Danny Amor's Online Library

home *** CD-ROM | disk | FTP | other *** search

/ Danny Amor's Online Library / Danny Amor's Online Library - Volume 1.iso / html / faqs / faq / sci-math-faq / part3 < prev

Wrap

Text File | 1995-07-25 | 16.1 KB | 436 lines

Subject: sci.math: Frequently Asked Questions [3/3] Newsgroups: sci.math,sci.answers,news.answers From: alopez-o@maytag.uwaterloo.ca (Alex Lopez-Ortiz) Date: Wed, 5 Oct 1994 15:36:48 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. The Extended Erdos Number applies to co-authors of Erdos. For People who have authored more than one paper with Erdos, their Erdos number is defined to be 1/#papers-co-authored. Ron Graham has the smallest, non-zero, Erdos number. 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 This FAQ is Copyright (C) 1994 by Alex Lopez-Ortiz. This text, in whole or in part, may not be sold in any medium, including, but not limited to electronic, CD-ROM, or published in print, without the explicit, written permission of Alex Lopez-Ortiz. -------------------------------------------------------------------------- Questions and Answers Edited and 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@neumann.UWaterloo.ca Department of Computer Science University of Waterloo Waterloo, Ontario Canada http://daisy.uwaterloo.ca/~alopez-o/home.html