THE MATH JOURNAL
TASIS ENGLAND
2014-2015
THE MATH JOURNAL
Faculty Advisor Mr. Robert Kennedy
Editor-in-Chief Arthur Schott-Lopes
Writers Dimitrios Gotsis
Rolf Grunner-Hegge
Ethan Holzberger
Francesco Insulla
Robert Kennedy
Cameron Lowdon
Graysen Phillips
ON THE COVER
Leonhard Euler (1707-1783), Swiss mathematician and physicist.
Picture from http://www.sfcc.edu.hk/academic_subjects/mathematics/web/1999_2000_projects/Peggy%20Cheung/Mathematica/images/Guys/euler.gif
CONTENTS
Cameron Lowdon
Are We Becoming Too Dependent on Technology?
1
Francesco Insulla
Differences
2
Rolf Grunner-Hegge
Ethan Holzberger
First Mathematical Analysis: Fancy Tricks
5
Graysen Phillips
Game Theory
6
Dimitrios Gotsis
The River Problem
8
Robert Kennedy
The Charge Problem
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2014-2015
ARE WE BECOMING TOO DEPENDENT ON TECHNOLOGY?
BY CAMERON LOWDON
Technology is a wonderful thing, but consider a scenario in which 28 American
soldiers are killed due to a computer miscalculation, this really happened and so did
the destruction of an oil platform, killing people, miscalculation cost America 7 billion
dollars. We hear about these tragedies but what do we do? We cling to our calculators
and our mathematical programs and shut out the possibility that our brilliant,
apparently fail proof technology is costing lives. I can’t say that I understand what has
happened and that I have had a change of heart and that I will be more careful
because I like many love having the use of calculating tools because it makes life
simple. Now simplicity plays a part… Have we all grown so lazy that we strive to find
the simplest route to finding an answer? Are we facing a future in which
supercomputers can overtake the infinite power of the human mind? Could technology
take jobs from the scientists of NASA? These are all admittedly extreme scenarios but
not scenarios which are out of the question completely. The way I see it maths is an
art form much like paintings or sculptures or music or theatre but the difference is
that maths is seen as less of a priority but all the arts I listed would be missed by
billions, but maths, as long as someone or something can make it easier it is a dying
form and this is the sad truth. In conclusion the expectation that technology holds in
our minds is foolish and unrealistic. Do we all really want to rely on a man-made
device not much bigger than our hands to save lives and keep people safe? Now you
feel the impact, right? I can guarantee that pretty much every building we enter on a
daily basis required the use of a calculator or calculation answering device to aid in
construction, now think back to another building, the oil platform in which people
were killed due to miscalculation, now think about your house, your maths classroom,
your safety is in the non-existent hands of a calculator. Reassured enough yet?
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DIFFERENCES
BY FRANCESCO INSULLA
In a linear equation with function
y = mx + b
there is a constant set of differences between the y of x and x + 1: m (shown in
Example 1).
EXAMPLE 1
y = 2x – 1
x
0
1
2
3
4
5
6
7
8
y
-1
1
3
5
7
9
11
13
15
+2
+2
+2
+2
+2
+2
+2
+2
The difference is always 2, or more generally m, and this can be proven easily:
[m(x + 1) + b] – (mx + b)= m
But is it true that the differences are constants for equations to other powers? Let’s
see. Let’s try with the simplest quadratic equation.
EXAMPLE 2
y = x2
x
0
1
2
3
4
5
6
7
y
0
1
4
9
16
25
36
49
+1
+3
+2
+5
+2
+7
+2
+9
+2
+11
+2
+13
+2
From Example 2, we can see that the differences become constants on the 2nd “level,”
as I like to think of it. But why? Well,
(x + 1)2 – x2 = x2 + 2x + 1 – x2 = 2x + 1 (1st-level difference)
[2(x + 1) + 1] – (2x + 1) = 2 (2nd-level difference which gives us our constant)
From this we can hypothesize that an equation of degree n has a constant difference at
its nth level.
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Before trying to prove it, let’s try doing the same thing again with a slightly different
type of equation, but it shouldn’t make a difference, just to see that it wasn’t just a
coincidence.
EXAMPLE 3
y = x4 – 1
x
0
1
2
3
4
5
6
y
-1
0
15
80
255
624
1295
+1
+15
+14
+65
+50
+36
+175
+110
+60
+24
+369
+194
+84
+24
+671
+302
+108
+24
From Example 3, the hypothesis that the constant is that the degree’s level still works.
So we should try proving it now.
y = xn (original equation, generalized with n as the power)
(x + 1)n – xn (1st-level difference; we cannot determine its exact value or simplify)
The expansion of (x + 1)n depends on what n is, and is can be visualized with Pascal’s
Triangle.
FIGURE 1
PASCAL’S TRIANGLE
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Since we are subtracting xn, we are reducing the degree of the equation. So, to
visualize it we are cutting off the diagonal column of ones on the left on Pascal’s
triangle.
For x4,
(x + 1)4 – x4 = 4x3 + 6x2 + 4x + 1
becomes the 1st-level difference equation.
If we try doing this again, a really cool thing happens. The process that happened for
the x4 occurs individually on each term.
[4(x + 1)3 + 6(x + 1)2 + 4(x + 1) + 1] – (4x3 + 6x2 + 4x + 1)
Grouping the terms gives
[4(x + 1)3 – 4x3] + [6(x + 1)2 – 6x2] + [4(x + 1) – 4x] + (1 – 1)
So we are reducing the degree of each term, and the one is cancelled, not just taking
away the highest term.
(12x2 + 12x + 4) + (12x + 6) + 4 = 12x2 + 24x + 14
And then again
[12(x + 1)2 – 12x2] + [24(x + 1) – 24x] + (14 – 14) = (24x + 12) + 24 = 24x + 36
And finally
[24(x + 1) – 24x] + (36 – 36) = 24
So all that counts is the highest degree term, because the others get cancelled out.
Going back to before, when I said that we are taking the first diagonal row of ones off
every time, when we do (x + 1)n – xn or even more deeper level differences, what we are
left with is the second diagonal row (and the other rows, but they don’t matter because
they will be cancelled). The numbers on the second diagonal rows are the counting
numbers (1, 2, 3, 4, 5, 6…). We are multiplying x by that, as we reduce degree each
time (going up the rows).
What I’m trying to say is that (x + 1)n – xn = nxn – 1 because we move up the triangle
and multiply by the power (2nd number of the row).
Therefore, because of this we can conclude that the constant we are looking for of xn is
n! because we move up, reducing until we hit the one… derivatives?
Figure 1 from http://gofiguremath.org/wp-content/uploads/2013/12/Rows-0-10-and-beyond-cropped.png !4
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FIRST MATHEMATICAL ANALYSIS: FANCY TRICKS
BY ROLF GRUNNER-HEGGE AND ETHAN HOLZBERGER
Materials: Calculator and cards
What will you learn?
You will learn an interesting trick you can use to amaze children or surprise
friends. It is a trick that uses math to get surprising results by telling the person you
are doing it with to choose a number and what to do after that, the number they chose
will be on the calculator 6 times.
Procedure: First, take your five cards, and label each with a number 3, 7, 11, 13, and
37. After that, ask someone to shuffle the cards, and for them to think of a number
between 1 and 9 inclusively. Ask the volunteer to pick one of the numbered cards and
multiply their secret number by the number on the card. Then tell them to pick
another card and multiply the previous result by the number on that chosen card.
Before they press the ‘=’ button, tell them that their secret number will appear six
times on the screen.
For Example: If they choose 6, the result would be 666,666. This is because, say they
chose the prime numbers from highest to lowest, it would be 6 x 37 = 222, then 222 x
13 = 2,886, then 2886 x 11 = 31,746, then 31746 x 7 = 222,222, then 222,222 x 3 =
666,666.
The reason this will always work for numbers from 1 to 9 is because each
number that is chosen will be multiplied by all of the cards. The product of all of the
numbers on the cards, 3, 7, 11, 13, and 37, will equal 111,111, which any whole
number from 1 to 9 will multiply by to become the chosen number six times. 3, 7, 11,
13, and 37 are all prime factors of 111,111.
11 Times Rules
The reason this works is because 3 x 7 x 11 x 13 x 37 = 111,111. This number is
important because if you multiply any one digit number, you get that number as a six
digit number. The reason this works is because whenever you multiply something by
1, you get that number. So when you multiply other numbers by 11, it is the same as
multiply a number a, by (10 + 1), so it is 10a + a. If a were 6, then it + would be 10(6)
+ 6 = 66. So when you multiply 6 by 111,111, it is 6(100,000 + 10,000 + 1,000 + 100 +
10 + 1). Distribute the six and you get 600,000 + 60,000 + 6,000 + 600 + 60 + 6 =
666,666. This is the product we were looking for, because if you multiply 6 by 3, 7, 11,
13, and 37, you get 6(111,111).
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GAME THEORY
BY GRAYSEN PHILLIPS
Game Theory is "the study of mathematical models of conflict and cooperation
between intelligent rational decision-makers.”* Several famous models of this exist,
such as The Prisoner’s Dilemma, The Traveler’s Dilemma, The Rendezvous Problem,
the Keynesian Beauty Contest, and Deal or No Deal. Important terms in Game Theory
include the Nash Equilibrium, Ambiguity Aversion, and Risk Aversion.
First I would like to introduce some of these principles. In The Prisoner’s
Dilemma, for example, the Nash Equilibrium comes into play. The Prisoner’s Dilemma
is as follows: Two gang members are arrested and convicted, and put into solitary
confinement, each with no means of communication with the other. The police plan to
sentence both to a year in prison on a lesser charge. However, the police offer each
prisoner another option. Each prisoner is given the opportunity either to betray the
other, by testifying that the other committed the crime, or to cooperate with the other
by remaining silent. Here's how it goes:
1. If A and B both betray the other, each of them serves three months in prison;
2. If A betrays but B remains silent, A will be set free and B will serve a full
year in prison (and vice versa);
3. If A and B both remain silent, both of them will only serve one month in
prison (on the lesser charge).
FIGURE 1
THE PRISONER’S DILEMMA
Prisoner A Choices
Prisoner B
Choices
Stay Silent
Confess and Betray
Stay Silent
Each serves 1 month in jail
Prisoner A goes free and
Prisoner B serves one year
Confess and
Betray
Prisoner A serves one year
and Prisoner B goes free
Each serves 3 months in jail
This is where the Nash Equilibrium ties in. In essence, the Nash Equilibrium is
where neither person can gain the upper hand solely by changing their strategy, but
instead makes decisions based on the other person’s best option. In this particular
instance, you have to anticipate the other prisoner’s probable choice. In this dilemma,
most people would probably opt to betray. While there are many valid reasons for a
person to remain silent, these stray from the topic of Game Theory, so I will not delve
into them.
Another classic is The Keynesian Beauty Contest. This is a fictitious newspaper
competition, in which the newspaper has published the pictures of six women. If you
select, as the most beautiful, the woman that most people select as such, then you will
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be eligible for a prize. This demonstrates the degrees of thinking. The first degree
would be to select the woman that you thought the most beautiful. The second degree
of thinking would be to select the person that you believe the majority of the
population would select as the most beautiful. Analysts suspect that the majority of
people today are on the third degree, and that some have even reached the fourth.
I would like to bring up Ellsberg paradox, which is a situation in which you are
given two pots. One consists of 50 red and 50 blue balls, while the other consists of an
unknown quantity of each. You have to place a bet on the color of the ball drawn from
the pot of your choice. If you are prone to ambiguity aversion, or the tendency not to
take unknown risks, you will select the pot with known quantities, and bet on either
color. However, you may figure that the other pot has a 50-50 chance of having better
odds, and bet on that. But then, going back the other way, they may be minimally
better, or even greatly worse. This is why it is a paradox, because either pot makes
some logical sense. Kids, don’t gamble.
The Nash Equilibrium is the essence of game theory. Without that principle,
game theorists would be at a loss. Game theory is an elaborate concept that I could not
elaborate on enough in just this small space. I recommend duping your siblings with
the Ellsberg paradox. You could make quite the pretty penny!
*Willy, Nethanel. Game Theory. London: Culp Press, 2012.
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THE RIVER PROBLEM
BY DIMITRIOS GOTSIS
!8
Dimitrios Gotsis
Question: If you are standing a certain distance d from a river, with a laser.
You begin with the laser pointing to the river, which creates a right angle with
the river. Assume now that you are slowly turning at an angular velocity of ! .
If the river stretches to infinity, and x is the distance of the river at any given
point, what will happen to dx
dt when x is infinitely long?
When doing this problem we must consider that light has a certain velocity
and will not cover R in any time. So, when solving this problem we must
consider that the time it takes light to go and come back to our eyes, for us to
realize.
t = t 1 + t2
d✓
=!
dt
d✓ = !dt
Z
Z
d✓ = !dt
✓ = !t
x
tan✓ =
d
x
arctan = ✓
d
x
t1 ! = arctan
d
x
arctan d
t1 =
!
vt = d
ct = R
R
t3 =
c
p
2
x + d2
t3 =
c
1
t 2 = t3
2
p
x2 + d2
t2 = 2t3 = 2
c
p
x
arctan d
x2 + d2
t=
+2
!
c
Texto
1
9
At this point the only way I could find to find dx
dt was to di↵erentiate
implicitly as to make x equal some form of t seems quite difficult.
1=
1
d!(1 +
1=
x2
d2 )
⇥
2x
dx
dx
+ p
⇥
dt
dt
c x2 + d2
dx
1
(
dt d!(1 +
dx
=
dt
x2
d2 )
1
2
d!(1+ x2 )
2x
+ p
)
c x2 + d2
1
+
d
dx
=
dt
p 2x
c x2 +d2
1
p
2
c x2 +d2 +2xd!(1+ x2 )
d
p
2
d!c(1+ x2 ) x2 +d2
d
p
d!c(1 +
x2 + d2
dx
= p
2
2
2
dt
c x + d + 2xd!(1 + xd2 )
x2
d2 )
To take the limit as x approaches infinity I used mathematica.
2 p
d!c(1 + xd2 ) x2 + d2
c
lim p
= = 1.50 ⇥ 108 m/s
2
x
2
2
x!1 c x + d + 2xd!(1 +
2
d2 )
Texto
2
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THE CHARGE PROBLEM
BY ROBERT KENNEDY
11
In the diagram below, 13 equal charges, q, are placed at the corners of a regular 13-sided polygon. What is
the force on a test charge Q at the centre?
The goal is to apply the principle of superposition and to sum the forces acting on F~Q . The model is
F~Q =
✓
13
X
1 qQ
2
4⇡✏
0 R
n=1
cos
✓
2n⇡
13
◆
ı̂ + sin
✓
First consider the sum of the forces in ı̂ . This amounts to computing
cos
✓
2⇡
13
◆
+ cos
✓
4⇡
13
◆
✓
16⇡
13
· · · + cos
+ cos
◆
✓
6⇡
13
+ cos
◆
✓
8⇡
13
◆
+ cos
+ cos
✓
20⇡
13
◆
cos
2n⇡
13
+ cos
18⇡
13
◆
✓
P13
We will use the following identities to compute
n=1
✓
2n⇡
13
P13
n=1
10⇡
13
◆
+ cos
+ cos
✓
22⇡
13
+ cos( + ↵) + cos( + 2↵) + · · · + cos( + n↵) =
N
X
cos(n✓) =
n=1
First apply (3) by letting
cos(0) + cos(0 +
= 0 and ↵ =
13
X
n=1
cos
✓
2n⇡
13
cos
, which is
12⇡
13
◆
+ cos
+ cos
✓
24⇡
13
◆
✓
14⇡
13
◆
+ ···
+1
⇣
(n+1)↵
2
⌘
· cos
+
n↵
2
(2)
sin ↵2
1 sin N + 12 ✓
+
2
2 sin ✓2
(3)
2⇡
13
2⇡
2⇡
2⇡
) + cos(0 + 2 ) + · · · + cos(0 + 13 ) =
13
13
13
1+
◆
✓
(1)
.
sin
cos
◆ ◆
ˆ⌘
2n⇡
13
◆
=
sin(⇡) cos
13
X
n=1
⇡
13
sin
cos
✓
2n⇡
13
12
1
◆
+ sin
⇡
13
=0
sin
⇡
13
⇣
(13+1)( 2⇡
13 )
2
cos(⇡)
⌘
sin
=1
⇣
· cos 0 +
⇣ 2⇡ ⌘
13
2
13( 2⇡
13 )
2
⌘
Second, applying (4) yields
13
X
cos
n=1
13
X
✓
cos
n=1
2n⇡
13
◆
✓
2n⇡
13
◆
1 sin 13 + 12
+
⇡
2
2 sin 13
=
1 sin 13 + 12
+
⇡
2
2 sin 13
=
2⇡
13
2⇡
13
⇡
+ sin
1 sin(2⇡) cos 13
+
⇡
2
2 sin 13
=
⇡
13
cos(2⇡)
=0
You may have been attempted to apply the cofunction identity which yields
cos
✓
2⇡
13
◆
+ sin
···
✓
cos
5⇡
26
✓
◆
⇣⇡⌘
+ sin
26
3⇡
13
◆
2 sin
✓
sin
✓
sin
3⇡
26
◆
✓
3⇡
26
◆
cos
✓
3⇡
13
◆
cos
⇣⇡⌘
13
cos
⇣⇡⌘
···
13
✓ ◆
✓ ◆
⇣⇡⌘
5⇡
2⇡
+ sin
+ cos
+1
26
26
13
sin
which simplifies to
1
2 sin
⇣⇡⌘
26
3⇡
26
◆
+ 2 sin
✓
5⇡
26
◆
2 cos
✓ ◆
⇣⇡⌘
2⇡
+ 2 cos
13
13
2 cos
✓
3⇡
13
◆
This answer is equal to zero but it is not obvious, in my opinion; however, interestingly enough, applying the
cofunction identity to the ˆ⌘ components does sum neatly to zero.
13
X
sin
n=1
✓
2n⇡
13
◆
= sin
⇣⇡⌘
sin
13
···
✓
◆
+ cos
✓
◆
2⇡
13
sin
3⇡
13
✓
5⇡
26
cos
◆
✓
✓ ◆
✓ ◆
⇣⇡⌘
⇣⇡⌘
3⇡
3⇡
+ cos
+ cos
+ sin
+ sin
+ ···
26
26
13
13
3⇡
26
◆
⇣⇡⌘
cos
26
cos
✓
5⇡
26
◆
sin
✓
2⇡
13
◆
+0=0
Therefore, this static arrangement is in equilibrium since the net force is equal to zero. What does this imply
about the geometry?
The previous two results suggest that it is possible to derive a general formula that demonstrates the net
force on a test charge located at the centre of a regular polygonal configuration of m charges, q, is zero.
F~Q =
m
X
1 qQ
4⇡✏0 R2
n=1
✓
cos
✓
2n⇡
m
◆
ı̂ + sin
✓
2n⇡
m
◆ ◆
ˆ⌘
(4)
Again consider the components independently and show that each sum is zero. Applying Lagrange’s identities
to each case as appropriate yields
m
X
cos
n=1
✓
2n⇡
m
◆
=
1 sin m + 12
+
⇡
2
2 sin m
2⇡
m
⇡
+ sin
1 sin(2⇡) cos m
+
⇡
2
2 sin m
=
⇡
m
cos(2⇡)
=0
similarly using the appropriate Lagrange identity for sine
m
X
n=1
sin
✓
2n⇡
m
◆
=
⇣⇡⌘
1
cot
2
m
cos
m + 12
⇡
2 sin m
2⇡
m
=
cos
2 sin
13
2
⇡
m
⇡
m
cos(2⇡) cos
⇡
m
sin
2 sin
⇡
m
⇡
m
sin(2⇡)
=0