Usenet 1994 January/usenetsourcesnewsgroupsinfomagicjanuary1994.iso/answers/sci-math-faq/part3

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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


--------------------------------------------------------------------------
Questions and Answers Edited and Compiled by:

Alex Lopez-Ortiz                              [e-mail]
Department of Computer Science                      University of Waterloo
Waterloo, Ontario                                                   Canada
--
Alex Lopez-Ortiz                             [e-mail]
Department of Computer Science                      University of Waterloo
Waterloo, Ontario                                                   Canada