From hip-hop!benjie@amdahl.com Thu Feb 17 13:07:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: Move completed
Attention all members of the Internet Amateur Mathematics Society:
The Internet Amateur Mathematics Society has officially moved to
iams@hh.sbay.org. The mailing list will be moderated and linked to
alt.math.iams. I will be the moderator.
Since I read my emails once everyday, most likely you will receive
IAMS messages once a day.
I have one request, to everybody: If possible, please unsubscribe from
^^^^^^^^^^^
the iams mailing-list (by emailing to listserv@hh.sbay.org with
'DELETE emailaddress iams' in the body of the message) and use the
alt.math.iams. This way the performance of the machine which IAMS is
^^^^^^^^^^^^^^
currently on will be improved.
Thanks for putting up with all the testing and things, but this is it.
Happy mathing from your moderator,
Benjie
From hip-hop!benjie@amdahl.com Fri Feb 18 07:49:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: POD: 2/18/94 Rectangles
Last September I received a dozen math problems from tulled@rpi.edu
(Thanks). Due to the lack of time I have not done most of them. From
now on I will post some of those questions as POD and hope everybody
will enjoy them.
POD 2/18/94
Does there exist a real number L such that, if m and n are integers
greater than L, then an m x n rectangle may be expressed as a union of
4 x 6 and 5 x 7 rectangles, any two of which intersect at most along
their boundaries?
Enjoy,
Benjie
From hip-hop!delphi.com!SOURCERER@amdahl.com Sat Feb 19 05:05:00 1994
From: SOURCERER@delphi.com
Subject: Problem Sourcerer 1.218
(Problem 1 for February 18, 1994 from Sourcerer to the Internet
Amateur Mathematics Society.)
Consider the two mutually tangent parabolas y = x^2 and y = -x^2.
[These have foci at (0, 1/4) and (0, -1/4), and directrices y = -1/4,
respectively.] The upper parabola rolls without slipping around the
fixed lower parabola. Find the locus of the focus of the moving
parabola.
--------------
Solution to problem 2.213:
"Consider an integer p > 1 with the property that the polynomial
x^2 - x + p takes prime values for all integers x in the range
0 L.E. x < p. (Examples: p = 5 and p = 41 have the property.)
Show that there is exactly one triple of integers a, b, c satisfying
the conditions:
b^2 - 4 * a * c = 1 - 4 * p,
0 < a L.E. c,
-a L.E. b < a."
One triple (a, b, c) satisfying the conditions is (1, -1, p); it
remains to show that this is the only solution. Clearly b must be odd
since b^2 is congruent with 1 (mod 4). Also, b^2 = (-b)^2, so write
|b| = 2 * x - 1. Then b^2 - 4 * c = 1 - 4 * p gives
x^2 - x + p = a * c.
If 0 L.E. x < p, the hypothesis tells us that a * c is prime; then
0 < a L.E. c implies that a = 1, it follows from -a L.E. b < a and
the oddness of b that b = -1, and
1 - 4 * p = b^2 - 4 * a * c = 1 - 4 * c gives us c = p. Since
x = (|b| + 1)/2 G.E. 0, it suffices to show that x < p. Since
|b| L.E. a L.E. c, b^2 - 4 * a * c = 1 - 4 * p, and p G.E. 2, one
sees that x < p using
3 * a^2 = 4 * a^2 - a^2 L.E. 4 * a * c - b^2 = 4 * p - 1,
|b| L.E. a L.E. SQRT[(4 * p - 1)/3],
x = (|b| + 1)/2 < SQRT(p/3) + (1/2) < p.
("L.E." means "less than or equal to," "G.E." means :greater than or
equal to.")
====
Note: as in this case, solutions to these problems appear attached
to new problems a few days after posting.
Bruce M. Bowden
Sourcerer@delphi.com
From hip-hop!alpha2.csd.uwm.edu!radcliff@amdahl.com Sun Feb 20 02:56:00 1994
From: David G Radcliffe
Subject: Construction of 17-gon with ruler and compass
CONSTRUCTION OF THE REGULAR 17-GON BY RULER AND COMPASS
The following is based on a talk by William Dunham, the author of
"Journey through Genius." This talk was in turn based on the proof
by Karl Gauss (at the age of 18!) that the regular 17-gon is
constructible by ruler and compass.
The proof requires only high school mathematics, but the calculations
are messy. I would be interested in seeing a more conceptual proof.
__________
Suppose we have a plane with some fixed unit of length. We wish
to describe the set of possible lengths which can be constructed
using ruler and compass.
THEOREM: If x,y are constructible, then the following lengths are
also constructible: x + y, x - y, xy, x / y, and sqrt(x).
The proof is included at the end of this note.
COROLLARY: Let x and y be real numbers. If x + y and xy
are constructible lengths, then so are x and y.
PROOF: Assume that x >= y. Let a = x + y and b = xy. Then
x = (a + sqrt(a^2-4b))/2 and y = (a - sqrt(a^2-4b))/2 by the
quadratic formula. The claim now follows from the previous result.
__________
In order to construct a regular 17-gon, it is sufficient to
construct a segment of length cos(2*pi/17). I will sketch
the proof that cos(2*pi/17) is constructible.
Let x = (cos(2*pi/17)) + i (sin(2*pi/17)). Then x^17 - 1 = 0 by
De Moivre's theorem. Divide both sides by x - 1, giving
x^16 + x^15+ ... x^2 + x + 1 = 0. Rewrite this equation as
(*) x + x^2 + x^3 + ... + x^15 = -1.
Let a = x + x^2 + x^4 + x^8 + x^9 + x^13 + x^15 + x^16
and b = x^3 + x^5 + x^6 + x^7 + x^10 + x^11 + x^12 + x^14.
Using (*), you can show that a+b = -1 and ab = -4. This
calculation is routine, but it is much too tedious to include
here! Now a and b are both real, because x^(17-j) is the
conjugate of x^j, so the terms can be grouped into conjugate
pairs. Therefore, a and b are constructible (by the corollary).
Let c = x + x^4 + x^13 + x^16, and d = x^2 + x^8 + x^9 + x^15.
You can check that c + d = a, and cd = -1. But c and d are
real; therefore, c and d are constructible.
Let e = x^3 + x^5 + x^12 + x^14, and f = x^6 + x^7 + x^10 + x^11.
Now e and f are real, e + f = b, and ef = -1; so e and f are
constructible.
Let g = x + x^16 and h = x^4 + x^13. Then g and h are real,
g + h = c, and gh = e. Therefore, g and h are constructible.
But g = 2*cos(2*pi/17), so we are done!
____________
APPENDIX
If x and y are constructible lengths, then the following lengths
are also constructible:
(a) x + y
(b) x - y
(c) xy
(d) x / y
(e) sqrt(x)
(a) and (b) are obvious.
Proof of (c): Draw perpendicular lines l and m which intersect at
a point O. Locate points P,Q on l,m respectively so that OP = x
and OQ = 1. Find a point R on m so that OR = y. Draw a line parallel
to PQ passing through R, and let S be the intersection of this line
with l. Then OS = xy (by similar triangles).
Proof of (d): By (c) it is enough to show that 1/x is constructible.
Use the same diagram as above, and draw a line n perpendicular to PQ
through Q. Let T be the intersection of n and l. Then OT = 1/x.
Proof of (e): Draw a semicircle with diameter ABC, where AB = 1 and
BC = x. Draw a line perpendicular to AC through B, and let D be the
point where this line intersects the semicircle. Then BD = sqrt(x).
(ADC is a right triangle, so ABD and DBC are similar triangles)
--
David Radcliffe
radcliff@alpha2.csd.uwm.edu
From hip-hop!benjie@amdahl.com Mon Feb 21 06:35:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: no subject (file transmission)
First let me remind everybody that if you could receive the newsgroup
alt.math.iams, please use the newsgroup and unsubscribe from the
mailing-list. I am trying to improve the performance of the machine
which IAMS is currently on.
Now, the Problem of the Day: (from tulled@rpi.edu)
POD: For which positive integers n is n^4+4^n prime?
Solution to the problem of the day 2/19, the rectangle problem follows.
SPOILER Alert
Problem:
Does there exist a real number L such that, if m and n are integers
greater than L, then an m x n rectangle may be expressed as a union of
4 x 6 and 5 x 7 rectangles, any two of which intersect at most along
their boundaries?
Solution:
From: "David W. Wilson"
A 12x12 square is tiled by 6 4x6 rectangles.
A 12x35 rectangle is tiled by 12 5x7 rectangles.
A 35x12 rectangle is tiled by 12 5x7 rectangles.
A 35x35 square is tiled by 35 5x7 rectangles.
Any rectangle with both sides of the form 12x+35y (x, y >= 0)
can clearly be tiled by 12x12, 12x35, 35x12, and 35x35
rectangles, and thus by 4x6 and 5x7 rectangles. The smallest
number not expressible in the form 12x+35y is 373 (** see
comment). Hence any rectangle with integer sides longer than
373 can be tiled by 4x6 and 5x7 rectangles. This answers the
question in the affirmative, with L = 373.
What is the smallest possible value LMIN of L? The above
clearly puts LMIN <= 373. To establish a lower bound on LMIN,
consider that any triangle tiled by 4x6 and 5x7 rectangles
must have an area of the from 24x+35y (x, y >= 0). In
particular, 676 is not of this form, and so a 26x26 square is
not tilable with 4x6 and 3x5 rectangles. We must therefore
have LMIN >= 26.
Moderator's Comment on the Problem:
This problem reminded me of the Chicken McNugget problem. Although
similar problems showed on various contests, the Chicken McNugget
problem was summarized in a math paper written by a high school
student Celia Liao. The paper was published in 1992 volume of
"mathematical buds", a math journal for the Mu Alpha Theta
organization (nationwide math club for high school students).
In the paper the author summarized what she called the Chicken
McNuggets problem:
Chicken McNuggets are sold in packages of 6, 9 and 20. Call
an integer buyable if you can purchase that exact number of
McNuggets. For example: 15 is, but 16 is not buyable. Find
the largest integer which is not buyable.
The author defined M(a_1, a_2, a_3 ..., a_n) to be the largest
non-buyable number. The author then derived the some theorems, and I
have included some of them here:
Theorem 1: M(1,b) = 0
Theorem 3: M(a,b) does not exist if a and b are not relative prime
Theorem 7: If (a,b)=1, then M(a,b) = (a-1)b-a = ab-b-a = ab-(a+b)
((**) From this theorem M(12,35)=373. However
it should be the largest number, not the smallest number, not
expressible in the form 12x+35y.
Theorem 8: M(a,b,c) does not exist if a,b,c have a common factor
greater than 1.
The author also made a conjecture that there is no closed formula for
M(a,b,c,d,...w) when the number of elements in {a,b,c,...,w} is
greater than 2.
The author proved most of the theorems by manipulating equations and
using modular arithmetics. It is also possible to prove the theorems
using linear algebra, I think. I will have to work on that a little
longer, it might be a good project. I believe if we could work out
the problem by linear algebra, the conjecture could be proved/
disproved.
If any of you knows more about the topic, please email me your
comments.
Benjie
From hip-hop!benjie@amdahl.com Tue Feb 22 02:38:00 1994
From: Sylvan Jacques
Subject: DISCUSSION: sum/product Problem
DISCUSSION PART1 - Tue Feb 22 04:54:04 1994
I have been having a email discussion with David Snook on the 1st step
of the sum/product problem--removing sums S which can be formed by
adding 2 primes (and all pairs (a,b) such that a+b = S, whether or not
a and b are prime). I argue, like Kermit Rose, that since S can say
that P does not know the solution, S must have a sum which contains no
possible pair (a,b) which are both primes.
David argues that this is a fallacy. I know several people on IAMS
were interested in this problem, and since this discussion is going
nowhere, I would like to get opinions from others.
Here is most of the initial post from David:
> From: ua532@freenet.victoria.bc.ca (David Snook)
> Subject: Product/Sum 1.212 ..... Fallacies
> (Stuff omitted).
> When any quantity of prime numbers are multiplied together they produce a
> product that can ALWAYS be factored to reveal the original prime numbers.
> Multiplication and factoring are inverse functions. This is because of the
> Fundamental Theorem of Arithmetic, which states that; ALL integers are either
> prime or a product of primes.
>
> There is NO functional inverse of the addition operation that will permit a
> sum to be decomposed into it's original components and to determine with
> certainty, that the resulting output matches exactly the input. (Exceptions
> are sum = 2 or 3) This is because a sum can be constructed in many ways.
>
> All of the rec.puzzle solutions that I have seen so far, have assumed that a
> SUM of primes is the same as a PRODUCT of primes and this, of course, is
> specious logic.
>
> (More omissions).
>
> By way of example, let us examine the Sv=18 within the the context of
> the puzzle contraints.
>
> Sv = 18 (a= 8 b= 10) Pv= 80
> (a= 7 b= 11) Pv= 77 [a=prime b=prime]
> (a= 6 b= 12) Pv= 72
> (a= 5 b= 13) Pv= 65 [a=prime b=prime]
> (a= 4 b= 14) Pv= 56
> (a= 3 b= 15) Pv= 45
> (a= 2 b= 16) Pv= 32
>
> Note that the sum=18 has seven(7) valid (a,b) pairs associated with it.
> Further, two(2) of these are of the form (a=prime, b=prime). It follows then,
> that the sum=18 is not uniquely associated to the sum of a (a=prime b=prime)
> pair. This is true for all integers except 2 & 3.
>
> The puzzle logic allows for the removal of the two(2) prime pairs based on
> the identifiablity of the products ... but only these.
>
> When the sum=18 is is removed on the basis of Sum=18=7+11 it, de facto,
> assumes that the remaining six(6) (a,b) pairs do not exist, and are therefore
> never accounted for. This will occur both in program algorithms, Math
> software, or formulae which directly or indirectly utilize the sum of primes
> parameter.
>
> Further, when the any sum is globally removed on the above basis, a
> cascading effect takes place. Note that each of the (a,b) pairs is associated
> with a product value (Pv). These are, in turn, associated to other sums ---->
> products ----> sums ---> products ----> ad nauseum! '-D This quickly
> detriorates into a classic "reductio ad absurdum".
>
> Now, the given solution may still be correct, of course, but it is very
> "INCORRECT", if it can only be derived by a method that involves the above
> "SUM of PRIMES" logic. 8-D
>
=============
Here is the last post I got from David:
>Here is some logic that deals with the given solution provided by
>the rec.puzzles FAQ. {P=52 (a=4 b=13) S=17}, based on auditable
>data in the accompanying tables. I think it clearly shows that
>it cannot be the correct answer.
>
>Remember, Mr P. starts with the product = 52. He factors it, and obtains
>exactly two(2) possibilities. Sum = 17 and Sum = 28. Now the sum = 28 can be
>formed by the sum of two(2) primes. (a=5 b=23). So by the sum of primes
>reasoning he can eliminate it. That leaves him with only one choice. Sum = 17
>a=4 b=13 ..... Before the conversation starts! 8-D
>
>_______________________________________________________
>Mr P has Pv = 52 ..... which he factors to obtain.
>----------------------------------
> 52 4 13 17 204 *
> 52 2 26 28 24 !!! cant be! (Sum of primes)(5,23)
>----------------------------------
>Mr P notes Sv = 17 and Sv = 28, which he "explodes" to obtain.
>----------------------------------
> 30 2 15 17 13
> 42 3 14 17 109
> 52 4 13 17 204 *
> 60 5 12 17 298
> 66 6 11 17 391
> 70 7 10 17 483
> 72 8 9 17 574
>----------------------------------
> 52 2 26 28 24 Note: Sum of primes logic says
> 75 3 25 28 120 that sum=28 is not possible.
> 96 4 24 28 215 Leaving only sum=17. Mr P
115 5 23 28 (added)
> 132 6 22 28 402 would then "know" (a,b) which
> 147 7 21 28 494 of course, is wrong!
> 160 8 20 28 585
> 171 9 19 28 675
> 180 10 18 28 764
> 192 12 16 28 939
> 195 13 15 28 1,025
>----------------------------------
>David J. Snook.................................ua532@freenet.victoria.bc.ca
DISCUSSION PART2 - Tue Feb 22 09:11:44 1994
I have a comment on one of David's arguments re the sum/product
problem. David says
>Here is some logic that deals with the given solution provided by
>the rec.puzzles FAQ. {P=52 (a=4 b=13) S=17}, based on auditable
>data in the accompanying tables. I think it clearly shows that
>it cannot be the correct answer.
>Remember, Mr P. starts with the product = 52. He factors it, and obtains
>exactly two(2) possibilities. Sum = 17 and Sum = 28. Now the sum = 28 can be
>formed by the sum of two(2) primes. (a=5 b=23). So by the sum of primes
>reasoning he can eliminate it. That leaves him with only one choice. Sum = 17
>a=4 b=13 ..... Before the conversation starts! 8-D
====
A table follows, which I previously posted. I don't understand the last
number in the table, or what it shows. Its beside the point here, IMO.
======
I say this is wrong. If P=52, the Mr. P know that (a,b) = (4,13) or
(2,26) as David says. He doesn't know which until Mr. S says he knew
that Mr. P couldn't know (a,b). From this statement of Mr. S, Mr. P,
like us, knows that the sum S cannot be a sum which can be formed from
2 primes, and since S = 28 = 5+23, he can eliminate S = 28 and thus
(a,b) = (2,26), so he knows (a,b) = (4,13) _AFTER_ S makes his
statement, just as he reponds to S in the problem.
Recall problem statements
<<2>>: Mr. S says to P: "I knew you didn't know (a,b)."
<<3>>: Mr. P says to S: "Now I do."
I would say that the above argument of David verifies, rather that
discredits the solution to the puzzle.
Van van@quack.kfu.com
From hip-hop!benjie@amdahl.com Mon Feb 21 04:28:00 1994
From: "David W. Wilson"
Subject: More puzzles (with answers)
All of the following assume base 10. Extra credit for proofs.
1. A palindrome is a number that is the same written in reverse
(e.g., 19891 is a palindrome). What is the smallest positive
palindrome divisible by 81?
2. Let a number be a twin if it is the same number written twice
in a row (e.g., 147147 and 58995899 are twins, whereas 08310831 is
not as we do not allow leading zeroes). What is the smallest
square twin?
3. A square number has leading digit 7. If its leading 7 is
replaced by 8, it remains a square. What is the smallest number it
could possibly be?
SPOILER ALERT: Answers follow
Answers:
1. 999999999
2. 183673469387755102041183673469387755102041
3. 7977708170172487211329625006796419620513916015625
From hip-hop!benjie@amdahl.com Mon Feb 21 04:36:00 1994
From: efedula@aol.com
Subject: Re: Construction of 17-gon with ruler and compass
--- In response to ---
CONSTRUCTION OF THE REGULAR 17-GON BY RULER AND COMPASS
The following is based on a talk by William Dunham, the author of
"Journey through Genius." This talk was in turn based on the proof
by Karl Gauss (at the age of 18!) that the regular 17-gon is
constructible by ruler and compass.
The proof requires only high school mathematics, but the calculations
are messy. I would be interested in seeing a more conceptual proof.
---
Well, this isn't much of a proof, but the construction goes something
like this (from a talk on the mathematical and social history of the
number 17 by Kelly from Hampshire College on July 17, 1993 at MIT):
Start with a circle centered at O with a diameter AB and radius OC
perpendicular to AB.
Construct the points D (on OB), Q (on OC) s.t. OD=(1/8)OB and CQ=OQ.
Construct E (on OA), F (on OB) s.t. QD=ED=DF.
Construct H (on OA), G (on the extension of OA past A) s.t. FH=FQ and
EG=QE.
Construct K (on OC) s.t. OK is the mean proportion (geometric mean)
between OH and OQ.
Construct line y parallel to AB at K.
The semicircle with diameter OG cuts y at M.
Construct N (on circle O) s.t. NM is perpendicular to y at M. It is
intuitively obvious that AN is 1/17 of the circumference of the
circle. :)
In case anyone's interested, cos(2*pi/17) = -1/16 + (1/16)SQRT(17) +
(1/16)SQRT(34-3*SQRT(17)) +
(1/8)SQRT(17+3*SQRT(17)-SQRT(34-2*SQRT(17))-2*SQRT(34+2*SQRT(17)))
-Gauss, March 30,1796
From hip-hop!benjie@amdahl.com Wed Feb 23 05:00:00 1994
From: Kermit Rose
Subject: DISCUSSION: followup to sum/product discussion
> From: ua532@freenet.victoria.bc.ca (David Snook)
> Subject: Product/Sum 1.212 ..... Fallacies
>
>Remember, Mr P. starts with the product = 52. He factors it, and obtains
>exactly two(2) possibilities. Sum = 17 and Sum = 28. Now the sum = 28 can be
>formed by the sum of two(2) primes. (a=5 b=23). So by the sum of primes
>reasoning he can eliminate it. That leaves him with only one choice. Sum = 17
>a=4 b=13 ..... Before the conversation starts! 8-D
The following should clarify what is happening.
Step 0: P knows that ab=52.
S knows that a+b = 17.
Step 1: P says to S, I do not know a and b.
Step 2: S says to P, I already knew that you did not know.
S can say this because a+b = 17
implies that
ab = (2)(15)
or ab = (3)(14)
or ab = (4)(13)
or ab = (5)(12)
or ab = (6)(11)
or ab = (7)(10)
or ab = (8)(9)
Thus S knows already that ab is not the product of two primes. The reason
that ab is not the product of two primes is that S is not the sum of two
primes.
Step 3: P says "Now I know a and b".
P knows that ab=52
that is ab = (2)(26) or ab=(4)(13).
If ab = (2)(26) and a+b=2+26, then S would not have been able to make his
claim that S knew that ab was not the product of two primes.
That is: P would reason: Suppose a+b = 28.
Then S would reason:
5+23 = 28, so ab might = 5*23 and in that case,P would know
immediately what a+b were.
P continues his reasoning: Since S was certain that I did not
know a and b, he could not have a number like 28 which might be the sum of
two primes.
Therefore the only possibility remaining is ab=(4)(13) and a+b=4+13.
Step 4: S says "Now I know a and b also".
S knew that a+b = 17
and that therefore
ab = (2)(15) = (3)(10) = (5)(6)
or ab = (3)(14) = (2)(21) = (7)(6)
or ab = (4)(13) = (2)(26)
or ab = (5)(12) = (2)(30) = (4)(15) =(6)(10) = (3)(20)
or ab = (6)(11) = (2)(33) = (3)(22)
or ab = (7)(10) = (2)(35) = (5)(14)
or ab = (8)(9) = (2)(36) = (3)(24) = (4)(18) = (6)(12)
S noted that P was able to figure out what a and b were. S knows that P
knows that a+b is not the sum of two primes. For S told P that he knew
that P does not know a and b.
(3)+(10) = 13 = 11 + 2 is the sum of two primes.
Therefore, S knows that ab could not equal (3)(10) = (2)(15).
(7) + (6) = 13 = 11+2 is the sum of two primes.
Therefore, S knows that ab could not equal (7)(6) = (3)(14).
(2) + (30) = 3 + 29 is the sum of two primes.
Therefore, S knows that ab could not equal (2)(30) = (5)(12).
(3)+(22) = 25 = 2 + 23 is the sum of two primes.
Therefore, S knows that ab could not equal (3)(22) = (6)(11).
(5)+(14) = 19 = 2+17 is the sum of two primes.
Therefore, S knows that ab could not equal (5)(14) = (7)(10)
(2)+(36)=38 = 31+7 is the sum of two primes.
Therefore, S knows that ab could not equal (2)(36) = (8)(9).
The only possibility left is that ab=(4)(17). Now S knows a and b.
Kermit.
From hip-hop!benjie@amdahl.com Tue Feb 22 02:21:00 1994
From: presiden@sfu.ca (Pat Presidente)
Subject: Common Volume
Find the volume common to two circular cones, each with radius r, if
the axes of the cylinders intersect at right angles.
From hip-hop!benjie@amdahl.com Wed Feb 23 09:02:00 1994
From: Gabriel Campuzano Trevino
Subject: Variational Problem
Problem: Given two points on the first quadrant, find the form of the
curve which joins both points and generates minimum surface area when
rotated around the x-axis.
Gabriel Campuzano
From hip-hop!benjie@amdahl.com Wed Feb 23 03:43:00 1994
From: Kermit Rose
Subject: algorithm puzzle
The following is an algorithm to find out whether or not a given integer n
has a particular property. The algorithm assigns 1 to the function f
if the integer n has the property, otherwise it assigns -1. Consider the
algorithm to be completed once a value has been assigned to f.
The puzzle is to find out what the property is, and why the algorithm works.
Also say something about what domain of n the algorithm is valid for.
define the function f(n) as follows:
The function int(x) is the integer part of x. ie. int(1.7)=1.
The function abs(m) is the absolute value of m. ie. abs(-1)=1.
for n < 0, f(n) = -1
f(0) = 1
f(1) = 1
if n is < 3 * int( (n+1)/3) then f(n) = -1
That is, add 1 to n, divide by 3, take the integer part, multiply by
3. If the result is greater than n, then f(n) = -1.
otherwise,
define m = n - 5 * int ( (n+2)/5)
if int( abs(m) / 2) > 0 then f(n) = -1
That is, take the absolute value of m, divide by 2, take the integer
part. If the result is greater than 0, then f(n) = -1.
otherwise,
define m = n - 7 * int( (n+3) /7)
if abs(m) = 3 then set m = -m. That is, change the sign of m.
if m < 0 then f(n) = -1
That is, if the absolute value of m is 3, change the sign of m.
Now if the sign of m is negative, then set f(n) = -1
otherwise,
define m = n - 11 * int( (n+5) / 11)
if abs(m) = 2 then set m = -m. That is, change the sign of m.
if m < 0 then f(n) = -1
That is, if the absolute value of m is 2, change the sign of m. Now
if the sign of m is negative, set f(n) = -1.
otherwise,
define m = n - 13 * int( (n+6) / 13)
if int((1+ abs( abs(m) - 2 )) /2) is not equal to 1 then f(n) = -1
That is, take the absolute value of m, subtract 2, take the absolute
value again, add 1, divide by 2, and take the integer part. If the result
is not 1, then set f(n) = -1.
otherwise,
define m = n - 17 * int( (n+8)/17)
define ma = abs(m)
if ma = 4*int(ma/4) then set ma = 0
if ma > 2 then f(n) = -1
That is, divide ma by 4, take the integer part, multiply by 4. If
the result is equal to ma, then reset ma = 0. Now if ma is greater
than 2, then f(n) = -1.
otherwise,
define m = n - 19 * int( (n+9)/19)
define ma = abs(m)
if int(ma/2) = 1 then set m = -m
if ma = 8 then set m = -m
if m < 0 then f(n) = -1
That is, divide ma by 2, and take the integer part. If the result
is equal to 1 then change the sign of m. Also if ma equals 8,
change the sign of m. Now if the sign of m is negative, set f(n) = -1.
otherwise,
define m = n - 23 * int( (n+11)/23)
define ma = abs(m)
if int(ma/2) = 5 then set m = -m
if abs(ma-6) = 1 then set m = -m
if m < 0 then set f(n) = -1
That is, divide ma by 2 and take the integer part. If the result is
5, then change the sign of m. Also, subtract 6 from ma, and take
the absolute value. If the result is 1, then change the sign of m.
Now if the sign of m is negative, set f(n) = -1.
otherwise,
define m = n - 29 * int( (n+14)/29)
define ma = abs(m)
define mb = int(ma/4)
if mb = 1 then set f(n) = 1.
if ma - 4*mb = 1 then set f(n) = 1.
if ma is not equal to 0 then set f(n) = -1
otherwise,
set f(n) = 1.
From hip-hop!benjie@amdahl.com Wed Feb 23 03:46:00 1994
From: pshinn@sas.upenn.edu (Paul Shinn)
Subject: proofs
Can anyone give me a proof why the product of two negative numbers is
positive?
Thanks.
Paul Shinn
pshinn@sas.upenn.edu
pshinn@stwing.resnet.upenn.edu (In a few days)
215-573-7314
From hip-hop!benjie@amdahl.com Wed Feb 23 04:55:00 1994
From: benjie@wales.sbay.org (Benjie Chen)
Subject: POD 2/23 (Maximization) and solution to the prime number POD
POD: (from tulled@rpi.edu)
Maximize 2^(-x)+2^(-1/x) over (0,inf).
SPOILER Alert
the solution to the prime number POD follows
Problem: For which positive integers n is n^4 + 4^n prime?
Solution: If n is an even number, n^4+4^n is even. Assume n is 2x+1.
Note that n^4 + 4^n = (n^2)^2 + (2^n)^2. So
n^4+4^n = (n^2+2^n)^2 - 2*n^2*2^n.
n^4+4^n = (n^2 + 2^n)^2 - [n^2 * 2^(n+1)]
n^4+4^n = (n^2 + 2^n)^2 - [n^2 * 2^(2x+2)]
n^4+4^n = (n^2 + 2^n)^2 - [n*2^(x+1)]^2
n^4+4^n = [(n^2 + 2^n) + n*2^(x+1)] [(n^2 + 2^n) - n*2^(x+1)]
If n^4+4^n was to be a prime, it has to be positive. Since n is a
positve integer, one of two factors has to be 1 and the other has to
be the prime itself. Note that the first factor is always greater than
1 because n is a positve integer. Thus the factor term must be the
prime and the second factor must be 1. So now we just have to find
the positive integer n that will result [(n^2 + 2^n) - n*2^(x+1)] to
be 1.
If n > 1, n = 2y+3 for all positive integer y. Since x+1 = (n+1)/2,
x+1 = y+2. Thus the second factor will be
(2y+3)^2 + 2^(2y+3) - (2y+3)*2^(y+2)
= (2y+3)^2 + [2^(y+2)]*[2^(y+1) - (2y+3)]
But for all y > 1, 2^(y+1)-(2y+3) > 0, so [2^(y+2)]*[2^(y+1) - (2y+3)]
is positive and (2y+3)^2 + [2^(y+2)]*[2^(y+1) - (2y+3)] > 1.
If y was 1, (2y+3)^2 + [2^(y+2)]*[2^(y+1) - (2y+3)] = 17 > 1.
Therefore for all n>1, [(n^2 + 2^n) - n*2^(x+1)] > 1.
Therefore for all n>1, the second factor
(2y+3)^2 + 2^(2y+3) - (2y+3)*2^(y+2) > 1
If n = 1, [(n^2 + 2^n) - n*2^(x+1)] = [3-2^(x+1)] = 1 when x=0.
Therefore only when n=1, the number n^4+4^n=5 is a prime.
A correct solution was sent in by csaba@teleport.com.
From hip-hop!benjie@amdahl.com Thu Feb 24 12:39:00 1994
From: Kermit Rose
Subject: DISCISSION: ongoing discussion of sum/product
Hello David.
Since I presume you are still on the listserve, I will send this to
the iams only.
>If you reread what was said in previous posts you may see that we actually
>ONLY use the product of primes logic. We use the product of primes logic
>to prove that S cannot be the sum of two primes. P does not know at the
>outset that a+b is not the sum of two primes. S knows that his number is
>not the sum of two primes.
>
>The fact that S could tell P that his product was not the product of two
>primes told P that a+b was not the sum of two primes.
. Kermit,
.
. What you state in the prior posts, is that ALL (a,b) pairs associated
. with a given sum may be excluded, IF a single (a=prime b=prime) pair was
. related to the sum.
From hip-hop!benjie@amdahl.com Thu Feb 24 14:10:00 1994
From: recos@cntrol.enet.dec.com ()
Reply-To: recos@cntrol.enet.dec.com ()
Subject: Question re: air resistance in kinematics
A math text illustrates a separation-of-variables differential equation
by setting up an equation of the form F = ma for a falling body starting
at rest, under the influence of gravity, and with a retarding force due
to air resistance, which is assumed to be directly proportional to velocity:
(W/g)dv/dt = W - Bv where F(air) = -Bv, and W= mg.
The math is trivial, so I won't present it here, but the solution for the
velocity as a fn. of time is
v = W/B(1 - e**(-Bgt/W))
My question is why doesn't this expression reduce to simply v = gt in the
absence of air resistance? Is it not valid to set B = 0 in the solution
and expect it to reduce to the ideal case?
Appreciate any helpful responses.
Rick
From hip-hop!benjie@amdahl.com Fri Feb 25 02:57:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: SPOILER to the Air resistance/Drag force problem
SPOILER follows, close your eyes......
Problem:
(W/g)dv/dt = W - Bv where F(air) = -Bv, and W= mg.
The math is trivial, so I won't present it here, but the solution for
the velocity as a fn. of time is
v = W/B(1 - e**(-Bgt/W))
My question is why doesn't this expression reduce to simply v = gt in
the absence of air resistance? Is it not valid to set B = 0 in the
solution and expect it to reduce to the ideal case?
Solution:
From: Leigh Blue Caldwell
Well, you can't always just set a variable equal to zero in a function to
get a special case. But what you _can_ usually do is take a limit - in
this case the limit of v as B tends to zero.
As B -> 0, -Bgt/W -> 0 also, so that:
e^(-Bgt/W) -> 1 + (-Bgt/W) [*]
= 1 - Bgt/W
If you substitute this into the original equation for v, you get:
v = W/B (1 - (1 - Bgt/W))
= W/B (Bgt/W)
= gt
which, spookily enough, is exactly what you would expect!
[*] The reason for this approximation is fairly easy to see. The
exponential function can also be represented as a power series:
exp(x) = e^x = 1 + x + x^2/2 + x^3/6 + ... + x^n/n! + ...
(notice this by differentiating the power series - you get the same
thing back again).
If x is very close to zero, then the x^2 and higher terms will be
so small as to make no difference - so the expression reduces to:
e^x = 1 + x
This is a _very_ useful result to know - it comes into all sorts
of approximation problems.
Leigh
p.s. is anyone else in Britain having trouble getting the newsgroup? It
hasn't shown up on my site, and I suspect it may be something to do with
the reputed 'censorship' of some alt. newsgroups at their feed into the UK.
If anyone else in the UK _can_ get alt.math.iams, could you let me know so
that maybe I can figure out how to do it myself.
>From quack.kfu.com!van
You have to take the limit as B --> 0, or you get the trivial v = 0 soln.
1 - e**(-Bgt/W) = 1 -(1-Bgt/W) = Bgt/W so we get v = gt as expected.
From hip-hop!benjie@amdahl.com Fri Feb 25 03:04:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: DISCUSSION: cont. on SUM/PROD problem
>From quack.kfu.com!van Fri Feb 25 04:31:34 1994
Subject: DISCUSSION: sum/product problem
RE: DISCUSSION: sum/product problem - Addendum
When Kermit's "SUM of PRIMES" logic is applied, there still remain,
145 valid (a,b) pairs, associated with the following sums:
11,17,23,27,29,35,37,41,47,53. These were not accounted for in his
overall reasoning, so I think they are worthy of mention.
>From quack.kfu.com!van Fri Feb 25 05:18:28 1994
Subject: DISCUSSION: Sum/product problem
>From: rcunniff@unix1.tcd.ie (ronan cunniffe)
I'm probably butting in here in a bad way....
I'm a 1st year uni. student, last two years I took part in Irish
Maths Olympics.... we got handed this problem both years. I know an
approach to a solution, but it requires a laborious finish, which I
haven't followed through.
First, I'll state the version I have:
Let the numbers be referred to as X and Y, the two people as S and P.
S knows that he has (X + Y), and that P has (X * Y)
P knows this also.
Both also know that X and Y are both in the range >1, <100,
exclusively.
P states: I don't know X and Y.
S answers: I don't either, but I knew automatically you couldn't know.
P laughs: Well, now I *do* know!
S grins: That just told me the answer as well!
As the post pointed out, P has a number with at least 3 factors (in
using the word factors, I am excluding the number itself and 1, unless
otherwise stated)
I found this an unprofitable place to start, I worked rather with the
second line, that S knew WITHOUT P SAYING ANYTHING that P *couldn't*
know. This implies that (X + Y) *cannot* be written as the sum of two
primes, or S couldn't be SURE, as he states he is.
This eliminates all even numbers, and as 2 is a prime, it elimates all
numbers which are 2 greater than a prime, ie 1+2=3, 3+2=5....19+2=21...
etc.
This allows the construction of a table containing all the possible
values for S consistent with the above argument. Obviously out of c.200
initial entries, something less than 100 will be left.
N.B. S's statement "I knew you couldn't know" allows P to construct this
table.
P now has both (X * Y), AND all the possible values for (X + Y). X and Y
are such that there is only one entry in S's table that matches his (X *
Y). Obviously, with both pieces of information, he knows X and Y, and
says so.
It is perhaps useful here to state my principal approach: I don't think
the problem can be approached from a sledgehammer technique, trying to
follow the logic, to tie down a specific number, generate a formula to
pop out the answer. I believe the correct way of tackling this is to
pose logical hurdles our numbers must pass, qualifications they must
meet to fit the criteria. Obviously, it is possible to pick X and Y such
that the question is insoluble... X=8, Y=12 (Off the top of my head) So
in this problem, we are dealing with very special creatures, let's try
and eliminate everybody else.
So, if P can tell X and Y, simply from gaining knowledge of S's table,
none of the factor pairs of (X * Y) (Given that at least 1 of X,Y are
composite) correspond to sums (X + Y) in S's table, bar one, the correct
one.
So: For every value in the table, write down all the possible additions,
eg: 23= 2+21, 3+18..... 11+12, and then the corresponding products:
42 54..... 132.
If P is to know X and Y, there must be only 1 value in the S's table
that can yield his product. (Examine this and you'll discover the
numbers are *very* special animals.)
This is the limit of where I have reached.... I have constructed S's
table, I have partly constructed the list of derived products, there are
lots of them. However, there are lots of duplicates, comparatively few
derived products that can come from only one source in S's table. These
are the only candidates, because P has to backtrack, to go from the
table to the unique, with only the knowledge of (X * Y) extra that we
don't have.
For us to approach the same knowledge, we have to construct a similar
table for P, all the possible products he could have. We eliminate all
primes and products of 2 primes, obviously. As I recall, we still get
left with more than 5000 numbers (I partly generated this table, it
didn't look pleasant.)
However, we're looking for those numbers P could have which
correspond to entries in S's table. (Correspond here does not mean
equal, rather that they share the same X and Y, or could do so.)
This yields a much smaller number of combinations, but probably not
just 1.
Now, just as S's statement in line 2 let P construct the S table, now
P's statement, line 3, lets S construct the P table. This, together with
the knowledge of X + Y, gives him a unique answer.
So, P is such that it can be generated by only one entry in the S table.
S is such that it can be generated by only one entry in the P table.
So, we have two shortlists, the possible values of X * Y, derived from
S's table, we know that P's number corresponds to a unique S. We know
the possible values of X + Y, derived from P's table, we know that S's
number corresponds to a unique P.
If my logic is correct, then there will be only one pair of values, one
in P, the other in S, that correspond.
However, the size of the tables, and the amount of drudgery involved,
suggests the use of a computer.
I state an overview of the logic:
Construct a table of all possible sums allowed by problem.
Construct a table of all possible products allowed by problem.
Derive from table S a list of all products generable from allowed sums.
Compare Table S with this list, create a shortlist of matches.
Derive from table P a list of all sums generable from allowed products.
Compare Table P with this list, create a shortlist of matches.
Compare the two final lists, finding all corresponding pairs.
My contention is that only one will exist.
This is probably not the most elegant solution, relying as it does on
number-bashing. If somebody has a simple & hard solution, or indeed any
data/comments/advice/misc, mail me. This logic is my own invention, so
can't be used to ridicule professors if it's flawed (B-P).
Ronan (rcunniff@unix1.tcd.ie)
=========
This logic looks fine to me. I have always thought that the 1st sentence
was unnecessary, and don't understand any argument that asserts that one
can't eliminate sums which are sums of primes, and all pairs (a,b)
associated with such sums.
From hip-hop!benjie@amdahl.com Fri Feb 25 04:28:00 1994
From: Sylvan Jacques
Subject: SPOILER: Common Volume
SPOILER:
>From: presiden@sfu.ca (Pat Presidente)
>Subject: Common Volume
>
>Find the volume common to two circular cones, each with radius r, if the axes
>of the cylinders intersect at right angles.
I assume the "circular cones" means circular cylinders.
The volume is a pyramid with curved
sides. Imagine the cylinders to be along the x and y axes. For any
z = const., the common area will be a square with area A(z).
The planes x = 0 and y = 0 intersect the common volume in the circles
y^2 + z^2 = r^2 and x^2 + z^2 = r^2 respectively.
Thus A(z) = 4(r^2 - z^2)
Integration from z = -r to +r gives the common volume
V = 2 Int{A(z);0;r} = 16*r^3/3 ; about 20 to 25% greater than that of
a sphere of radius r, and 2/3 the volume of a cube with edge 2r.
Van van@quack.kfu.com
From hip-hop!benjie@amdahl.com Sat Feb 26 12:25:00 1994
From:
Subject: Problem: Parallel Algorithm
"Given an array of length n containing a legal sequence of
parentheses, determine for each left parenthesis in the sequence its
matching right parenthesis. Develope an algorithm running in O(log n)
time using a total of O(n) operations. Hint: start by assigning a
weight of +1 to each left parenthesis and a weight of -1 to each right
parenthesis. Compute the prefix sums."
The algorithm shoule be an parallelism. Hope you who are interested
in parallel computing will enjoy it.
Thank you all.
Sincerely,
David Tsai
From hip-hop!benjie@amdahl.com Mon Feb 28 11:09:00 1994
From: Kermit Rose
Subject: complete analysis of sum,product puzzle.
p: 1
Subject: slight variation of sum-product puzzle.
If we assume that S and P do NOT know that a and b are restricted to the
range 2 to 99, but know only that a and b are both > 1, then in the range
2 to 99, there are 4 solutions.
s = 17 p = 52 a = 13 b = 4
s = 65 p = 244 a = 61 b = 4
s = 89 p =1168 a = 73 b = 16
s = 137 p =4672 a = 73 b = 64
Kermit.
From hip-hop!benjie@amdahl.com Thu Mar 3 13:28:00 1994
From: "David A. Wagner"
Subject: Slick proofs
I thought I'd share two slick proofs I saw this past few
weeks - they talk about algebraic numbers. A complex number x is
algebraic if there are integers a_0, a_1, ..., a_n so that
a_0 + a_1 x + ... + a_n x^n = 0.
[You can also let the a_j's be rational; it's equivalent.] Let
A denote the set of all algebraic numbers. If you're bored, try
to prove that A is (a) countable and (b) a field.
I'll post a spoiler in a couple of days in the unlikely
event that noone posts a solution...
-------------------------------------------------------------------------------
David Wagner dawagner@princeton.edu
From hip-hop!benjie@amdahl.com Thu Mar 3 13:34:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: General stuff
Here are some general stuff:
I will post a general bulletin containing info posts to IAMS once a
week or two weeks. This way some general information could be past
out at once.
Also, I bet you that most of you have some useful tricks when asked to
solve a special kind of problem, or some of you have special problem
solving techniques for special kind of math problems. Could we share
with each other? Post your tricks......
Here is one of my tricks, the Heronian formula for area of a triangle.
Most of you probably already know that the area of a triangle with
sides a b and c could be calculated as
Area = sqrt(s(s-a)(s-b)(s-c)) where s = 0.5(a+b+c)
I've found this formula very effective and useful. For example when a
triangle is inscribed in a circle, and the center of the circle is
with in the triangle, what is the radius of the circle? We could draw
lines from center of the circle to vertexes of triangle and thus
dividing the triangle in three.
We could then use the fact that two tangents to a single circle (at
two points) have the same length if we drew them from the same point
to derive an equation for the area of the triangle, using the formual
area = base * height / 2. Then we use the heron formula to get
another equation for the area, then we could solve for the radius of
the circle......
There are other uses for heronian triangle, such as for calculating
the heronian triangle, calculating the sides of the triangle inscribed
in a circle, etc. In all, I found the formula helpful and effective.
Enjoy,
Benjie
From hip-hop!benjie@amdahl.com Sat Mar 5 08:06:00 1994
From: luis@math.psu.edu (Luis Manuel Barreira)
Subject: Ellipses inscribed in triangles
Suppose that we have a triangle with sides a,b,c
and try to inscribe ellipses in the triangle with
axis say s,t with s/t fixed.
Does anyone know of an explicit formula for s (and t)?
This question is connected with the estimation of
Hausdorff dimension of general Cantor sets. Of course
that for circles the answer is known. It also must be
somehow easier to express s and t if the triangle is
isosceles.
Luis Barreira
From hip-hop!benjie@amdahl.com Sat Mar 5 08:07:00 1994
From: Jin-Chong Tsai
Subject: Proof of Clos permutation network
Hi there,
I have a proof of Clos permutation network, and I hope someone who can help me to modify this proof more clear.
Thank you.
-------- -------- --------
-|1 |---------|1 |---------|1 |-
-|2 |- -|2 |- -|2 |-
.| |.\ /.| |.\ /.| |.
.| CL(m)|. \ / .| CL(k)|. \ / .| CL(m)|.
.| |. \ / .| |. \ / .| |.
-|m |- / -|k |- / -|m |-
-------- \ / \ / -------- \ / \ / --------
-------- / \ -------- / \ --------
-|1 |-/| |\-|1 |-/| |\-|1 |-
-|2 |---------|2 |---------|2 |-
.| |. | | .| |. | | .| |.
.| CL(m)|. | | .| CL(k)|. | | .| CL(m)|.
.| |. | | .| |. | | .| |.
-|m |- | | -|k |- | | -|m |-
-------- \| |/ -------- \| |/ --------
. \ / . \ / .
. |\ /| . |\ /| .
. | / | . | / | .
-------- |/ \| -------- |/ \| --------
-|1 |-// \\-|1 |-// \\-|1 |-
-|2 |-/ \-|2 |-/ \-|2 |-
.| |. .| |. .| |.
.| CL(m)|. .| CL(k)|. .| CL(m)|.
.| |. .| |. .| |.
-|m |---------|k |---------|m |-
-------- -------- --------
( Cl(m) and Cl(k) are permutation networks on m and k inputs, respectively. )
(Clos 3-stage permutation network on n inputs where n = m*k)
Given n = m * k, where m and k are integers, tells that the inputs and the outputs can be divided into k subsets of cardinality m or m subsets of cardinality k. Given such subsets, let each of the input subsets have their own assignment list which telling which outputs the corresponding inputs should connect to. The entries of this list, one for each input, gives first the output subset, then the output number within that subset. Therefore, for an arbitrary bijection a subassignment list corresponding to input subset number g ( g = 1, ..., k) can be formed F*(g) = ( (i1, j1), ... ,(im,jm) ) where
F : a family,
i1, ..., im : output subset number, and
j1, ..., jm : output terminal number (within each subset).
Since the totality of all the F*(g)'s describes a bijection, all pairs in the F*(g)'s must be distinct.
The output subsets of an assignement list F(g) are only the i's:
F(g) = { i1, i2, ..., im }. the elements are no longer necessarily distinct.
Prove the Clos 3-stage network is a permutation network:
(1) Hall's condition:
Let a family divided in k parts. If the union of any c ( c = 1, ..., k ) arbitrary chosen different parts contains at least c distinct elements, the family satisfy Hall's condition for a family to have at least one transversal.
(2) Looking at c arbitrarily chosen F*(g)'s ( with c distinct g's), they shows c*m bijections that can be made between c subsets of inputs and an unknown number, u, of subsets of outputs.
However, the maximum number of assignments to a single subset of outputs is m. The number u of different subsets pointed to by the union of c arbitrary chosen F*(g)'s is therefore limited by the inequality
u G.E. (c*m) / m = c /* G.E. means great than or equal to */
This means that c arbitrary chosen F*(g)'s contain at least c different i values. This fact must also be true for the union of the corresponding F(g)'s so that Hall's condition is satisfied. Therefore, the family F has a transversal.
(3) A canonical reduction of the family F is defined by removing one element in each subfamily corresponding to the found transversal. This leaves a reduced family F where every subfamily has the cardinality m-1.
The rest of the bijection concerns n-k inputs partioned in sets of cardnality m-1. By taking n-k as a new n and m-1 as a new m, we have:
n-k = (m-1)*k (old value)
n = m*k (new value)
so that n is m times k as before.
Therefore, repeat the argument given in point (2) above. This proves the existence of another transversal, and the procedure can then be repeated until the assignment family F is emptied. This completes the proof.
David Tsai
From hip-hop!benjie@amdahl.com Sun Mar 6 04:32:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: bulletin
Hello everybody,
Here is the bulletin for the past week. There are lots of interesting
ads. etc.
To AOL subscribers, American On Line installed their USENET feed the
past week. This means you will be able to get alt.math.iams newsgroup
from AOL. I think the keyword is "newsgroup" (I am not sure because I
heard all this on the radio). If you could receive the newsgroup,
please unsubscribe from the mailing-list (the mailing-list and the
newsgroup are the same thing) by mailing to listserv@hh.sbay.org with
"delete YOURADDRESS iams" in the body of the message.
This goes for everyone. If you could receive alt.math.iams, it would
be nice if you unsubscribe from the mailing-list.
=========================================================================
From: wyman@math.ohio-state.edu (Bostwick F. Wyman)
Subject: Summer Workshops for College Faculty
SUMMER OPPORTUNITY FOR ALL COLLEGE FACULTY IN MATHEMATICS DEPARTMENTS
AND RELATED AREAS.
Technology Strategies Workshops at The Ohio State University Summer
1994
Supported by the National Science Foundation
Session 1: June 20 - July 1, 1994
Session 2: July 11- July 22, 1994
Community college, two-year college, four-year college, and university
faculty are all welcome. Teams from a single school are encouraged to
apply together.
**** APPLICATION DEADLINE APRIL 1, 1994. FOR APPLICATION MATERIALS CONTACT
BOSTWICK WYMAN DIRECTLY BY EMAIL TO wyman@math.ohio-state.edu *****
* Introduction to uses of technology in undergraduate mathematics
instruction in areas ranging from precalculus through calculus to advanced
topics.
* Hands on experience and laboratory work with graphing calculators,
computer algebra systems, computer graphing programs, linear algebra
packages, differential equation packages. Maple, Mathematica, Matlab,
maybe some statistics. Offerings will be very flexible and will depend on
the audience.
* Discussions and reports on pedagogical implications of the new
technologies.
* Opportunities to work on course and curriculum development for use at
home institutions. Follow-up conference support and publication
opportunities will be available.
* Room, board, instruction, and materials paid by the National Science
Foundation. Travel costs will be the responsibility of participants.
*Faculty of community colleges, two-year colleges, four-year colleges and
universities are eligible for these workshops. Members of
underrepresented groups, and those deeply involved in the teaching of
underrepresented groups are especially encouraged to apply. Applications
from teams of participants from a single institution are welcomed.
For more information and application materials, please contact:
Professor Bostwick F. Wyman
Technology Strategies Workshops
Ohio State Mathematics Department
231 West 18th Avenue
Columbus, OH 43210
wyman@math.ohio-state.edu
fax: (614) 292-1479
office phone: (614)292-4901
math dept office (614)292-4975
Applications due April 1, 1994. We hope to send acceptance letters by
April 15. If this deadline is a problem, please contact Wyman as soon as
possible.
From: Pertsel Vladimir
Subject: Russian (in English) journal of the school math.
Dear sir!
I think, that it would be interesting for the subscribers
to You newsgroup to get an information about the Russian
journal of the school and recreational mathematics and
physics that is nowdais available in English. I have sent
a request to the managing editor, and here is the response:
/
/From: Tim Weber <72030.3162@CompuServe.COM>
/
/Thank you for your inquiry about Quantum, the English-language version of
/Kvant. Orders outside of North America should be placed through Springer-
/Verlag, Postfach 31 13 40, D-10643 Berlin, FRG. The cost of an individual
/subscription is US$20, but you may want to inquire about air mail rates.
/
/We are in currently negotiating with a publisher in Athens regarding a
/Greek edition of Quantum. There are no other authorized translations of
/Quantum magazine.
/
/Please feel free to contact me again if you have any problems subscribing.
/
/Sincerely,
/
/Tim Weber
/Managing Editor, Quantum
/
I, myself, highly recommend it. It is dealt with the same sort of
problems, that You discuss in the newsgroup.
From: efedula@aol.com
Subject: 1994 AHSME
I don't know how many high school students are here, but I was hoping
to get a little bit of national feedback on how the AHSME went. It was
obviously the easiest AHSME ever, but I'd appreciate some unofficial
results to see how it went at other schools.
The UNOFFICIAL results from my school (Science Academy at LBJ in
Austin, TX) are that we have a team score of 444 with two 150's (one
of which is mine) and a 144. Since we won't get the official answer
key until the second week of March, we're just using the answers that
we agree on (I'll post them tomorrow if anyone's interested). We
haven't counted how many people got 100 or better yet, so I won't even
try to guess.
Anybody else?
From: roconnel@ccvax.ucd.ie
Subject: Plasma Physics Summer School
T H E 3 1 s t C U L H A M P L A S M A P H Y S I C S
###########################################################
S U M M E R S C H O O L
#########################
1 1 - 2 2 J U L Y 1 9 9 4
C u l h a m L a b o r a t o r y, A b i n g d o n,
O x f o r d s h i r e, U K
An International Summer School intended for students near the start
of postgraduate courses. No previous knowledge of plasma physics is
assumed. Since 1985, the Summer School series has been attended by
over 600 students from 47 countries, more than two thirds coming from
outside the UK.
Culham Laboratory is the primary centre for plasma physics & nuclear
fusion research in the UK; it is located close to the city of Oxford,
and shares a site with the world's largest magnetic fusion
experiment, the Joint European Torus (JET).
The School covers a broad curriculum :-
* Plasma particle dynamics * Plasma waves * Kinetic theory * MHD
* Computational techniques * Astrophysical plasmas * Laser plasmas
* Magnetically confined plasmas * Solar plasmas * Poster session
* Space plasmas * Laboratory visits * Industrial plasmas
* Turbulence & chaos * Diagnostics * Gravitational plasmas
A copy of the course textbook "Plasma Phyics: An Introductory Course"
(Cambridge University Press,1993) is given to each student.
ACCOMMODATION WILL BE IN A HISTORIC COLLEGE OF OXFORD UNIVERSITY.
CLOSING DATE FOR APPLICATIONS : 13th MAY 1994
(Late applications may be accepted, depending on accommodation.)
Further details / application forms are available from :-
Mrs Joan Stimson,
Culham Laboratory,
Abingdon,
Oxfordshire OX14 3DB, Tel: 44 235 463293
UK. FAX: 44 235 463288
or e-MAIL enquiries to :- geoff.maddison@aea.orgn.uk
From hip-hop!benjie@amdahl.com Sun Mar 6 13:15:00 1994
From: Kermit Rose
Subject: Re: Ellipses inscribed in triangles
>Suppose that we have a triangle with sides a,b,c
>and try to inscribe ellipses in the triangle with
>axis say s,t with s/t fixed.
You should be able to inscribe an ellipses at every angle of the long axis.
>Does anyone know of an explicit formula for s (and t)?
I do not know an explicit formula, but have a suggestion for how you may
derive a formula in terms of the angle of the long or short axis.
Choose the direction of the long axis. Then define r = min(s/t,t/s).
That is, r is the ratio of the smaller of s and t to the larger. Then
shorten distances in the direction of the long axis by the factor r.
This will give new lengths to the sides of the triangle. Call these
lengths x,y,z. Inscribe a circle in the new triangle. Apply the
inverse transformation to lengthen distances in the direction of the long
axis by the factor 1/r. The circle has become an elipse in the triangle
with sides a,b,c.
>This question is connected with the estimation of
>Hausdorff dimension of general Cantor sets.
I do not know the meaning of Hausdorff dimension of general Cantor sets.
>Of course
>that for circles the answer is known.
So following the above, we should be able to make explicit formulae.
Perhaps you could even find the best ellipse of the family of ellipses
according to any one of several criteria.
>It also must be
>somehow easier to express s and t if the triangle is
>isosceles.
I do not know why this should be. If it is so, then perhaps it is because the
formulae become simplier because two sides of the triangle are equal.
However, the transformed triangle would be isosceles only if you choose
the direction of the long or short axis of the ellipse to be along the line
dividing the isosceles triangle into two equal parts.
>Luis Barreira
Kermit Rose
From hip-hop!benjie@amdahl.com Tue Mar 8 13:57:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: POD Solving for X
Here is a short little problem, the solution is pretty slick.
There is only one real k for which
log(3x)log(5x)=k
has only one real solution for x. What is this value of x?
Enjoy,
Ben
From hip-hop!benjie@amdahl.com Fri Mar 11 00:19:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: SPOILER to the POD
SPOILER Alert
Problem: There is only one real k for which
log(3x)log(5x) = k has only one real solution for x, what
is the value of x?
Solution:
log(3x)log(5x)=(logx)^2 + (log3+log5)x + log3log5=k
So
(logx)^2 + (log3+log5)log(x) + constant term = 0.
Let y = logx, then
y^2 + (log(3)+log(5))y + C = 0
If the solution to this equation, call it y0, was to be unique,
(y-y0)^2 = 0
so
-2y0 = log(3)+log(5)
-2y0 = log15
2y0 = -log15 = log(1/15)
sub y0 = log(x) back,
2(log(x)) = log(1/15)
log(x^2)=log(15) and so
x^2 = 1/15,
x = 1/sqrt(15)
From hip-hop!benjie@amdahl.com Thu Mar 10 13:36:00 1994
From: mfhlh@uxa.ecn.bgu.edu (Howard L. Hansen)
Subject: Horizontal Asymptotes
I have a question which was posed by a student while we were
investigating the derivatives of functions with horizontal asymptotes.
I am unable to give an satisfactory answer. Given a real-valued
continuous function f(x) which has a horizontal asymptote and is
differentiable, then its dervative is asymptotic to the x-axis. This
much is clear and intuitively obvious. The question which arose was,
can we look at the asymptotic behavior of the derivative of a function
and determine whether the function has horizontal asymptotes. We came
up with examples for which the derivative is asymptotic to the x-axis,
but the function had no asymptotes (e.g. sqrt(x), etc.). This still of
course does not eliminate the possibility that within some bounds,
inspection of the derivative can be used to determine the existence of
asymptotes in the function. That's where I'm at "Are there sufficient
conditions based on the derviativbe of a function for the function to
have horizontal asymptote?"
Thanks for any light you can shed.
Howard Hansen aka H^2
Mathematics Education
Western Illinois University
From hip-hop!benjie@amdahl.com Sat Mar 12 11:18:00 1994
From: tom@wg.icl.co.uk (Tom Thomson)
Subject: Re: SPOILER to the POD
the solution as given has a flaw. we are given that there is a unique
real solution x, but this does not make log(x) unique and real; the
quadratic equation for log(x) can have two solutions which differ by
2ni, where n is any integer. Choosing k so that n is even (eg n=0,
which your solutiion assumes) gives x = 1/sqrt(15). Choosing k so that
n is odd (eg n=1) still gives a unique solution for x, but this solution
is x = -sqrt(1/15).
Tom
From hip-hop!benjie@amdahl.com Sun Mar 13 08:15:00 1994
From: Andrew Wilcock
Subject: Probability problem.
Here is a little probability problem for you all ...
Let X,Y be independent random variables, each geometrically distributed
with parameter p. Thus
P(X=x)=P(Y=y)=p*q^x for x=0,1,2,...
i) Find P(min(X,Y)=X).
ii) Find P(Y=y|X+Y=z) for y,z=0,1,2,...
From hip-hop!benjie@amdahl.com Sun Mar 13 08:16:00 1994
From: allenk@gluttony.ugcs.caltech.edu (Allen Knutson)
Subject: Re: Horizontal Asymptotes
mfhlh@uxa.ecn.bgu.edu (Howard L. Hansen) writes, in part:
>Given a real-valued
>continuous function f(x) which has a horizontal asymptote and is
>differentiable, then its dervative is asymptotic to the x-axis.
Nope. Take sin(x^2)/x.
>"Are there sufficient conditions based on the derviativbe of a function
>for the function to have horizontal asymptote?"
As you said and I snipped, f' -> 0 is not enough. It needs to go to 0 fast
enough. I take it you're not doing convergence of series stuff in this class.
One standard sufficient condition is that x^{1+\epsilon} f'(x) has to
go to zero (\epsilon any positive constant). This condition is not necessary.
Allen K.
From hip-hop!benjie@amdahl.com Thu Mar 17 13:27:00 1994
From: benjie@hh.sbay.org (Benjie Chen)
Subject: Vector space and transformations
While doing a project (the Chicken McNugget problem) I got to the
following problem:
Assume I have a transformation matrix T,
- -
| 4 |
| 7 |
- -
that transforms (x,y) into a single positive integer.
If I call the image space of T (Im(T)) S, then S is
a subspace (subset) of the set of positive integers I+.
That is, S (= I+.
How would I find out the max number in (I+) - S? Is it
possible?
Benjie
From hip-hop!benjie@amdahl.com Sat Mar 19 12:19:00 1994
From: efedula@aol.com (EFEDULA)
Subject: Diophantine equation
Find all integral solutions (x,y,z), ignoring permutations, of:
x^3 + y^3 + z^3 - 3xyz = 2^32
I've found 17 solutions, but I don't know if there are others. If you don't
want any help, stop reading now...
*
*
Hint1: To find my 17 solutions, let x = y.
*
Hint2: To factor the left side, let a=y-x, and b=z-x, then substitute x+a for y
and x+b for z, expand completely, cross out a bunch of stuff, and factor what's
left.
From hip-hop!benjie@amdahl.com Sat Mar 19 12:22:00 1994
From: csaba@teleport.com (Csaba Gabor)
Subject: Re: Probability problem.
In article <2m0abh$sss@hip-hop.hh.sbay.org> Andrew Wilcock
writes:
[edited]
>Let X,Y be independent random variables with distributions such that
> P(X=x)=P(Y=x)=p*q^x for x=0,1,2,... and p = 1-q is in (0,1)
>i) Find P(min(X,Y)=x).
>ii) Find P(Y=y|X+Y=z) for 0 <= y <= z for z in {0,1,2,...}
SPOILER FOLLOWS:
i) P(min(X,Y)=x) = P(X=x & Y>x) + P(Y=x & X>x) + P(X=Y=x)
= p*q^x*q^(x+1) + p*q^x*q^(x+1) + p*q^x*p*q^x
= 2pq^(2x+1) + ppq^(2x)
= (1+q)pq^(2x)
ii) P(Y=y|X+Y=z) = P(Y=y)*P(Z=z-y)/P(X+Y=z)
= p*pq^z/P(X+Y=z)
now to find the denominator, we sum the numerator z+1 times
= 1/(z+1)
Csaba
From hip-hop!benjie@amdahl.com Mon Mar 21 09:18:00 1994
From: ca102@fim.uni-erlangen.de (Ray Girvan)
Subject: Square roots - long division method
Can ayone advise?
I've just learnt the 'long division' method for calculating
square roots, the one that looks like this:
eg for SQRT (45657049)
6 7 5 7
-------------------
) 4 5 6 5 7 0 4 9
3 6
---
12(7) ) 9 6 5
8 8 9
------
134(5) ) 7 6 7 0
6 7 2 5
--------
1350(7) ) 9 4 5 4 9
9 4 5 4 9
-----------
0
I can't work out *why* it works, though. Can anyone who
recgonises the method tell me what's going on, or point
me to a reference? Thanks in advance.
Ray Girvan
rgirvan@cix.compulink.co.uk
From hip-hop!benjie@amdahl.com Tue Mar 22 13:22:00 1994
From: rossum@di.unito.it (Peter van Rossum)
'EFEDULA' recalled a message from 'ivoigt@rz.uni-leipzig.de',
stating that
- a number abcdefgh is divisible by 7 if and only if abcdefg - 2h is
- a number abcdefgh is divisible by 11 if and only if abcdefg - h is
- a number abcdefgh is divisible by 13 if and only if abcdefg + 4h is
and asked
(a) Why does this work?
(b) How can you generalise this to higher primes?
First (a), for the prime number 7, but I guess it will be clear how
to do the same for the others:
7 divides abcdefg - 2h
if and only if
7 divides 10 * (abcdefg - 2h) = 10*abcdefg - 20h
(multiplying the right hand side by 10)
(since 7 and 10 do not have a divisor in common)
which holds if and only if
7 divides 10*abcdefg - 20h + 21h = 10*abcdefg + h = abcdefgh
(adding 21h to the right hand side)
(since 7 divides 21h)
Note that it is necessary that 7 and 10 do not have a divisor in
common, so there would be problems if you'd try to generalise this
to 2 and 5, but for all other primes it works (and even for
not primes, provided that your chosen number and 10 do not
have a divisor in common)
For the reasoning above to work, for an arbitrary prime p (except
2 or 5), you need a multiple of p that is 1 more than a
multiple of 10, e.g.
3 --> 7 * 3 = 21 --> 3 div abcdefgh iff 3 div abcdefg - 2h
7 --> 3 * 7 = 21 --> 7 div abcdefgh iff 7 div abcdefg - 2h
11 --> 1 * 11 = 11 --> 11 div abcdefgh iff 11 div abcdefg - h
13 --> 7 * 13 = 91 --> 13 div abcdefgh iff 13 div abcdefg - 9h
(or for 13, alternatively
13 --> -3 * 13 =-39 --> 13 div abcdefgh iff 13 div abcdefg + 4h)
and now for higher primes:
17 --> 3 * 17 = 51 --> 17 div abcdefgh iff 17 div abcdefg - 5h
19 --> 9 * 19 = 171 --> 19 div abcdefgh iff 19 div abcdefg - 17h
(here it is easier to use
19 --> -1 * 19 = 19 --> 19 div abcdefgh iff 19 div abcdefg + 2h)
23 --> 7 * 23 = 161 --> 23 div abcdefgh iff 23 div abcdefg -16h
(or
23 --> -3 * 23 =-69 --> 23 div abcdefgh iff 23 div abcdefg + 7h)
etc.
For every prime p (execpt 2 and 5) you can always find such a
multiple of p that is 1 more than a multiple of 10:
if p ends in a 1, take p itself
if p ends in a 3, take 7*p (or -3*p)
if p ends in a 7, take 3*p (or -7*p)
if p ends in a 9, take 9*p (or -1*p)
+--------------------+----------------------------------------------------+
| Peter van Rossum | int ll(n)int n;{int m=1,i=0,r=4;for(;i
Subject: Palindrome numbers
I am doing a project for my number theory class on Palindrome
numbers and was looking for any information which may not appear
in most texts or references books (little known theorems, interesting
applications, "fun facts", etc.). Any help would be greatly
appreciated.
Please respond by E-mail (at #s45@utmartn.bitnet) since I don't use
this very often. Thanks..........Pat Riley
From hip-hop!benjie@amdahl.com Thu Apr 3 08:01:00 1994
From: Jim Locum
Subject: Integral of x^x...
My Real analysis book says that integral of x^-x from 0 to 1 is the sum
of n^-n from 1 to infinity. How do we show this?
Jim Locum