MEI Conference
Friday, 3rd July 2008
Mathematics: an historical perspective
Notes
Stephen Lee
A remarkable formula: summation being equal to an Integral (Bernoulli).
Historical
Snippets
Stephen Lee
MEI/FM Network
Images from www.mathematik.ch and wikipedia
Sources of information
• Professional Bodies ‘free’ Publications
– Institute of Maths and its Applications
• Mathematics Today
– Maths Association
• Mathematics in School
• Mathematical Gazette
• Mechanism for gaining students interest in
mathematics
Historical perspective
1 of 11
MEI conference 2008
A remarkable formula (or two)
• Mathematical Gazette
– March 2008
– Vol 92
– No. 523
– Pgs 94-96
Discovered by Bernoulli in 1697:
∞
1 1
1
=
dx
∑
n
∫
0 xx
n
n =1
Investigate
∫
1
0
1
dx
x
x
=
= 1.29129 (to 5 d.p. using CAS system)
Historical perspective
2 of 11
MEI conference 2008
Investigate
n
∞
1
∑
n
n =1 n
1 1 1 1
= 1 + 2 + 3 + 4 + ...
1 2 3 4
Summation
1
1
2
1.25
3
1.28703703703704
4
1.29094328703704
5
1.29126328703704
6
1.29128472050754
7
1.29128593477322
8
1.29128599437787
9
1.29128599695904
10
1.29128599705904
11
1.29128599706255
12
1.29128599706266
13
1.29128599706266
14
1.29128599706266
Proof
(Dealing with the interchange of summation and Integration)
∞
1
∑n
n =1
n
=∫
1
0
1
dx
xx
(1)
n
∞
1
n ( x ln x )
exp(
x
ln
x
)
(
1)
=
−
=
−
∑
xx
n!
n =0
(2)
The convergence of this series being uniform on [0,1] we get that
∫
1
0
∞
1
( −1) n
dx
=
∑
xx
n!
n =0
1
∫ ( x ln x)
n
dx
0
(3)
Proof (cont.)
To evaluate the integral of ( x ln x) it is more convenient, but
not essential to look at the more general function, x n (ln x) m
where m, n are positive integers, since it is easy to get a general
recurrence formula:
n
I m,n =
∫
1
0
x n (ln x) m dx = −
m
I m −1,n
n +1
m!
m!
I 0,n = (−1) m
(n + 1) m
(n + 1) m +1
n!
= (−1) n
(n + 1) n +1
I m,n = (−1) m
I n,n
Historical perspective
3 of 11
MEI conference 2008
Proof (cont.)
I n ,n = (−1) n
n!
(n + 1) n +1
Substituting this result in (3)
∫
1
0
∞
1
(−1) n
dx
=
∑
xx
n!
n =0
∞
Gives (1)
1
∑n
n =1
n
1
∫ ( x ln x)
0
=∫
1
0
1
dx
xx
n
(3)
dx
(1)
1
(−1) n
= ∫ x x dx
n
0
n =1 n
∞
∑
Same argument
Proof
Same argument proves the equally
interesting formula:
1
(−1) n
=
x x dx
∑
n
∫
0
n =1 n
∞
Proof…is irrational
Common knowledge of proof that
using proof by contradiction.
2 is irrational,
Equally interesting, but perhaps slightly less
common is proof by contradiction that e is irrational.
If you search for it there are several webpages that
show this, including for example the excellent
meikleriggs webpages (meikleriggs.org.uk)
Historical perspective
4 of 11
MEI conference 2008
Andrew Rogers
Frank Nelson Cole and his 1903 lecture, “On the Factorisation of Large Numbers”, in
which he proved that 2^67 – 1 was composite.
“On the Factorisation of Large Numbers” (MEI Conference, 2008)
Starting point
Core 3, January 2008, Question 5
These are Mersenne primes. More on Mersenne in a minute.
If I remember correctly from awarding, in part (i) candidates identified the primes p
less than 11 correctly, and produced 3, 7, 31 and 127 which are all prime. In part (ii),
they could calculate 23 × 89 correctly (!) but misunderstood the last part, often
attempting to evaluate 22047 − 1. Anyway, 211 − 1 is composite.
Fr Marin Mersenne, 1588-1648
He was a French monk, philosopher and mathematician. He was friendly with Fermat,
Pascal and Galileo. There were no mathematical journals in his time and he acted as a
clearing house for mathematical ideas: people sent him details of their discoveries and
he passed them on. But he had many ideas of his own. One of them concerned
numbers of the form 2n − 1, and if any such numbers were prime.
It is easy to see that 2n − 1 is not prime if n is not prime, e.g. by using the binary
representation (although I doubt if Mersenne did this!). So consider the case where n
is prime. Mersenne, in a letter to Frenicle de Bessy, proposed that 2n − 1 was prime
for a list of prime values of n, not including 11 (see C3 W08) but including 31, 67,
127 and 257. The smallest of these numbers has 10 digits and it was beyond the
capacity of anybody at the time to prove or disprove this assertion. In 1750 Euler
succeeded to prove that 231 − 1 was prime.
Edouard Lucas, 1842-1891
He devised a simple test for checking the “primeness” of these Mersenne numbers,
and showed that 2127 − 1 was prime (it was the largest known prime for more than 70
years after this discovery) and that 267 − 1 was not prime. However, his method did
not exhibit the factors. He does not seem to have said anything about 2257 − 1.
Subsequent searching showed that Mersenne had missed a few: 2n − 1 is prime for n =
61, 89 and 107.
Frank Nelson Cole, 1861-1926
Cole was born in Massachusetts and educated at Harvard. In 1903 he gave a
presentation to the American Mathematical Society called “On the Factorisation of
Large Numbers”. Obviously he had no modern technology to help him deliver his
lecture. Without saying a word, he wrote up on the blackboard the calculation for
raising 2 to the 67th power, then carefully subtracted 1. (More carefully than
Wikipedia or the BBC web site, which have the answer wrong.) On the other side of
the board, he then wrote out 193,707,721 × 761,838,257,287 and worked out the
multiplication by hand. The two answers agreed. “Cole returned to his seat, not
having uttered a word during the hour-long presentation.” (Wikipedia). He received a
standing ovation.
Historical perspective
5 of 11
MEI conference 2008
Three years of Sundays
How had Cole found the factors? He admitted that it had taken him “three years of
Sundays”. But these days the problem, and its history, can be presented to pupils and
the answer worked out before their eyes.
Do you know the program Derive? It can work out calculations to integer accuracy,
and the most truculent Year 7 class can be entertained by asking Derive to work out
100000! and then asking them to see if they can see their phone number in the sea of
digits. It disposes of the primeness of 267 − 1 in 0.04 seconds. The Texas TI-92 has a
version of the Derive software on it, and can evaluate the expression to integer
accuracy, but cannot do the factorisation. Derive on a PC can factorise 2257 − 1, and it
takes about one minute. The smallest prime factor has 15 digits.
Frank Nelson Cole
He lectured at Harvard, the University of Michigan and Columbia University, was the
secretary of the American Mathematical Society and editor of its Bulletin. There is a
Frank Nelson Cole prize for work in number theory. He died suddenly of heart failure
in New York City in 1926. A Google Image Search produced the following
photograph (Nat King Cole and Frank Sinatra).
Frank Nelson Cole
“On the Factorisation of
Large Numbers”
Andrew Rogers
King Edward VI Camp Hill Boys’ School
Core 3, January 2008
Historical perspective
6 of 11
MEI conference 2008
Fr Marin Mersenne, 1588-1648
2n – 1 prime for
n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257
Edouard Lucas, 1842-1891
Proved that 2127 – 1 is prime
(largest known prime for over 70 years)
Showed 267 – 1 wasn’t, but couldn’t find
the factors
Frank Nelson Cole, 1861-1926
Advertised a lecture called “On the
Factorisation of Large Numbers” (1903)
He had a blackboard
He showed that
267 – 1 = 147,573,952,589,676,412,927
(spot the error on Wikipedia)
Then he worked out
193,707,721 × 761,838,257,287
The two answers agreed
Three years of Sundays
Children like large numbers
Use Derive to work out 267 – 1 and
factorise the result
It takes 0.04 seconds
The TI-92 can’t do it
I tried 2257 – 1 which has 78 digits
Derive took 1 minute to factorise it
The smallest prime factor has 15 digits
Historical perspective
7 of 11
MEI conference 2008
Google Image Search
Historical perspective
8 of 11
MEI conference 2008
Peter Mitchell
The Königsberg Bridge Problem.
Reference
http://www.contracosta.edu/math/konig.htm
Gerald Goodall
Experimental design in Wales.
The 1920s and 1930s must have been an incredibly exciting time for applied
statisticians. Wonderful new advances in the design and analysis of experiments
came out all the time. Nowadays we teach this as classical material - not least, in
MEI Statistics 4 and in the AS Statistics - and it is in absolutely standard practical use
in all application areas everywhere all round the world.
Including in Wales.
References
http://www.york.ac.uk/depts/maths/histstat/images/latin_square.gif
A key to the layout of this Latin square is at
http://www.york.ac.uk/depts/maths/histstat/images/latin_square_key.gif.
Some information is available by Googling for something like "beddgelert forest latin
square". Note the correct spelling of "beddgelert" rather than the incorrect
"bettgelert" which is often quoted for this material. Note for example that it appears
in the key given above. Googling using "bettgelert" will obtain some of the
information but not all of it. Several of the internet items that can be found make
interesting comparisons between Latin Squares and Sudoku puzzles.
Historical perspective
9 of 11
MEI conference 2008
Bernard Murphy
How mathematicians before Newton and Leibniz attempted to find the gradients of
tangents to various curves.
Alternative methods for finding the gradient of a tangent
Mathematicians before Newton and Leibniz attempted to find the gradients of
tangents to various curves. Here are two methods, using the curve y = x 2 as an
illustration.
Fermat’s method
y
y = x2
( x + e)
2
x2
x
s
e
By similar triangles,
x2 ( x + e)
≈
s
s+e
2
Making s the subject:
x2
s≈
2x + e
So as e → 0 , s →
x2 x
=
2x 2
Therefore, the slope of the tangent approaches
Historical perspective
10 of 11
x2 2 x2
=
= 2x
s
x
MEI conference 2008
y
Descartes’ method
y = x2
The circle, centre ( v, 0 ) , radius r ,
touches the curve y = x 2 at the point
(α , α ) . Therefore the tangent to the
2
circle is also the tangent to the curve at
this point.
r
x
α
v
The equation of the circle is ( x − v ) + y 2 = r 2 . The circle touches, rather than crosses,
2
the curve y = x 2 . Therefore we have a double root of x = α when we try to solve the
equations simultaneously.
One way to solve the equations simultaneously is to substitute y = x 2 into
( x − v)
2
+ y 2 = r 2 . This gives
( x − v)
2
+ ( x2 ) = r 2
2
We need values of v and r so that this equation has a double root of x = α .
In other words, we need ( x − v ) + x 4 − r 2 to be of the form ( x − α ) ( x 2 + px + q )
2
2
Comparing coefficients (in the expanded forms) gives:
p − 2α = 0
α − 2α p + q = 1
2
α 2 p − 2α q = −2v
α 2q = v2 − r 2
These simplify to p = 2α , q = 1 + 3α 2 , v = α + 2α 3
α2
and the gradient of the
v −α
3
v − α (α + 2α ) − α
= 2 =
= 2α
α
α2
Therefore, the gradient of the radius shown is −
−1
(perpendicular) tangent is
−
Historical perspective
α2
v −α
11 of 11
MEI conference 2008