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Archive-Name: sci-math-faq Version: $Id: sci-math-faq,v 4.5 93/07/19 15:55:00 $ C A L L F O R C O N T R I B U T I O N S ***This FAQ list is about to undergo a major overhaul.*** It will be split in several parts, with a short introduction being posted weekly and the full version being posted monthly (instead of posting 70K biweekly). If there's a topic you think should be included send e-mail to alopez-o@maytag.uwaterloo.ca Topics to be included: Questions 22, 24 and 25. This is a list of Frequently Asked Questions for sci.math (version 4.5). Any contributions/suggestions/corrections are most welcome. Please use * e-mail * on any comment concerning the FAQ list. Changes and additions are marked with a # on the table of contents. This FAQ list (and most others, for that matter) is available via anonymous ftp at rtfm.mit.edu (18.70.0.224). The list of contributors to this FAQ list is too large to include here; but thanks are due to all of them (you know who you are folks!). Table of Contents ----------------- 1Q.- Fermat's Last Theorem, status of .. # 2Q.- Four Colour Theorem, proof of .. 3Q.- Values of Record Numbers 4Q.- General Netiquette 5Q.- Computer Algebra Systems, application of .. 6Q.- Computer Algebra Systems, references to .. 7Q.- Fields Medal, general info .. 8Q.- 0^0=1. A comprehensive approach 9Q.- 0.999... = 1. Properties of the real numbers .. 10Q.- Digits of Pi, computation and references 11Q.- There are three doors, The Monty Hall problem, Master Mind and other games .. 12Q.- Surface and Volume of the n-ball 13Q.- f(x)^f(x)=x, name of the function .. 14Q.- Projective plane of order 10 .. 15Q.- How to compute day of week of a given date 16Q.- Axiom of Choice and/or Continuum Hypothesis? 17Q.- Cutting a sphere into pieces of larger volume 18Q.- Pointers to Quaternions 19Q.- Erdos Number 20Q.- Odd Perfect Number 21Q.- Why is there no Nobel in mathematics? 22Q.- General References and textbooks... 23Q.- Formula for prime numbers... # 24Q.- Interest Rate... 25Q.- Euler's formula e^(i Pi) = - 1 ... 1Q: What is the current status of 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. ... (wlog we assume x,y,z to be relatively prime) 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 Andrew Wiles, a researcher at Princeton 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 parametrizes 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 sqaure 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: Has the Four Colour Theorem been solved? (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). 3Q: 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 4650828 * 1001 * 10^3429 +/- 1. They were found by H. Dubner For an article by the previous record holders see: 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 bf04@UTMartn.bitnet (preferred) or kvax@utkvx.UTK.edu, 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. (Please send updates to alopez-o@maytag.UWaterloo.ca) 4Q: I think I proved <insert big conjecture>. OR I think I have a bright new idea. What should I do? A: Are you an expert in the area? If not, please ask first local gurus for pointers to related work (the "distribution" field may serve well for this purposes). If after reading them you still think your *proof is correct*/*idea is new* then send it to the net. 5Q: 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. 6Q: 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, 160 Columbia Street West, Waterloo, Ontario, Canada N2L 3L3 Phone: (519) 747-2373 wmsi@daisy.uwaterloo.ca wmsi@daisy.waterloo.edu 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 pari@ceremab.u-bordeaux.fr and for the Macintosh versions to bernardi@mathp7.jussieu.fr 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 info@wri.com 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: info-macsyma@macsyma.com 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 info@mathworks.com 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 cayley@maths.su.oz.au 7Q: Let P be a property about the Fields Medal. Is P(x) true? A: There are a few gaps in the list. If you know any of the missing information (or if you notice any mistakes), please send me e-mail. 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. 8Q: 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. 9Q: 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. 10Q: 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 20,000-500,000 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. References : (This is a short version for a more comprehensive list contact Juhana Kouhia at jk87377@cc.tut.fi) 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 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 11Q: 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, where the problem is solved and carefully explained. (The Monty Hall problem). MANY OTHER "MATHEMATICAL" GAMES ARE EXPLAINED IN THE REC.PUZZLES FAQ. 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 frequently asked questions list for rec.puzzles, send to the address archive-request@questrel.com an email message consisting of the text send switch 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. 12Q: 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. 13Q: 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. 14Q: The existence of a projective plane of order 10 has long been an outstanding problem in discrete mathematics and finite geometry. A: More precisely, the question is: 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. 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. See Am. Math. Monthly, recent issue. 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. 15Q: Is there a formula to determine the day of the week, given the month, day and year? A: 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, 1990. K. Rosen, "Elementary Number Theory", p. 156. 16Q: 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. 17Q: 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. Also read in the Mathematical Intelligencer an article on the Banach-Tarski Paradox. 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?' '' 18Q: 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. 19Q: 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. 20Q: 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. 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... # [a list of general references and most commonly used textbooks] [ ] 23Q.- 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? If so, then yes, 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). 24Q.- Interest Rate... >Below is the standard equation for calculating monthly payments (m) >given the periodic interest rate (i), the principal (p), and the number >of payments (n). I need to rearrange the equation to solve for the >periodic interest rate, given the other variables, but my math is sadly >no longer up to it. > >If someone would be so kind as to solve for (i), or look up the >equation, or whatever, and respond by email, I would be eternally >grateful (well, a long time, anyway). I ask for email since this site >does not receive sci.math. Thanks in advance. > > i >m = p * ------------ > -n > 1 - (i + 1) > > 25Q.- Euler's formula e^(i Pi) = - 1 ... -1 = e^(ip) where i = sqrt(-1), p = pi ... -------------------------------------------------------------------------- Questions and Answers _Compiled_ by: Alex Lopez-Ortiz alopez-o@maytag.UWaterloo.ca Department of Computer Science University of Waterloo Waterloo, Ontario Canada -- Alex Lopez-Ortiz alopez-o@maytag.UWaterloo.ca Department of Computer Science University of Waterloo Waterloo, Ontario Canada