DC Mathematica 2016

DC M ATHEMATICA 2016

E DITORS

Minghao Zhang

Jack Kurtulus

Charles Cheung

Arthur Cheung

Johnnie Baggs

S UPERVISOR

Dr Purchase

C OVER D ESIGNED B Y

Tate Sun

A RTICLE C ONTRIBUTORS

Arthur Cheung

Jay Connor

Toby Evans

Harry Goodwin

Thomas Kuijlaars

Joseph Lane

Theo Macklin

Park Jun Sang

Shawn Shen

Stanley Traynor

Minghao Zhang

Zhengyuan Zhu

Table of Contents

The Sum of All Natural Numbers? ................................................................ 1 By Theo Macklin (Yr 11)

Globalization and Cointegration – Study of Stock Markets in China and USA ................................................................................................................. 5 By Zhengyuan Zhu (Yr 13)

The Mathematics inside the optical fibre .................................................... 14 By Shawn Shen (Yr 13)

The Fourth Dimension and Those Above It................................................ 23 By Stanley Traynor (Yr 7)

Marble on a turntable ................................................................................... 25 By Minghao Zhang (Yr 12)

Mr Ottewill - How he won a hundred pounds with ease............................. 27 By Arthur Cheung (Yr 12)

Pythagoras and His Cult .............................................................................. 28 By Park Jun San (Yr 9)

Sir Roger Penrose, mathematician and physicist ........................................ 31 By Joseph Lane (Yr 8)

Maths Problems from History...................................................................... 33 By Harry Godwin (Yr 10)

Much Ado about Nothing: the History of Zero ........................................... 35 By Jay Connor (Yr 8)

Phi - not to be confused with pie!................................................................. 37 By Thomas Kuijlaars (Yr 9)

Square Numbers – I bet you didn’t know this! ............................................ 38 By Toby Evans (Yr 8)

The Sum of All Natural Numbers?

1 + 2 + 3 + 4 + 5 … = −1

12⁄

By Theo Macklin (Y11)

This succinct, self-confident statement is a reason to hate Euler. Unfortunately, the maths that leads us to this arrogant but strangely compelling summation as simple as it comes. The idea that the infinite sum of positive whole numbers should give a negative fraction is, frankly, unimaginable; however, once again, the sly, mischievous talons of infinity delight in ruining the fundamental bedrock of nursery addition. Despite this initital mental impasse, there are in fact two ways to prove that the sum of positive integers does reach that ridiculous -1/12. There is the scary sounding proof by Leonhard Euler that utilises zeta function regularisation and the more accessible proof by Srinivasa Ramanujan. It should be obvious which one I intend to start with.

Ramanujan’s Proof:

Ramanujan’s proof requires two realisations: these require us to utilise algebra to represent the constants of series, let’s call them C, C 1 and C 2 . The first is such:

1 = 1 − 1 + 1 − 1 + 1 − 1 + 1 … = 1 2⁄

This innocent equation is the premise of this proof and while it seems initially unintuitive, on further inspection it makes logical sense. Where n is odd in this series the value of the sequence up to that point is 1. Conversely, where n is even the value is 0. Since ∞ ∉ 𝑂 ∪ the logical value for the sequence at the infinite point is the average of the two: 1 2⁄ . The next stage tackles another new sequence with the knowledge gained from the last. Here the sequence C 2 is used:

= 1 − 2 + 3 − 4 + 5 …

2 2 = (1 − 2 + 3 − 4 + 5 … ) + (1 − 2 + 3 − 4 + 5 … ) = 1 − 1 + 1 − 1 + 1

2 2

= 1

2⁄

2 = 1 4⁄ This sequence also causes unease with a strange blend of positive and negative integers culminating in a fraction. Fortunately, this brings us to a position from which we can prove our titular summation:

= 1 + 2 + 3 + 4 + 5 …

− 2

= (1 + 2 + 3 + 4 + 5 … ) − (1 − 2 + 3 − 4 + 5 … ) = 4 + 8 + 12 +

16 + 20 …

− 2

= 4 = − 1

4⁄

3 = −1

4⁄

∴ = −1

12⁄

And there we have it: a concise and intuitive proof of what is a deeply unintuitive sum. For all that work, however, something feels wrong: incomplete. This foundation-rocking piece of mathematics was all so simple! Surely there must be more to it that that; I, like that part of you deep inside, crave intelligible squiggles littering the page. Fear not, however, for I mentioned Euler’s proof: a forest of differentiation, sigmas and Riemann-Zeta functions. Despite this, the mathematics itself is quite rational and very satisfying.

Euler’s Proof:

Euler’s proof begins with a seemingly unrelated series and the proof of its constant, K :

= 1 + + 2 + 3 …

+ 1 = 1 + + 2 + 3 … =

− = 1 = (1 − )

∴ = 1 + + 2 + 3 … = 1

; < 1

(1 − ) ⁄

Following this, we differentiate K:

𝑥 = 1 + 2 + 3 2 + 4 3 … = 1

(1 − ) 2 ⁄

And let = −1 :

1 − 2 + 3 − 4 + 5 … = 1

2 2 ⁄ = 1

4⁄ = 2

Straight away we can see a similarity to Ramanujan’s method in Euler’s except he has reached his equivalent of C 2 via stronger mathematical logic. Reaching this point I feel a quick pause is in order while we quickly go over the Riemann-Zeta function. This is a function that was first introduced by Euler and then generalised by Bernhard Riemann. It behaves as follows:

∞

3 ⁄ … = 1 − + 2 − + 3 − …

⁄

() = ∑ 1

= 1

1 ⁄ + 1

2 ⁄ + 1

In practical terms, the Riemann-Zeta function is the infinite sum of 1, divided by integers beginning at 1, which have been placed to the power of the argument, s . Returning to the problem at hand, Euler’s proof continues:

(2 − )() = 2 − + 4 − + 6 − …

(1 − 2(2 − ))() = (1 + 2 − + 3 − … ) − 2(2 − + 4 − + 6 − … ) = 1 − 2 − + 3 − …

Set = −1

1 − 2(2 −−1 ) = −3

(−1) = 1 −−1 + 2 −−1 + 3 −−1 … = 1 + 2 + 3 …

1 − 2 −−1 + 3 −−1 − 4 −−1 … = 1 − 2 + 3 − 4 …

∴ −3(1 + 2 + 3 + 4 … ) = 1 − 2 + 3 − 4 … = 1 4⁄

1 + 2 + 3 + 4 + 5 … = −1

12⁄

It is notable that along the path of Euler’s proof there were close parallel at points to Ramanujan’s proof despite the fact that they were separated by around 250 years and Ramanujan worked, for much of his career, in isolation from the mathematical world, having taught himself and developed his own system of maths. Other than for the pleasure in maths, however, the sceptics are asking: “But what use is it?” and leaning back with a smug look smeared across their faces confident that a series reliant on infinity has no application in the real world. And these neigh-sayers would be correct if it weren’t for the great mathematical receptacle of quantum field theory that applies the -1/12 result in computing the Casimir effect. This result also has applications in bosonic string theory as it allows calculation of the number of physical dimensions and

rectifies the failure of string theory to be consistent in dimensions other than 26.

Overall, despite these pertinent usages, it is unlikely that this summation will be used in day to day life, but I feel that this concise reality of mathematics is truly under-appreciated. It was for this reason that I felt I should put it under spotlight so that the idea of the innate logic of numbers can be dispelled in favour of their actual mind-bending reality. However, I think that this brief line of numbers truly sums up the majesty in pure maths

Bibliography:

‘1 + 2 + 3 + 4 +…’ (7/04/16) Wikipedia . Available at: https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF [Accessed: 10/04/16]

‘ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12’ (9/01/14) YouTube: Numberphile. Available at: https://www.youtube.com/watch?v=w-I6XTVZXww

‘Sum of Natural Numbers (second proof and extra footage)’ (11/01/15) YouTube: Numberphile. Available at: https://www.youtube.com/watch?v=E- d9mgo8FGk&feature=youtu.be

Cargal, J.M. (1988) Discrete Mathematics for Neophytes: Number Theory, Probability, Algorithms, and Other Stuff.

Maths Puzzle 1 Each of the Four Musketeers made a statement about the four of them, as follows. a. Artagnan: “Exactly one is lying.” b. Athos: “Exactly two of us are lying.” c. Porthos: “An odd number of us is lying.” d. Aramis: “An even number of us is lying.” How many of them were lying (with the others telling the truth)?

Globalization and Cointegration – Study of Stock Markets in China and USA

By Zhengyuan Zhu (Yr 13)

Abstract This paper investigates the integrating and long term casual underlying relationship of share prices in Shanghai Stock Exchange (China) and S&P 500(USA) 1 . On the basis of the Dickey Fuller unit root and the Durbin Watson test, my results suggest that both shares have an integration of order one (unit root). More importantly, a long term relationship measured by cointegration does not exist between these two indices, they are statistically weakly cointegrated.

1. Introduction

a. Impact of globalization

By the end of last century, the international economy started moving towards a single market, due to the liberalization of trade and production, which allowed better and more efficient allocation of resources. 2 The growth and integration of the world capital markets is in fact one of the engines of globalization. As foreign exchange and bond markets become more integrated, the law of one price begins to apply throughout the world. The national stock exchanges are moving towards increasing linkages to other international stock exchanges. The linkages are extended from developed stock markets to other emerging stock markets, as well as from stock markets to other financial and banking systems. 3 The Shanghai Stock Exchange (SSE) was founded on Nov. 26th, 1990 and in operation on December 19th the same year. “Shares in mainland China have recorded their biggest one-day fall for more than eight years following a sell-off towards the end of the trading day . The Shanghai Composite closed down 8.5% at 3,725.56 after more weak economic data raised concerns about the health of the world's second largest economy.” From BBC News 27 July 2015. b. Brief introduction to Chinese and USA stock markets

1 I am grateful to Wenda Zhang who gives me valuable supervision in writing this paper. 2 Armanious, A. N. R. (2007) Globalization Effect On Stock Exchange Integration . 3 O.Lerche III, C. (1995) The Conflicts Of Globalization .

Standard & Poor's (S&P 500) introduced its first stock index in 1923. The S&P 500 index in its present form began on March 4, 1957. The open prices of S&P 500 on 24 th and 27 th of July 2015 were 2102.24 and 2078.19 respectively where a sharp drop on Index could be clearly seen.

SSE VS S&P

6000

5000

4000

3000

2000

1000

SSE Adjusted

S&P Adjusted

Figure 1 Adjusted close share prices of SSE and S&P 500 which have been rebased, 2001:12-2015:4, monthly observations.

From visually expectations, it is clear that there is no obvious relationship between these two shares during this period of time so we might expect the cointegration statistic measured is low.

2. Methodology

a. Integration: basic introduction

i. Stationarity and Non-stationarity

Why do we need to test for stationarity?

 Spurious regressions. If two variables are trending over time, a regression of one on the other could have high 𝑅 2 even if the two are totally unrelated.  If the variables in the regression model are not stationary, then it can be proved that the standard assumptions for asymptotic analysis will not be valid. In other words, the usual ‘t-ratios’ will not follow a t-distribution,

so we cannot validly undertake hypothesis tests about the regression parameters. 4

What are the conditions for a stationary series:

) = 𝜇 ()

 E(

) = ()

 Var(

) = () ≠ ()

 Covarience(

, −

ii. Mathematical Modelling

This is the model which has been frequently used to characterise non- stationarity: the random walk with drift: = + −1 + ~(0, 𝜎 2 )

(1)

where −1 < < 1

is independent identically distributed.

Consider the above autoregressive model, the error process usual assumptions of the classical model, which are:

retains all the

) = 0

Zero mean: E(



 Normality: the error terms follow a normal distribution  Homoscedasticity: the variance of the error term is always the same Var( ) = 𝜎 2  Independence: the error terms , − associated with two different settings of the predictor variables are statistically independent Covarience( , − ) = () ≠ () We can write: −1 = + −2 + −1 −2 = + −3 + −2 Substitute −1 into (∗) yields: = + ( + −2 + −1 ) + = (1 + ) + 2 −2 + −1 + Then substitute −2 into (∗) yields: = (1 + ) + 2 ( + −3 + −2 ) + −1 + = (1 + + 2 ) + 3 −3 + 2 −2 + −1 +

Successive substitutions of this type lead to:

= (1 + + 2 + 3 + ⋯ ) +

+ 2 −2

+ 3 −3

+ −1

+ ⋯

4 Brooks, C. (2008) Introductory Econometrics for Finance . 2nd edn. New York: Cambridge University Press.

where −

= 0 as n → ∞

so that

) = (1 + + 2 + 3 + ⋯ ) = 1−

(



) =

Var(



1−

 , ≥ 0 This model satisfies the conditions of weak stationarity as long as || < 1 so that all mean, variance and autocovariances are constant over time. Cov( , − ) = 𝜎 2 (1 + 2 + 4 + 6 + ⋯ ) = 𝜎 2 𝑠 1− 2

Therefore there three different situations for different values of β:

1. < 1 ⇒ → 0 𝑇 → ∞ So the shocks to the system gradually die away. 2. > 1 ⇒ = 1 ∀ 𝑇 So shocks persist in the system and never die away. We obtain: = 0 + ∑ ∞ =0 (2) So just an infinite sum of past shocks plus some starting value of 0 3. > 1 , now shocks become more influential as time goes on since if > 1 , 3 > 2 > etc. Now consider what happens to the properties of the model if β = 1 , the process started at t = 0 with fixed initial condition 0 = 0 . Then, = + −1 + , = 1,2, … , 𝑇 (3) Similarly, substitute for lagged ( −1 , −2 , −3 , ) = + ∑ ( 2 =1 ) + 0 (4) Therefore in this model E(Y t ) = tc so it is not constant over time. Thus this process is no longer stationary and is known as a random walk with drift. Note that the random walk process can be stationary by transforming the dependent variable by first differencing since ∆ = − −1 = + (5)

is a stationary process

A series that can be made stationary by differencing once is said to be integrated of order one, or to possess a unit root.

b. Dickey Fuller integration test

The early and pioneering work on testing for a unit root in time series was done by Dickey and Fuller. The reason why we focus on DF test is that it is simple

and there is no uniformly better alternative. The basic objective of this test is to test the level of integration of the time series using autoregressive model: 5 = −1 + (6) The majority if economic and financial series contain a single unit root, although some are stationary and consumer prices have been argued to have two unit roots. Here to simplify the situation we have the hypothesises: 0 : 1 :

We usually use the regression in the form:

∆

= −1

(7)

So that a test of β=1 is equivalent to a test of δ=0 (since δ-1=β)

The null and alternative hypothesises are:

0 1

: :

= −1

= −1 , < 1 This is a test for a random walk against a stationary autoregressive process of order one(AR(1)) +

i. Computing the DF test statistic

Fuller (1976) and Dickey and Fuller (1979, 1981) proposed a test based on the t-ratio t(α) in the Ordinary Least Squares (OLS) regression.

The test statistics are defined as:

𝛿̂ (𝛿̂ )̂

test statistic =

Where SE means standard deviation. The test statistic does not follow the usual t-distribution under the null since the null is one of non-stationarity, but rather follows a non-standard distribution. Critical values are derived from Monte Carlo experiments, the critical values are shown in the table 4.1.

c. Durbin Watson test for Cointegration

i. The Meaning of Cointegration

If I(1) variables are cointegrated, this means that although they are individually non-stationary, they are moving together so that there is some long run

5 Armstrong, J. S. (ed.) (2001) Principles of Forecasting: A Handbook for Researchers and Practitioners . 1st edn. Boston, MA: Kluwer Academic Publishers.

relationship between them. We use the static equation between two I(1) variables which now may possibly be cointegrated: = +

(8)

Where

is monthly index of S&P 50 0

is monthly index of SSE

If and are not cointegrated then there is no possible value of the parameters can be stationary. If they are cointegrated however, then there is a single value for the two parameters such that the linear combination − ( + ) is stationary. If a set of variables are cointegrated, then the residuals from a static regression will be stationary. If not, then the residuals will be integrated. The unit root Durbin Watson test can be used to test cointegration in the residuals from a static regression and is described in Sargan and Bhargava (1983). 6 and such that

ii. Testing cointegration

0 0

: :

= (1)

= (0) Here we could again employ the Dickey Fuller test again to determine the order of integration of whereas we would use an alternative method which is Durbin Watson test.

Durbin Watson test statistic:

       t

     



e e )

(



 t t 1





n 2











, the sample error terms, also known as residuals, could be simply computed. Let and be possible estimated of the regression parameters and , and fit the regression line to describe the relationship between expected value of the dependent variable and the independent variable. ̂ = + (9) The difference between the observed and predicted values of the dependent variable is

̂ =

−

− −

Under the null hypothesis that

is a random walk and that = 0 , so there is no

cointegration, and

becomes a random walk with theoretical first order

6 Pierse, R. . (no date) Lecture 8: Nonstationality, Unit Roots and Cointegration . .

autocorrelation equal to unity. Under the null of no integration, the DW statistic value will not be significantly from zero. Therefore, a Cointegrating Regression Durbin-Watson (CRDW) statistic different from zero implies cointegration.

iii. Slight problems with DW test for cointegration

This test suffers from two major problems, it is extremely sensitive to the assumption of being a true random walk and, the critical values of the test statistic are not consistent as the number of regressors increases over the sample size. The practical use of this test is therefore extremely limited. 7

3. Data

We select monthly data of adjusted close share prices of SSE and S&P 500 from 2001.12 to 2015.04, this is because China joint the World Trade Organization (WTO) on 2001.11.12 from which China began to globalize with the world’s financial system. (Notes: all data are in natural logarithms) Also, the reason why we do not use daily data is because of the difference between Chinese and US stock markets open dates, they are always mismatched due to different national holidays as well as time lag. Therefore it is difficult to judge which dates should be paired.

4. Results

a. Unit Root Test

The first stage of the analysis is to determine whether the time series data possess a unit root by using Dickey-Fuller test.

Unit root test for SSE Index

 

X Variable β coefficient: -0.00863

Test statistic: -1.24184

Significance Level

10%

C.V for constant but no trend C.V for constant and trend

-2.57

-2.86

-3.43

-3.12

-3.41

-3.96

Table 2: Critical Values for Dickey Fuller test (Fuller, 1976, p373)

7 Sjo, B. (2008) Testing for Unit Roots and Cointegration .

From the test statistic calculated, reject 1 that SSE Index time series has a unit root.

, there is sufficient evidence showing

Similarly, the test statistic is calculated for S&P 500 Index:



Test Statistic: -0.21946

Again, reject 1

, there is sufficient evidence supporting that S&P 500 possess a

unit root.

To conclude, both hypotheses of 0 cannot be rejected at any significance level, therefore the original data do follow a random walk process, which are both I (1) time series.

1.1Durbin Watson test for Cointegration

∑ ( =2

) 2

− −1

∑ =1

  

Numerator: 0.821091 Denominator: 16.24786

d=0.050535 Finally, we need to use the CRDW critical value table to test the statistic calculated.

Number of variables (incl. )

Number of observations

50 100

200

0.72 0.89 1.05 1.19

0.38 0.48 0.58 0.68

0.20 0.25 0.30 0.35

3 4 5

Table.3: 5% critical values CRDW tests for no

cointegration

In our sample there are 2 variables and totally 164 observations therefore, from the table, the critical values are between 0.38 and 0.20, however, the 𝒹 we’ve calculated (0.050535) is way lower which strongly suggests that there is no cointegration between these two time series.

5. Conclusion

The object of this paper was to examine long-term cointegration between S&P Index and SSE Index for the period December 2001 to April 2015. We first use the Dickey Fuller test to examine the order of integration of each time series as it is a necessary step for setting up an econometric model and do inference later on. The test has the null hypothesis that the series is (1) which

is general might be hard to reject. Our results of the test is as expected, both series possess a unit root. On the basis of having two (1) series, we then employ the Durbin-Watson test to investigate the cointegration relationship. Then I found valid evidence that cointegrating relations have yet emerged between the SSE and S&P 500 markets. A possible explanation of this result could be that cointegration is long run property of variables. In the short run, the variables can be moving in different ways, driven by dynamic processes. However, cointegration ties the variables together in the long run. Also, the Chinese stock market is a bit special as being not absolutely efficient where government intervention plays an important role. “With the party cheerleading the market’s inexorable rise, it became even easier to imagine that the government was, in effect, declaring an informal debt obligation on the stock market’s future – essentially covering any bets with its own considerable assets. How could you lose? And so the bubble grew and grew: price-to-earnings ratios for Chinese stocks averaged an astounding 70-to-1, against a worldwide average of 18.5 to 1; the value of the A-shares inside China grew to be nearly double the equivalent shares of the same companies on the Hong Kong exchange. Ordinary Chinese people had become so intoxicated by bull-market euphoria that stories began to proliferate about people leaving their jobs, and even their families, to become day traders, often using funds borrowed from high-interest rate “shadow banks” or loans taken out against their homes.”- from the Guardian. However, a reasonable prediction would be that globalization is an inevitable trend and in the foreseeable future the Chinese stock market would be further integrated with the world financial system.

Maths Puzzle 2 Find all possible solutions to the ‘word sum’ on the right. Each letter stands for one of the digits 0−9 and has the same meaning each time it occurs. Different letters stand for different digits. No number starts with a zero. O D D

+ O D D E V E N

The Mathematics inside the optical fibre

By Shawn Shen (Yr 13)



Introduction of Optical Fibres Developing a highly efficient communication system has always been one of the principal interests of human beings since ancient time. In ancient China, the army used smoke which was to light the wolves’ dung on the smoke tower along the great walls to send the message in order to defence or defend, furthermore, fireworks were used to send messages and pigeon were used to send letters. However, these kinds of communication systems was either time consuming or limited by short distance. Two hundred years ago, Samuel F.B Morse devised the first commercial telegraph service using wire cables which ushered in a new era in communication – electrical communication. The electrical communication is commonly used throughout the world and an increasing portion of the electromagnetic spectrum was utilized for conveying information from one place to another in the following years. However, entering the century which is called the century of explosion of information. The copper wire is no longer capable of transmitting such tremendous amount of information. Therefore, a great interest in optical communication was created in 1960 since optical frequencies are on the order of 5 × 10 14 Hz, which means the information capacity of lasers exceeds that of microwave systems by approximately 10 million TV channels. (Keiser, 1991) Charles Kuen Gao, who is known as the father fibre optic communication, discovered some certain physical properties of glass which laid the groundwork for high-speed data communication in the information age. Since then, more than 10 billion kilometre optical fibres were installed to make internet connection faster. Optical fibre, nowadays, has become the most widely used system to transfer information. To attain a more detailed understanding of the optical power propagation mechanism in the fibre, there are two methods. The first method is to use the geometrical optics such as concepts of light refraction and reflection to provide a clear picture of the propagation mechanisms. The second method, which involves solving Maxwell’s equations subject to the cylindrical boundary conditions of the fibre, treats light as an electromagnetic wave which propagates along the optical fibre waveguide. However, the geometrical approach is only valid when the ratio of the fibre radius to the wavelength is large. So when it comes to the analysis of single- mode fibres or problems involving interference or coherence phenomena, they must be dealt with by using electromagnetic theory.

An optical fibre, by definition, is a dielectric waveguide that operates at optical frequencies. This fibre waveguide is normally in cylindrical form and it guides the light in a direction parallel to its axis. The structural characteristics determines the transmission properties of the optical waveguide, which is how an optical signal is affected as it propagates along the fibre. The propagation of light along an optical waveguide can be described in terms of a set of guided electromagnetic waves called the modes. Therefore, a mode is a pattern of electric and magnetic field lines that is repeated along the fibre at interval equal to the wavelength. Not every modes can propagate along the waveguide, only those modes which satisfy the homogeneous wave equation –Maxwell’s equations can be guided along the core of the fibre. Therefore, it is essential to solve Maxwell’s equations that give the relationships between the electric and magnetic fields to understand the optical power propagation mechanism in a fibre.  From Maxwell’s equation to Bessel function Assuming a linear, isotropic dielectric material having no currents and free charges, these equations take the form 1 ∇ × = − (1) ∇ × H = (2) ∇ ∙ = 0 (3) ∇ ∙ = 0 (4) D =∈ E and = 𝜇 . The parameter ∈ is the permittivity (or dielectric constant) and 𝜇 is the permeability of the medium. (Keiser, 1991)

∇ × 𝑉 is called the curl of the vector filed V which is a measure of the angular velocity of the fluid in the neighbourhood of that point .

∇ ∙ is called the divergence of the vector field V which is can be considered as a quantitative measure of how much a vector field diverges (spread out) or converges at any given point. (K.F.RILEY, 2006)

A relationship defining

∇ × (∇ × ) = ∇ × (−

) = −∇ × (

) = − 𝜎 𝜎

(∇ × ) (5)

Substitute Eq.2 into Eq.5 yields

2 2

𝜎 𝜎

∈ E

(∇ × ) = −

(

) = −

(

) = −∈

−

(6)

And using Eq.6 and this vector identity ∇ × (∇ × ) = ∇(∇ ∙ ) − ∇ 2 E

Since ∇(∇ ∙ ) = 0

2 2

∇ 2 E =

(7)

Taking the curl of Eq.2, we can get

2 2

∇ 2 H =

(8)

The equations 7 and 8 are the standard wave equations.

Electromagnetic light field that propagates along a cylindrical fibre can be represented by a superposition of bound or trapped modes (which are two kinds of modes). Therefore, a cylindrical coordinate system {r, ∅, z} can be established in the optical fibre. It is defined with the z axis lying along the axis of the waveguide. If the electromagnetic waves are to propagate along the z axis, the will have a functional dependence form = 0 (, ∅) (−𝑧) (9) = 0 (, ∅) (−𝑧) (10) The electromagnetic waves are harmonic in time t with radian frequency , the parameter is the z component of the propagation vector and is the main parameter in describing wave modes. There are only certain values for which will be determined from the mode fields that satisfy Maxwell’s equations and the electric and magnetic field boundary conditions at the core-cladding interface. (Keiser, 1991)

When Eq.9 and 10 are substituted into Maxwell’s curl equations, we have

Using the determinant curl ∇ × = | 1

∅ ∅

𝑧

| (Keiser, 1991)

∅

𝑧

)

𝑧 ∅

𝑧

( ∅

∅

(

)

) ∅

∇ × =

+ ∅

+ (

(

−

) 𝑧

−

= −𝜇

+ 𝜇 ∅

∅

+ −𝜇 𝑧

Equating the coefficients we get 1 ( 𝑧 ∅

) = −𝜇

+ ∅

(11)

𝑧

= 𝜇 ∅

(12)

)

( ∅

∅

(

) = −𝜇 𝑧

−

(13)

Similarly, from Eq.1

)

𝑧 ∅

𝑧

( ∅

∅

(

)

) ∅

∇ × =

+ ∅

+ (

(

−

) 𝑧

− ∅

∅

+ 𝑧

𝑧

Equating the coefficients, we get 1 ( 𝑧 ∅

) =

+ ∅

(14)

𝑧

（ 15 ）

= − ∅

)

( ∅

∅

(

) = 𝑧

−

(16)

By eliminating variables these equations can be rewritten such that, when 𝑧 and 𝑧 are known, the remaining transverse components , ∅ , , ∅ can be determined. For example, ∅ or ca be eliminated from Eqs 11 and 14 so that the component ∅ or , respectively, can be found in terms of 𝑧 and 𝑧 = − 2 ( 𝑧 + 𝜇 𝑧 ∅ ) (17) = − 2 ( 𝑧 ∅ − 𝜇 𝑧 ) (18) = − 2 ( 𝑧 − 𝑧 ∅ ) (19) ∅ = − 2 ( 𝑧 ∅ + 𝑧 ) (20) Where 2 = 2 𝜇 − 2 = 2 − 2 Substitution of Eq. 19 and 20 into Eq. 16 results in the wave equation in cylindrical coordinates 2 𝑧 2 + 1 𝑧 + 1 2 𝑧 ∅ 2 + 2 𝑧 = 0 (21)

And substitution of Eq.17 and 18 into Eq.13 leads to 2 𝑧 2 + 1 𝑧 + 1 2 2 𝑧 ∅ 2 + 2 𝑧 Then we can use the separation of variables method 𝑧 = 1 () 2 (∅) 3 ( ) 4

= 0 (22)

() (23)

As was already assumed, the time and z-dependent factors are given by 3 ( ) 4 () = (−𝑧) (24)

Since the wave is sinusoidal in time and propagates in the z direction. In addition, because of the circular symmetry of the waveguide, each field component must not change when the coordinate ∅ is increased by 2𝜋 . We thus assume a periodic function of the form 2 (∅) = 𝑣∅ (25)

Substituting Eq.24 and 25 into Eq.23, we get 𝑧 = 1

() (−𝑧) 𝑣∅ (26)

Then substitute Eq. 26 into the wave equation for 𝑧

Eq. (21) yields

2 1 2

2 2

+ ( 2 −

) 1

= 0 (27)

This is the well-known differential equations for Bessel functions. Considering the processes of solving the Eq.27 is too long which will affect the understanding of the whole picture in solving Maxwell’s equations, I will put the process of getting the solution of Bessel functions in the last part. The configuration of the step-index fibre is basically a homogeneous core of refractive index n1 and radius a , which is surrounded by an infinite cladding of index n 2 ,The reason for assuming an infinitely thick cladding is that the guided modes have exponentially decaying fields outside the core and the harmonically varying fields inside the core. For the inside region, < so the solutions to Eq. (27) are Bessel functions of the first kind 𝑣 () where 2 = 2 − 1 2 with 1 = 2𝜋 2 , these expressions are 𝑧 = 𝑣 () 𝑣∅ (−𝑧) 𝑧 = 𝑣 () 𝑣∅ (−𝑧) While outside the core the solutions to Eq. (27) are Bessel functions of the second kind 𝑣 () , where 2 = 2 − 2 2 with 2 = 2𝜋 2 . These expressions are 𝑧 = 𝑣 () 𝑣∅ (−𝑧) 𝑧 = 𝑣 () 𝑣∅ (−𝑧) The A, B, C, D are all arbitrary constants. These four short and elegant equations describe the propagation mechanism of waveguide in optical fibres.

 The process of solving Bessel equation Now, I would like to show how to solve the Bessel functions. Bessel equation is in this form 2 ′′ + ′ + ( 2 − 2 ) = 0 , v ≥ 0 The solution of y will be = ∑ + Differentiate y, we get ′ = ∑( + ) +−1 ′′ = ∑( + )( + − 1) +−2 Substitute y, y’ and y’’ in to the Bessel equation, we get

2 ∑( + )( + − 1)

+−2 + ∑( + )

+−1

+ ( 2 − 2 )∑

+ = 0

+ +∑( + )

⇒ ∑( + )( + − 1)

+ ( 2 − 2 )∑

+ = 0

+ +∑( + )

+ +∑

++2

⇒ ∑( + )( + − 1)

−∑ 2

+ = 0

⇒ ∑(( + ) 2 − 2 ) ⇒ (∑(( + ) 2 − ∞ =0

+ +∑

++2 = 0

∞

2 )

+∑

) = 0

Now we want the power of x both be the same, so we let n = k and n+2=k respectively

Then, we get

(∑(( + ) 2 − ∞ =0

∞

2 )

+∑ −2

) = 0

Write down the case of k=0 and k=1

1 +∑((( + ) 2 − 2 ) ∞ =2

{( 2 − 2 ) 0

0 + ((1 + ) 2 − 2 ) 1

) )} = 0

+ −2

Since the whole equation equals to zero, is a viable, we get 2 − 2 = 0, 1 = 0, ((( + ) 2 − 2 ) + −2 ) = 0

From 2 − 2 = 0 , 2 = 2 so

= −

Consider the case that = We get

1 +∑(( 2 + 2) ∞ =2

) = 0

(1 + 2v) 1

+ −2

Since the equation equals 0 again, so 1 + 2v = 0, ( + 2) + −2

= 0 Then we get

1 −2

= − 𝑘−2 (+2𝑣)

And

Let k = 2

− 0 (2 + 2)2 − 1 (3 + 2)3 − 2 (4 + 2)4

Let k = 3

Let k = 4

− 0 (2 + 2)2

−

→ 4

(4 + 2)4

0 2 ∗ 4 ∗ (2 + 2) ∗ (4 + 2)

→ 4

− 0 2 ∗ 4 ∗ 6 ∗ (2 + 2) ∗ (4 + 2) ∗ (6 + 2)

Then according to the pattern

(−1) 0 2 2 ∗ ! ∗ (1 + ) ∗ (2 + ) ∗ (3 + ) ∗ ( + )

Let k = 5

− 3 (5 + 2)5

− 1 (3 + 2)3

−

(5 + 2)5

− 1 3 ∗ 5 ∗ (3 + 2) ∗ (5 + 2)

According to the pattern 2+1 =

(−1) 1 1 ∗ 3 ∗ 5 ∗ … ∗ (2 + 1) ∗ (1 + ) ∗ (2 + ) ∗ (3 + ) ∗ … ∗ (2 + 1 + ) Let 1 = 0 So 2+1 = 0 Define 0 = 1 2 𝑣 ∗ Γ(1 + v) Where Γ(1 + a) = aΓ(a)

Therefore, we get

Γ(1 + v) = vΓ(v)

Substitute v=v+1

Γ(2 + v) = (v + 1) ∗ v ∗ Γ(v) Γ(3 + v) = (2 + v) ∗ (v + 1) ∗ v ∗ Γ(v) Γ(n + 1 + v) = (n + v) ∗ (n − 1 + v) ∗ … .∗ (2 + v) ∗ (v + 1) ∗ v ∗ Γ(v)

Substitute the above equation into 2

1 2 𝑣 ∗ Γ(1 + v)

(−1)

Γ(n + 1 + v) v ∗ Γ(v)

2 2 ∗ ! ∗

1 2 𝑣 ∗ vΓ(v)

(−1)

Γ(n + 1 + v) v ∗ Γ(v)

2 2 ∗ ! ∗

(−1) 2 2+ ∗ ! ∗ Γ(n + 1 + v) (−1) 2 2− ∗ ! ∗ Γ(n + 1 − v)

=𝑣 =

=−𝑣 =

Since 2+1 = 0 in the case r=v , then we get

= ∑

∞

+𝑣

= ∑

𝑁=0

∞

(−1) 2 2+𝑣 ∗ ! ∗ Γ(n + 1 + v)

2+𝑣 = ∑

2+𝑣

= ∑ 2

𝑁=0

∞

(−1) ( ) 2+𝑣 ! ∗ Γ(n + 1 + v)

𝑣

= ∑

𝑁=0

Let this equation be J v In the case = −

∞

(−1) ( ) 2−𝑣 ! ∗ Γ(n + 1 − v)

−𝑣

= ∑

𝑁=0

Let this equation be J −v Therefore, the above two equations J v

and J −v

are called the Bessel equation of

the first kind of the order v and –v respectively y = 1 𝑣 () + 2 𝑣 ()

Where

() − −𝑣

cos(𝜋) 𝑣

()

() =

𝑣

sin(𝜋)

For example 2 ′′ + ′ + ( 2 − 1 3 ) = 0 v=-3 or v=3 then all you have to do is to substitute v= 1 3

and v= − 1 3

into 𝑣

, −𝑣

and

y = 1

𝑣

() + 2

𝑣

()

1 3

(−1) 𝑛 ( 𝑥 2

) 2𝑛+

∞ =0

E.g.: Bessel function of first kind of order 3 1 3 Bessel function of first kind of order -3

= ∑

!∗Γ(n+ 4 3

)

1 3

∞

) 2−

(−1) (

−1 3

= ∑

2 3

)

! ∗ Γ (n +

Bessel function of second kind

y = 1

𝑣

() + 2

𝑣

()

1 3

() −

cos (

𝜋) 1 3

()

1 3

−

() =

1 3

sin(

𝜋)

1 2

() −

() − 2

1 3

()

1 3

()

1 3

−

√3 2

√3

Therefore,

−

1 3 ∗ Γ(

1 3

4 3

∞

) 2+

(−1) (

)

∑

4 3

16 3

)

! ∗ Γ (n +

as 1

= 0

Bibliography K.F.RILEY, M. a. (2006). Mathematical Methods for physcs ad Engineering. Keiser, G. (1991). Optical Fiber Communications.

The Fourth Dimension and Those Above It

By Stanley Traynor (Y7)

Whenever I think of dimensions, the book ‘Flatland’ springs to mind. From the perspective of a storywriter it is not a very good book, but to a mathematician it is supreme. The story is about a square who lives in a 2D world and one night has a vision of a 1D world which is populated by lines and dots. He is amazed that the inhabitants think that it is all there is to the world, and terrifies them by passing through the ‘line’ that is their 1D world and out again. Later the square is visited by a sphere who amazes him by passing through the plane of the 2D world. Then the sphere knocks the square out of the 2D world, and he can now see the 3D world. Then the square has a brainwave and asks the sphere if there are other dimensions beyond 3. The sphere silences him and tells him that the rulers of the 3D world would kill to stop the inhabitants finding out about a 4D world. The point is that there are an infinite number of dimensions, but we cannot imagine any above us. The idea of another plane than width, length, and depth is unimaginable. We can only see 4D shapes as

a 3D shadow of themselves, just as the 1D people could only see the square as a line. The 4D shapes are represented as 3D shapes as they pass through our domain. If a hypercube (4D cube) were to pass through 3D space face first, we would just see a normal 3D cube appearing and disappearing. Edge first would produce a much wider variety of shapes. It would start as a triangular orientation.

With this new knowledge, can you imagine a hypercube? No. You can’t because no human can. Our brains are simply not made for 4D space. However, we can find out about 4D space through 3D shadows of 4D shapes. The shadows, however, are so complicated that only professional mathematicians can understand them. Think of a sheet of paper. You can display the X and Y axis easily but when you try to put a Z axis into the diagram it all goes wrong,

as the paper you are drawing on is (effectively) 2D. You can easily put a Z axis onto a 3D diagram, but the W axis you cannot, as long as the diagram you are trying to put it on has less than 4 dimensions. Venturing into 4D space would be nearly impossible, as we would have to move along the W axis to get there, but to us the W axis does not exist. We cannot move along the W axis; we are fixed at 0. Another force would have

to move us along the W axis and would also have to move us back. To a 2D being, moving out of 2D space into 3D space would be impossible, and once

they had been moved by an external force they would be stuck on another 2D plane, able to move along X and Y but not Z. We would also be stuck if we were moved along the W axis, not able to see the 3D cube that we live in. What if one day, a 4D being thought of a fifth dimension? They would try to put the V axis on a 4D diagram, but it would not work. A 5D world to a 4D being is like a 4D world to us. What shapes would live in the fifth dimension? Shapes that we can only see the shadows of the shadows of. If imagining a 4D shape is impossible, the merest glimpse of a 5D shape would have two hidden dimensions. The dimensions are infinite, there is always a being that can see into your house, planet, car, and even see what you had for lunch. The world of dimensions is a puzzling one. It is best not to think about the 4D beings looking at your property and secrets. We can’t hide from the 4D creatures- they can see us wherever we go. More intelligent 4D beings might realise that there are an infinite amount of beings watching them, as there are an infinite number of dimensions. A 2364-dimensional creature might be watching you right now.

Marble on a turntable

By Minghao Zhang (Yr 12)

Just as everyone has a thousand reasons to love or hate maths, I have my own. I appreciate maths, because it can provide even the most ridiculously counter- intuitive phenomenon with a solid solution. Hardly any other forms of explanations could be as convincing as a mathematical expression. Here in this article, I would like to discuss with you one example of such problems, the bizarre behaviour of a marble ball, which turns out to fall right into the mathematical prediction. Imagine a large, flat turntable, which is rotating at constant a speed. A naughty boy placed a marble ball on it, allowing it to start rotating before letting go. What would you expect to happen? Will the marble be expelled immediately? Or will it start to orbit around the centre resembling a planet? I shall not reveal the answer right away, because that way I would ruin your interests in taking out a record player and carrying out your own ‘experiment’. While you are setting up the gadgets, we will begin our theoretical approach - first by modelling the situation and analysing the dynamics. Let the angular velocity of the turntable be Ω⃗ , and , , 𝑅⃗ and ⃗ denote the mass, moment of inertia, radius (pointing from centre of ball to surface of contact) and angular velocity of the ball bearing respectively. Since the turntable is a plane, we use a 2-D vector to represent the ball bearing’s position. We assume the force is some friction force exerted on the ball by the surface, which should be the only horizontal force that concerns us. Our goal is to work out the equation of motion of the ball bearing.

First, apply Newton’s Second Law,

2 2

= ⃗⃗⃗

Then Newton’s second law for rotation,

= 𝑅⃗ × ⃗⃗⃗ There is no slipping between the ball and the turntable, so 𝛺⃗ × = + ⃗ × 𝑅⃗

Job done for all the mechanics. With three equations and three unknowns, we should be able to acquire our solution. However, there is no division rules for vectors and this will bring us trouble in manipulating the algebra. At this critical point, the flexibility of maths comes onto stage. While Cartesian coordinate system, vector space and Argand diagrams are essentially the same thing, the quantities they carry are very different. Therefore, we are able to make one final facilitation to our equations before solving them - mapping all vector quantities onto an Argand diagram. Let = ∙ ̂ + ∙ ̂ , = ⃗ ∙

̂ + ⃗ ∙ ̂ , etc. This is viable because every one of these quantities, some after taking vector products, are in the same horizontal plane. Nevertheless, signs need to be taken good care of, for example, ‘ Ω⃗ × ’ will become ‘ × Ω ’, while ‘ 𝑅⃗ × ’ is ‘ × (−𝑅) ’. After all these preparation work, we are left with the following set of complex equations ̈ = ̇ = −𝑅 𝛺 = ̇ + 𝑅 By differentiating (3) we can substitute into (2), eliminate ̇ , and obtain an expression for , which could in turn be eliminated. Finally, we reached an ordinary differentiation equation for z.

Ω 𝑅 2 +

̈ − ̇ = 0 It is obvious that the solution will take the form = + Ω + 2 In the case of a solid ball, = 2 5

𝑅 2 . If we substitute in initial location 0

, and

initial velocity 0 are looking for.

, we can eventually complete the equation of motion that we

= ( 0

) −

where = 2 7

Ω . This is an amazing result! Substituting with Euler’s formula, = cos + sin , we can return to the vector space. The marble ball traces a new circle on the turntable! More surprisingly, it has a constant period, which is 3.5 times the period of the turntable, regardless of the mass or radius of the ball, or its position or velocity on the surface. Now switch on your record player and pick a marble, and you will discover almost exactly what is described above, provided the resistive forces play a negligible part. The result is so unexpected that barely anyone could make the correct prediction in the beginning. Although the phenomenon is counter-intuitive, mathematics does not feel it this way. Maths has its own strict disciplines for us to follow, avoiding any potential biases; while it is flexible enough to provide shortcuts to the solution. People often recognize maths as a slave for sciences, and yet in this very physical problem, it is maths who takes the dominance. Without mathematics, science could hardly turn anything descriptive into definitive.

Above all, there lies the sense of joy and relief when a convincing mathematical solution brings a problem to a satisfactory conclusion.

References Behavior of a ball on the surface of a rotating disc Am. J. Phys. 62, 151 (1994)

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