1
The birth of modern science
“Philosophy is written in this grand book the universe, which stands continually open to our gaze. But the
book cannot be understood unless one first learns to comprehend the language and to read the alphabet
in which it is composed. It is written in the language of mathematics, and its characters are triangles,
circles and other geometric figures, without which it is humanly impossible to understand a single word of
it; without these, one wanders about in a dark labyrinth.”
In the early 17th century, the Italian Galileo and the Frenchman Descartes, although radically different in
their approaches, between them re-defined the goals of science by the interaction of their ideas – the critical
rationalism of Descartes, and the practical empiricism of Galileo. They selected the concepts that should
be used in good scientific thinking, and they transformed the methodology – the styles and techniques of
observation, theory-making and experiment. The one thing they totally agreed about was the importance
of mathematics. Their legacy was a new way of looking at and analysing the world mathematically, and a
dramatically greater power to explain the world scientifically. That’s where Isaac Newton started. (He was
born in the year Galileo died.)
Galileo represents the pioneering spirit of the new experimental science, with its novel concentration upon
measuring things quantitatively, thinking mathematically about the world, and submitting all hypotheses to the verdict of Nature. We should not ask ‘Why do things happen?’ which is usually beyond our
brief, so much as ‘How do they happen?’ We aim at describing the ways of Nature in mathematical terms,
thus talking of mathematical ‘laws of Nature’. We can expect to comprehend the world this way, said
Galileo, because God has designed the world in mathematical terms, and has created us in his image to
make (mathematical) sense of the world. Descartes’s contribution to the fruitful mix was to conceive the
world as constituted primarily by mechanical particles acting upon one another and so giving rise to all other
‘secondary’ qualities that our senses perceive. He combined this with a resolute refusal to simply believe
what pretended authority of any kind says, but rather to practise systematic doubt about everything, and
to use one’s own reason, validated, said Descartes, by the existence of a perfect Creator-God who gives us
reason as a good gift to lead us to truth.
What Descartes achieved, in radically limiting and re-focusing scientific thought in mechanistic terms and,
mathematically, in uniting the respective powers of geometry and algebra in the new coordinate geometry,
and what Galileo achieved, in expressing, as variables in geometric diagrams and also as solutions of algebraic
equations, the position and velocity of a body moving under gravity, together laid down the essential preparation for the majestic edifice of Newton’s mechanics. Today, using Descartes’s coordinate geometry, and
Newton’s (and Leibniz’s) calculus, and starting with what we now call Newton’s laws of motion and his law
of universal gravitation, we have a magnificent mathematical model of the world. We can deduce Galileo’s
laws of falling bodies and projectiles and also Kepler’s laws of planetary motion. In this grand Newtonian
synthesis, terrestrial motion and celestial motion are wonderfully brought together in one comprehensive,
unified theory. But let’s go back to Galileo’s simple yet ground-breaking combination of observation, algebra
and geometry, applied to a basic human experience: an object falling under gravity.
2
Galileo’s falling stone
Galileo’s assumption/hypothesis: “When ... I observe a stone initially at rest falling from an elevated
position and continually acquiring new increments of speed, why should I not believe that such increases
take place in a manner which is exceedingly simple? ... A motion is said to be uniformly accelerated when
starting from rest, it acquires, during equal increments of time, equal increments of speed. That is the
concept of accelerated motion that is most simple and easy, and it is confirmed to be that actually occurring
in Nature, by exact correspondence with experimental results based upon it.”
1
Galileo’s deduction: Distances are to each other as the squares of the times.
Galileo’s experiment: The ball and the inclined plane ...
time
1
2
3
4
5
distance
1
4
9
16
25
Galileo’s other assumption:
“In order to handle this matter in a scientific way, it is necessary to cut loose from such irksome difficulties
[as friction or air resistance], otherwise it is not possible to give any exact description. Once we have
discovered our theorems about the stone, assuming there is no resistance, we can correct them or apply
them with limitations as experience will teach. I can show by experiments that these external and incidental
resistances are scarcely observable for bodies travelling over short distances.”
Our modern version of Galileo’s simple mathematical model:
Acceleration
so velocity
hence distance
d2 y
= g,
dt2 Z
dy
v(t) =
= g dt = gt,
dt
Z
1
y(t) =
gt dt = gt2 .
2
a(t)
=
Ball rolling on an inclined plane:
If we roll the ball, as Galileo said he did, down a plane inclined
at angle θ then the gravitational force mg can be composed into (so replaced by) a force mg cos θ acting
perpendicularly to the plane, together with a force mg sin θ acting downwards parallel to the plane.
N
F
mg sin θ
mg cos θ
θ
2
The first force on the ball is exactly counterbalanced by an equal and oppositely directed reaction force N
exerted by the plane itself on the ball, so no movement takes place perpendicularly to the plane. And it is
the other component of force that causes the ball to accelerate in the downward direction along the plane.
If it were a stone sliding instead of a ball rolling, we would do well to take a retarding frictional force F into
account, acting upwards along the plane, and depending on the magnitude of that perpendicular reaction
force as well as the coefficient of friction of stone on planar surface. But with a rolling ball we can afford to
ignore the minor effects of friction. All the equations and graphs of distance rolled, velocity and acceleration,
would have the same form as with the stone falling freely, with the new (smaller) acceleration g sin θ in place
of g. The effect of the inclined plane is just what Galileo meant it to be: a viewing in slow motion.
Falling stone with air resistance: Taking into account the air resistance should improve our real-life
modelling. It will be proportional to the number of collisions per second with air molecules, hence to the
product of the velocity v and the cross-sectional area of the ball presented to the resisting medium. Therefore
we assume that the resistance is −kmv, where m is the mass of the ball (included in there for convenience
to allow cancellation) and k is a constant to be found by experiment. Its value is proportional to the density
of the medium as well as to the cross-sectional area, and it will vary according to local altitude and weather
conditions. As the velocity increases so does the resisting force; the minus sign is because resistance is
always in the opposite direction to the velocity. Using Newton’s law, we equate the product of mass and
acceleration to the total force downward, that due to gravity minus that due to air resistance acting upward.
The resulting differential equation is second order, but if we express it in terms of the velocity it becomes a
first order equation, easy to solve.
mass × accel.
d2 y
so ma = m 2
d t
dv
hence ma = m
dt
dv
and therefore a(t) =
dt
=
force,
=
mg − km
=
mg − kmv,
=
g − kv.
dy
,
dt
We could integrate this equation as a first order linear equation, or directly by “separating the variables”,
and we obtain equations for velocity v and for distance y, in terms of time t. The equation for v(t) is very
illuminating. It tells us that in the long run, if the fall continued through the resisting air, the velocity would
approach a constant limiting value g/k, called the terminal velocity.
Z
Z
dv
=
dt,
g − kv
1
hence, integrating,
− ln(g − kv) = t + c1 ,
k
so that
ln(g − kv) = −kt + c2 , where c2 = −kc1 .
c3 e−kt , where c3 = ec2 ,
g
c3
v(t) =
− c4 e−kt , where c4 = ,
k
k
g
→
as t → ∞.
k
g − kv
Taking exponentials,
so
=
But note that we could have found the terminal velocity much more easily from our fourth equation above,
by putting a(t) = 0. Note also that we have made crucial assumptions – about the acceleration and about
the resisting medium, that are not exactly true. What are these assumptions, precisely? When might it be
necessary to improve our model? How could we do this?
3
3
Galileo’s simple pendulum
Galileo’s first observation
“One must observe that each pendulum has its own time of vibration so definite and determinate that it is
not possible to make it move with any other period than that which nature has given it. For let any one take
in his hand the cord to which the weight is attached and try, as much as he pleases, to increase or decrease
the frequency of its vibrations; it will be time wasted.”
Galileo’s assumptions: That the angle of swing is small, and that friction/resistance may be ignored.
Galileo’s experimental conclusion:
“As to the times of vibration of bodies suspended by threads of different lengths, they bear to each other
the same proportion as the square roots of the lengths of the thread; or one might say that the lengths are
to each other as the squares of the times. ...
Our modern modelling of this situation:
θ
L
sS
M y
F
mg
From the geometry,
y
≈ arc M S = Lθ.
By Newton’s law,
F
= mg sin θ
≈ mgθ
mg
≈
y
L
Let the displacement at time t of the bob from the middle point M be y(t), and the corresponding rope
angle be θ (the Greek letter theta). Because the angle is small, y is approximately the arc of swing, which
is Lθ. The force F tending to restore the bob to the middle equilibrium position M , is the component of
the gravitational force acting perpendicularly to the string, which for small angles is approximately in the
direction SM . Its value is mg sin θ, where m is the mass of the bob. For small angles this is approximately
mgθ, which is approximately mgy/L. This means we have a restoring force proportional (approximately) to
the displacement.
To demonstrate that this results in simple harmonic motion (given by a sine curve),we write down the
differential equation of the motion, equating the force F to the product of the mass m and the acceleration
towards the middle point M . The mass cancels out – it is irrelevant to our problem, just as the mass of the
falling stone was irrelevant to its acceleration. We have a second order differential equation to solve, with
4
two “initial conditions”, say initial displacement given, and initial velocity zero.
Force = mass × acceleration,
d2 y
mg
y = F =m 2,
so that
−
L
dt
d2 y
g
giving
= − y.
dt2
L
p
This leads to a general solution y(t) = A sin( Lg t + α), where A (the “amplitude”) and α (the “phase”)
q
are constants to be determined by the initial conditions. The period of this sinusoidal function is 2π Lg , in
harmony with Galileo’s experimental conclusion.
Now, what assumptions have we made in this model? When might it fail to model vibrations accurately?
How could we improve it?
4
Fibonacci’s rabbits, trees, and bees
In 1202 one of the great books in the history of science was published in Italy: Liber Abbaci (Book of
Counting). It was written by Leonardo of Pisa, called Leonardo Fibonacci – son of Bonacci. His father
Bonacci was a merchant who travelled widely around the coast of the Mediterranean Sea, and the bright
young Leonardo picked up a great deal of learning from the Arab lands – in particular, the so-called IndianArabic base-ten place-value numeral system we still use today. In fact it’s his book that is credited with
introducing the Europeans to the brilliant new numeral system, and ultimately persuading the merchants
and bankers to stop using the old, clumsy Roman numerals. However, Leonardo is more famous 800 years
later for one problem in his book – about the way rabbits increase in number.
Problem: A certain person places one pair of rabbits in a certain place surrounded on all sides by a wall.
We want to know how many pairs can be bred from that pair in one year, assuming that it is in their nature
that each month they give birth to another pair, and in the second month after birth each new pair can also
breed.
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, · · ·
Fibonacci’s answer is 377 pairs of rabbits, and his sequence of numbers is now known as the Fibonacci
sequence. This is an example of a problem in population dynamics, which has dominated mathematical
biology over much of the 200 years the subject has existed. However, you can see that someone was making
a mathematical model over 800 years ago! How did he solve it?
“After the first month there will be two pairs, after the second, three. In the third month, two pairs will
produce, so at the end of that month there will be five pairs. In the fourth month, three pairs will produce,
so there will be eight pairs. Continuing thus, in the sixth month there will be five plus eight equals thirteen,
in the seventh month, eight plus thirteen equals twenty-one, etc. There will be 377 pairs at the end of the
twelfth month. For the sequence of numbers is as follows, where each is the sum of the two predecessors,
and thus you can do it in order for an infinite number of months.”
5
B
?
Y
?
A -B
?
R
A -B
Y
?- R R A B
Y
A B
(B, Y, A)
Total
(1, 0, 0)
1
(0, 1, 0)
1
(1, 0, 1)
2
(1, 1, 1)
3
(2, 1, 2)
5
(3, 2, 3)
8
After n months, let there be (Bn , Yn , An ) (pairs of) babies, young rabbits, adults, respectively. Then
An+1
= An + Yn
Bn+1
= An + Yn
Yn+1
= Bn
We deduce
An+1 = Bn+1 = Yn+2 .
Let the total number of (pairs of) rabbits at n months be given by
Fn = An + Bn + Yn .
Then we deduce that
for n ≥ 0,
Fn+2 = Fn+1 + Fn ,
F0 = F1 = 1.
This recursion formula will generate “Fibonacci numbers” as far as we want, as Fibonacci pointed out. He
seems to have got there with less fuss and no symbols, and indeed, if you think about that recursion formula,
you will see how it becomes almost obvious: Everyone added to the last generation is a baby born to a
member of the last-but-one generation! As Fibonacci might have put it, in the 100th month, F98 pairs will
produce.
But we have a more general method, and our notation and technique can be applied to a host of other
problems. The number of pairs of rabbits after Fibonacci’s 12th month is F13 , as he starts with F1 , not F0 ,
with a pair of young rabbits, not babies. Is there a quick way of arriving at this number or higher Fibonacci
numbers like F100 ? A simple computer programme will produce recursively generated numbers very fast.
Otherwise we can programme the computer to generate successive pairs if Fibonacci numbers with powers
of a matrix:
Observe that
Fn+2
Fn+1
=
1
1
1
0
6
Fn+1
Fn
,
for n ≥ 0.
Hence
F2
F1
1
1
1
0
1
1
1
0
1
1
1
0
1
1
1
0
=
=
F3
F2
F13
F12
=
=
F1
F0
1
,
1
2 1
,
1
12 1
377
=
.
1
233
There is in fact a lovely closed-form expression for Fn in terms of the golden ratio φ, which you can do some
research about.
Before leaving the rabbits we must point out that this same model for population growth can be applied to
other situations. Assuming that each branch of a tree gives rise to a new branch, but only after skipping a
season’s maturation period, we obtain a visually very plausible model:
8
5
3
2
1
And in the ancestry of the male bee of certain species, the forebears follow a Fibonacci law, because, while
female bees are born from the mating of male and female, male bees are born asexually to single females.
M
M
F
F
3 great grandparents
M
2 grandparents
F
1 parent
M
1 male bee
Try to prove for yourself that the number Bn of bee ancestors in the nth generation backward, satisfies the
Fibonacci generating equation. (Let Bn = Mn + Fn , the sum of the number of males and females, and write
down the equations connecting Mn , Fn , Mn+1 , Fn+1 .)
Fibonacci numbers are also the solution to the following problem. Can you see why?
Problem:
Taking either one or two steps at a time, how many different ways are there to climb n steps?
Questions: How accurate is the Fibonacci model for the biological situations we are modelling? What
were the assumptions we made? What could be done to improve the model?
7
For the male bee’s ancestors, the numbers are exact. But for rabbits and trees, the model will only ever be
an approximate fit, for we assumed a fixed interval of time between being born and reproducing, and we
assumed that exactly one entity is produced without fail. Real life is never so predictable. The next step in
modelling is to test the model against real data. Perhaps we can improve the estimate of that fixed interval;
a better model will not be deterministic – we need to think in terms of probabilities.
5
Red blood cells
These give your blood its red colour. Their function is to carry oxygen around the body. When there are
too few in your body you suffer from anaemia. They are manufactured in your bone marrow, and each cell
lives for about 4 months before dying, probably in your kidney or spleen.
You should normally have about 2 × 1013 red blood cells in your body at any moment, unless you live in
Lesotho, where people need more (why?).
Your bone marrow produces about 2 × 106 red blood cells per second, or 1728 × 108 per day.
Let’s make a mathematical model. We assume that a fixed number b of cells are produced every day, and
that a fixed proportion m of existing cells die each day, and we want to find: What percentage of cells
survive each day?
Problem 1: What percentage of cells normally survive each day in the equilibrium situation?
Problem 2: Is this equilibrium stable? On our assumptions above, what will normally happen if the
red blood cell count is suddenly different from normal?
Let Cn be the number of red blood cells present at day n.
Let m be the proportion of red blood cells that die each day.
Let b be the number of red blood cells produced each day in the bone marrow.
Then
Cn+1 = (1 − m)Cn + b.
Normally, there is a constant situation: Cn = C, say, for all n. Substituting in the equation we have
C = (1 − m)C + b, hence mC = b,
m=
1728 × 108
b
=
= 0.00864.
C
2 × 1013
Thus, the percentage of red blood cells that die each day is 0.864%, so that 99.136% of them survive each
day.
Next, suppose Cn 6= C, and let Dn = Cn − C, so Dn is the deviation from the normal. Then
8
Dn+1
Hence
6
Dn
=
Cn+1 − C
=
(1 − m)Cn + b − C
=
(1 − m)Cn + mC − C
=
(1 − m)(Cn − C)
=
(1 − m)Dn .
=
(1 − m)n D0 → 0, so Cn → C.
Crop yields and bacteria cultures
Over 2000 years ago, in the Chinese classic Jiu zhang suan shu (The Nine Chapters of the Mathematical
Art), there appeared an amazing feat of what might, at a stretch, be called mathematical biology. The
following problem is solved on the Chinese counting board, by manipulating red and black rods in a manner
equivalent to what we call Gaussian elimination in the theory of matrices today.
Problem: Now there are 3 classes of paddy: top, medium and low grade. Given 3 bundles of top grade
paddy, 2 bundles of medium grade paddy and 1 bundle of low grade paddy, the yield is 39 dou. For 2 bundles
of top grade, 3 bundles of medium grade and 1 bundle of low grade, the yield is 34 dou. And for 1 bundle
of top grade, 2 bundles of medium grade and 3 bundles of low grade, the yield is 26 dou.
How much does one bundle of each grade yield?
Answer: Top grade paddy yields nine and a quarter dou per bundle; medium grade paddy four and a
quarter dou; and low grade paddy two and three quarters dou.
high grade
medium grade
low grade
yield
Exercise:
1
2
3
26
2
3
1
34
3
2
1
39
3x+2y+z
2x+3y+z
x+2y+3z
=
=
=
39
34
26
(1)
(2)
(3)
Solve the Chinese problem by doing row operations on the matrix of coefficients.
Now here is a modern problem, of very similar structure. Three species of bacteria B1 , B2 , B3 are mixed together in the same culture in a laboratory, and it is known that each requires three types of food, F1 , F2 , F3 ,
and that they consume the food at the following rates:
B1 consumes F1 , F2 , F3 at the rates (1, 2, 2);
B2 consumes F1 , F2 , F3 at the rates (3, 2, 1);
B3 consumes F1 , F2 , F3 at the rates (8, 4, 1).
It is found experimentally that when 2750 units of F1 , 1500 units of F1 , and 500 units of F3 are supplied per
day, a constant situation is reached in which all of the food is consumed. Consider the following problems.
1. Find the amounts of the bacteria present in this constant situation.
9
2. Which set of constant states is possible, and which is not, regardless of the food supply?
Solution: Suppose that there are amounts X1 , X2 , X3 , respectively, of the three bacteria present. Then

1
 2
2
3
2
1

 

8
X1
2750
4   X2  =  1500 
1
X3
500
which we may write as AX = F . Note that det(C) = 0, so there are many possible solutions:
(X1 , X2 , X3 ) = (t − 250, 1000 − 3t, t), t ∈ R.
Notice that for positive values of the Xi we must have 250 ≤ t ≤
1000
3 .
For the second question, consider the map f : R3 → R3 defined by f (X) = AX. The rank is easily found to
be 2, so the kernel of f has dimension 1 (it’s a line) and the range is the 2-dimensional plane through the
origin, spanned by the two image vectors:
   
   
1
1
0
3
f  0  =  2 , f  1  =  2 .
0
2
0
1
Taking the vector product of these two, we get a vector in direction (−2, 5, −4) perpendicular to the required
plane, hence the plane is the set of all vectors whose scalar product with that vector is 0; the equation of
the plane is
2x − 5y + 4z = 0.
One point on this plane is that experimentally discovered equilibrium state (2750, 1500, 500). But all
possible equilibrium states must lie on this plane.
7
Age-stratified populations & a butterfly’s life cycle
Consider a population of individuals each of which gives birth, lives for a while, then dies. We assume that
there is a maximum life span, and we divide the population into k age-groups, numbering a1 (n), a2 (n), . . . , ak (n)
individuals at time n, measured in years or breeding seasons. We suppose that
1. everyone in stage k dies at the end of the year/season;
2. the probability that an individual in class j survives another year/season is pj ;
3. the probability that an individual in class j gives birth to exactly one individual is fj (the fertility
rate).
For example, take k = 4. Then

 
a1 (n + 1)
f1
 a2 (n + 1)   p1

 
 a3 (n + 1)  =  0
a4 (n + 1)
0
f2
0
p2
0
10
f3
0
0
p3

f4
a1 (n)
 a2 (n)
0 

0   a3 (n)
0
a4 (n)


.

This is called the Leslie matrix. After N years we will have
 
f1
a1 (N )
 a2 (N )   p1

 
 a3 (N )  =  0
0
a4 (N )

f2
0
p2
0
N 
f4
a1 (0)
 a2 (0)
0 
 
0   a3 (0)
0
a4 (0)
f3
0
0
p3


.

Consider the four stages of a butterfly’s life cycle: egg, caterpillar, pupa (or chrysalis) and adult butterfly.
EGG
p
~
CATERPILLAR
>
BUTTERFLY
o
r
q
/
PUPA
(CHRYSALIS)
Of course, of the four stages, only the butterfly lays eggs, so f1 = f2 = f3 = 0. Assume that a butterfly
succeeds in surviving to lay eggs with probability f , and then always lays N eggs. Then f4 must be replaced
by f N . Let En , Cn , Pn , Bn be the numbers of individual eggs, caterpillars, pupa, and butterflies, in a
population after the nth season. And let the respective probabilities of an egg, caterpillar and pupa surviving
to move into the next class be p, q, r. Then

 
En+1
f N Bn
 Cn+1   pEn

 
 Pn+1  =  qCn
Bn+1
rPn


0
  p
=
  0
0

En
0 0 fN
 Cn
0 0
0 

q 0
0   Pn
0 r
0
Bn


.

The parameters f, p, q, r could be estimated from experimental data. The long term fate of the population
will depend on their values. For the butterfly, it is easy to see what will happen. We are interested in high
powers of the matrix – so it’s natural to consider the eigenvalues. And this leads to:

0
 p

 0
0
4

0 0 fN

0 0
0 
 = f N pqr 

q 0
0 
0 r
0
1
0
0
0
0
1
0
0
0
0
1
0

0
0 

0 
1



En+4
 Cn+4 


 = f N pqr 
 Pn+4 

Bn+4

En
Cn 

Pn 
Bn
It’s easy to check this, but where does it come from? The Cayley-Hamilton theorem says that the matrix will
satisfy its characteristic equation, and the eigenvalues of the Leslie matrix are easily seen to be the solutions
of the cubic λ3 = f N pqr
Thus
if f N pqr
<
1,
ultimate extinction occurs;
if f N pqr
=
1,
there is unstable equilibrium;
if f N pqr
>
1,
there is unsustainable expansion.
How realistic is this model? What were our assumptions? How can we test it? How can we improve it?
What can we learn from it anyway?
11