Subject: Re: RE: Differentiation Question
From: Bill Dubuque
Date: 31 Dec 1996 05:42:08 -0500
Rod writes:
>
> ... My original question was ment to ask how would one take the derivative
> with respect to x of the *function* 2x^(sqrt of 5). I realize that the
> power rule applies, but I thaught that the sqrt of 5 was to be
> differentiated without reducing it to decimal form. I suppose that this
> can't be done based on the majority of the responses I recieved.
It would appear that in the power differentiation formula
c c-1
(x )' = c x if c is a constant
you are mistakenly assuming that "constant" means "real number in
decimal representation". In fact, whenever one speaks of a constant
with respect to a derivation such as ( )' = derivative wrt x, it simply
means any expression whose derivative is 0. That this is the appropriate
condition for the above formula to hold can be seen by examining the
general formula for differentiation of an arbitrary exponential, say
h = f^g, where f and g may depend upon x
g g log(f) g g log(f)
(f )' = (e )' since f = e
g log(f) u u
= e (g f'/f + g' log(f)) since (e )' = e u'
g-1 g
= g f f' + g' f log(f)
Note that in the special case g' = 0 the second term vanishes and with
f = x the formula thus simplifies to the usual power differentiation rule
g g-1
(x )' = g x if g' = 0
So in this context the meaning of "g is a constant" is simply
that "g' = 0". It matters not what particular constant is the
value of g, or how it is represented or denoted -- whether an
integer, rational number, algebraic number such as sqrt(5), or
transcendental number such as pi, or even some symbolic constant
parameter c with unspecified value. All that matters is that g
is a constant with respect to the variable of differentiation --
i.e. the value of g remains constant however x varies.
In fact, when one studies differentiation from an algebraic
point of view -- so called "differential algebra" -- constants
are defined to be simply those elements whose derivative
vanishes. In this abstract setting one is working over a ring
or field whose elements are abstract objects, not necessarily
functions, and a derivation is simply a linear map satisfying
the product rule D(a*b) = a*D(b) + D(a)*b. For example
consider the field of univariate rational expressions consisting
of quotients of polynomials P(x)/Q(x) with integer coefficients,
where such expressions are considered as abstract objects,
independent of any possible interpreation as functions (e.g. as
they might be represented in a computer algebra system such as
Macsyma, Maple or Mathematica).
For further info see the beautiful little book by Kaplansky
titled Differential Algebra. This is the field of study underlying
the algorithms employed in computer algebra systems for closed
form symbolic integration in finite terms (for which see e.g.
the textbook "Algorithms for Computer Algebra" by Geddes et. al.).
-Bill Dubuque
Subject: Re: Help: a practical statistics problem
From: girod@niktow.canisius.edu (Don Girod)
Date: 31 Dec 1996 13:59:05 GMT
Chris (cpy_DeleteMe@world.std.com) wrote:
: I have a few large data sets (daily stock prices) and a reliable
: source against which to check the accuracy of the data (the
: newspaper).
:
: Is it possible to sample the data in such a way that I can say with X
: degree of certainty that the data is Y% accurate.
:
: Is there some non-arbitrary way to determine an appropriate sample
: size? Is it better to sample a block of consecutive elements or to
: spread the sample evenly across the set?
:
: This may be too complicated to answer in a post, but I'd appreciate
: any comments that could point me in the right direction, i.e., a
: relevant theorem, the relevant branch of statistics--something to
: search on. Many thanks.
:
: --Chris
:
You have to have a probability model for the situation before you can
decide what you want to do. The simplest interpretation of what you
wrote is that each individual stock price is either right or wrong,
and you want to estimate in some way the proportion of data values
(prices) which are correct. On the other hand, it is possible for
things like prices to be nearly correct. I can't imagine a process
which would generate a few hundred prices in independent ways (that
is, the prices in the newspaper are produced in one way, and your
data is produced independently) and would not generally produce results
which are close to each other if not exact. For example, if you gathered
prices for your data at 2:00 pm and the newspaper quotes the closing price,
the differences would not be large in the vast majority of cases.
Anyway, if what you have is just a right-wrong decision on each price,
all you are asking to do is to estimate the proportion of items in
a population posessing a given characteristic (that of being correct).
You do this by taking a RANDOM sample of the population (your data),
determine the fraction correct in the sample, and use this as an estimate
for the fraction correct in the whole population. The reliability of
this estimate will depend on your sample size in a standard way.
Complications arise if your sample is "large" in relation to the size
of the population. If you have 500 stock prices to sample from, and you
decide to take a sample of size 200, you have a different statistical
problem than if you took a sample of size 20.
You can learn about this in any elementary statistics book. The chapter
will say something like "Estimating proportions" or "Confidence intervals
for a proportion". The statistical thing you want is a confidence
interval; you get to say "The proportion of correct prices in the population
is between .78 and .86, and I am 95% confident that this is a true statement."
The actual methods you use will depend on how you choose your sample,
how big the sample is, and also on how large the proportion of correct
prices is. Probability tools include using the binomial distribution
(relatively small samples), the normal distribution (larger samples,
proportion of correct prices not too large or too small), and the Poisson
distribution.
If you could get this data into electronic form it would be much more
sensible just to determine the exact number of correct values. If
all you have is a piece of paper with a list of numbers on it, even the
process of making a random sample is going to be difficult and tedious,
unless it is quite small. But it depends on what you are trying to do.
If I had a list of 1000 values and I wanted to know about how good it
was, I would tape it into a circle, put it on the floor, and spin the
bottle 20 times inside the circle, check the 20 selected values, and
use the binomial distribution. If I didn't find any wrong values,
the probability that more than .01 of the values are bad is less than
5%, so I would say "At least .99 of the values are ok and I am 95%
sure I am right".
Hope this helps. I might be able to say something more useful if I know
what you are actually trying to do and why.
Subject: Re: Best way to study maths?
From: hrubin@b.stat.purdue.edu (Herman Rubin)
Date: 31 Dec 1996 09:53:08 -0500
In article <5a8fb7$jm0@metro.ucc.su.OZ.AU>,
Peter wrote:
>Hi, everybody.
>Could anyone give me the answer of my subject?
One of the real problems with giving you an answer is that the
foundations are only studied by those who are already using
mathematics. The entire pedagogical sequence is from the
non-basic to the more basic.
To some extent, this must be so. If you really want to start
with the basics, start with a book on mathematical logic.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (317)494-6054 FAX: (317)494-0558
Subject: Re: Hypergeometric Probability Function
From: hrubin@b.stat.purdue.edu (Herman Rubin)
Date: 31 Dec 1996 10:20:36 -0500
In article <5a94rp$cvr5@chinook.generation.net>,
Gilles Lehoux wrote:
>I must program to evaluate the hypergeometric probability
>distribution,
>I have obtained from NETLIB and other sources some relevant source
>code examples. These examples are for the "U-confluent Hypergeometric"
>or the "Confluent Hypergeometric".
>What is the meaning of this "confluent" qualifier?
>Are "Hypergeometric" and "Confluent Hypergeometric" equivalent
>functions?
>How can I go from the "Hypergeometric" input variables to the
>"Confluent Hypergeometric" input variables?
>Note:
>The confluent hypergeometric function is the solution to the
>differential equation: zf"(z) + (a-z)f'(z) -
>bf(z) = 0
>The Hypergeometric function is:
> n m
>C * C
> k r-k
>------------- = Hypergeometric(n,m,k)
> n+m
>C
> r
This is not the hypergeometric function, but the hypergeometric
distribution. It received this name because its probability
generating function is P(0)F(-n, -r; m-r+1; x), where F is the
hypergeometric function as generally used in mathematics.
The general hypergeometric function F(a,b; c; x) gives rise to
a distribution if the terms are all non-negative and F(a,b; c; 1)
is finite. For the confluent hypergeometric function, the series
always converges.
Many distributions are named in this way, or for the use of
their probabilities or densities in connection with useful
functions. This includes the binomial, negative binomial,
logarithmic series, and Zeta distribuions for the integer case,
and the exponential, uniform, Beta, and Gamma distributions in
the continuous case.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu Phone: (317)494-6054 FAX: (317)494-0558
Subject: Re: x^y+y^x>=1
From: "William E. Sabin"
Date: Tue, 31 Dec 1996 09:40:52 -0800
sfly wrote:
>
> who can poof that x^y+y^x>=1 with x,y>=0?
> it's seem to be easy,but not..
Let A = x^y + y^x
If y=0 then A = x^0 + 0^x = x^0 + 0
limit, as x approaches 0+, of x^0 = 1
If x=0 then A = 0^y + y^0 = 0 + y^0
limit, as y approaches 0+, of y^0 = 1
If x=0+ and y=0+ then A > 1
This covers all possibilities for real x, y if they are not both
exactly zero. If x any are both exactly zero then the hypothesis is not
correct.
Bill W0IYH
Subject: Re: circle algorithm
From: martind@veblen.acns.carleton.edu (Daniel T. Martin)
Date: 31 Dec 1996 15:52:30 GMT
In article <01bbf5fd$34038ec0$8a40aace@default>, "Robert E Sawyer" writes:
>
>In the orthogonal LS case (OLS), the full minimization involves finding the
>center of the best fitting circle, as well as its radius, but I think it's
>clear that the center must turn out to be the point (xbar,ybar), i.e. the
>"center of mass" of the data points. Therefore, transform the data from
>(x_i,y_i) to (X_i,Y_i) with X_i=x_i-xbar, Y_i=y_i-ybar, so these (X_i,Y_i)
>are centered on (0,0) and only the radius needs to be found.
>
No. Think of the special case of the three data points (1,0), (-1,0), and
(0,1). The circle which minimizes the OLS case is then clearly the unit
circle centered at the origin, yet your center of mass argument would center
the circle around (0, 0.5), resulting in a worse fit.
However, your argument about which radius to use, given a certain center, I
find sound; that is:
>radius is the mean radius Rbar of the "centered data" coordinates.
So, given that the optimal radius is just the average of the distances to
each point from the center, one should then be able to derive a formula (or
at least an approximation algorithm) for the location of the center itself.
I have an idea how an approximation algortihm might work, but I'm not
certain yet whether or not my idea generates an algorithm that always
converges, so I'll hold off on posting it. Deriving a formula seems
possible, though the algebra seems to get a little messy for my taste.
<_|_> DANIEL
Subject: Re: x^y+y^x>=1
From: "E.M."
Date: Tue, 31 Dec 1996 10:29:05 -0500
sfly wrote:
>
> who can poof that x^y+y^x>=1 with x,y>=0?
> it's seem to be easy,but not..
Why, for at least one of x, y >1 the theorem is obviuos. When
both x,y<1 we have A = x^y + y^x = (1/p)^(1/q) + (1/q)^(1/p),
where p = 1/x, q=1/y. Then A^(pq) > 1/p + 1/q > 1., Q.E.D.
Subject: Re: x^y+y^x>=1
From: b841039@math.ntu.edu.tw (sfly)
Date: Tue, 31 Dec 1996 04:38:22 GMT
"William E. Sabin" wrote:
>sfly wrote:
>>
>> who can poof that x^y+y^x>=1 with x,y>=0?
>> it's seem to be easy,but not..
>Let A = x^y + y^x
>If y=0 then A = x^0 + 0^x = x^0 + 0
>limit, as x approaches 0+, of x^0 = 1
>If x=0 then A = 0^y + y^0 = 0 + y^0
>limit, as y approaches 0+, of y^0 = 1
>If x=0+ and y=0+ then A > 1
~~~~~~~~~~~~~~~~~~~~
????why?
please proof it!
>This covers all possibilities for real x, y if they are not both
>exactly zero. If x any are both exactly zero then the hypothesis is not
>correct.
>Bill W0IYH
Subject: Re: Ugly Mathematics?
From: marnix@worldonline.nl (Marnix Klooster)
Date: Tue, 31 Dec 1996 18:02:40 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."
And I agree. As E.W. Dijkstra once put it: "Beauty is our
business."
> So, I ask you professional mathematicians, what do you
> consider a beatiful proof to a theorem? What are your
> parameters? Can you describe?
Not really. But I'll try.
One book you might like to read is A.J.M. van Gasteren, "On
the Shape of Mathematical Arguments", Springer 1990, Lecture
Notes in Computer Science 445. This book is on the design and
presentation of proofs. It consists of a series of case studies,
where often clear and concise proofs are given as alternatives
for wieldy and/or vague proofs from the literature. The second
half tries to distill some general principles from these case
studies.
As an example, in Chapter 5 she presents two proofs for the
theorem that "[a] natural number is a multiple of 3 iff the sum
of the digits in its decimal representation is a multiple of 3."
The proof given by Arbib, Kfoury, and Moll in "A Basis for
Theoretical Computer Science" -- they consider it exemplary -- is
quite complicated, making many case distinctions. Conversely,
here is Van Gasteren's proof, which in fact proves r(n) = n +
multiple of 9; r(n) is the sum of the digits of n in its decimal
representation.
"We adopt the decimal positional system, dropping, however
the constraint that each digit be less than 10. We start
with the decimal number having n as its only significant
``digit'' --and, hence, having digit sum n--, and by
repeated carries transform it into n's standard decimal
representation --which has digit sum r(n)--. A carry
consists in subtracting 10 from a digit >= 10 and adding,
or creating, 1 at the position to the immediate left.
Each carry, hence, changes the digit sum by -9. This
process terminates because the digit sum is at least 0
and decreases at each carry. Hence r(n) = n + multiple
of 9."
This, IMHO, is an example of an elegant proof. Why? For one
thing, because it is so much simpler than the original proof. It
is about the simplest proof possible.
I agree with Van Gasteren that the following things are among
those that make a proof more elegant:
* It is presented in a readable way, so that the reader can
easily understand and check it.
* It treats separate points separately.
* Neither too much nor too little terms (names, abbreviations,
labels) are used.
To be fair, Van Gasteren's book is not about beauty or
elegance, but about the design and presentation of proofs. It is
my experience, however, that following her advice results in more
elegant proofs. On several occasions it allowed me to rewrite
proofs from literature in a much more readable way. A good side
effect of spending time on proof presentation is that it becomes
more obvious what the crucial points in a proof are. I find that
highlighting such points is another thing that makes a proof
better.
> Sometimes I think a proof is amazing and the author of
> the text I'm reading says that "there are much nicer
> proofs to this theorem" and sometimes the author says:
> "See how this is beautiful" to something that seems
> trivial to me.
But why do you consider a proof amazing? Because one of its
steps surprises you? Or because you say to yourself "this must
be the simplest proof possible for this theorem"? Or because the
theorem proved is beautiful? Or something else altogether?
Personally, I think I don't like a proof if it is more
complex than necessary.
Also, generally I prefer a proof which relies on formula
manipulation as opposed to one described using mostly text. For
example, proving that (Ex:: (Ay:: P)) => (Ay:: (Ex:: P)) can be
done as follows:
(Ex:: (Ay:: P))
== "(Ex:: (Ay:: P)) does not contain y free"
(Ay:: (Ex:: (Ay:: P)))
=> "(Ay:: P) => P"
(Ay:: (Ex:: P))
(Here (E_:: _) and (A_:: _) stand for existential and universal
quantification, respectively, == is equivalence, and => is
implication. The text within "..." are hints.) A mostly-text
proof of this theorem is long-winded and incomprehensible, in my
experience.
Note how in a certain sense the above proof is trivial.
Still I consider it elegant due to its simplicity.
This formal style of proof is the basis of David Gries and
Fred B. Schneider, "A Logical Approach to Discrete Math",
Springer 1993. There is information on this on the web too, at
http://www.cs.cornell.edu/Info/People/gries/Logic/Introduction.html
Two other titles: E.W. Dijkstra and C.S. Scholten, "Predicate
Calculus and Program Semantics", Springer 1990; and (I think)
"Beauty is our business -- a birthday tribute to Edsger W.
Dijkstra".
> So, since different people have different tastes, how do
> you agree that a proof is elegant?
I don't know. All I know is that often we just agree, at
least about the beauty of the basic structure of a proof. There
may very well be disagreement on its presentation. But that is
just my limited experience.
> Just curious, Roger...
And I'm curious, too. Is there more literature on this,
perhaps?
Groetjes,
<><
Marnix
--
Marnix Klooster | If you reply to this post,
marnix@worldonline.nl | please send me an e-mail copy.
Subject: Re: The Power Of Numbered Words Revealed.
From: bob jacobs
Date: Tue, 31 Dec 1996 11:47:52 -0800
William Long wrote:
>
> David Kaufman (davk@netcom.com) wrote:
> : A numbered word is usually a number next to the left of
> : a word (or abbreviation). The number tells how many, while
> : the word reveals what items are under consideration. For
> : example, "5 pounds" (5 lb) or "4 fruit" are numbered words.
> --
> Ah, a definition of "numbered word." Thanks. But, the sentence
> "A numbered word is usually a number ..." seems to me to say that the
> numbered word is the NUMBER. Then the examples seem to say that the
> numbered word consists of BOTH the number and the word. Shouldn't a
> numbered word be a special kind of WORD? (Sorry to be so picky, but I
> really feel the need for a clear precise definition.)
>
> William Long
> a018379t@bcfreenet.seflin.lib.fl.us
I came in on this thread just today and so I don't have the background
posts other than this one. I'm not sure that I understand, but...
I like the phrase, "numbered word" because it emphasizes the word and
reduces the number to the status of adjective. My chemistry students
often will solve problems without units and are indignant when I do not
give them credit for "correct answers" even though they only get the
adjectives right and not the "nouns." An understanding of the process of
handling units, dimensional analysis, is so important to mathematical
reasoning and a respect for process so important to scienctific studies
that I have now adopted the phrase, "numbered word" as an intimate part
of my science-teaching vocabulary. Have I got it right or am I missing
the point?
Thanks.
Bob
Subject: Re: here's a program to calculate pi
From: jasonp@Glue.umd.edu (Jason Stratos Papadopoulos)
Date: 31 Dec 1996 08:24:46 GMT
Simon Read (nfic@internetmci.com) wrote:
: jasonp@Glue.umd.edu (Jason Stratos Papadopoulos) wrote:
: >'
: >' 1 1 1 1 1 2 3 4 5
: >' 1 - - + - - - * - (1 + - (1 + -- (1 + --(1 + --(1 + --)))))
: >' 3 5 2 7 9 11 13 15 17
: >'
: >'Gregory's terms take less time to compute than Euler's, and overall
: >'the error drops by a factor of 3 per term instead of 2 per term.
: Starting with 1/9 we have the ratios of successive terms as n/(2n+7)
: which (for n=1, 2, 3, 4, 5, ...) gives us
: 1/9, 2/11, 3/13, 4/15, 5/17, .... as you quote above.
: The +7 eventually becomes less noticeable and each term approaches the
: limit of 1/2 x (previous term)
: which means the error after each term is half what it
: would have been without that term.
: I am unclear when you say
: "the overall error drops by a factor of 3 per term"
: if you mean the series you quote or some other series.
That's the strange part! The standard Euler acceleration starts with
a sequence of partial sums
S0
S1
S2
S3
...
and forms more columns; each entry is the average of the entry to its
left and the entry above and to the left:
S00
\
S01 - S10
\ \
S02 - S11 - S20
\ \ \
S03 - S12 - S21 - S30
The standard Euler series for pi/4 is just the sequence of diagonal
elements. What's strange (and what the program takes advantage of) is that
if you look at the error from pi/4 along a given row, the minimum error
occurs 2/3 of the way out and NOT all the way out. This corresponds to
adding up a few terms of the original series (going straight down the
first column) and then accelerating (forming a smaller triangle whose
diagonal is the sequence of approximations actually used)
Not only are the errors smaller 2/3 of the way out on any row, but they
decrease much faster than those of the diagonals...in effect by a factor
of 3 per term instead of 2 per term! The reason has to do, I
think, with linear programming (I'm not at all sure why, it seems like
magic) but check the code for yourself: the number of terms the program
adds together should not be enough to calculate pi to all the decimal
places shown, but by some tiny miracle it works and is the major
time-saver in the code.
For more rigor and some references see "The Lag-Averaged Euler Method"
(or something close), Applied Mathematics and Computation vol 72 #2,
Sept 1995
jasonp
Subject: Re: REAL NUMBERS (STANDARD)
From: abian@iastate.edu (Alexander Abian)
Date: 31 Dec 1996 17:39:28 GMT
In article <851972914.785@dejanews.com>, wrote:
>In article <5a72m9$lmn@cymbal.aix.calpoly.edu>,
> jplee@cymbal.aix.calpoly.edu (Jason Lee) wrote:
>>
>> And then abian@iastate.edu (Alexander Abian) quote:
>> >
>> > DEFINITION (Abian). A real number is a picture with a decimal point and
>> > with finitely many digits to the left of the decimal point and with
>> >infinitely many digits to the right of the decimal point (and the
>> >picture is preceded with + or minus sign ( + is usually omitted).
>>
>>
>> JLee
>
>
> Actually it's a perfectly reasonable definition of "real number".
>A tad ambiguous, (and it needs a little addendum to handle the case where
>one real number has two decimal expansions)
>..............
>David Ullrich
Abian answers:
No, there is no ambiguity, because (please see my previous posting)
After proving the fundamental Theorem of Completeness, I introduce
subtraction, and I define a = b iff a - b = 0 in which case clearly, say,
3.4699999999..... = 3.4700000... and 0.99999..... = 1.0000000....
The entire power of the above Definition is that it shows how to
pass from kindergarten arithmetic of "finite pictures" such as
3.65, 787.9876, etc to the PH.D arithmetic of "infinite pictures". The
transition from kindergarten "Finite arithmetic" to PH.D "Infinite
arithmetic" is performed via considering the SET of TRUNCATIONS (of
addition, multiplication,....etc) and assigning the LUB or GLB of
those sets (with proper precautions to handle negative numbers, etc) of
truncations (as explained in my previous postings).
Obviously in the above DEFINITION
..... and with infinitely many digits to the right of the decimal point...
is an abbreviation of:
....a (countable infinite) sequence of ordinal type omega of digits
0 to 9 to the right of the decimal point "
Alexander Abian
--
--------------------------------------------------------------------------
ABIAN MASS-TIME EQUIVALENCE FORMULA m = Mo(1-exp(T/(kT-Mo))) Abian units.
ALTER EARTH'S ORBIT AND TILT - STOP GLOBAL DISASTERS AND EPIDEMICS
ALTER THE SOLAR SYSTEM. REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT
TO CREATE A BORN AGAIN EARTH (1990)
Subject: Re: The Inferiority of Cydonian "Math" in Comparison to the Earthly Science-Art
From: caj@sherlock.math.niu.edu (Xcott Craver)
Date: 31 Dec 1996 18:55:51 GMT
Oh, God. It seems we have a KotM Clash of the Titans in
the works. Brace yerselves!
Actually, there might be good in this after all: Jiri, why
don't you provide a concise debunking [it's not that hard, just a
little time-consuming] of the TETET stuff, and Angel, you can debunk
Jiri's "Science-Art" [again, not hard].
Good luck one and all....
Caj
Subject: Re: EXTRAORDINARY PI
From: caj@sherlock.math.niu.edu (Xcott Craver)
Date: 31 Dec 1996 18:40:25 GMT
*SIGH*
BLStansbury wrote:
>>
>> No, not quite. If we want to write pi down in some
>>positional notation (i.e., decimal) or use it in computation,
>>THEN we must approximate it.
>Oh, so we do not approxiamte it at other times. I did not realize
>this.
Correct. We only approximate PI when doing real-world
computation with it. In mathematics, we use its exact value.
>> It is a lie that the best one can do is approximate.
>>Mathematics is exact
>I understand. Could you please send me the exact value of pi.
Sure. How about 6*arcsin(1/2)? See, Mr. Stansbury, I
think you are not apprehending the distinction between the VALUE
of a constant and its decimal representation. Decimal
representations are things we write down, and do arithmetic with.
But while PI can be written as "approximately 3.14159...," that
is not the value of PI, any more than the name "Mr. Stansbury"
*is* Mr. Stansbury.
Imagine that, one day, I decide to change my last name
to an infinitely long string, whose letters form no recognizable
pattern. You then have me, Scott Craverthreepointonefouronefive...
and the actual name "Scott Craverthreepointonefouronefive...."
You'll never write my name down completely [lots of fun at
graduation!] but that's just a name. If you respond to "I
bumped into Scott Craverpi the other day" with, "NO YOU DIDN'T!!
YOU JUST BUMPED INTO AN APPROXIMATION OF Scott Craverpi!!", then
you are probably not clear on the use-mention distinction,
between the name of something and its value. This is what you
seem to be displaying here.
We will never be able to do base-10 arithmetic with the
exact value of PI, because we would have to write PI in base-10,
which would require an infinite number of digits. But mathematics
IS NOT COMPUTATION. We use the exact value of PI in mathematics.
We just don't restrict ourselves to writing answers entirely in
decimal, get it? A circle of radius 2 has area 4pi, exactly.
>1/3 * 300 = 100 and 100/pi * pi = 100
>1/pi * 3.14159265359... does not = 1
I assume that by "3.14159265359..." you mean only a finite
number of digits, yes? Sure, then. But what does that show?
1/(1/3) * 0.33333333333 does not = 1 either. Again, just because
we can't write it in base-10 doesn't mean we can't use its exact
value. Similarly, just because we can't draw a perfect circle
doesn't mean we can't use perfect circles in mathematics.
>> Um, no. What we can do in mathematics does not depend
>>on how many digits of PI a machine can crank out, or how sharp a
>>pencil we can build.
>?
Do you want me to elaborate on this?
>> The squaring a circle is impossible even
>>with perfectly accurate tools [meaning, of course, an unmarked
>>straightedge and compass].
>Or by any other means.
Not true. Squaring the circle *is* possible if we
slightly modify those perfectly accurate tools. With an umarked
straightedge and collapsing compass, it cannot be done.
>BLS
,oooooooo8 o ooooo@math.niu.edu -- http://www.math.niu.edu/~caj/
o888' `88 ,888. 888
888 ,8'`88. 888 "This year's Summer fasions are simple yet
888o. ,oo ,8oooo88. 888 vibrant, as I will prove using the following
`888oooo88 o88o o888o 888 lemma." -Cindy Crawford, _Gauss_of_Style_
____________________8o888'_________________________________________________
Subject: Re: 0^0
From: The Universal Heretic
Date: Tue, 31 Dec 1996 11:43:59 -0800
Michael Cohen has correctly shown that
lim (x->0+) x^x = 0.
However, the following must be noted (for it is the reason that
0^0 is undefined generally).
Let f(x,y)= y^x.
Then f(x,x) = x^x and we have already shown that as x=>0+, we arrive
at 1 for the value of 0^0. This is equivalent to approaching the
origin along the line y=x.
However, approaching along the line y=0, we have
f(x,0)= 0^x = 0 for all x.
Thus, lim (x->0) 0 = 0.
Therefore, the value of 0^0 is undefined generally.
Frank
Subject: Re: Diophanto Eq
From: rusin@olympus.math.niu.edu (Dave Rusin)
Date: 31 Dec 1996 19:26:49 GMT
sfly (b841039@math.ntu.edu.tw) wrote:
> how to solve the eq.......x^4-d*y^2=1 where d=41*73
> and , is ther there rational numbers a,b,c such that
> a^2-b^2=b^2-c^2=d (d=41*73)
There are no nontrivial solutions in either case, although this
conclusion is harder to come by than is typical in these questions.
Hauke Reddmann wrote:
>Elliptic Curves (move x=1,y=0 to infinity)
>a+b=d/l
>a-b=l
>b+c=d/m
>b-c=m
>d/l-l=d/m+m
>Which is again an elliptic curve after a bit
>variable rewriting.
This is the way to go.
Let's look at the second system of equations first. As Hauke noted,
the equations may be rewritten d/l-l=d/m+m ;
upon multiplying by l*m we obtain a cubic equation which may be
expressed as an elliptic curve in standard form. Indeed, using the
invertible transformation {l=d*(X-d)/Y, m=X*(X-d)/Y} the equation becomes
2
Y = X (X - d) (X + d)
This equation clearly has solutions with Y=0 (namely X=0,
X= +- d, and the "point at infinity"). But these don't correspond to
solutions (l,m) of Hauke's pair of equations, nor to solutions
(a,b,c) of the original pair.
Now, the points on an elliptic curve form a finitely-generated abelian
group; these four points are merely the torsion subgroup. What's really
interesting is the question of whether this group has positive rank.
There are a number of algorithms which enable us to estimate the rank
of a general elliptic curve over the rational field, but these don't
give sufficiently strong upper bounds for our particular curve.
However, curves of the form described above are precisely those
involved in Tunnell's resolution (sort of) of the "congruent number
problem". Tunnell proves that there is a more-or-less complete
algorithm which can tell us at least whether the rank of the curve is
positive or not.
Here's the test (for d square-free). Count the number of integer solutions
(x,y,z) to the equation
2x^2 + y^2 + 8z^2 = d.
Let c1 be the number of such solutions. Let c2 be the number of
such solutions with z even. Then if the elliptic curve y^2=x(x^2-d^2)
has positive rank, we must have c1 = 2 c2. (The converse is probably true,too).
Unless I miscounted, there are for d=43*71 exactly c1=96 solutions but
only c2=40 with z even, so by Tunnell's theorem, the rank of this elliptic
curve is zero. The only points are the torsion points, and so there are no
solutions to the original equations.
Now we can address sfly's other equation. Again, Hauke's suggestion is fine:
using the invertible transformation {y=4*Y/(X-2*d)^2, x=(X+2*d)/(X-2*d)}
the equation becomes
2 2 2
Y = X (X + (2 d) )
While this curve is not of the type covered directly by Tunnell's theorem,
it happens to be isogenous to the first curve we considered. (That is,
there are homomorphisms E -> E' and E' -> E whose composites are the
doubling maps on each curve). In particular, these two elliptic curves
have the same rank, which we have shown to be zero. Thus again the only
points on this curve are the torsion points, which can be checked to be
only the point (X,Y)=(0,0) and the point at infinity (unless d=1).
In terms of the original coordinates, this means the only rational solutions
are (x,y) = (+-1, 0).
Of course, if you start with a different value of d, then the situation
can be quite different. I'm glad you didn't suggest d=157.
dave
Subject: Re: EXTRAORDINARY PI
From: juanvp@impsat1.com.ar (JuanVP)
Date: Tue, 31 Dec 1996 19:51:26 GMT
On Tue, 31 Dec 1996 19:43:50 GMT, bstan@datasync.com (BLStansbury)
wrote:
>If I ask 100 of my students to give me the area of a pi x pi square,
>and each one of them gives me a right but different answer, which
>value(s) should I accept? All of them right? And none of their answers
>would be exactly the same.
Well-defined maths problems like this have only 1 answer. You cannot
have "right but different answers", if that were the case no relation
containing products would be a function(!!). The only right answer is
a pi^2 area, but if you consider that the answer to this problem is
right if it has for instance 3 significant digits or more, then any
number in the interval [9.869, 9.870] would be considered correct, but
bear in mind that every number in the interval but 1 (Pi^2) is
only an approximation of the answer with an error less than 10^-3.
Juan
Subject: Re: Adv Calc Question: Bounded Sets
From: mlerma@pythagoras.ma.utexas.edu (Miguel Lerma)
Date: 31 Dec 1996 20:15:01 GMT
Jjo31420 (jjo31420@aol.com) wrote:
[...]
> Must all bounded non empty sets have a non empty boundary?
In what topological space?
> The problem I'm having is with the boundary part. How is a boundary
> defined for an arbitrary set? Obviously, a closed set has a boundary
> (the inverse of the set is open), but what conclusions can I draw
> from a bounded, non empty set?
There are several (equivalent) definitions of "boundary" of a
subset S of a topological space (X,T). One of them says that x
is in the boundary of S iff every neighborhood of x intersects
both S and X\S.
The answer to the original question depends on the topological
space you are working with. Assuming that it is R (the real line)
with its usual topology, a hint could be to look at sup S, which
must exist since S is non-empty and bounded. If you are in R^n,
try to find some similar idea. However, if you are in Q (rational
numbers) the answer is negative...
Miguel A. Lerma
Subject: Re: Adv Calc Question: Bounded Sets
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Date: 31 Dec 1996 14:57:49 -0500
In article <19961231142400.JAA09990@ladder01.news.aol.com>,
Jjo31420 wrote:
>
> Hello
>
> I'm taking an advanced calc class and cannot seem to get an angle on
>the following question:
>
> Must all bounded non empty sets have a non empty boundary?
>
>
> The problem I'm having is with the boundary part. How is a boundary
>defined for an arbitrary set? Obviously, a closed set has a boundary
>(the inverse of the set is open), but what conclusions can I draw
>from a bounded, non empty set?
>
> Thanks!
> Joel O.
> jjo31420@aol.com
You are right in one respect: to answer the question, you need a
definition of the boundary. (Scan the textbook - the definition should be
there.) One possible definition: it is the set of boundary points, and
x is a boundary point of S if every neighborhood of x contains points
both from S and from the complement of S.
Equivalently in metric spaces,
x is a boundary point of S if there are two sequences, both convergent
to x, one consisting of points from S and the other consisting of points
from the complement of S.
You see that the boundary depends on the surrounding space (in which the
complements are taken), and boundedness depends on how we define the
distance in the surrounding space. To give anyone a chance to answer the
question, this information must be specified.
The introductory scenario is, I presume, that the surrounding space is R,
the set of all real numbers, with abs(x-y) as the distance between x and
y. In that case, given a non-empty bounded S, consider b = sup S. You do
need the concept of the supremum, or a concept equally powerful (to
formulate the completeness of R), to answer the question.
In spaces R^n with Euclidean distance, take sequences of points where the
first coordinate approaches its supremum, and subsequences of those.
In other spaces, the situation may be a jungle. If the surrounding space
is Q (the rationals), for example, then there are non-empty bounded sets
without a boundary: take the set of all rationals whose square is less
than 2. It is a non-empty bounded set which is both open and closed, and
has no boundary.
Hope it helps, ZVK (Slavek).
Subject: Re: circle algorithm
From: Rick Decker
Date: Tue, 31 Dec 1996 15:05:37 +0000
Daniel T. Martin wrote:
>
> In article <01bbf5fd$34038ec0$8a40aace@default>, "Robert E Sawyer" writes:
[...]
> >I think it's
> >clear that the center must turn out to be the point (xbar,ybar), i.e. the
> >"center of mass" of the data points.
[...]
> >
> No. Think of the special case of the three data points (1,0), (-1,0), and
> (0,1). The circle which minimizes the OLS case is then clearly the unit
> circle centered at the origin, yet your center of mass argument would center
> the circle around (0, 0.5), resulting in a worse fit.
^^^
You mean 1/3 here, though it doesn't affect your argument.
> However, your argument about which radius to use, given a certain center, I
> find sound; that is:
> >radius is the mean radius Rbar of the "centered data" coordinates.
>
Nope. Consider the three points (-1, 0), (0, 0), (1, 0). In this case
we have Rbar = 2/3. If we put the center at (c, 0) and w.l.o.g. c >= 0,
we find that the c which minimizes the sum of (the squares of) the
orthogonal distances is c = 2/9. This gives a total orthogonal distance
of 14/27. However, if we do the same calculation and minimize over both
c and r (the radius), we find that the minimal orthogonal distance
occurs when c = 1/4 and r = 3/4, giving a distance of 1/2, which is less
than
the 14/27 we found using r = Rbar.
In fact, the symmetry of this instance is misleading--you can make the
orthogonal distance as close to zero as you wish in this case. Put the
center on the y-axis and fit a circle through (-1, 0) and (1, 0). Those
two points will then contribute zero to the orthogonal distance and you
can move the center (adjusting the radius appropriately) away from the
origin (upwards, say), to make the lower part of the circle arbitrarily
close to the remaining point (0, 0).
Regards,
Rick
-----------------------------------------------------
Rick Decker rdecker@hamilton.edu
Department of Comp. Sci. 315-859-4785
Hamilton College
Clinton, NY 13323 = != == (!)
-----------------------------------------------------
