| Subject: sci.math: Frequently Asked Questions [1/3]
Originator: [email protected]
Organization: University of Waterloo
Date: Fri, 19 Nov 1993 14:03:31 GMT
Archive-Name: sci-math-faq/part1
Last-modified: October 26, 1993
Version: 5.0
This is the list of Frequently Asked Questions for sci.math (version 5.0).
Any contributions/suggestions/corrections are most welcome. Please use
* e-mail * on any comment concerning the FAQ list.
Section 1 of 3, questions 1Q to 5Q.
Table of Contents
-----------------
1Q.- Fermat's Last Theorem, status of ..
2Q.- Values of Record Numbers
3Q.- Formula for prime numbers...
4Q.- Digits of Pi, computation and references
5Q.- Odd Perfect Number
6Q.- Computer Algebra Systems, application of ..
7Q.- Computer Algebra Systems, references to ..
8Q.- Fields Medal, general info ..
9Q.- Four Colour Theorem, proof of ..
10Q.- 0^0=1. A comprehensive approach
11Q.- 0.999... = 1. Properties of the real numbers ..
12Q.- There are three doors, The Monty Hall problem, Master Mind and
other games ..
13Q.- Surface and Volume of the n-ball
14Q.- f(x)^f(x)=x, name of the function ..
15Q.- Projective plane of order 10 ..
16Q.- How to compute day of week of a given date
17Q.- Axiom of Choice and/or Continuum Hypothesis?
18Q.- Cutting a sphere into pieces of larger volume
19Q.- Pointers to Quaternions
20Q.- Erdos Number
21Q.- Why is there no Nobel in mathematics?
22Q.- General References and textbooks...
23Q.- Interest Rate...
24Q.- Euler's formula e^(i Pi) = - 1 ...
1Q: What is the current status of Fermat's last theorem?
and
Did Fermat prove this theorem?
Fermat's Last Theorem:
There are no positive integers x,y,z, and n > 2 such that x^n + y^n = z^n.
I heard that <insert name here> claimed to have proved it but later
on the proof was found to be wrong. ...
A: The status of FLT has remained remarkably constant. Every few
years, someone claims to have a proof ... but oh, wait, not quite.
UPDATE... UPDATE... UPDATE
Andrew Wiles, a researcher at Princeton, Cambridge claims to have
found a proof.
The proof was presented in Cambridge, UK during a three day seminar
to an audience including some of the leading experts in the field.
The proof is long and cumbersome, but here are some of the first
few details:
*From Ken Ribet:
Here is a brief summary of what Wiles said in his three lectures.
The method of Wiles borrows results and techniques from lots and lots
of people. To mention a few: Mazur, Hida, Flach, Kolyvagin, yours
truly, Wiles himself (older papers by Wiles), Rubin... The way he does
it is roughly as follows. Start with a mod p representation of the
Galois group of Q which is known to be modular. You want to prove that
all its lifts with a certain property are modular. This means that the
canonical map from Mazur's universal deformation ring to its "maximal
Hecke algebra" quotient is an isomorphism. To prove a map like this is
an isomorphism, you can give some sufficient conditions based on
commutative algebra. Most notably, you have to bound the order of a
cohomology group which looks like a Selmer group for Sym^2 of the
representation attached to a modular form. The techniques for doing
this come from Flach; you also have to use Euler systems a la
Kolyvagin, except in some new geometric guise.
If you take an elliptic curve over Q, you can look at the
representation of Gal on the 3-division points of the curve. If you're
lucky, this will be known to be modular, because of results of Jerry
Tunnell (on base change). Thus, if you're lucky, the problem I
described above can be solved (there are most definitely some
hypotheses to check), and then the curve is modular. Basically, being
lucky means that the image of the representation of Galois on
3-division points is GL(2,Z/3Z).
Suppose that you are unlucky, i.e., that your curve E has a rational
subgroup of order 3. Basically by inspection, you can prove that if it
has a rational subgroup of order 5 as well, then it can't be
semistable. (You look at the four non-cuspidal rational points of
X_0(15).) So you can assume that E[5] is "nice." Then the idea is to
find an E' with the same 5-division structure, for which E'[3] is
modular. (Then E' is modular, so E'[5] = E[5] is modular.) You
consider the modular curve X which parameterizes elliptic curves whose
5-division points look like E[5]. This is a "twist" of X(5). It's
therefore of genus 0, and it has a rational point (namely, E), so it's
a projective line. Over that you look at the irreducible covering
which corresponds to some desired 3-division structure. You use
Hilbert irreducibility and the Cebotarev density theorem (in some way
that hasn't yet sunk in) to produce a non-cuspidal rational point of X
over which the covering remains irreducible. You take E' to be the
curve corresponding to this chosen rational point of X.
*From the previous version of the FAQ:
(b) conjectures arising from the study of elliptic curves and
modular forms. -- The Taniyama-Weil-Shmimura conjecture.
There is a very important and well known conjecture known as the
Taniyama-Weil-Shimura conjecture that concerns elliptic curves.
This conjecture has been shown by the work of Frey, Serre, Ribet,
et. al. to imply FLT uniformly, not just asymptotically as with the
ABC conj.
The conjecture basically states that all elliptic curves can be
parameterized in terms of modular forms.
There is new work on the arithmetic of elliptic curves. Sha, the
Tate-Shafarevich group on elliptic curves of rank 0 or 1. By the way
an interesting aspect of this work is that there is a close
connection between Sha, and some of the classical work on FLT. For
example, there is a classical proof that uses infinite descent to
prove FLT for n = 4. It can be shown that there is an elliptic curve
associated with FLT and that for n=4, Sha is trivial. It can also be
shown that in the cases where Sha is non-trivial, that
infinite-descent arguments do not work; that in some sense 'Sha
blocks the descent'. Somewhat more technically, Sha is an
obstruction to the local-global principle [e.g. the Hasse-Minkowski
theorem].
*From Karl Rubin:
Theorem. If E is a semistable elliptic curve defined over Q,
then E is modular.
It has been known for some time, by work of Frey and Ribet, that
Fermat follows from this. If u^q + v^q + w^q = 0, then Frey had
the idea of looking at the (semistable) elliptic curve
y^2 = x(x-a^q)(x+b^q). If this elliptic curve comes from a modular
form, then the work of Ribet on Serre's conjecture shows that there
would have to exist a modular form of weight 2 on Gamma_0(2). But
there are no such forms.
To prove the Theorem, start with an elliptic curve E, a prime p and let
rho_p : Gal(Q^bar/Q) -> GL_2(Z/pZ)
be the representation giving the action of Galois on the p-torsion
E[p]. We wish to show that a _certain_ lift of this representation
to GL_2(Z_p) (namely, the p-adic representation on the Tate module
T_p(E)) is attached to a modular form. We will do this by using
Mazur's theory of deformations, to show that _every_ lifting which
'looks modular' in a certain precise sense is attached to a modular form.
Fix certain 'lifting data', such as the allowed ramification,
specified local behavior at p, etc. for the lift. This defines a
lifting problem, and Mazur proves that there is a universal
lift, i.e. a local ring R and a representation into GL_2(R) such
that every lift of the appropriate type factors through this one.
Now suppose that rho_p is modular, i.e. there is _some_ lift
of rho_p which is attached to a modular form. Then there is
also a hecke ring T, which is the maximal quotient of R with the
property that all _modular_ lifts factor through T. It is a
conjecture of Mazur that R = T, and it would follow from this
that _every_ lift of rho_p which 'looks modular' (in particular the
one we are interested in) is attached to a modular form.
Thus we need to know 2 things:
(a) rho_p is modular
(b) R = T.
It was proved by Tunnell that rho_3 is modular for every elliptic
curve. This is because PGL_2(Z/3Z) = S_4. So (a) will be satisfied
if we take p=3. This is crucial.
Wiles uses (a) to prove (b) under some restrictions on rho_p. Using
(a) and some commutative algebra (using the fact that T is Gorenstein,
'basically due to Mazur') Wiles reduces the statement T = R to
checking an inequality between the sizes of 2 groups. One of these
is related to the Selmer group of the symmetric square of the given
modular lifting of rho_p, and the other is related (by work of Hida)
to an L-value. The required inequality, which everyone presumes is
an instance of the Bloch-Kato conjecture, is what Wiles needs to verify.
He does this using a Kolyvagin-type Euler system argument. This is
the most technically difficult part of the proof, and is responsible
for most of the length of the manuscript. He uses modular
units to construct what he calls a 'geometric Euler system' of
cohomology classes. The inspiration for his construction comes
from work of Flach, who came up with what is essentially the
'bottom level' of this Euler system. But Wiles needed to go much
farther than Flach did. In the end, _under_certain_hypotheses_ on rho_p
he gets a workable Euler system and proves the desired inequality.
Among other things, it is necessary that rho_p is irreducible.
Suppose now that E is semistable.
Case 1. rho_3 is irreducible.
Take p=3. By Tunnell's theorem (a) above is true. Under these
hypotheses the argument above works for rho_3, so we conclude
that E is modular.
Case 2. rho_3 is reducible.
Take p=5. In this case rho_5 must be irreducible, or else E
would correspond to a rational point on X_0(15). But X_0(15)
has only 4 noncuspidal rational points, and these correspond to
non-semistable curves. _If_ we knew that rho_5 were modular,
then the computation above would apply and E would be modular.
We will find a new semistable elliptic curve E' such that
rho_{E,5} = rho_{E',5} and rho_{E',3} is irreducible. Then
by Case I, E' is modular. Therefore rho_{E,5} = rho_{E',5}
does have a modular lifting and we will be done.
We need to construct such an E'. Let X denote the modular
curve whose points correspond to pairs (A, C) where A is an
elliptic curve and C is a subgroup of A isomorphic to the group
scheme E[5]. (All such curves will have mod-5 representation
equal to rho_E.) This X is genus 0, and has one rational point
corresponding to E, so it has infinitely many. Now Wiles uses a
Hilbert Irreducibility argument to show that not all rational
points can be images of rational points on modular curves
covering X, corresponding to degenerate level 3 structure
(i.e. im(rho_3) not GL_2(Z/3)). In other words, an E' of the
type we need exists. (To make sure E' is semistable, choose
it 5-adically close to E. Then it is semistable at 5, and at
other primes because rho_{E',5} = rho_{E,5}.)
2Q: What are the values of:
largest known Mersenne prime?
A: It is 2^756839-1. It was discovered by a Cray-2 in England in 1992.
It has 227,832 digits.
largest known prime?
A: The largest known prime is the Mersenne prime described above.
The previous record holder, and the largest known non-Mersenne prime,
is 391581*2^216193-1. See Brown, Noll, Parady, Smith, Smith, and
Zarantonello, Letter to the editor, American Mathematical Monthly,
vol. 97, 1990, p. 214. Throughout history, the largest known prime
has almost always been a Mersenne prime; the period between Brown
et al's discovery in Aug 1989 and Slowinski & Gage's in March 1992
is one of the few exceptions.
largest known twin primes?
A: The largest known twin primes are 1691232 * 1001 * 10^4020 +- 1,
which is a number with 4030 digits, found by H. Dubner.
The second largest known twin primes are 4650828 * 1001 * 10^3429 +- 1.
They were found by H. Dubner
References:
B. K. Parady and J. F. Smith and S. E. Zarantonello,
Smith, Noll and Brown.
Largest known twin primes, Mathematics of Computation,
vol.55, 1990, pp. 381-382.
largest Fermat number with known factorization?
A: F_11 = (2^(2^11)) + 1 which was factored by Brent & Morain in
1988. F9 = (2^(2^9)) + 1 = 2^512 + 1 was factored by
A.K. Lenstra, H.W. Lenstra Jr., M.S. Manasse & J.M. Pollard
in 1990. The factorization for F10 is NOT known.
Are there good algorithms to factor a given integer?
A: There are several that have subexponential estimated
running time, to mention just a few:
Continued fraction algorithm,
Class group method,
Quadratic sieve algorithm,
Elliptic curve algorithm,
Number field sieve,
Dixon's random squares algorithm,
Valle's two-thirds algorithm,
Seysen's class group algorithm,
A.K. Lenstra, H.W. Lenstra Jr., "Algorithms in Number Theory",
in: J. van Leeuwen (ed.), Handbook of Theoretical Computer
Science, Volume A: Algorithms and Complexity, Elsevier, pp.
673-715, 1990.
List of record numbers?
A: Chris Caldwell maintains "THE LARGEST KNOWN PRIMES (ALL KNOWN
PRIMES WITH 2000 OR MORE DIGITS)"-list. Send him mail to
[email protected] (preferred) or [email protected], on any new
gigantic primes (greater than 10,000 digits), titanic primes
(greater than 1000 digits).
What is the current status on Mersenne primes?
A: Mersenne primes are primes of the form 2^p-1. For 2^p-1 to be prime
we must have that p is prime. The following Mersenne primes are
known.
nr p year by
-----------------------------------------------------------------
1-5 2,3,5,7,13 in or before the middle ages
6-7 17,19 1588 Cataldi
8 31 1750 Euler
9 61 1883 Pervouchine
10 89 1911 Powers
11 107 1914 Powers
12 127 1876 Lucas
13-14 521,607 1952 Robinson
15-17 1279,2203,2281 1952 Lehmer
18 3217 1957 Riesel
19-20 4253,4423 1961 Hurwitz & Selfridge
21-23 9689,9941,11213 1963 Gillies
24 19937 1971 Tuckerman
25 21701 1978 Noll & Nickel
26 23209 1979 Noll
27 44497 1979 Slowinski & Nelson
28 86243 1982 Slowinski
29 110503 1988 Colquitt & Welsh jr.
30 132049 1983 Slowinski
31 216091 1985 Slowinski
32? 756839 1992 Slowinski & Gage
The way to determine if 2^p-1 is prime is to use the Lucas-Lehmer
test:
Lucas_Lehmer_Test(p):
u := 4
for i from 3 to p do
u := u^2-2 mod 2^p-1
od
if u == 0 then
2^p-1 is prime
else
2^p-1 is composite
fi
The following ranges have been checked completely:
2 - 355K and 430K - 520K
More on Mersenne primes and the Lucas-Lehmer test can be found in:
G.H. Hardy, E.M. Wright, An introduction to the theory of numbers,
fifth edition, 1979, pp. 16, 223-225.
3Q.- Formula for prime numbers...
Is there a polynomial which gives all the prime numbers?
No, there is not. This is a simple exercise to prove.
Is there a non-constant polynomial that only takes on prime values?
It has been proved that no such polynomial exists.
The proof is simple enough that an high school student could probably
discover it. See, for example, Ribenboim's book _The Book of Prime
Number Records_.
Note, however, by the work of Jones, Sato, Wada, and Wiens, there *is* a
polynomial in 26 variables such that the set of primes coincides with
the set of *positive* values taken by this polynomial. See Ribenboim,
pp. 147-150.
But most people would object to the term "formula" restricted to mean
polynomial. Can we not use summation signs, factorial, and the floor
function in our "formula"? If so, then indeed, there *are* formulas
for the prime numbers. Some of them are listed below.
If we can't, then exactly what operations do you allow and why?
Indeed, as I have previously argued, a reasonable interpretation of
the word "formula" is simply "Turing machine that halts on all inputs".
Under this interpretation, there certainly are halting Turing machines
which compute the n'th prime number. However, nobody knows how to
compute the n'th prime in time polynomial in log n. That's still
an open question.
Herb Wilf has addressed the question, "What is a formula?" in his
delightful article, "What is an answer?" which appeared in the
American Mathematical Monthly, 89 (1982), 289-292. He draws the
distinction between "formula" and "good formula". Anyone who claims
"there is no formula for the prime numbers" should read this article.
Here are just a few articles that discuss "formulas" for primes. Almost
all of these do *not* require computation of the primes "ahead of time".
Most of them rely on standard mathematical functions such as summation,
factorial, greatest integer function, etc.
C. Isenkrahe, Math. Annalen 53 (1900), 42-44.
W. H. Mills, Bull. Amer. Math. Soc. 53 (1947), 604.
L. Moser, Math. Mag. 23 (1950), 163-164.
E. M. Wright, Amer. Math. Monthly 58 (1951), 616-618. (Correction,
59 (1952), 99.)
E. M. Wright, J. Lond. Math. Soc. 29 (1954), 63-71.
B. R. Srinivasan, J. Indian Math. Soc. 25 (1961), 33-39.
C. P. Willans, Math. Gazette 48 (1964), 413-415.
V. C. Harris, Nordisk Mat. Tidskr. 17 (1969), 82.
U. Dudley, Amer. Math. Monthly 76 (1969), 23-28.
C. Vanden Eynden, Amer. Math. Monthly 79 (1972), 625.
S. W. Golomb, Amer. Math. Monthly 81 (1974), 752-754.
For more references see
J.O. Shallit, E. Bach, _Algorithmic Number Theory_ (to be published,
MIT Press).
4Q: Where I can get pi up to a few hundred thousand digits of pi?
Does anyone have an algorithm to compute pi to those zillion
decimal places?
A: MAPLE or MATHEMATICA can give you 10,000 digits of Pi in a blink,
and they can compute another 20,000-1,000,000 overnight (range depends
on hardware platform).
It is possible to retrieve 1.25+ million digits of pi via anonymous
ftp from the site wuarchive.wustl.edu, in the files pi.doc.Z and
pi.dat.Z which reside in subdirectory doc/misc/pi.
New York's Chudnovsky brothers have computed 2 billion digits of pi
on a homebrew computer.
How is pi calculated to many decimals ?
There are essentially 3 different methods.
1) One of the oldest is to use the power series expansion of atan(x)
atan(x)=x-x^3/3+x^5/5-... together with formulas like
pi=16*atan(1/5)-4*atan(1/239). This gives about 1.4 decimals per term.
2) A second is to use formulas coming from Arithmetic-Geometric mean
computations. A beautiful compendium of such formulas is given in the
book of Borwein and Borwein: Pi and the AGM, Canadian Math. Soc. Series,
John Wiley and Sons, New York, 1987. They have the advantage of converging
quadratically, i.e. you double the number of decimals per iteration.
For instance, to obtain 1 000 000 decimals, around 20 iterations are
sufficient. The disadvantage is that you need FFT type multiplication
to get a reasonable speed, and this is not so easy to program.
3) A third one comes from the theory of complex multiplication of elliptic
curves, and was discovered by S. Ramanujan. This gives a number of
beautiful formulas, but the most useful was missed by Ramanujan and
discovered by the Chudnovsky's. It is the following (slightly modified
for ease of programming):
Set k1=545140134;k2=13591409;k3=640320;k4=100100025;k5=327843840;k6=53360;
Then in AmsTeX notation
$\pi=\frac{k6\sqrt(k3)}{S}$, where
$$S=\sum_{n=0}^\infty (-1)^n\frac{(6n)!(k2+nk1)}{n!^3(3n)!(8k4k5)^n}$$
The great advantages of this formula are that
1) It converges linearly, but very fast (more than 14 decimal digits per
term).
2) The way it is written, all operations to compute S can be programmed
very simply since it only involves multiplication/division by single
precision numbers. This is why the constant 8k4k5 appearing in the
denominator has been written this way instead of 262537412640768000.
This is how the Chudnovsky's have computed several billion decimals.
Question: how can I get a C program which computes pi?
Answer: if you are too lazy to use the hints above, you can use the
following 160 character C program (who is the author of this?) which
computes pi to 800 decimal digits. If you want more, it is easy to modify,
but you have to understand how it works first.
int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;
for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a)for(b=c;d+=f[b]*a,
f[b]=d%--g,d/=g--,--b;d*=b);}
References :
(This is a short version for a more comprehensive list contact
Juhana Kouhia at [email protected])
J. M. Borwein, P. B. Borwein, and D. H. Bailey, "Ramanujan,
Modular Equations, and Approximations to Pi", American Mathematical
Monthly, vol. 96, no. 3 (March 1989), p. 201 - 220.
P. Beckman
A history of pi
Golem Press, CO, 1971 (fourth edition 1977)
J.M. Borwein and P.B. Borwein
The arithmetic-geometric mean and fast computation of elementary
functions
SIAM Review, Vol. 26, 1984, pp. 351-366
J.M. Borwein and P.B. Borwein
More quadratically converging algorithms for pi
Mathematics of Computation, Vol. 46, 1986, pp. 247-253
J.M. Borwein and P.B. Borwein
Pi and the AGM - a study in analytic number theory and
computational complexity
Wiley, New York, 1987
Shlomo Breuer and Gideon Zwas
Mathematical-educational aspects of the computation of pi
Int. J. Math. Educ. Sci. Technol., Vol. 15, No. 2, 1984,
pp. 231-244
David Chudnovsky and Gregory Chudnovsky
The computation of classical constants, Columbia University,
Proc. Natl. Acad. Sci. USA, Vol. 86, 1989.
Y. Kanada and Y. Tamura
Calculation of pi to 10,013,395 decimal places based on the
Gauss-Legendre algorithm and Gauss arctangent relation
Computer Centre, University of Tokyo, 1983
Morris Newman and Daniel Shanks
On a sequence arising in series for pi
Mathematics of computation, Vol. 42, No. 165, Jan 1984,
pp. 199-217
E. Salamin
Computation of pi using arithmetic-geometric mean
Mathematics of Computation, Vol. 30, 1976, pp. 565-570
D. Shanks and J.W. Wrench, Jr.
Calculation of pi to 100,000 decimals
Mathematics of Computation, Vol. 16, 1962, pp. 76-99
Daniel Shanks
Dihedral quartic approximations and series for pi
J. Number Theory, Vol. 14, 1982, pp.397-423
David Singmaster
The legal values of pi
The Mathematical Intelligencer, Vol. 7, No. 2, 1985
Stan Wagon
Is pi normal?
The Mathematical Intelligencer, Vol. 7, No. 3, 1985
J.W. Wrench, Jr.
The evolution of extended decimal approximations to pi
The Mathematics Teacher, Vol. 53, 1960, pp. 644-650
5Q: Does there exist a number that is perfect and odd?
A given number is perfect if it is equal to the sum of all its proper
divisors. This question was first posed by Euclid in ancient Greece.
This question is still open. Euler proved that if N is an odd
perfect number, then in the prime power decomposition of N, exactly
one exponent is congruent to 1 mod 4 and all the other exponents are
even. Furthermore, the prime occurring to an odd power must itself be
congruent to 1 mod 4. A sketch of the proof appears in Exercise 87,
page 203 of Underwood Dudley's Elementary Number Theory, 2nd ed.
It has been shown that there are no odd perfect numbers < 10^300.
Copyright Notice
Copyright (c) 1993 A. Lopez-Ortiz
--------------------------------------------------------------------------
Questions and Answers Edited and Compiled by:
Alex Lopez-Ortiz [email protected]
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada
|
| From: [email protected] (Alex Lopez-Ortiz)
Subject: sci.math: Frequently Asked Questions [2/3]
Organization: University of Waterloo
Date: Fri, 19 Nov 1993 14:03:38 GMT
Expires: Fri, 31 Dec 1993 14:03:28 GMT
Archive-Name: sci-math-faq/part2
Last-modified: October 26, 1993
Version: 5.0
This is a list of Frequently Asked Questions for sci.math (version 5.0).
Any contributions/suggestions/corrections are most welcome. Please use
* e-mail * on any comment concerning the FAQ list.
Section 2 of 3, questions 6Q to 18Q.
Table of Contents
-----------------
1Q.- Fermat's Last Theorem, status of ..
2Q.- Values of Record Numbers
3Q.- Formula for prime numbers...
4Q.- Digits of Pi, computation and references
5Q.- Odd Perfect Number
6Q.- Computer Algebra Systems, application of ..
7Q.- Computer Algebra Systems, references to ..
8Q.- Fields Medal, general info ..
9Q.- Four Colour Theorem, proof of ..
10Q.- 0^0=1. A comprehensive approach
11Q.- 0.999... = 1. Properties of the real numbers ..
12Q.- There are three doors, The Monty Hall problem, Master Mind and
other games ..
13Q.- Surface and Volume of the n-ball
14Q.- f(x)^f(x)=x, name of the function ..
15Q.- Projective plane of order 10 ..
16Q.- How to compute day of week of a given date
17Q.- Axiom of Choice and/or Continuum Hypothesis?
18Q.- Cutting a sphere into pieces of larger volume
19Q.- Pointers to Quaternions
20Q.- Erdos Number
21Q.- Why is there no Nobel in mathematics?
22Q.- General References and textbooks...
23Q.- Interest Rate...
24Q.- Euler's formula e^(i Pi) = - 1 ...
6Q: I have this complicated symbolic problem (most likely
a symbolic integral or a DE system) that I can't solve.
What should I do?
A: Find a friend with access to a computer algebra system
like MAPLE, MACSYMA or MATHEMATICA and ask her/him to solve it.
If packages cannot solve it, then (and only then) ask the net.
7Q: Where can I get <Symbolic Computation Package>?
THIS IS NOT A COMPREHENSIVE LIST. There are other Computer Algebra
packages available that may better suit your needs. There is also
a FAQ list in the group sci.math.symbolic. It includes a much larger
list of vendors and developers. (The FAQ list can be obtained from
math.berkeley.edu via anonymous ftp).
A: Maple
Purpose: Symbolic and numeric computation, mathematical
programming, and mathematical visualization.
Contact: Waterloo Maple Software,
450 Phillip Street
Waterloo, Ontario
N2L 5J2
Phone (519)747-2373
FAX (519)747-5284
email: [email protected]
A: DOE-Macsyma
Purpose: Symbolic and mathematical manipulations.
Contact: National Energy Software Center
Argonne National Laboratory 9700 South Cass Avenue
Argonne, Illinois 60439
Phone: (708) 972-7250
A: Pari
Purpose: Number-theoretic computations and simple numerical
analysis.
Available for most 32-bit machines, including 386+387 and 486.
This is a copyrighted but free package, available by ftp from
math.ucla.edu (128.97.4.254) and ftp.inria.fr (128.93.1.26).
Contact: questions about pari can be sent to [email protected]
and for the Macintosh versions to [email protected]
A: Mathematica
Purpose: Mathematical computation and visualization,
symbolic programming.
Contact: Wolfram Research, Inc.
100 Trade Center Drive Champaign,
IL 61820-7237
Phone: 1-800-441-MATH
[email protected]
A: Macsyma
Purpose: Symbolic numerical and graphical mathematics.
Contact: Macsyma Inc.
20 Academy Street
Arlington, MA 02174
tel: 617-646-4550
fax: 617-646-3161
email: [email protected]
A: Matlab
Purpose: `matrix laboratory' for tasks involving
matrices, graphics and general numerical computation.
Contact: The MathWorks, Inc.
21 Prime Park Way
Natick, MA 01760
508-653-1415
[email protected]
A: Cayley
Purpose: Computation in algebraic and combinatorial structures
such as groups, rings, fields, modules and graphs.
Available for: SUN 3, SUN 4, IBM running AIX or VM, DEC VMS, others
Contact: Computational Algebra Group
University of Sydney
NSW 2006
Australia
Phone: (61) (02) 692 3338
Fax: (61) (02) 692 4534
[email protected]
8Q: Let P be a property about the Fields Medal. Is P(x) true?
A: Institution is meant to be the Institution to which the researcher
in question was associated to at the time the medal was awarded.
Year Name Birthplace Age Institution
---- ---- ---------- --- -----------
1936 Ahlfors, Lars Helsinki Finland 29 Harvard U USA
1936 Douglas, Jesse New York NY USA 39 MIT USA
1950 Schwartz, Laurent Paris France 35 U of Nancy France
1950 Selberg, Atle Langesund Norway 33 Adv.Std.Princeton USA
1954 Kodaira, Kunihiko Tokyo Japan 39 Princeton U USA
1954 Serre, Jean-Pierre Bages France 27 College de France France
1958 Roth, Klaus Breslau Germany 32 U of London UK
1958 Thom, Rene Montbeliard France 35 U of Strasbourg France
1962 Hormander, Lars Mjallby Sweden 31 U of Stockholm Sweden
1962 Milnor, John Orange NJ USA 31 Princeton U USA
1966 Atiyah, Michael London UK 37 Oxford U UK
1966 Cohen, Paul Long Branch NJ USA 32 Stanford U USA
1966 Grothendieck, Alexander Berlin Germany 38 U of Paris France
1966 Smale, Stephen Flint MI USA 36 UC Berkeley USA
1970 Baker, Alan London UK 31 Cambridge U UK
1970 Hironaka, Heisuke Yamaguchi-ken Japan 39 Harvard U USA
1970 Novikov, Serge Gorki USSR 32 Moscow U USSR
1970 Thompson, John Ottawa KA USA 37 U of Chicago USA
1974 Bombieri, Enrico Milan Italy 33 U of Pisa Italy
1974 Mumford, David Worth, Sussex UK 37 Harvard U USA
1978 Deligne, Pierre Brussels Belgium 33 IHES France
1978 Fefferman, Charles Washington DC USA 29 Princeton U USA
1978 Margulis, Gregori Moscow USSR 32 InstPrblmInfTrans USSR
1978 Quillen, Daniel Orange NJ USA 38 MIT USA
1982 Connes, Alain Draguignan France 35 IHES France
1982 Thurston, William Washington DC USA 35 Princeton U USA
1982 Yau, Shing-Tung Kwuntung China 33 IAS USA
1986 Donaldson, Simon Cambridge UK 27 Oxford U UK
1986 Faltings, Gerd 1954 Germany 32 Princeton U USA
1986 Freedman, Michael Los Angeles CA USA 35 UC San Diego USA
1990 Drinfeld, Vladimir Kharkov USSR 36 Phys.Inst.Kharkov USSR
1990 Jones, Vaughan Gisborne N Zealand 38 UC Berkeley USA
1990 Mori, Shigefumi Nagoya Japan 39 U of Kyoto? Japan
1990 Witten, Edward Baltimore USA 38 Princeton U/IAS USA
References :
International Mathematical Congresses, An Illustrated History 1893-1986,
Revised Edition, Including 1986, by Donald J.Alberts, G. L. Alexanderson
and Constance Reid, Springer Verlag, 1987.
Tropp, Henry S., ``The origins and history of the Fields Medal,''
Historia Mathematica, 3(1976), 167-181.
9Q: Has the Four Colour Theorem been proved?
Four Color Theorem:
Every planar map with regions of simple borders can be coloured
with 4 colours in such a way that no two regions sharing a non-zero
length border have the same colour.
A: This theorem was proved with the aid of a computer in 1976.
The proof shows that if aprox. 1,936 basic forms of maps
can be coloured with four colours, then any given map can be
coloured with four colours. A computer program coloured this
basic forms. So far nobody has been able to prove it without
using a computer. In principle it is possible to emulate the
computer proof by hand computations.
References:
K. Appel and W. Haken, Every planar map is four colourable,
Bulletin of the American Mathematical Society, vol. 82, 1976
pp.711-712.
K. Appel and W. Haken, Every planar map is four colourable,
Illinois Journal of Mathematics, vol. 21, 1977, pp. 429-567.
T. Saaty and Paul Kainen, The Four Colour Theorem: Assault and
Conquest, McGraw-Hill, 1977. Reprinted by Dover Publications 1986.
K. Appel and W. Haken, Every Planar Map is Four Colourable,
Contemporary Mathematics, vol. 98, American Mathematical Society,
1989, pp.741.
F. Bernhart, Math Reviews. 91m:05007, Dec. 1991. (Review of Appel
and Haken's book).
10Q: What is 0^0 ?
A: According to some Calculus textbooks, 0^0 is an "indeterminate
form". When evaluating a limit of the form 0^0, then you need
to know that limits of that form are called "indeterminate forms",
and that you need to use a special technique such as L'Hopital's
rule to evaluate them. Otherwise, 0^0=1 seems to be the most
useful choice for 0^0. This convention allows us to extend
definitions in different areas of mathematics that otherwise would
require treating 0 as a special case. Notice that 0^0 is a
discontinuity of the function x^y.
Rotando & Korn show that if f and g are real functions that vanish
at the origin and are _analytic_ at 0 (infinitely differentiable is
not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from
the right.
From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):
"Some textbooks leave the quantity 0^0 undefined, because the
functions x^0 and 0^x have different limiting values when x
decreases to 0. But this is a mistake. We must define
x^0 = 1 for all x,
if the binomial theorem is to be valid when x=0, y=0, and/or x=-y.
The theorem is too important to be arbitrarily restricted! By
contrast, the function 0^x is quite unimportant."
Published by Addison-Wesley, 2nd printing Dec, 1988.
References:
H. E. Vaughan, The expression '0^0', Mathematics Teacher 63 (1970),
pp.111-112.
Louis M. Rotando & Henry Korn, "The Indeterminate Form 0^0",
Mathematics Magazine, Vol. 50, No. 1 (January 1977), pp. 41-42.
L. J. Paige, A note on indeterminate forms, American Mathematical
Monthly, 61 (1954), 189-190; reprinted in the Mathematical
Association of America's 1969 volume, Selected Papers on Calculus,
pp. 210-211.
11Q: Why is 0.9999... = 1?
A: In modern mathematics, the string of symbols "0.9999..." is
understood to be a shorthand for "the infinite sum 9/10 + 9/100
+ 9/1000 + ...." This in turn is shorthand for "the limit of the
sequence of real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000,
..." Using the well-known epsilon-delta definition of limit, one
can easily show that this limit is 1. The statement that
0.9999... = 1 is simply an abbreviation of this fact.
oo m
--- 9 --- 9
0.999... = > ---- = lim > ----
--- 10^n m->oo --- 10^n
n=1 n=1
Choose epsilon > 0. Suppose delta = 1/-log_10 epsilon, thus
epsilon = 10^(-1/delta). For every m>1/delta we have that
| m |
| --- 9 | 1 1
| > ---- - 1 | = ---- < ------------ = epsilon
| --- 10^n | 10^m 10^(1/delta)
| n=1 |
So by the (epsilon-delta) definition of the limit we have
m
--- 9
lim > ---- = 1
m->oo --- 10^n
n=1
An *informal* argument could be given by noticing that the following
sequence of "natural" operations has as a consequence 1 = 0.9999....
Therefore it's "natural" to assume 1 = 0.9999.....
x = 0.99999....
10x = 9.99999....
10x - x = 9
9x = 9
x = 1
Thus
1 = 0.99999....
References:
E. Hewitt & K. Stromberg, Real and Abstract Analysis,
Springer-Verlag, Berlin, 1965.
W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976.
12Q: There are three doors, and there is a car hidden behind one
of them, Master Mind and other games ..
A: Read frequently asked questions from rec.puzzles, as well as
their ``archive file'' where the problem is solved and carefully
explained. (The Monty Hall problem).
MANY OTHER MATHEMATICAL GAMES ARE EXPLAINED IN THE REC.PUZZLES
FAQ AND ARCHIVES. READ IT BEFORE ASKING IN SCI.MATH.
Your chance of winning is 2/3 if you switch and 1/3 if you don't.
For a full explanation from the rec.puzzles' archive, send to the
address [email protected] an email message consisting
of the text
send monty.hall
Also any other FAQ list can be obtained through anonymous ftp from
rtfm.mit.edu.
References
American Mathematical Monthly, January 1992.
For the game of Master Mind it has been proven that no more than
five moves are required in the worst case. For references look at
One such algorithm was published in the Journal of Recreational
Mathematics; in '70 or '71 (I think), which always solved the
4 peg problem in 5 moves. Knuth later published an algorithm which
solves the problem in a shorter # of moves - on average - but can
take six guesses on certain combinations.
Donald E. Knuth, The Computer as Master Mind, J. Recreational Mathematics
9 (1976-77), 1-6.
13Q: What is the formula for the "Surface Area" of a sphere in
Euclidean N-Space. That is, of course, the volume of the N-1
solid which comprises the boundary of an N-Sphere.
A: The volume of a ball is the easiest formula to remember: It's r^N
times pi^(N/2)/(N/2)!. The only hard part is taking the factorial
of a half-integer. The real definition is that x! = Gamma(x+1), but
if you want a formula, it's:
(1/2+n)! = sqrt(pi)*(2n+2)!/(n+1)!/4^(n+1)
To get the surface area, you just differentiate to get
N*pi^(N/2)/(N/2)!*r^(N-1).
There is a clever way to obtain this formula using Gaussian
integrals. First, we note that the integral over the line of
e^(-x^2) is sqrt(pi). Therefore the integral over N-space of
e^(-x_1^2-x_2^2-...-x_N^2) is sqrt(pi)^n. Now we change to
spherical coordinates. We get the integral from 0 to infinity
of V*r^(N-1)*e^(-r^2), where V is the surface volume of a sphere.
Integrate by parts repeatedly to get the desired formula.
It is possible to derive the volume of the sphere from ``first
principles''.
14Q: Does anyone know a name (or a closed form) for
f(x)^f(x)=x
Solving for f one finds a "continued fraction"-like answer
f(x) = log x
-----
log (log x
------
...........
A: This question has been repeated here from time to time over the
years, and no one seems to have heard of any published work on it,
nor a published name for it (D. Merrit proposes "lx" due to its
(very) faint resemblance to log). It's not an analytic function.
The "continued fraction" form for its numeric solution is highly
unstable in the region of its minimum at 1/e (because the graph is
quite flat there yet logarithmic approximation oscillates wildly),
although it converges fairly quickly elsewhere. To compute its value
near 1/e, use the bisection method which gives good results. Bisection
in other regions converges much more slowly than the "logarithmic
continued fraction" form, so a hybrid of the two seems suitable.
Note that it's dual valued for the reals (and many valued complex
for negative reals).
A similar function is a "built-in" function in MAPLE called W(x).
MAPLE considers a solution in terms of W(x) as a closed form (like
the erf function). W is defined as W(x)*exp(W(x))=x.
An extensive treatise on the known facts of Lambert's W function
is available for anonymous ftp at daisy.uwaterloo.ca in the
maple/5.2/doc/LambertW.ps.
15Q: Does there exist a projective plane of order 10?
More precisely:
Is it possible to define 111 sets (lines) of 11 points each
such that:
For any pair of points there is precisely one line containing them
both and for any pair of lines there is only one point common to
them both?
A: Analogous questions with n^2 + n + 1 and n + 1 instead of 111 and 11
have been positively answered only in case n is a prime power.
For n=6 it is not possible, more generally if n is congruent to 1
or 2 mod 4 and can not be written as a sum of two squares, then an
FPP of order n does not exist. The n=10 case has been settled as not
possible either by Clement Lam. As the "proof" took several years of
computer search (the equivalent of 2000 hours on a Cray-1) it can be
called the most time-intensive computer assisted single proof. The
final steps were ready in January 1989.
References
R. H. Bruck and H. J. Ryser, "The nonexistence of certain finite
projective planes," Canadian Journal of Mathematics, vol. 1 (1949),
pp 88-93.
C. Lam, Amer.Math.Monthly 98 (1991), 305-318.
16Q: Is there a formula to determine the day of the week, given
the month, day and year?
A: First a brief explanation: In the Gregorian Calendar, over a period
of four hundred years, there are 97 leap years and 303 normal years.
Each normal year, the day of January 1 advances by one; for each leap
year it advances by two.
303 + 97 + 97 = 497 = 7 * 71
As a result, January 1 year N occurs on the same day of the week as
January 1 year N + 400. Because the leap year pattern also recurs
with a four hundred year cycle, a simple table of four hundred
elements, and single modulus, suffices to determine the day of the
week (in the Gregorian Calendar), and does it much faster than all the
other algorithms proposed. Also, each element takes (in principle)
only three bits; the entire table thus takes only 1200 bits, or 300
bytes; on many computers this will be less than the instructions to do
all the complicated calculations proposed for the other algorithms.
The Julian Calendar has a four year cycle of leap years; after four
years January 1 has advanced by 5 days. Since 5 is relatively prime
to 7, a table of 35 elements is necessary.
Incidental note: Because 7 does not divide 400, January 1 occurs more
frequently on some days than others (in the Gregorian calendar, of
course)! Trick your friends!
In a cycle of four hundred years, January 1 and March 1 occur on the
following days with the following frequencies:
Sun Mon Tue Wed Thu Fri Sat
Jan 1: 58 56 58 57 57 58 56
Mar 1: 58 56 58 56 58 57 57
Of interest is that (contrary to most initial guesses) the occurrence
is not maximally flat.
Here is the standard method.
A. Take the last two digits of the year.
B. Divide by 4, discarding any fraction.
C. Add the day of the month.
D. Add the month's key value: JFM AMJ JAS OND
144 025 036 146
E. Subtract 1 for January or February of a leap year.
F. For a Gregorian date, add 0 for 1900's, 6 for 2000's, 4 for 1700's, 2
for 1800's; for other years, add or subtract multiples of 400.
G. For a Julian date, add 1 for 1700's, and 1 for every additional
century you go back.
H. Add the last two digits of the year.
Now take the remainder when you divide by 7; 1 is Sunday, the first day
of the week, 2 is Monday, and so on.
Another formula is:
W == k + [2.6m - 0.2] - 2C + Y + [Y/4] + [C/4] mod 7
where [] denotes the integer floor function (round down),
k is day (1 to 31)
m is month (1 = March, ..., 10 = December, 11 = Jan, 12 = Feb)
Treat Jan & Feb as months of the preceding year
C is century ( 1987 has C = 19)
Y is year ( 1987 has Y = 87 except Y = 86 for jan & feb)
W is week day (0 = Sunday, ..., 6 = Saturday)
This formula is good for the Gregorian calendar
(introduced 1582 in parts of Europe, adopted in 1752 in Great Britain
and its colonies, and on various dates in other countries).
It handles century & 400 year corrections, but there is still a
3 day / 10,000 year error which the Gregorian calendar does not take.
into account. At some time such a correction will have to be
done but your software will probably not last that long :-) !
References:
Winning Ways by Conway, Guy, Berlekamp is supposed to have it.
Martin Gardner in "Mathematical Carnival".
Michael Keith and Tom Craver, "The Ultimate Perpetual Calendar?",
Journal of Recreational Mathematics, 22:4, pp. 280-282, 19
K. Rosen, "Elementary Number Theory", p. 156.
17Q: What is the Axiom of Choice? Why is it important? Why some articles
say "such and such is provable, if you accept the axiom of choice."?
What are the arguments for and against the axiom of choice?
A: There are several equivalent formulations:
-The Cartesian product of nonempty sets is nonempty, even
if the product is of an infinite family of sets.
-Given any set S of mutually disjoint nonempty sets, there is a set C
containing a single member from each element of S. C can thus be
thought of as the result of "choosing" a representative from each
set in S. Hence the name.
>Why is it important?
All kinds of important theorems in analysis require it. Tychonoff's
theorem and the Hahn-Banach theorem are examples. Indeed,
Tychonoff's theorem is equivalent to AC. Similarly, AC is equivalent
to the thesis that every set can be well-ordered. Zermelo's first
proof of this in 1904 I believe was the first proof in which AC was
made explicit. AC is especially handy for doing infinite cardinal
arithmetic, as without it the most you get is a *partial* ordering
on the cardinal numbers. It also enables you to prove such
interesting general facts as that n^2 = n for all infinite cardinal
numbers.
> What are the arguments for and against the axiom of choice?
The axiom of choice is independent of the other axioms of set theory
and can be assumed or not as one chooses.
(For) All ordinary mathematics uses it.
There are a number of arguments for AC, ranging from a priori to
pragmatic. The pragmatic argument (Zermelo's original approach) is
that it allows you to do a lot of interesting mathematics. The more
conceptual argument derives from the "iterative" conception of set
according to which sets are "built up" in layers, each layer consisting
of all possible sets that can be constructed out of elements in the
previous layers. (The building up is of course metaphorical, and is
suggested only by the idea of sets in some sense consisting of their
members; you can't have a set of things without the things it's a set
of). If then we consider the first layer containing a given set S of
pairwise disjoint nonempty sets, the argument runs, all the elements
of all the sets in S must exist at previous levels "below" the level
of S. But then since each new level contains *all* the sets that can
be formed from stuff in previous levels, it must be that at least by
S's level all possible choice sets have already been *formed*. This
is more in the spirit of Zermelo's later views (c. 1930).
(Against) It has some supposedly counterintuitive consequences,
such as the Banach-Tarski paradox. (See next question)
Arguments against AC typically target its nonconstructive character:
it is a cheat because it conjures up a set without providing any
sort of *procedure* for its construction--note that no *method* is
assumed for picking out the members of a choice set. It is thus the
platonic axiom par excellence, boldly asserting that a given set
will always exist under certain circumstances in utter disregard of
our ability to conceive or construct it. The axiom thus can be seen
as marking a divide between two opposing camps in the philosophy of
mathematics: those for whom mathematics is essentially tied to our
conceptual capacities, and hence is something we in some sense
*create*, and those for whom mathematics is independent of any such
capacities and hence is something we *discover*. AC is thus of
philosophical as well as mathematical significance.
It should be noted that some interesting mathematics has come out of an
incompatible axiom, the Axiom of Determinacy (AD). AD asserts that
any two-person game without ties has a winning strategy for the first or
second player. For finite games, this is an easy theorem; for infinite
games with duration less than \omega and move chosen from a countable set,
you can prove the existence of a counter-example using AC. Jech's book
"The Axiom of Choice" has a discussion.
An example of such a game goes as follows.
Choose in advance a set of infinite sequences of integers; call it A.
Then I pick an integer, then you do, then I do, and so on forever
(i.e. length \omega). When we're done, if the sequence of integers
we've chosen is in A, I win; otherwise you win. AD says that one of
us must have a winning strategy. Of course the strategy, and which
of us has it, will depend upon A.
From a philosophical/intuitive/pedagogical standpoint, I think Bertrand
Russell's shoe/sock analogy has a lot to recommend it. Suppose you have an
infinite collection of pairs of shoes. You want to form a set with one
shoe from each pair. AC is not necessary, since you can define the set as
"the set of all left shoes". (Technically, we're using the axiom of
replacement, one of the basic axioms of Zermelo-Fraenkel (ZF) set theory.)
If instead you want to form a set containing one sock from each pair of an
infinite collection of pairs of socks, you now need AC.
References:
Maddy, "Believing the Axioms, I", J. Symb. Logic, v. 53, no. 2, June 1988,
pp. 490-500, and "Believing the Axioms II" in v.53, no. 3.
Gregory H. Moore, Zermelo's Axiom of Choice, New York, Springer-Verlag,
1982.
H. Rubin and J. E. Rubin, Equivalents of the Axiom of Choice II,
North-Holland/Elsevier Science, 1985.
A. Fraenkel, Y. Bar-Hillel, and A. Levy, Foundations of Set Theory,
Amsterdam, North-Holland, 1984 (2nd edition, 2nd printing), pp. 53-86.
18Q: Cutting a sphere into pieces of larger volume. Is it possible
to cut a sphere into a finite number of pieces and reassemble
into a solid of twice the volume?
A: This question has many variants and it is best answered explicitly.
Given two polygons of the same area, is it always possible to
dissect one into a finite number of pieces which can be reassembled
into a replica of the other?
Dissection theory is extensive. In such questions one needs to
specify
(A) what a "piece" is, (polygon? Topological disk? Borel-set?
Lebesgue-measurable set? Arbitrary?)
(B) how many pieces are permitted (finitely many? countably? uncountably?)
(C) what motions are allowed in "reassembling" (translations?
rotations? orientation-reversing maps? isometries?
affine maps? homotheties? arbitrary continuous images? etc.)
(D) how the pieces are permitted to be glued together. The
simplest notion is that they must be disjoint. If the pieces
are polygons [or any piece with a nice boundary] you can permit
them to be glued along their boundaries, ie the interiors of the
pieces disjoint, and their union is the desired figure.
Some dissection results
1) We are permitted to cut into FINITELY MANY polygons, to TRANSLATE
and ROTATE the pieces, and to glue ALONG BOUNDARIES;
then Yes, any two equal-area polygons are equi-decomposable.
This theorem was proven by Bolyai and Gerwien independently, and has
undoubtedly been independently rediscovered many times. I would not
be surprised if the Greeks knew this.
The Hadwiger-Glur theorem implies that any two equal-area polygons are
equi-decomposable using only TRANSLATIONS and ROTATIONS BY 180
DEGREES.
2) THM (Hadwiger-Glur, 1951) Two equal-area polygons P,Q are
equi-decomposable by TRANSLATIONS only, iff we have equality of these
two functions: PHI_P() = PHI_Q()
Here, for each direction v (ie, each vector on the unit circle in the
plane), let PHI_P(v) be the sum of the lengths of the edges of P which
are perpendicular to v, where for such an edge, its length is positive
if v is an outward normal to the edge and is negative if v is an
inward normal to the edge.
3) In dimension 3, the famous "Hilbert's third problem" is:
"If P and Q are two polyhedra of equal volume, are they
equi-decomposable by means of translations and rotations, by
cutting into finitely many sub-polyhedra, and gluing along
boundaries?"
The answer is "NO" and was proven by Dehn in 1900, just a few months
after the problem was posed. (Ueber raumgleiche polyeder, Goettinger
Nachrichten 1900, 345-354). It was the first of Hilbert's problems
to be solved. The proof is nontrivial but does *not* use the axiom
of choice.
"Hilbert's Third Problem", by V.G.Boltianskii, Wiley 1978.
4) Using the axiom of choice on non-countable sets, you can prove
that a solid sphere can be dissected into a finite number of
pieces that can be reassembled to two solid spheres, each of
same volume of the original. No more than nine pieces are needed.
The minimum possible number of pieces is FIVE. (It's quite easy
to show that four will not suffice). There is a particular
dissection in which one of the five pieces is the single center
point of the original sphere, and the other four pieces A, A',
B, B' are such that A is congruent to A' and B is congruent to B'.
[See Wagon's book].
This construction is known as the "Banach-Tarski" paradox or the
"Banach-Tarski-Hausdorff" paradox (Hausdorff did an early version of
it). The "pieces" here are non-measurable sets, and they are
assembled *disjointly* (they are not glued together along a boundary,
unlike the situation in Bolyai's thm.)
An excellent book on Banach-Tarski is:
"The Banach-Tarski Paradox", by Stan Wagon, 1985, Cambridge
University Press.
Robert M. French, The Banach-Tarski theorem, The Mathematical
Intelligencer 10 (1988) 21-28.
The pieces are not (Lebesgue) measurable, since measure is preserved
by rigid motion. Since the pieces are non-measurable, they do not
have reasonable boundaries. For example, it is likely that each piece's
topological-boundary is the entire ball.
The full Banach-Tarski paradox is stronger than just doubling the
ball. It states:
5) Any two bounded subsets (of 3-space) with non-empty interior, are
equi-decomposable by translations and rotations.
This is usually illustrated by observing that a pea can be cut up
into finitely pieces and reassembled into the Earth.
The easiest decomposition "paradox" was observed first by Hausdorff:
6) The unit interval can be cut up into COUNTABLY many pieces which,
by *translation* only, can be reassembled into the interval of
length 2.
This result is, nowadays, trivial, and is the standard example of a
non-measurable set, taught in a beginning graduate class on measure
theory.
References:
In addition to Wagon's book above, Boltyanskii has written at least
two works on this subject. An elementary one is:
"Equivalent and equidecomposable figures"
in Topics in Mathematics published by D.C. HEATH AND CO., Boston. It
is a translation from the 1956 work in Russian.
Also, the article "Scissor Congruence" by Dubins, Hirsch and ?,
which appeared about 20 years ago in the Math Monthly, has a pretty
theorem on decomposition by Jordan arcs.
``Banach and Tarski had hoped that the physical absurdity of this
theorem would encourage mathematicians to discard AC. They were
dismayed when the response of the math community was `Isn't AC great?
How else could we get such counterintuitive results?' ''
Copyright Notice
Copyright (c) 1993 A. Lopez-Ortiz
--------------------------------------------------------------------------
Questions and Answers Edited and Compiled by:
Alex Lopez-Ortiz [email protected]
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada
|
| From: [email protected] (Alex Lopez-Ortiz)
Subject: sci.math: Frequently Asked Questions [3/3]
Organization: University of Waterloo
Date: Fri, 19 Nov 1993 14:03:43 GMT
Expires: Fri, 31 Dec 1993 14:03:28 GMT
Archive-Name: sci-math-faq/part3
Last-modified: October 12, 1993
Version: 5.0
This is a list of Frequently Asked Questions for sci.math (version 5.0).
Any contributions/suggestions/corrections are most welcome. Please use
* e-mail * on any comment concerning the FAQ list.
Section 3 of 3, questions 19Q to 24Q.
Table of Contents
-----------------
1Q.- Fermat's Last Theorem, status of ..
2Q.- Values of Record Numbers
3Q.- Formula for prime numbers...
4Q.- Digits of Pi, computation and references
5Q.- Odd Perfect Number
6Q.- Computer Algebra Systems, application of ..
7Q.- Computer Algebra Systems, references to ..
8Q.- Fields Medal, general info ..
9Q.- Four Colour Theorem, proof of ..
10Q.- 0^0=1. A comprehensive approach
11Q.- 0.999... = 1. Properties of the real numbers ..
12Q.- There are three doors, The Monty Hall problem, Master Mind and
other games ..
13Q.- Surface and Volume of the n-ball
14Q.- f(x)^f(x)=x, name of the function ..
15Q.- Projective plane of order 10 ..
16Q.- How to compute day of week of a given date
17Q.- Axiom of Choice and/or Continuum Hypothesis?
18Q.- Cutting a sphere into pieces of larger volume
19Q.- Pointers to Quaternions
20Q.- Erdos Number
21Q.- Why is there no Nobel in mathematics?
22Q.- General References and textbooks...
23Q.- Interest Rate...
24Q.- Euler's formula e^(i Pi) = - 1 ...
19Q: Is there a theory of quaternionic analytic functions, that is, a four-
dimensional analog to the theory of complex analytic functions?
A. Yes. This was developed in the 1930s by the mathematician Fueter.
It is based on a generalization of the Cauchy-Riemann equations,
since the possible alternatives of power series expansions or
quaternion differentiability do not produce useful theories. A number
of useful integral theorems follow from the theory. Sudbery provides
an excellent review. Deavours covers some of the same material less
thoroughly. Brackx discusses a further generalization to arbitrary
Clifford algebras.
Anthony Sudbery, Quaternionic Analysis, Proc. Camb. Phil. Soc.,
vol. 85, pp 199-225, 1979.
Cipher A. Deavours, The Quaternion Calculus, Am. Math. Monthly,
vol. 80, pp 995-1008, 1973.
F. Brackx and R. Delanghe and F. Sommen, Clifford analysis,
Pitman, 1983.
20Q: What is the Erdos Number?
Form an undirected graph where the vertices are academics, and an
edge connects academic X to academic Y if X has written a paper
with Y. The Erdos number of X is the length of the shortest path
in this graph connecting X with Erdos.
What is the Erdos Number of X ? for a few selected X in {Math,physics}
Erdos has Erdos number 0. Co-authors of Erdos have Erdos number 1.
Einstein has Erdos number 2, since he wrote a paper with Ernst Straus,
and Straus wrote many papers with Erdos.
Why people care about it?
Nobody seems to have a reasonable answer...
Who is Paul Erdos?
Paul Erdos is an Hungarian mathematician, he obtained his PhD
from the University of Manchester and has spent most of his
efforts tackling "small" problems and conjectures related to
graph theory, combinatorics, geometry and number theory.
He is one of the most prolific publishers of papers; and is
also and indefatigable traveller.
References:
Caspar Goffman, And what is your Erdos number?, American Mathematical
Monthly v. 76 (1969), p. 791.
21Q: Why is there no Nobel in mathematics? #
Nobel prizes were created by the will of Alfred Nobel, a notable
swedish chemist.
One of the most common --and unfounded-- reasons as to why Nobel
decided against a Nobel prize in math is that [a woman he proposed
to/his wife/his mistress] [rejected him beacuse of/cheated him
with] a famous mathematician. Gosta Mittag-Leffler is often claimed
to be the guilty party.
There is no historical evidence to support the story.
For one, Mr. Nobel was never married.
There are more credible reasons as to why there is no Nobel prize
in math. Chiefly among them is simply the fact he didn't care much
for mathematics, and that it was not considered a practical
science from which humanity could benefit (a chief purpose
for creating the Nobel Foundation).
Here are some relevant facts:
1. Nobel never married, hence no ``wife''. (He did have a mistress,
a Viennese woman named Sophie Hess.)
2. Gosta Mittag-Leffler was an important mathematician in Sweden
in the late 19th-early 20th century. He was the founder of the
journal Acta Mathematica, played an important role in helping the
career of Sonya Kovalevskaya, and was eventually head of the
Stockholm Hogskola, the precursor to Stockholms Universitet.
However, it seems highly unlikely that he would have been a
leading candidate for an early Nobel Prize in mathematics, had
there been one -- there were guys like Poincare and Hilbert around,
after all.
3. There is no evidence that Mittag-Leffler had much contact with
Alfred Nobel (who resided in Paris during the latter part of his
life), still less that there was animosity between them for whatever
reason. To the contrary, towards the end of Nobel's life
Mittag-Leffler was engaged in ``diplomatic'' negotiations to try to
persuade Nobel to designate a substantial part of his fortune to the
Hogskola. It seems hardly likely that he would have undertaken this
if there was prior bad blood between them. Although initially Nobel
seems to have intended to do this, eventually he came up with the
Nobel Prize idea -- much to the disappointment of the Hogskola,
not to mention Nobel's relatives and Fraulein Hess.
According to the very interesting study by Elisabeth Crawford,
``The Beginnings of the Nobel Institution'', Cambridge Univ. Press,
1984, pages 52-53:
``Although it is not known how those in responsible positions
at the Hogskola came to believe that a *large* bequest was forthcoming,
this indeed was the expectation, and the disappointment was keen when
it was announced early in 1897 that the Hogskola had been left out of
Nobel's final will in 1895. Recriminations followed, with both
Pettersson and Arrhenius [academic rivals of Mittag-Leffler in the
administration of the Hogskola] letting it be known that Nobel's
dislike for Mittag-Leffler had brought about what Pettersson termed the
`Nobel Flop'. This is only of interest because it may have contributed
to the myth that Nobel had planned to institute a prize in mathematics
but had refrained because of his antipathy to Mittag-Leffler or
--in another version of the same story-- because of their rivalry for
the affections of a woman....''
4. A final speculation concerning the psychological element.
Would Nobel, sitting down to draw up his testament, presumably
in a mood of great benevolence to mankind, have allowed a mere
personal grudge to distort his idealistic plans for the monument
he would leave behind?
Nobel, an inventor and industrialist, did not create a prize in
mathematics simply because he was not particularly interested
in mathematics or theoretical science. His will speaks of
prizes for those ``inventions or discoveries'' of greatest
practical benefit to mankind. (Probably as a result of this
language, the physics prize has been awarded for experimental work
much more often than for advances in theory.)
However, the story of some rivalry over a woman is obviously
much more amusing, and that's why it will probably continue to
be repeated.
References:
Mathematical Intelligencer, vol. 7 (3), 1985, p. 74.
Elisabeth Crawford, ``The Beginnings of the Nobel Institution'',
Cambridge Univ. Press, 1984.
22Q: General References and textbooks...
Full references and/or title suggestions will be appreciated:
Algebra:
Lang, Serge. Algebra
Birkhoff, McLane, Algebra
Analysis:
Rudin
Hewitt, Stromberg
23Q: Here's a formula which can be used in 123, Excel, Wings and
Dynaplan:
------- Input this data -------------------------------
principal amount = E9 ( in dollars )
Amortization Period = d10 ( in years ie 6 mon = .5 )
Payments / year = D11 ( 12 = monthly, 52 = weekly )
Published Interest rate = D12 ( ie 9 % = 0.09 )
Times per year Int calculated = d13 ( CDN mortgage use 2
US mortgage use 12
all other loans use 12 )
----- Calculate the proper rate of interest -----------
e14 = Effective annual rate = EXP(D13*LN(1+(D12/D13)))-1
e15 = Interest rate per payment = (EXP(LN(E14+1)/(D10*D11))-1)*D10*D11
e17 = Payments = APMT(E9,E15/D11,D10*D11) ( both these functions are
= PMT (E9,E15/D11,D10*D11) ( indentical,diff spreadsheet)
APMT( principal amount,interest rate per period,# periods )
( this is a standard function on any true commercial spreadsheet)
OR use the following if done using a calulator
= Payments = P*I/[1-(I+1)^-T]
= E9*(E15/D11)/(1-((E15/D11) +1)**(-1*D10*D11))
Total interest cost = E17*D10*D11-E9
-- Use these formulas if you wish to generate an amortization table --
always add up to 'Payments (e17)'
Interest per payment = current balance * ( E15 / D11 )
Principal per payment = current balance - Interest per payment
new current balance = current balance - Principal per payment -
(extra payment)
keep repeating until 'new current balance' = 0
Derivation of Compound Interest Rate Formula
Suppose you deposited a fixed payment into an interest bearing
account at regular intervals, say monthly, at the end of each month.
How much money would there be in the account at the end of the nth
month (at which point you've made n payments)?
Let i be the monthly interest rate as a fraction of principle.
Let x be the amount deposited each month.
Let n be the total number of months.
Let p[k] be the principle after k months.
So the recursive formula is:
p[n] = x + ( (1 + i) * p[n-1] ) eq 1
This yields the summation:
n-1
---
\
p[n] = / x * (1 + i)^k eq 2
---
k=0
The way to solve this is to multiply through by (1 + i) and
subtract the original equation from the resulting equation.
Observe that all terms in the summation cancel except the
last term of the multiplied equation and the first term of
the original equation:
i * p[n] = x * ( (1 + i)^n - 1) eq 3
or
p[n] = x * ( (1 + i)^n - 1) / i eq 4
Now suppose you borrow p at constant interest rate i.
You make monthly payments of x. It turns out that this problem
is identical to taking out a balloon loan of p (that is it's
all due at the end of some term) and putting payments of x
into a savings account. At the end of the term you use the
principle in the savings account to pay off the balance of
the loan. The loan and the savings account, of course, must
be at the same interest rate. So what we want to know is:
what monthly payment is needed so that the balance of the
savings account will be identical to the balance of the balloon
loan after n payments?
The formula for the principal of the balloon loan at the
end of the nth month is:
p[n] = p[0] * (1 + i)^n eq 5
So we set this expression equal to the expression for the
the savings account (eq 4), and we get:
p[0] * (1 + i)^n = x * ( (1 + i)^n - 1) / i eq 6
or solving for x:
x = p[0] * (1 + i)^n * i / ( (1 + i)^n - 1) eq 7
If (1 + i)^n is large enough (say greater than 5), here
is an approximation for determining n from x, p, and i:
n ~= -ln( ln(x/(i*p) ) ) / ln(1+i) eq 8
The above approximation is based upon the following approximation:
ln(y - 1) ~= ln y - 1/y
eq 9
Which is within 2% for y >= 5.
For example, a $100000 loan at 1% monthly, paying $1028.61
per month should be paid in 360 months. The approximation
yields 358.9 payments.
If this were your 30 year mortgage and you were paying $1028.61
per month and you wanted to see the effect of paying $1050
per month, the approximation tells you that it would be paid
off in 303.5 months (25 years and 3.5 months). If you stick
304 months into the equation for x (eq 7), you get $1051.04, so
it is fairly close. This approximation does not work, though,
for very small interest rates or for a small number of payments.
The rule is to get a rough idea first of what (1 + i)^n is.
If that is greater than 5, the approximation works pretty
well. In the examples given, (1 + i)^n is about 36.
Finding i given n, x, and p is not as easy. If i is less than
5% per payment period, the following equation approximately
holds for i:
i = -(1/n) * ln(1 - i*p/x) eq 10
There is no direct solution to this, but you can do it by
Newton-Raphson approximation. Begin with a guess, i[0].
Then apply:
x*(1 - i[k]*p/x) * (n*i[k] + ln(1 - i[k]*p/x))
i[k+1] = i[k] - ----------------------------------------------
x*n*(1 - i[k]*p/x) - p
eq 11
You must start with i too big, because eq 10 has a solution
at i=0, and that's not the one you want to end up with.
Example: Let the loan be for p=$10000, x=$50/week for
5 years (n=260). Let i[0] = 20% per annum or 0.3846%
per week. Since i must be a fraction rather than a percent,
i[0] = 0.003846. Then, applying eq 11:
i[1] = 0.003077
i[2] = 0.002479
i[3] = 0.002185
i[4] = 0.002118
i[5] = 0.002115
The series is clearly beginning to converge here.
To get i[5] as an annual percentage rate, multiply by 52 weeks
in a year and then by 100%, so i[5] = 10.997% per annum.
Substituting i[5] back into eq 7, we get x = $50.04, so it
works pretty well.
The theory of interest, by Stephen G. Kellison. Homewood, Ill., R. D.
Irwin, 197o-.
24Q.- Euler's formula e^(i Pi) = - 1 ...
e^(ip) = -1
where i = sqrt(-1), p = pi ...
Copyright Notice
Copyright (c) 1993 A. Lopez-Ortiz
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Questions and Answers Edited and Compiled by:
Alex Lopez-Ortiz [email protected]
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada
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