math quotes

pi/4 = 1-1/3+1/5-1/7+1/9 ...
-- Wilhelm von Leibniz

2000 BC Babilonians use pi=25/8, Egyptians use pi=256/81
1100 BC Chinese use pi=3
200 AC Ptolemy uses pi=377/120
450 Tsu Ch'ung-chih uses pi=255/113
530 Aryabhata uses pi=62832/20000
650 Brahmagupta uses pi=sqrt(10)
1593 Romanus finds pi to 15 decimal places
1596 Van Ceulen calculates pi to 32 places
1699 Sharp calculates pi to 72 places
1719 Tantet de Lagny calculates pi to 127 places
1794 Vega calculates pi to 140 decimal places
1855 Richter calculates pi to 500 decimal places
1873 Shanks finds 527 decimal places
1947 Ferguson calculates 808 places
1949 ENIAC computer finds 2037 places
1955 NORC computer computes 3089 places
1959 IBM 704 computer finds 16167 places
1961 Shanks-Wrench (IBM7090) find 100200 places
1966 IBM 7030 computes 250000 places
1967 CDC6600 computes 500000 places
1973 Guilloud-Bouyer (CDC7600) find 1 Mio places
1983 Tamura-Kanada (HITACM-280H) compute 16 Mio places
1988 Kanada (HITAC M-280H) computes 16 Mio digits
1989 Chudnovsky finds 1000 Mio digits
1995 Kanada computes pi to 6000 Mio digits
1996 Chudnovsky computes pi to 8000 Mio digits
1997 Kanada determines pi to 51000 Mio digits
--David Blatner, the joy of pi

However successful the theory of a four dimensional world may be,
it is difficult to ignore a voice inside us which whispers:
"At the back of your mind, you know a fourth dimension is all
nonsense". I fancy that voice must have had a busy time in the
past history of physics. What nonsense to say that this solid
table on which I am writing is a collection of electrons
moving with prodigious speed in empty spaces, which relative to
electronic dimensions are as wide as the spaces between the
planets in the solar system! What nonsense to say that the thin
air is trying to cursh my body with a load of 14 lbs. to the
square inch! What nonsense that the star cluster which I see
through the telescope, obviously there NOW, is a glimpse into a
past age 50'000 years ago! Let us not be beguiled by this voice.
It is discredited...
-- Sir Arthur Eddington

In a forest a fox bumps into a little rabbit, and says,
"Hi, junior, what are you up to?"
"I'm writing a dissertation on how rabbits eat foxes," said the rabbit.
"Come now, friend rabbit, you know that's impossible!"
"Well, follow me and I'll show you."
They both go into the rabbit's dwelling and after a while the rabbit
emerges with a satisfied expression on his face.
Along comes a wolf. "Hello, what are we doing these days?"
"I'm writing the second chapter of my thesis, on how rabbits devour wolves."
"Are you crazy? Where is your academic honesty?"
"Come with me and I'll show you." ......
As before, the rabbit comes out with a satisfied look on his face and
this time he has a diploma in his paw. The camera pans back and into the
rabbit's cave and, as everybody should have guessed by now, we see an
enormous mean-looking lion sitting next to the bloody and furry remains
of the wolf and the fox. The moral of this story is:
It's not the contents of your thesis that are important --
it's your PhD advisor that counts.
- Unknown Usenet Source

How to catch a lion:
- THE HILBERT METHOD. Place a locked cage in the desert.
Set up the following axiomatic system.
(i) The set of lions is non-empty
(ii) If there is a lion in the desert, then there is a lion in the cage.
Theorem. There is a lion in the cage
- THE PEANO METHOD. There is a space-filling curve passing through every
point of the desert. Such a curve may be traversed in as short a time as
we please. Armed with a spear, traverse the curve faster than the
lion can move his own length.
- THE TOPOLOGICAL METHOD. The lion has a least the connectivity of a torus.
Transport the desert into 4-space. It can now be deformed in such a way as
to knot the lion. He is now helples.
- THE SURGERGY METHOD. The lion is an orientable 3-manifold with boundary
and so may be rendered contractible by surgery.
- THE UNIVERSAL COVERING METHOD. Cover the lion by his simply-connected
covering space. Since this has no holes, he is trapped.
- THE GAME THEORY METHOD. The lion is a big game, hence certainly a game.
There exists an optimal strategy. Follow it.
- THE SCHROEDINGER METHOD. At any instant there is a non-zero probability
that the lion is in the cage. Wait.
- THE ERASTOSHENIAN METHOD. Enumerate all objects in the desert: examine
them one by one; discard all those that are not lions. A refinement
will capture only prime lions.
- THE PROJECTIVE GEOMETRY METHOD. The desert is a plane. Project
this to a line, then project the line to a point inside the cage. The
lion goes to the same point.
- THE INVERSION METHOD. Take a cylindrical cage. First case: the lion
is in the cage: Trivial. Second case: the lion is outside the cage.
Go inside the cage. Invert at the boundary of the cage. The lion is
caught. Caution: Don't stand in the middle of the cage during the
inversion!

All science requires Mathematics. The knowledge of mathematical things is
almost innate in us... This is the easiest of sciences, a fact which is
obvious in that no one?s brain rejects it; for laymen and people who are
utterly illiterate know how to count and reckon.
-- Roger Bacon

The first million decimal places of pi are comprised of:
99959 0's
99758 1's
100026 2's
100229 3's
100230 4's
100359 5's
99548 6's
99800 7's
99985 8's
100106 9's
--David Blatner, the joy of pi

Today, it is not only that our kings do not know mathematics,
but our philosophers do not know mathematics and - to go a step
further - our mathematicians do not know mathematics.
-- J.R. Oppenheimer

THEOREM (A. Katok) There exists a measurable set E of area one in
the unit square (0,1) x [0,1] together with a family of disjoint
smooth real analytic curves G(y) which fill out this square, so that
each curve G(y) intersects E in at most one single point.
PROOF. Define for p in (0,1) the piecewise linear map T on [0,1]
by T(x)=x/p for x in A=[0,p) and f(x)=(x-p)/(1-p) for x in B=[p,1).
It is easy to see that T is measure-preserving. Denote by T^n(x) the
n'th iterate of the map, that is T^n(x)=T(T^(n-1)(x)). For fixed p, code
x by an infinite sequence b(n)=0 if x(n)=T^n(x) in A and b(n)=1 else.
In terms of this coding, T corresponds to the shift map. By the strong
law of large numbers, for Lebesgue almost every x in [0,1], the
frequency of 1's in the associated symbol space is defined and equal
to (1-p). Let E be the subset of (p,x) in (0,1) x [0,1] such that
the frequency of 1's is equal to 1-p. It is a measurable set. Because
the intersection of E with each line {p} x [0,1] has full Lebesgue
measure, Fubini's theorem implies that E has Lebesgue area 1.
For x in [0,1] define y(p,x) = sum b(n) 2^n, where b(n) is the coding
of x. The sets G(y) = { (p,x) | y(p,x)=y } are disjoint and each G(y)
is a smooth real analytic curve.
[Proof: set p(0)=p,p(1)=1-p. From x(n)=b(n) p(0)+x(n+1) p(b(n)) follows
x=x(p,y)=p(0)(b(1)+p(b(1))(b(2)+p(b(2))(b(3)+...) ...) ...)
=p(0)(b(1)+b(2) p(b(1)) +b(3) p(b(1)) p(b(2))
+b(4) p(b(1)) p(b(2)) p(b(3)) +...)
Set p(0)=p=(1+t)/2, p(1)=1-p=(1-t)/2. If |t| the above series has absolute value at most [(1+c)/2]^n so that the
series converges uniformly. By Weierstrass uniform convergence theorem,
for each y, the series defines x as an analytic function of t for |t|<1
and so as an analytic function of p for p in (0,1).
]
Now, G(y) is the graph of the function p -> x(p,y). Since a given symbol
sequence b(n) can have at most one limiting frequency
lim (b(1)+ ... + b(n))/n = 1-p, it follows that each G(y) can intersect
E in at most a single point (p,x(p,y)).
-- John Milnor, Mathem. Intelligencer, Vol 19, 1997

It is true that a mathematician who is not also something of a
poet will never be a perfect mathematician.
- K. Weierstrass,
Quoted in D MacHale,
Comic Sections (Dublin 1993)

II III V VII XI XIII XVII XIX XXIII XXIX ...

Let no one ignorant of mathematics enter here.
-- Plato, Inscription written over the
entrance to the academy

After a few years at MIT, the Mathematician Norbert Wiener moved to a larger house.
His wife, knowing his nature, figured that he would forget his new address and
be unable to find his way home after work. So she wrote the address
of the new home on a piece of paper that she made him put in his shirt pocket.
At lunchtime that day, the professor had an inspiring idea. He pulled the
paper out of his pocket and used it to scribble down some calculations. Finding
a flaw, he threw the paper away in disgust. At the end of the day he realized
he had thrown away his address, he now had no idea where he lived.
Putting his mind to work, he came up with a plan. He would go to his old
house and await rescue. His wife would surely realize that he
was lost and go to his old house to pick him up. Unfortunately, when he
arrived at his old house, there was no sign of his wife, only a small girl
standing in front of the house. "Excuse me, little girl" he said "but do you
happen to know where the people who used to live here moved to?" "It's okay,
Daddy," said the little girl, "Mommy sent me to get you".
Moral 1. Don't be surprised if the professor doesn't know your name by the end
of the semester.
Moral 2. Be glad your parents aren't mathematicians. if your parents are
mathematicians, introduce yourself and get them to help you through the
course.
- From the introduction of "How to ace calculus" by
C. Adams, A. Thompson and J. Hass

"Can you do addition?" the White Queen asked.
"What's one and one and one and one and one and one
and one and one and one and one?"
"I don't know," said Alice, "I lost count.".
-- Lewis Carrol alias Charles Lutwidge Dodgson,
Alice's Adventures in Wonderland

It is still an unending source of surprise for me to see
how a few scribbles on a blackboard or on a sheet of paper
could change the course of human affairs.
-- Stanislaw Ulam.

David Hilbert was one of the great European mathematicians at the turn of the
century. One of his students purchased an early automobile and died in one of
the first car accidents. Hilbert was asked to speak at the funeral. "Young Klaus"
he said, "was one of my finest students. He had an unusual gift for doing
mathematics. He was insterested in a great variety of problems, such as..."
There was a short pause, follwed by "Consider the set of differentiable functions
on the unit interval and take their closure in the ..."
Moral 1. Sit near the door.
Moral 2. Some mathematicians can be a little out of touch with reality. If your professor
falls in this category, look at the bright side. You will have lots of
funny stories by the end of the semester.
- From the introduction of "How to ace calculus" by
C. Adams, A. Thompson and J. Hass

THEOREM: All triangles are equilateral.
PROOF:
1) Given an arbitrary triangle ABC. Construct the middle orthogonal
on AB in D and cut it with the line dividing the angle at C. Call the
intersection E. Form the normal from E to AC in F and from E to BC
in G. Draw the lines AE und BE.
C *
/ \
/ \
*F *G
/ E* \
/ | \
/ | \
/ |D \
A*---------*------------*B
2. The angles ECF and ECG are gleich. The angles EFC and EGC are both
right angles. Because the triangles ECF and ECG have also EC common,
they myust be congruent. Therefore CF=CG and EF=EG.
3. The sides DA and DB are equal. The angle EDA and EDB are both
right angles. Because the triangles EDA and EDB have also ED in common,
they have to be congruent and EA=EB.
4. The angle EGB and EFA are both right angle. Also, EF=EG and EA=EB.
Therefore both triangles EGB and EFA are congruent. Therefore FA=GB.
5. Since CF=CG and FA=GB, addition of the sides gives also CA=CB.
6. Having proved that two arbitrary sides are equal, all are equal.

Euler's formula: A connected plane graph with n vertices, e edges and f faces
satisfies n - e + f = 2
Proof. Let T be the edge set of a spanning tree for G. It is a subset of the
set E of edges. A spanning tree is a minimal subgraph that connects all the
vertices of G. It contains so no cycle. The dual graph G* of G has a vertex
in the interior of each face. Two vertices of G* are connected by an edge if
the correponding faces have a common boundary edge. G* can have double edges
even if the original graph was simple. Consider the collection T* of edges
E* in G* that correspond to edges in the complement of T in E. The edges of
T* connect all the faces because T does not have a cycle. Also T* does not
contain a cycle, since otherwise, it would seperate some vertices of G
contradicting that T was a spanning subgraph and edges of T and T* don't
intersect. Thus T* is a spanning tree for G*. Clearly e(T)+e(T*)=2.
For every tree, the number of vertices is one larger than the number of
vertices. Applied to the tree T, this yields n = e(T)+1, while for the tree
T* it yields f=e(T*)+1. Adding both equations gives n+f=(e(T)+1)+(e(T*)+1)=e+2.
-- from M.Aigner, G. Ziegler "Proofs from THE BOOK"

There is keen competition to produce the largest pair of twin primes.
On October 9, 1995, Dubner discovered the largest known pair of twin
primes p,p+2 where p=570918348*10^5120 - 1. It took only one day
with 2 crunchers. The expected time would be 150 times longer! What
luck!
-- from P. Ribenboim,
'The new book of prime number records'

I knew a mathematician, who said 'I do not know as much as God. But I
know as much as God knew at my age'.
-- Milton Shulman, Candian writer