FIRST YEAR OF MATHS SOCIETY

MATHS SOCIETY

Abstract. In the first year of maths society, we travelled through

various topics. Saga 1 covers the basics of dynamical systems, from
the Collatz conjecture to the Mandelbrot set. In saga 2, we look at
a more number theoretical topic: the Fibonacci numbers. We un-
cover connections between the Fibonacci numbers and the golden
ratio through continued fractions, and then generalise this. We also
looked briefly at transcendental numbers after this. We then got
deep and philosophical in saga 4, looking at topics that ask ques-
tions about the foundations of mathematics itself from Russel’s
Paradox and how that motivates ZFC set theory, to the incom-
pleteness theorems. We then finished off the year with saga 5 by
going back to a more number theoretic theme, but this time look-
ing at various functions in analytic number theory, most notably
the Zeta function and the Gamma function and some applications.
Finally, and most importantly, Gregor Sanfey (leader of maths soci-
ety) would like to give a huge thanks to Wren Shakespeare, Prasan
Patel, Alex Dowling, Rowan Oliviera, Saik Rudad, Akshayan and
Danica for their excellent talks, all of the regular attendees (you
know who you are) and everyone reading this right now. I hope
that you enjoy reading this feast of mathematical knowledge before
your eyes!

Contents

Part 1. Saga 1- Dynamical Systems 4

1. Collatz Conjecture 4
1.1. Introduction 4
1.2. Arguments for the conjecture being true 4
1.3. Arguments Against 5
1.4. Further Reading 5
2. Introduction to Dynamical Systems and the Mandelbrot Set 5
2.1. Introduction 5
2.2. A simple example 5
2.3. Chaos and the Julia Set 6
2.4. A closer look at periodic points 6
Date: 2023-2024.
1
2 MATHS SOCIETY

2.5. Mandelbrot Set 7

2.6. Summary 8
3. Newton’s Fractal 8
3.1. Introduction 8
3.2. The Newton-Raphson Method 8
3.3. The Newton Fractal 9
3.4. The Connection to the Mandelbrot Set 9
3.5. Summary 10

Part 2. Saga 2- Generalising Connections Between the

Golden ratio and Fibonacci Numbers 11
4. Continued Fractions 11
4.1. Introduction 11
4.2. Application to Quadratics 11
4.3. Two Shocking Connections 12
5. On the Fibonacci Numbers and Golden Ratio 13
5.1. Introduction 13
5.2. Deriving the General Formula for Fn 13
5.3. Developing a Test For Fibonacci Numbers 14
6. Pell Numbers, The Silver Ratio and More Generalisations 15
6.1. Introduction 15
6.2. Pell Numbers and The Silver Ratio 15
6.3. The Maths Society Ratio 17
6.4. Further Reading 18

Part 3. Saga 3- Transcendental Numbers 18

7. Transcendental Numbers Exist- A Look At Liouville
Numbers 18
7.1. Introduction 18
7.2. Liouville Numbers 19
7.3. Liouville Constant 20
8. e and π are transcendental 20

Part 4. Saga 4: Set Theory and Logic 21

9. Russel’s Paradox and ZFC Set Theory 21
9.1. Introduction 21
9.2. The Paradox 21
9.3. A look towards ZFC set theory 21
9.4. Various Interpretations 22
9.5. The Von Neumann Universe 23
10. A Closer Look at the Natural Numbers 24
10.1. Introduction 24
 FIRST YEAR OF MATHS SOCIETY 3

10.2. The Peano Axioms 24

10.3. Defining Addition and Multiplication 25
10.4. Set Theoretic Definition of Natural Numbers 26
11. Gödel’s Incompleteness Theorems 26
11.1. Introduction 26
11.2. What are Gödel’s Incompleteness Theorems? 26

Part 5. Saga 5: Functions in Analytic Number Theory

(featuring Prasan) 27
12. The Basel Problem 27
12.1. Introduction 27
12.2. The Solution 27
13. The Euler Product for the Zeta Function: Prasan Patel 28
13.1. Introduction 28
13.2. Setting up the Product 29
13.3. The Euler Product 29
14. The Gamma Function: 12 !


30
14.1. Introduction 30
14.2. The Gamma Function 30
14.3. The Integral 31

Part 6. Guest Talks 33

15. The Riemann Zeta Function, or, How to Win £1000000-
Wren Shakespeare 33
15.1. On the Real Line 33
15.2. Complex Numbers 33
16. Introduction to Taylor Series- Prasan Patel 34
16.1. Introduction 34
16.2. Example: ex 34
16.3. Generalising This Process 36
17. Differentiation under the integral sign: Wren Shakespeare 37
18. The logistic map: Rowan Oliveira 39
18.1. Introduction 39
18.2. Studying what happens as r varies and the Feigenbaum
Constant 40
18.3. Connection with the Mandelbrot Set 41
19. Derivatives and Integrals of Displacement: Saik Rudad 42
19.1. Introduction 42
19.2. The Derivatives 42
19.3. The integral 42
4 MATHS SOCIETY

Part 1. Saga 1- Dynamical Systems

1. Collatz Conjecture
1.1. Introduction. The Collatz conjecture is an infamous problem
which was posed in 1937. It’s simplicity has lured many mathemati-
cians into thinking they can solve it, yet nobody has done so so far
(there was even a conspiracy that the Soviets posed this problem to
the West to deliberately slow down their mathematicians’ progress!).
The problem is as follows: Pick a number n.
• If n is odd, multiply by 3 and add 1.
• If n is even, divide it by 2.
For example, if we start with n = 5 we would get the sequence:
5 → 16 → 8 → 4 → 2 → 1.
Conjecture 1.1 (Collatz Conjecture). No matter which number n you
start with, your sequence will always go down to 1.
1.2. Arguments for the conjecture being true.
(1) The conjecture has been tested for the first 268 numbers, and
all of those numbers have satisfied the conjecture.
(2) It could be the case that there is another loop of numbers in the
collatz conjecture other than 1 → 4 → 2 → 1, which would dis-
prove it, but any such loop would have to be at least 17,000,000
numbers long!
(3) The sequence where you multiply odd numbers by 3 and add
1 seems to be special. For example, when we add 5 instead of
1 (sending odd n to 3n + 5) we get all sorts of other loops of
integers, for example 5 → 20 → 10 → 5, yet when you multiply
the odd numbers by 3 and add 1 we still haven’t found any
counterexamples for the first 268 digits.
(4) In 2003, Krasikov and Lagarias proved that for any set of num-
bers 1, 2, 3, . . . , x, at least x0.84 of them will satisfy the collatz
conjecture.
(5) “The probabilistic argument”. Every even number will ob-
viously be decreased by the sequence (because it is divided by
2). However it seems that odd numbers increase, so possibly
they could diverge off to infinity. However “on average” this is
not true either- it will be multiplied by a factor of 34 “on aver-
age”. More precisely, an odd integer n will be sent to 3n + 1.
This will always be even, so this goes to 3n+1 2
. Now this has a
50% chance of being even so 25% of the time the odd integer
will be sent to 3n+1 4
and so 12.5% of the time it will be sent to
 FIRST YEAR OF MATHS SOCIETY 5

3n+1
8
.
Thus the expected growth of this odd integer n will end
up being:
  21   14   18
3 3 3 3
··· = .
2 4 8 4
Don’t worry if that last one went a bit over your head, essentially all it
was saying is that odd numbers get multiplied by 43 on average so they
should decrease to 1- I just wanted to add the precise detail for those
who were interested.

1.3. Arguments Against.

(1) They still haven’t proven it! It’s nearly been 90 years since the
problem was posed, and nobody has solved it yet so perhaps
that is simply because it is not true; you can’t prove something
that is actually false.
(2) Despite the fact that they have computed the first 268 numbers,
that is still not convincing- some conjectures have broken only
at the 21000 th number!
(3) Even though it would be at least 17, 000, 000 numbers long, the
3
4
argument doesn’t counter the possibility of there being a loop.

1.4. Further Reading.

• In 2019, mathematician Terrence Tao made huge progress in
the problem, which you can look into if interested.

2. Introduction to Dynamical Systems and the

Mandelbrot Set
2.1. Introduction. In today’s talk we talk about dynamical systems.
Roughly speaking, this is the study of what happens when you iterate
F (z)
some rational function ϕ (meaning it is of the form ϕ(z) = G(z) where
F, G are polynomials) a bunch of times. For example, the Collatz con-
jecture is a question about a dynamical system, which should already
show how interesting this field of maths can get very quickly. Today
we look in more depth into some interesting properties, and some more
unsolved questions in this mysterious, but beautiful field of maths.

2.2. A simple example. Let us begin by looking at the iterations of

ϕ(z) = z 2 . We can ask if any numbers will go in a loop, and of course
the answer is yes.
• The orbit of z is the set {ϕ(z), ϕ2 (z), ϕ3 (z), . . . }
6 MATHS SOCIETY

• For example 0 will go in the sequence {0, 0, 0, . . . } and similarly

1 will continue off as {1, 1, 1, . . . }. These numbers are called
fixed points, because they satisfy the equation ϕ(a) = a.
• The number −1 is called preperiodic, because it will eventually
end up in a loop which doesn’t actually have the number −1 as
follows: {−1, 1, 1, 1, . . . }. √
• The complex number −1+2 −3 is called periodic, because it goes
√ √ √
in a loop as follows: { −1+2 −3 , −1−2 3 , −1+2 −3 , . . . }.
• Points such as 2 or 13 are called wandering points, because they
will never go in a loop; the number 2 diverges off to infinity and
the number 13 gets closer and closer to 0.
• Note that we can have “points at infinity”. For example if we
2 +1
study the function ϕ(z) = zz2 −1 , then if we allow the point ∞,
1 is periodic with the following orbit: {1, ∞, 1, ∞, . . . }.
2.3. Chaos and the Julia Set. If we look again at the function
ϕ(z) = z 2 , we notice something interesting. For numbers α between
1 and 0, the orbits will go towards zero and for any number β that is
close to α, their orbits will stay very similar too. Similarly, for numbers
greater than 1, if α and β are close to each other then the orbits will
also stay close to each other. However, when α = 1, because it is a
fixed point, no matter what β we pick, no matter how close it is to 1,
will always have an orbit that goes further and further away from 1.
This is an example of a chaotic point.
Definition 2.1. We can define the Fatou set F (ϕ) and Julia set J(ϕ)
as follows:
J(ϕ) = {α such that α is chaotic.}
F (ϕ) = {α such that α is not chaotic} = J(ϕ)′

2.4. A closer look at periodic points. Recall that a periodic point

of a rational function ϕ is a number z such that ϕn (z) = z for some in-
teger n. We will spend the rest of the talk examining rational periodic
points specifically. It is known that a rational function will have infin-
itely many complex periodic points, and many times infinitely many
real ones as well. However Northcott’s theorem states the following:
Theorem 2.1 (Northcott, 1949). A rational function ϕ has only finitely
many rational periodic points.
Let us look at an example. ϕc (z) = z 2 + c. When c = 0, this has a
couple of rational numbers that have period 1 (aka fixed points). Those
 FIRST YEAR OF MATHS SOCIETY 7

Figure 1. Julia set of ϕ(z) = z 2 − 0.8 + 0.156i

are z = 0 and z = 1. Now let’s take c = 41 . Then z = 12 is also a fixed

point. What about larger loops? Well when c = −1, then z = 0 will
be a point of period 2 as follows:
−1 → 0 → −1 → 0 → . . .
−29 −7
Can we go higher? Yes. Take c = 16
and z = 4
. Then the loop
becomes:
−7 5 −1 −7
→ → → .
4 4 4 4
At this point you might be thinking that we can always keep making
loops that are longer and longer, but here comes a shock:
• Morton proved that ϕc can’t have a rational periodic point of
period 4.
• Flynn, Poonen and Schaefer proved that ϕc can’t have a rational
periodic point of period 5.
• Poonen has conjectured that there are no rational periodic points
of φc that have a period greater than 5 either.

2.5. Mandelbrot Set. Now we will look at a famous set that has
come out of the study of dynamic systems: the Mandelbrot set. We
look again at the function ϕc (z) = z 2 +c. The Mandelbrot set is defined
as the set of all complex numbers c such that the function doesn’t go
off to infinity when we start at z = 0. For example, c = 1 is not in the
Mandelbrot set because the sequence will be:
1 → 11 + 1 = 2 → 22 + 1 = 5 → 26 → . . .
8 MATHS SOCIETY

But −2 is clearly not going to explode to infinity because the sequence

is:
−2 → 2 → 2 → 2 → . . .
To finish the talk we played around with the Mandelbrot set on geoge-
bra here and watched a cool animation of it, because it is simply so
aesthetically pleasing.
2.6. Summary. In this talk we went through a tour in the basics of
dynamical systems. We started by looking at the basic example of
ϕ(z 2 ) to define some of the important terms in dynamical systems, and
then went through an interesting conjecture about rational periodic
points, before finally ending by looking at the Mandelbrot set, which
is widely considered one of the most beautiful mathematical objects in
the world.

3. Newton’s Fractal
3.1. Introduction. This talk can be split roughly into three parts:
(1) First, look at the Newton-Raphson method, which is a method
for approximating solutions to polynomials.
(2) Look at how we can make it into a fractal on the complex plane
(known as the Newton fractal).
(3) Third, look at the connection between the Newton fractal and
the Mandelbrot set (a set which was introduced in the last talk)
and generalise it (based on the paper entitled “The Mandelbrot
set is universal”), which is a truly remarkable result.
3.2. The Newton-Raphson Method. First we look at the Newton-
Raphson method. For solving quadratics, we have an easy method:
simply use the quadratic formula. For cubics, it’s not so simple but
there is a cubic formula one can use if one wishes to get an exact
solution. For quartics, there is an absolute mess of a formula that will
give you solutions. However, there is no quintic (or above) formula,
so we can’t get the exact solutions in this way. Thus, we approximate
it instead (and also we do the same for quartics and cubics, because
their formulas are so clunky it’s better to just use this method instead).
Since it is not the focus of the talk, I shall simply state the formula
used to approximate the solutions without saying why it works. For a
polynomial p(z), we guess an answer z0 and then we use the recursive
formula to get a more and more accurate estimate:
p(z)
zn+1 = zn − .
p′ (z)
 FIRST YEAR OF MATHS SOCIETY 9

Example 3.1. If we want to approximate the square root of 2, we can

use the Newton-Raphson method on the equation z 2 − 2. If we guess
z0 = 1.5 then we have:
z0 = 1.5
1.52 − 2
z1 = 1.5 − = 1.42
2 × 1.5
z2 − 2
z2 = z1 − 1 = 1.41
2z1
z3 = . . .
√
here we see how quickly the Newton-Raphson approximated 2,
which is a good sign that it is a good way to approximate roots. Now
we will move to the complex plane and see how to make a fractal out
of it.
3.3. The Newton Fractal. Now we will look at how this makes a
fractal. Clearly, since there are multiple roots of most polynomial
equations, different starting guesses will be attracted to different roots.
Thus, we will colour the points in the plane that are attracted to differ-
ent roots in different colours and this process will produce some beau-
tiful fractals. It also highlights the sensitivity of the Newton-Raphson
method in certain areas of the complex plane.

Figure 2. Newton’s Fractal for p(z) = z 3 − 1

3.4. The Connection to the Mandelbrot Set. Now we look at

a shocking connection to the Mandelbrot set that Newton’s fractal
reveals. We will now look at the cubic polynomial
1 1
p(z) = (z + 1)(z − − λ)(z − + λ).
2 2
Now start off the Newton-Raphson method for this polynomial with
z0 = 0. If the Newton-Raphson method converges to a certain root,
colour that λ a certain colour. But here’s where it gets exciting: when
you colour the points λ such that it doesn’t converge to any root,
10 MATHS SOCIETY

we get a picture which looks exactly like the Mandelbrot set! I

cannot exaggerate enough how incredible this is: a seemingly unrelated
fractal given by the Newton-Raphson method gives us a fractal (already
surprising), which also reveals the Mandelbrot set too! This is a prime
example of the beauty of mathematics. This connection was further
explored in a paper entitled “The Mandelbrot Set is Universal” by
McMullen, if you are curious about how deep this goes.

Figure 3. Zooming in on the points that fail to con-

verge for p(z) = (z + 1)(z − λ − 21 )(z + λ − 21 ), taken from
“On the Dynamics of Polynomial-Like Mappings”

3.5. Summary. In conclusion, we looked at the Newton-Raphson method,

and how it surprisingly generates fractals which produces some of the
best images in mathematics. Then, we proceeded to look at an ex-
tremely interesting connection between this fractal and the Mandelbrot
set, which highlights the beauty of mathematics.
 FIRST YEAR OF MATHS SOCIETY 11

Part 2. Saga 2- Generalising Connections Between the

Golden ratio and Fibonacci Numbers
4. Continued Fractions
4.1. Introduction. In this talk we looked at the idea of continued fractions.
We begin with a simple example
43 5
=2+
19 19
1 1
= 2 + 19 = 2 + 4
5
3+ 5
1 1
= 2 + 19 = 2 + 1
5
3+ 5
4

1
=2+
3 + 1+1 1
4

Thus, we see that even a simple fraction admits an interesting continued

fraction expansion. But the theory of continued fractions, as we shall
see provides us with some deep connections to number theory...

4.2. Application to Quadratics. Now consider a quadratic equa-

tion, for example x2 − 5x − 1 = 0. Normally, to solve this we would use
the quadratic formula, but here we shall use the language of continued
fractions:

x2 = 5x + 1
1
x=5+
x
1
x=5+ 1
5+ x
1
x=5+ 1
5+ 5+ 1
1
5+ ...

And so we’ve expressed the solution to this quadratic as an infinite

continued fraction! But now let us dive deeper and look at a connection
to the golden ratio. Recall that the golden ratio is the solution to the
equation

φ2 = φ + 1.
12 MATHS SOCIETY

And so we shall rearrange it:

1
φ=1+
φ
1
φ=1+ 1
1+ φ
1
φ=1+ 1
1+ 1
1+ ...

Or in other notation, φ = [1; 1, 1, 1, . . . ]. It turns out that this contin-

ued fraction has a shocking connection to the Fibonacci numbers...

4.3. Two Shocking Connections.

Definition 4.1. Remember that the Fibonacci numbers Fn are defined

by the relation Fn = Fn−1 + Fn−2 with F0 = 0 and F1 = 1. The first
few numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, . . .

We observe that the convergents of the golden ratio is precisely the

ratio of the Fibonacci numbers. Let us the first few incidents:

1 1 1
1, 1 + , 1 + 1 ,1 + ,...
1 1 + 1+1 1 + 1+1 1
1+1

which becomes:

1 3 5 8
, , , ,...
1 2 3 5

As we can see, as we go through the convergents of the infinite con-

tinued fraction of φ, we keep getting ratios of two Fibonacci numbers.
From this we can also infer the cool fact that as we take this process
to infinity, we have:

Fn+1
lim =φ
n→∞ Fn

We now look at a second shocking connection to number theory which

we observed- a connection to Pell numbers.
√ First, we have to find
the continued fraction expansion for 2 + 1. This is the solution to
 FIRST YEAR OF MATHS SOCIETY 13

(x − 1)2 = 2 and so we obtain:

x2 = 2x + 1
1
x=2+
x
1
x=2+ 1
2+ x
1
x=2+ 1
2+ 1
2+ ...

Now we look at how this connects to Pell numbers.

Definition 4.2. The Pell numbers are defined by Pn = 2Pn−1 + Pn−2
with P0 = 0, P1 = 1. The first few Pell numbers are: 0, 1, 2, 5, 12, 29, 70, . . . .
√
The convergents of x = 2 + 1 are the following:
1 1 1
2, 2 + , 2 + 1,2 + ,...
2 2+ 2 2 + 2+1 1
2

which become
2 5 12 29
, , , ,...
1 2 5 12
and again we see how the convergents of a certain special continued
fraction are deeply related√to a famous sequence in number theory.
Fun fact: the number 1 + 2 is called the silver ratio. Next time we
shall look in more depth into the Fibonacci numbers and the golden
ratio.

5. On the Fibonacci Numbers and Golden Ratio

5.1. Introduction. In the last talk we unveiled a shocking connection
between the continued fraction of the golden ratio and the Fibonacci
numbers (check it out if you haven’t seen the notes from that talk).
This week we delve deeper into the connection between the golden
ratio and the Fibonacci numbers, and uncover some deep results about
the Fibonacci numbers.

5.2. Deriving the General Formula for Fn . We will now use an

ingenious argument to find a general formula for the nth Fibonacci
numbers. First recall that the golden ratio satisfies φ2 = φ + 1 and so
we have:
φn = φn−1 + φn−2 .
14 MATHS SOCIETY

We also have that the other solution to the quadratic ψ := 1 − φ =

−1
φ
satisfies this too. Thus in general, any sequence that satisfies the
recursion Un = Un−1 + Un−2 will be of the form:
Un = aφn + bψ n
since
Un = aφn + bψ n
= a(φn−1 + φn−2 ) + b(ψ n−1 + ψ n−2 )
= Un−1 + Un−2
as required. Now we apply this to Fn . Since F0 = 0 and F1 = 1, we
have that:
a+b=0
aφ + bψ = 1
√1 , −1
And once the algebra is done, you have: a = 5
b = √
5
and so we
have:
φn − (−φ)−n
(1) Fn = √
5
(using ψ = −φ−1 ).
5.3. Developing a Test For Fibonacci Numbers. Using the equa-
tion we derived in the previous section, we can develop a test to see
if a random integer x is a Fibonacci number, meaning you don’t have
to go through all of the Fibonacci numbers to test it! First we must
prove a preliminary lemma:
Lemma 5.1. φn = φFn + Fn−1
Proof. We do this by induction which means we prove it for the first
case (n = 1) and then we assume that it is true for the nth case and
show that this implies that the statement is true for the (n + 1)th case.
So let us proceed:
The base case: φ1 = φ(1) + 0 which is clearly true.
The inductive step: Let us assume that φn = φFn + Fn−1 . Then:
φn+1 = φ2 Fn + φFn−1
= (φ + 1)Fn + φFn−1
= φ(Fn + Fn−1 ) + Fn
= φFn+1 + Fn
as required. □
 FIRST YEAR OF MATHS SOCIETY 15

Now we can go about making this test. Using 1 we have that:

√
5Fn = φn − (−φ)−n
√
φn 5Fn = φ2n − (−1)n
√
φ2n − φn 5Fn − (−1)n = 0
√ p
5Fn + 5Fn2 + 4(−1)n
φn = ,
2

where the last step comes from the quadratic formula (if look closely,
it’s a quadratic of φn ). Now using the lemma we derived before, we
have:
√ p
5Fn + 5Fn2 + 4(−1)n
φFn + Fn−1 =
√ 2p
= 2φFn + 2Fn−1 = 5Fn + 5Fn2 + 4(−1)n
√ √ √
2φ=1+ 5 p
=⇒ (1 + 5)Fn + 2Fn−1 = 5Fn + 5Fn2 + 4(−1)n
p
=⇒ Fn + 2Fn−1 = 5Fn2 + 4(−1)n
=⇒ (Fn + 2Fn−1 )2 = 5Fn2 + 4(−1)n .

And here we are at the crux of the argument! Since the right hand side
is always a perfect square, we can say that x is a Fibonacci number iff
5x2 − 4 is a perfect square or 5x2 + 4 is a perfect square. Pretty nifty,
eh!

6. Pell Numbers, The Silver Ratio and More

Generalisations
6.1. Introduction. This week we finish the saga on all that we’ve
been looking at in number theory for the last few weeks. And this will
hopefully be an exciting end!

6.2. Pell Numbers and The Silver Ratio. The silver ratio is de-
fined in a very similar way to the golden ratio. Whilst the golden ratio
is defined as the solution to x2 − x − 1 = 0, the silver ratio shall be
defined as the positive√solution to x2 − 2x − 1 = 0. In other words,
the silver ratio is 1 + 2. We can observe that the silver ratio enjoys
16 MATHS SOCIETY

a very similar continued fraction expansion to the golden ratio:

x2 − 2x − 1 = 0
x2 = 2x + 1
1
x=2+
x
1
=2+
2 + x1
1
=2+
2 + 2+ 1 1
2+...

which looks exactly like the golden ratio, except for that one the 2s were
replaced with 1s (see the notes from that legendary talk last week for a
refresher). However, here the fun has just begun! Because recall that
last time we observed a connection between the golden ratio and the
Fibonacci numbers (which were defined with the recurrance relation
Fn = Fn−1 + Fn−2 .) So this time, the silver ratio will have a shocking
connection with the very similar Pell numbers which are defined by
Pn = 2Pn−1 + Pn−2 (with P0 = 0 and P1 = 1). So the first few Pell
numbers would be: 0, 1, 2, 5, 12, 29, 70, . . . . Again there is a shocking
connection: the convergents of the silver ratio are the following:
1 1 1
2, 2 + , 2 + 1,2 + ,...
2 2+ 2 2 + 2+1 1
2

which become
2 5 12 29
, , , ,...
1 2 5 12
Furthermore, just as we did with the Fibonacci numbers, we can now
use the silver ratio to get a formula for the nth Pell number. Because
we can simply multiply x2 = 2x + 1 by xn−2 on both sides, we √ have
n n−1 n−2
that x = 2x +x and so since the solutions are x = 1 ± 2, we
know the formula for the nth Pell number will be of the form
√ √
Pn = a(1 + 2)n + b(1 − 2)n
since
√ √
Pn = a(1 + 2)n + b(1 − 2)n
√ √ √ √
= 2a(1 + 2)n−1 + 2b(1 − 2)n−2 + a(1 + 2)n−1 + b(1 − 2)n−2
= 2Pn−1 + Pn−2
as required. So now we must find a and b. We have set P0 = 0
and P1 = 1, so we just plug these two values in and get simultaneous
 FIRST YEAR OF MATHS SOCIETY 17

equations and then we should be fine and dandy. More specifically, we

get:
√ √
P0 = a(1 + 2)0 + b(1 − 2)0 = a + b = 0
√ √
P1 = a(1 + 2) + b(1 − 2) = 1.
After you do the algebra, you get that a = 2√1 2 and b = 2−1
√ . So we
2
obtain a glorious formula for the nth Pell number being:
√ √
(1 + 2)n − (1 − 2)n
Pn = √ .
2 2
So again we see that there is a connection between these special ra-
tios and some very nice sequences in number theory. Thus, as any
mathematician would seek to do, we generalised it and saw the full
story!
6.3. The Maths Society Ratio. Yes, I know these are called the
metallic means, but we generalised it ourselves and the results agreed
with what mathematicians had inevitably already done so I called it
the maths society ratio: sue me! Anyway, remember that the quadratic
equations defining the golden and silver ratios ,respectively, were x2 −
x − 1 = 0 and x2 − 2x − 1 = 0. So naturally we then looked at
x2 − nx − 1 = 0, and called this the maths society ratio (which we
denoted δ because it was√Mithush’s favourite Greek letter). Anyway, let
2
us observe that δ(= n+ 2n +4 ) enjoys a very similar continued fraction
expansion to the golden and silver ratios:
x2 − nx − 1 = 0
x2 = nx + 1
1
x=n+
x
1
=n+
n + x1
1
=n+
n + n+ 1 1
n+...

We now see how this links to the sequence Mi = nMi−1 + Mi−2 , with
M0 = 0 and M1 = 1. For example, when n = 3 the first few terms
would be: 0, 1, 3, 10, 33, 109, . . . . And the convergents of the continued
fraction would be:
1 1 1
3, 3 + , 3 + 1,3 + ,...
3 3+ 3 3 + 3+1 1
3
18 MATHS SOCIETY

which become
3 10 33 109
, , , ,...
1 3 10 33
And in general the convergents of the maths society ratio for any n are
1 1 1
n, n + , n + 1 ,n + ,...
n n+ n n + n+1 1
n

which become
n n2 + 1 n(n2 + 1) + n n(n(n2 + 1) + n) + n2 + 1
, , , ,...
1 n n2 + 1 n(n2 + 1) + n
which is precisely just the first few terms of our sequence in a fraction.
The last thing we did was find a general formula for Mi . Using the
same
√
reasoning again it will be of the form Mi = aδ i + bδ̄ i , where δ̄ is
2
n− n +4
2
. And again, doing the algebra on the simultaneous equations
a+b=0
aδ + bδ̄ = 1
yields a = √ 1 and b = √ −1 and so our general formula is:
n2 +4 n2 +4

δ i − δ̄ i
Mi = √
n2 + 4
(fun exercise: check that this agrees with the formulae for the nth
Fibonacci and nth Pell number.)

6.4. Further Reading. Thank you to Saik for pointing out this link
which had some more amazing results on the metallic means- we barley
scratched the surface in this saga!

Part 3. Saga 3- Transcendental Numbers

7. Transcendental Numbers Exist- A Look At Liouville
Numbers
7.1. Introduction. Last time it was revealed that our next saga would
be a look into the world of transcendental numbers, which are numbers
that are not a root of any equation an xn +an−1 xn−1 +· · ·+a0 where ai ∈
Z. In this talk we establish the fact that these numbers exist by looking
at the first example that was discovered in 1844- the Liouville constant,
which is an example of a class of transcendental numbers known as
Liouville numbers
 FIRST YEAR OF MATHS SOCIETY 19

7.2. Liouville
P Numbers. The goal of this talk is to prove that the
number L = ∞ 1
n=1 10n! is irrational. We do this in the following way:
(1) Prove that all Liouville numbers are transcendental
(2) Show that L is a Liouville number.
Sadly, the definition of a Liouville number is pretty technical.
Definition 7.1. A Liouville number α is a number such that for all
n ∈ N, there exists a rational number ab (with b > 1) such that:
a 1
0< α− < n.
b b
Proposition 7.1. If α is a Liouville number, then α is irrational.
Proof. Let us assume that α is a rational Liouville number pq . Then for
some rational number ab ̸= pq , we have:
a p a pb − aq
0< α− = − =
b q b qb
Now we pick a natural number n such that 2n−1 > q. Then we have:
pb − aq 1 1
> n−1 ≥ n
qb 2 b b
where the last inequality sign comes from the fact that b > 1. Thus
we have shown that for any ab we try to choose, there will be an n ∈ N
such that α − ab > b1n which contradicts the assumption that α was a
Liouville number. Thus, all Liouville numbers are irrational. □
Now that we have established that all Liouville numbers are irra-
tional, we can now move forward and try to prove that they are also
transcendental too.
Theorem 7.2. Liouville numbers are transcendental.
Proof. Let f (x) = an xn +an−1 xn−1 +. . . a0 and assume by contradiction
that f (α) = 0. We now define a few things: M = max[α−1,α+1] |f ′ (x)|,
A < {1, M1 , |α − α1 |, . . . |α − αm |} where f (αi ) = 0, ∀i ≤ m. Then we
pick some r ∈ N such that 2r ≥ A1 . Since α is a Liouville number, we
have some ab ∈ Q such that:
a 1 1 A
(2) 0< α− < n+r ≤ r n ≤ n < A.
b b 2b b
a
Now, because α − b < A, we have that:
(1) ab ∈
[α − 1, α + 1]
(2) f ab ̸= 0.
20 MATHS SOCIETY

Thus we can

use the mean value theorem to say that there exists some
x0 ∈ α, ab such that
f (α) − f ab −f ab



′
f (x0 ) = =
α − ab α − ab
So we can say that:
f ab f ab f ab




′ a
(3) |f (x0 )| = =⇒ α − = ′ ≥
α − ab b |f (x0 )| M
Now note that we have:
a a1 a an an 1 1
f = a0 + + · · · + n = n |bn a0 + a1 abn−1 + · · · + an an | ≥ n .
b b b b b
So finally we are done because now we plug this into 3 to obtain:
f ab


1 A a
(4) ≥ n > n > α− .
M b M b b
But 2 and 4 imply that α − ab > α − ab which is a clear contradiction
so there could not have existed some f such that f (α) = 0, so α is
transcendental as required! □
P∞ 1
7.3. Liouville Constant. We now show that L = n=1 10n! is a Li-
ouville number.
Proposition 7.3. L is a Liouville number.
P∞
Proof. Let us write L as m 1 1
P
n=1 10n! + n=m+1 10n! . Then this first part
can be collapsed into a single fraction of the form 10an! . We shall pick
our b in the fraction ab to be 10n! . Then:
∞ ∞
!
a X 1 1 X 1 1
0< L− = n!
< n n
= n
b n=m+1
10 b n=1
2 b

where the last inequality comes from simply comparing the denomina-
tors of the fractions on the sums and observing that they are bigger in
the sum on the left and so we have shown that L satisfies the conditions
being a Liouville number and so L is transcendental! □

8. e and π are transcendental

I was lacking that week, so we just watched a YouTube video. The
notes are in this link: here
 FIRST YEAR OF MATHS SOCIETY 21

Part 4. Saga 4: Set Theory and Logic

9. Russel’s Paradox and ZFC Set Theory
9.1. Introduction. Today we shall begin our journey through set the-
ory and logic by discussing one of the most famous paradoxes around-
Russel’s paradox. This paradox sparked a whole branch of discussion
about set theory and logic, which then prompted mathematicians Ernst
Zermelo and Abraham Fraenkel to come up with a set of axioms to re-
solve this, which seems to fix Russel’s paradox and has worked very
well so far. We conclude by looking at the Von Neumann universe,
which describes what all of the sets in ZFC set theory should look like.

9.2. The Paradox. We shall jump straight into the paradox itself.
Firstly, we shall talk about the barber of Seville. The barber of Seville
shaves everyone who doesn’t shave themselves. Then does he shave
himself? Well, if he doesn’t then he does, but if he does then he
doesn’t- a paradox! Whilst this is an interesting paradox, you may be
wondering what relation this has to set theory. Well naively, one may
say that a set is any definable collection of elements. Well, in that case
we can define the set of all sets that don’t contain themselves:
S = {X|X ̸∈ X}
Well then the question is: is S ∈ S. Well if that were the case, then it
contains itself, so its not in S, but if it doesn’t contain itself, then it is
in S, which is a paradox!

9.3. A look towards ZFC set theory. Since naive set theory was
shown to lead to contradictions, mathematicians had to try to find
a satisfactory way to rebuild the foundations of mathematics, or else
much of mathematics would be under threat. This lead to a set of
axioms known as the Zermelo–Fraenkel axioms (ZFC set theory). We
shall roughly outline them here, but they will be a topic for further
discussion in the following talks.
(1) The axiom of extensionality
This states that two sets are equal if they have the same ele-
ments. For example, the set of all numbers of the form 2n + 2
and the set of all numbers of the form 2(n + 1) are the same
set.
(2) The axiom of regularity
This states that every non-empty set A contains a member B
such that A ∩ B = ∅. Note that this means that we cannot have
a set that contains itself, which eliminates Russel’s paradox.
22 MATHS SOCIETY

(3) The axiom of seperation

Any “subset” of a set is itself a set (we need to define what
precisely we mean by subset).
(4) The axiom of pairing
Given two sets a, b we can form a new set which contains a and
b as elements.
(5) The axiom of union
Given some sets a, b, c, . . . we can form their union a∪b∪c∪. . . .
(6) The axiom of infinity
There is a set X which has infinitely many elements.
(7) The axiom of powerset
Given a set a, we can form a set P(a) which is the set of all
subsets of a.
(8) The axiom of replacement
The image of a function is itself a set.
(9) The axiom of choice
This axiom says that if we’re given a bunch of sets, we can take
one element of each set and together those elements will form
a new set. Whilst this seems controversial, this was the most
controversial axiom, but it is now accepted by most mathemati-
cians. ZF set theory refers to just axioms 1 to 8, whilst ZFC
set theory includes the axiom of choice (C stands for choice).

9.4. Various Interpretations. We now move onto the slightly more

philosophical question of how to interpret these axioms. There are
many ways of doing this:
(1) Platonic
The axioms are describing the collection of all sets, and the
properties they have. The problem with this interpretation is
that we can’t really justify why if we can form this collection of
all sets, why shouldn’t it be a set itself?
(2) Formalism
There’s no point in discussing the meaning of the axioms, we
should just investigate their consequences and see what hap-
pens. The issue with this is if you ignore the meaning of the
axioms completely, then why bother studying them in the first
place? How do you know that they are interesting at all?
(3) Pragmatism
Assume that the axioms are true and just get on with proving
theorems without worrying about this stuff in the first place.
(4) Finitism
The axioms are nonsense anyway because infinite sets don’t
 FIRST YEAR OF MATHS SOCIETY 23

actually exist so there’s no point studying these axioms because

they’re not actually describing anything anyway. The problem
with this is the question of: why do these axioms work so well?
If we simply reject these axioms, then we would lose most of
mathematics!
9.5. The Von Neumann Universe. The Von Neumann universe is
a way of describing what all of the sets in ZFC set theory look like.
The idea behind the Von Neumann universe is that we shall build it up
from the starting point V0 = ∅. Let us proceed by taking the powerset
(set of all subsets) at each stage:
V0 = {} = ∅
V1 = {∅}
V2 = {∅, {∅}}
V3 = {∅, {∅}, {∅, {∅}}, {{∅}}}
V4 = . . .
One can see that this is already getting complicated, but we’ve just
begun- we now form:
Vω = V0 ∪ V1 ∪ V2 ∪ . . .
But we can still keep going: we take
Vω+1 = P(Vω )
Vω+2 = P(Vω+1 )
..
.
but we’re still not done! We now can form
V2ω = Vω+ω = Vω ∪ Vω+1 ∪ Vω+2 ∪ . . .
and then ,naturally, we can keep going for V2ω+1 and so on. These sub-
scripts are known as ordinal numbers. Then we form the Von Neumann
universe:
[
V = Vα
α

where α ranges through all the ordinal numbers. Note that this isn’t
a set itself, since if there were a set of all sets we would run into a
Russel’s paradox type of situation. The interpretation of this is that
all of the members of a set are themselves sets which are built up from
smaller sets. We shall see next time how this works for the natural
numbers.
24 MATHS SOCIETY

Figure 4. A visualisation for the Von Neumann uni-

verse, also giving the reason for why the letter “V ” is
used.

10. A Closer Look at the Natural Numbers

10.1. Introduction. Last time we looked at the axioms of ZFC set
theory and we finished with the Von Neumann universe. Today we
look closer at the natural numbers, and the set of rules that we use
as axioms for the natural numbers called the “Peano axioms”. We
finish by seeing the set theoretic definition of the natural numbers,
which should hopefully shed some insight into how the Von Neumann
universe works, with the idea of building up sets from smaller sets.
10.2. The Peano Axioms. The first axioms are to do with the rela-
tion “x = y”. The first Peano axiom (in some literature) is that this is
an equivalence relation. This means that:
(1) For all x ∈ N, we have x = x. This is called reflexivity.
(2) For all x, y ∈ N, we have x = y ⇐⇒ y = x. This is called
symmetry.
(3) For all x, y, z ∈ N. we have x = y and y = z implies x = z.
This is called transitivity.
With this out of the way, we can move onto the main Peano axioms.
The idea is to start with the first natural number 0 and to use a function
succ(x), called the successor function, in order to construct the rest of
them. The first two axioms are:
(1) 0 ∈ N
(2) If x ∈ N, then succ(x) ∈ N.
 FIRST YEAR OF MATHS SOCIETY 25

Now intuitively, we know that succ(x) = x + 1, so one might think that

we’re done. However, we don’t even know what “+” is, so as of now
we could have S(0) = 0 and the set of natural numbers would be {0}
and this would satisfy the axioms right now. Thus we have to add:
• For all x ∈ N, succ(x) ̸= 0.
Now we run into the issue that we could now define N = {0, 1} where
succ(0) = 1 = succ(1) and all the axioms are satisfied. Thus we also
add:
• For all x, y ∈ N, if S(x) = S(y), then x = y.
Now we’re very close, since we have that succ(0) = 1, but then succ(1)
must be some other number, which we call 2 and then succ(2) must
be a different number called 3 so on forever. However, we could still
define:
N = {0, 1, 2, . . . } ∪ {a, b}
with S(a) = b and S(b) = a and all the axioms are satisfied. Thus we
have the last axiom, called the axiom of induction:
• If there is a set S that satisfies the previous axioms, then it
contains all of the natural numbers.
So the final list of Peano axioms are the following, given a function
succ:
(1) 0 ∈ N
(2) If x ∈ N, then succ(x) ∈ N
(3) For all x ∈ N, succ(x) ̸= 0
(4) For all x, y ∈ N, if succ(x) = succ(y), then x = y
(5) If there is a set S that satisfies the previous axioms, then it
contains all of the natural numbers.

10.3. Defining Addition and Multiplication. Now, this is all well

and good, but we still want to define addition and multiplication in the
natural numbers. We define addition as follows:
For all a ∈ N, a + 0 = 0
a + succ(b) = succ(a + b)
For example, we have
1 + 1 = 1 + succ(0) = succ(1 + 0) = succ(1) = 2.
Then we can use this to do
1 + 2 = 1 + succ(1) = succ(1 + 1) = succ(2) = 3,
26 MATHS SOCIETY

and so addition is defined recursively as such.

We also define multiplication recursively by saying:
a·0=0
a · succ(b) = a + a · b
Then we can show, for example, that a · 1 = a, because:
a · 1 = a · succ(0) = a + a · 0 = a.

10.4. Set Theoretic Definition of Natural Numbers. Again, we

define the natural numbers recursively. This time, we define: 0 = ∅
and succ(n) = n ∪ {n}. Then we have:
0=∅
1 = 0 ∪ {0} = ∅ ∪ {∅} = {∅} = {0}
2 = 1 ∪ {1} = {∅} ∪ {{∅}} = {∅, {∅}} = {0, 1}
3 = 2 ∪ {2} = {∅, {∅}} ∪ {{∅, {∅}}} = {∅, {∅}, {∅, {∅}}} = {0, 1, 2}
..
.
We can see that this interpretation satisfies the Peano axioms. This is
very nice, a bit too nice in fact, so I shall throw a spanner in the works
next time!

11. Gödel’s Incompleteness Theorems

11.1. Introduction. Now that we have constructed some axioms that
seem to work very well in set theory, mathematics seems to be back
on steady grounding. In today’s talk, we examine a very unnerving set
of theorems known as Gödel’s incompleteness theorems, which tell us
some scary things about mathematics.

11.2. What are Gödel’s Incompleteness Theorems? At the start

of the 20th century, David Hilbert proposed a set of 23 problems which
he thought would be very important in 20th century mathematics.
The second one of these problems was to provide a finite list of axioms
to start out with and to prove that these axioms don’t lead to any
contradictions (such as Russel’s paradox). Ideally, one would want to
be able to prove every true statement in mathematics, building up from
these rules.
Theorem 11.1 (Gödel’s first incompleteness theorem). No consistent
system of axioms is capable of proving all true statements about the
arithmetic of natural numbers. That is, no matter what system of rules
 FIRST YEAR OF MATHS SOCIETY 27

you work with, there will always be statements about numbers that we
cannot prove with those rules.

This is disappointing! It would be nice to say that all true statements

about natural numbers can flow from a certain set of rules, but it turns
out that this is not the case. Perhaps the reason that the Collatz
conjecture (first ever talk at maths society) or the Riemann hypothesis
(Wren’s talk) are so difficult is because they literally cannot be proven
with the system of rules that we have in place. In fact, weirdly enough,
if one showed that it was impossible to prove these statements with our
rules, that itself would be a proof because if the statements were false,
then our rules would certainly allow us to find a counterexample (just
let a computer run for long enough). You may be thinking: wouldn’t
that be a proof, so surely it can be proven? Well remember that this
hypothetical proof strategy would use rules outside of our system.

Theorem 11.2 (Gödel’s second incompleteness theorem). No system

can prove it’s own consistency.

This means that, given any system of rules that we want to start out
with, we cannot prove that no contradictions arise within this system,
using the system itself. For example, the Peano axioms for the natural
numbers can only be proven to be consistent using ZFC set theory.
And ZFC set theory cannot prove that ZFC set theory is consistent;
one must assume another rule in order to do this!

Part 5. Saga 5: Functions in Analytic Number Theory

(featuring Prasan)
12. The Basel Problem
12.1. Introduction. The Basel problem is a famous problem in math-
ematics, originally solved by Euler in 1734. The problem asks what the
following sum converges to:
1 1 1 1
2
+ 2 + 2 + 2 + ...
1 2 3 4
In this talk we looked at a very clever solution to the problem, but
there are many alternative ways to do it, if you are interested.

12.2. The Solution. We start by considering the function sin(x). This

function has roots at 0, π, −π, 2π, −2π, etc. Thus, by the Weierstrass
28 MATHS SOCIETY

factorisation theorem, we can write:

 x  x  x  x
sin(x) = x 1 − 1+ 1− 1+ ...
 π
 π   2π 2π

 x 2  x 2  x 2
=x 1− 1− 1− ...
π 2π 3π
On the other hand, we can look at the Taylor series of sin(x) (thanks
to Prasan’s excellent talk on the subject!) which is:
x3 x5 x7
sin(x) = x − + − + ...
3! 5! 7!
Now if we divide both of these by x, we have:
  x 2    x 2    x 2 
sin(x)
(5) = 1− 1− 1− ...
x π 2π 3π
sin(x) x2 x4 x 6
(6) =1− + − + ...
x 3! 5! 7!
Now it doesn’t look like it, but we’re actually now in business! If we
expand the top equation, we have:
 
sin(x) 2 1 1 1
=1−x + + + . . . + O(x4 )
x π 2 (2π)2 (3π)2
where the big O notation roughly means: “junk that we don’t care
about”. Now, all that’s left to do is compare the coefficients of x2 in
both series and we obtain:
 
−1 1 1 1
=− + + + ...
3! π 2 (2π)2 (3π)2
and so multiplying both sides by −π 2 , we have the result in all of it’s
glory:

1 1 1 π2
+ + + · · · =
12 22 32 6

13. The Euler Product for the Zeta Function: Prasan

Patel
13.1. Introduction. In this talk we looked at a stunning connection
between the Zeta function and the prime numbers, known as the Euler
product, by considering a special probability distribution.
 FIRST YEAR OF MATHS SOCIETY 29

13.2. Setting up the Product. We shall begin by considering a prob-

ability distribution given by:
1
P (X = n) =
ns ζ(s)
for some s > 1. First we have to check that all of the probabilies add
up to 1:
∞ ∞
X X 1
P (X = n) =
n=1 n=1
ns ζ(s)
∞
1 X 1
=
ζ(s) n=1
ns
ζ(s)
= =1
ζ(s)
as required. Now, we look at the probability of picking some number
divisible by k. More precisely, this is:

P (X = k) + P (X = 2k) + P (X = 3k) + . . .
1 1 1
= s + + + ...
k ζ(s) (2k)s ζ(s) (3k)s ζ(s)
∞
X 1
=
n=1
ζ(s)(nk)s
∞
1 X 1
=
ζ(s) n=1 (nk)s
∞
1 X 1
= s
k ζ(s) n=1 ns
ζ(s)
=
k s ζ(s)
1
= s.
k

13.3. The Euler Product. Now, the probability of being divisible by

a prime p is p1s , so the probability of not being divisible by p (denoted
later as p′ ) is 1 − p1s . Now, since being divisible by any two numbers are
independent events, we have (denoting the probability of being divisible
by n as P (n)): P (n ∩ m) = P (n)P (m). Therefore the probability of
30 MATHS SOCIETY

not being divisible by any prime is given by:

!
\ Y  1

′
P p = 1− s .
prime prime
p

where the big Π just denotes: “product of”.

But what is the only number that is not divisible by any prime? 1. So
therefore we have:
1 Y  1

P (X = 1) = = 1− s
ζ(s) prime p

and thus we have the Euler product:

Y 1
ζ(s) =
prime
1 − p1s

1


14. The Gamma Function: 2
!
14.1. Introduction. In this

talk we ponder a question which at first
1
seems bizarre: what is 2 ! Recall that for integers, we define the
factorial as follows:
n! = n(n − 1)(n − 2) . . . 2 · 1
So for example we have 5! = 5 · 4 · 3 · 2 · 1 = 120. It is unclear, however,
how to extend this for non integers like 12 . The answer lies within the
Gamma function.

14.2. The Gamma Function. Notice we can define the factorial func-
tion inductively for natural numbers: once we have 0! = 1, we can use
the relation n! = n(n − 1)! to do the rest for us. So therefore if we
can construct a function such that f (1) = 1 and f (n + 1) = nf (n),
then f (n) = (n − 1)! for all natural numbers n. We now introduce the
Gamma function, which obeys these properties:
Theorem 14.1. The Gamma function
Z ∞
Γ(n) = tn−1 e−t dt
0

satisfies:
(1) Γ(1) = 1
(2) Γ(n + 1) = nΓ(n)
 FIRST YEAR OF MATHS SOCIETY 31

Proof. For the first property, we just plug in x = 1 and are left with
an elementary integral:
Z ∞ Z ∞
1−1 −t
Γ(1) = t e dt = e−t dt
0 0
∞
= −e−t 0


= −e−∞ − −e0 = 1
For the second property we can use integration by parts to obtain:
Z ∞
Γ(n + 1) = tn e−t dt
0
Z ∞
n −t ∞
= [−t e ]0 − ntn−1 (−e−t )dt
0
Z ∞
=0+n tn−1 e−t dt
0
= nΓ(n)
□

Therefore, we can say that the Gamma function is
an extension of

n! and so now it makes sense in a way to compute 12 !- it’s just Γ 32 .

14.3. The Integral. Let’s not dilly dally and get straight down to the
computation. We wish to compute:
Z ∞
1
t 2 e−t dt.
0

Firstly, by integration by parts we have:

Z ∞ h 1 i∞ 1 Z ∞ 1
1
t 2 e dt = −t 2 e−t +
−t
t− 2 e−t dt
0 0 2 0
1 ∞ − 1 −t
Z
= t 2 e dt
2 0
−1
We now employ the u substitution u = t 2 to transform this into the
following integral:
Z ∞
2
e−u du.
0

This is exactly half of a very famous integral known as the Gaussian

integral, and we will get stuck into a satisfying method for computing
32 MATHS SOCIETY
R∞ 2
it. We say that I = e−x dx and so we have:
−∞
Z ∞ Z ∞
−x2 2
2
I = e dx e−y dy
Z ∞−∞Z ∞ −∞
2 2
= e−(x +y ) dxdy.
−∞ −∞

We now have to transform into polar co-ordinates, which is a method

of expressing any complex number in the form
z = r cos θ + ir sin θ
where r ranges from 0 to ∞ and θ ranges from 0 to 2π. We can
see from the formula that the x-coordinate (real part) is r cos θ and
similarly the y-coordinate (imaginary part) is r sin θ and so we employ
the substitutions x = r cos θ and y = r sin θ. Now, because we are now
working in 2d, doing this substitution is more complicated (for 1d you
just see what dx would be in terms of du and plug that in), but since
we’re in 2d here we have to take all the partial derivatives and put
them into a Jacobian matrix and take the determinant of that and in
the end we obtain r so overall we get:
Z 2π Z ∞
2 2
2
I = re−((r cos θ) +(r sin θ) ) drdθ
0 0
Z 2π Z ∞
2
= re−r drdθ
0 0
R∞ 2
Now the integral 0 re−r dr can be done with elementary calculus:
notice that the derivative of r2 is 2r, so if you do the reverse of the
chain rule in your head (otherwise just do u sub with u = r2 ), you can
obtain that:
Z ∞  ∞
−r2 −1 −r2 1
re dr = e = .
0 2 0 2
So overall our integral becomes:
Z 2π
2 1
I = dθ = π
0 2
√ 1


And so the Gaussian integral I = π. And since 2
! was half of that
we have:
  √
1 π
!= .
2 2
Neat, eh?
 FIRST YEAR OF MATHS SOCIETY 33

Part 6. Guest Talks

15. The Riemann Zeta Function, or, How to Win
£1000000- Wren Shakespeare
15.1. On the Real Line. We begin by considering two related prob-
lems from the earlier days of maths. The first is the harmonic series:
1 1 1 1
+ + + ...
1 2 3 4
This series was proved to equal ∞ by Nicole Oresme in 1350. The
second is the Basel problem:
1 1 1 1
+ + + ...
1 4 9 16
This series has a finite value that is now quite well-known:
1 1 1 1 π2
+ + + ··· =
1 4 9 16 6
This was proven, if very unrigorously, by Leonhard Euler, who is also
credited with the first use of the modern-day Riemann Zeta function,
which is a generalisation from both of these ideas:
∞
X 1
ζ(s) =
n=1
ns

This infinite series only produces a finite value in a traditional sense for
s > 1, as can be seen by comparison with the harmonic series - s < 1
means all the terms will be bigger than the harmonic series, so its
value will be greater than the harmonic series, which is infinite. Euler
also proved, again slightly unrigorously in some cases, two interesting
properties of the zeta function:
ζ(2n) = kπ 2n , k ∈ Q, n ∈ N
1 1 1 1
ζ(s) = −s
× −s
× −s
× ...,s > 1
1−2 1−3 1−5 1 − 7−s
By contrast, very little is known about the zeta function on odd num-
bers - the most we know is that ζ(3) is irrational.

15.2. Complex Numbers. The zeta function was most famously con-
sidered as a complex function by Bernhard Riemann, which is why it
bears his name today. It can be extended to a complex function for
Re(s) > 1 very easily. However, what if we want to define it for other
34 MATHS SOCIETY

numbers?
∞
X (−1)n−1
Taking η(s) = :
n=1
ns
η(s)
We can show that ζ(s) = , Re(s) > 1
1 − 21−s

However, η also converges for 0 ≤ Re(s) ≤ 1, giving us a handy way to

extend the zeta function to this region, which is exactly the region we
want to study.
The Riemann Hypothesis asserts that if ζ(s) = 0, then Re(s) = 21 .
Initially discarded as a mere curiousity by Riemann himself, it became
increasingly apparent that RH implies a number of other things in var-
ious fields of maths. Notably, it implies a fairly regular distribution of
prime numbers.

So, how far have we gotten with proving it? We can prove there are
infinitely many zeroes of the form Re(s) = 21 . We also know, if there
are zeroes with other real parts, they cannont be too close to 0 or 1
(how close depends on their exact location). The strongest result that
we currently have is that 40% of the zeroes are of the form Re(s) = 12 .
Interestingly, the monetary prize is not for a resolution of the Rie-
mann Hypothesis one way or another, it’s for a proof (so if it was false,
whoever showed that would not be able to collect the prize).

16. Introduction to Taylor Series- Prasan Patel

16.1. Introduction. In this talk we were introduced to the concept of
a Taylor series, which was named after English mathematician Brook
Taylor who was born in 1685 here in Edmonton! So we were blessed to
see the magic of Taylor series delivered in its hometown. The idea be-
hind Taylor series is to convert functions that are not polynomials into
“infinite polynomials”, so that they are easier to compute; a calculator
doesn’t actually know what e2 is- it is just plugging in values into the
Taylor series to approximate it.

16.2. Example: ex . Our goal is to write ex in the form p(x) = c0 +

c1 x + c2 x2 + . . . . To do this we shall look at the derivatives. It is well
d x
known that dx e = ex and so when x = 0, the derivative will be 1, no
 FIRST YEAR OF MATHS SOCIETY 35

matter how many times we take the derivative. Thus we need:

p(0) = 1
p′ (0) = 1
p′′ (0) = 1
..
.

Plugging this information in gives us that c0 = 1, since all of the rest of

the terms vanish when plugging in 0 into the polynomial. Furthermore
p′ (x) = c1 + 2c2 x + 3c3 x2 + . . . so plugging in 0 into this and setting it
equal to one leaves us with c1 = 1. Taking the second derivative leaves
us with 2c2 + 6c3 x + 12c4 x2 + . . . and so, again, when x = 0 we are left
with just the c2 term and so we have 2c2 = 1 =⇒ c2 = 21 . In general,
we see that because of the power rule and the fact that when x = 0 all
higher terms cancel, we have that

p(n) (0) = n!cn

1
and so since we want p(n) (0) = 1, we have that cn = n!
and so our
function actually looks like the following:

∞
X
ex = cn x n
n=0
∞
X xn
=
n=0
n!

We observe on desmos how well this can approximate ex . The first

2 2 3
three terms are 1 + x + x2 , the first three terms are 1 + x + x2 + x6 .
36 MATHS SOCIETY

Figure 5. The approximations get closer and closer!

(Red curve is ex )

The purple (really close) line is the first 5 terms of the polynomial,
the green (still pretty close) is the first 4 terms and so on. So really
your calculator is just adding up the first 10-15 terms of this polynomial
when you plug in ex for some x in your calculator. Pretty neat, eh.

16.3. Generalising This Process. Now, considering that this is maths

society, we want a general formula to do this process with any function
f (which isn’t a polynomial). Again, let us say that we want to write it
in the form p(x) = ∞ n
P
c
n=0 n x . Then again we would take derivatives
and set them equal to the derivative of the original function, except we
need not take the derivative at 0, we can pick any point a (for example
if we were to do this process with ln(x), we couldn’t pick x = 0 to take
our derivative on since it isn’t defined there). However, like in the ex
example, we want that when we take the nth derivative at a, all of the
higher terms cancel out so instead we shall write:
∞
X
2
p(x) = c0 + c1 (x − a) + c2 (x − a) + · · · = cn (x − a)n .
n=0

Now, just like last time we take the nth derivative of p(x) at a and we
find that:

p(n) (a) = n!cn

 FIRST YEAR OF MATHS SOCIETY 37

pn (a)
since we’ve set p(n) (a) = f (n) (a), we find that cn = n!
and so we
have:
∞
X p(n) (a)
f (x) = (x − a)n
n=0
n!

which is called the Taylor series of f .

17. Differentiation under the integral sign: Wren

Shakespeare
Here we consider a genius method for integrating functions which
have no elementary anti derivative. This method, popularised by Feyn-
man, goes as follows:
(1) Introduce a new variable inside the integral a and call the inte-
gral I(a). We want there to be a value of a for which we know
what I(a) is, and also we want there to be some a such that
I(a) gives us our original integral.
(2) Then, differentiate with respect to a, hopefully making the in-
tegral now easier to solve.
(3) Once you have found I ′ (a), integrate that to find out I(a) and
then plug in your value of a that gives you back the original
integral.
We shall, of course, consider some examples.

R∞ sin(x)
17.0.1. 0 x
dx.

Example 17.1.
Z ∞
sin(x)
dx.
0 x

The thing in this integral that’s really annoying us is the x on the

bottom of that fraction. Therefore, we shall introduce a new variable
as follows:
Z ∞
sin(x)
I(a) = e−ax dx.
0 x
38 MATHS SOCIETY

Note that I(0) is our original integral. Now, we shall differentiate both
sides with respect to a:
Z ∞
′ ∂ −ax sin(x)
I (a) = e dx
0 ∂a x
Z ∞
−x sin(x)
= e−ax dx
0 x
Z ∞
=− e−ax sin(x)dx.
0
ix −ix
We shall use the fact that sin(x) = e −e 2i
to make this integral easier,
although it can also be done with integration by parts.
1 ∞ −ax ix
Z
′
I (a) = − e (e − e−ix )dx
2i 0
i ∞ (i−a)x
Z
= e − e−(a+i)x dx
2 0
 ∞
i 1 (i−a)x 1 −(a+i)x
= e + e
2 i−a a+i 0
 
−i 1 1
= +
2 i−a i+a
 
−i i + a + i − a
=
2 1 − a2
 
−i 2i
=
2 −1 − a2
−1
= 2 .
a +1
Now that we have I ′ (a) = a−1
2 +1 , we can use standard results from

calculus to deduce that I(a) = −arctan(x) + c. Since I(a) → 0 as

a → ∞, we have: 0 = − π2 + c =⇒ c = π2 . Therefore:
π
I(a) = −arctan(x) +
2
Thus:
Z ∞
sin(x) π
I(0) = =
0 x 2
R1 x−1
17.0.2. 0 ln(x)
dx.
 FIRST YEAR OF MATHS SOCIETY 39

Example 17.2.
1
x−1
Z
dx.
0 ln(x)
How do we get rid of the ln(x) on the denominator? Well we know
that the derivative of ax is ln(a)ax , so we shall utilise this to our full
advantage!
Z 1 a
x −1
I(a) = dx
0 ln(x)
Z 1
′ ∂ xa − 1
I (a) = dx
0 ∂a ln(x)
Z 1
ln(x)xa
= dx
0 ln(x)
Z 1
= xa dx
0
 1
1 a+1
= x
a+1 0
1
= .
a+1
Success! We have found a nice form for I ′ (a) and we can easily deduce
that I(a) = ln(a + 1) + c. Since we also know that I(0) = 0, we can also
say that 0 = ln(1) + c and so c = 0. Therefore we have I(a) = ln(a + 1)
and so:
Z 1
x−1
I(1) = dx = ln(2)
0 ln(x)

18. The logistic map: Rowan Oliveira

18.1. Introduction. In this talk we look at a fascinating iterative map
known as the logistic map, given by:

xn+1 = rxn (1 − xn ),

where we take xn to be between 0 and 1. This map originally finds its

applications in biology in the paper ”Simple mathematical models with
very complicated dynamics” by Robert M. May, due to the fact that
for varying values of r it can capture different types of phenomena.
40 MATHS SOCIETY

18.2. Studying what happens as r varies and the Feigenbaum

Constant. When r is between 0 and 1, the iterative sequence will
always just die off and get closer to zero, for example if we have r = 0.5
and we pick our starting point to be x0 = 0.25, then our sequence
becomes:
0.25 → 0.5(0.25)(0.75) = 0.09375 → 0.04248 → . . .
Now, when r is between 1 and 3, the sequence will converge to r−1 r
,
but the map sequence will fluctuate more when r is between√2 and
3. Now, things get interesting: when r is between 3 and 1 + 6, the
sequence will eventually start bouncing between two separate points,
which are given by some √complicated surd which isn’t important. Then
when r is between 1 + 6 and approximately 3.54409, the sequence
will bounce between four different points. Now this pattern contin-
ues, with the number of points doubling very rapidly until we reach
r ≈ 3.56995, when we have most points above this point becoming
completely chaotic. However, the beauty is that there still exist islands
of stability above 3.56995 where the r value gives us stable behaviour.
We can see this in a bifurcation diagram which nicely shows exactly
what’s going on.

Figure 6. The little white strips in the midst of the

chaos are the islands of stability.

Furthermore in the area before the chaos, the ratio of the distances
between each new r value that doubles the amount of points that the
sequence converges to approaches a number δ which is known as Feigen-
baum’s constant, which is given by 4.669201609 . . . . We can see this
with a table:
 FIRST YEAR OF MATHS SOCIETY 41

rn −rn−1
n period rn δ → rn−1 −rn−2
1 2 3√
2 4 1+ 6
3 8 3.5440903 4.7514
4 16 3.5644073 4.6562
5 32 3.5687594 4.6683
6 64 3.5696916 4.6686

18.3. Connection with the Mandelbrot Set. It turns out that

(this is hardly even surprising anymore!) the Mandelbrot set is con-
nected with the logistic map as well. Indeed, it turns out that the
“islands of stability” in the logistic map are deeply connected to the
Mandelbrot set. Recall that the equation that the Mandelbrot concerns
itself with is:

zn+1 = zn2 + c.

When we make the change of variables z = r 12 − x , c = 2r 1 − 2r ,

we can observe a stunning connection between the real parts of the
Mandelbrot and the logistic map.

Figure 7. The islands of stability line up perfectly with

the Mandelbrot set, under the change of variables
42 MATHS SOCIETY

This actually makes sense when you think about it, because the Man-
delbrot set measures exactly where the iterative sequence is stable and
both the logistic map and the Mandelbrot set are given by quadrat-
ics, so it is feasible that they should have some connection under a
change of variables. That being said, it doesn’t make the result any
less remarkable!
19. Derivatives and Integrals of Displacement: Saik
Rudad
19.1. Introduction. In this talk we looked at the various concepts in
physics that arise when you take repeated integrals and derivatives of
displacement with respect to time.
19.2. The Derivatives. Of course, when we take the first derivative
of displacement ds
dt
, we get velocity v in ms−1 , which measures how fast
(and the direction) something is going. Next, we take the derivative
of velocity to obtain acceleration, a in ms−2 , which measures the rate
of change of velocity. Now here things get interesting: when we take
the third derivative of displacement is called jerk, j, measured in ms−3 .
Jerk has many applications in engineering, for example it is considered
when designing lifts and cars. This is because one must limit not only
the maximum force, but the rate at which the force changes. Since
F = ma, this means that we not only need to look at acceleration,
but also the rate of change of acceleration- that “jerky” feeling- which
engineers spend a lot of time to try to minimise when designing differ-
ent vehicles. The fourth derivative of displacement (equivalently, the
derivative of jerk) is called snap. Snap is minimised in the design of
railway tracks and is measured in ms−4 . The fifth and sixth deriva-
tives of displacement are called crackle and pop respectively and are
measured in ms−5 and ms−6 respectively.
19.3.
R The integral. We shall look at the integral of displacement
sdt, which is called absement and is measured in ms. What this
measures is the sustained displacement of an object from its initial po-
sition. Absement has been used in studying fluid flow, for example in
music instruments like the hydraulophone.