Subject: Re: Ugly Mathematics?
From: margot@cnwl.igs.net (David)
Date: Tue, 31 Dec 1996 06:50:35 GMT
rbrito@ime.usp.br (Rogerio Brito) wrote:
> Hi.
>
> Once I saw a quotation with credits due to Hardy that
> said something like: "There's no place to ugly
> Mathematics."
> So, I ask you professional mathematicians, what do you
> consider a beatiful proof to a theorem? What are your
> parameters? Can you describe?
>
One parameter must be "elegance". That is; neat, concise, and to the
point. So it must be with the written word. in mathematics what is
saught is "necessary AND sufficient."
Some proofs are "good" in the sense that they do the job very well
indeed. However, some show deep insight, and may inspire others to
further study. (I believe that Gauss was sos inspired?)
One example, perhaps:
To prove that (x-a)(x-b)/(c-a)(c-b) + (x-b)(x-c)/(a-b)(a-c) +
(x-c)(x-a)/(b-c)(b-a) = 1.
1. One could slog it out by finding a common denominator. It might
be seen that a "good" solution would involve seeing that (a-b) and
(b-a) etc. will provide the simple denominator (a-b)(a-c)(b-c) and
that appropriate signs are cared for easily.
2. A more "elegant " solution depends upon wider knowledge. Gauss's
Fundamental Theorem of Algebra states something like "Every nth degree
Polynomial Equation has exactly n roots." An "Equation with more than
the required number of roots is then an identity. It can be seen
easily that the original statement is satisfied for x = a, b, or c.
Also, expanded it is of 2nd degree. Therefore it must be an identity.
Q.E.D. [Source of solution... Hall & Knight; Higher Algebra.]
Subject: Re: Best way to study maths?
From: margot@cnwl.igs.net (David)
Date: Tue, 31 Dec 1996 07:02:07 GMT
Peter wrote:
>Hi, everybody.
>Could anyone give me the answer of my subject?
Hi,
Read, read, read. Practice, practice, practice. ...until *you* know
it is right. Then read, read read some more. [Really!]
Live long and prosper.
Subject: Re: function question , need help
From: Fred Galvin
Date: Tue, 31 Dec 1996 03:28:25 -0600
On 28 Dec 1996, LSC wrote:
> Hi, everyone, I need help on a question. It's very tough to me. I found
> a solution that I think is correct but I cannot give a satisfying proof.
> I can only give some example to show it is correct. I hope that someone
> can help me solve the problem or give a counter example to show my
> solution is wrong. Ok, here is the question:
>
> question:Does there exist a function f(n) from natural number to natural
> number such that
> (i) f(1)=2
> (ii) f((f(n))=f(n)+n for all natural number n
> (iii) f(n+1)>f(n) for all natural number n
What a neat homework problem! Would you please tell me what textbook it's
from?
> solution:Yes, there is such a function. We need to use the series u1, u2,
> u3,... where u1=1, u2=2, u(n+2)=u(n+1)+u(n)
I.e., the sequence of Fibonacci numbers.
> first we expressed n as the sum of terms from the series,
> n=u(a)+u(b)+u(c)+...
> we define f(n)=u(a+1)+u(b+1)+u(c+1)+...
> for example: 25=21+3+1=u7+u3+u1, then f(25)=u8+u4+u2
More precisely, you want to express n as the sum of *distinct* and, what's
more, *nonconsecutive* terms from the series. I suppose you have already
shown (or been shown) that every natural number has a unique
representation as the sum of nonconsecutive Fibonacci numbers; this is
needed for your definition of f(n) to make sense, and also to verify
condition (ii). And I guess you must know that this representation is
obtained by the greedy algorithm, i.e., you start by taking u(a) to be the
greatest Fibonacci number that's <= n, and so on. This is called the
Zeckendorf representation, if I remember right. As for your function f(n),
I seem to recall that it was mentioned in a book called Concrete
Mathematics, by Knuth et al., as a rough and ready way of converting miles
to kilometers in your head.
> My problem is, I can prove condition (i) and (ii), but I cannot prove
> condition (iii)f(n+1)>f(n).I can only show it is correct by some example.
O.K., suppose n is the least number for which (iii) fails (so, n > 1).
Let u(a) be the greatest Fibonacci number <= n (so, a > 1); then your
definition of f(n) says that
f(n) = u(a+1)+f(n-u(a));
and we know that f(n-u(a)) < f(n+1-u(a)), since n-u(a) < n.
Now we consider two cases.
Case I. u(a+1) > n+1
So u(a) is also the greatest Fibonacci number <= n+1, whence
f(n+1) = u(a+1)+f(n+1-u(a)) > u(a+1)+f(n-u(a)) = f(n).
Case II. u(a+1) = n+1
Then f(n-u(a)) < f(n+1-u(a)) = f(u(a+1)-u(a)) = f(u(a-1)) = u(a), and so
f(n) = u(a+1)+f(n-u(a)) < u(a+1)+u(a) = u(a+2) = f(n+1).
> Please help.Thanks for any help.
You're welcome. Of course, you may prefer to use the simpler (but less
fun) solution posted by Brent Hetherwick:
On 31 Dec 1996, Brent Hetherwick wrote:
> LSC (liausc@pl.jaring.my) wrote:
> : question:Does there exist a function f(n) from natural number to natural
> : number such that
> : (i) f(1)=2
> : (ii) f((f(n))=f(n)+n for all natural number n
> : (iii) f(n+1)>f(n) for all natural number n
>
> I didn't try to follow your solution, but try looking at the following f:
>
> Let f(1) = 2
>
> For n > 1, define f(n) = f(k) + k if k < n and f(k) = n.
> If no such k exists, let f(n) = f(n-1) + 1.
>
> I haven't verified all the details, but I'm fairly confident that this
> works. Draw a picture, if you're not convinced.
Looks fine to me. Might be a tad easier if you use the following simpler
(but equivalent) definition:
For n > 1, define f(n) = n+k(n) where k(n) is the greatest natural number
k < n such that f(k) <= n.
On Mon, 30 Dec 1996, Jim Hunter wrote:
> No, there is no such function. You can use the pigeonhole
> princple to show it.
>
> First observe that from ii and iii you can conclude
> f(n) > n for all n, otherwise they are contradictory.
>
> So we know f(1) > 1 for any such function.
>
> Therefore using n = 1 in ii, f(n) must satisfy
>
> f(f(1)) = f(1) + 1. But, there aren't enough "holes"
> open in the range to satisfy this.
>
> i.e.
> f(1) < f(2) <= f(1) + 1 <-- You have at least these
> f(1) < f(3) <= f(1) + 1 <-- two to satisfy.
> and f(3) =/= f(2)
> ...
> f(1) < f(f(1)) = f(1) + 1.
What makes you think that f(3) <= f(1) + 1 ?
Subject: Re: Best way to study maths?
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 31 Dec 1996 04:22:07 -0500
In article <5aa396$k0n$2@nntp.igs.net>, David wrote:
:Peter wrote:
:
:>Hi, everybody.
:
:>Could anyone give me the answer of my subject?
:Hi,
:
:Read, read, read. Practice, practice, practice. ...until *you* know
:it is right. Then read, read read some more. [Really!]
:
:Live long and prosper.
:
Also: If all people intending to study maths thought alike, there might
be an answer to your question. But they don't.
What has helped me:
a pile of scrap paper to work out the details which are often "left to
the reader";
writing a lot of notes in general; sometimes the presentation is to the
author's liking but not to yours, so you can improve on it in this
direction;
seek different sources if the one you're studying from is not well
written for your needs;
it's a definite plus if you find a buddy to study with: compare notes,
ask questions, etc. Often I got closer to the answer just by asking the
question out loud (i.e. realizing what to ask about);
if it is problem solving that gives you a hard time, read Polya's little
book "How to solve it" and look for newer books on this topic;
if abstract definitions or very general theorems are the problem, make up
your own examples or counterexamples, as the situation requires; some
also prefer graphical illustrations;
to help remember things, employ more of your body: sight, as you read the
facts, your muscles as you write them down, hearing, as you read them
back aloud.
don't get frustrated when you don't get the "feeling" for a theory or a
technique instantly; like in learning a language, there is a barrier
that waits to be broken by practice (in languages, it marks the
transition from speaking in separate words to speaking in whole
sentences).
Hope some of it helps, ZVK (Slavek).
Subject: Re: Monty Hall Trap
From: eleong@netcom.com (Eric Leong)
Date: Tue, 31 Dec 1996 08:57:10 GMT
Cheung Koon Tung, Kent (ktcheung@cs.cityu.edu.hk) wrote:
: Hello all,
: I have just finished reading a book, called "For Experts Only". One of
: the chapters in this book discussed about probablities and has caused me
: much confusion. I hope anyone who like bridge and good at mathematics
: would not mind to give some explanations on this problem.
: In the chapter that has the same name as the subject title, the author
: wrote a story to introduce his idea:
: In a game show, one audience is invited to play the game. There are
: three doors. Behind one of these doors, there is a prize of value
: $100,000. The host, Monty Hall, asked the audience to choose a door to
: see whether he wins the prize.
: "No.1", said the audience.
: "Before I open the door, I would like to buy anything behind door 1
: with $20,000.", said Monty Hall.
: The audience answered confidently, "Of course not. My expected value is
: $33,333, why should I accept $20,000?"
: "Before you see what is behind door 1, let use see what is behind door
: 2?", annouce Monty Hall, "right, it is not the great prize."
: Monty Hall continued to urge the audience to sell his rights on door 1.
: "I 'd like to give a last chance, would you sell anything behind door 1
: to me with $40,000?"
: "I accept.", said the audience. The author explained that if the door
: to open is randomly chosen, the probability that door 1 is the great
: prize will be 0.5 and it is better not to accept the offer. However,
: since the door to open is not randomly chosen, the opened door DOES NOT
: AFFECT THE PROBABILITY THAT DOOR 1 IS THE GREAT PRIZE. IT IS STILL 1/3.
: I really don't understand the above argument. Since the book I read was
: a Chinese edition. Maybe I have misunderstanded the original meaning.
: Could anyone give me an explanation to this argument, or tell me the
: original statement in English?
: Thank you very much.
: Rgds,
: Kent.
You could have selected any of two doors which doesn't have the prize so
you have a 2/3 chance of not picking the door with the prize. If that is
the case then you will always win by switching given the extra information
provided by the host. Your decision only loses if you selected the one door
out of the three with the prize originally.
As an example, say the prize is distributed as follows:
Door 1: No prize
Door 2: No prize
Door 3: Prize
There are three equally likely cases:
Case 1: You pick door 1, the host opens door 2
Case 2: You pick door 2, the host opens door 1
Case 3: You pick door 3, the host opens either door 1 or door 2
If you switch, case 1 and case 2 is favorable to you. If you decide to
stay only case 3 is favorable to you.
Eric Leong
Subject: Re: Polynom with prime number values
From: haha@astral.magic.ca (Hans Havermann)
Date: Tue, 31 Dec 1996 00:28:52 -0500
Vladimir Volovich wrote:
> (k+2)*(1-(w*z+h+j-q)^2-((g*k+2*g+k+1)*(h+j)+h-z)^2-(2*n+p+q+z-e)^2-
> (16*(k+1)^3*(k+2)*(n+1)^2+1-f^2)^2-(e^3*(e+2)*(a+1)^2+1-o^2)^2-
> ((a^2-1)*y^2+1-x^2)^2-(16*r^2*y^4*(a^2-1)+1-u^2)^2-
> (((a+u^2*(u^2-a))^2-1)*(n+4*d*y)^2+1-(x+c*u)^2)^2-
> (n+l+v-y)^2-((a^2-1)*l^2+1-m^2)^2-(a*i+k+1-l-i)^2-
> (p+l*(a-n-1)+b*(2*a*n+2*a-n^2-2*n-2)-m)^2-
> (q+y*(a-p-1)+s*(2*a*p+2*a-p^2-2*p-2)-x)^2-
> (z+p*l*(a-p)+t*(2*a*p-p^2-1)-p*m)^2)
I don't have the original but Ribenboim's "Book of Prime Number Records"
mentions it. Comparing it with yours, I've found two variances...
Your second line (above) might be:
(16*(k+1)^3*(k+2)*(n-1)^2+1-f^2)^2-(e^3*(e+2)*(a+1)^2+1-o^2)^2-
i.e., (n-1) instead of (n+1).
Your fourth line (above) might be:
(((a+u^2*(u^2-a))^2-1)*(n+4*d*y)^2+1-(x-c*u)^2)^2-
i.e., (x-c*u) instead of (x+c*u).
--
Nature requires five,
Custom allows seven,
Idleness takes nine,
And wickedness eleven.
