Queen of Scots to her co- conspirator Anthony Babington. Every symbol stands for a letter of the alphabet.
Letters in a language are pretty unusual because some get used more often than other letters. The graph below shows the average frequency of letters in English. To compile the information, people looked through thousands of books, magazines and more and counted them all up to come to a conclusion.
And it's this information that can help you to crack codes. All Elizabeth the First's Spy- Master had to do to crack Mary's code, was to look through the coded message and
count the number of times each symbol came up. The symbol that came up the most would probably stand for the letter 'E'. When you crack codes like this, by looking for the most common letter, it's called 'frequency analysis'. In conclusion, we can see not only has coding been around for a long time, it has very close relevance to mathematics as a majority of code breakers use the system. Other examples could be things such as cracking the enigma, code in 1915 or Julius Caesar sending code languages 2000 years ago in battle.
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Reflections on Reflections
----Mr. Ottewill
An old conundrum runs as follows. Suppose you stand in front of a full size mirror and hold both hands out horizontally to each side. Suppose you shake your left hand. Which hand of your ‘image’ shakes? If you picture yourself where your image is, then the answer is their right hand that shakes. Similarly, if you lean forward and move your hands in towards the mirror, eventually your left hand joins with ‘their’ right hand. This is not hard to understand, after all, they are your ‘mirror image’ and so it seems natural for things to be reversed. The question arises however: if everything is reversed, then why when you shake your head don’t the image’s feet shake? In other words, why are left and right swapped but not top and bottom? The beauty of this question is that it is very simple to state, yet actually quite hard to pin down exactly what the answer is. I believe that there are a surprisingly large number of ‘wrong’ answers which are often given. I will look at what I think are two possible wrong answers before giving what I believe is the correct answer. The first wrong answer is that left and right change, but not up and down, because we have two eyes side by side, not one above the other. I think that this answer is most easily seen to be incorrect simply by shutting one eye and noticing that this does not change anything – left and right still swap but not up and down. The reply could come that we are so used to having two eyes side by side that we still think like this when one is shut. A simple reply is that someone unfortunate enough to have only one eye from birth presumably sees things the same as us, i.e. their feet still don’t shake when they look in a mirror and shake their head. (Another alternative answer is to turn one’s head on one side, i.e. so that one’s eyes are now above each other – nothing odd happens, i.e. the image’s left hand doesn’t suddenly morph into their right hand, or the image’s head to their feet.) A second possible wrong answer, one given to me a few years ago by a well- respected Biology teacher (no longer at the College I should add) is that the effect is due to the image on a retina being upside down, our brains turning it the ‘right way up’ when we process the image. This sounds plausible until one realises that the reason for the image on the retina being upside down, namely that light coming in from above passes through the lens at the front of the eye and ends up towards the bottom of the retina, applies equally well to light coming in from the left, i.e. such light ends up on the right hand side of the retina and vice versa for light coming in from the right. The image being upside down on the retina
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seems stranger than left and right swapping, but both happen and so this again fails to explain why left and right swap but top and bottom do not.
There are other possible answers, i.e. it is perhaps worth the reader considering what they think the answer is before reading on. However, rather than discuss them all, I will instead state what I believe is the correct answer, which involves symmetry and rotations, plus, not surprisingly, reflections and also, perhaps more surprisingly, a little bit of gravity. I believe that the key insight is that when we are asked to consider which is the image’s ‘left hand’ we picture ourselves moving around to where the image appears to be, possibly picturing ourselves literally walking around to it, and standing where the imagine is, that is, we effectively picture ourselves rotating around a vertical axis which runs down the middle of the mirror. We do this for two reasons. The first is due to the very simple fact that a human body, roughly speaking, has a vertical line of symmetry. This means that when we picture ourselves having rotated around to where the image is standing, we match up pretty well with the image. Of course, we don’t match up exactly, e.g. if we are wearing a ring on our left hand and we picture ourselves moving around, then the ring ends up on the right hand. The reply may come back: okay, so suppose we had both a vertical and horizontal line of symmetry, e.g. suppose that we were like a starfish, but with only four arms in the shape of a cross, and just one round eye in the middle. Wouldn’t we still look at ourselves in the mirror and see left going to right but not top to bottom? This is where gravity plays a part too – we see things in the way just described because when we line ourselves up with the image we still picture ourselves moving around as before, rotating around a vertical line, doing this now because this is what we have to do on earth. If we picture ourselves like a cross shaped starfish but in the sea we might actually find it just as easy to swim ‘over the top and down’ to line ourselves up with the image facing back, i.e. to rotate around a horizontal line through the mirror. We see now that in this case, our top and bottom have swapped, but not our left and right. It is hard to picture this because we are unused to turning around like this, and we end up literally ‘upside down’, which we don’t usually encounter. However, considering this gives what I believe is the correct answer to the original question, namely that left and right swap first of all because, due to our vertical line of symmetry, we fit ourselves so much better if we picture ourselves rotating around a vertical line to match up with our image, and secondly because this is much a much more natural movement on earth. If neither of these are the case, e.g. the cross shaped starfish in the sea, then it is just as easy to swap top and bottom as it is to swap left and right.
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Shackleton, Sextant and the…quadratic regression
----Jan Rybojad (Y12)
A brief history of nautical navigation 10
Ever since the ancient times, sailors have been wondering about the way to navigate. Firstly, the travellers used maps – which are equally old as the scripture. However, it was impossible to make a map of a sea as the sailors could see there nothing but water. The 3 rd century BC brought 2 new inventions that have revolutionised the world of navigation. One of them is the magnetic compass, which was invented by the Chinese. 11 Firstly used by the fortune tellers, a magnetized needle suspended in a liquid was discovered to point always in the same direction. The other invention was the idea of the geographic coordinate system invented by Eratosthenes of Cyrene in the similar point in time. This unified system enables all travellers to communicate easily about any position on Earth. Later on, in 1884, the conference in Washington decided that the Prime Meridian (0°) is the one determined by the Greenwich Observatory. 12
Longitudes and latitudes 13
But how can one determine their longitude and latitude having no GPS available. Well, already in the past, scientists have discovered that some celestial bodies appear roughly in the same place on the sky every day on a particular hour or they even do not move at all! Thanks to the equal velocity of Sun’s rotation
around Earth (relative to us; as perceived by people on Earth), one can predict Sun’s position on the sky for every day of the year. The angle by which the sunlight shines on the equator is known as the declination. Knowing the declination at the noon for the particular day of the year, and the angle between the horizon and the
Figure 1: The declination of Sun [12]
10 A. J. Bratcher, “History of Navigation at Sea,” [Online]. Available: http://www.waterencyclopedia.com/Mi- Oc/Navigation-at-Sea-History-of.html. [Accessed 15 April 2015]. 11 M. Bellis, “The Compass and other Magnetic Innovations,” [Online]. Available: http://inventors.about.com/od/cstartinventions/a/Compass.htm. [Accessed 15 April 2015]. 12 “The Prime Meridian at Greenwich,” [Online]. Available: http://www.rmg.co.uk/explore/astronomy-and- time/astronomy-facts/history/the-prime-meridian-at-greenwich. [Accessed 15 April 2015]. 13 “How to navigate by the Sun?,” [Online]. Available: https://www.youtube.com/watch?v=CycmCFb-6VU.
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Sun in our position (the altitude), one can calculate their latitude (as I will show later). At night, a good way to find the latitude is to measure the altitude of Polaris, whose declination is always approximately 90°. The problem was how to measure the altitude. Many inventions tried to deal with this problem: astrolabes, quadrants, octants, but none of those was accurate enough to rely on completely. However, in 1731 two men from different parts of the world John Hadley in England and Thomas Godfrey in Philadelphia created a revolutionary device: a sextant. 14 In 1914-1916 Sir Ernest Shackleton OA made one of the world most famous marine journeys when he was planning to reach the South Pole on his ship ‘Endurance’. Unfortunately, she got trapped in ice, making the rest of the journey impossible. The crew decided to escape on 3 little boats, so that they reached Elephant Island. This is when Shackleton with 5 of his crew members sailed 1300 miles on his boat ‘James Caird’ (placed in our school’s new lab building) to get help. Of course neither of the crew members had an access to Google Maps or other satellite technology (GPS wasn’t here until 1960s). This is why Frank Worsely [6] (the skipper) was obliged to use the old good sextant and the Nautical Almanac (a book containing all the astronomical data published every year) to find their way on the sea. This was not an easy task as the horizon and the sky must be clear so that the measurement was accurate enough. A mistake of just one minute could be as bad as 4 kilometres on the sea! Another extremely useful piece of equipment is the chronometer, invented by a British mathematician John Harrison in 1737, (or a precise clock) that provided the sailors with the Coordinated Universal Time (UTC – the solar time on the Prime Meridian). This should remain unchanged even on the Elephant Island, where the Time Zone is GMT -3. Shackleton’s journey 15,16
Using a sextant 17,18
Sextant, presented on the Figure 2 below, consists of a telescope, 2 mirrors, frame, a couple of shade glasses and an index bar. The horizon mirror in the
14 P. Iflant, “The History of Sextant,” [Online]. Available: http://www.mat.uc.pt/~helios/Mestre/Novemb00/H61iflan.htm. [Accessed 15 April 2015]. 15 D. TV, “Shackleton: Death Or Glory: Navigation With Sextant,” [Online]. Available: https://www.youtube.com/watch?v=9zzYxdWGda8. [Accessed 16 April 2015]. 16 “Ernest Shackleton,” [Online]. Available: http://www.bbc.co.uk/history/historic_figures/shackleton_ernest.shtml. [Accessed 15 April 2015]. 17 “How to use a sextant,” [Online]. Available: http://www.robinsdocksideshop.com/how_to_use_a_sextant.htm. [Accessed 14 April 2015]. 18 “ Your position with a sextant,” [Online]. Available: http://www.idea2ic.com/Manuals/YOUR%20POSITION%20WITH%20A%20SEXTANT%20.pdf. [Accessed 16 April 2015].
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