The Axiom - Issue 17 - St Andrew's Day 2024

THE
AXIOM
THE
AXIOM
THE
AXIOM
Issue 17
St Andrew’s Day ’24

1 | The Axiom
Editors’ Note
Dear reader,
At long last, the wait is over: The Axiom’s St. Andrew’s Day edition is here.
We believe that Maths is not just a rigorous and precise discipline, but also an
elegant field that requires profound creativity. It is this beauty of Maths that we
aim to display to you in this edition. The selection of articles is wide-ranging and
contains some of the most fascinating topics within the subject. From a ranking of
proofs of the Pythagorean Theorem; to how Maths is used to compose music; to
what speedcubers are really thinking about, and so much more, there is certainly
something for everyone.
The Axiom provides an opportunity to express this mathematical creativity outside
the classroom. Be it through drawing up an elegant solution to a challenging problem;
introducing readers to an interesting area of Maths; or editing articles for the
publication, it is a great way to engage with the subject on a supercurricular level and
constitutes a respectable paragraph on a personal statement for any STEM course. To
get involved, contact the Editors-in-Chief below.
Without further ado, get stuck in! Of course, this has only been possible because
of the hard work of our writers, editors, and designers, whose efforts we greatly
appreciate. As ever, we are grateful for Dr Moston’s ongoing support for this
publication.
Happy reading,
The Editors-in-Chief
Rajas Nanda KS  
Jacob Potter KS
Ralph Matta ma OS
Designer-in-Chief
Ahnaf Kabir OS
Master-in-Charge
Dr Moston - JM
Editors
Aaryan Agnihotri KS
Akaash Agnihotri OS
Giacomo Amiti
Joshua Fernandes
Mungo Forbes-Cable
Charles Marsden
William Moorhouse OS
Gaoyang Qiu OS
Giacomo Rubino ma
Leo Sun
Rayaan Vyas KS

St Andrew’s Day ’24 | 2
Contents
G. Qiu OS
Ranking Proofs of the Pythagorean Theorem 03
Forbes-Cable
Prime Numbers: The Foundation of Internet
Security 13
A. Agnihotri OS
Electron Geometry 35
L. Sun
Pascal’s Wager and the Interconnectedness of
Philosophy and Mathematics 09
Vansh ma MS
Composing with Cubes 27
Amiti
The Beauty of Chaos: Exploring the Butterfly Effect
in Mathematics 20
Marsden
How Often Do the Planets Align? 07
Zeng MS
Unsettled Relations between Correlation and
Causation 17
Tjandramaga
3x + 1 11
A. Agnihotri KS
Functional Equations 38
Moorhouse OS
Gauss’ Law: Differentials and Integrals 42
Vyas KS
‘The Holy Grail of Mathematics’ — The Riemann
Hypothesis 31
Rubino ma
Cracking the Cube: Maths Behind The Rubik’s Cube 23

3 | The Axiom
Ranking Proofs of the Pythagorean Theorem
G.Qiu OS
1 Introduction
a2
+ b2
= c2
I’m sure that the vast majority of our distinguished readers will recognise this formula, even if they
do not study maths as a specialist, or even if they left maths behind a long long time ago. The
Pythagorean Theorem is one of the most recognisable and iconic theorems in all of maths. Named
after the Greek philosopher Pythagoras, born around 570 BC, it has been proved through numerous
di!erent methods. However, Pythagoras became such a figure of legend after his death, that we
are sceptical about whether he actually proved this. It is also reputed that he was so delighted
with his proof that he sacrificed one hundred cows to the gods. We do know that a version of
the theorem was being used in what is now Iraq, over twelve centuries before Pythagoras’ lifetime.
As such, many proofs have been developed for this theorem. I will take you through some of my
favourites and give my opinion on each.
2 Proof by similar triangles
Einstein is reputed to have discovered this proof in his boyhood.
Let us construct a right angled triangle, with shorter sides a and b, and with the longest side
opposite the right angle being c. The angles are 90→
, ω, and 90→
→ ω. Then, draw a straight line
from the right angle to the opposite side c, so that the angle between the drawn line and c is a
right angle.
As seen from the diagram, we have constructed three similar triangles: ”ABC, ”ADB, and
”BDC (AA similarity). Then,
CB
AC
=
CD
CB
and
AB
AC
=
AD
AB
So,
CB2
= AC ↑ CD and AB2
= AC ↑ AD
Further,
CB2
+ AB2
= AC ↑ CD + AC ↑ AD = AC(CD + AD) = AC2
Thus,
CB2
+ AB2
= AC2
(Q.E.D.)
My opinion: This proof relies on similar triangles, and thus is accessible to most people who
have done a degree of maths to a secondary level. It is very carefully constructed, so that one

can cross-multiply to get a squared side. However, it took me great pain to find out which sides
would do this, through meticulously finding which sides are similar and then finding the right
combination. Props for simplicity, yet it was a lot more challenging than I first expected. 3/5
3 Experiment Time
Suppose that you had a right-angled triangle, with sides a, b, and h. h is the longest side and is
opposite the right angle. Attached to each side is a square. Thus, the square attached to a has
area a2
, the square attached to b has area b2
, and the square attached to h has area h2
.
We then carry out the following experiment:
Fill up the square attached to side a and the square attached to side b with liquid. The area
that the liquid occupies is a2
+ b2
. Then, take out all the liquid, and pour it into the square at-
tached to side h. You will find that it fits exactly. Thus, we have proved that the sum of the square
of the two shorter sides is equal to the square of the longest side, and thus the Pythagorean theorem.
My opinion: The only positive thing I have to say about this is that this can, in fact, be
verified by someone who has no capacity to do maths, and has access to this apparatus. However
overall, I am not a fan. I am very wary of the reference to ’the area that the liquid occupies’. In
addition, the only way that this proof works is if carried out empirically, with this very specific set
of apparatus. Although we may be able to feel this intuitively, there is no way for us to prove it
through maths alone. 1/5
4 Algebraic Proof
Once again, let us construct a right angled triangle, with sides of length a, b, and c. The side c is
the longest and is opposite the right angle. Now, let us make three more identical copies of this
triangle, and arrange it so that we get two squares. It is clear that the area of the larger square is
equal to the sum of the area of the smaller square and the area of the four triangles. Thus,
(a + b)2
= c2
+ 4 ↑
1
2
ab
a2
+ 2ab + b2
= c2
+ 2ab
a2
+ b2
= c2
My opinion: I like this proof; very little algebra is required. It is low e!ort: you do not need to
grope around looking for which sides are similar or for apparatus. Although it takes a slight leap
of imagination to construct the 2 squares initially, I am sure that most people would be able to.
After that, the equation derives itself very nicely. 5/5

5 | The Axiom
5 Euclid’s Proof
This is taken from Euclid’s Elements, written around 300BC, around two centuries after Pythago-
ras’ death. The first chapter ends with this geometric proof.
Again, we have a right angled triangle. It has points A, B, and C. We then build squares on each
side. In order to prove the theorem, we need to show that the sum of the areas of the two smaller
squares is equal to the area of the largest square, attached to the hypotenuse.
We drop down a perpendicular from the right angle at C to the hypotenuse and across to the
far edge of the square, constructing the straight line CKH. Now, we are going to draw some lines
in the diagram to help us spot some congruent triangles. Although it is possible to use similar
triangles, that has already been used in a previous proof, and Euclid does not actually introduce
similarity until book six of ‘Elements’, so it is better if we do without.
First draw in BD and CE.
Area of ”DAB =
1
2
DA ↑ AC = Half the area of the square DACL
”CAE and ”DAB are congruent (S.A.S.) as DA = CA and BA = EA and ↭DAB = ↭CAE =
90→
+ ↭CAB. We also have:
Area of ”CAE =
1
2
AE ↑ EH
Thus, the area of the triangle CAE is half that of the rectangle AKEH. Equally,
Area of ”DAB =
1
2
DA ↑ AC
The area of the triangle DAB is half that of the square LCAD.
Since the triangles are congruent, the area of the rectangle AKEH and the square LCAD are
equal. The equivalent line of reasoning can be applied to the square and rectangle on the right,
to give that the area of the rectangle and the square are equal. Since each rectangle has the same
area as the square on its side, it is evident that the area of the squares sum to give the area of the
largest square, which is comprised of the two rectangles. Thus, we have proved that the sum of
the squares of the two shorter sides sum to give the square of the largest side.

My opinion: Although it is rather nice to be walking the same path as one of the ancient
mathematicians, I feel that this proof is quite clunky. It makes use of the area of a triangle for-
mula, although it takes some time to identify which lines to construct and which sides are the
base and height, it is simple and original. It has the same set up as the proof which requires an
experiment, and accomplishes the same in a much nicer way. 4/5
6 Di!erentiation
Let us consider a right angled triangle ABC with shorter lengths a and b, and hypotenuse of length
c. Then, let us construct a coordinate system so that the vertices of the triangle have coordinates
(0, 0), (a, 0), (a, b). Finally, draw a semicircle centred at the origin with radius c, passing though
the points (a, b) and (0, c).
Since point P lies on the semicircle, the line running through P is perpendicular to the radius from
the origin and is a tangent to the semicircle. It follows that the semicircle must be the graph of a
solution to:
dy
dx
= →
x
y

y dy =

→x dx
y2
= →x2
+ C
Since the circle passes through (0, c), C = c2
. But, the circle also passes through (a, b), so we get:
a2
+ b2
= c2
My opinion: This proof is quite intuitive, although it can be bettered if we use proper limits
instead of dy and dx. However, to use A Level maths on Year 6 maths seems like overkill. Indeed,
I doubt that this is the easiest, most aesthetic, or neatest proof of Pythagoras’s Theorem. 2/5
7 Conclusion
So, my ranking, from worst to best, goes as follows: Experiment Time, Di!erentiation, Similar
Triangles, Euclid’s Proof, and finally Algebraic Proof. Although originality is always to be ap-
plauded, the best proofs will be concise, require maths alone, and not require bounds of assumed
knowledge in order to understand it.
References
[1] Umb.edu. (2021). Pythagorean Theorem and its many proofs. [online] Available at:
https://www.faculty.umb.edu/gary zabel/Courses/Phil%20281b/Philosophy%20of%20Magic/
Arcana/Neoplatonism/Pythagoras/index.shtml.html.
[2] Head, A. (n.d.). Proofs of the Pythagorean Theorem. [online] jwilson.coe.uga.edu. Avail-
able at: https://jwilson.coe.uga.edu/EMT668/EMT668.Student.Folders/HeadAngela/essay1/
Pythagorean.html.
[3] brilliant.org. (n.d.). Proofs of the Pythagorean Theorem — Brilliant Math Science Wiki.
[online] Available at: https://brilliant.org/wiki/proofs-of-the-pythagorean-theorem/.

7 | The Axiom
How Often Do the Planets Align?
Marsden
1 Introduction
Throughout history, the alignment of planets has grabbed the attention of scientists and philoso-
phers alike. In early Mesopotamia, it was commonly believed that gods indicated future events to
mankind through omens, which could be read through astronomy and astrology. A primary source
of our knowledge of the relationship Mesopotamians had with omens is the Omen Compendia, a
Babylonian text that was first started at the beginning of the second millennium BC. In Assyrian-
Babylonian religion, Ea was the god of water, knowledge, and creation, and was thought to be the
cause of these foreshadowing events. The Compendia contains information on rites to avert evil
or “namburbu,” including events that took place before and during planetary alignments. Since
omens arising from the planets were produced without any human interaction, they were seen as
especially powerful, so being able to predict their frequency was vital.
On 27 February 1953 BC, Chinese astronomers observed the conjunction of 5 planets in the sky:
Mercury, Venus, Mars, Saturn, and Jupiter, the first four spread out over an angular diameter of
just half a degree. Planetary alignments were associated with significant changes in China’s power
dynamics. Given how important these events were, you might ask, “how often do they occur?”
2 Derivation
We’ll start by using Earth and Mars as an example. Kepler’s third law of planetary motion says
that for Solar System planets,
T2
= a3
(1)
where T is the time period of a planet’s orbit around the Sun (measured in years) and a is the
length of the orbit’s semi-major axis (measured in Astronomical Units).
Angular velocity, ω, is a measure of how the angular position of an object changes with respect to
time. Since the period T is the time taken for one complete orbit, we can say that
ω =
2ε
T
. (2)
Since we want to find the time between successive alignments, it would be helpful to know the
daily angular velocities. For Earth, this is 2ε/365.25 radians per day. For Mars, we need to work
out its period using Kepler’s third law. Its distance from the Sun is 1.52368055 AU, so its period
T = 1.52368055(3/2)
→ 365.25 = 687 days. Our two expressions for angular velocity are therefore
ωE =
2ε
365.25
and ωM =
2ε
687
.
The rate of angular separation is simply ωE ↑ ωM , and using equation (1), we can work out the
time it takes for one full orbit at this angular velocity, referred to as the synodic time period.
Tsyn =
2ε
ωE ↑ ωM
= 779 days. (3)

(a) Initial aligned position of Earth and Mars (in op-
position)
(b) Position of Earth and Mars after a time interval t
Figure 1: Illustration of Earth and Mars positions over time.
We can generalize this to find a formula for the synodic period where ω1 ω2:
1
Tsyn
=
1
T1
↑
1
T2
. (4)
3 Conclusion
Congratulations! You can now predict oppositions and learn when the next omens are coming,
just as various civilizations have been doing for thousands of years.
References
[1] H. Hunger and D. Pingree. “Astral Sciences in Mesopotamia”. In: Handbook of Oriental Studies
44 (1999).
[2] H. Hunger and D. Pingree. “Astral Sciences in Mesopotamia”. In: Handbook of Oriental Studies
44 (1999), pp. 1–33.
[3] D. F. Nivison. Library of Sinology, Volume 1: The Nivison Annals. De Gruyter, 2018.
[4] D. W. Pankenier. “Astralogical Origins of Chinese Dynastic Ideology”. In: Vistas in Astronomy
39 (1995), pp. 503–516.
[5] D. W. Pankenier. “Mozi and the Dates of Xia, Shang, and Zhou: A Research Note”. In: Early
China 9/10 (1983–1985), pp. 175–183.

9 | The Axiom
Pascal’s Wager and The Interconnectedness of
Philosophy and Mathematics
L.Sun
1 Introduction
Although perceived as a humanity and a science respectively and thus disparate, in mod-
ern times, the studies of Philosophy and Mathematics have always been interconnected.
For example, Pythagoras’s school of philosophical thought, most notably his teachings
about the material world have also had strong influence throughout history. In this
article, we will explore Pascal’s Wager, which links theology with probability theory.
2 Pascal’s Wager
Proposed by Blaise Pascal in his posthumously published Pensées, Pascal’s Wager is
an argument for why one should believe in the existence of God from a perspective of
personal gains and losses. First, the Law of Excluded Middle states that God either exists
or does not exist. Furthermore, Pascal assumes that we are uncertain of whether God –
and, crucially, thus the afterlife – exists.
Then, Pascal o!ers a wager on the existence of God. If God does not, in reality, exist,
then belief in God’s existence is irrelevant and the outcome is invariant: no life after
death. Yet, if God exists, belief in God’s existence will grant eternal life in heaven, but
refusal to do so will result in misery (see Romans 1:18-20). As we live in uncertainty, we
must wager, for any middle ground is not wagering for God and is, therefore, wagering
against God. Thus, we can generate the following table of outcomes.
God exists God doesn’t exist
For God Heaven Nothing
Against God Misery Nothing
The potential outcomes of wagering for God are always at least as good as the potential
outcomes of wagering against God. Note that the probability of God’s existence does not
actually a!ect the argument, as even if it approaches 0, wagering for God will still have
a better expected gain. Therefore, through a basic probability argument, Pascal presents
why one should believe in the existence of God.
Let us now look at the expected gain in an algebraic sense. Our table becomes:
where we assume y to be finite (though the true nature of y is beyond the scope of this
article). As rational beings, we try to maximise the expected gain E(X) =

xP(X = x),

God exists God doesn’t exist
For God → 0
Against God y 0
where X is our random variable defined above, with probability p of God existing, for some
fixed value p. In the case of wagering for God, the expected utility is →(p)+0(1↑p) = →.
When wagering against God, the expected utility is y(p) + 0(1 ↑ p) = py →. Thus,
wagering for God produces the maximum utility and thus is the optimal choice.
3 Criticisms of the wager
3.1 The ’many Gods’ argument
Pascal makes the mistaken assumption that the Catholic conception of God is the only
possible way in which a God can exist. Other deities existing is also an outcome that
must be considered. Thus, the table of outcomes becomes increasingly complex, for if
we wager for God and God does not exist, we have to take into account the possibility
(and probabilities) of other Gods’ existences and their attitudes towards heathenism.
Therefore, Pascal’s argument breaks down.
3.2 p as undefined or 0
The first premise of the argument relies on the probability of God’s existence being
defined. However, it might be the case that it is not possible to assign this quantity a
value. Unlike other probabilistic experiments, such as a coin being flipped, the existence
of God might be transcendental. That is, God’s nature may not be governed by the laws
of Mathematics we have invoked. The wager fails.
On the other hand, atheists might use arguments such as the presence of evil and the
inconsistent triad to arrive at the conclusion that the p = 0. If so, then wagering for or
against God will give us an expected utility of 0, which is trivial.
4 Conclusion
Through the problem of Pascal’s Wager, the interconnectedness between theological and
philosophical questions and Mathematics is displayed, and hopefully you can appreciate
that they both concern fundamental truth and employ systems of logical reasoning and
deduction.
References
[1] Wikipedia Contributors (2019). Plato. [online] Wikipedia. Available at:
https://en.wikipedia.org/wiki/Plato.
[2] Wikipedia Contributors (2024). Pascal’s wager. Wikipedia.
[3] Hájek, A. (2017). Pascal’s Wager (Stanford Encyclopedia of Philosophy). [online]
Stanford.edu. Available at: https://plato.stanford.edu/entries/pascal-wager/.

11 | The Axiom
3x + 1
Tjandramaga
The Collatz Conjecture, often referred to as the hardest problem in the world, at first may seem
trivial and benign due to the simple arithmetic operation given in the title. However, it is a
problem that has tormented mathematicians since 1937, when Lothar Collatz first introduced it.
Since then, the world of mathematics has been deemed ‘not ready for such problems’ by Paul Erdős,
and the mathematician Je!rey Lagarias even went as far as to say that it ‘is an extraordinarily
di”cult problem, completely out of reach of present-day mathematics.’
Many of you at this point are probably wondering what makes this problem so hard. To start o!,
the conjecture asks whether repeating two simple arithmetic operations would eventually transform
every integer into 1. These operations di!er depending on whether the integer is odd or even. If
even, the integer is to be divided by two; if odd, the number is to be multiplied by 3 and then
added by 1 (hence 3x + 1). These two operations are then repeated infinitely, and ultimately (to
our knowledge) the integer always ends up being 1 after however many operations.
Thus it becomes obvious that anyone can understand this simple concept. However, the more
challenging aspect is trying to prove it. Mathematicians have put the conjecture to the test,
examining integers up to 2.95→1020
(a little under a sextillion) using computers. Despite the wide
range of numbers that have proven the conjecture true, no general proof has been found.
Mathematicians have approached this problem in di!erent ways, with some dedicating years to
finding a proof for this seemingly impossible problem.
Figure 1: Visualization of the Collatz conjecture
However, what I find perhaps even more interesting than the attempts to solve this complex
problem are the visualizations mathematicians have created of the conjecture.
This organic, feather-like visualization of the conjecture is produced like this: each line follows
a sequence, with each operation causing the line to bend up or down depending on whether the
number is being divided by 2 or multiplied by 3 with 1 added on.
Another interesting observation about the conjecture is that if you take some large number n at
random, the first digits of the numbers produced by the sequence roughly follow Benford’s law.
The law uses empirical data to state the probability and frequency with which the numbers 1
through 9 should appear in the natural world. Many real-life large datasets follow this law to
detect fraud or tampering in things such as tax returns and financial records.

Figure 2: Benford’s Law
However, again, no one has been able to explain why this is true.
All in all, the Collatz Conjecture has been a mathematical mystery for years, haunting hundreds
of talented mathematicians for this past century. Newcomers to the mathematical world are even
warned not to tread down the slippery and dangerous slopes of the conjecture. However, despite all
of this stigma as the ‘world’s hardest problem,’ Terence Tao (recognized by many as the greatest
mathematician of the century) confronted this problem and made a breakthrough in 2019. He
managed to prove that at the very least, the conjecture is almost true for almost all numbers.
This may not seem to reveal much, but it is still considered a major breakthrough in the Collatz
conundrum and is a much bigger breakthrough than what Tao had originally expected to achieve,
simply highlighting the di”culty of this conjecture and solidifying its place as the ‘world’s hardest
problem’.
References
[1] Wikipedia. (2020). Collatz conjecture. [online] Available at: https://en.wikipedia.org/wiki/
Collatz conjecture.
[2] Muller, D. (2021). The Simplest Math Problem No One Can Solve - Collatz Conjecture. [online]
www.youtube.com. Available at: https://www.youtube.com/watch?v=094y1Z2wpJg.

13 | The Axiom
Prime Numbers: The Foundations of Internet Security
Forbes-Cable
Every time you log in to Instagram, make an online purchase, or even send a text, your personal
information is kept safe by a wall of encryption, constructed on a foundation of prime numbers.
The idea of a prime number may seem quite simple; however, their unique nature makes them an
integral part of the digital world we live in. To understand just why they are so vital, we need to
look at the qualities of a prime number and why they are significant in the field of encryption
known as RSA cryptography.
1 Encryption: What is it and how does it work?
To understand why primes are important to cryptography, we first need to understand how
cryptography works.
Imagine you want to send a secret message to your friend, so you lock it in a box with a special
lock and send it. This lock is special in that it can be locked by anyone, but once it is locked, it
can only be opened by a unique key. In cryptography, this “key” represents a private key, and
the “lock” represents what is known as a public key. The message represents the transfer of data.
The process of sending this message can be split into two parts: Encryption and Decryption.
1.1 Encryption
Imagine everyone has access to this box; anyone can place a message in it and lock it. The box
becoming locked represents encryption—it now cannot be accessed without the private key, and so
it is safe from anyone else.
Decryption
Once the locked box is received, it can be opened using the unique key. This represents
decryption. This can only be done by the desired recipient who has the private key. The beauty of
this is that it is secure, simple, and scalable. However, problems arise when actually making this
“lock.” It needs to be accessed by everyone in one direction (locked) and only one person in the
other (being opened). This is where prime numbers come in.
2 Prime Numbers: Building Blocks of Mathematics
A prime number is defined as a number that has factors of only 1 and itself. The
Fundamental Theorem of Arithmetic states that every integer greater than 1 can be factored

into prime numbers. Therefore, when a prime is multiplied by another prime, its only factors are
those two prime numbers. No other number has the same prime factorisation. It is this intrinsic
quality of primes that makes them so valuable to cryptographers and the RSA algorithm.
3 RSA Cryptography
RSA cryptography is one of the most widely used methods of online encryption. It is based on the
fact that multiplying two very large prime numbers together is relatively easy, but factorising the
result back into prime factors is incredibly di!cult. The first step of RSA encryption is creating a
public key:
Step 1 - Pick Two Very Large Prime Numbers
Let these primes be p and q (each is usually hundreds of digits long). Multiply them together to
create n, which makes up half of the public key and will serve as the modulus for both the
encryption and decryption process.
n = p → q
Example:
Let p = 23 and q = 41:
n = 23 → 41 = 943
Step 2 - Compute Euler’s Totient Function ω(n)
This function calculates the number of integers less than n which are coprime to n.
ω(n) = (p ↑ 1)(q ↑ 1)
In this example:
ω(943) = (23 ↑ 1)(41 ↑ 1) = 880
Step 3 - Choose a Value e Which is Coprime to ω(n)
Select an integer e such that 0 e ω(n) and that e and ω(n) do not share any prime factors
except 1, so they are coprime.
In our example, let e = 7.
It is necessary that e is coprime to ω(n) so that there is an inverse of e modulo ω(n). This will be
required in the decryption of the message.
This value e makes up the other half of the public key. Therefore, e and n are publicly available
and are all that is necessary to send a message to the owner of the private key.

15 | The Axiom
Step 4 - Calculate the Encrypted Message c
Now for simplicity, say you want to send your friend a number, for example, 35. The encoded
message (m) can be calculated by:
c = me
mod n
In our example:
357
mod 943 = 545
So, c = 545 is what is sent.
This formula ensures the message is su!ciently encoded so that it cannot be reversed without the
private key. It is, therefore, safe from interception. (If you have any doubt, try solving 545 = x7
mod 943.)
4 Decryption
Step 1 - Calculate the Private Key d
This is calculated using the formula:
de ↓ 1 mod ω(n)
For our example:
7d ↓ 1 mod 880
d = 503. This value can only be calculated with knowledge of the numbers p and q (as
ω(n) = (p ↑ 1)(q ↑ 1)) and is thus private. It also allows us to decrypt c, making it a private key.
Step 2 - Decrypt c back into m
To decrypt c back into m, the formula used is:
m = cd
mod n
For our example:
545503
mod 943 = 35
5 Proof of RSA
The modular arithmetic exponentiation law states that when A, K, n ↔ Z and A n, then:
AK
mod n = (A mod n)K
mod n
Euler’s theorem states that if a and n are coprime positive integers, then:

aω(n)x
mod n = 1
Rearranged for the RSA encryption:
ed = 1 mod ω(n)
ed = ω(n)x + 1, x ↔ Z
As me
mod n = c, cd
mod n is equal to:
(me
mod n)d
mod n
Using the modular arithmetic exponentiation law, this can be rewritten as:
med
mod n
Now, using Euler’s theorem, ed can be substituted for ω(n)x + 1, so the equation becomes:
mω(n)x+1
mod n
m → mω(n)x
mod n
Substituting aω(n)x
mod n = 1 into the equation leaves just m, proving that:
(me
mod n)d
mod n = m
Therefore:
me
mod n = c =↗ cd
mod n = m
(This is only true when m is coprime to n. The chance of this not occurring is negligible.)
6 Conclusion: The Future
As previously highlighted, RSA encryption relies on the extreme di!culty of factorising the
product of two very large primes back into its prime factors. It has been key to the safety of the
online world since it was created in 1977. However with the prospect of improvements in quantum
computing, RSA becomes vulnerable. In response to this threat cryptographers are developing
new methods of protecting data which do not rely on prime factorisation such as algorithms which
rely on multivariate polynomials instead.
References
[1] Brilliant (2010). RSA Encryption — Brilliant Math Science Wiki. [online] Brilliant.org.
Available at: https://brilliant.org/wiki/rsa-encryption/.
[2] Wikipedia. (2019). RSA (cryptosystem). [online] Available at:
https://en.wikipedia.org/wiki/RSA (cryptosystem).
[3] Tom Rocks Maths (2023). How does RSA Cryptography work? YouTube. Available at:
https://www.youtube.com/watch?v=qph77bTKJTM.

17 | The Axiom
Unsettled Relations between Correlation and
Causation
Zeng MS
1 Introduction
Correlation and causation have been entangled for centuries over a number of fields, in-
cluding philosophy, maths, statistics, economics, etc. From Aristotle’s causal pluralism,
Karl Pearson’s discovery of correlation, to Judea Pearl’s causal diagrams, the heated
discussion between their relationship is ongoing. However, the argument becomes in-
creasingly scientific when statistical and mathematical methods are used to examine
their relationship.
2 Equations and diagrams
Both the scatter plots in Figure 1 show positive correlation between two variables. Can
we truly draw the conclusion of causation? Of course not. Well, why is the number of
ice cream sales per month linked to shark attacks then? Perhaps eating ice cream makes
you taste better. It is clear, not mathematically, but in a more broad rational sense, that
one does not cause another. It is quite reasonable to suggest that, instead, they are both
caused by a third variable. However, there exists a well-researched causal relationship
between smoking and lung cancer, wherein smoking directly increases the likelihood of
developing lung cancer, according to medical study.
Figure 1: Scatter plots showing correlation
A correlation analysis does not necessarily lead to a causation conclusion. Correlation
does not imply causation. In other words, mathematical relationships between quantities

do not express anything about whether the quantities are causally related. Let us examine
the two equations below.
(1.a) : Y = 4X, (1.b) : Z = Y + 1 (1)
(2.a) : X =
1
4
Y, (2.b) : Y = Z → 1 (2)
Equations 1 and 2 represent the relations of X ⊋ Y and Y ⊋ Z, and are equivalent
to each other, from the mathematical viewpoint. Meanwhile, we can use a represen-
tation like circuit diagrams to show their relationship. For example, we could use the
multiplier (ω), adder and subtracter gates to represent the two equations in Figure 2 (a)
and (b) respectively. They are able to predict the output from the input through the
arrow directions. When we set Y = 0, both equations result in the identical solutions
X = 0 and Z = 1. However, in the diagram of Figure 2 (a), Y = 0 leads to Z = 1 and
no constraint is placed on X; while the diagram of Figure 2 (b) leads to X = 0 and un-
constrained Z. Obviously, the diagram o!ers the capability to interpret the impact of Y
when it is manipulated, which is conducted in an interventional study. The intervention
is to execute operations in the equations using di!erent values, but has to be guided by
the diagram. The causality is to predict the consequences of such interventions. Unfor-
tunately, diagrams are not considered to be formal mathematical language and are not
most welcomed by statisticians. But the strength of diagrammatic representations along
with formal equations do pave the way to explore causality.
* 4 + 1
X
Y
Z
(a)
* 1/4 - 1
X
Y
Z
(b)
Smoking
Lung
Cancer
Genes
Figure 2: Diagrammatic representations of equations to encode causality
In a similar vein, we can represent the causal relation between smoking and lung cancer
through a directed graph, as shown in the right panel of Figure 2. The arrow points the
cause (smoking) to the e!ect (lung cancer). The magnitude of the arrow is quantified
as the correlation degree, which comes from statistical calculation of the data shown in
Figure 1. The directed graph could become complicated when more variables, as well
as their causal relations marked by arrows, are added into the model. For example,
a person’s genes could cause lung cancer, so the red node can be included. With the
consideration of a causal direction and correlative quantity, probability theory emerges
as a powerful concept to equip a causal model. For example, we can write the conditional
probability of getting lung cancer, given personal genetic factors and smoking habits, as
P(lung cancer | smoking, genes).
The graph with the directed structure and probabilistic strength is often coined as a
causal model for a problem domain of interest. A causal model becomes a means to

19 | The Axiom
examine the causality among variables, which is very useful for mathematicians and com-
puter scientists. To build such a causal model, we require both knowledge of causal
relations (deciding the arrow direction) and correlations that formally quantify the ar-
row’s magnitude. Note that the knowledge is not necessarily drawn from data, but from
our understanding of the world, science, and rationality and common sense. For exam-
ple, the correlation between genes and lung cancer is common medical knowledge. Health
consultants can draw upon their expertise to identify this causal relationship. Fundamen-
tally, causality can only occur when equations are supplemented with knowledge of the
world which has been encapsulated into a causal model. This seems to be a harmonious
union among statisticians, mathematicians, computer scientists, and philosophers.
3 Not at peace yet
An interesting connection with artificial intelligence (AI) arises. First, since correlation is
a cornerstone of modern statistics, which tends to rely heavily on knowledge being derived
from data, statistics must also concern the philosophy of knowledge on causality. In
particular, modern AI programs which seem able to draw knowledge from data suddenly
opens up the potential for statisticians to understand the world through mathematical
knowledge. However, one fundamental component is missing: data-driven AI cannot
predict the e!ect of a cause which it has not met. Hence it needs a causal model to
supplement its partial understandings of the world from the data on which it has been
trained.
Immediately, one question is raised: where does a causal model come from? Causality
indeed benefits deep understanding of the world, but it, of course, depends on a correct
causal model. As the true model of the world cannot be known, we are led to another
question. A comprehension of causality is not merely to understand how things have
behaved, but also how things will behave when interventions are placed into the world.
How can you tell whether you have the right causal model? We can never know for sure.
Whilst we can falsify models on account of being patently incorrect, we cannot know with
certitude whether a causal model is correct. This is crucial to the scientific method.
4 Conclusion
In conclusion, we can see that correlation does not necessarily imply causation. In cases
such as the link between smoking and lung cancer, the cause and e!ect are linked by both
correlation and causation. Both correlation and causality are important to statistics, but
we must be careful with the relationship between the two and mindful not to assume that
correlation implies causality. So, next time you are at the beach, licking an ice-cream,
don’t worry about sharks.
References
[1] UCLSPP (2017).https://github.com/UCLSPP/datasets/blob/master/data/shark attacks.csv
[2] Statcrunch.com.(2024) Available at:https://www.statcrunch.com/reports/view?reportid=
36865tab=preview.

The Beauty of Chaos: Exploring the Butterfly E!ect
in Mathematics
Amiti
1 Introduction
Mathematics often o!ers a comforting predictability, yet it also concerns systems so sen-
sitive that even the smallest nudge can alter their fate dramatically. The Butterfly E!ect,
an idea popularised by Edward Lorenz, epitomises this concept. While initially exploring
weather prediction, Lorenz found that minuscule variations in starting conditions could
make long-term forecasts impossible, revealing an inherent unpredictability within deter-
ministic systems. This phenomenon has since become synonymous with Chaos Theory
and serves as a window into the complex behaviour of non-linear systems. In this article,
we explore the mathematical underpinnings of the Butterfly E!ect, using it as a gateway
to understanding chaotic systems.
2 The birth of the Butterfly E!ect
In 1961, Lorenz was refining a simplified model of atmospheric convection, described by
a set of non-linear equations. One day, he re-entered a value into the system, rounded to
three decimal places instead of six, expecting a minor di!erence. The result was startling:
the weather patterns his model generated became wildly di!erent. What seemed to be a
trivial rounding error had upended the entire trajectory of the system, marking the birth
of what is now known as sensitive dependence on initial conditions.
At the heart of the Butterfly E!ect lies non-linearity. In non-linear systems, the relation-
ship between inputs and outputs is not proportional—small inputs can have dispropor-
tionately large e!ects. A prime example of this behaviour is found in the logistic map, a
simple recurrence relation that models population growth:
xn+1 = rxn(1 → xn)
where xn represents the population at time step n, and r is a growth rate parameter that
controls the system’s dynamics. The equation’s deceptively simple form belies its rich
and often chaotic behaviour.
3 Exploring the logistic map
The logistic map behaves predictably for certain values of r, while for others, its behaviour
becomes chaotic. For instance:

21 | The Axiom
• For 0 r 1, the system stabilises at a fixed point, and we say it converges.
• For 1 r 3, it enters a periodic cycle, oscillating between two or more values.
• For r 3.57, the system enters full chaos – a “chaotic regime”. Even the tiniest
change in initial conditions causes the system to diverge rapidly.
Take r = 3.9 as an example. Starting with two initial values x0 = 0.500 and x0 =
0.501, the di!erence between them seems insignificant. Yet, after only 10 iterations, they
produce these significantly di!erent outcomes:
x0 = 0.500 x0 = 0.501
x10 = 0.907 x10 = 0.748
This stark divergence illustrates the very essence of chaos: small variations compound
over time, rendering long-term prediction futile.
4 Measuring chaos: Lyapunov exponent
The Lyapunov exponent quantifies the sensitivity to initial conditions, measuring how
divergent trajectories of initially nearby values are. For a system to be classified as
chaotic, its Lyapunov exponent ω must be positive, where
ω := lim
n→↑
1
n
n↓1

i=0
ln

d

f(xi)

dxi

For the logistic map, the derivative, by the product rule, is
d
dx
(rx(1 → x)) = r(1 → 2x)
If ω 0, nearby trajectories diverge exponentially, indicating chaotic behaviour. The
Lyapunov exponent o!ers a precise metric for how unpredictability manifests in systems
like the logistic map.
5 Attractors: chaos with structure
Despite the apparent randomness, chaotic systems often exhibit underlying patterns
known as strange attractors. These attractors, typically fractal in nature, represent the
set of states toward which a system tends to evolve. The Lorenz attractor, arising from
Lorenz’s famous system of equations for atmospheric convection, provides a striking visual
example:
dx
dt
= ε(y → x),
dy
dt
= x(ϑ → z) → y,
dz
dt
= xy → ϖz
For certain values of ε, ϑ, and ϖ, the Lorenz system traces out an intricate, looping shape
in phase space, representing the system’s long-term behaviour. Even though the system is
deterministic, its sensitive dependence on initial conditions makes its precise future state
unpredictable. The attractor demonstrates how chaotic systems, while unpredictable, are
not entirely without structure.

6 Applications and implications
The Butterfly E!ect and Chaos Theory extend far beyond weather models. In fields
like economics, biology, and even engineering, chaotic behaviour is frequently observed.
In population dynamics, for example, the logistic map is used to model species popula-
tions under limited resources. In neuroscience, Chaos Theory has helped to explain the
seemingly erratic yet structured patterns of brain activity.
However, chaos is not synonymous with randomness. Chaotic systems are deterministic,
governed by specific, fixed rules, whilst random systems are probabilistic and the same
experiment with the same initial conditions may produce di!erent outcomes each time it
is done. It is the sensitivity to initial conditions in chaotic systems that gives the illusion
of randomness. This distinction is crucial, as it suggests that, while chaotic systems may
be di”cult to analyse at a high level, for a particular initial condition, they are entirely
predictable, whereas outcomes on a random system can never be predicted with certainty.
7 Conclusion
The Butterfly E!ect o!ers a profound insight into the nature of complex systems. Through
simple mathematical models like the logistic map, we can see how chaos emerges from
deterministic equations, revealing an underlying sensitivity that challenges our ability to
predict the future. Far from undermining the role of mathematics, Chaos Theory enriches
our understanding, showing that even the most unpredictable systems follow discernible
patterns. The beauty of chaos lies not in any randomness, but in the intricate and of-
ten fractal structures it produces, structures that defy our conventional expectations of
mathematical order.
References
[1] www.sciencedirect.com. (n.d.). Chaos Theory - an overview — ScienceDirect
Topics. [online] Available at: https://www.sciencedirect.com/topics/chemical-
engineering/chaos-theory.
[2] Wikipedia Contributors (2019). Butterfly e!ect. [online] Wikipedia. Available at:
https://en.wikipedia.org/wiki/Butterfly e!ect.
[3] Wikipedia Contributors (2019). Logistic map. [online] Wikipedia. Available at:
https://en.wikipedia.org/wiki/Logistic map.
[4] Elert, G. (2016). Glenn Elert. [online] The Chaos Hypertextbook. Available at:
https://hypertextbook.com/chaos/strange/.

23 | The Axiom
Cracking The Cube: Maths Behind The Rubik’s Cube
Rubino ma
1 Introduction
The Rubik’s Cube, invented by Ernő Rubik in 1974, has since become one of the most
iconic puzzles worldwide. With its vibrant colors and seemingly limitless combinations,
the cube is not just a toy, but rather a gateway to understanding deep mathematics.
When you pick up a Rubik’s Cube, you may wonder in how many ways it can be arranged.
The answer is a staggering 43 quintillion. In spite of this, the cube can be solved in a
mere 20 moves, a concept that has gained fame as “God’s Number”.
In this article, I will delve deeper into the three underlying mathematical concepts behind
solving the Rubik’s Cube: Group Theory, combinatorics and permutations, and parity.
2 Group Theory: Defining the Rubik’s Cube Group
Group Theory is the mathematical study of symmetry, focusing on sets and operations
that follow specific rules. The Rubik’s Cube is a classic example of a non-abelian group,
meaning that the order in which operations are applied matters – that is, the group
operation is not commutative.
A group G is defined as a set of elements combined with a binary operation → that satisfies
the following four properties.
• Closure: For any two elements g, h ↑ G, the operation g → h results in another
element in G.
g → h ↑ G ↓g, h ↑ G
• Associativity: For any three elements f, g, h ↑ G, where we write the brackets
does not matter.
(f → g) → h = f → (g → h) ↓f, g, h ↑ G
• Identity Element: There exists an identity element e ↑ G.
↔ e ↑ G, g → e = e → g = g ↓g ↑ G
• Inverses: For any element g ↑ G, there exists an inverse g→1
.
↓ g ↑ G ↔ g→1
, g → g→1
= g→1
→ g = e

A simple example of a group, demonstrating the above properties, is G = (Z, +). That
is, the set is the integers, and the operation is addition.
• Closure: Adding integers always results in an integer. The set of integers is closed
under the operation of addition.
• Associativity: Adding three values always gives the same result, regardless of
the order in which each of the two individual additions is done. Addition is an
associative operation.
• Identity Element: Since n + 0 ↗ 0 + n ↗ n, and 0 is an integer, there is an
identity element e = 0 in our set.
• Inverses: Since n + (↘n) ↗ (↘n) + n ↗ 0 = e, every element of the set has an
inverse.
In the case of the Rubik’s Cube, the group still consists of a set and an operation, satis-
fying the above properties. Here, the elements of the set are the di!erent configurations
of the cube, the operation is a rotation applied to the faces of the cube, the identity
element e corresponds to the solved state of the cube, and every rotation has an inverse
that undoes it.
The Rubik’s Cube group, therefore, includes all transformations, and is crucial to analysing
the puzzle, as each move sequence (to get to a certain state) is an application of the group
operation on group elements.
3 Mathematical representation and permutation par-
ity
The Rubik’s Cube can be described using permutations, which rearrange smaller cubes
(let’s call them cubies) to achieve various configurations. We categorise the cubies into:
• Corners: 8 corner cubies, each with 3 possible orientations.
• Edges: 12 cubies, each with 2 possible orientations.
To calculate the total number of configurations of the Rubik’s Cube, we use the formula:
total configurations =
8! ≃ 37
≃ 12! ≃ 211
2 ≃ 3
The 8 corner cubies can be arranged in 8! ways. Each corner cubie has 3 orientations,
giving 37
possibilities (the 8th
is determined by the others).
The 12 edge cubies can be arranged in 12! ways. Each edge has 2 orientations, lead-
ing to 211
possibilities (the 12th
is determined by the others).
We divide by 2 to account for the edge flip parity, and by 3 to account for the cor-
ner twist parity.
This calculation results in approximately 43 quintillion configurations. Note that not
all configurations are legal due to the cube’s constraints.

25 | The Axiom
A critical property in solving the Rubik’s Cube is parity, referring to the even or odd
nature of a permutation. In the context of the Rubik’s Cube, any legal configuration
must have an even permutation of the corner cubies. Similarly, the edge cubies must also
follow an even permutation.
To illustrate this mathematically, consider that swapping two corner cubies or two edge
cubies individually would constitute an odd permutation, which is not possible on the
cube without additional swaps to maintain overall parity. This parity constraint helps in
developing algorithms that strategically swap and reorient pieces without violating the
cube’s structure.
3.1 Cycle notation and permutations
Permutations are best visualised using cycle notation. For example, if a face rotation
cycles four edge pieces, such as (A, B, C, D), it means cubie A moves to B’s position, B
moves to C’s, and so on. Each cycle length contributes to the order of the permutation.
The order of a permutation is the number of times the cycle must be applied before all
elements return to their starting positions. For instance, a 3-cycle has an order of 3
because applying it 3 times returns the elements to their original configuration.
3.2 Proof of even parity in the Rubik’s Cube
Parity refers to whether a permutation involves an even or odd number of swaps. The
Rubik’s Cube adheres strictly to even parity, meaning it is impossible to swap only 2
pieces without a!ecting others. Understanding and proving parity is crucial in designing
algorithms for solving the cube.
To prove that the Rubik’s Cube has even parity for both edge and corner swaps, we can
use induction on the number of moves n.
Base Case: The solved state has no swaps, representing an even parity (0 swaps).
Induction Step: Suppose that after n moves, the cube configuration has even parity
(an even number of swaps). We must show that applying an algorithm or move sequence
preserves the even parity. Any valid move rotates a set of 4 pieces in a cycle, which is
equivalent to 3 swaps. Since 3 is an odd number, repeating such cycles maintains even
parity in the overall cube configuration.
Therefore, after every valid move sequence, the cube must remain in an even parity state.
This is why it is impossible to solve the cube if only 2 pieces (either corners or edges) are
swapped. Such a configuration is unsolvable because it violates the parity rule.
4 Algorithms for solving the cube
While Group Theory and permutation parity provide a mathematical framework for
understanding the Rubik’s Cube, solving it requires practical strategies and algorithms,
which manipulate specific sections of the cube while leaving others unchanged.
The most common method for solving the Rubik’s Cube is the CFOP method (Cross,
F2L, OLL, PLL), used by many speedcubers:

• Cross: Solving a cross on one face by aligning edge pieces with their corresponding
centres.
• F2L (First Two Layers): Solving the first two layers by pairing corner and edge
pieces.
• OLL (Orientation of the Last Layer): Aligning the colours of the pieces on the
final layer, even if they are not yet in the correct positions.
• PLL (Permutation of the Last Layer): Permuting the pieces of the final layer
into their correct positions without disturbing their orientations.
Each stage of this process is driven by specific algorithms, sequences of rotations designed
to manipulate small sets of pieces while maintaining the overall structure of the cube.
These algorithms are carefully crafted based on the Rubik’s Cube group properties and
permutation constraints.
5 God’s Number
In 2010, researchers, aided by computing power from Google, discovered that any config-
uration of a Rubik’s Cube can be solved in 20 moves or fewer. Another way of thinking
about this number, known as “God’s Number”, is as the upper bound of the minimum
number of moves required to solve any scrambled cube. While most configurations can
be solved in fewer than 20 moves, the existence of a universal upper bound demonstrates
the power of mathematical analysis in solving complex problems.
6 Conclusion
The Rubik’s Cube is much more than a simple puzzle; it is a physical representation
of deep mathematical principles. By understanding Group Theory, permutations, and
parity, we can not only appreciate the complexity of the cube, but also develop e”cient
strategies for solving it. Whether you are a casual solver or a speedcuber, the cube o!ers
a fascinating intersection of Mathematics, logic, and creativity.
In the end, the Rubik’s Cube teaches us that, even in a world of 43 quintillion possibilities,
mathematical structure and logic can guide us to the solution, one twist at a time.
References
[1] MIT (2009). The Mathematics of the Rubik’s Cube.
[2] www.youtube.com. (n.d.). Groups — Mathematics of Rubik’s Cube. [online] Available
at: https://www.youtube.com/watch?v=Cxko6UcArL4.
[3] J Perm (2021). How Algorithms ACTUALLY Work on the Rubik’s Cube. YouTube.
Available at: https://www.youtube.com/watch?v= Zv3YcQeNVI.
[4] Using maths to solve the Rubik’s Cube. (2017). YouTube. Available at:
https://www.youtube.com/watch?v=UqQaqbvDZUA.

27 | The Axiom
Composing with Cubes
Kumar ma MS
1 Introduction
Iannis Xenakis was a Greek-French contemporary composer and music-theorist, with a
background in engineering and architecture. He is mostly known for composing the orchestral
piece ‘Metastasis’, the score of which ended up serving as the architectural blueprint for The
Philips Pavilion.
Perhaps one of Xenakis’s most interesting pieces is Nomos Alpha, a 15 minute piece for solo cello,
which is entirely based o! of something called Group Theory.
2 Group Theory
Sometimes, shapes have multiple lines which, when reflecting through said lines (or in the case of
a 3D shape, axes), the resulting shape is the same as the original shape. In other words, they
have multiple lines of symmetry - an incredibly fundamental mathematical concept.
A great example to give a basic idea of group theory is a snowflake. Its Group theory number is
D6 (i.e. there are 12 ways in which it can be rotated or reflected resulting in the same snowflake).
Figure 1: The 12 lines of symmetry of a snowflake.
In order to compose Nomos Alpha, Xenakis applied Group theory on a cube. There can be up to
188,743,680 ways of rotating and reflecting a cube and its individual faces within its axes, but for
less complexity, Xenakis focused on the 24 lines of symmetry.
3 Setup
The vertices of each cube were numbered 1 to 8, each corresponding to a basic sound. Joining
vertex 1 to vertex 4 and vertex 5 to vertex 8, 2 tetrahedra are formed. Each rotation gives a
permutation of both groups (e.g. 1234 to 2314 and 6578 to 7586).
The 24 rotations can be classified into 2 groups, labelled A and Q:
A: The rotation starts with a permutation of the 1-4 group.
Q: The rotation starts with a permutation of the 5-8 group.

Figure 2: The diagram drawn by Xenakis as a plan for the piece.
In group A, there are 12 transformations, each named by a letter;
A, B, C, D, D2
, E, E2
, G, G2
, I, L, L2
In group Q, there are 12 transformations:
Q1 to Q12
For both groups, there are 3 subsets of the transformations:
A: V1- {A, B, C, I}, V2- {D, E2
, G, L2
}, V3- {D2
, E, G2
, L}
Q: V4 - {Q6, Q12, Q7, Q1}, V5 - {Q8, Q10, Q3, Q5}, V6 - {Q2, Q4, Q11, Q9}
Figure 3: Table containing the subsets and their products which Xenakis used for determining
certain parameters.
By multiplying subsets together, starting with V2 and V6, then multiplying the product with V6,
we form a loop of the following sub-groups:
V2, V4, V6, V3, V5, V6.
Given that each subset contains 4 di!erent transformations, Xenakis was able to choose a
di!erent one from each group for each time for variety within the piece.
4 Execution
Xenakis set parameters for the length and intensity for sound, with 8 dynamic levels and 8
lengths of notes (figure 4).

29 | The Axiom
Figure 4: Parameters of intensity and length.
Figure 5: Configuration
We can now take an example of the piece itself. For example, when writing the notes for Q12, the
order of sounds is 5, 6, 8, 7 - 1, 4, 2, 3. Hence, the configuration ends up as shown in figure 5.
After combining the 3 parameters of number, length and intensity, Xenakis ended up with the
score shown in figure 6.
The end result is a 15 minute mathematical and musical wonder. There are many idiosyncrasies
within the piece, and, due to the non-repeating nature of each permutation, there are constant
surprises thrown at the listener.
References
[1] Josquin Buvat (2019). Nomos Alpha of Iannis XENAKIS for cello solo : guide for performers
[NOMOS ALPHA of Iannis XENAKIS for cello solo: guide for performers - 2019-09-17 13:08]
by Josquin Buvat. [online] Researchcatalogue.net. Available at:
https://www.researchcatalogue.net/view/686273/686426.
[2] RNCM PRiSM (2022). Professor Marcus du Sautoy discusses Iannis Xenakis’ ‘Nomos Alpha’
for solo cello. [online] YouTube. Available at:
https://www.youtube.com/watch?v=-i hok9TQjU.
[3] VEDANTU. (n.d.). Group Theory in Mathematics. [online] Available at:
https://www.vedantu.com/maths/group-theory-in-mathematics.
[4] Anon, (n.d.). Iannis Xenakis. [online] Available at: https://www.iannis-xenakis.org/en/.

Figure 6: The final score.

31 | The Axiom
‘The Holy Grail of Mathematics’ – The Riemann
Hypothesis
Vyas KS
1 Introduction
The Riemann Hypothesis, first proposed in 1859 by Bernhard Riemann, is one of the
seven Millennium ‘Millionaire’ Problems. Considered by many to be the greatest
unsolved problem on the list, a proof of the Riemann Hypothesis would have
far-reaching consequences in number theory, especially in our understanding of prime
numbers. Nonetheless, it remains one of the hardest questions to solve to date, with
many even arguing that our tools for complex analysis and analytic number theory are
simply not capable of providing a solution.
2 The Zeta Function
The hypothesis concerns the Euler-Riemann zeta (ω) function, and, as with the greatest
conjectures in mathematics, it attempts to solve what appears to be a simple question:
how many values of the analytical continuation of the function sum to zero?
The zeta function is defined as follows:
ω(s) =
→

n=1
1
ns
(1)
Essentially, the function raises each term in the harmonic series to the power of s.
When s is defined as a real number greater than 1, the infinite sum has a finite answer,
as the sum is convergent. Some obvious patterns can be spotted upon first observation,
such as when s = →1: the function is divergent and equals the sum of all natural
positive integers. After analyzing the Prime Harmonic series, Euler was the first to
provide a defined solution to the function when s 1 and to find the limit that the sum
approached for each of these values of s.
3 The Function on the Complex Plane
In the function, s is not actually restricted to a real number – one can input s as any
complex value on the complex plane. The value of s will be convergent if it is to the
right of the red line below, which represents 1 + yi.

Figure 1: Complex plane, with line x = 1 + yi
Whilst the answer will be a complex sum, it is still defined as long as it is within this
region.
4 Holomorphic Functions and Analytical
Continuation
Currently, our domain is quite restricted, as we only have values for the function in a
limited region of the plane. However, the zeta function by nature is a holomorphic
function – a complex-valued function that is complex di!erentiable. A benefit of
holomorphic functions is that they can undergo a process called ‘analytical
continuation,’ which extends the function so that the domain of defined solutions is
increased.
Hence, Riemann calculated the analytical continuation of the zeta function, giving him
a new functional equation. Using this, he could now find the defined solution for every
complex and real value for s except for 1, which is a singularity in this case and is still
undefined.
5 Finding Zero
The main question regarding the zeta function is as follows: for what values of s will
ω(s) = 0? Already found are the ‘trivial’ solutions: s = →2n. -2, -4, -6, etc., are all
trivial values where ω(s) = 0.
The key task is to find non-trivial values for ω(s) = 0. We know for certain that all
non-trivial values must lie in the ‘critical strip,’ shown below in green.
Every non-trivial value that sums to 0 is contained within this strip. The problem arises
when trying to find these values.

33 | The Axiom
Figure 2: Red region where Re(s) 1, Blue region where Re(s) 1, Green region where
0 Re(S) 1. The bold black line is the ’critical line’, at x = 1
2
+ yi
6 Riemann’s Hypothesis and the Search for Proof
Within this critical strip, there is the ‘critical line’ where s = 1
2
+ yi. Riemann
hypothesized that every non-trivial solution within the critical strip lies on the critical
line. Currently, mathematicians have found trillions of non-trivial solutions, all of which
lie on the critical line.
The task arises to prove or disprove Riemann’s Hypothesis. Many have attempted proof
by counterexample – trying to find a non-trivial solution not on the critical line.
However, none have been found so far, and the hypothesis remains unsolved.
7 Impact of the Hypothesis
The reason the hypothesis is so important is due to the impact it would have on
number theory if proven true.
Since Euclid proved there are an infinite number of primes, Prime Number Theory
(PNT) has devised a way to estimate the number of prime numbers between 1 and n. If
we let n = 100, PNT would estimate the number of prime numbers as ↑ 100/ ln(100),
or 22. In reality, there are 25 prime numbers between 1 and 100, so the estimation isn’t
perfect, with a deviation of 3 in this case. Riemann’s Hypothesis provides us with a way
to estimate this deviation. Hence, if the hypothesis were true, one could have a
relatively accurate idea of how many prime numbers there were between 1 and n.
The hypothesis also posits the location of prime numbers on a number line. Essentially,
as you go down the number line, while PNT states that primes decrease in density,
there is still roughly representative coverage of the distribution. As Scientific American
puts it, in the same way that air is denser on the floor than the ceiling, but the particles

are still spread evenly across the density distribution, the hypothesis poses that there
are no large areas with no primes or with a large number of primes; instead, areas have
an approximate distribution of primes.
8 Conclusion
To this day, many mathematicians are still on the hunt for zeta function zeros and a
proof of the Riemann Hypothesis. A proof would mean so much for mathematics, and
particularly for the field of number theory. As is the case with many of the hardest
conjectures, the solution can come at any time, from anyone; perhaps it’s yours to find
in the future.
References
[1] Riemann (2014). Riemann Hypothesis - Numberphile. [online] YouTube. Available
at: https://youtu.be/d6c6uIyieoo?si=7akNP0TJtHbo4wS6.
[2] The (2016). The Key to the Riemann Hypothesis - Numberphile. [online] YouTube.
Available at: https://youtu.be/VTveQ1ndH1c?si=rcaZrVzkcPY-VtJt.
[3] the, B. (2023). 23% Beyond the Riemann Hypothesis - Numberphile. [online]
YouTube. Available at: https://youtu.be/dwe4-OiRw7M?si=ObgQ Ceg OsnwOF8.
[4] Bombieri, E. (n.d.). The Riemann Hypothesis. [online] Available at:
https://citeseerx.ist.psu.edu/document?repid=rep1type=pdfdoi=e3e9a513f0422acb
0e49ac1395fa50f054040894#page=98.
[5] Conrey, J. ed., (n.d.). The Reimann Hypothesis. [online] Available at:
https://www.ams.org/journals/notices/200303/fea-conrey-
web.pdf?adat=March202003trk=200303fea-conrey-
webcat=featuregalt=feature.
[6] Sarnak, P. (n.d.). Problems of the Millennium: The Riemann Hypothesis. [online]
Available at:
https://homepages.math.uic.edu/ libgober/math592/export/Sarnak.pdf.
[7] Miller, G. (n.d.). Riemann’s Hypothesis and tests for primality. [online] Available at:
https://dl.acm.org/doi/abs/10.1145/800116.803773.
!

35 | The Axiom
Electron Geometry
A.Agnihotri OS
1 Introduction
If you have done some A level chemistry, you will know about molecular and electron geometry.
Electron densities form as far apart on average as possible; 3 electron densities form a trigonal
planar structure (equilateral triangle, bond angle 60→
if electron densities are the same), and 4
electron densities form a tetrahedral structure (regular tetrahedron, bond angle 109.5→
if electron
densities are the same).
These shapes seem intuitive, but I wondered about a proof. Is the proof as simple as these shapes?
I wanted to find out, so I came up with some questions that translate the chemistry into maths.
Let there exist a central point in a 3D field. 3 outer points surround this
central point, each equidistant from the central point. What is the optimal
arrangement such that the mean of the distances between the outer points is
maximised? What about for 4 outer points?
2 Proofs
2.1 3 outer points angles proof
Let me start by labeling the central point O and the outer points A, B and C. Since the points
A, B and C are equidistant from O, they lie on the circumference of the sphere with centre O. It
becomes clear that the points must all lie on the same 2D plane (for brevity, these proofs won’t
be complete), and so A, B, C are points on the circumference of the circle with centre O. Before
we get into the proofs, what are we quantitatively trying to find? We are trying to maximise
the expression |AB|+|AC|+|BC|
3 , and since 1
3 is just a constant, it would be su!cient to maximise
|AB| + |AC| + |BC|.
My first idea was to look at angles. How can we intertwine angles and lengths? Trigonometry.
Using the cosine rule:
|AB|
2
= |OA|
2
+ |OB|
2
→ 2|OA||OB| cos ω = 2r2
→ 2r2
cos ω
|AC|
2
= |OA|
2
+ |OC|
2
→ 2|OA||OC| cos ε = 2r2
→ 2r2
cos ε
|BC|
2
= |OB|
2
+ |OC|
2
→ 2|OB||OC| cos ϑ = 2r2
→ 2r2
cos ϑ

Where r is the radius. Let S = |AB|
2
+ |AC|
2
+ |BC|
2
. Since all lengths are positive reals, max-
imising S would also maximise our goal. So now we have:
S = 2r2
→ 2r2
cos ω + 2r2
→ 2r2
cos ε + 2r2
→ 2r2
cos ϑ
= r2
(6 → 2(cos ϑ + cos ε + cos ω))
Since r is independent of ϑ, ε and ω, S is maximised when cos ϑ+cos ε +cos ω (= s) is minimised.
I am going to leave out the case on the right. We can minimise s by finding the partial derivative
with respect to ϑ, ε and ω, and equating them to 0.
substituting ω = 360→
→ (ϑ + ε) into s:
s = cos ϑ + cos ε + cos(360→
→ (ϑ + ε))
= cos ϑ + cos ε + cos(ϑ + ε)
ϖs
ϖϑ
= → sin ϑ → sin(ϑ + ε) = 0, when s is minimised.
ϖs
ϖε
= → sin ε → sin(ϑ + ε) = 0, when s is minimised.
↭ sinϑ = → sin(ϑ + ε) = sin ε =↑ sin ϑ = sin ε
Doing a similar process (with ϑ = 360→
→ (ε + ω)), you will get sin ϑ = sin ε = sin ω, but since
ϑ + ε + ω = 360→
, and observing the graph of sin from 0→
to 360→
, you will obtain the angles
(180→
, 180→
, 0→
) and (120→
, 120→
, 120→
). The former is a phantom result from the case on the right,
and is clearly not the right answer (it would be if there was only two outer points), so the angles we
get (the bond angles) are (120→
, 120→
, 120→
), which gives the arrangement of an equilateral triangle.
2.2 4 outer points coordinates proof
So what about for 4 outer points? Notice how we could only exploit angles because we had three
which summed to 360→
. There is no such standard sum for the angles in 3D. I had to look back at
3 points and ponder a proof which was more general and didn’t necessarily use fixed angle sums.
Coordinates.
Again, let the central point be O and the outer points be A, B, C and D. Just to make the
algebra easier, let r = 1, which shouldn’t be an issue since the arrangement will scale with r. Let
A have coordinates (0, 1, 0), and let B have coordinates (xB, ±
↓
1 → xB
2, 0) – I have made A and
B be in only the x → y plane, so B has coordinates of a point on the circle x2
+ y2
= 1. Let C
and D be the generic points (xC, yC, ±

1 → xC
2 → yC
2) and (xD, yD, ±

1 → xD
2 → yD
2). You
can try all the combinations for positive and negative square roots using the method I am about
to show you for yourself... or you can trust that this combination is the optimal one (there are
other combinations, but they just give rotations of this one):
A = (0, 1, 0)
B = (xB, →

1 → xB
2, 0)
C = (xC, yC, →

1 → xC
2 → yC
2)
D = (xD, yD,

1 → xD
2 → yD
2)

37 | The Axiom
Now, using the distance formula:
|AB|
2
= (0 → xB)
2
+ (1 → (→

1 → xB
2))
2
= 2 + 2

1 → xB
2
|AC|
2
= (0 → xC)
2
+ (1 → yC)
2
+ (0 → (→

1 → xC
2 → yC
2))
2
= 2 → 2yC
|AD|
2
= (0 → xD)
2
+ (1 → yD)
2
+ (0 →

1 → xD
2 → yD
2)
2
= 2 → 2yD
|BC|
2
= (xB → xC)
2
+ (→

1 → xB
2 → yC)
2
+ (0 → (→

1 → xC
2 → yC
2))
2
= 2 → 2xBxC + 2yC

1 → xB
2
|BD|
2
= (xB → xD)
2
+ (→

1 → xB
2 → yD)
2
+ (0 →

1 → xD
2 → yD
2)
2
= 2 → 2xBxD + 2yD

1 → xB
2
|CD|
2
= (xC → xD)
2
+ (yC → yD)
2
+ →

1 → xC
2 → yC
2 →

1 → xD
2 → yD
2
2
= 2 → 2xCxD → 2yCyD + 2

1 → xC
2 → yC
2 ·

1 → xD
2 → yD
2
Let S = |AB|
2
+|AC|
2
+|AD|
2
+|BC|
2
+|BD|
2
+|CD|
2
. Like before, maximising S also maximises
our goal. To find the optimal values of xB, xC, xD, yC, yD, we equate the partial derivatives to 0,
to get a system of 5 simultaneous equations.
ϖS
ϖxB
= →2xC → 2xD →
2xB + 2xByC + 2xByD
↓
1 → xB
2
= 0
dS
dxC
= →2xB → 2xD →
2xC

1 → xD
2 → yD
2

1 → xC
2 → yC
2
= 0
ϖS
ϖxD
= →2xB → 2xC →
2xD

1 → xC
2 → yC
2

1 → xD
2 → yD
2
= 0
ϖS
ϖyC
= →2 → 2yD + 2

1 → xB
2 →
2yC

1 → xD
2 → yD
2

1 → xC
2 → yC
2
= 0
ϖS
ϖyD
= →2 → 2yC + 2

1 → xB
2 →
2yD

1 → xC
2 → yC
2

1 → xD
2 → yD
2
= 0
Doing some “o”-camera” simultaneous equations solving, the values we obtain are: xB = 2
↑
2
3 , xC =
xD = →
↑
2
3 , yC = yD = →1
3 . Substituting these values back into the coordinates, they do in fact
make a regular tetrahedron, and the bond angle is arccos →1
3 .
2.3 Geometric proofs
The coordinates proof works for any number of outer points; it is general (I think), but it is quite
tedious and not very intuitive. I don’t have the space to go through all my findings, but for 3 outer
points, the general idea is to consider two generic points on a circle, and then find the point on the
circle such that the total distance between this new point and the other 2 points each is maximised.
You’ll find that the point lies on the intersection of the circle with perpendicular bisector of the line
segment connecting the two original points. From there, you can “slide” the two original points
up and down the circle simultaneously to find the optimal arrangement.
A similar process can be done in the 4 point system, but in 3D. Consider a triangle with ver-
tices being three points on a sphere. Similarly to the 3 point system, the point which maximises
the total distance between this new point and the original 3 points each lies on the intersection of
the sphere with the orthogonal lines passing through the triangle’s circumcentre, though I am not
entirely certain.

Functional Equations
Agnihotri KS
1 Introduction
Functional equations are equations in which you are tasked with finding all possible solutions for
the function(s). They are a fun section of algebra which can produce some quite ingenious solutions
using clever tricks. Examples of functional equations might include:
f : N → N, f(1) = 2, f(n + 1) = 2f(n)
f : Q → Q, f(1) = 2, f(xy) = f(x)f(y) ↑ f(x + y) + 1
f : Z → Z, 2f(f(n)) = 5f(n) ↑ 2n
1.1 Introductory Example
Cauchy’s Functional Equation Over the Rationals
f : Q → Q, f(x + y) = f(x) + f(y)
The first two things that should go through your mind when approaching a functional equation
are:
• Guess the solution.
• Try some values.
Here it seems like f(x) = kx is a solution, and playing around with a few substitutions might reveal
useful properties of the function or give hints as to how to go about proving the solution. Putting
x = 0, y = 0 reveals that f(0) = 0. From this, putting y = ↑x reveals that f(x) = ↑f(↑x). So
we have shown that f is odd, hence we only need to work with positive values.
It is not so obvious how to proceed, so let’s try some more substitutions and see what happens.
Putting x = 1, y = 1 we see that f(2) = 2f(1). This is promising; we have something of the form
f(x) = kx (k = f(1)). Continuing subbing x = 2, y = 1, we might begin to get an idea of how
to solve the problem. Noticing that we are getting new values from previous values, we begin to
suspect an induction is at hand. For reasons which will become apparent later, we will try to prove
that f(mn) = mf(n) for m ↓ Z, n ↓ Q and then substitute n = 1.
Base case, m = 0: f(0n) = 0f(n)
Assume true for m = k: f(kn) = kf(n)
Consider m = k + 1:
f

(k + 1)n

= f(kn + n)
= f(kn) + f(n) (definition of the function)
= kf(n) + f(n) (induction hypothesis)
= (k + 1)f(n)
True for m = 0, true for m = k =↔ true for m = k + 1, hence true for m ↓ Z+
.
And since f is odd, this is also true for m ↓ Z. So we have proved that f(mn) = mf(n) for
m ↓ Z and hence that f(x) = xf(1), so f(x) = kx for x ↓ Z. Now we want to extend this over
the rationals. Let p, q be co-prime integers where q ↗= 0.
f

p
q

=
q
q
f

p
q

=
1
q
f

pq
q

=
1
q
f(p) = k
p
q

39 | The Axiom
Step two is true because we proved f(mn) = mf(n) where m ↓ Z and n ↓ Q. Here, m = q and
n = p
q . And so we have proved that f(x) = kx for x ↓ Q. Now we put it back into the original
equation to make sure that our answer works, since we made the subtle assumption that a solution
exists, and we only proved that if there is a solution, then our answer is the only solution. Clearly,
our answer works and we are done.
2 Substituting Functions
A useful trick is to substitute a function which might be easier to solve for. For example:
Example 1
f : Q → R++
, f(x + y) = f(x)f(y)
This is really close to the Cauchy’s functional equation, except there is a product instead of a sum.
We know that logarithms and exponents are helpful for going between products and sums, and a
guess at the solution tells us that f(x) = ax
is a solution. So let’s define a new function g(x) such
that:
g : Q → R, bg(x)
= f(x), b 0
and put this into the original equation, giving:
bg(x+y)
= bg(x)
bg(y)
bg(x+y)
= bg(x)+g(y)
=↔ g(x + y) = g(x) + g(y)
This is just Cauchy’s functional equation, which we already know. So g(x) = kx. So
f(x) = bg(x)
= bkx
= (bk
)x
= ax
We check that this works, and we are done. Now see if you can solve these problems:
f : N → R, f(xy) = f(x) + f(y)
f : N → R++
, f(xy) = f(x)f(y)
BMO Round 2 2024, Question 2
f : Z → Z, 2f(f(n)) = 5f(n) ↑ 2n
As before, we first try to guess the solution. Since this is from a BMO2, it is likely that the solution
is linear. Putting f(n) = kn + c, we obtain:
2(k(kn + c) + c) = 5(kn + c) ↑ 2n
2k2
n + 2kc + 2c = 5kn + 5c ↑ 2n
2k2
n + (2kc + 2c) = (5k ↑ 2)n + (5c)
Equating the coe!cients, we get:
2k2
= 5k ↑ 2 (1)
2kc + 2c = 5c (2)
The first is a quadratic which we can solve to get that k = 1
2 or k = 2. But f is over the integers,
so this forces k = 2 as the only solution. This means that c = 0, so our guess is that f(x) = 2x.
Trying to prove this directly seems di!cult, so we employ the trick we just learnt in order to solve
a potentially easier problem. We might try g(x) = f(x)
x and prove that g is constant. This doesn’t

really help, so we might try g(x) = f(x)↑2x and prove that g ↘ 0. We get some nice cancellations
if we do this, so it looks promising. Subbing in f(x) = g(x) + 2x:
2f(g(n) + 2n) = 5(g(n) + 2n) ↑ 2n
2

g

g(n) + 2n

+ 2

g(n) + 2n

= 5g(n) + 10n ↑ 2n
2g

g(n) + 2

+ 4g(n) + 8n = 5g(n) + 8n
g(n) = 2g

g(n) + 2

g(n) = 2g(a1)
Where a1 = g(n) + 2 is some integer. But g(a1) is itself equal to 2g(a2), where a2 = g(a1) + 2 is
another integer. So g(n) = 4g(a2). And g(a2) is itself equal to 2g(a3), so g(n) = 8g(a3). And we
can keep doing this indefinitely. This means 2 | g(n) and 4 | g(n) and 8 | g(n) and so on forever.
So g(n) has infinite factors. The only integer with infinite factors is 0, so g(n) = 0, so f(x) = 2x.
We check this with the original equation and we are done.
3 Useful Properties
The useful properties of functions are:
• Oddness, f(x) = ↑f(↑x)
• Evenness, f(x) = f(↑x)
• Injectivity, f(a) = f(b) ≃↔ a = b
• Surjectivity, for all elements, y, in the range of f, there exists at least one element, x, in the
domain such that f(x) = y
When solving functional equations, be on the lookout for these properties since they can be in-
credibly useful.
Example 2
f : R → R, f

x2
+ yf(z)

= xf(x) + zf(y)
It is not hard to see that f ↘ 0 is a solution. We will then assume f is not 0 everywhere, and try
to find all other solutions. Subbing in y = z = 0, and x = 0, we get:
f(x2
) = xf(x)
f

yf(z)

= zf(y)
Both equations here are almost symmetric. The slight asymmetry can be exploited. For the first
equation, we can exploit the evenness of the square function, by subbing x = ↑x,
f(x2
) = ↑xf(↑x)
= xf(x)
=↔ f(x) = ↑f(↑x)
And so we have established that f is odd (and that f(0) = 0). This property doesn’t turn out to be
particularly useful on this occasion, but this is a good illustration of the mindset when approaching
functional equations.
In the second equation, one z is a ‘free’ variable; it is not in a function, but the other z is.
This is a telling sign that we can prove injectivity. Setting y = y0, and subbing in z = z1 and
z = z2 such that f(z1) = f(z2) and f(y0) ↗= 0,
f

y0f(z1)

= z1f(y0)
f

y0f(z2)

= z2f(y0)
=↔ z1f(y0) = z2f(y0)
=↔ z1 = z2

41 | The Axiom
Thus we have proven injectivity. So it makes sense to try and exploit this by substituting y = t,
z = t, t ↗= 0 into the second equation,
f

tf(t)

= tf(t)
f

tf(t)

= f(t2
)
tf(t) = t2
(f is injective)
f(t) = t
So f ↘ 0 or f is the identity function. Checking this with the original equation, we see that both
work, and we are done.
IMO 2022, Question 2
Let R+
denote the set of positive real numbers. Find all functions f : R+
→ R+
such that
for each x ↓ R+
, there is exactly one y ↓ R+
satisfying
xf(y) + yf(x) ⇐ 2.
Since each y is unique to each x, it is most likely that y is a function of x. So we guess that the
inequality can be rewritten as:
xf(g(x)) + g(x)f(x) ⇐ 2
Subbing x = g(x) into this, we get:
g(x)f(g(g(x))) + g(g(x))f(g(x)) ⇐ 2
And so it appears that g(g(x)) = x. So g is an involution and our educated guess is that y = x
and f(x) = 1
x , which clearly works.
In order to prove we have found the only solution, we employ a proof by contradiction. Let us
assume that x ↗= y. Due to the uniqueness of y, this implies that xf(x)+xf(x) 2 =↔ f(x) 1
x
and yf(y) + yf(y) 2 =↔ f(y) 1
y . So (noting that ω + 1
ω ⇒ 2 due to the AM-GM inequality),
2 ⇒ xf(y) + yf(x)
x
y
+
y
x
⇒ 2
which is clearly a contradiction, so our assumption that x ↗= y was wrong. So x = y.
So the inequality becomes f(x) ⇐ 1
x , and we have that xf(x→
) + x→
f(x) 2 for each x→
↗= x, which
implies
2 xf(x→
) + x→
f(x) ⇐
x
x→
+ x→
f(x)
=↔ f(x)
2x→
↑ x
(x→)2
As x→
approaches x, f(x) approaches
2x ↑ x
x2
=
1
x
And since we already have that f(x) ⇐ 1
x , this becomes an equality:
f(x) =
1
x
Upon checking this with the original restrictions, we see that it works and thus we are done.
References
[1] Chen E. (2016). Introduction to Functional Equations. Available at:
https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf.

Gauss’ Law: Di!erentials and Integrals
Moorhouse
1 Introduction
In 1865, James Clerk Maxwell published A Dynamical Theory of the Electromagnetic
Field, which unified electric and magnetic phenomena through quantitative descriptions
given by four di!erential equations – Maxwell’s equations. The theoretical framework
was built on newly defined mathematics founded on Sir William Hamilton’s theory of
quaternions, which, when later formalised by Gibbs and Heaviside, became known as
vector calculus. This branch of mathematics is extensively used in physics in descriptions
of field theories and fluid flow, such as with electromagnetism. Here, we will focus on the
first of Maxwell’s equations, Gauss’ Law for Electrostatics, and develop its integral form
to the more common appearance as a di!erential equation.
2 The Electric Field
Gauss’ Law is a statement that relates the flux through a surface to the charge it encloses.

S
(E · n̂) dS =
qenc
ω0
Focusing on the left-hand side of the equation, we see the mathematical expression that
defines flux. This is a surface integral of the component of the vector function, which
here is the electrostatic field, that is normal to the surface. In essence, recalling that
integrals are simply the limit of sums, we can think of electric flux ”E, or indeed flux, as
the total number of field lines that penetrate the given surface. (Discussing flux as the
flow of a vector field would not be much help here since an electrostatic field does not
‘flow’, as we would say a fluid does, from a charge through a closed surface). This form
of Gauss’ Law relies heavily on the system having some property of symmetry that can
be exploited to help us find the electrostatic field.
1. There is a spherically symmetric distribution of charge.
2. There is a cylindrically symmetric distribution of charge (i.e. an infinite rod).
3. There is a planar symmetry in the charge distribution.
However, it can be shaped into a more useful form for when symmetry is not available.
Going back to the idea of integrals as the limit of a sum, the problem with a system
where symmetry is not present, is that the resulting equation has an infinite number of
unknowns (the value of E·n̂ at all points along the surface). In other words, it is hopeless
for finding the electrostatic field in these cases. Instead, we want to deal with the flux at

43 | The Axiom
a single point (whatever this might mean) which would reduce the infinite values of E · n̂
to one unknown.
To do this, we consider a general point P(x, y, z), at which we centre a series of con-
centric spherical shells around. It is tempting at this point to use the current form of
Gauss’ Law to define the flux at P to be the limiting value approached by the sequence
of concentric spherical shells. If we did this, we would end up with a zero value of flux
every time, since we are stating that as the surface tends to a point, the volume tends
to zero and hence, there is no enclosed charge to account for. Instead, we are inclined to
define a new property, the average charge density ε, in some volume #V .
Rewriting Gauss’ Law by noting that the enclosed charge is the product of the aver-
age charge density and the volume it applies to, we confirm that as S → 0, #V → 0 and
hence ”E → 0 as expected. Charge density, unlike the enclosed charge, is a non-vanishing
property as #V → 0, and in fact we can state that at a point P(x, y, z) the density func-
tion returns ε(x, y, z). Isolating this property and considering now the limit as #V → 0,
we arrive at quite a terrifying expression that appears to be even more useless than what
we started with. Luckily, the dreadful-looking left-hand side can be made more workable
by introducing a new operator.
lim
!V →0
about(x,y,z)
1
#V

S
(E · n̂) dS =
ε(x, y, z)
ω0
3 Divergence
The divergence of a vector function is defined to be the ratio of the surface integral of that
vector function to the volume enclosed by the surface as the volume tends to zero about
some point. This new property of a vector function is clearly a scalar, as, by design, it
is equivalent to the right-hand side of the terrifying equation above. We would therefore
say it is a scalar function of position. How useful this new operator is depends on whether
we can derive a simpler form, as currently it provides nothing more than cosmetic value.
div F = lim
!V →0
about(x,y,z)
1
#V

S
(E · n̂) dS
To derive this form, we will consider a rectangular parallepiped (a cuboid) with sides of
length #x, #y, #z parallel to their corresponding axes. Further, the point at the centre
of the cuboid will have coordinates (x, y, z). Due to the linearity of integration, the net
flux through the cuboid’s surface can be computed as the sum of the flux through each of
the six faces. We will first consider the face S1. Recall, that the flux of a vector function
is the surface integral of the component of the vector that is normal to the surface.

S
(F · n̂) dS
The unit normal vector pointing outwards from the surface S1 is i and hence, we can
simplify the scalar product to F · i = Fx, which is simply the projection of the vector
function along the x-axis direction. As such, the flux integral becomes:

S1
Fx(x, y, z) dS

Figure 1: Rectangular parallepiped positioned with sides parallel to the axes. (H. M.
Schey)
Further simplification occurs as a result of the mean value theorem and our assumption
that the surface S1 is small. It states that the integral of a function over a surface is
equivalent to the area of that surface multiplied by the function evaluated somewhere on
it. We will evaluate the function at the centre of the surface S1, because as we consider
the limit where the volume of the cuboid tends to zero, our approximation here will return
the exact value of the flux through S1 by the mean value theorem. At the centre of S1,
the coordinates are (x + !x
2
, y, z) and hence the flux through S1 is approximately:

S1
Fx(x, y, z) dS ↑ Fx(x +
#x
2
, y, z)#y#z
It would now be of worth to evaluate the flux through the opposite face labelled S2, which
has the unit normal vector pointing outwards of i and coordinates at the centre of the
face of (x ↓ !x
2
, y, z). The same line of reasoning returns:

S
(F · n̂) dS =

S2
Fx(x, y, z) dS ↑ Fx(x ↓
#x
2
, y, z)#y#z
From these two results, we can calculate the total flux of the vector function as projected
along the i direction by simply adding the two together.

S1+S2
(F · n̂) dS ↑

Fx(x +
#x
2
, y, z) + Fx(x ↓
#x
2
, y, z)

#y#z
Recall that our expression for the divergence of a vector function involved the limit as
the volume approached zero. By multiplying and dividing the expression by #x, we
introduce the volume of the cuboid #V = #x#y#z.

S1+S2
(F · n̂) dS ↑
Fx(x + !x
2
, y, z) + Fx(x ↓ !x
2
, y, z)
#x
#x#y#z
↑
Fx(x + !x
2
, y, z) + Fx(x ↓ !x
2
, y, z)
#x
#V
=↔
1
#V

S1+S2
(F · n̂) dS ↑
Fx(x + !x
2
, y, z) + Fx(x ↓ !x
2
, y, z)
#x
In the limit as #V approaches zero (and hence as #x, #y, #z → 0), we arrive at a result
involving partial derivatives. These are derivatives where we are only interested in how

45 | The Axiom
a multi-variable function changes with respect to one of its variables, and so we hold
any others constant. The right-hand side of the above equation therefore represents the
derivative of Fx with respect to x - we have held any changes in y or z constant. We
write:
lim
!V →0
1
#V

S1+S2
(F · n̂) dS =
ϑFx
ϑx
From the symmetry of the cuboid system we assumed, the contributions from the other
two pairs of faces are simply partial derivatives with respect to the other two variables.
Our final result is therefore the limit as #V → 0 of the surface integral of our vector
function as projected onto a unit normal to the surface expressed in terms of partial
derivatives.
lim
!V →0
1
#V

S
(F · n̂) dS =
ϑFx
ϑx
+
ϑFy
ϑy
+
ϑFz
ϑz
Hence, we have derived a simpler form of the divergence of a vector function. This result
holds true for all volumes, but we shall leave this as a research task for the reader.
Returning to our expression for Gauss’ Law, we can now see that we have a relation
between a property of the electrostatic field (the divergence) and a known quantity (the
charge density) at a point.
ϑEx
ϑx
+
ϑEy
ϑy
+
ϑEz
ϑz
=
ε(x, y, z)
ω0
However, despite being a di!erential equation, it contains three unknowns and so is still
not in its most workable form. We therefore introduce a new notation and operator that
has considerable use in vector calculus - the del operator. The del operator, given symbol
↗, is defined by the following equation.
↗ = i
ϑ
ϑx
+ j
ϑ
ϑy
+ k
ϑ
ϑz
By taking the scalar product of ↗ and a vector function F = iFx + jFy + kFz, we arrive
at:
↗ · F =

i
ϑ
ϑx
+ j
ϑ
ϑy
+ k
ϑ
ϑz

· (iFx + jFy + kFz)
=
ϑ
ϑx
Fx +
ϑ
ϑy
Fy +
ϑ
ϑz
Fz
=
ϑFx
ϑx
+
ϑFy
ϑy
+
ϑFz
ϑz
= div F
Thus, Gauss’ Law for Electrostatics can be rewritten in its di!erential form, as it appears
in Maxwell’s equations.
↗ · E =
ε
ω0
4 Di!erential or integral?
We now have two di!erent, but entirely equivalent, forms of Gauss’ Law that allow us to
describe the relationship between a given charge distribution and the resultant electro-
static field. But how would you decide which form to apply to a situation? Revisiting

our derivation, the motivation for finding a new form of Gauss’ Law came from being
unable to e!ectively deal with non-symmetric charge distributions. Since the integral
form provided an ‘equation’ with an infinite number of unknowns, we decided to consider
the electric flux locally, about a single point. The key distinction between the two forms
is therefore that the integral form describes the relationship globally, across a surface,
whilst the di!erential form is only concerned with local behaviour of the electrostatic
field. Maxwell’s equations, as presented in A Dynamical Theory of the Electromagnetic
Field, were written in di!erential form and his subsequent work that showed electromag-
netic waves are the result of perpendicular oscillations of the electric and magnetic fields
with propagation limited to the speed of light, came from manipulation in di!erential
form. Whilst the integral form of Gauss’ Law may seem more intuitive, the point-wise
analysis of the di!erential form is fundamental to many physical theories (not to mention
its arguably greater cosmetic value).
Our ‘physicist’s rough-and-ready proof’, as H. M. Schey (author of Div, Grad, Curl
and All That: An Informal Text on Vector Calculus) would describe it, of the di!erential
form of Gauss’ Law beginning with it in integral notation, can be continued to arrive
at a powerful mathematical statement known as ‘The Divergence Theorem’. Through
our work with electrostatics and Gauss’ Law, we have already made some connections
between surface and volume integrals; the theorem formerly relates the two.

S
(F · n̂) dS =

V
(↗ · F) dV
As an aside, it is left to the reader to use the theorem to derive the di!erential form of
Gauss’ Law.
References
[1] M. J. Crowe, (2002) A History of Vector Analysis.
[2] H. M. Schey, (1997) Div, Grad, Curl and All That: An Informal Text on Vector
Calculus.
[3] Sequential Math, (n.d.) A Brief History of: Vector Calculus.
[4] Wikipedia, (2024) Divergence theorem.
[5] Wikipedia, (2024) Mean value theorem

The Axiom - Issue 17 - St Andrew's Day 2024

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The Axiom - Issue 17 - St Andrew's Day 2024