mm-371
===
Subject: Lebesgue number, Lie algebras question
Let G be a connec matrix lie group, and let A be a matrix in
G. Thus,
we
can find a continuous path A(t) in G with A(0) = I and A(0) =
A.
Let V be a neighborhood of I so that every element of V is the
exponential
of an element of g, the Lie algebra of G. (We can do this by a
theorem I
state at the bottom of this message)
Then how can we show that using the compactness of the
interval [0,1], we
can pick a sequence of numbers t_0, ..., t_m with 0 = t_0 <
t_1 < ... < t_m
= 1 such that
((A_{t_{k-1}})^(-1))(A_{t_k}) is in V? (sorry for the excessive
parenthesis!)
(I don't have enough experience with the Lebesgue number lemma
to think of
a
way to prove this, but I assume that this is used)
Theorem : If G is a matrix Lie group with Lie algebra g, then
there exists
a
neighborhood U of 0 in g and a neighborhood V of I in G such
that the
exponential mapping takes U homeomorphically onto V.
===
Subject: Binary Quadratic Forms
I'm looking for primes of the form
a^2 - ab + b^2, where b is congruent to 0 (mod 3) and a is not
congruent to
0(mod 3)
Anyone know if there's an infinite amount of these?
(Without using Dirichlet's theorem...which I don't think you
can use
anyway,
with the b^2 in there...)
If so, a book or paper on the subject would be helpful.
GREG
===
Subject: Re: Binary Quadratic Forms
There is a large theory about primes representable by binary
quadratic
forms, especially positive definite ones such as the one you're
looking at, which is associa with the field Q(sqrt(-3)).
A nice book with lots of material on this subject is
Primes of the Form x^2 + ny^2: Fermat, Class Field Theory, and
Complex
Multiplication by David A. Cox
You're quadratic form isn't quite of this form, since it's
(a - b/2)^2 + 3 (b/2)^2,
but I suspect that Cox's book will have a lot of relevant
stuff in it.
Joe Silverman
> I'm looking for primes of the form
> a^2 - ab + b^2, where b is congruent to 0 (mod 3) and a is
not congruent
to
> 0(mod 3)
> Anyone know if there's an infinite amount of these?
> (Without using Dirichlet's theorem...which I don't think you
can use
anyway,
> with the b^2 in there...)
> If so, a book or paper on the subject would be helpful.
 GREG
===
Subject: Re: Binary Quadratic Forms
> I'm looking for primes of the form
> a^2 - ab + b^2, where b is congruent to 0 (mod 3) and a is
not congruent
to
> 0(mod 3)
> Anyone know if there's an infinite amount of these?
> (Without using Dirichlet's theorem...which I don't think you
can use
anyway,
> with the b^2 in there...)
> If so, a book or paper on the subject would be helpful.
 GREG
If I haven't made a mistake they appear to be the primes of
the form 6n+1.
That is, my computations of these primes produced a sequence
of primes
agreeing with the following sequence in the OEIS: I checked
that the two
sequences of primes (yours and the one below) agree up to
around 3000
terms.
So there may be a theorem here. But I don't know about such
matters. You
might inquire at the number theory mailing list:
http://listserv.nodak.edu/archives/nmbrthry.html
ID Number: A002476 (Formerly M4344 and N1819)
URL: http://www.research.att.com/projects/OEIS?Anum=A002476
Sequence:
7,13,19,31,37,43,61,67,73,79,97,103,109,127,139,151,157,163,
181,193,199,211,223,229,241,271,277,283,307,313,331,337,349,
367,373,379,397,409,421,433,439,457,463,487,499,523,541,547,
571,577,601,607,613,619
Name: Primes of form 6n + 1.
Comments: -3 is a quadratic residue mod a prime p iff p is in
this
sequence.
Primes p dividing sum(k=0,p,C(2k,k)) -3 = A006134(p)-3 - Benoit
Primes p such that tau(p)==2 (mod 3) where tau(x) is the
Ramanujan tau
function (cf. A000594). - Benoit Cloitre
References M. Abramowitz and I. A. Stegun, eds., Handbook of
Mathematical
Functions, National Bureau of Standards Applied Math. Series
55, 1964
(and various reprintings), p. 870.
K. G. Reuschle, Tafeln Complexer Primzahlen. K{o}nigl.
Akademie
der Wissenschaften, Berlin, 1875, p. 1.
Links: C. Banderier, Calcul de (-3/p)
Maple: a:=[ ]: for n from 1 to 400 do if isprime(6*n+1) then
a:=[
op(a),
n ]; fi;
od: A002476:=n->a[n];
See also: For values of n see A024899. Primes of form 3n-1
give A003627.
Sequence in context: A088513 A004611 A038590 this_sequence
A040079
A038160 A040038
Adjacent sequences: A002473 A002474 A002475 this_sequence
A002477
A002478 A002479
These are the primes arising in A024892, A024899, A034936;
A091178
gives prime index.
Clark
Subject: Re: A mathematical proof of the Cantors goof?
===
>Thanks for you complete refutation. However in the whole
refutation
>I only agree with you in two things:
>1) Effectively, the proper word is addition and not addiction
>2) Proof 3 is clearly a Nicolas' goof
>About Proof 1, I have never seen to someone to complicate
things in a
>so unnecessary way.
Nonsense. You claimed to have a bijection between N and the
set of
terminating decimals in [0,1). You did not. I merely supplied a
bijection such as you required for the rest of your proof. I
also poin out that your proof does not in fact prove that
either
[0,1) or R is uncountable. You need a more rigourous proof
than you
supplied.
>You just have to carry out a simple combinatory,
>as the positional systems of numeration do,
I reiterate: you did not have a bijection between N and the
terminating
decimals between [0,1). I supplied a bijection. With regard to
your
attempt to describe your nonexistent bijection, you had an
initial
bijection between {0,1,...,9} and {0,0.1,...,0.9}, given by 0
<-> 0,
1 <-> 0.1, 2 <-> 0.2, ..., 9 <-> 0.9. You then had a bijection
between
{0,1,2,...,99} and {0,0.01,0.02,...,0.99}, given by 0 <-> 0, 1
<-> 0.01,
..., 99 <-> 0.99. The next bijection between {0,1,2,...,999}
and
{0,0.001,0.002,...,0.999} would be given by 0 <-> 0, 1 <->
0.001,
2 <-> 0.002, ..., 999 <-> 0.999. And it is obvious what the
other
bijections in your sequence of bijections look like. The first
thing to
notice about your bijections is that they are
order-preserving, so it
looks like you expect your limiting bijection between N and
the set of
terminating decimals in [0,1) to be order-preserving. It is
very simple
to show that no bijection between N and the set of terminating
decimals
in [0,1) can be order-preserving, so that your limiting
bijection cannot
exist. Specifically, N has a successor function, so that there
exist
pairs of distinct elements of N with no intervening element of
N, i.e. the
ordering of N is not dense. On the other hand, the ordering of
the set of
terminating decimals in [0,1) is dense, and so since denseness
is
preserved by order-preserving bijections, no such
order-preserving
bijection exists. As a result, ALL bijections between N and
the set of
terminating decimals in [0,1) are non-order-preserving.
Alternatively, 1 corresponds to 0.1 in your first bijection,
to 0.01 in
your second bijection, and in subsequent bijections, it will
correspond to
0.001, 0.0001, 0.00001, 0.000001, and so on. What would 1
correspond to
in the limiting bijection, should the limiting bijection exist?
Similarly, 0.1 corresponds to 1 in your first bijection, 10 in
your second
bijection, and in subsequent bijections, it will correspond to
100, 1000,
10000, 100000, etc. What would 0.1 correspond to in the
limiting
bijection, should the limiting bijection exist?
>and you will get all the
>real numbers. By the way, the decimal expansion of e is the
decimal
>expansion of e. So, why do you insist in take into account the
>integer part (2)?
The decimal expansion of e is 2.718281828..., not
0.718281828...
The integer part IS part of the decimal expansion.
>Finally, the diagonal argument is a transfinite PROCESS.
Cantor
>synthesises each digit of the diagonal number one at a TIME.
It is NOT a transfinite process over a period of time. That is
just
your fiction. The number S is determined by its decimal
digits. Cantor
described a means of determining what each digit in the
decimal expansion
of S is. In other words, Cantor allowed us to determine S as
the limit
of a series. In other words, Cantor allows us to determine a
specific
real number by allowing us to unambiguously determine all the
digits of
the number. Let f be a fuction from N to [0,1). Define g : N x
N ->
{0,1,2,...,9}, by the conditions that
f(m) = sum_{n=0}^infty g(m,n)/10^{n+1},
and for all m, there does not exist k such that for all n > k,
g(m,n) = 9,
i.e. Am Ak En > k (g(m,n) != 9), where Am means for all m, En >
k
means there exists n > k, and g(m,n) != 9 means g(m,n) does not
equal 9. Then f uniquely determines g. This means that g
determines
the digits in the decimal expansion of f(m) for all m, where
the decimal
expansion of f(m) does not end in a repeating sequence of 9's.
Define the
function h : N -> {0,1,2,...,9} by h(n) = 5 if g(n,n) != 5,
and h(n) = 6
if g(n,n) = 5. Let S = sum_{n=0}^infty h(n)/10^{n+1}, so that h
determines the digits in the decimal expansion of S. Then S is
not equal
to f(m) for any m.
>Nicolas de la Foz
David McAnally
Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills.
>A SUMMARY OF CONCLUSIONS OF THE THREAD
>1st CONCLUSION:
> * The transfinite construction N <-> R is not possible (See
PROOF
>1 below)
>>I note that your PROOF 1 below is identical to the flawed
proof
that
>>you initially offered in order to demonstrate the existence
of a trivial
>>proof that there is no bijection between N and R. You have
repea the
proof in spite of the fact that I have already debunked it by
>>demonstrating that your so-called proof relies on an unsta
assumption
>>which is KNOWN TO BE FALSE. This means that your PROOF 1
below is
>>actually worthless. I also note that there was no attempt by
you to show
>>anything wrong with my debunking of your proof.
>That means that the naturals fail in counting the reals
because they
>cannot undertake the infinity of R.
>>Noting that your PROOF 1 is worthless, and that something
more
>>sophistica will have to be adop.
>Then, we cannot assume N <-> R as
>possible.
>>If you want to prove it by reductio ad absurdum, then it is
legitimate
>>to assume that there is such a bijection, and then to show
that the
>>assumption leads to a contradiction.
>>Note that you have not yet proven that no such bijection
exists, so you
>>will have to look around for a legitimate proof.
>On the other hand, senders have sent at the moment three
kinds of
>proofs, trying to refute my arguments
>a) Proofs by contradiction o reductio ad absurdum.
>>Which is a legitimate method of proof. If the assumption ~P
leads to
>>a contradiction, then P must be true.
>b) Direct proof (supposedly the Cantor's first proof).
>c) Others (supposedly the Cantor's first proof too).
>a) As far as I've seen, all the proofs by contradictions need
the
>premise *let's assume that f: N <-> R is true*.
>>And if the assumption that there exists a bijection between
N and
>>R leads to a contradiction, the only logical conclusion is
that there
>>is no such bijection. No problem so far.
>If we carry on with a
>proof by contradiction in spite of the 1st conclusion,
>>Since your PROOF 1 is worthless, this statement holds no
water.
>then the
>proof's outcome must be proved that is true, since the
premise (raw
>material in the proof) is false. In fact it is a silly thing
to go
>with a false premise to a proof,
>>It is not a silly thing to assume a false premise and to
demonstrate
>>that it leads to a contradiction. This technique allows one
to
>>prove that the premise must be false because it cannot
possibly be true.
>>That is the essence of reductio ad absurdum.
>but if you insists, then you must
>prove that your result is reliable.
>>Since you are the only person who knows what you mean by
reliable,
>>might as well be hand waving.
>2nd CONCLUSION:
>* The results obtained by contradiction are not reliable,
>>Yes, they are. If assuming P leads to a contradiction, then
that
>>can only happen if P is false, and so it can only happen if
~P is
>>true. So demonstrating that the assumption of P leads to a
>>contradiction IS a proof of ~P.
>if the
>conclusion has not been obtained by the method of the diagonal
>argument.
>>What is so special about the method of the diagonal argument?
>b) Direct proof
>This proof does not assume anything in advance, but the
mapping f: N -
>> R, trying to prove that it is not a surjection. So, there
is not
>false premise, and then it goes directly to its goal applying
the
>diagonal argument, getting a wrong conclusion.
>>False.
>Why? Have a look to my
>own impartial version of this argument,
>>Your own impartial version being the thoroughly discredi
PROOF
1,
>>and therefore being absolutely worthless.
>and compare it with the
>partial version given by Cantor. (See PROOF 2 below)
>>This PROOF 2 relying on the furphy that time must come into
any
>>infinite process in mathematics (or any process at all), so
that
>>the generation of Cantor's diagonal number supposedly takes
time.
>>Cantor's argument has NO dependence on time.
>3rd CONCLUSION:
>As this proof uses the diagonal argument, the result is false.
>>So why does making use of the diagonal argument make the
proof
>>false?
>c) Others
>The only proof sent to the thread that fits in this section
states
>that R is uncountable because N <-> P(N) it is not possible,
>>By which you mean that there is no bijection between N and
P(N).
>>Please use the proper terminology, rather than not bothering.
>since
>the cardinality of P(N) is bigger than the cardinality of N
(Cantor's
>theorem). On the other hand, it can be rigorously proved that
it is
>possible the bijection P(N) <-> R.
>>By which you mean that there is a bijection between P(N) and
R.
>The logical conclusion is that R
>is uncountable, and that there are more reals than naturals.
However,
>Proof 3 shows that N <-> P(N) is possible. (See PROOF 3 below)
>>Your PROOF 3 is worthless for the reasons detailed below.
>4th CONCLUSION:
>The reals are countable.
>>Only because you want them to be. None of your proofs are
valid. All
>>of your proofs are flawed. It is easy to prove that there is
no
>>surjection from N to P(N), by explicitly constructing, for
any
>>function g : N -> P(N), a subset of N which is not in the
range of g.
>5th FINAL CONCLUSION
>>Since PROOF 1 is discredi, all conclusions drawn from it are
>>worthless.
>it is obvious that the naturals cannot undertake the
>counting of the reals using the decimal expansion. Why?
Because in
>the list the reals clearly show a two-dimensional infinity,
one of
>them horizontal with infinitely many digits, and the other
vertical
>with infinitely many reals. On the other side, naturals just
show a
>one-dimensional infinity (the vertical) and therefore they
cannot
>deal with the horizontal infinity of the reals.
>>The above is just a lot of meaningless jargon, designed to
befuddle,
>>rather than provide any clear thought.
>As a consequence of this fact, all the decimal-expansion
based proofs
>do not work properly, as we have already seen.
>>Maybe, if you spent some time learning logic, you might
actually learn
>>what is a legitimate proof and what is not. That would be
better for
>>you than proclaiming many legitimate proofs are
nonlegitimate, while
>>trying to force all your nonlegitimate proofs on us.
> Finally, although the naturals fail counting the reals by
means of
>the decimal expansion, it does not imply that R may not be
coun by
>other mathematical means.
>>Make up your mind. You claim to have a proof that there is
no bijection
>>between N and R, and also a proof that there is a bijection
between N
>>and R. You can't have both.
>Effectively, the elements of the power set
>of N *P(N) * can do it,
>>No, it can't. It is very easy to prove that there is no
bijection
between
>>N and P(N).
>because it is possible the one-to-one
>correspondence between P(N) <-> R, as it has been rigorously
proved
>by several mathematicians and in different ways. Perhaps the
simpler
>is the one that considers the binary expansion of the reals.
>On the other side, the Cantor's theorem proof fails because it
>analyzes a counting process (the bijection N <-> P(N)) taking
a
>timeless approach. To count is a process, the one-to-one
>correspondence is a kind of counting process, and time is
inherent to
>any kind of process.
>>No. To count is to establish a bijection with a subset of N.
Time does
>>not enter into it.
>So, it is very dangerous, if possible, to carry
>out any process into a timeless context.
>>Justify this claim rigourously. All that I have seen from
you is hand
>>waving.
>As we have seen in PROOF 3,
PROOF 3 is wrong. I show this below.
>P(N) is countable with naturals because after doing P(N) <->
R we get
>rid of the horizontal infinity of the reals, i. e., P(N) has
just a
>one-dimensional infinity, as N, so that it works the
bijection N <-P(N).
>>Since PROOF 3 is wrong, the above statements are unsuppor.
>I think it is a conclusive summary for everybody,
>>Including the fact that you are still trying to use an
argument in
PROOF 1 which I have discredi, and you did not address my
>>discrediting.
>except for some
>mathematicians that, I am completely sure, they will try to
discredit
>my arguments, but not to refute them, because simply they
cannot.
>>This is the arrogance of the crank who is so thoroughly
convinced that he
>>is right that he refuses to even consider the possibility
that he may be
>>wrong, except to dismiss it immediately. The question is: Is
it possible
>>to get such a crank to see the error of their ways, or are
they so fixed
>>on how right they are that even the most rigourous rebuttals
will not
sway
>>them?
>****************
>PROOF 1
> A METHOD TO PROVE THAT NATURALS CANNOT UNDERTAKE THE
COUNTING OF THE
>REALS DIRECTLY FROM ITS DECIMAL EXPANSION
>It is possible to represent any real number using a given
positional
>system of numeration. For ease we will use the decimal
notation.
>Proposition:
>The transfinite construction N <-> R is not possible
>Proof:
>Every real number within the interval (0, 1) has as first
decimal
>>As can be seen from below, 0 should be an element of the
interval since
>>you are listing it below. The correct interval is [0,1).
>digit one of the ten digits of the decimal system of
numeration, i.
>e. it must be of the form 0.0, 0.1, 0.2, ..., 0.9. As you can
see for
>this first digit there are 10 ^ 1 possibilities. For the
second digit
>we have 0.00, 0.01, 0.02,..., 0.99. Consequently, there are
10 ^2
>possible combinations for the two first digits, 10 ^3 for the
first
>three digits, and so on. When the number of digits is
infinite we get
>R.
>>When the number of digits is infinite, we get the elements
of R with a
>>nonterminating decimal expansion.
>Now, when we have only a digit of the decimal expansion of
the reals,
>we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2
<-0.2, ..., 9 <-> 0.9. Next, with two digits we begin the 1-1
>correspondence 0 <-> 0.00, 1 <-> 0.01,..., 10 <-> 0.10, 11
<-> 0.11,
...,
>99 <-> 0.99. With three digits, we continue doing the same,
and so on.
>And now is when the impossible transfinite construction
appears.
>While we keep doing the 1-1 correspondence within the finite
decimal
>expansion, everything will go fine.
>>As I poin out previously, what you have written above will
NOT go
fine.
>>You will not end up with a bijection between N and the
terminating
>>decimals in the interval [0,1). Think carefully about it,
and ask
>>yourself what natural number corresponds to 0.1 in your
intended limiting
>>bijection between N and the terminating decimals in the
interval [0,1),
or
>>what terminating decimal corresponds to 1 in your intended
limiting
>>bijection. The answer is that neither of these questions has
a
legitimate
>>answer since there is no limiting bijection. Did you even
bother reading
>><-> 0, 1 <-> 0.1, ..., 9 <-> 0.9, 10 <-> 0.01, 11 <-> 0.11,
12 <-> 0.21,
>., 254987631 <-> 0.136789452, .... Note that I am REVERSING
the order
>>of the digits, and placing them after the decimal place.
THIS will
>>determine a bijection such as you require. Your way
certainly won't.
>>And note that I can immediately tell you what natural number
corresponds
>>to a given terminating decimal, and vice versa.
>But, what natural number would
>correspond to the decimal expansion of the number e?
>>NO natural number would correspond to e. e is between 2 and
3, and so e
>>is not an element of [0,1). It follows that e is not in the
range. A
>>better question would have been: What natural number would
correspond to
>>the decimal expansion of the number e-2? The reason why this
would have
>>been a better question is that e-2 DOES fall in the interval
[0,1).
>By induction it
>is obvious that it should be 71828182...,
>>With your ill-defined, and nonexistent, bijection. With my
correc
>>bijection, the answer would be ...828182817, i.e. not a
natural number.
>i. e., a number with infinite
>nonzero digits, which doesn't belong to N (Point 1).
>>This is true, but does not lead where you think it does.
>Conclusion:
>The transfinite construction N <->R is not possible, i. e.
the naturals
>cannot undertake the counting of the reals directly from its
decimal
>expansions.
>>Here, you would use the fact that there exists a bijection
between N and
>>the terminating decimals (specifically, one specific
bijection supplied
by
>>ME, by the way, since you supplied no such bijection at
all), to assert
>>that there is no bijection between N and R. Your argument is
based on
the
>>statement that there cannot be a bijection between N and a
subset of
[0,1)
>>which properly contains as a subset the set of all
terminating decimals
in
>>[0,1). In other words, if A is a subset of [0,1), and A
contains all the
>>terminating decimals in [0,1) as elements, and A contains at
least one
>>nonterminating decimal as an element, then you would have us
believe that
>>the existence of my bijection above forbids the existence of
a bijection
>>between N and A. So let us take a look at a set B whose
elements are the
>>terminating decimals in [0,1), and also e-2. Can I find a
bijection
>>between N and B? Since you expect us to buy your argument,
then you
would
>>presumably answer No. But the correct answer is Yes. Take the
>>correspondence 0 <-> e-2, 1 <-> 0, 2 <-> 0.1, ..., 9 <->
0.8, 10 <-> 0.9,
>>11 <-> 0.01, 12 <-> 0.11, 13 <-> 0.21, ...., 12356987 <->
0.68965321, ....
>>Note the rule: 0 corresponds to e-2, and for natural numbers
greater than
>>zero, subtract 1, reverse the order of the digits, and place
after the
>>decimal point.
>>The point that I am making is that just because I can come
up with a
>>bijection between N and the terminating decimals in [0,1),
that does not
>>of itself forbid any bijections between N and [0,1), since
we have not
>>elimina the possibility that I can manipulate the bijection
so that it
>>is a bijection between N and [0,1). Of course, I cannot so
manipulate
the
>>bijection in this manner, but more sophistica tools are
required to
>>prove it than you have used here.
>>In short, you relied on the unsta assumption that N is not
bijective
>>with a proper subset of itself, an assumption which is known
to be false.
>>Your PROOF 1 is now worthless.
>PROOF 2
>AN IMPARTIAL VERSION OF THE DIAGONAL ARGUMENT
>uses the one that it is useful for his proof. Here we will
take all
>the information the diagonal argument supplies.
>Proposition:
>is not possible to conclude whether the synthesised number is
in the
>list or not.
>Proof:
>Let's suppose we have the mapping f: N -> R, being then an
arbitrary
>list of reals in the form
>f(1) =0.a1 a2 a3 a4 ...
>f(2) = 0.b1 b2 b3 b4 ...
>f(3) = 0.c1 c2 c3 c4 ...
>f(4) = 0.d1 d2 d3 d4 ...
>.
>.
>.
>Moreover, let S be the synthesized number, being S = 0.s1 s2
s3 ... In
>addiction,
>>That's in addition, not in addiction, which has a completely
different
>>meaning.
>*t1* will be the moment when s1 is genera, *t2*, when
>s2 is genera and so forth.
>>Time again. What is this obsession about time that you have?
>Working in decimal notation, a simple combinatory tell us
that in t1
>there can be in the list 10^1 sets of reals beginning with a
>different decimal digit. Obviously f(1) necessarily belongs
to one of
>these sets, and S to another. So, in t1 there is one set that
no
>contains S and (10^1 - 1) sets to which S could belong. In t2
we can
>be sure that S can't belong to 2 sets of reals out of the 10^2
>possible sets that we can form with the elements beginning
with two
>specific digits; so, there are (10^2 - 2) sets left where S
could be
>an element. Finally, in tn it will be n sets where S cannot
be a
>member, and (10^n - n) sets where S could be. Therefore, if we
>suppose that in every moment tk the sets are more or less of
the same
>size (cardinality), how we can be so sure that S is not a
member in
>one of the (10^n - n) sets?
>>That is not how the argument works. Take f(n). After the
n-th decimal
>>place is established, then S and f(n) disagree in the n-th
place, and are
>>therefore not equal.
>In other terms, in t1 we can be sure that S != f(1), but we
cannot
>guess whether S is in the list or not, because there are
infinitely
>many reals that begin with s1.
>In t2 we can be sure that S != f(1) and S != f(2) but we
cannot find
>out whether S is in the list, because there are infinitely
many reals
>that begin with s1 s2, and so on.
>Therefore, may someone tell us in what precise moment (tn)
and why we
>begin to be sure that S is not in the list?
>>The argument has nothing to do with your artificial
impositions of time.
>>The argument is simple enough for anybody except the most
wilfully
>>obstinate to comprehend.
>Conclusion:
>It is not possible to conclude from the diagonal argument
whether the
>synthesised diagonal number is in the list or not.
>>This is because you have introduced the illogical idea that
time must
>>play a part in the proof. If you had ANY understanding, you
would
>>understand what the proof is REALLY saying.
>PROFF 3
>A TRANSFINITE METHOD TO COUNT THE POWER SET OF N
>Proposition:
>The power set of N is countable.
>Proof:
>Being P(N) the set of the subsets of N (power set of N), we
will
>count its elements in an orderly way. First we will count the
subset
>with 0 elements, and then the subsets with one element
(singletons),
>then the subsets with two elements, and so on.
>1) 1 <-> {}
>2) In order to count the subsets with 1 element {n}, we will
use the
>naturals that are multiple of 2. We have 2 <-> {1}, 4 <->
{2}, 6 <-{3} ... 2k <-> {k}
>3) In order to count the subsets with 2 elements {n1, n2} we
will use
>the naturals multiple of 3 that they are not divisible by 2.
We have
>3 <-> {1, 2}, 9 <-> {1, 3}, 15 <-> {1, 4}, ...
>>What numbers correspond to {2,3}, {4,11}, {5,10}, etc.? At
present,
>>you have not accoun for ANY of these subsets of N to appear
in
>>your list.
>4) In order to count the subsets with 3 elements {n1, n2, n3}
we will
>use the naturals multiple of 5 that they are not divisible by
2 and
>3. We have 5 <-> {1, 2, 3}, 25 <-> {1, 2, 4}, 35 <-> {1, 2,
5}, ...
>>What numbers correspond to {1,4,10}, {1,8,9},
{65,478,10000}, etc.?
>>At present, you have not accoun for ANY of these subsets of
N to
>>appear in your list.
>4) In order to count the subsets with 4 elements {n1, n2, n3,
n4} we
>will use the naturals multiple of 7, that they are not
divisible by
>2, 3 and 5, and so on.
>>And similar finite subsets of N of all finite cardinalities
greater than
3
>>will be left out of your list.
>Conclusion:
>As the naturals are infinite and the prime numbers too, the
bijection
>N <-> P(N) is possible, so P(N) is countable.
>>You have not proven this since you have left many subsets of
N out of your
>>list. Specifically, for any finite cardinality greater than
1, you have
>>left a large number of subsets of N of that cardinality off
your list.
>>Further, you have left ALL infinite subsets of N off your
list. It
>>follows that you have NOT proven a bijection between N and
P(N).
>>The standard proof that there is no bijection between N and
P(N) still
>>holds, and you proof has not challenged the statement that
there
>>is no such bijection.
>>David McAnally
>> Despite anything you may have heard to the contrary,
>> the rain in Spain stays almost invariably in the hills.
===
Subject: Re: Socrates' Nothing is Everything
In sci.logic, Garry Denke
<garrydenke@catholic.org>
<4e63857.0401220627.6600e37c@posting.google.com>:
>> Sadly yes, that seems clear even to me with his last post.
>> Never mind, I could do with plenty of lessons in patience.
>> Love and respect
>> Chris
> Hello Chris,
> Because every number multiplied by 0 equals 0, every number
divided by
> 0 equals every number. One of those numbers is -1. Your
problem, as
> others' problem here in sci.logic,sci.math,sci.physics, is
_limiting_
> the solution.
> Don't be sad, it's only a fact.
Well, you're right that x = 0/0 can solve every equation.
You're wrong in assuming that this is even remotely useful. :-P
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Socrates' Nothing is Everything
> In sci.logic, Garry Denke
>
<garrydenke@catholic.org<4e63857.0401220627.6600e37c@
posting.google.com>:
>> Sadly yes, that seems clear even to me with his last post.
Never mind, I could do with plenty of lessons in patience.
Love and respect
>> Chris
Hello Chris,
Because every number multiplied by 0 equals 0, every number
divided by
> 0 equals every number. One of those numbers is -1. Your
problem, as
> others' problem here in sci.logic,sci.math,sci.physics, is
_limiting_
> the solution.
Don't be sad, it's only a fact.
> Well, you're right that x = 0/0 can solve every equation.
> You're wrong in assuming that this is even remotely useful.
:-P
Concerning the sqrt(-1) = 0/0
(x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
-Doctor Garry Whilhelm Denke, 1655
solving every equation, it was of 0 use, and it made 0 sense.
===
Subject: Re: Socrates' Nothing is Everything
In sci.logic, Garry Denke
<garrydenke@catholic.org>
<4e63857.0401230604.60fe67d9@posting.google.com>:
>> In sci.logic, Garry Denke
>> <garrydenke@catholic.org> <4e63857.0401220627.6600e37c@posting.google.com>:
> Sadly yes, that seems clear even to me with his last post.
Never mind, I could do with plenty of lessons in patience.
Love and respect
> Chris
Hello Chris,
Because every number multiplied by 0 equals 0, every number
divided by
>> 0 equals every number. One of those numbers is -1. Your
problem, as
>> others' problem here in sci.logic,sci.math,sci.physics, is
_limiting_
>> the solution.
Don't be sad, it's only a fact.
>> Well, you're right that x = 0/0 can solve every equation.
>> You're wrong in assuming that this is even remotely useful.
:-P
> Concerning the sqrt(-1) = 0/0
(x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.
> -Doctor Garry Whilhelm Denke, 1655
> solving every equation, it was of 0 use, and it made 0 sense.
Let a*x^2 + b*x + c = 0 be an equation we want to solve.
Since a*(0/0)^2 + b*(0/0) + c = 0 = 0*0/0 (obviously true if
one multiplies through by the denominator, namely 0), it is
clear that x = 0/0 is a solution of the above equation.
It is also clear that this result is of no practical value,
and violates at least one law of arithmetic.
One of those equations, of course, is a = 1, b = 0, c = 1,
as a special case. Therefore, x = 0/0 solves x^2 + 1 = 0.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
Subject: Re: Action Device Tragedy
===
> http://www.geocities.com/inertial_propulsion
> 17 years ago, in 1986, when I was 17 year old, I ran away
from one of
> the most prestigious engineering college in India, VJTI
Bombay. Reason
> was, I don't want to be Engineer. I want to become Pilot.
Ran away? Or did they throw you out?
> through which we can hang the things in air. We can build
air vehicles
> and any one can become Pilot.
Almost anyone can become a pilot now, all it takes is damned
hard work.
Apparently you are afraid of hard work.
> But I am back to square 1. I will have to become Engineer to
build
> this action device. There are no mechanical engineers in any
workshop
> in this town. All you people in world are forcing me to
build this
> device myself even though you are aware that at least one
person has
> built one of the parts of this device and it works. Shame!
Shame!
> Shame!
> I don't want to be Engineer. I want to be Pilot, you Morons!
And you are! You pile it as deep as anyone.
> I am not going to build this action device. That spring in
hardware
> shop struck to my finger very hard.
> -Abhi.
Oh, poor baby! Did Mommy kis the booboo and make it all better?
===
Subject: Pi 3.14
pi
pie ( P ) Pronunciation Key (p)
n.
1.A baked food composed of a pastry shell filled with fruit,
meat,
cheese, or other ingredients, and usually covered with a
pastry crust.
2.A layer cake having cream, custard, or jelly filling.
3.A whole that can be shared: That would... enlarge the
economic pie
by making the most productive use of every investment dollar
(New
York Times).
Idiom:
pie in the sky
An empty wish or promise: To outlaw deficits... is pie in the
sky
(Howard H. Baker, Jr.).
===
Subject: Re: Pi 3.14
> pi
> pie ( P ) Pronunciation Key (p)
> n.
> 1.A baked food composed of a pastry shell filled with fruit,
meat,
> cheese, or other ingredients, and usually covered with a
pastry crust.
> 2.A layer cake having cream, custard, or jelly filling.
> 3.A whole that can be shared: That would... enlarge the
economic pie
> by making the most productive use of every investment dollar
(New
> York Times).
> Idiom:
> pie in the sky
> An empty wish or promise: To outlaw deficits... is pie in
the sky
> (Howard H. Baker, Jr.).
Good thing a pie is usually circular in shape otherwise we
never could
have figured out how pie ties to 3.1415926... Also,
mathmeticians are
basically lazy, always looking for the simple solutions; they
even
leave the e off pie!
===
Subject: Re: Pi 3.14
> pi
pie ( P ) Pronunciation Key (p)
> n.
> 1.A baked food composed of a pastry shell filled with fruit,
meat,
> cheese, or other ingredients, and usually covered with a
pastry crust.
> 2.A layer cake having cream, custard, or jelly filling.
> 3.A whole that can be shared: That would... enlarge the
economic pie
> by making the most productive use of every investment dollar
(New
> York Times).
Idiom:
> pie in the sky
> An empty wish or promise: To outlaw deficits... is pie in
the sky
> (Howard H. Baker, Jr.).
> Good thing a pie is usually circular in shape otherwise we
never could
> have figured out how pie ties to 3.1415926... Also,
mathmeticians are
> basically lazy, always looking for the simple solutions;
they even
> leave the e off pie!
So they can have their pi and e it too! Ha ha ha ha!
Jason
===
Subject: Re: Pi 3.14
>pi
>pie ( P ) Pronunciation Key (p)
>n.
>1.A baked food composed of a pastry shell filled with fruit,
meat,
>cheese, or other ingredients, and usually covered with a
pastry crust.
>2.A layer cake having cream, custard, or jelly filling.
>3.A whole that can be shared: That would... enlarge the
economic pie
>by making the most productive use of every investment dollar
(New
>York Times).
>Idiom:
>pie in the sky
>An empty wish or promise: To outlaw deficits... is pie in the
sky
>(Howard H. Baker, Jr.).
>>Good thing a pie is usually circular in shape otherwise we
never could
>>have figured out how pie ties to 3.1415926... Also,
mathmeticians are
>>basically lazy, always looking for the simple solutions;
they even
>>leave the e off pie!
> So they can have their pi and e it too! Ha ha ha ha!
LOL. Shame on you, Jason.
Rick
===
Subject: Re: Pi 3.14
> pi
pie ( P ) Pronunciation Key (p)
> n.
> 1.A baked food composed of a pastry shell filled with fruit,
meat,
> cheese, or other ingredients, and usually covered with a
pastry crust.
> 2.A layer cake having cream, custard, or jelly filling.
> 3.A whole that can be shared: That would... enlarge the
economic pie
> by making the most productive use of every investment dollar
(New
> York Times).
Idiom:
> pie in the sky
> An empty wish or promise: To outlaw deficits... is pie in
the sky
> (Howard H. Baker, Jr.).
> Good thing a pie is usually circular in shape otherwise we
never could
> have figured out how pie ties to 3.1415926...
That continues to be a source of confusion. People keep saying
Pie are square
when everyone knows that it's cornbread that are square.
> Also, mathmeticians are
> basically lazy, always looking for the simple solutions;
they even
> leave the e off pie!
===
Subject: universal property of semidirect products?
Does the semidirect product of groups have a universal
property?
===
Subject: Re: universal property of semidirect products?
>Does the semidirect product of groups have a universal
property?
Well, suppose that K is an abelian group and H is group with a
given action
on K, written as k^h. If H.K is the semidirect product using
this
action, then H.K acts on K via conjugation, and we have maps
p:H.K -> H,
q:H.K -> K given by p(h,k) = h, q(h,k) = k, such that p is a
group
homomorphism, the actions of (h,k) and p(h,k) on K are the
same, and q
is a derivation; i.e. q(g1 g2) = q(g1)^g2 q(g2) for g1, g2 in
H.K.
Now suppose we are given any triple (G,p,q) such that G is a
group with
a given action on K, p:G -> H is a group homomorphism, the
actions of g
and p(g) on K are the same for all g in G, and q:G -> K is a
derivation.
Then the map f: G -> H.K given by f(g) = (p(g),q(g)) is a group
homomorphism and the resulting diagram clearly commutes.
I can't see how to make this work when K is not abelian.
Derek Holt.
===
Subject: Re: universal property of semidirect products?
|>Does the semidirect product of groups have a universal
property?
|
|Well, suppose that K is an abelian group and H is group with
a given
|action on K, written as k^h. If H.K is the semidirect product
using
|this action, then H.K acts on K via conjugation, and we have
maps
|p:H.K -> H, q:H.K -> K given by p(h,k) = h, q(h,k) = k, such
that p
|is a group homomorphism, the actions of (h,k) and p(h,k) on K
are the
|same, and q is a derivation; i.e. q(g1 g2) = q(g1)^g2 q(g2)
for g1,
|g2 in H.K.
|
|Now suppose we are given any triple (G,p,q) such that G is a
group
|with a given action on K, p:G -> H is a group homomorphism,
the
|actions of g and p(g) on K are the same for all g in G, and
q:G -> K
|is a derivation.
|
|Then the map f: G -> H.K given by f(g) = (p(g),q(g)) is a
group
|homomorphism and the resulting diagram clearly commutes.
|
|I can't see how to make this work when K is not abelian.
hmm, apparently you decided to try to describe the
_right_-universal
property of the semi-direct product, which strikes me as a
somewhat
peculiar thing to do considering that the semi-direct product
is most
simply characterized by its _left_-universal property.
nevertheless
it sounds like you may have actually gotten something
interesting (and
which would generalize pretty straightforwardly to the case
where k is
check for sure.
anyway, the semi-direct product of a group h acting on a group
k is
simply the homotopy colimit of the obvious functor f:h->groups
assigning the group k to the unique object of h. (the ordinary
colimit is obtained by starting with k and then universally
compelling
the autofunctor of k associa with each element of h to become
equal
to the identity autofunctor, whereas the homotopy colimit is
obtained
by starting with k and then universally compelling the
autofunctor of
k associa with each element of h to become naturally
isomorphic to
the identity autofunctor, subject to a certain straightforward
coherence property of this system of natural isomorphisms.)
--
[e-mail address jdolan@math.ucr.edu]
===
Subject: Re: A gift from Ilya Shambat
X-Who-Cares: Who cares?
Wanker's credentials at:
http://groups.google.com/groups?selm=QXbC8.4591%24Mp1.541193%
40n
ews.shore.net&oe=UTF-8&output=gplain
Hey, imbecile, have you noticed how many bad poetry contests
there are on the web? Since you have no hope of ever getting a
job, you should earn a living by entering them. The public has
no obligation to feed and shelter useless wankers like you
until
you die of old age. They don't pay much, but it would be the
first time you ever earned anything, and you're assured of
winning.
===
Subject: Re: A gift from Ilya Shambat
> See DR. ROCKET's web page at
May I suggest that in the future, even when the budget gets
tight, next
time
don't scrimp on the antipsychotic meds.
Clear Skies
Chuck Taylor
Do you observe the moon?
Try the Lunar Observing Group
http://groups.yahoo.com/group/lunar-observing/
************************************
===
Subject: Re: A gift from Ilya Shambat
In sci.math, Ilya Shambat
<aa347590.0401220917.482bb374@posting.google.com>:
> See DR. ROCKET's web page at
> http://www.geocities.com/drr0cket.
> Features:
> Snapshots of America
> Cats in various compromising positions
> DR. ROCKET's family recipes
> Poetry by the ROCKETS
> Anthology of posts by DR. ROCKET
> Anthology of posts by Layo Lehmann
> Penetrating analysis of romance, culture and psychology of
gringos
> Links to the best of the best web sites
> And an enjoyable meditation that everyone here will love!
> A radioactive potato in every pot, a ROCKET in every garage!
So be (a
> man, a woman, whatever you are) and check out the home of
DR. ROCKET!
Sorry, this web site is temporarily unavailable because it
ran out of bandwidth for the month. Please view this ad
with a smiling attractive woman selling Yahoo! webhosting
instead.
:-)
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: JSH: Rick Decker's example 2
> I've wondered and wondered how simple algebra could be such a
> difficult thing for so many of you, but I'm at a loss.
> What gives?
>
It is probably your false assumption that, in the ring of
algebraic
integers 2 and 7 are primes, so 14 has essentially only one
factorization.
In the ring of algebraic integers, both 2 and 7 factor in a
lot of ways
that you have not considered.
And their product, 14, factors in a lot of ways, too.
Note that every root of x^2 + n*x + 14, for every natural
number n, is a
quadratic integer which is a factor of 14, and only for |n| =
9 are tey
rational integers. Then there are roots of monic cubics with
constant
term 14 or - 14, and so on, that we haven't even looked at.
Subject: Re: JSH: Rick Decker's example 2
===
>>I've wondered and wondered how simple algebra could be such a
>>difficult thing for so many of you, but I'm at a loss.
>>What gives?
> It is probably your false assumption that, in the ring of
algebraic
> integers 2 and 7 are primes, so 14 has essentially only one
> factorization.
> In the ring of algebraic integers, both 2 and 7 factor in a
lot of ways
> that you have not considered.
> And their product, 14, factors in a lot of ways, too.
> Note that every root of x^2 + n*x + 14, for every natural
number n, is a
> quadratic integer which is a factor of 14, and only for |n|
= 9 are tey
> rational integers. Then there are roots of monic cubics with
constant
> term 14 or - 14, and so on, that we haven't even looked at.
Add |n| = 15 to that list.
--
Will Twentyman
email: wtwentyman at copper dot net
Subject: re:Hilbert space problems
===
Anyone try it??
----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet
News==----
http://www.newsfeed.com The #1 Newsgroup Service in the World!
>100,000
Newsgroups
---= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
=---
===
Subject: when NaturalNumbers = p-adics what alters in the
Riemann Hypothesis
Re: proof of the Riemann Hypothesis
(snip my old post of 1996) And below is a re-post of Riemann
Hypothesis that was on my website (now moving to a new
location and
will be up and running in the next several months).
COMMENTS about RH: If I had stuck with doing and concentra on
math
to reconcile RH with the idea that the NaturalNumbers were the
p-adics. And that precarious situation would have forced me
patch up
or mend the notion that the NaturalNumbers lie on the 1/2 Real
line.
I would have had to consider two extreme cases: (1) that the
RH is
completely false just as FLT is completely false when
NaturalNumbers =
p-adics. If taking that extreme route would have had me find
out what
was flawed in my claimed two proofs below.
(2) the second extreme case is to say that something lies on
the 1/2
Real line but not the NaturalNumbers of the illdefined notion
of
finite-integers but rather instead the p-adics. Only the line
is not a
straight line. And my second *alleged proof below* using a
spiral sort
of touches or hints of a curved line.
As far as I know from 1993-1996 in conversation with Karl
Heuer about
my RH proofs the number 2 as N in the encoding of ((N+N)^1/N) =
((NxN)^1/N) = N remains the same whether we think of 2 as a
p-adic or
as a finite integer. So that encoding formula is not corrup.
Then there are loads of possibilities between those two
extreme cases.
My gut feeling is that my 2 alleged proofs of RH below are
flawed
because I discovered them before I discovered the idea
NaturalNumbers
= p-adics.
Flawed because if I really believe NaturalNumbers = p-adics
means FLT
is false and so that demands the *nonexistence of a pattern
ascribed
by RH that the NaturalNumbers all lie on a 1/2 Real line.* If
NaturalNumbers are truly the p-adics, then the acute
regularity that
Riemann Hypothesis projects cannot be regular. NaturalNumbers
as the
same set as p-adics cannot be so regular as to be on a
straight line
that is 1/2 Real line.
mess of the RH had to be straightened out by me. Meaning that
I would
have had to teach myself a deeper understanding of RH. I would
have
perhaps had to learn as much about RH as the genius Riemann
himself
knew of RH and then been able to make the patches.
My gut feeling is that RH is as false as FLT is false where
p-adics
yield millions of counterexamples. And this gut feeling is
suppor
by the fact that thousands of mathematicians have battered
their
brains out trying to prove RH just as thousands battered their
brains
out trying to prove FLT. None will succeed with the framework
of
NaturalNumbers = FiniteIntegers because that is the root cause
of the
entire problem. So the long history of never a proof for RH or
for FLT
(Wiles is a sham) is all due to the fact that you have to fix
NaturalNumbers before you make a FLT-like statement or before
you make
a RH-like statement.
So I believe my below 2 alleged proofs of RH are false because
I have
not fully incorpora the tenet that NaturalNumbers = p-adics.
So, If I can show p-adic counterexamples to RH, then RH is
conquered.
There is a P-adic Riemann Hypothesis but I have never spent
the time
to learn it.
> I am pleased that someone can repost one of my old posts as
if it were
> yesterday. And I find it somewhat refreshing to see a 1996
post of
> mine as if it were pos yesterday, even a 1993 post of mine
because
> the idea that the Natural Numbers set is ill-defined and
that these
> are in fact the p-adics was an idea of mine dating back to
1993.
> The website that contained all of the math proofs no longer
exists:
> http://www.newphys.se/elektromagnum/physics/LudwigPlutonium/
> I am in the process of getting a new website to house all of
the
> science I have done, and please be patient for it will take
months to
> get this accomplished.
> I have not been active in mathematics after 1996 and the
reason being
> is that I eventually realized that mathematics is the least
satisfying
> of the sciences. Not anything personal. The history of
science
> compared to mathematics is my testiment proof of the claim
that
> mathematics is the least satisfying of all the sciences.
What I mean
> by that is that Mathematics does not have that arm of
Experimental
> proof that all the other sciences have, especially the
hardcore
> sciences. And thus, a person who becomes creatively involved
with
> mathematics easily ends up with frustration rather than joy
of
> accomplishment. That creative math usually ends in a shouting
> argumentative nasty tirade. Example: back in 1988 was news
of a
> test-tube water fusion called cold fusion and circa 1993 was
news of
> Wiles claiming proof of FLT. It took only 7 years for the
physics
> community to debunk the cold-fusion claim and that is
because so many
> Experiments were rapidly conduc. On the other hand,
mathematics is
> mostly hot air talk and gibberish scribbling on paper
lacking the
> unbiased, unjudgemental arm of Experiment. It may take 700
years to
> debunk Wiles FLT, but in physics it takes a matter of a few
years and
> in many cases as fast as a few months to debunk false claims.
> abandoned mathematics simply because the other sciences are
far
> superior in making progress. Since 1996 I have discovered
the Fusion
> Barrier Principle, discovered the Unification of the Forces
of
> Physics, discovered the true theory of the Solar System--
CellWell1
> vice the false NebularDust Cloud theory. And recently in the
past few
> years discovered the Stonethrowing theory as the basis for
the species
> Homo sapiens and why we are bipeds and why we have a
civilization. And
> just last night I pos looking for a bone signature of
balance for
> Orrorin, just as Pickford found a signature for bipedalism
in the
> groove region of the femur.
> In the years of 1993 to 1996 I did alot of mathematics, but
I found no
> one except for Karl Heuer and a few others serious and
willing to
> discuss or think of these new ideas. I am not surprised that
it would
> take 10 years lag in Internet post before more than one or
two people
> understand and take serious a new idea in mathematics.
> Later I am going to repost the Riemann Hypothesis proof of
mine in the
> early 1990s. Partly because that website mentioned above is
down where
> most of the material on p-adics used to be housed.
> I have come to believe that the Natural Numbers are a
fictional set,
> just as a mirage in the desert when thirsty is a fiction.
Natural
> Numbers as finite entities is contradictory to them being an
infinite
> set. The minute you make Natural Numbers infinite you lose
the
> characterization that each number is 100% finite. The set is
a
> contradiction in terms. You make each number 100% finite
then the
> entire set cannot extend to infinity. You allow each number
to be
> infinite in decimal places such as 34.00000.......... then
it is no
> longer finite but a p-adic.
> I have said that physics will prove the Natural Numbers are
p-adics
> faster than all of the mathematics community will, because
all it
> takes is another glance at the Quantized Hall Effect or
other aspects
> of Quantum Mechanics in that QM is not NaturalNumbers but
that all of
> physics is based on p-adics.
> But there is something nagging me about the Riemann
Hypothesis for if
> the NaturalNumbers are a fictional set that is ill-defined,
then my
> proof of the Riemann Hypothesis appears to be the
conventional
> foundation of NaturalNumbers. That is to say, I have not
sterilized my
> RH proof to use only p-adics. And it would reflect on the RH
idea that
> the NaturalNumbers lie on the 1/2 line. So, you see, I am
not settled
> in my own mind that I have a proof of RH. You see, FLT is a
false
> statement because p-adics solve all equations of form. So
the Fermat
> conjecture was false and no proof exists for p-adics are
> counterexamples. But that leaves the question as to what is
ill-sta
> in the Riemann Hypothesis?. Can we say the p-adics lie on
the 1/2
> line? No, somewhat absurd. So I have been wrestling in my
mind, on and
> off, as to how to clean up the Riemann Hypothesis. I know
there is a
> p-adic Riemann Hypothesis but that one was never with the
> foundational-idea that the NaturalNumbers were a bogus set.
Proofs of RH and PC first pos to Internet sci.math 1993 and
many
times
thereafter
Below are the hearts of 2 proofs of the Riemann Hypothesis.
Both are
copyrigh and I hold priority rights. So, I should not expect
to see
anyone use the log spiral or variants and equivalents thereof,
or, use
the fact that 2 is the only Real or Complex number N which has
the
encoding ((N+N)^1/N) = ((NxN)^1/N) = N or variants and
equivalents
thereof, in a proof of Riemann Hypothesis, without my name
involved,
for that is theft of intellectual property. The Plutonium Atom
Foundation does not sit still for a minute on cases of theft
of its
intellectual properties. And these properties are vast and
numerous
and
exquisitely beautiful.
TWO PROOFS OF THE RIEMANN HYPOTHESIS
PROOFS: Two proofs of the Riemann Hypothesis follows as A
and B.
Proof (A) is a geometrical proof. It was proved that the
Riemann
Hypothesis is equivalent to the following-- the Moebius
function mu of
x, m(x), and adding-up the values of m(x) for all n less than
or equal
to N giving M(N). Then M(N) grows no faster than a constant
multiple k
of (N^1/2)(N^E) as N goes to infinity (E is arbitrary but
greater than
0). Figure1, by setting-up a logarithmic spiral in a rectangle
of
whirling squares where the squares are the sequences:
1,1,2,3,5,8,13,21,34,55,89, . . . 2,2,4,6,10,16,26, . . .
3,3,6,9,15,24,39, . . . then every number appears in at least
one of
these sequences because every number will start a sequence.
Since all
numbers are represen uniquely by prime factors (the unique
prime
factorization theorem or called the fundamental theorem of
arithmetic)
and The Prime Number Theorem: the distribution of prime
numbers is
governed by a logarithmic function, where (An/n)/(1/Ln of n)
tends to
1
as n increases, where An denotes the number of primes below the
positive integer n, and where An/n is called the density of
the primes
in the first n positive integers. The density of the primes,
An/n, is
approxima by 1/(Ln of n), and as n increases, the approximation
gets
better. The distribution of prime numbers is governed by a
logarithmic function where these two math concepts-- one of
prime
numbers, and the other, logarithms seem unconnec at first
appearance, but in reality they are totally connec.
Geometrically,
the logarithmic spiral exhausts every positive integer, see
figure 1.
The area of the rectangles containing the logarithmic spiral
is always
greater, since the spiral is always inside the rectangles.
Thus the
Moebius function k (N^1/2)(N^E) is satisfied since the area of
the
logarithmic spiral is less than the rectangle whose area
represents
the
number N, and whose sides represent its factors. The area of a
logarithmic spiral is represen by A=(r)(e^(Hj)) , and so
depending
on where the point of origin for the spiral is taken rsubO
determines
k, and depending on the value of H, H determines the E value
for N,
when H=0 then the curve is a circle. The logarithmic spiral
inside
rectangles of whirling squares implies that for any number N
then
N^1/2
is the limit of the factors for N, for example, given the
number 28,
then 28^1/2 = 5.2915. . and so looking for the factors of 28,
it is
useless to try beyond 5 because the factors repeat, 4x7 then
repeats
as
7x4. But if the Moebius function was false then there must
exist a
number M such that M^1/2 is not the limit of the factors for M
and the
spiral is outside of the square, which is impossible, hence the
Moebius
function is true. Therefore the Riemann Hypothesis is proved.
Q.E.D.
My second proof (B) of the Riemann Hypothesis uses a reductio
ad absurdum argument. Euler proved that a formula encoding the
multiplication of primes was equal to the zeta function.
Euler's
formula in complex variable form is as follows:
(1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1
/(7^c))))x
(1/(1-(1/(11^c))))x . . . , where c is a complex variable,
c=u+iv. The
Riemann zeta function is as follows. Re(c) =
1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex
variable,
c=u+iv. Euler's formula involves multiplication of terms and
the
Riemann zeta function involves addition of terms of a
sequence. Taking
Re(c) > 0, suppose the Riemann Hypothesis is false then there
is a 0
such that Re(c)=0 and c not equal 1/2 +iy, which implies there
is
another 0 which is not on the 1/2 real line. Which means
another real
number other than 1/2 works as an exponent resulting in a zero
for the
Riemann zeta function, and a zero in the Euler formula. Thus,
Riemann
zeta function subtract Euler formula must equal zero. This
implies
for
any other real number exponent, either rational or irrational
numbers,
such as for example the rational exponents: 1/3,1/4,1/5, . . .
(Note:
any other exponent y/x , where y and x are Real numbers and
where the
Real number of A^(y/x) such that y not equal 1, immediately
transforms
to a number A^y(1/x), so that exponents with a 1 in the
numerator
entail all of the Real exponents). To make clear of the above,
for
example, 2^2/3 is 4^1/3. So then back to the proof. Then for
exponent
1/3 there has to exist a number M not equal 0 where
(M+M+M)^1/M =
(MXMXM)^1/M = M. Then for exponent 1/4 there has to exist a
number M
not equal 0 where (M+M+M+M)^1/M = (MXMXMXM)^1/M = M, and so on.
Including the infinite number of cases where the x denominator
is
irrational are impossible. Only the real number 1/2 works
since 2 does
not equal 0, and (2+2)^1/2 = (2X2)^1/2 = 2, and so (2+2)^1/2
- (2X2)^1/2 = 0. In all of Reals and the Complex numbers, 2 is
the
only number N which has the encoding ((N+N)^1/N) = ((NxN)^1/N)
= N.
Unlike 0, the number 2, its sum equals its product and where
the sum
and product is a new number 4. If RH were false, then another
number
other than 2 would satisfy a generalized encoding formula
((N+N)^1/N)
= ((NxN)^1/N) = N. False, hence the proof. QED (Quantum
Electrodynamics)
-------------------------------------------------------------
> ....0002 is the one and only one adic Integer (take any
> adic) which solves this encoding---- (2+2)^1/2 = (2X2)^1/2 =
2
------------------------------------------------------------
Subject: Yes, the Riemann Hypothesis is True
Plutonium)
Message-Id: (5an0kr$t9s@dartvax.dartmouth.edu>
Distribution: world
X-Authentica: Archimedes.Plutonium on DND host dartmouth.edu
> The proof of Riemann Hypothesis as a true theorem depends on
2.00...
> being the unique solution to (N+N) = (NxN) = N^N = M. If
there are no
> p-adic unique solution means that RH was false all along.
> The Euler formula is a multiplication and use of prime
integers. IN
> the P-adics there are an infinitude of primes , and for
2-adics it is
> 2, for 3-adics it is 3 and 5-adics it is 5 and so on ad
infinitum.
I posed this question to [ ] before start of the holidays,
> and I pose it again. Can you adequately define exponential
and
> logarithm in p-adics?
What solutions exist for (N+N) = (NxN) = N^N = M in p-adics?
> I believe the answer is in. The answer is that Riemann
Hypothesis is
> true after all. I was informed that: Yes, 2 is still a
unique solution.
> In the n-adics the equation x+x = x*x has the obvious
solutions 0 and
> 2 and also (when n is composite) some hybrids like the
10-adic
> ...3574218752, which are partly like 0 and partly like 2.
But the
> exponentiation constraint eliminates 0 and also the 0-like
part of the
> hybrids, so 2 is the only solution. For each n >= 2, the
equation in
> the n-adic ring (or field) has exactly one solution. (Which
can be
> written ...0010 when n = 2, and ...0002 when n > 2).
> And I have also been told that there is a p-adic Riemann
Hypothesis.
> I have never heard of this and do not know exactly what they
are
> talking about. I can be sure though that what they are not
talking
> about is the program that Naturals = P-adics
The above one paragraph is actually a proof of RH. It is
explained in
more detail in my website, and there I also give a geometrical
proof
of
RH.
But the pattern is this: If FLT were a true statement of math,
then
the proof of FLT would be that if false then there exist a
number N
such that N+N+N = N*N*N = M to form a building block number
for a
P-triple. But of course with Naturals being the p-adics there
does
exist these numbers that satisfy kN= N^k.
Now the pattern can be used for RH. Look at RH and ask
yourself why
is 1/2 so special? And what sort of pattern is connec with the
specialness of the Real 1/2 and the Euler encoding and the Zeta
function encoding? What is the connection between 1/2 and
Euler and
Zeta? And the answer is that those three are connec by the
addition
of Naturals equaling the multiplication of Naturals. The
linkage, or
the connection is that addition equals multiplication and is
equal to
this special number of 1/2. Here is the pattern of linkage.
((N+N)^1/N) = ((NxN)^1/N) = N
reduces to (N+N) = (NxN) = N^N = M
There, in that encoding is the Riemann Hypothesis distilled to
its
primal foundation. Just as kN=N^k distills FLT to its primal
foundation.
Sorry to say that FLT is false because in Naturals = P-adics
that
primal foundation collapses.
But look at RH. RH in primal foundation is true in both
Reals+i+j
since 2 is the unique solution and also in P-adics where
....0002 is
the unique solution.
There, Riemann Hypothesis is proved.
I do not know if the primal foundation above with the P-adic
solution
also answers the Riemann Hypothesis for negative numbers?
Anyone
know? I am busy at the moment to give it my full attention.
===
Subject: Re: god, I'm dumb
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MDTxL31752;
if 1=6N then what property is being used and what does n equal?
===
Subject: Re: unit of intelligence
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MDU1F31785;
Truth be told, the difference IS the same, as measured in
standard
deviations. Generally, IQ tests have a standard deviation of
15 or 16 points,
depending on which test it is. Assuming a 15 point standard
deviation here,
the differnce between 100IQ and 130IQ is two standard
deviations, as is the
difference between 130IQ and 160IQ.
The RARITY of a 160 IQ, however, is much much higher than that
of a 130IQ.
The rates at which these IQs can be expec to be found (and by
which the IQ
tests are centered) are shown by your everyday bell curve, the
properties
of which are not terribly complica, but far beyond the scope
of THIS
particular reply.
Any questions, moans, gripes, complaints -- feel free to email
me directly.
><preWhatever your opinion about IQ as a measurement of
intelligence, I think
>one can hardly call it a unit of intelligence. For instance I
very
much
>doubt
>that the scale is linear. (Is the distance between an IQ of
100
>and one of 130 the same as between 130 and 160? I don't think
so.)
>richard
>las P. McNutt skrev i meldingen ...
>>In Article <34D498B4.E6415D10@service.raksnet.com.tr>, mim
eke
><HTMLthere are units of length mass time and of many other
properties.
><BR>is there any unit of intelligence.
><BR>does anybody have any information or idea.</HTML>In spite of the funny replies our friends in the social
sciences talk
>>about an intelligence quotient. Standard tests produce a
grade which is
>>compared to results on the same test for a population of the
same age as
>the
>>testee with a maximum of 15 years. The ratio is the IQ and is
>dimensionless.
>>Let's NOT fill this NG with discussions of the validity of
IQ!
===
Subject: Some good news and some bad news (ver. 2)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MI3aZ22727;
Claim, below, was originally made without stating
explicitely that x, y, z should be pure quaterions.
Otherwise, one immediately has infinetly many counterexamples,
for example: x = x, y = 1, z = 1.
fairly big claims about the binary operation $ and about the
determinant, namely:
1. $ associative/distributive binary operation
2. det(res(x), res(y), res(z)) = tes(res(x*y*z))
After all the proofs, both claims are incorrect.
(the proofs involving determinants were wrong because (1)
was supposed true)- and my day had already been
going badly *before* I realized this.
Define the surjection: tes: Q(A)-> R
tes(x) = tes( (x_1)i + (x_2)j + (x_3)k + (x_4)1 ) = x_4
Remeber that res: Q(A) -> R^3 was defined as
res(x) = res( (x_1)i + (x_2)j + (x_3)k + (x_4)1 ) = (x_1, x_2,
x_3)
Thus, informally speaking, tes and res together insure that no
information is lost.
Claim:
det(res(x), res(y), res(z)) = - tes(x*y*z)
where * is multiplication on the quaternion algebra and (!)
x, y and z pure quaternions.
Initial reaction: strange but pretty cool
Proof: Please check for yourself, maybe I made
another mistake?
===
Subject: Rationals are countable
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MI3a422734;
Yeah, and I know a lot of other properties, too. Alas,
I only have time for this one.
Proof:
Let x be any rational. Then it can be expressed as
n/m where both n and m are naturals. First count to n, i.e.
{1, 2, 3, 4, ..., 25, 26, 27, ..., 89, 90, ....1001, 1002,
1003, ..., 4500232, n)
Now divide that sequence by m and create an equivilance class
or something. Then, n/m will be the last member of this
sequence and
since you coun it the first time, you will have still coun
it after you divided by m. Thus, you coun it! Q.E.D.
===
Subject: Re: Please read my preprint for a proof of Goldbach
Conjecture
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MI4h522892;
>It is a big deal. You claim you have a proof; in a proof I
don't have to
>guess what values of a and b you are using. In a proof of
Goldbach's
>conjecture, everybody stops reading as soon as the author
cannot express
>himself clearly. No point trying to guess what you mean.
You would not have to even guess if N(3, q) is given; indeed
in the hypothesis of Lemma, it is sta.
>It doesn't, therefore no proof.
Okay. My point of view would be rather axiomatic and so
that's the reason I can say no more.
The point of this repetitive postings is that I can not see
why I am wrong. I would not like to stop defending my thesis
until I am totally convinced that I was wrong. Your
counterargument was
mainly the illegality of defining the number of elements in a
covering,
and I am saying that using converings with one element or not
has
nothing to do with contradictions, having admit that according
to the definition of coverings in Definition 2.2., covering
with one
element can exist. Whatelse could I speak of?
Would my argument be independent of, has nothing to do with
your remark? If independent, then why? This would be all I
would
be asking for.
As I said earlier, unless the questions above are not
answered properly, we must continue this postings.
Thanks for your strictness, though.
Hisanobu Shinya
===
Subject: Re: Please read my preprint for a proof of Goldbach
Conjecture
> The point of this repetitive postings is that I can not see
> why I am wrong. I would not like to stop defending my thesis
> until I am totally convinced that I was wrong.
It is not me who has to convince you that you are wrong. It is
you who
has to convince me that you are right. That's how proofs work:
The
person writing the proof has to convince the people reading it
that it
is correct. I am not convinced, therefore it is not a proof.
By the way, this was just the first problem in the proof, and
that is
when I stopped reading. Fix it, and I may read on to the next
problem.
===
Subject: Re: Any news on odd perfect numbers?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MIrJd26905;
I published my own naive geometrical argument that there are
no odd perfect
numbers on Jan 14th here
http://mathquest.com/discuss/sci.math/m/569936/570941
and my non-standard notation irrita a reader or two -
apologies.
My outline argument is that since Euler showed an odd perfect
number must be
a multiple of a square, then if we consider how summing the
factors must
fill in an array of squares equalling the odd prime, a
contradiction
emerges when we look at the square holding some or all of the
subfactors of
the factor which is a square [which I helpfully name 'sq'].
If anyone friendly has the patience to wade through three
pages of that, I'd
be grateful for an opinion, and will gladly answer any
questions.
Mark G.
===
Subject: Re: basketball parabola
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MIrJs26914;
>I'm 6'2. When I shoot a basketball, my release point is 7'
high. I
>want to practice my three point shot, but I dont have a
basketball
>goal. Instead, I want to place a mark on a wall, which
represents the
>top of my shot's arc.
>I read somewhere that an initial trajectory of 45 degrees
will yield
>the best results. The 3 point arc in college is 19'9.....call
it
>20. A goal would be 10 high.
>How far from the wall should I stand, and how high up is the
target
>mark such that if my shot strikes this mark at the top of my
shooting
>parabola, the ball would fall into the goal from 20.
In real life the parabola doesn't hold. Besides the effect of
the
backboard, you would also need to take into account air
resistance,
the Magnus effect, and other real world effects.
Advice: get a hoop
===
Subject: Re: Fundamental basics of unilaterality
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MJjO431326;
>> As an engineer, I do not need at all the fictitious notion
of an
>> infinitely small value zero. Its neighbours 0- and 0+ do
the same job.
>> Why not declaring the exact zero Cantors pipe dream?
>If you have 10 euros and then a mugger takes away those 10
euros
>from you, how many euros have you left? 0+? 0-? or just 0?
0+ if you were at least able to remember what he looked like
when going to the police. Otherwise 0-.
>--
>
Needless to say, I had the last laugh.
> Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Connecness in the plane
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MK3MC00674;
Hi everybody,
on the real line we know exactly which sets are connec-
the intervals.Is something similar true for the plane?
I mean,is there some characterization of the connec sets in
RxR?
Even for relatively harmless sets in RxR such as
A={(x,y):x and y are rational}
B={(x,y):x or y is rational}
C={(x,y):x and y are irrational}
D={(x,y):x or y is irrational}
I don't know if they are connec.Which of them are?Why?
Any help welcomed.
===
Subject: Re: Connecness in the plane
> on the real line we know exactly which sets are connec-
> the intervals.Is something similar true for the plane?
> I mean,is there some characterization of the connec sets in
RxR?
In R^2 and in any normed vector space, connec <=> path-connec.
Now it
becomes clear that:
> A={(x,y):x and y are rational}
Not C
> B={(x,y):x or y is rational}
Not C
> C={(x,y):x and y are irrational}
Not C
> D={(x,y):x or y is irrational}
Not C,
Using the mean-value theorem applied to each component of
those vectors
(and
the fact that if X, Y, Z are three topological spaces, f:
X->YxZ,
f(x)=(h(x),i(x)) is continuous <=> h:X->Y and i:X->Z are
continuous).
--
Julien Santini
===
Subject: Re: Connecness in the plane
>>on the real line we know exactly which sets are connec-
>>the intervals.Is something similar true for the plane?
>>I mean,is there some characterization of the connec sets in
RxR?
> In R^2 and in any normed vector space, connec <=>
path-connec. Now
it
> becomes clear that:
Umm... that's not true at all. The standard example of a
connec space
which isn't path connec works. (The topologist's sine curve).
An *open* connec set in R^n is path connec, but none of the
sets
he asked about were open.
David
E-mail address spam blocked. Reverse the first part to e-mail
me.
===
Subject: Re: Connecness in the plane
>>on the real line we know exactly which sets are connec-
>>the intervals.Is something similar true for the plane?
>>I mean,is there some characterization of the connec sets in
RxR?
> In R^2 and in any normed vector space, connec <=>
path-connec. Now
it
> becomes clear that:
>>A={(x,y):x and y are rational}
> Not C
>>B={(x,y):x or y is rational}
> Not C
Wrong. B contains all vertical lines with rational
x-coordinate, as well
as all horizontal lines with rational y-coordinate. A path
connecting
one point to another can be construc by moving along the
appropriate
set of horizontal and vertical lines.
>>C={(x,y):x and y are irrational}
> Not C
>>D={(x,y):x or y is irrational}
> Not C,
Wrong. D contains all vertical lines with an irrational
x-coordinate,
as well as all horizontal lines with an irrational
y-coordinate.
> Using the mean-value theorem applied to each component of
those vectors
(and
> the fact that if X, Y, Z are three topological spaces, f:
X->YxZ,
> f(x)=(h(x),i(x)) is continuous <=> h:X->Y and i:X->Z are
continuous).
> --
> Julien Santini
Dale
===
Subject: Re: Egyptian-Fraction Expansions Of REALS
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MN7mg16255;
>An obvious idea I have not seen *much* about before (which
means
>nothing...).
>If we have a positive REAL, possibly irrational, x,
>then we can find a sequence (many many sequences..., I bet) of
>positive integers
>{n(k)}
>such that
>sum{k=1 to M} 1/n(k) = x,
>where M is infinity if x is irrational.
>For instance, after dealing with the integer-part
differently, we can
>apply the Greedy-Algorithm to x.
>x = pi, as an example:
>pi = 3 + 1/8 + 1/61 +...
>Or x = (sqrt(5)+1)/2 = phi:
>phi = 1 + 1/2 + 1/9 + 1/145 +...
>(I am FAR from certain about the last terms in the above two
>expansions, since my calculator is low-precision.)
>We might want to, in order to be strict with our definition of
Egyptian-Fraction, rewrite the 3 in the pi-expansion as 1+1+1,
>to get:
>pi = 1 + 1 + 1 + 1/8 + 1/61(?) +...
>In any case, I do not believe these sequences are in the
Encyclopedia
>of Integer Sequences, although this idea seems basic to me.
>I also wonder about alternative EF-expansions, both with all
positive
>terms and with the possibility of having negative terms.
>For instance, aside from:
>pi^2/6 = sum{k=1 to oo} 1/k^2,
>what other expansions exist?
>We could consider:
>pi^2/6 = sum{k=1 to oo} (4/3)/(2k-1)^2,
>but 4 does not divide (2k-1)^2, so we can rewrite this as:
>pi^2/6 =
>1/3 + 1/3 + 1/3 + 1/3 +
>1/27 + 1/27 + 1/27 + 1/27 +
>1/75 + 1/75 + 1/75 +1/75 +...
>So, therefore, if duplicate terms are allowed, then it is
easy to see
>that it is possible for an x to have multiple expansions.
>Anything more to be said about EFs of reals??
>(I am sure there *must* be more to be said!...)
>Leroy
> Quet
One (perhaps not-so-useful) way to start would be to write
down an
irrational number (at least the decimal part) as the sum of
inverse powers of
ten. So
.1726 ... becomes
(1/10) + (7/100) + (2/1000) + (6/10000) + ...
Since you seem familiar with egyptian fractions you should
easily be able to
find a way to convert each rational part into a corresponding
egyption
fraction and then take your infinite sequence from this sort of
construction.
Jacob Woolcutt
===
Subject: Re: Some good news and some bad news (ver. 2)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MNV5J18403;
Well, I called that ver. 2 because Id sent one just before
which apparently didn`t go /hasn't gone through.
Basically, despite the new definition, the proofs
to the questions such as: why does the determinant change
its sign if two columns are exchanged?, etc. don't have to be
changed much,
if at all.
This is because, again
(a) tes, res linear
(b) x, y pure quaternions -> x*y = -y*x
(c) x, y pure quaternions -> exists lambda in R: (x^2)*y =
lambda y
(d) x*(lambda y) = lambda(x*y)
Ex.
(a) and (d) show why the determinant is linear in
each slot
(a) and (b) immediately answer why the determinant changes
its sign after columns are exchanged.
(a), (b), (c) and (d) answer why the determinant is zero if
res(x), res(y), res(z) are linear dependant: since, for ex.
tes(x^2*y) = 0 by (c).
===
Subject: Is this a field or a borel field?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MJjON31318;
Suppose that a=[A1, A2, ... ,An], where each of these is a
subset of
the Sample Space.
Let us form the the collection of sets consisting of A1,
A2,..., An,
and the complements A1', A2', ..., An' and all possible
intersections
which can be formed between the 2n sets.
Then it can be seen that the family consisting of these sets
and all
finite unions which can be formed among them constitute a
Borel field
deno by B(a).
*************************************************************
Q1. I am thinking that since we are having finite union and
intersection here, we can actually constructing a field, not a
Borel-field (sigma-field). Borel-field closed under countable
unions
and intersection.
Is there anything wrong with my reasoning?
Thanks a lot!
===
Subject: Re: Is this a field or a borel field?
>Suppose that a=[A1, A2, ... ,An], where each of these is a
subset of
>the Sample Space.
>Let us form the the collection of sets consisting of A1,
A2,..., An,
>and the complements A1', A2', ..., An' and all possible
intersections
>which can be formed between the 2n sets.
>Then it can be seen that the family consisting of these sets
and all
>finite unions which can be formed among them constitute a
Borel field
>deno by B(a).
>*************************************************************
>Q1. I am thinking that since we are having finite union and
>intersection here, we can actually constructing a field, not a
>Borel-field (sigma-field). Borel-field closed under countable
unions
>and intersection.
>Is there anything wrong with my reasoning?
If your point is that the field genera by finitely many
sets is that same as the sigma-field genera by those
sets then that's exactly correct.
>Thanks a lot!
************************
===
Subject: Re: Intersection between a line and circle
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MN7mZ16247;
>I know:
> - origin of circle
> - line x,y
> - line gradient
>I would like to know whether the line and circle intersect
and how far
>down the line the intersection occurs.
>Someone sugges that there is a simultaneous equation that
could solve
>this. However, as this is forming part of a computer program,
a
>simultaneous equation is not ideal.
>Any help would be apprecia.
Let's say your line is y = m*x + b and your circle is
y = sqrt (R^2 - x^2) and also y = - sqrt (R^2 - x^2)
where R is the radius.
You can equate them, i.e.
m*x + b = sqrt (R^2-x^2) and m*x + b = - sqrt (R^2-x^2)
Now solve each for x_0 (if it exists), then get y_0 from
y_0 = m * x_0 + b , where (x_0, y_0) is the intersection point
Good luck
===
Subject: Re: Sum (A(k)*sin(kx) = Constant => How to determine
A(k)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MLrdi10110;
>I am dealing with the following problem :
>Sum of (A(k)*sin(kx) = B from k=0 to infinity
>where A(k) is a constant and B is a constant.
>Supposing I know the value of B,how do I find the coefficients
>A(k) as a function of B ??
>How is it rela to Fourier integrals.
>thanks in advance for your help
Look up Fourier Series in your textbook.
===
Subject: Re: goldbach's and fermat's proofs
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MIrIX26893;
>You might like to know that I am rewriting some segments of my
>Goldbach proof. Meanwhile there is a pdf listing on
>www.mathprints.com. Kerry
I copied and pas the given address, but my computer
says that that address could not be found. Well, let me know
if you finished your rewriting. My email is
hshinya2002@yahoo.com
A fan of Britney Spears
===
Subject: Re: goldbach's and fermat's proofs
>You might like to know that I am rewriting some segments of my
>Goldbach proof. Meanwhile there is a pdf listing on
>www.mathprints.com. Kerry
> I copied and pas the given address, but my computer
> says that that address could not be found. Well, let me know
> if you finished your rewriting. My email is
> hshinya2002@yahoo.com
> A fan of Britney Spears
Thanks for pointing out me address error. Correctly, it is
www.mathpreprints.com
Thanks for your indulgement
Subject: Re: A Certain Random Maze-Generating Algorithm
===
>Even though the resulting mazes LOOKED interesting,
>the mazes' paths would practically always just pass through
>the centers of the mazes...
>Very simple to solve, each of these mazes were...
In a wall adding algorithm where you pick a random cell each
time, the
Maze will tend to grow inward from the four boundary walls at
equal rates,
always finishing up in the center. The passage layout will
roughly be a
bunch of passages radiating from the center to the edges,
meaning the
solution indeed always goes through the center. Passage carved
versions of
this algorithm have the same issue, where all passages will
radiate from
the first cell visi.
This is very similar to Prim's Algorithm. The one subtle
difference is
Prim's keeps track of frontier cells, or unoccupied cells
adjacent to
an occupied cell, and chooses randomly among the frontier
cells. Prim's
Algorithm results in Mazes that are similar, but look more
random. Your
algorithm tends to smooth away differences in the wave of
occupied cells
that move toward the center, like throwing sand on a rough
surface. For
example, have a smooth wall with a one unoccupied cell
indentation inward,
and there's triple the chance that unoccupied cell will get
visi, since
it's adjacent to three occupied cells. Prim's Algorithm
however tends to
accentuate and grow any differences, like a crystal in a
solution. For
example, have a smooth wall with a one occupied cell
indentation outward,
and there's triple the chance that occupied cell will get
extended, since
it's adjacent to three frontier cells.
>(And, I must stress, the program never connec two unoccupied
>vertexes, nor did it connect two occupied vertexes.)
Adding a wall segment between two unoccupied vertices would
create a loop.
Adding a wall segment between two occupied vertices would
create an
inaccessible section or make the Maze unsolvable (unless there
are already
loops, in which case it may reconnect a detached wall). Hence
your Mazes
are perfect, or form a minimal spanning tree over the cells,
with
exactly
one solution and one path between any pair of points.
>But I believe that we can improve the mazes' difficulty by
>using a variable v, where 0 <= v <= 1.
v is the probability that the program, instead of simply
>picking an occupied vertex to connect to at-random,
>will connect to the most recently conver-to-occupied vertex
>in the maze, if this vertex has at least one currently
>unoccupied neighbor.
What you describe here is the Growing Tree Algorithm. Growing
Tree is a
generalized algorithm that grows the passages or walls in the
Maze like
other algorithms, where all occupied cells are appended to a
list. You have
total freedom how to select the next cell from the list to try
to grow
from. Always select the most recent cell (i.e. v=1) and this
turns into the
Recursive Backtracking Algorithm. Always select a random cell
(i.e. v=0)
and this turns into your algorithm. You can do things like
select among the
n most recent cells, or always select the most recent cell but
a certain
percentage of the time select a random cell, or have a
variable v like you
have, for different textures.
>On the other extreme, v=1 would make certain the path passes
>along the maze's outer wall, at least for some part of the
path.
When v=1, your algorithm becomes very similar to a wall added
version of
the Recursive Backtracking Algorithm. The only difference is
that when you
can no longer grow a wall in any direction, you go to a random
vertex,
instead of following the stack back to the first available
vertex. This
indeed has a potential problem in that the solution path may
follow the
boundary wall. In general wall added Maze algorithms are
harder to get
right than passage carved ones. If v is too low, all paths go
through the
center. If v is too high, the solution goes around the outside
edge.
The Mazes are indeed simple, but this is actually one of the
faster
algorithms if implemen right. Have a list of all occupied
vertices,
and a mapping from each cell coordinate pair to its index in
the list. When
a new cell is visi, add it to the end of the list. This allows
picking
a random occupied vertex in constant time. If an occupied
vertex being
considered has no unoccupied neighbors, remove it from the
list (by
swapping
it with the end). This works regardless of the value of v and
is just as
fast at the start and at the end of the Maze creation process.
>So, somewhere between v=0 and v=1 is a value for v which leads
>to the most interesting mazes.
>Also, interesting refers to the aesthetic appeal of the mazes.
>But in either case, the definition of interesting is
subjective.
Having v near 1, or the Recursive Backtracking Algorithm,
works well for
passage carved Mazes, and will result in long convolu
solutions.
Interesting is indeed subjective, however I can say for the
best Mazes,
create it by carving passages, and have a v at least 0.5
All the algorithms mentioned: Recursive Backtracking, Prim's
Algorithm, and
Growing Tree, and many other types (including perfect Mazes
that don't
grow the Maze like a tree, and methods of non-perfect Mazes
with loops) are
described at http://www.astrolog.org/labyrnth/algrithm.htm.
That page also
links to a free program called Daedalus that implements each
algorithm,
in both passage carved and wall added versions, where you can
watch the
Maze get drawn if you like. :)
O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*
O*O*O*O*O*O*O
* Walter D. Cruiser1 Pullen :) ! Astara@msn.com
*
O Get lost in my Labyrinth Web page:
http://www.astrolog.org/labyrnth.htm O
*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O
*O*O*O*O*O*O*
===
Subject: Re: A Certain Random Maze-Generating Algorithm
Originator: jgamble@ripco.com (John M. Gamble)
>maze-generating program (in BASIC).
>Even though the resulting mazes LOOKED interesting,
>the mazes' paths would practically always just pass through
>the centers of the mazes...
>Very simple to solve, each of these mazes were...
For what it's worth, see the perl module description at
<http://search.cpan.org/search?query=Maze&mode=module>.
This is, if i'm reading correctly, the dual of what you're
doing. It breaks down walls to the next unoccupied cell,
until there are no unoccupied cells left. If a cell had
more than one choice of cells to break into, it would save
that location in an array, and return to it if it ran into
a dead end.
Originally i used a stack for this, and it produced fairly
uninteresting mazes. When i changed it to a queue, the
paths branched out with longer alternate branches. The
reason for this was that a stack would pop off the most
recently visi cell, which would be one close to the
dead end. This would in turn usually produce a short
branch that dead-ended quickly. Yawn. Queues produced
much more interesting choices, since the cell to revisit
would usually be farther away from the dead end.
>Any comments?
I'm fascina by the fact that you came up with the dual
of this method first. This is the opposite of what most
people, including myself, do. It was only after mulling
over the possibilities for change that i thought of the
wall-drawing method that you described. Someday i'll
add it as an option.
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: Google's rankings
> We perfectly agree on this: it doesn't matter the intrinsic
value of the
> website, which makes Google's ranking totally useless and
their
> proprietary ranking algorithm a joke.
Experiment proves you wrong: millions of people have made
Google the first
place they go to search for information, usually find what
they want on the
first page of results, and rarely find a need to go to other
search
engines.
People have done this because Google *works*, and works much
*better* than
any of the previous search engines.
--
===
Subject: Re: Google's rankings
> [M]illions of people have made Google the first
> place they go to search for information, usually find what
they want on
the
> first page of results, and rarely find a need to go to other
search
engines.
> People have done this because Google *works*, and works much
*better*
than
> any of the previous search engines.
No. People have done this because, while Google *doesn't
work*, it
*doesn't work* much *less* than any of the previous search
engines.
Colin Percival
===
Subject: Re: Cantor's diagonal proof
>1. Its easy to prove that the sythesized number S doesn't
appear in the
>list. If it does appear in the list, it must be in some
position in the
>list - say the nth position. So S=f(n). But by definition of
S, the nth
>digit of S is different to the nth digit of f(n), so its not
f(n), a
>contradiction. Therefore it cannot appear at any position in
the list, and
>so isn't in the list. Pretty straightforward, huh?
I've got a question and I can't quite get the answer. Suppose
you make
a list of all the Natural Numbers, but you represent them in
binary
and write them down in reverse (So 2 is a 01 with infinitely
many 0
behind it). If you now constuct a new number S using the
proposed
method (flipping the nth digit of the nth number from 0 to 1
or vice
versa), that number won't be in the list (it simply can't be).
Does
that mean that the construc number S is not a Natural Number?
What
am I missing here (because it sure seems to me that the number
S would
be a Natural Number)?
===
Subject: Re: Cantor's diagonal proof
> I've got a question and I can't quite get the answer.
Suppose you make
> a list of all the Natural Numbers, but you represent them in
binary
> and write them down in reverse (So 2 is a 01 with infinitely
many 0
> behind it). If you now constuct a new number S using the
proposed
> method (flipping the nth digit of the nth number from 0 to 1
or vice
> versa), that number won't be in the list (it simply can't
be). Does
> that mean that the construc number S is not a Natural
Number? What
> am I missing here (because it sure seems to me that the
number S would
> be a Natural Number)?
It is not. What you are missing is that each number in your
set has a last
position with the digit 1. The numbers S does not have such a
last
position,
so it is not in the set.
--
===
Subject: Re: Cantor's diagonal proof
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i0MNi2N19291;
>I have no idea why anybody replies to cranks writing about
Cantor. (If you
>try and prove with a straight face that the Cantor diagonal
proof is
wrong,
>you are implicitly saying that every serious mathematician of
the 20th
>century is also wrong, but you are right. This is a good
enough enough
>definition of crank for me).
The most common and amusing characteristic of the anti-crank
people
is their ability to use the rules of the mathematical game in
the way
they prefer and always in their own profit, whenever they
want. Their
motto is: The rules are mine, so you are a crank because you
don't
use them in the way I want.
The method used to synthesize the diagonal number in my proof
is
exactly the same that the one used by Cantor. Therefore, if my
method
works in a finite time (only because you like that it be this
way)
then the Cantor's diagonal method also works in a finite time,
so it
doesn't conclude anything.
The only difference between my method and the diagonal
argument is
that I show ALL THE INFORMATION that the diagonal argument
provides
and Cantor just show and use the information that is useful
for his
reasoning.
>But for the record:
>1. Its easy to prove that the sythesized number S doesn't
appear in the
>list. If it does appear in the list, it must be in some
position in the
>list - say the nth position. So S=f(n). But by definition of
S, the nth
>digit of S is different to the nth digit of f(n), so its not
f(n), a
>contradiction. Therefore it cannot appear at any position in
the list, and
>so isn't in the list. Pretty straightforward, huh?
The mistake in your reasoning is the following. When you
synthesize
the nth digit of S, the combinatory tells us that there are
10^n
possible real numbers with n digits (even if n is infinite).
However,
Cantor and you get the conclusion that S is not in the list
because
S is different to the first n reals in the list. What's the
matter
with the other (10^n  n) reals in the list?
>2. So where is the error in yor proof? Insofar as it proves
anything,
it
>proves that you cannot find S in the list in finite time
using the process
>you propose. This is hardly surprising, as S doesn't appear
in the list.
>Too bad! Looks like you will not go down in history as the
man who turned
>mathematics on its head!
It is a pity if your mathematical castle is falling down due
to a
Cantor's mistake, but mathematics is a science, not a
religion. Don't
try to defend what it is not defendable.
Nicolas de la Foz
>> AN IMPARTIAL VERSION OF THE DIAGONAL ARGUMENT
>> uses the one that it is useful for his proof. Here we will
take all
>> the information the diagonal argument supplies.
>> Proposition:
>> is not possible to conclude whether the synthesised number
is in the
>> list or not.
>> Proof:
>> Let's suppose we have the mapping f: N -> R, being then an
arbitrary
>> list of reals in the form
>> f(1) =0.a1 a2 a3 a4 .
>> f(2) = 0.b1 b2 b3 b4 .
>> f(3) = 0.c1 c2 c3 c4.
>> f(4) = 0.d1 d2 d3 d4 .
>> .
>> .
>> .
>> Moreover, let S be the synthesized number, being S = s1 s2
s3 . In
>> addiction, *t1* will be the moment when s1 is genera, *t2*,
when
>> s2 is genera and so forth.
>> Working in decimal notation, a simple combinatory tell us
that in t1
>> there can be in the list 10^1 sets of reals beginning with a
>> different decimal digit. Obviously f(1) necessarily belongs
to one of
>> these sets, and S to another. So, in t1 there is one set
that no
>> contains S and (10^1 - 1) sets to which S could belong. In
t2 we can
>> be sure that S can't belong to 2 sets of reals out of the
10^2
>> possible sets that we can form with the elements beginning
with two
>> specific digits; so, there are (10^2 - 2) sets left where S
could be
>> an element. Finally, in tn it will be n sets where S cannot
be a
>> member, and (10^n - n) sets where S could be. Therefore, if
we
>> suppose that in every moment tk the sets are more or less
of the same
>> size (cardinality), how we can be so sure that S is not a
member in
>> one of the (10^n - n) sets?
>> In other terms, in t1 we can be sure that S != f(1), but we
cannot
>> guess whether S is in the list or not, because there are
infinitely
>> many reals that begin with s1.
>> In t2 we can be sure that S != f(1) and S != f(2) but we
cannot find
>> out whether S is in the list, because there are infinitely
many reals
>> that begin with s1 s2, and so on.
>> Therefore, may someone tell us in what precise moment (tn)
and why we
>> begin to be sure that S is not in the list?
>> Conclusion:
>> It is not possible to conclude from the diagonal argument
whether the
>> synthesised diagonal number is in the list or not.
>><pre> Apart from the question of whether such pairs of numbers
(x_0,
a_l)
>> even exist, we _do_ have a notation to express such
potential
number
>> pairs: ..01... and ..10...
>> How is that a question? How can there be a least c with b <
c?
>>Do you doubt b < c -> 2b < b+c < 2c -> b < b+c /2 < c ?
>> Note that it is pointless to take an x, b < x, form y = b+x
/2,
>>then b+y /2, ... maybe take a limit, or whatever. It doesn't
matter;
>>any purpor least c , however arrived at, fails as above.
>> .01 x_0
>> .11 y_0
>> .011 x_1
>> .101 y_1
>> ...lots of numbers between 0.0 and .011 which are =< x_1;
not
>selec
>> ...lots of numbers between 1.0 and .101 which are >= y_1;
not
>selec
>> .0111 x_2
>> .1001 y_2
>> ...lots of numbers between 0.0 and .0111 which are =< x_2;
not
>selec
>> ...lots of numbers between 1.0 and .1001 which are >= y_2;
not
>selec
>> .01111 x_3
>> .10001 y_3

>> Where does .1 appear in your list?
>> [...] You tell me.
That particular sequence of digits (blanks are 0) is not in
the list.
>> Right. I don't know why you go back to binary, though; this
was
>>notation-independent. The sequence starts with x_0. First
time it
goes
>>to the right, you have y_0. Then the first time it goes to
the
left
>(and
>>between), that's x_1. And so on; at every step, the sequence
has not
yet
>>visi the current in between interval. All this has nothing to
do
>with
>>what representation one might use for the x_i's and y_i's.
>_Two_ sequences of digits of equal value _do_ appear in the
list,
>wherever you choose to set that particular limit on the
procedure
>by which you evaluate equal or not from sequences of digits.
>> No, the proof is not affec by this... or any other issue
regarding
>>how reals are expressed.
>> In particular: yes, the proof does work for binary
notation. There
>>is no problem one digit-sequence for x is not on the list,
but
another
>is
> for this proof, as well as for the usual diagonalization.
>> That's easy; ambiguity has no mystery. The eventually-1
sequence
>>s = A0111... denotes the same real as the eventually-0 one
s* =
>A1000...;
>>that is all. (Let * be this flip in either direction; if s
is not
>eventu-
>>ally constant let s* = s).
>> So just double up your list: given <a_i>, define a new list
by
>>b_2n = a_n, b_2n+1 = a_n*. E.g. your list above, L,
>>.01
>>.11
>becomes L*,
>>.01
>>.00111...
>>.11
>>.10111...
>Given L, L* is uniquely defined; diagonalization of L*
produces a
result
>d
>>that is not in L... and if d has a flip version, that's not
in L
>either.
>>So we get rid of this triviality.
>> Of course the x_0, y_0 ... proof proceeds the same way in
either
>>L or L*... since they list the same reals! And no, x = .1 =
.0111...
>>is not on the list.
>> What does setting the limit mean, by the way? There is a
crite-
>>rion for whether a_k, qua binary sequence, denotes the same
real as a
z
>>(i.e. s, t denote different reals iff both {s, t} and {s, t*}
>differ at
>>some digit; i.e. for some m, s(m) != t(m) and for some n,
s(n) !=
>t*(n).).
>>Either z shows up as an a_k, for some k, or it doesn't.
>(If you like to call that limit infinity then I'll call the
>corresponding entry the infinitieth entry.)
>> Eh?... Lists have no such thing; they have an a_k for each
>>finite k, and nothing else.
>> a) No loss is involved; all finite k is enough for
countable.
>>Having an w-th entry, or going up to w+6, or w^3, makes no
difference;
>one
>>can rearrange to an <a_k>. It takes a_u for all countable
ordinals
u
>to
>>get a higher cardinality, w_1 (aleph-one).
>> b) The list a_k = .111...1 (k times) does _not_ contain y =
.11111...
>>Sure, lim a_k = y; that doesn't make y an w-th entry of L
(nor
would
>it
>>help if it did -- per above). You can move everything up and
put in
y
>at
>>position 0; but that's a different list L'. In no way does L
now
>contain y
>>-- by some miracle.
>> c) Saying L could have contained this v is illusory. There
are
>>L' that contain v; so what? Be it L or L', the proof yields,
for any
>>list, an f(list) not on the list. Switching to some L'' is in
vain.
>> d) Limits of (sets/sequences of) a_k are whatever they are;
e.g.
>>list Q, and every real will be such a limit. Again... so
what?!
>> Whereever it is, it eventually gets picked, either as an
x_i or as a
>> y_i for some i. Thereafter, your pattern breaks down.
What does break down then is the proof idea.
>Thereafter there's no number x with x_i < x < y_i left in the
list,
>because: _you_ will evaluate that x as equal.
>Perhaps there are still many of those x and of the ...lots of
numbers
>left, or perhaps not.
>> The proof doesn't break down. If .1 is on L and gets picked
>>as x_7, then x_8, etc will be greater than it... and will
generate an
>x
>>that is not .1 .
>The point is that, yes, I construct a list, but only after
>_you tell me_ how you're going to decide whether numbers are
pairwise
>equal, larger than, or smaller than. That's the big
difference
>between reals and Richman reals, IMHO.
>> Only the fact that one of <, = or > holds is needed -- not
>>how you decide it. In any case, I did specify; I gave the
usual rule
>>for base 2. But it has no bearing; one can follow r such
that r >without reference to bases and digits.
>> If x_i = .1 for some i, then only finitely many of the
numbers you
>name
>> will be found among the x's.
I construct the i from the way you propose to evaluate
equality.
>That's most likely not a finite procedure.
>> Right; a real is not a finite object (nor is a list L).
Verifying
>>x = y means checking x(n) = y(n) for all n. Even if you
restrict
>this
>>to the recursive reals (i.e. n |--> x(n) being genera by a
program),
>>which _are_ a countable set by the way, there is no
algorithm that,
given
>>the programs of x and y , will decide x = y (in essence, the
Halting
>>Problem).
>> But this is beside the point. We are talking about all x,
not
>>just the recursive ones!... and looking for any function
from w onto
R,
>>whether computable in some sense or not. And there is none.
>> If y_i = .1 for some i, a similar comment applies to the
y's.
>> If neither, then you forgot to include .1 in your list to
start
with,
>> or you didn't follow the selection rules correctly.
The selection rules are yours to propose.
>What selection rules??
>> Defining x_0, y_0, ... in terms of a_k.
>> Ilias
>
===
Subject: Re: Cantor's diagonal proof
>>I have no idea why anybody replies to cranks writing about
Cantor. (If
you
>>try and prove with a straight face that the Cantor diagonal
proof is
wrong,
>>you are implicitly saying that every serious mathematician
of the 20th
>>century is also wrong, but you are right. This is a good
enough enough
>>definition of crank for me).
>The most common and amusing characteristic of the anti-crank
people
>is their ability to use the rules of the mathematical game in
the way
>they prefer and always in their own profit, whenever they
want. Their
>motto is: The rules are mine, so you are a crank because you
don't
>use them in the way I want.
You got it. It's perfectly valid to think of it all as a big
game -
these things you've been posting simply do not follow the
rules,
so by definition they're not mathematical proofs.
Whether you like it or not. Someone does get to say what the
rules are, and it's not you.
************************
===
Subject: Re: Cantor's diagonal proof
>>I have no idea why anybody replies to cranks writing about
Cantor. (If
you
>>try and prove with a straight face that the Cantor diagonal
proof is
>>wrong, you are implicitly saying that every serious
mathematician of the
>>20th century is also wrong, but you are right. This is a
good enough
>>enough definition of crank for me).
> The most common and amusing characteristic of the anti-crank
people
> is their ability to use the rules of the mathematical game
in the way
> they prefer and always in their own profit, whenever they
want. Their
> motto is: The rules are mine, so you are a crank because you
don't
> use them in the way I want.
The most common and amusing characteristic of the crank
is its ability to twist the rules of language in the peculiar
way
it prefers and always in its own profit, whenever it wants. Its
motto is: language is are mine, so I am right because the rest
of the world
doesn't twist it the way I do.
> The method used to synthesize the diagonal number in my
proof is
> exactly the same that the one used by Cantor. Therefore, if
my method
> works in a finite time (only because you like that it be
this way)
time? (not a mathematical concept)
> The only difference between my method and the diagonal
argument is
> that I show ALL THE INFORMATION that the diagonal argument
provides
> and Cantor just show and use the information that is useful
for his
> reasoning.
Que?
> The mistake in your reasoning is the following. When you
synthesize
Synthesize?
> the nth digit of S, the combinatory tells us that there are
10^n
combaintory? (I alsway though that was a verb).
> possible real numbers with n digits (even if n is infinite).
However,
> Cantor and you get the conclusion that S is not in the list
because
> S is different to the first n reals in the list. What's the
matter
> with the other (10^n n) reals in the list?
10^n n?
The list is the range of a function f: N -> R.
The point is that there is a number x such that x =/= f(n) for
all n in N.
The reason being that x and f(n) differ in the n-th digit.
(You can easily ensure that |x - f(n)| > 10^{-n}.)
>>2. So where is the error in yor proof? Insofar as it proves
anything,
it
>>proves that you cannot find S in the list in finite time
using the
process
>>you propose. This is hardly surprising, as S doesn't appear
in the list.
>>Too bad! Looks like you will not go down in history as the
man who turned
>>mathematics on its head!
> It is a pity if your mathematical castle is falling down due
to a
> Cantor's mistake, but mathematics is a science, not a
religion. Don't
> try to defend what it is not defendable.
Mathematics is not a science or a religion; mathematics is
mathematics.
--
===
Subject: Re: theorem vs. proposition
>What is the difference between a theorem and a proposition?
>>All statements that are true or false are propositions.
> In a given language which can assign truth values
exclusively of true
> or false to certain sentences of the language, to those
sentences
> which are assumed to be strictly either true or false are
called
> propositions. The truth value of the proposition may be
unknown but
> not provably unknowable (undecidable), since that would
violate its
> definition.
But in the context of presenting mathematical material (rather
than in
logic) one wouldn't call such a thing a proposition. If you
present
such a statement followed by a proof, then it should be called
a
theorem.
If you present it without a proof, it should be called a
conjecture
or hypothesis (in my opinion). I don't see, outside of logic,
why
anything would be called a proposition.
--
Mitch Harris
(remove q to reply)
===
Subject: Re: theorem vs. proposition
>What is the difference between a theorem and a proposition?
>>All statements that are true or false are propositions.
In a given language which can assign truth values exclusively
of true
> or false to certain sentences of the language, to those
sentences
> which are assumed to be strictly either true or false are
called
> propositions. The truth value of the proposition may be
unknown but
> not provably unknowable (undecidable), since that would
violate its
> definition.
> But in the context of presenting mathematical material
(rather than in
> logic) one wouldn't call such a thing a proposition. If you
present
> such a statement followed by a proof, then it should be
called a
theorem.
> If you present it without a proof, it should be called a
conjecture
> or hypothesis (in my opinion). I don't see, outside of
logic, why
> anything would be called a proposition.
In the first place, the term theorem is applied -- for better
or for
worse -- to a conjecture even if it has no proof attending it.
It's
also true that the term theorem is generally assumed to be
used only
for proven conjectures, but this is not strictly adhered to
(e.g.
FLT). So, one can make definitions in accordance with the way
things
are done or according to how things ought to be done. I chose
to
ballance both extremes.
Of course one can just adopt a convention to not use the term
proposition in mathematics, but I think this is too
restricting. (In
a similar manner I suppose that we can get by without labeling
equations with equation numbers of any kind, but at a loss of
clarity
and efficiency.) The term proposition has a stronger meaning
than
any of theorem, conjecture, or hypothesis.
For example, consider the case of this statement: For a given
N there
are M number of prime numbers in the first 10^{10^N} positive
integers. This statement is either true or false, definitely
NOT
undecidable, though perhaps for the moment unknown. It's
perfectly
decidable as true or false given enough time to test the
primeness of
all the numbers in the interval up to 10^{10^N}. Now, we could
certainly call such a statement a conjecture or a hypothesis,
or even
a theorem, yet the first two descriptors are not as
informative as the
term proposition, and the theorem descriptor is likely too
pretentious for such a statement, though that's also a matter
of
opinion.
For example, say a reader of a mathematics subject is interes
for
some reason in all the claims that are decidable in a given
field. To
such claims as propositions. But if you're asking if we can
get by
without the distinction available through the term
Proposition, the
answer is yes, but at at cost I think.
===
Subject: Re: theorem vs. proposition
>> What is the difference between a theorem and a proposition?
> All statements that are true or false are propositions.
>> In a given language which can assign truth values
exclusively of true
>> or false to certain sentences of the language, to those
sentences
>> which are assumed to be strictly either true or false are
called
>> propositions. The truth value of the proposition may be
unknown but
>> not provably unknowable (undecidable), since that would
violate its
>> definition.
What we are doing here is semantics, right?
So:
A proposition is a proposition.
A theorem should enlight the structure of a theory.
Frank
===
Subject: Re: Equations of type: cx^n - kabx - a^n - b^n = 0
> I've noticed, that equations of type:
> cx^n - kabx - a^n - b^n = 0
> should not have integer solutions for a;b
> natural numbers and of gcd=1 and for optiable
> fixed natural values of c and k.
> Are there some contra examples and for what
> c and k ?
> I'm not familiar with the word optiable.
I guess it's supposed to mean 'chosen' or 'given', not that
there's
any such word (mentioning that for the benefit of the original
poster).
> Your equation has scads of integer solutions with
> a, b, c, k, x natural and gcd(a, b) = 1. For example,
> a = 2, b = 3, c = 281, k = 1, n = 5, x = 1.
> You must have some conditions in mind that you are not
stating clearly.
===
Subject: Re: Practical problems with correlation dimension
> Hallo!
> I want to calculate the correlation dimension of a time
serie.
> What I have done
> I calcula the correlation integral C(r) (number of point
having a
> distance smaller than r) for different embedding dimensions.
Taking
> the slopes of the curve of log C(r) against log r for the
different
> embedding dimensions and plotting them against the embedding
dimension
> should result in a limes of the slopes: the correlation
dimension.
> My problem
> Which slope shall I take?
> In examples I saw in text books there is a nice limit of the
slopes
> with higher embedding dimensions. In my data I do not know
which slope
> I should take because the slope of the curve varies. If I
take the
> slope at a certain value of log r I can not get a limes.
> My curves (log C(r) against log r) can be seen in
> http://karlknoblich.4t.com/korrdim.jpg
> What to do? Does anybody knows such data and how to handle
it?
> Hope somebody can help!
> Karl
Value of C.D. depends on several factors: level of data
stationarity,
the length of a stationary segment, external noise, sampling
rate, and embedding dimension.
For example, if we choose stationary region of time series,
the length
of
this
region must be about 10^(D+2) points. Therefore to
determine D=10 we need about 10^6 points of suficiently
stationary
observable.
I guess you are trying apply the method of C.D. to EEG series,
if so, you'll be disappoin unless you can find a subject with
sufficiently (see above)
long segment of quasi-stationary data.
Some time ago I got similar results for EEG series.
If you want, you can check the paper at:
http://mindspring.com/~nldap/publ/ind2000.pdf
Hope this help,
D.Gribkov
===
Subject: Re: rotations w quaternions and eulers
o it..
>>Maybe the problem lies in the fact that the euler angles
describe a
>>series of rotations.. and thus I perhaps miss something when
converting
>>from Euler to quaternion.
> What do you do when you convert them? You should multiply
the matrices
> in the correct order, for example Rz*Ry*Rx for the order
x-axis, y-axis
> then z-axis (if you multiply column vectors on the right
side), then
> convert from this matrix to quaternion form. Otherwise
you'll have to
> give more details.
Let's say wx(t),wy(t),wz(t) are the angular velocities at time
t. Then I
create one quaternion for each angular velocity, resulting in
Wx(t),
Wy(t), Wz(t) and I multiply to get an overall quaternion W(t)
= Wx(t)
Wy(t) Wz(t)
quaternion A(t) = Ax(t) Ay(t) Az(t). Then, to get the new
orientation
after time dt I just use Euler integration and I have:
A(t+dt) = A(t) + A(t) W (t) dt
After that I extract Euler angles from the new quaternion.
This seems to
work when my all angles are initially 0 and there is a single
non-zero
angular velocity, but fails in other cases. Maybe the problem
is that I am
concatenating all the angular velocites together instead of
applying them
separately. Any ideas?
--
Christos Dimitrakakis
IDIAP (http://www.idiap.ch/~dimitrak/main.html)
===
Subject: irreducibility of polynomial
How do I check irreducibility of polynomial x^4+6x^2+2x+1 over
Z/5Z?
Can I apply theorem I found on Internet:
Let D be a unique factorization domain, let Q be the quotient
field
of D, and let f(x) be a primitive polynomial in D[x]. Then
f(x) is
irreducible in D[x] if and only if f(x) is irreducible in Q[x].
How is this theorem in accordance whith reducibility of x^2+1
over
Z/2Z?
===
Subject: Re: irreducibility of polynomial
>How do I check irreducibility of polynomial x^4+6x^2+2x+1
over Z/5Z?
First you show that it doesn't have linear factors - just try
all
elements of Z/5Z. Then you try to factor it as (x^2 + ax
+b)(x^2 + cx + d)
and try to solve for a,b,c and d. Since Z/5 is quite small,
this
is easily doable by hand.
(Actually, this polynomial *does* factor over Z/5Z; it even has
linear factors)
>Can I apply theorem I found on Internet:
> Let D be a unique factorization domain, let Q be the
quotient field
>of D, and let f(x) be a primitive polynomial in D[x]. Then
f(x) is
>irreducible in D[x] if and only if f(x) is irreducible in
Q[x].
No. Or rather, you can apply it, but it doesn't get you
anywhere.
Z/5Z is a field and hence its oqn quotient field.
Am I right in guessing that this is an exercise and that it
really asks to
show the irreducibility of x^4 + 6x^2 + 2x + 1 over Q? (The 6
seems weird
if it is really over Z/5Z). Because then you can apply this
theorem to
reduce it to the irreducibility of the polynomial over Z and
you can prove
that by proving irreducibility modulo some prime.
>How is this theorem in accordance whith reducibility of x^2+1
over
>Z/2Z?
Z/2Z is a field, so its quotient field is itself.
Peter
--
Peter van Rossum, <petervr@nmsu.edu> | Universal law of
linearity: for
all
Dept. of Mathematics, New Mexico | f : R -> R and for all x, y
in
R:
State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y)
===
Subject: Re: irreducibility of polynomial
>How do I check irreducibility of polynomial x^4+6x^2+2x+1
over Z/5Z?
 First you show that it doesn't have linear factors - just
try all
> elements of Z/5Z.
As a first step I would say that x^4 = 1 for all x != 0 in
Z/5Z.
The polynomial you get *must* have the same linear factors as
the original
(although possibly with different multiplicity).
--
===
Subject: Re: irreducibility of polynomial
> How do I check irreducibility of polynomial x^4+6x^2+2x+1
over Z/5Z?
Berlekamp's algorithm?
(But actually it isn't irreducible).
--
===
Subject: Re: irreducibility of polynomial
> How do I check irreducibility of polynomial x^4+6x^2+2x+1
over Z/5Z?
> Berlekamp's algorithm?
> (But actually it isn't irreducible).
Maple says...
> Factor(x^4+6*x^2+2*x+1) mod 5;
2
(x + 4) (x + 3 x + 3) (x + 3)
--
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: irreducibility of polynomial
> Can I apply theorem I found on Internet:
> Let D be a unique factorization domain, let Q be the
quotient field
> of D, and let f(x) be a primitive polynomial in D[x]. Then
f(x) is
> irreducible in D[x] if and only if f(x) is irreducible in
Q[x].
It seems to me to be true, since you may use the same
demonstration than
Gauss in Z, using the gcd of the coefficients of the
polynomial in D and
showing it is multiplicative.
But here it's useless, since D=Q.
--
Maxi
===
Subject: Re: Rational Limits
<4eCdnaUn0MJfppLdRVn-sQ@comcast.com>
<2MqdnaPRRLZIA5LdRVn-gQ@comcast.com>
<8765f46qhl.fsf@phiwumbda.org>
<AYmdnV-KyJmJ643dRVn-tw@comcast.com>
Discussion, linux)
>> A = {x | x < 0 or x = 0}
>> B = {x | exists n. x > 1/n}
>> Then A and B satisfy the requirements, and clearly 0 is the
witness
>> for the gaplessness of R for these sets. If x < 0 then x is
in A and
>> if x > 0 then x is in B.
>> The definition of gapless does *not* say that every
intersection of
>> nes open intervals is non-empty. It is not a consequence of
>> gaplessness, either.
>> Now, a consequence of gaplessness is the following (hope I
got it
>> right).
>> Theorem. Suppose we have two sequences (a_n) and (b_n)
satisfying the
>> following:
>> (1) (a_n) is increasing and (b_n) is decreasing.
>> (2) For all n, a_n < b_n.
>> (3) For all x, there exists n such that x > b_n or x < a_n.
>> Then there exists an x such that for all y, if x < y, then
y > b_n for
>> some n and if x > y, then y < a_n for some n.
>> It's not a particularly interesting theorem, but I believe
it's
>> correct and that it's close to what you had in mind above.
According
>> to this theorem, there must be an x such that for all y,
>> * if x < y, then y > 1/n for some n
> For all n.
>> * if x > y then y < 0.
> if x > y then y <= 0
Only trivially. x = 0 and so if x > 0, then y < 0 and hence y
<= 0.
>> But this is obviously true and the witness is x = 0.
> Except 0 is in set A.
Russell, can you read? Really?
Here's the definition that you have given repealy.
R has no gaps, i.e., if it is partitioned into two sets A and
B in
such a way that every member of A is less than every member of
B,
then there is a boundary point c, so that every point less
than c is
in A and every point greater than c is in B.
Nowhere in this definition does it claim that c is neither an
element
of A or B. It says that c is a boundary point, but that does
not
imply that c is not in A.
Why do you insist that my witness cannot be a witness? It
satisfies
exactly the criteria it ought.
--
Jesse Hughes
Yes, I'm one of those arrogant people who tries to be quotable.
There is actually at least one person who quotes me often.
--
===
Subject: Re: Rational Limits
>> I'm curious.
>> What does this set of intervals converge to?
>> Category mistake: seqences of points (in a topological
space)
>> may converge; sets of intervals don't.
> Can you suggest a better way to express the question?
> Is confusing enough without making category mistakes.
Wow! It's up to me to make sense of your ramblings since
you are unable to.
--
===
Subject: Re: Rational Limits
> I'm curious.
>> What does this set of intervals converge to?
>> Category mistake: seqences of points (in a topological
space)
>> may converge; sets of intervals don't.
>>
Can you suggest a better way to express the question?
> Is confusing enough without making category mistakes.
> Wow! It's up to me to make sense of your ramblings since
> you are unable to.
Of course. That is why I write them.
Why else would you read them?
Russell
- 2 many 2 count
===
Subject: Re: Rational Limits

>> I'm curious.
>> What does this set of intervals converge to?
>> Category mistake: seqences of points (in a topological
space)
>> may converge; sets of intervals don't.
>>
Can you suggest a better way to express the question?
> Is confusing enough without making category mistakes.
Wow! It's up to me to make sense of your ramblings since
> you are unable to.
 Of course. That is why I write them.
> Why else would you read them?
Watching you make stupid errors helps me avoid making them.
Even the worst of us can serve, as bad examples.
===
Subject: Re: Rational Limits
I'm curious.
> What does this set of intervals converge to?
Category mistake: seqences of points (in a topological space)
> may converge; sets of intervals don't.
 Can you suggest a better way to express the question?
> Is confusing enough without making category mistakes.
Nes sequences of intervals have an intersections, which is as
close
to convergence as you are going to get.
===
Subject: Re: Rational Limits
 Let A_1 be a non-empty open interval.
> Let A_n be a subset of A_(n-1).
> The intersection of A_n and A_(n+1) is:
> A_n x A_(n+1) = A_(n+1)
> We are assuming that every A_n is non-empty.
> The intersection of A is equal to some A_n.
So the intersection of the set { (0, 1/n) | n in N } is the
interval
> (0, 1/n) for some n? And hence, I suppose, we have a proof
that
there
> exists a smallest rational. And nothing here begs any
questions,
> nosiree.
Lame, Russell, bloody lame and stupid.
I'm curious.
> What does this set of intervals converge to?
It can't be 0 because 0 is one of the bounds.
> Assuming it converges to a real number,
> it would have to be the smallest real
> greater than 0. Does it converge to the empty set?
Yes, in the sense that the intersection of all such intervals
is empty.
> But note that the lower bounds do not form a (strictly)
increasing
> sequence, as is required by the theorem.
> That should make it easier to find c,
> the number that proves the reals are uncountable.
> When the lower bounds of the intervals are strictly
increasing and the
> upper bounds are strictly decreasing and X = R, then the
intersection
is
> never empty.
> This is part of the wikipedia prrof:
The two monotone sequences a and b move toward each other.
> By the gaplessness of R, some point c must lie between them.
> The claim is that c cannot be in the range of the sequence
x, and that is
> the contradiction.
> Are you saying that c must exist if the two sequences move
toward each
> other,
> but it doesn't exist if one of the sequences is constant?
I am saying that it need not exist if one sequence is
constant, as the
example showed ( It can, but doesn't have to) whereas, at
least for the
reals, if both sequences are strictly monotone, there must
always be at
least one real c between them.
But for the rationals, even for strictly monotone sequences of
rationals, there sometimes is no rational between such
sequences.
Example: Two such sequences of rationals squeezing in on
sqrt(2) have no
rational between them.
Thus the hypotheses of the theorem are not met.
Thus there is no such proof that the set of rationals is
uncountable.
Got it yet, Russal?
===
Subject: Re: Rational Limits
> Are you saying that c must exist if the two sequences move
toward each
> other,
> but it doesn't exist if one of the sequences is constant?
> I am saying that it need not exist if one sequence is
constant, as the
> example showed ( It can, but doesn't have to) whereas, at
least for the
> reals, if both sequences are strictly monotone, there must
always be at
> least one real c between them.
The real numbers might have gaps if one sequence is constant,
but they don't have gaps if both sequences are monotonic?
I am supposed to believe this?
It would be nice if someone gave a definition of what gapless
means.
The only definition I have seen so far is that its a property
the
real numbers have. I still think it is a novel way to say the
set is dense.
> But for the rationals, even for strictly monotone sequences
of
> rationals, there sometimes is no rational between such
sequences.
Sometimes there is an open interval with rational endpoints
that doesn't contain a rational.
Because you took an infinite number of intersections.
And all of these intersections were between an infinite open
set
with an infinite open subset of itself.
I would still like to know when is
A n B = B not equal to B?
Why would something that is true for any finite number
of intersections not be true for an infinite number
of intersections?
Russell
- 2 many 2 count
===
Subject: Re: Rational Limits
 Are you saying that c must exist if the two sequences move
toward
each
> other,
> but it doesn't exist if one of the sequences is constant?
I am saying that it need not exist if one sequence is
constant, as the
> example showed ( It can, but doesn't have to) whereas, at
least for the
> reals, if both sequences are strictly monotone, there must
always be at
> least one real c between them.
> The real numbers might have gaps if one sequence is constant,
> but they don't have gaps if both sequences are monotonic?
> I am supposed to believe this?
Where do you get that peculiar idea? It certainly is not part
of the
Wikipedia statements, nor of mine.
Do you know what a LUB (least upper bouund) is?
Do you know that R has the LUB property?
(Every non-empty set of reals which has any UB has a unique
LUB)
Do you know that Q does not have the LUB property?
> It would be nice if someone gave a definition of what
gapless means.
Gapless means having the LUB property (and teh corresponding
GLB
property).
> The only definition I have seen so far is that its a
property the
> real numbers have. I still think it is a novel way to say the
> set is dense.
You thing wrongly. Again.
> But for the rationals, even for strictly monotone sequences
of
> rationals, there sometimes is no rational between such
sequences.
> Sometimes there is an open interval with rational endpoints
> that doesn't contain a rational.
Only for an empty open interval whose endpoints are equal, and
in this
case a real inverval would also be empty.
> Because you took an infinite number of intersections.
> And all of these intersections were between an infinite open
set
> with an infinite open subset of itself.
You make it sound like only two sets, instead of infinitely
many, were
intersec.
> I would still like to know when is
> A n B = B not equal to B?
An equation is rarely equal to a set.
For intersections of any finite family of nes sets, meaning
each is
a proper subset of all its predecessors, that intersection
equals the
last, and therefore smallest, set in the family.
For intersections infinite families of nes sets, this cannot
be the
case because there is no last or smallest.
Do you insist that the intersection shall be equl to something
that does
not exist? You might by accident end up with a Pegasus that
way.
> Why would something that is true for any finite number
> of intersections not be true for an infinite number
> of intersections?
Finite sequences have a last member, infinite sequences do not.
That is the whole point of infinite sequences!
Why do you keep wanting to have something that does not exist?
Just greedy, I guess.
===
Subject: Re: Rational Limits
> Sometimes there is an open interval with rational endpoints
> that doesn't contain a rational.
> Only for an empty open interval whose endpoints are equal,
and in
this
> case a real inverval would also be empty.
> Because you took an infinite number of intersections.
> And all of these intersections were between an infinite open
set
> with an infinite open subset of itself.
> You make it sound like only two sets, instead of infinitely
many, were
> intersec.
I would still like to know when is
> A n B = B not equal to B?
> An equation is rarely equal to a set.
> For intersections of any finite family of nes sets, meaning
each is
> a proper subset of all its predecessors, that intersection
equals the
> last, and therefore smallest, set in the family.
> For intersections infinite families of nes sets, this cannot
be the
> case because there is no last or smallest.
There must have been a smallest if you are claiming it was the
empty set.
It doesn't matter if it was the last set.
Russell
- 2 many 2 count
===
Subject: Re: Rational Limits
> The real numbers might have gaps if one sequence is constant,
> but they don't have gaps if both sequences are monotonic?
> I am supposed to believe this?
The definition of gaps does not mention sequences in any way
whatsoever. The definition simply says every cut is genera by a
member of the set.
> It would be nice if someone gave a definition of what
gapless means.
You gave the definition yourself at the beginning of the
thread, quoting
wikipedia:
> Suppose a set R is
> linearly ordered, and
> densely ordered, i.e., between any two members there is
another, and
> has no endpoints, i.e., smallest or largest members, and
> has no gaps, i.e., if it is partitioned into two sets A and
B in such a
> way
> that every member of A is less than every member of B, then
there is a
> boundary point c, so that every point less than c is in A
and every
> point
> greater than c is in B.
> The only definition I have seen so far is that its a
property the
> real numbers have.
You are wrong twice in that statement. You have indeed seen a
definition
of gaplessness (it's quo just above from one of your own
postings),
and this definition does not even mention the real numbers.
The set R
mentioned there is intended to be any set that has the sta
properties.
>I still think it is a novel way to say the
> set is dense.
Density and gaplessness are different concepts, as has been
explained to
you many times. The integers are gapless, but not dense. The
rationals
are dense, but not gapless.
> Sometimes there is an open interval with rational endpoints
> that doesn't contain a rational.
Name one.
> Because you took an infinite number of intersections.
No, there is only one intersection. That intersection happens
to have an
infinite number of sets as its operands, but it is still a
single
operation.
Intersections and unions are quite unlike sums and products;
the set
operations can take as many operands as you like, even
infinitely many,
all in a single operation. The numeric operations are restric
to
working on pairs of numbers at a time. When we talk about an
infinite
sum, we are actually talking about a limit. Not so with
infinite
intersections. There is no limit involved.
> And all of these intersections were between an infinite open
set
> with an infinite open subset of itself.
No. You are talking about intersections of just two sets.
There is no
intersection like that in the proof. The only intersection
that appears
in the proof is a single intersection that involves an
infinite number of
sets, all at once.
> I would still like to know when is
> A n B = B not equal to B?
> Why would something that is true for any finite number
> of intersections not be true for an infinite number
> of intersections?
We are talking about just one intersection, not an infinite
number.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Rational Limits
<4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <HrvG26.Ms4@cwi.nl>
<O7OdnYSerPBP1ZLdRVn-hQ@comcast.com>
<877jzkxzdf.fsf@phiwumbda.org>
<jaqdnfu-6OAOAJLdRVn-jg@comcast.com>
<OuSdnWCa2ai95I3dRVn-ug@comcast.com>
<z8WdnQXNSaT4SI3dRVn-tA@comcast.com>
Discussion, linux)
>> Are you saying that c must exist if the two sequences move
toward each
>> other,
>> but it doesn't exist if one of the sequences is constant?
>> I am saying that it need not exist if one sequence is
constant, as the
>> example showed ( It can, but doesn't have to) whereas, at
least for the
>> reals, if both sequences are strictly monotone, there must
always be at
>> least one real c between them.
> The real numbers might have gaps if one sequence is constant,
> but they don't have gaps if both sequences are monotonic?
> I am supposed to believe this?
No, of course not. The real numbers have no gaps. You've
repealy
made the error of assuming that the witness c in the
definition of
without gaps does not lie in either A or B. It does. Indeed, it
must, because A and B are assumed to be a *partition*.
R has no gaps, i.e., if it is partitioned into two sets A and
B in
such a way that every member of A is less than every member of
B,
then there is a boundary point c, so that every point less
than c is
in A and every point greater than c is in B.
A pair of sets A and B partition R if (1) A and B are disjoint
(a n B = emptyset) and (2) A u B = R.
*Every* x in R, then, must lie in exactly one of the two sets
A and
B. This includes the boundary point c above.
,----
| Theorem. Suppose we have two sequences (a_n) and (b_n)
satisfying the
| following:
|
| (1) (a_n) is increasing and (b_n) is decreasing.
|
| (2) For all n, a_n < b_n.
|
| (3) For all x, there exists n such that x > b_n or x < a_n.
|
| Then there exists an x such that for all y, if x < y, then y
> b_n for
| some n and if x > y, then y < a_n for some n.
`----
Note that the theorem does *not* require that the witness x
satisfies
a_n < x < b_n for all n. However, in the case that (a_n) is
*strictly* increasing and (b_n) is *strictly* decreasing, one
can
prove that, indeed, a_n < x < b_n for all n.
There is no contradiction to be found here. Proofs available on
request.
> It would be nice if someone gave a definition of what
gapless means.
> The only definition I have seen so far is that its a
property the
> real numbers have. I still think it is a novel way to say the
> set is dense.
You've given it a million times. I've pas it in half a million.
Regardless of whether you think it's a novel way to say the
set is
dense, you haven't proven so. In fact, there are a quarter of a
million proofs that the dense set Q is *not* gapless floating
about
these threads.
>> But for the rationals, even for strictly monotone sequences
of
>> rationals, there sometimes is no rational between such
sequences.
> Sometimes there is an open interval with rational endpoints
> that doesn't contain a rational.
No.
> Because you took an infinite number of intersections.
An infinite intersection of nes open intervals need not be
open.
> And all of these intersections were between an infinite open
set
> with an infinite open subset of itself.
So, give us a proof that an infinite intersection of nes open
intervals need not be open. It would be swell if the proof
didn't
depend on the existence of a greatest natural number or a least
positive rational or any of your other dubious claims.
> I would still like to know when is
> A n B = B not equal to B?
You're babbling son.
> Why would something that is true for any finite number
> of intersections not be true for an infinite number
> of intersections?
Because not everything that is true of finite constructions
holds for
infinite constructions.
Here's an example that even you should agree to (but probably
won't)
Let A_n = {0,...,n}.
For any finite n, the union of A_0,..., A_n is finite, in fact
it is
A_n.
Do you conclude that the (infinite) union of A_0,A_1,... is
*also*
finite? (Please don't claim that it is A_{largest natural
number}.)
--
Jesse Hughes
How lucky we are to be able to hear how miserable Willie
Nelson could
imagine himself to be. -- Ken Tucker on Fresh Air
===
Subject: Re: More Accessible Algorithm-Maze
> Come on, this is easy!...
> I will post the answer, if no one else gets it sooner, in
the next few
days.
> Leroy
Yeah, it was easy - I solved by hand and there is only one
solution
that I could find. As for having a reliable internet
connection to
post the answer sooner than this, that's a different problem.
spoiler space
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
sequence of moves = #R3A58*RIGHT%%347*DOWN*F
We end up at F in the botttom right hand corner.
===
Subject: Re: More Accessible Algorithm-Maze
> Come on, this is easy!...
> I will post the answer, if no one else gets it sooner, in
the next few
days.
Leroy
> Yeah, it was easy - I solved by hand and there is only one
solution
> that I could find. As for having a reliable internet
connection to
> post the answer sooner than this, that's a different problem.
> spoiler space
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> .
> sequence of moves = #R3A58*RIGHT%%347*DOWN*F
> We end up at F in the botttom right hand corner.
YES!....Correct, of course...
Besides being a source of mindless fun (FUN!), I made this
maze (as
well as the other algorithm mazes I have pos) as an
inspiration for
others to make such puzzles themselves, perhaps making them a
lot more
difficult and creating them using a lot more creativity
(inspired by,
say, Game's Calculatrivia, for example, or inspired by actual
computer algorithms, or by mathematics or by word-puzzles, or
by image
puzzles, or by...).
;)
Leroy
Quet
===
Subject: Re: Factoring Conjecture: Triplets
<bupg09$jsehr$1@ID-206850.news.uni-berlin.de>
<Hrx89q.933@cwi.nl>
Discussion, linux)
> ...
> Now JSH want to start on Nora Baron because she has been so
effective
in
> showing him up for what he really is.
I guess I better be careful. The place where I work is easy to
find
online.
> For me it does not matter. Both the place where I work and
the place
> where I live are easy to find. They are in my signature. But
I wonder
> what the FBI will do when James puts them onto me.
Don't forget that James has the US Army, too. Generals *like*
him.
--
I'm talking about mathematics--hard, brutal, extreme ...
pushing your
mind beyond the limits to understand what no one else can
because
they're afraid to risk it all, to lose their freaking
worthless minds
in the push to know. --, for the Nike Derivator
===
Subject: Re: Factoring Conjecture: Triplets
...
> For me it does not matter. Both the place where I work and
the place
> where I live are easy to find. They are in my signature. But
I
wonder
> what the FBI will do when James puts them onto me.
 Don't forget that James has the US Army, too. Generals
*like* him.
O, dear me. But I think I will be safe from them until we put
a US
soldier in jail.
--
===
Subject: Re: Factoring Conjecture: Triplets
<87k73ix41a.fsf@phiwumbda.org> <Hry26G.7qA@cwi.nl>
Discussion, linux)
> ...
> For me it does not matter. Both the place where I work and
the
place
> where I live are easy to find. They are in my signature. But
I
wonder
> what the FBI will do when James puts them onto me.
Don't forget that James has the US Army, too. Generals *like*
him.
> O, dear me. But I think I will be safe from them until we
put a US
> soldier in jail.
You never know. The US Army has Yankee spies *everywhere*. I
hear
that one will be at CWI next Tuesday.
--
Conservative, n:
A statesman who is enamored of existing evils, as distinguished
from the Liberal who wishes to replace them with others.
-- Ambrose Bierce
===
Subject: Re: Factoring Conjecture: Triplets
> ...
> For me it does not matter. Both the place where I work and
the
place
> where I live are easy to find. They are in my signature. But
I
wonder
> what the FBI will do when James puts them onto me.
 Don't forget that James has the US Army, too. Generals
*like* him.
 O, dear me. But I think I will be safe from them until we
put a US
> soldier in jail.
 You never know. The US Army has Yankee spies *everywhere*. I
hear
> that one will be at CWI next Tuesday.
I will go in hiding.
--
===
Subject: Re: Factoring Conjecture: Triplets
> ...
> For me it does not matter. Both the place where I work and
the
place
> where I live are easy to find. They are in my signature.
But I
wonder
> what the FBI will do when James puts them onto me.
> Don't forget that James has the US Army, too. Generals
*like*
him.
O, dear me. But I think I will be safe from them until we put
a US
> soldier in jail.
You never know. The US Army has Yankee spies *everywhere*. I
hear
> that one will be at CWI next Tuesday.
> I will go in hiding.
But don't pick a 'spider's hole' in your home town cos they'll
look there.
Eventually. After one of your old friends sells you out.
===
Subject: Re: Final Rout of Synchronization Clocks in Relativity
 For instance, it MUST be wrong for an orbit 1 centimetre
above the
Earth's
> surface since the gravitational potential is the same there
as o the
ground and
> since we know that clock rates are not affec by movement. (to
discourage
> smartarse replies from the idiots here - you know who I mean
-
assume the earth
> is perfectly round and has no atmosphere)
So a clock circling the Earth at that height would not run at a
different rate
> from one at rest and so it woud remain always in synch with
the GC.
>>Beautiful, Henry.
>>Thanks for yet another great demo!
>> Oh thank you, Paul. Yes, I thought it was quite clever
myself.
>> It completely annihilates your argument and brings down
your religion.
Of course you assertion of what the outcome of an impossible
>experiment would be is fatal to GR.
>After all, you KNOW!
>Henry Wilson knows that the outcome of an unfeasible
experiment
>>never done MUST support Henry Wilson's idea of reality,
>>and thus will falsify GR.
>> Do you not agre that the gravitational potential 1cm
(better still, 1
um) ABOVE
>> the surface is near enough to that ON the surface, for the
purpose of
this
>> experiment?
Sure, Henry.
>As I said, since you know that the outcome of this experiment
>MUST falsify GR, there is nothing more to discuss.

>This is simply too stupid to be funny, Henry.
>I am bored by this game.
Paul
 This message typifies your tactics when you find yourself
well and truly
> cornered. You resort to ridicule, jokes, anything.... that
will divert
> attention from the question you cannot answer.
Well and truly cornered, indeed.
A guy thinking he can falsify a theory by asserting what
he think MUST be the result of an impossible experiment,
isn't amenable to reason.
So why shouldn't I ridicule the ridiculous?
But I am getting bored of doing that as well.
I think you definitely are deteriorating.
Keep it up, and you can feel safe from me.
It isn't fun to ridicule an idiot.
Paul
===
Subject: Sierpinski's curve
some one help me ?Please respond to email.THX for imformation.
===
Subject: Re: Sierpinski's curve
> some one help me ?Please respond to email.THX for
imformation.
Try this:
[ http://ecademy.agnesscott.edu/~lriddle/ifs/ifs.html ]
Sierpinski's curve = Sierpinski carpet.
Sierpinski's other curve = Sierpinski gasket.
--
http://www.math.ohio-state.edu/~edgar/
Subject: Topological genus of the human body
===
If we treat the human body as a topological surface, male and
female, what
is its genus? How many holes do we have in our bodies; what
counts as
a hole?
===
Subject: Re: Topological genus of the human body
> If we treat the human body as a topological surface, male
and female,
what
> is its genus? How many holes do we have in our bodies; what
counts as
> a hole?
Genus 3 to first order. However, if one counts the rupture of
an ear drum
then the Eustachian
tube could add a genus +1.
Now remember that this is gross topology at the macro scale.
===
Subject: Re: Topological genus of the human body
Originator: grubb@lola
>If we treat the human body as a topological surface, male and
female, what
> is its genus? How many holes do we have in our bodies; what
counts as
>a hole?
This may depend on whether your mouth is open or closed.
--Dan Grubb
===
Subject: Re: Topological genus of the human body
Originator: dwildstr@euclid.ucsd.edu (Jake Wildstrom)
The Prophet Daniel Grubb known to the wise as
grubb@math.niu.edu, opened the
Book of Words, and read unto the people:
>>If we treat the human body as a topological surface, male
and female,
what
>> is its genus? How many holes do we have in our bodies; what
counts as
>>a hole?
>This may depend on whether your mouth is open or closed.
Mathematics made difficult had a chapter on topology, which
drew a
distinction between a 'Quiet Man' (closed orifices, so a
surface of
genus 0) and a 'Loud Man' (constantly flatulent and speaking,
so, with
an oversimplified view of the digestive system, a surface of
genus
1). They made the rather disturbing point that two loud men
could be
interlocked, which is probably best not visualized.
+-------------------------------------------------------------+
| D. Jacob Wildstrom -- Math monkey and freelance thinker |
| Graduate Student, University of California at San Diego |
| A mathematician is a device for turning coffee into |
| theorems. -Alfred Renyi |
+-------------------------------------------------------------+
The opinions expressed herein are not necessarily endorsed by
the
University of California or math department thereof.
===
Subject: Re: Topological genus of the human body
charset=utf-8
===
Stuart Anderson <stuart@unsound.tv>
[CapitalEth][EDouble
Dot][Micro]
.b3
.b9[EDo
ubleDot].b9
> If we treat the human body as a topological surface, male
and female,
what
> is its genus? How many holes do we have in our bodies; what
counts as
> a hole?
Ignoring the nasal cavity altogether, it's a torus with a
single hole
(running from mouth to anus). So in this case, it's genus 1.
Taking the nasal cavity into account, you have an additional
hole (nose)
which is furthermore divided into two separate ones by the
nasal diaphragm.
So in this case the (actual) genus is 3.
All that, ignoring any artificial, pathological or otherwise
induced
passages (such as doubly communicating internal cysts) that
communicate
with
each other.
(My own body, for example, was at least genus 300, since on my
back's skin
I
have cysts which have crea self communicating corridors,
because of
cystic acne. At least 300, because at least this many were
removed by a
skin
surgeon)
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Topological genus of the human body
> (My own body, for example, was at least genus 300, since on
my back's
skin
> I have cysts which have crea self communicating corridors,
because of
> cystic acne. At least 300, because at least this many were
removed by a
> skin surgeon)
Fascinating... have you a picture on your website?
--
===
Subject: Re: Topological genus of the human body
>>(My own body, for example, was at least genus 300, since on
my back's
skin
>>I have cysts which have crea self communicating corridors,
because of
>>cystic acne. At least 300, because at least this many were
removed by a
>>skin surgeon)
> Fascinating... have you a picture on your website?
EWWWW! One hopes not.
Rick
p.s. Counting pores, I'd suggest that the number is far higher
than
300 for most, if not all, people.
===
Subject: Re: Topological genus of the human body
charset=utf-8
===
Rick Decker <rdecker@hamilton.edu>
[CapitalEth][EDouble
Dot][Micro]
.b3
.b9[EDo
ubleDot].b9
[snip]
> Rick
> p.s. Counting pores, I'd suggest that the number is far
higher than
> 300 for most, if not all, people.
Wrong. Pores are all dead ends, so they are essentially skin
folds, so they
don't alter the genus count. All orifices on the human body
with the
exception of the three main cavities (nasal, mouth, rectal)
eventually dead
end. Including urethral cavities, which eventually dead end on
the kidneys
and ear cavities which dead end inside the cochlea.
Makes sense, too, since if you look at the blastula while the
embryo grows,
it only forms a single fold, which eventually wraps around to
form the
digestive tract. All other organs are formed by folding
surface cells. The
zygote starts as genus 0, becomes genus 1 when it folds once
upon itself to
form the digestive system and changes to genii 2 and 3 when
the nasal
cavity
and diaphragm form. The formation of the rest of the organs do
not change
the genus, which stays 3, unless some external procedure
changes it. For
example a tracheaectomy (given to free aerial passages), would
change the
genus to 4, while shooting one on the head, would change the
genus to 4
also.
The rest about closed and open orifices is nonsense, since no
matter
how
tight, the openings are still there, unless somebody
cauterizes the opening
to form a skin/muscle joint that seals the cavity permanently.
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Topological genus of the human body
> ? Rick Decker <rdecker@hamilton.edu> ?????? ??? ??????
> [snip]
Rick
p.s. Counting pores, I'd suggest that the number is far higher
than
> 300 for most, if not all, people.
> Wrong. Pores are all dead ends, so they are essentially skin
folds, so
they
> don't alter the genus count. All orifices on the human body
with the
> exception of the three main cavities (nasal, mouth, rectal)
eventually
dead
> end. Including urethral cavities, which eventually dead end
on the
kidneys
> and ear cavities which dead end inside the cochlea.
> Makes sense, too, since if you look at the blastula while
the embryo
grows,
> it only forms a single fold, which eventually wraps around
to form the
> digestive tract. All other organs are formed by folding
surface cells.
The
> zygote starts as genus 0, becomes genus 1 when it folds once
upon itself
to
> form the digestive system and changes to genii 2 and 3 when
the nasal
cavity
> and diaphragm form. The formation of the rest of the organs
do not change
> the genus, which stays 3, unless some external procedure
changes it. For
> example a tracheaectomy (given to free aerial passages),
would change the
> genus to 4, while shooting one on the head, would change the
genus to 4
> also.
> The rest about closed and open orifices is nonsense, since no
matter
how
> tight, the openings are still there, unless somebody
cauterizes the
opening
> to form a skin/muscle joint that seals the cavity
permanently.
I wonder how to designate rectal Kleins like or Peter Brown?
===
Subject: Re: Topological genus of the human body
charset=utf-8
===
Robin Chapman
<rjc@ivorynospamtower.freeserve.co.uk>
[CapitalEth][EDouble
Dot][Micro]
.b3
.b9[EDo
ubleDot].b9
(My own body, for example, was at least genus 300, since on my
back's
skin
> I have cysts which have crea self communicating corridors,
because
of
> cystic acne. At least 300, because at least this many were
removed by a
> skin surgeon)
> Fascinating... have you a picture on your website?
No, but I have a nice picture of something else there. Wanna
see it?
> --
>
Needless to say, I had the last laugh.
> Alan Partridge, _Bouncing Back_ (14 times)
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Lie algebras, matrix exponential help!
Given a function f:R^n->R, we have that
d/dt(f(tX)) (at t=0) = Df . X
where Df is the Jacobian, and . is the matrix product (X is a
column
vector).
Thus from knowing d/dt(f(tX)) (at t=0) for all X (or indeed
just a selection
of
X that spans R^n), we can find Df.
===
Subject: Re: Lie algebras, matrix exponential help!
> Given a function f:R^n->R, we have that
> d/dt(f(tX)) (at t=0) = Df . X
> where Df is the Jacobian, and . is the matrix product (X is
a column
vector).
> Thus from knowing d/dt(f(tX)) (at t=0) for all X (or indeed
just a
selection of
> X that spans R^n), we can find Df.
Thanks!
===
Subject: Re: A plane geomerty problem
In-reply-to: Stephen J. Herschkorn
<herschko@rutcor.rutgers.eduThe bisector of angle ABC intersects segment AC at D. The
bisector
>of angle ACB intersects segment AB at E. Segments BD and CE
>are congruent. True or false: triangle ABC is isosceles?
Suppose we have a triangle, ABC. Let the side AB be the base
of the
triangle. Place a point D on the line containing A and B so
that AD is
perpindicular to CD; that is, ACD is a right triangle. The
length of
side CD, |CD|, is the altitude of ABC. Since all right
triangles which
share a common angle are similar, |CD|/|AC| is a function
solely of the
angle at A, call it sin(A). Thus, the altitude of ABC is
|AC|sin(A).
Since the area of a triangle is 1/2 of the base times the
altitude, we
get that the area of ABC is equal to |AB||AC|sin(A)/2.
The area of triangle ABC in the problem above is
|AB||BC|sin(B)/2.
Since BD bisects angle ABC, we get that the area of triangle
BCD is
|BC||BD|sin(B/2)/2 and the area of triangle ABD is
|AB||BD|sin(B/2)/2.
Thus, the area of triangle ABC is (|AB|+|BC|)|BD|sin(B/2)/2.
Therefore,
|AB||BC|sin(B)/2 = (|AB|+|BC|)|BD|sin(B/2)/2
which yields
|BD| |BD| sin(B)
---- + ---- = -------- [1]
|AB| |BC| sin(B/2)
Equation [1] says that the sum of the bisector divided by each
of the
adjacent sides is a function solely of the angle between the
adjacent
sides. Consider a triangle in which |AB| = |BC| = 1, it is
easy to see
that |BD| decreases monotonically from 1 to 0 as angle B
increases from
0 to pi. Thus, the function sin(B)/sin(B/2) decreases
monotonically
from 2 to 0 as B increases from 0 to pi.
The analogue of [1] for angle ACB is
|CE| |CE| sin(C)
---- + ---- = -------- [2]
|AC| |BC| sin(C/2)
Since |BD| = |CE| we can combine [1] and [2] to get
1 1 sin(C) 1 1 sin(B)
( ---- + ---- ) -------- = ( ---- + ---- ) -------- [3]
|AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
If angle ABC is less than angle ACB, then |AC| < |AB| and,
from the
discussion above, sin(C)/sin(C/2) < sin(B)/sin(B/2). Therefore,
1 1 sin(C) 1 1 sin(B)
( ---- + ---- ) -------- < ( ---- + ---- ) --------
|AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
If angle ABC is greater than angle ACB, then |AC| > |AB| and,
from the
discussion above, sin(C)/sin(C/2) > sin(B)/sin(B/2). Therefore,
1 1 sin(C) 1 1 sin(B)
( ---- + ---- ) -------- > ( ---- + ---- ) --------
|AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
Thus, angle ABC must equal angle ACB and the triangle is
isosceles.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: A plane geomerty problem
>>The bisector of angle ABC intersects segment AC at D. The
bisector
>>of angle ACB intersects segment AB at E. Segments BD and CE
>>are congruent. True or false: triangle ABC is isosceles?
>Suppose we have a triangle, ABC. Let the side AB be the base
of the
>triangle. Place a point D on the line containing A and B so
that AD is
>perpindicular to CD; that is, ACD is a right triangle. The
length of
>side CD, |CD|, is the altitude of ABC. Since all right
triangles which
>share a common angle are similar, |CD|/|AC| is a function
solely of the
>angle at A, call it sin(A). Thus, the altitude of ABC is
|AC|sin(A).
>Since the area of a triangle is 1/2 of the base times the
altitude, we
>get that the area of ABC is equal to |AB||AC|sin(A)/2.
[etc.]
A somewhat intuitive but much easier proof is to draw a circle
around
the triangle. Suppose there is a triangle that satisfies the
condition and that line BC and angle BAC are specified. Then,
the
point A can rotate around the circle and, as long as it does
not cross
the line BC, will retain the same value, no matter what the
shape of
the resulting triangle, for the angle BAC. (Length BC =
diameter times
sine of angle BAC.)
When A is close to B, the bisectors of the two other angles, as
described in the problem, will obviously be unequal, and one
much
greater than the other. As point A approaches the point P on
the
diameter perpendicular to BC, one of the bisectors increases,
and the
other decreases. It is clear that the two bisectors can have
equal
lengths only when point A reaches point P.
Perhaps you can take this intuitive proof and make it more
rigorous.
===
Subject: Re: A plane geomerty problem
In-reply-to: Stephen J. Herschkorn
<herschko@rutcor.rutgers.edu>The bisector of angle ABC intersects segment AC at D. The
bisector
>>of angle ACB intersects segment AB at E. Segments BD and CE
>>are congruent. True or false: triangle ABC is isosceles?
>Suppose we have a triangle, ABC. Let the side AB be the base
of the
>triangle. Place a point D on the line containing A and B so
that AD is
>perpindicular to CD; that is, ACD is a right triangle. The
length of
>side CD, |CD|, is the altitude of ABC. Since all right
triangles which
>share a common angle are similar, |CD|/|AC| is a function
solely of the
>angle at A, call it sin(A). Thus, the altitude of ABC is
|AC|sin(A).
>Since the area of a triangle is 1/2 of the base times the
altitude, we
>get that the area of ABC is equal to |AB||AC|sin(A)/2.
>The area of triangle ABC in the problem above is
|AB||BC|sin(B)/2.
>Since BD bisects angle ABC, we get that the area of triangle
BCD is
>|BC||BD|sin(B/2)/2 and the area of triangle ABD is
|AB||BD|sin(B/2)/2.
>Thus, the area of triangle ABC is (|AB|+|BC|)|BD|sin(B/2)/2.
Therefore,
> |AB||BC|sin(B)/2 = (|AB|+|BC|)|BD|sin(B/2)/2
>which yields
> |BD| |BD| sin(B)
> ---- + ---- = -------- [1]
> |AB| |BC| sin(B/2)
>Equation [1] says that the sum of the bisector divided by
each of the
>adjacent sides is a function solely of the angle between the
adjacent
>sides. Consider a triangle in which |AB| = |BC| = 1, it is
easy to see
>that |BD| decreases monotonically from 1 to 0 as angle B
increases from
>0 to pi. Thus, the function sin(B)/sin(B/2) decreases
monotonically
>from 2 to 0 as B increases from 0 to pi.
>The analogue of [1] for angle ACB is
> |CE| |CE| sin(C)
> ---- + ---- = -------- [2]
> |AC| |BC| sin(C/2)
>Since |BD| = |CE| we can combine [1] and [2] to get
> 1 1 sin(C) 1 1 sin(B)
> ( ---- + ---- ) -------- = ( ---- + ---- ) -------- [3]
> |AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
>If angle ABC is less than angle ACB, then |AC| < |AB| and,
from the
>discussion above, sin(C)/sin(C/2) < sin(B)/sin(B/2).
Therefore,
> 1 1 sin(C) 1 1 sin(B)
> ( ---- + ---- ) -------- < ( ---- + ---- ) --------
> |AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
>If angle ABC is greater than angle ACB, then |AC| > |AB| and,
from the
>discussion above, sin(C)/sin(C/2) > sin(B)/sin(B/2).
Therefore,
> 1 1 sin(C) 1 1 sin(B)
> ( ---- + ---- ) -------- > ( ---- + ---- ) --------
> |AB| |BC| sin(C/2) |AC| |BC| sin(B/2)
>Thus, angle ABC must equal angle ACB and the triangle is
isosceles.
formatting a couple of the lines. Hopefully, the quoting will
improve
the ASCII-art formatting.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: Fibonacci PHI, PI, e Relation
But the 99th Fibonacci number won't be divisible by 99 - in
fact,
> it's not divisible by 3; you can work out that a Fibonacci
number
> is a multiple of 3 if and only if its position is a multiple
of 4,
> which 99 isn't.
> Also, the 12th position is *not* the first time the Fibonacci
> number is exactly divisible by its position; this happens at
> 0, 1, and 5 before it happens at 12.
I stand correc. I should have said position 12 is the first
time
the position^2 = Fibonacci.
I don't like position 0 because I question whether the sequence
actually starts with 0. You get the same resulting sequence if
you
start with 1. In other posts here we discussed the derivation
of left
truncatable primes from the fibonacci sequence, which only
worked if
you elimina the 0 from the sequence.
Brings up the interesting osophical question: did everything
(the
universe) start from nothing (0) or was something (1) there to
start
the process? This always seems to come back to 0 1, + -, black
white,
open closed, good bad, God Satan, ... on and on. Sorry got WAY
off
the point there.
===
Subject: Re: Fibonacci PHI, PI, e Relation
> The decimal positions out to 10,000 digits that are all the
same in
> Pi, E and Phi are:
> {12, 99, 169, 395, 499, 595, 606, 693, 824, 827, 840, 940,
1282, 1291,
> 1384,
> 1594, 1705, 1742, 1905, 2020, 2060, 2153, 2257, 2302, 2359,
2367,
> 2507, 2546,
> 2557, 2710, 2724, 2791, 2832, 2857, 3036, 3051, 3280, 3309,
3429,
> 3497, 3518,
> 3591, 3651, 3709, 3867, 4210, 4292, 4390, 4493, 4719, 4826,
4859,
> 4862, 4892,
> 4934, 4940, 5087, 5315, 5427, 5480, 5488, 5653, 5699, 6155,
6426,
> 6617, 6838,
> 6854, 7113, 7155, 7202, 7358, 7390, 7659, 7685, 7721, 7761,
7816,
> 7833, 7867,
> 7923, 8417, 8570, 8611, 8653, 8731, 8914, 9051, 9077, 9133,
9283,
> 9286, 9310,
> 9704, 9717, 9724, 9805, 9897}
Jason, thanks. I will use your post as a source for playing
with
these numbers a bit further. Outside of my excel spreadsheet I
currently have no other means to develop these numbers on my
own so I
depend on internet sources. Do you use any particular program
language or application to develop these numbers?
===
Subject: Re: Fibonacci PHI, PI, e Relation
>The decimal positions out to 10,000 digits that are all the
same in
>Pi, E and Phi are:
>{12, 99, 169, 395, 499, 595, 606, 693, 824, 827, 840, 940,
1282, 1291,
>1384,
>1594, 1705, 1742, 1905, 2020, 2060, 2153, 2257, 2302, 2359,
2367,
>2507, 2546,
>2557, 2710, 2724, 2791, 2832, 2857, 3036, 3051, 3280, 3309,
3429,
>3497, 3518,
>3591, 3651, 3709, 3867, 4210, 4292, 4390, 4493, 4719, 4826,
4859,
>4862, 4892,
>4934, 4940, 5087, 5315, 5427, 5480, 5488, 5653, 5699, 6155,
6426,
>6617, 6838,
>6854, 7113, 7155, 7202, 7358, 7390, 7659, 7685, 7721, 7761,
7816,
>7833, 7867,
>7923, 8417, 8570, 8611, 8653, 8731, 8914, 9051, 9077, 9133,
9283,
>9286, 9310,
>9704, 9717, 9724, 9805, 9897}
Huh - in particular there's 98 such places, which agrees quite
well
with the 100 predic if they're independent normal numbers...
************************
===
Subject: Re: Fibonacci PHI, PI, e Relation
> Huh - in particular there's 98 such places, which agrees
quite well
> with the 100 predic if they're independent normal numbers...
shucks, we coulda been famous if it turned out different
--
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Fibonacci PHI, PI, e Relation
>> Huh - in particular there's 98 such places, which agrees
quite well
>> with the 100 predic if they're independent normal numbers...
>shucks, we coulda been famous if it turned out different
There's still hope - could be there are no matches at all
between
position 10,000 and 10,000,000.
************************
===
Subject: Re: Fibonacci PHI, PI, e Relation
> Huh - in particular there's 98 such places, which agrees
quite well
>> with the 100 predic if they're independent normal numbers...
shucks, we coulda been famous if it turned out different
> There's still hope - could be there are no matches at all
between
> position 10,000 and 10,000,000.
I'm just thrilled there's a match at the 2020th position.
Carlson2020
is my web site! I guess if you look hard enough you can find
coincidences everywhere.
===
Subject: Re: Left Truncatable Primes Derived from Fibonacci
> Here are the next few.
> {1,1,1}
> {2,2,1}
> {3,3,1}
> {21,7,3}
> {377,13,29}
> {2584,17,152}
> {46368,23,2016}
> {39088169,37,1056437}
> {701408733,43,16311831}
> {4807526976,47,102287808}
> {86267571272,53,1627690024}
>[sniped remainder of sequence]
Jason, thanks. Do you (or anyone reading this) use a
programming
language or special application to generate these numbers. I am
looking for a math package (beyond excel) to do things similar
to what
you provided above.
Thanks for your input!
Bob
===
Subject: Re: Left Truncatable Primes Derived from Fibonacci
> Here are the next few.
{1,1,1}
{2,2,1}
{3,3,1}
{21,7,3}
{377,13,29}
{2584,17,152}
{46368,23,2016}
{39088169,37,1056437}
{701408733,43,16311831}
{4807526976,47,102287808}
{86267571272,53,1627690024}
[sniped remainder of sequence]
> Jason, thanks. Do you (or anyone reading this) use a
programming
> language or special application to generate these numbers. I
am
> looking for a math package (beyond excel) to do things
similar to what
> you provided above.
I used Python when I replied to your original post.
http://www.python.org
It's free and it can handle the kind of math I play around with
(mostly really large integers). And there are a lot of add-ons
available, such as the GMPY module referenced below which has
neat
functions like is_prime(), next_prime() and fib(). It makes a
great
step-up from Excel.
Some things to keep in mind once you get your hands on such a
package:
Although you can easily work with individual numbers that have
20,000
digits,
you can't loop through _all_ 20,000 digit numbers.
Outputting the numbers can be horribly slow. In the program
below, I just
settled for knowing how many digits there were, didn't need to
actually
see them.
If you actually _do_ need to see them, say, because you're
looking for
patterns in the binary representation (which GMPY has a
function for)
of the nth fibonacci number (which GMPY has a function for),
you'll
find it more efficient to write the results to a file.
The program below tests all the left truncatable primes up to
9986113
and prints any that fail your test (the left truncatable
primes actually
continue up to 357686312646216567629137, but as mentioned
above, I can't
loop through all the integers up to that, so I cut it off at
10 million).
# start of fib_ltp.py
import gmpy
ltps = [ 2, 3, 5, 7, 13, 17, 23, 37, 43, 47,
53, 67, 73, 83, 97, 113, 137, 167, 173, 197,
223, 283, 313, 317, 337, 347, 353, 367, 373, 383,
397, 443, 467, 523, 547, 613, 617, 643, 647, 653,
673, 683, 743, 773, 797, 823, 853, 883, 937, 947,
953, 967, 983, 997,
1223, 1283, 1367, 1373, 1523, 1613, 1823, 1997, 2113, 2137,
2347, 2383,
2467, 2617, 2647, 2683, 2797, 2953, 3137, 3167, 3313, 3347,
3373, 3467,
3547, 3613, 3617, 3643, 3673, 3797, 3823, 3853, 3947, 3967,
4283, 4337,
4373, 4397, 4523, 4547, 4643, 4673, 4937, 4967, 5113, 5167,
5197, 5347,
5443, 5647, 5653, 5683, 5743, 5953, 6113, 6173, 6197, 6317,
6337, 6353,
6367, 6373, 6397, 6547, 6653, 6673, 6823, 6883, 6947, 6967,
6983, 6997,
7283, 7523, 7547, 7643, 7673, 7823, 7853, 7883, 7937, 8167,
8317, 8353,
8443, 8467, 8647, 9137, 9173, 9283, 9337, 9397, 9467, 9547,
9613, 9643,
9743, 9883, 9967,
12113, 12347, 12647, 12953, 13313, 13613, 13967, 15443, 15647,
15683, 16547,
16673, 16823, 16883, 18353, 18443, 21283, 21523, 21613, 21997,
23167, 24337,
24373, 24547, 24967, 26113, 26317, 26947, 27283, 27673, 27823,
27883, 29137,
29173, 31223, 32467, 32647, 32797, 33347, 33547, 33613, 33617,
33797, 33967,
34283, 34337, 34673, 36353, 36373, 36653, 36947, 36997, 37547,
37643, 37853,
38167, 38317, 39397, 39883, 42467, 42683, 42797, 42953, 43313,
43613, 43853,
45197, 45953, 46337, 46997, 48353, 48647, 49547, 49613, 51283,
51613, 53617,
54547, 54673, 56113, 56197, 56983, 57283, 57853, 59467, 59743,
61223, 61283,
61613, 62137, 62347, 62383, 62467, 62617, 62683, 63313, 63347,
63467, 63617,
63823, 63853, 64283, 64373, 64937, 65167, 65647, 66173, 66337,
66373, 66653,
66883, 66947, 67523, 67547, 67853, 67883, 68443, 69337, 69467,
72383, 72467,
72617, 72647, 72797, 72953, 73547, 73613, 73643, 73673, 73823,
75167, 75347,
75653, 75683, 75743, 76367, 76673, 76883, 78167, 78317, 78467,
79283, 79337,
79397, 79613, 79967, 81223, 81283, 81373, 83137, 83617, 84523,
84673, 84967,
86113, 86197, 86353, 87523, 87547, 87643, 87853, 89137, 91283,
91367, 91373,
91823, 91997, 92347, 92383, 92467, 92647, 92683, 93967, 94397,
94547, 95443,
96337, 96353, 96823, 96997, 97283, 97523, 97547, 97673, 97883,
98317, 98443,
98467, 99137, 99173, 99397, 99643,
121283, 121523, 121997, 124337, 126317, 132647, 133967,
136373, 139397,
139883, 162683, 163853, 181283, 184523, 184967, 186113,
187547, 192347,
192383, 195443, 196337, 213613, 215443, 231223, 233347,
233617, 234673,
236653, 236947, 237547, 242467, 242797, 243613, 261223,
264283, 266947,
267523, 272383, 273613, 273643, 275167, 276673, 276883,
279337, 279397,
279613, 279967, 291367, 291373, 291997, 294397, 296353,
297523, 299137,
313613, 318443, 326113, 326947, 327673, 327823, 332467,
336353, 336373,
336653, 336997, 337853, 338167, 342467, 343313, 345953,
346337, 348353,
356113, 356197, 357283, 361223, 362137, 362347, 363313,
364373, 364937,
366173, 367547, 367853, 367883, 368443, 372797, 373613,
373823, 375743,
378167, 378317, 378467, 379283, 379397, 381223, 381373,
384673, 387853,
391283, 391367, 391373, 391823, 392347, 392383, 392467,
392647, 395443,
396353, 396997, 397283, 397547, 397673, 398467, 399137,
399173, 399643,
421997, 424547, 424967, 427283, 427883, 429137, 432797,
433967, 439883,
453617, 454547, 454673, 459467, 462467, 463313, 463823,
465167, 466373,
481373, 492467, 492647, 493967, 496997, 498467, 499397,
513313, 516673,
516883, 534283, 536353, 536947, 537547, 537853, 542467,
542683, 542797,
543313, 543853, 549547, 563467, 564373, 564937, 566173,
566653, 566947,
567883, 573673, 576883, 578167, 578317, 578467, 579283,
579613, 579967,
594397, 597523, 597673, 612113, 613967, 616547, 616673,
621997, 626113,
626317, 626947, 627673, 629137, 631223, 632647, 633613,
633797, 633967,
636353, 636653, 636947, 636997, 638317, 642683, 642797,
642953, 649613,
653617, 656113, 659467, 661613, 662617, 663823, 663853,
666173, 667547,
667883, 672953, 673613, 673643, 675347, 675743, 676883,
686197, 686353,
687523, 691997, 692347, 692467, 692647, 693967, 696823,
697523, 697673,
721283, 721613, 721997, 723167, 724547, 724967, 727673,
727823, 729173,
732467, 738317, 739397, 751613, 753617, 759467, 763823,
766373, 781283,
783137, 786197, 787547, 789137, 792383, 792647, 793967,
796337, 798443,
812347, 813613, 816547, 816883, 818353, 833347, 833617,
834283, 837853,
843613, 845197, 846997, 848647, 861613, 866653, 867547,
869467, 872383,
872647, 872953, 873643, 875683, 878167, 878467, 879283,
891823, 891997,
894547, 896353, 912647, 912953, 915683, 918353, 918443,
921523, 924337,
924967, 926113, 932647, 933613, 933797, 933967, 934673,
946997, 951283,
956113, 959467, 961283, 961613, 962617, 962683, 964283,
964373, 965647,
966337, 966373, 966653, 966883, 969467, 973547, 973823,
975743, 976883,
979283, 979337, 981283, 981373, 983617, 986113, 986197,
987523, 995443,
997547, 998443,
1237547, 1261223, 1279337, 1297523, 1326947, 1327673, 1332467,
1336997,
1338167, 1356197, 1368443, 1384673, 1516883, 1537853, 1549547,
1563467,
1564373, 1629137, 1632647, 1633967, 1636997, 1686197, 1686353,
1812347,
1813613, 1818353, 1833347, 1872953, 1875683, 1891997, 1896353,
1965647,
1966337, 1987523, 1998443, 2126317, 2184967, 2186113, 2195443,
2313613,
2336353, 2345953, 2364937, 2366173, 2373823, 2375743, 2379283,
2379397,
2396353, 2421997, 2424547, 2424967, 2427883, 2433967, 2463313,
2465167,
2616673, 2636353, 2642797, 2649613, 2667883, 2675347, 2723167,
2729173,
2738317, 2759467, 2763823, 2796337, 2912953, 2961613, 2969467,
2973547,
2979337, 3136373, 3192347, 3233617, 3243613, 3272383, 3276883,
3279337,
3291367, 3291373, 3294397, 3299137, 3327673, 3348353, 3364937,
3372797,
3378317, 3381223, 3396353, 3396997, 3398467, 3399173, 3433967,
3454673,
3513313, 3549547, 3563467, 3564937, 3573673, 3578167, 3578467,
3579613,
3597673, 3616673, 3626113, 3626947, 3633797, 3636947, 3673613,
3675743,
3692347, 3692467, 3724967, 3732467, 3739397, 3751613, 3786197,
3787547,
3834283, 3837853, 3872647, 3879283, 3912647, 3915683, 3918353,
3918443,
3924337, 3926113, 3932647, 3946997, 3961283, 3964283, 3966883,
3969467,
3986113, 3987523, 3995443, 3997547, 4215443, 4233347, 4234673,
4261223,
4279337, 4279967, 4296353, 4297523, 4326947, 4327823, 4338167,
4384673,
4392383, 4398467, 4536947, 4537853, 4542467, 4543313, 4549547,
4563467,
4578467, 4579283, 4594397, 4597673, 4626113, 4627673, 4632647,
4696823,
4816883, 4833617, 4837853, 4873643, 4891823, 4921523, 4933967,
4966373,
4981373, 4986197, 4987523, 5181283, 5187547, 5196337, 5337853,
5345953,
5367853, 5379397, 5391367, 5391823, 5396353, 5397283, 5397673,
5424967,
5427283, 5432797, 5439883, 5462467, 5493967, 5499397, 5613967,
5661613,
5667547, 5673643, 5675347, 5697673, 5724967, 5729173, 5753617,
5766373,
5792383, 5915683, 5934673, 5961283, 5964373, 5979283, 6126317,
6132647,
6139883, 6184967, 6231223, 6234673, 6237547, 6266947, 6273643,
6275167,
6279397, 6336373, 6342467, 6346337, 6367547, 6367853, 6367883,
6368443,
6373613, 6379283, 6387853, 6391373, 6395443, 6396997, 6397283,
6398467,
6399137, 6399643, 6421997, 6424547, 6463823, 6492467, 6499397,
6516673,
6516883, 6536947, 6537547, 6543853, 6576883, 6579283, 6597673,
6621997,
6626947, 6631223, 6633967, 6666173, 6673613, 6676883, 6687523,
6692347,
6721283, 6738317, 6813613, 6878167, 6878467, 6891823, 6912953,
6915683,
6933967, 6934673, 6946997, 6961613, 6962617, 6975743, 6981283,
6986197,
6987523, 7233617, 7236947, 7237547, 7266947, 7291367, 7368443,
7392647,
7399643, 7542797, 7543853, 7564373, 7567883, 7573673, 7594397,
7621997,
7626947, 7629137, 7636997, 7659467, 7662617, 7663823, 7692647,
7693967,
7696823, 7812347, 7813613, 7816547, 7816883, 7818353, 7861613,
7867547,
7869467, 7894547, 7896353, 7924967, 7932647, 7981283, 7983617,
7986113,
7986197, 7995443, 7998443, 8121283, 8133967, 8184523, 8192383,
8373823,
8391283, 8397283, 8397547, 8427283, 8432797, 8454673, 8463313,
8481373,
8493967, 8498467, 8613967, 8616547, 8666173, 8672953, 8675743,
8676883,
8721997, 8724967, 8727673, 8729173, 8739397, 8759467, 8787547,
8966653,
8979337, 9121523, 9139397, 9139883, 9162683, 9184523, 9187547,
9192347,
9243613, 9267523, 9279337, 9313613, 9318443, 9326113, 9336373,
9345953,
9356197, 9362137, 9364373, 9366173, 9391373, 9427883, 9433967,
9454547,
9516673, 9516883, 9536353, 9537547, 9542683, 9563467, 9564937,
9566173,
9576883, 9579967, 9594397, 9616547, 9633797, 9636947, 9636997,
9642683,
9662617, 9663853, 9687523, 9693967, 9729173, 9751613, 9759467,
9789137,
9812347, 9813613, 9818353, 9833347, 9833617, 9834283, 9848647,
9861613,
9866653, 9869467, 9875683, 9878467, 9879283, 9891823, 9891997,
9915683,
9918353, 9933613, 9951283, 9956113, 9961283, 9961613, 9962683,
9966883,
9973547, 9979283, 9979337, 9981373, 9986113]
i = 1
while i<9986114:
if gmpy.is_prime(i):
if ltps.count(i)>0:
f = gmpy.fib(i+1)
d = divmod(f,i)
if d[1]>0:
print lt-prime: %4d d[1]: %4d %
(i,d[1])
i += 1
print position:,i-1,f(position):,gmpy.numdigits(f),digits
# end of fib_ltp.py
> Thanks for your input!
> Bob
===
Subject: Axioms and Reducibility
Question for all:
If I have a mathematical system (A) based on 5 axioms and
another
mathematical system (B) that is based on 3 of A's 5 axioms,
does it
logically follow that B is reducible to A?
-Steve
===
Subject: Re: The reals are countable
> By the way, we don't know whether there is a bijection
between P(N) and
R
> because no one has yet resolved the continuum hypothesis.
Sorry to
point
> out another error you've made. But no doubt you will thank
me.
> More likely he will look puzzled and wonder why you think
the continuum
> hypothesis has anything to do with this.
> Here is a bijection between P(N) and R:
> Let [a0, a1, a2, ...] be the continued fraction for |r|. For
r >= 0,
> the bijection is
> r <--> { 2+a0, 3+a0+a1, 4+a0+a1+a2, ... }
> and for r < 0, the bijection is
> r <--> { 1, 2+a0, 3+a0+a1, 4+a0+a1+a2, ... }
Thank you for correcting me.
-Leonard
===
Subject: Re: The reals are countable
>> Final conclusion: As we have proved that f: N <-> P(N) is a
>> bijection, and as we already know that P(N) <-> R is also a
>> bijection,
By the way, we don't know whether there is a bijection
between P(N) and
>R because no one has yet resolved the continuum hypothesis.
> That's not CH. Schroeder-Bernstein says that since there is
both an
> injection P(N) -> [0, 1] and an injection from [0, 1] ->
P(N) then
> there exists a bijection P(N) <-> [0, 1] <-> R.
> CH says that there is no cardinality between |N| and |R|.
Thank you for correcting me. You are of course correct.
You see Nicalas de la Foz? It is nice when people point out
your errors.
-Leonard
===
Subject: ECDLP cracked mathematically??!
Name: James Wanless DONE
Address: 30 Christ Church Road, East Sheen, London, SW14 7AA,
UK DONE
Email: james@grok.ltd.uk DONE
Phone: +44 (020) 8876 8443 DONE
Exercise or Challenge: All challenges still open DONE
Solution: Since your document
[http://www.certicom.com/download/aid-111/cert_ecc_
challenge.pdf] does not
list the specific necessary data (ie public keys and
parameters) no
specific
results can be given here (ie private keys) - instead a
general method is
described to quickly and easily solve any applicable problem
DONE
Method: Using Wanless' Theorem [aka Fermat's Big Theorem]
http://www.bearnol.pwp.blueyonder.co.uk/Math/wanless.html
Discrete log problem: B^x==y [mod p]
Find x given B,y,p
Wanless' Theorem: a^(p2^n)==a^(2^n) [mod p2^n]
Therefore look for B^x==y [mod p2^n], this will also satisfy
original
problem.
Find smallest n s.t. 2^n > B^x (in order to convert the
problem from one in
Fp to one in reals)
Then, x==log-to-base-B[y*2^n] - 2^n [mod p]
QED
And analogously, for the other type of problem, namely in
F2^m, convert to
reals, by finding smallest p > B^x
DONE
cc:certicom-ecc-challenge@certicom.com
===
Subject: Re: ECDLP cracked mathematically??!
> Method: Using Wanless' Theorem [aka Fermat's Big Theorem]
Dear James,
Do you ever find it incredible that you are able to solve so
many
famous, difficult mathematical problems in about half a page?
Take Fermat's Last Theorem, for example. For you, it's trivial.
Any yet it takes mathematical geniuses like Andrew Wiles and
James
Harris pages and pages of cleverly defined constructs and
intricate
arguments to reach the same conclusion.
Your work tacitly implies that Wiles and Harris are bunglers
to be
giving such convolu proofs instead of the simple ones you've
come
up with. It's as if their deep insight into the structure of
mathematics, such as the fundamental flaw in the algebraic
integers,
is totally irrelevant.
So what do you think of the monumental proofs of mathematics?
Is
Gauss's proof of quadratic reciprocity comical to you? Is the
classification of finite simple groups a hideous waste of
paper?
Which of the Hilbert problems can be solved in one line, as
opposed
to half a dozen? Does Harris's invention of the prime counting
function do nothing to alter the landscape of number theory?
If any of the famous, repuly difficult problems you've solved
were as easy as you suggest, don't you think that somewhere in
the
history of mathematics that one of its pantheon of geniuses
would
have noticed that there is such a simple solution? How could
Euler
miss a solution so simple? And even if he did, how could Gauss
miss
it too? And hundreds of other of mathematician? And, how can
you
sit there and suggest that even , who has spent years of
effort closely scrutinizing Fermat's Last Theorem, looking for
a
simple solution -- how can you suggest that even he has missed
what
you claim to have found so easily, so effortlessly, like it
was a
shiny penny sitting on the sidewalk waiting for you to pick it
up?
Well?
-Jim Ferry
===
Subject: sqrt(-1)=0/0
Concerning the sqrt(-1)=0/0
(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.
-Doctor Garry Whilhelm Denke, 1655
solving every equation; it was of 0 use, making 0 sense.
-/-
===
Subject: Re: sqrt(-1)=0/0
> Concerning the sqrt(-1)=0/0
(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.
> -Doctor Garry Whilhelm Denke, 1655
> solving every equation; it was of 0 use, making 0 sense.
> -/-
Are you some sort of idiot?
sqrt(-1) = 0/0
Square both sides.
-1 = 0/0
How can -1 be 0? The real numbers form a field and its trivial
that
there can only be one number such that a * 0 = 0, for any a.
But you
are saying -1 * 0 = 0. Contradiction.
So are you using some different definition of real numbers,
one that
isn't a field? I think the Completeness Axiom has something to
say
about that.
Jason
===
Subject: Re: sqrt(-1)=0/0
> Concerning the sqrt(-1)=0/0
(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.
> -Doctor Garry Whilhelm Denke, 1655
> solving every equation; it was of 0 use, making 0 sense.
> -/-
Evidently you are at least 500 years behind the times.
Why do you insist on retardation.
===
Subject: System of conservation Laws
===
Hi!
I have a system of conservation laws of this kind:
ro1_t + (f1(x) ro1)_x = -a ro1 + b ro2
ro2_t + (f2(x) ro2)_x = a ro1 - b ro2
Does anyone know how can I solve this analytically?
I can reduce it to an hyperbolic second order equation but
it's still too
complica.
Thanks
Tiago
===
Subject: Re: Fundamental basics of unilaterality
> In other words, the paradox is that the theory doesn't
explain
the real
> world.
It's not a matter of unhappyness. Models are models and not
the real
world. You need a better model? Build it don't whine.
> As an engineer, I do not need at all the fictitious notion
of an
> infinitely small value zero. Its neighbours 0- and 0+ do the
same job.
> Why not declaring the exact zero Cantors pipe dream? He and
most
> mathematicans after him were perhaps not aware of our
unability to have
> the cake and eat it. Zero is a value that is smaller than
any value. So
> it cannot be a value at all. It might nonetheless be of
value for pipe
> dreams.
Zero is a value that cannot be a value... Hey! thats a
paradox, and
there are no donkeys in it! So the people who put years of
learning
into building mathematics where having pipe dreams, but you
the self
acknowledge layman are going to put things right. Would you be
at
least so kind as to try and briefly explain what it is you are
after?
What is this unilaterality you are talking of. Is it the kind a
moebius band has got? Is it the condition of having just one
functioning brain half? or is it rather something like that
german
saying about pleasure being totally on ones side?
> Many people expect there to be a useful theory of everything
> I do not see a compelling reason for not creating
mathematial basics
> with less arbitrariness.
well then just do.
> Let's not quarrel about words. Please read my reasons and
then try to
> falsify them or benefit from them as did I.
Then state those reasons if they be in fact reasons.
> Do not feel in a too strong position just because you were
trained in
> mathematics. The matter is quite fundamental.
> Eckard Blumschein
Yeah, I guess consulting your local fishmonger should do as
well. When
unhappy or in doubt, proclaim the new paradigm, and the world
will
follow...
Regards
===
Subject: Re: Fundamental basics of unilaterality
> If you have 10 euros and then a mugger takes away those 10
euros
> from you, how many euros have you left? 0+? 0-? or just 0?
You are on the right truck. Euros are countable entities, even
multiples
of smaller units. A number (but not a real number) is
something designed
to count with.
If chapmans are selling portions of butter, cheese whatsoever,
they are
cutting pieces approaching a desired countable size. So
approaching a
value only comes close to it. In other words, engineers like
me have got
a notion of continuity that sees a non-bridgeable though
unfinitely
small difference between anything that stems from and can be
reduced to
numbers on the one hand and continuous measures on the other
hand.
Englishmen distinguish between how many and how much in a
similar
manner. For my understanding, crossing the non-bridgeable gap
requires
what I learned in German as Grenzuebergang (limes) denotes a
transition
in the sense of crossing any limitation.
More imaginably speaking, outside Cantor's paradise, the
continuum can
be zoomed infinitely. Let's consider a line marked with
points. Look at
the night at the open sky with increasing magnification. You
will see an
apparently infinite number of stars. Points can never be packed
densely., and if they could this would not make sense.
Division by zero is forbidden as if a king was necessary to
decide
something quite natural in an unnatural arbitrary manner. If we
restrict to a continuous world, thought to exist besides the
discrete
one, then we will earn a lot. We may consider it impossible to
exactly
measure identities and abstract numbers like 0, 4, or even
3.14. Instead
our measures will merely come as close to them as we like. In
practical
terms, it does not make sense to exclude division by zero by
tolerating
division by almost zero. If Matlab warns me: 'division by
zero', then
this is strictly speaking a lie.
Approprietly dealing with what non-mathematicians regard the
continuum
requires chapmens' cheese-knife. It is able to separate R into
two equal
halves wqith no neutral zero in between: R+ and R-, both open
to both
sides. Anything else is arbitrary and suffers from the
self-deceptive
readiness for an unjustified compromise.
Eckard Blumschein
===
Subject: Re: Fundamental basics of unilaterality
>> If you have 10 euros and then a mugger takes away those 10
euros
>> from you, how many euros have you left? 0+? 0-? or just 0?
> You are on the right truck.
Or the left truck?
> Euros are countable entities, even multiples
> of smaller units.
Are you sure that euros are not infinitesimals.
> A number (but not a real number) is something designed
> to count with.
At last. Numbers are for counting. A glimmer of sense at last.
> If chapmans are selling portions of butter, cheese
whatsoever, they are
> cutting pieces approaching a desired countable size. So
approaching a
> value only comes close to it. In other words, engineers like
me have got
> a notion of continuity that sees a non-bridgeable though
unfinitely
> small difference between anything that stems from and can be
reduced to
> numbers on the one hand and continuous measures on the other
hand.
In other words you are fetishizing one particular application
of numbers: approximation of quantities in the real
(phenomenal)
world and imagining that mathematics should be subordina to
that.
I reverse the question. Is engineering of use or of application
to mathematics?
> More imaginably speaking, outside Cantor's paradise, the
continuum can
> be zoomed infinitely.
zoom-zoom-zoom
> Let's consider a line marked with points. Look at
> the night at the open sky with increasing magnification. You
will see an
> apparently infinite number of stars. Points can never be
packed
> densely., and if they could this would not make sense.
I don't see very many stars at all, 'cos of the brightness of
street
lighting. But it seems your stargazing is rotting your
intellect.
You are so dazzled by your infinite number of stars that you
suddenly
pontificate about points being packed densely and imagine
that not making sense, rather than you not making sense.
> Division by zero is forbidden as if a king was necessary to
decide
Verboten!
> something quite natural in an unnatural arbitrary manner. If
we
> restrict to a continuous world, thought to exist besides the
discrete
> one, then we will earn a lot.
Great, more than 100,000 a year?!
> We may consider it impossible to exactly
> measure identities and abstract numbers like 0, 4, or even
3.14.
I never knew that one could measure the number 4.
> Instead
> our measures will merely come as close to them as we like.
In practical
> terms, it does not make sense to exclude division by zero by
tolerating
> division by almost zero. If Matlab warns me: 'division by
zero', then
> this is strictly speaking a lie.
Ach! A worshipper of the MACHINE!
> Approprietly dealing with what non-mathematicians regard the
continuum
> requires chapmens' cheese-knife. It is able to separate R
into two equal
> halves wqith no neutral zero in between: R+ and R-, both
open to both
> sides. Anything else is arbitrary and suffers from the
self-deceptive
> readiness for an unjustified compromise.
So when you were mugged, were you left with 0+ or 0- euros.
--
===
Subject: Re: Fundamental basics of unilaterality
Robin Chapman unfortunately merely joked without deeper
understanding:
>>You are on the right truck.
> Or the left truck?
Maybe.
> Are you sure that euros are not infinitesimals.
???
> At last. Numbers are for counting. A glimmer of sense at
last.
Measures are perhaps not equally applicable for counting.
>>If chapmans are selling portions of butter, cheese
whatsoever, they are
>>cutting pieces approaching a desired countable size. So
approaching a
>>value only comes close to it. In other words, engineers like
me have got
>>a notion of continuity that sees a non-bridgeable though
unfinitely
>>small difference between anything that stems from and can be
reduced to
>>numbers on the one hand and continuous measures on the other
hand.
> In other words you are fetishizing one particular application
> of numbers: approximation of quantities in the real
(phenomenal)
> world and imagining that mathematics should be subordina to
that.
Well, math. is roo within the real world, and it still tries
to serve
it best. Ancient geometry relates to geo and meter, and it is
not
identical with a science of counting numbers. I reverse the
reproach. It
is my gut feeling that analysis tends to be wrongly domina by
'generalized'numbers instead of being balanced between means
for
counting and means for measuring.
>>More imaginably speaking, outside Cantor's paradise, the
continuum can
>>be zoomed infinitely.
> zoom-zoom-zoom
Yes, and you will never resolve a single point.
>> Points can never be packed
>>densely, and if they were, this would not make sense.
> You are so dazzled by your infinite number of stars
number means that it is not a number as isn't zero.
>>Division by zero is forbidden as if a king was necessary to
decide
> Verboten!
I would rather appreciate a traffic sign: reconstruction work
(in math.)
>> something quite natural in an unnatural arbitrary manner.
If we
>>restrict to a continuous world, thought to exist besides the
discrete
>>one, then we will earn a lot.
> Great, more than 100,000 a year?!
Or course, insight first.
>>We may consider it impossible to exactly
>>measure identities and abstract numbers like 0, 4, or even
3.14.
> I never knew that one could measure the number 4.
>>Instead
>>our measures will merely come as close to them as we like.
In practical
>>terms, it does not make sense to exclude division by zero by
tolerating
>>division by almost zero. If Matlab warns me: 'division by
zero', then
>>this is strictly speaking a lie.
> Ach! A worshipper of the MACHINE!
Not at all. However some machines represent more geniuis than
an average
analyst.
>>Approprietly dealing with what non-mathematicians regard the
continuum
>>requires chapmens' cheese-knife. It is able to separate R
into two equal
>>halves with no neutral zero in between: R+ and R-, both open
to both
>>sides. Anything else is arbitrary and suffers from the
self-deceptive
>>readiness for an unjustified compromise.
> So when you were mugged, were you left with 0+ or 0- euros.
Only a (British) mug might count his negative or zero euros. I
didn't
mug up to much mandatory mathematical nonsense.
Disclaimer: Those who might feel offended might blame nothing
else than
my poor command of English for that.
Eckard Blumschein
===
Subject: Re: Fundamental basics of unilaterality
>> At last. Numbers are for counting. A glimmer of sense at
last.
> Measures are perhaps not equally applicable for counting.
>If chapmans are selling portions of butter, cheese
whatsoever, they are
>cutting pieces approaching a desired countable size. So
approaching a
>value only comes close to it. In other words, engineers like
me have got
>a notion of continuity that sees a non-bridgeable though
unfinitely
>small difference between anything that stems from and can be
reduced to
>numbers on the one hand and continuous measures on the other
hand.
>> In other words you are fetishizing one particular
application
>> of numbers: approximation of quantities in the real
(phenomenal)
>> world and imagining that mathematics should be subordina to
that.
> Well, math. is roo within the real world,
The real world. Always the favourite rhetorical device
of anti-mathematics. As I say, it's just a fetishization of
one's pet obsessions.
Of course I try to serve the real world of mathematics.
> Yes, and you will never resolve a single point.
Are points to be resolved?
>> You are so dazzled by your infinite number of stars
number means that it is not a number as isn't zero.
Que?
>We may consider it impossible to exactly
>measure identities and abstract numbers like 0, 4, or even
3.14.
>> I never knew that one could measure the number 4.
It's a category mistake.
--
===
Subject: Re: Fundamental basics of unilaterality
> I never knew that one could measure the number 4.
>It's a category mistake.
Not sure about that!
According to the definition from set theory, 4 = {0,1,2,3}
which is (or
can identified with if you prefer) a measurable subset of R.
The measure of 4 is 0.
Derek Holt.
===
Subject: Re: Fundamental basics of unilaterality
>> I never knew that one could measure the number 4.
>>It's a category mistake.
> Not sure about that!
> According to the definition from set theory, 4 = {0,1,2,3}
which is (or
> can identified with if you prefer) a measurable subset of R.
> The measure of 4 is 0.
Pah! Set theoretic fundamentalism
--
===
Subject: Re: Stupid question on notation (norm of a matrix)
>> I am having some issues with the norm of a matrix. Let A be
an n x n
^^^^^^^^^^^^^
>> matrix.
[...]
>That is called the operator norm
[...]
>That is called the Hilbert-Schmidt norm
[...]
>These are different norms. Any given space has many norms on
it,
>perhaps even many useful norms.
But then the good news are that they are all equivalent,
thanks to the
finite-dimensionality of the space.
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Stupid question on notation (norm of a matrix)
>> I am having some issues with the norm of a matrix. Let A be
an n x n
> ^^^^^^^^^^^^^
>> matrix.
> [...]
>That is called the operator norm
> [...]
>That is called the Hilbert-Schmidt norm
> [...]
>These are different norms. Any given space has many norms on
it,
>perhaps even many useful norms.
> But then the good news are that they are all equivalent,
thanks to the
> finite-dimensionality of the space.
Under the operator norm, the identity n x n matrix has norm
equal to 1,
under the other norm, it equals square root of n.
How do you mean by equivalent?
> Michele
> --
> Comments should say _why_ something is being done.
> Oh? My comments always say what _really_ should have
happened. :)
> - Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Stupid question on notation (norm of a matrix)
>>These are different norms. Any given space has many norms on
it,
>>perhaps even many useful norms.
>> But then the good news are that they are all equivalent,
thanks to the
>> finite-dimensionality of the space.
>Under the operator norm, the identity n x n matrix has norm
equal to 1,
>under the other norm, it equals square root of n.
>How do you mean by equivalent?
I mean they are equivalent in the sense of the definition of
*equivalent* norms that is not to say that they take the same
values
on the same matrices, in which case they would be
*identical*...
It should suffice to you to know that they generate the same
topologies.
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Stupid question on notation (norm of a matrix)
> I am having some issues with the norm of a matrix. Let A be
an n x n
>> ^^^^^^^^^^^^^
> matrix.
>> [...]
>>That is called the operator norm
>> [...]
>>That is called the Hilbert-Schmidt norm
>> [...]
>>These are different norms. Any given space has many norms on
it,
>>perhaps even many useful norms.
>> But then the good news are that they are all equivalent,
thanks to the
>> finite-dimensionality of the space.
> Under the operator norm, the identity n x n matrix has norm
equal to 1,
> under the other norm, it equals square root of n.
> How do you mean by equivalent?
norms ||_1 and ||_2 are equivalent iff there are A and B > 0
with
|x|_1 <= A |x|_2 and |x|_2 <= B |x|_1 for all x.
Two norms on a finite-dimensional space must be equivalent.
Two norms on an infinite-dimensional space need not be
equivalent.
--
===
Subject: Re: JSH: Push for Java
> So what I thought about this is probably totally wrong, and
I'm
> not going to be able to ask the question very efficiently:
There
> certainly exist standard automatic parsers, that accept a
formal
> description of a grammar. Some programming languages do
> not admit description in these standard grammar-definition
> languages, but my impression was that Python does. Yes?
> Yes. C does too. GCC's parser for C is amazingly complex, as
is
> few thousand lines long. A parser for Scheme can be written
in about
> a hundred lines.
> Thomas
Questions about formal parsing are of theoretical interest and
of interest to language implementors. They are only indirectly
of interest to language users, and then only if parsing is so
difficult that it affects the quality or number of
implementations.
To a user the term parse is more or less equivalent to
decipher. Under this meaning, Python is usually considered
easy to parse.
- William Hughes
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
SHHHH!! I hear SIX TATTOOED TRUCK-DRIVERS tossing ENGINE
BLOCKS into
empty OIL DRUMS..
> Questions about formal parsing are of theoretical interest
and
> of interest to language implementors. They are only
indirectly
> of interest to language users, and then only if parsing is so
> difficult that it affects the quality or number of
implementations.
Right, but questions about syntax complexity are not. FORTRAN
is a
bugger to parse, but isn't any harder in syntax complexity
(which I
use as an impressionistic tag for what users care about) than
C or
Pascal or Python.
However, experience in teaching the language to people who are
brand
new to programming is that the trivial syntax of Lisp is a
boon over
the complex syntax typical of Algol-like languages.
Thomas
===
Subject: Re: JSH: Push for Java
> I repeat my question. Have you been taking gnomic lessons
from Harris?
> We now know that you think it is possible to come up with
useful
> criteria for why a programming language would be suitable
for large
> systems. We know you don't consider C or Python suitable for
large
> systems. You have not actually given an example of a
language that
> you do consider suitable for large systems, although I
assume that
> Scheme fits into this catagory. However, we are still
completely
> in the dark as to what criteria you are using. I hope its
something
> more than functional good, procedural bad.
> I think Scheme is good, Common Lisp is as well. I could
imagine
> Python evolving into something that would be good.
> I have already identified some specific features that I
think are
> missing in Python, and other factors that are quite
specifically
> relevant as well.
It is hard to believe that you think the list
of weaknessess you identified in Python have anything
to do with its suitability for large systems.
However, if this is what you mean, say so!
You have not identified ...other factors that are quite
specifically
relevant as well.
I take it you have been taking evasion lessons from Harris as
well. Standard prodcedure:
Question: X
Answer: Talk about R
Question: X
Answer: Talk about S
Question: X
Answer: Talk about T
...
Question: X
Answer: I already answered this
I will repeat the question.
What are your criteria for a programming language
to be suitable for large systems?
No, you have not answered this question, directly
or indirectly.
Maybe an example will help.
To be suitable for programming a large system
a language needs the following properties:
(i) Fully modular: It is very difficult to deal with a large
problem except in small chunks. It must be possible
to divide the problem up. It must be possible
to work on more than one section of the program
at the same time (this is necessary even if there
is only a single developer)
(ii) Supports ADT's: I find it difficult to concieve of a
design
for a large system that doesn't include ADT's
(abstract data types).
(iii) Exception Handling: Needed for clean implementation and
very useful in debugging.
(Iv) Sufficiently High Level. It is possible to argue that
assembler satisfies (i)-(iii) (it just takes
a little work).
(v) Tool set: The language needs a good set of support
tools. As a reductio ad absurdum, a language
for which there is no interpreter or compiler
is not suitable for large systems. However,
if your editing, compilation and or interpretation
and error checking (both static and dynamic)
tools are deficient, it will be difficult to
implement a large system.
Many languages fail (i) (E.g. BASIC [1]). Fortran (at least
some earlier varients) is deficient in (ii). C is deficient in
(iii)
(it is possible to have good exception handling in C, just not
easy). Condition (iv) is a judgement call. In the end you
will probably end up implementing your system, not in a
language
but in a set of tools writen in the language. If the set of
tools
is very complete, it becomes problematic to claim that
you are working in the original language. In my opinion, the
dividing
point comes somewhere near C.
By these criteria, Python is suitable for progamming
large systems.
Of course the choice of a language for a project depends on
many, many other factors (e.g. is the language well sui
to the problem, what existing resources are there, what
about third party support, what about execution speed, etc.
etc. etc.).
However, these questions are mostly orthogonal to the question
of whether
the language is suitable for a large system.
- William Hughes
[1] Well, there are many many BASIC's. Some of them
(e.g. I have been told of at least one BASIC with
a full set of loop contructs, subroutines and structures)
may be suitable for large systems. However, anything
depending on line numbers and GOSUB is not.
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
> What are your criteria for a programming language
> to be suitable for large systems?
My criterion is one of looking at it, with the experience of
having
written multiple large and medium sized systems in different
languages. Based upon that, I consider how easy or difficult
it would
be to use it to write a large system in it; what problems I
know come
up frequently and which do not. Further, I consider the
question of
whether there have been large systems written in the language,
most
especially by its strongest advocates, and whether it is
demonstrably
their language of choice for writing large systems.
I do not claim to have some algorithm for making this
judgment, it is
a judgment call, and I do not pretend anything else.
Sure, modularity, ADTs, exception handling, and so forth are
important. They are necessary conditions, but not sufficient
ones.
Note that language stability is not on your list, but seems
obvious to
me: the definition of the langauge should not change, at least
not in
core areas, over the lifetime of the project. Large systems
have
larger lifetimes, making stability more important. Python
fails here,
in large part because issues that the designers hand-waved
away (like
good variable scoping rules) come back to bite them, and they
decide
to add the real feature in. And typically, they do so with an
eye to
what is easiest to implement, and so they get it wrong and
have to
come back and redefine it a second time (as happened with
variable
scoping).
Likewise, Python evolved an exception system and a module
system over
time. Both are passable at the current time; but this is an
area
where precious few languages do a really good job (including
most
Scheme or Lisp implementations). But they are not stable, and
this is
a serious problem.
Also not on your list is memory management; any system without
garbage
collection and a decent safe memory system is going to suck,
and the
steady run of security disaster buffer-overrun bugs show that
this
deficiency clearly marks C as not suitable for large systems.
Python
gets this one right.
Thomas
===
Subject: Re: JSH: Push for Java
>> What are your criteria for a programming language
>> to be suitable for large systems?
>My criterion is one of looking at it, with the experience of
having
>written multiple large and medium sized systems in different
>languages. Based upon that, I consider how easy or difficult
it would
>be to use it to write a large system in it; what problems I
know come
>up frequently and which do not.
But you haven't _mentioned_ any problems that are going to come
up when one attempts to use Python in a large system! If you've
considered them you've been keeping your considerations to
yourself, in spite of the fact that I asked specifically what
sort of
problem was going to arise.
No, Python is not self-hosting does not seem to anyone but
you as an answer to the question What sort of problems are
going to arise in using Python in a large system?
>Further, I consider the question of
>whether there have been large systems written in the
language, most
>especially by its strongest advocates, and whether it is
demonstrably
>their language of choice for writing large systems.
>I do not claim to have some algorithm for making this
judgment, it is
>a judgment call, and I do not pretend anything else.
>Sure, modularity, ADTs, exception handling, and so forth are
>important. They are necessary conditions, but not sufficient
ones.
>Note that language stability is not on your list, but seems
obvious to
>me: the definition of the langauge should not change, at
least not in
>core areas, over the lifetime of the project. Large systems
have
>larger lifetimes, making stability more important.
Ok, I lied, you did mention one. But it says _nothing_ about
why,
say, Python 1.5.2 is not suitable for large systems. A person
working on a large system is not required to upgrade as soon
as a new version comes out - in fact that Python on _my_
system has been 100% stable, it's exactly the same today
as it was many years ago when I first installed Python.
> Python fails here,
>in large part because issues that the designers hand-waved
away (like
>good variable scoping rules) come back to bite them, and they
decide
>to add the real feature in. And typically, they do so with an
eye to
what is easiest to implement, and so they get it wrong and
have to
>come back and redefine it a second time (as happened with
variable
>scoping).
>Likewise, Python evolved an exception system and a module
system over
>time. Both are passable at the current time; but this is an
area
>where precious few languages do a really good job (including
most
>Scheme or Lisp implementations). But they are not stable, and
this is
>a serious problem.
>Also not on your list is memory management; any system
without garbage
>collection and a decent safe memory system is going to suck,
and the
>steady run of security disaster buffer-overrun bugs show that
this
>deficiency clearly marks C as not suitable for large systems.
Python
>gets this one right.
>Thomas
************************
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
> I take it you have been taking evasion lessons from Harris as
> well.
We're discussing something which is not susceptible of proof;
it's an
impressionistic question, it's one that depends on conceptions
of
aesthetics, it's one that reasonable people will disagree
about.
Harris argues about settled facts which are susceptible of
mathematical proof.
When you treat me with respect, I'll address your points.
===
Subject: Re: JSH: Push for Java
In his first post in the thread, in reply to a post reading in
part
Just for giggles I whipped up a _totally_ brainless PrimePi
in Python just now - it's nothing but a sieve, and not a
clever sieve, a totally idiotic sieve, not particularly well
written either, just top-of-the-brainless-head code:
'I don't think you can prove programming skill by saying crazy
things
like Python is an extremely fabulous language. Sorry.'
And then somewhat later he said
>When you treat me with respect, I'll address your points.
That _is_ Harresque, complaining when people treat you
the same way as you treat them. Sorry.
************************
===
Subject: Re: JSH: Push for Java
> The point being to make it clear that there are _other_
> examples of large systems.
> Sure. The reason one asks about self-hosting is because one
assumes
> that the implementors of a language are fans of it--perhaps
the
> greatest fans, as well as those who know more about it than
anyone
> else. So if it were susceptible to self-hosting, wouldn't
they do so
> When they don't, an initial assumption with some merit (but
> rebuttable, yes) is that the language's greatest fans and
the people
> who know more about it than anyone else don't think it's
suitable for
> such a project.
> Thomas
Right, if one of the criteria for suitable for large systems
is suitable for self hosting then this is relevant.
And if one of the criteria for suitable for large systems
is name of language starts with 'B', then BASIC is suitable
for large systems and Lisp is not.
-William
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
X-Tom-Swiftie: My terminal is completely screwed up, Tom cursed
> Right, if one of the criteria for suitable for large systems
> is suitable for self hosting then this is relevant.
For a general purpose language, such as Python claims to be,
it's a
glaring ommission that the large system on which the Python
maintainers most work, and which they most care about, is not
written
in Python.
Another way to put it: ask the implementors why Python is not
written
in Python. The answer is your list of things that Python
cannot do
well enough (yet, at least). There are other possible answers
too, of
course, which is why non-self-hosting is only suggestive and
not
probitive.
===
Subject: Re: JSH: Push for Java
>> Right, if one of the criteria for suitable for large systems
>> is suitable for self hosting then this is relevant.
>For a general purpose language, such as Python claims to be,
it's a
>glaring ommission that the large system on which the Python
>maintainers most work, and which they most care about, is not
written
>in Python.
>Another way to put it: ask the implementors why Python is not
written
>in Python. The answer is your list of things that Python
cannot do
>well enough (yet, at least). There are other possible answers
too, of
>course, which is why non-self-hosting is only suggestive and
not
>probitive.
Right. It's only suggestive but not probitive.
This star with your statement that Python is
extremely fabulous was a crazy thing to say,
indicating a lack of programming expertise (in
a context where the person saying the crazy
thing had made it very clear that programming
expertise was _not_ what he was trying to
demonstrate, but never mind that.)
By far the most significant of your complaints about
Python, and the _only_ one you offered for quite a
while, was that it's not suitable for large systems.
It seems plenty suitable for large systems to many
of us - the _only_ reason you've given for its
unsuitability there is the fact that Python has
not been implemen in Python. Now it turns
out that that's only suggestive. But it's suggestive
enough to make Python is an extremely fabulous
language a crazy thing to say. Right.
A lot of this has been interesting. But if I've
been showing a lack of programming expertise,
I do think that you've been showing exactly the
sort of [can't come up with the right adjective]
debating style that William has been commenting on.
************************
===
Subject: Re: JSH: Push for Java
 I said that it *would* have been a nightmare in C and C++,
and that I
> suspec it would have been similar in Scheme or Lisp. But I
do not
> know enough about the latter two to be sure about it.
Anyhow, the
> multiplexer is about 300 lines of Python code.
 Good grief, that's not even a small program!
Did I say it was large?
--
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
I said that it *would* have been a nightmare in C and C++, and
that
I
> suspec it would have been similar in Scheme or Lisp. But I do
not
> know enough about the latter two to be sure about it.
Anyhow, the
> multiplexer is about 300 lines of Python code.
Good grief, that's not even a small program!
> Did I say it was large?
I thought it was offered as an example of a large system
written in
Python.
Nobody disputes that Python can run tiny programs well.
===
Subject: Re: JSH: Push for Java
>> So what I thought about this is probably totally wrong, and
I'm
>> not going to be able to ask the question very efficiently:
There
>> certainly exist standard automatic parsers, that accept a
formal
>> description of a grammar. Some programming languages do
>> not admit description in these standard grammar-definition
>> languages, but my impression was that Python does. Yes?
>Yes. C does too. GCC's parser for C is amazingly complex, as
is
>few thousand lines long. A parser for Scheme can be written
in about
>a hundred lines.
Hmm. If I sugges you should have written it in Python you
might not realize it was a joke...
I wonder how many lines of C code the parser takes in the
standard Python implentation? The code is freely available,
presumably easy to find starting at www.python.org ; I'm not
gonna bother looking at it since it would be Greek to me.
The tokenizer.py that comes with Python 1.5.2 is less than
200 lines, including comments and blank lines. It's not clear
to
me how much more work a parser needs to do after the
tokenization; for the record here's what tokenizer.py does:
Tokenization help for Python programs.
This module exports a function called 'tokenize()' that breaks
a
stream of
text into Python tokens. It accepts a readline-like method
which is
called
repealy to get the next line of input (or for EOF) and a
token-eater
function which is called once for each token found. The latter
function is
passed the token type, a string containing the token, the
starting and
ending (row, column) coordinates of the token, and the
original line.
It is
designed to match the working of the Python tokenizer exactly,
except
that
it produces COMMENT tokens for comments and gives type OP for
all
operators.
_My_ probably totally wrong impression is that that's most of
what
we need to do to parse Python code. (Presumably there's a clear
split between the tokenizing and the rest of it in that Python
parser
post-tokenizating part?)
>Thomas
************************
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
> The tokenizer.py that comes with Python 1.5.2 is less than
> 200 lines, including comments and blank lines. It's not
clear to
> me how much more work a parser needs to do after the
> tokenization; for the record here's what tokenizer.py does:
Huh? All the hard work is in the parsing. Tokenization (what is
actually called scanning, no need to invent a new word) is
typically
easy.
> _My_ probably totally wrong impression is that that's most
of what
> we need to do to parse Python code. (Presumably there's a
clear
> split between the tokenizing and the rest of it in that
Python parser
> post-tokenizating part?)
Nearly all. I must apologize, the numbers I gave before came
from
looking at the wrong files. My scanner is 2248 lines, but it
does
everything, including numeric conversion and so forth.
The expression parser is 5542, and the rest of the parser is
about the
same size again. So the total size is something like 13000
lines.
Not all of this can be attribu to parsing per se, because
several
other things are going on in that program at the same time; I
would
say that about half of it is strictly parsing code.
Thomas
===
Subject: Re: JSH: Push for Java
>> The point being to make it clear that there are _other_
>> examples of large systems.
>Sure. The reason one asks about self-hosting is because one
assumes
>that the implementors of a language are fans of it--perhaps
the
>greatest fans, as well as those who know more about it than
anyone
>else. So if it were susceptible to self-hosting, wouldn't
they do so?
>When they don't, an initial assumption with some merit (but
>rebuttable, yes) is that the language's greatest fans and the
people
>who know more about it than anyone else don't think it's
suitable for
>such a project.
Right. We seem to be caught in a loop here: I've never dispu
the idea that Python is not suitable for writing an optimizing
Python interpreter. That's not the same as the statement that
it's not suitable for large projects...
>Thomas
************************
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
> Right. We seem to be caught in a loop here: I've never dispu
> the idea that Python is not suitable for writing an
optimizing
> Python interpreter. That's not the same as the statement that
> it's not suitable for large projects...
Compiler, not interpreter.
Regardless, it claims to be a general purpose language. If the
problem is that the claim is incorrect, then so be it. I
dispute the
need for non-general-purpose languages in problem spaces where
existing general purpose languages already do the job.
Thomas
===
Subject: Re: JSH: Push for Java
<874quvyfjt.fsf@becket.becket.net>
<4d5e4663.0401161825.6e65b7ce@posting.google.com>
<87ektu8leo.fsf@becket.becket.net>
<eccr00tprl2htrpp7ijv44nj4m02asiihn@4ax.com>
<87n08i55ut.fsf@becket.becket.net>
<tsqs005nsnd4g873rr1stj74o4s6amqk54@4ax.com>
<87vfn4dhh0.fsf@becket.becket.net>
<ercv0093vfj9uuhrvr646s0nnrc0kstoj6@4ax.com>
<87y8rzmjt7.fsf@becket.becket.net>
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<87smi61o55.fsf@becket.becket.net>
Discussion, linux)
>> Right. We seem to be caught in a loop here: I've never dispu
>> the idea that Python is not suitable for writing an
optimizing
>> Python interpreter. That's not the same as the statement
that
>> it's not suitable for large projects...
> Compiler, not interpreter.
> Regardless, it claims to be a general purpose language. If
the
> problem is that the claim is incorrect, then so be it. I
dispute the
> need for non-general-purpose languages in problem spaces
where
> existing general purpose languages already do the job.
This seems a rather stronger criticism than your previous ones.
(1) All general purpose languages are (capable of)
self-hosting.
(2) There is no need for any language which is not general
purpose.
You may dispute (2) since you qualified it: there's no need
for any
non-general-purpose language in problem spaces where existing
general
purpose languages already do the job. I don't really know what
that
qualification means or how it applies.
--
Jesse F. Hughes
[Lancelot] sighed, defea. 'It is as practical to hurry an acorn
toward treeness as to urge a damsel when her mind is set.'
-- John Steinbeck, /The Acts of King Arthur and His Noble
Knights/
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
X-Tom-Swiftie: Time to test this new formula, Tom said
experimentally
> This seems a rather stronger criticism than your previous
ones.
> (1) All general purpose languages are (capable of)
self-hosting.
> (2) There is no need for any language which is not general
purpose.
> You may dispute (2) since you qualified it: there's no need
for any
> non-general-purpose language in problem spaces where
existing general
> purpose languages already do the job. I don't really know
what that
> qualification means or how it applies.
Suppose we have a non-general-purpose language FOO which is
good for
job X, and the general-purpose languages aren't good at job X.
(The
name general-purpose is not necessarily synonymous with
all-purpose.) In that case, FOO is needed.
But suppose that FOO does X, that FOO is not general purpose,
and that
there are existing general purpose languages which also do X
and do it
just as well as FOO does. In that case, FOO is not needed.
Thomas
===
Subject: Re: JSH: Push for Java
>> Huh. Of course dynamic typing is a great boon for many
reasons,
>> in particular it's crucial in the sort of things that I find
>> fascinating (long description of an example elsewhere), but
I'm
>> surprised that you say this - I thought the standard wisdom
>> was that !Q's point about what the Python guys call bondage
and
>> discipline languages was exactly correct. Could you give a
>> hypothetical example of how dynamic typing can increase
>> reliability?
>The B&D languages remark is actually one that the Lisp
community did
>first too. :)
Wow, those guys are impressive.
>> If so, is there some reason there could not exist a strongly
>> typed language with garbage collection, theoretically
>> immune from memory errors? Hmm, Java is such a language,
>> is it not?
>Sure; they are more or less orthogonal issues.
Right. Then this leaves my question unanswered: Can you
explain how it is that dynamic typing increases reliability?
I'm certainly not disputing that, just don't see why it would
be so.
(Alternately, you could clarify that you didn't really mean
that
the dynamic typing increases reliability per se, just that
various dynamically typed languages have been found to
be more reliable than various B&D languages - _if_ that's
what you meant I don't find that puzzling at all.)
>Thomas
************************
===
Subject: Re: JSH: Push for Java
permission for an emailed response.
> Right. Then this leaves my question unanswered: Can you
> explain how it is that dynamic typing increases reliability?
> I'm certainly not disputing that, just don't see why it would
> be so.
Many large systems end up with things that are dynamically
typed
anyway. For example, the GNU Hurd (all written in C) uses Mach
ports
to represent system facilities. A Mach port is like a Unix file
descriptor: it's an integer index into a kernel table of
resources
owned by the task, which can be used to communicate to other
tasks.
Some ports refer to files, some to processes, some to regions
of
memory, some to ...: they are, in fact, generic system
pointers.
We pretend these are different types, with typedefs like io_t,
process_t, memory_object_t, etc. But they are all really just
32-bit
integers, and the compiler doesn't care beans if you pass an
io_t
where the system expects a memory_object_t, but of course,
you'll get
a crash nearly any time you do something like that.
The details are different in different systems, but things
like this
come up all the time. So perhaps a different way to try and
express
why I think a dynamic language is better for reliability is
that it
doesn't lull you into a false sense of security. I know that
doesn't
sound very powerful, but I admit that:
> (Alternately, you could clarify that you didn't really mean
that
> the dynamic typing increases reliability per se, just that
> various dynamically typed languages have been found to
> be more reliable than various B&D languages - _if_ that's
> what you meant I don't find that puzzling at all.)
I have usually preferred dynamic typing on this basis and not
bothered
trying to think through the ways that dynamic typing is
independently
wonderful for reliability, as well as in other ways too.
Thomas
===
Subject: Trigonometric Integral
Dear all,
How to solve the following integral
Int (cos nx /(a - b cos x), x = 0.. Pi ), n =
0,1,2,3,4,........,
a, b are real numbers
Thank you,
N. Karjanto
===
Subject: Re: Trigonometric Integral
> Dear all,
> How to solve the following integral
> Int (cos nx /(a - b cos x), x = 0.. Pi ), n =
0,1,2,3,4,........,
> a, b are real numbers
> Thank you,
> N. Karjanto
First, reduce it to a one-parameter problem: after handling
exceptional
cases (division by zero), you are looking for
A(n) = Int (cos(n*x) / (1 - q * cos(x)) , x = 0 .. pi )
and you'd better have -1 < q < 1 (otherwise the integral is
divergent).
(This corresponds to abs(b) < abs(a) .)
The result is classical: introduce
r = q / (1 + (1 - q^2)^(1/2))
and then
A(n) = r^n * pi / (1 - q^2)^(1/2) .
Curious about the method? It's the generating function trick.
For -1 < r < 1, the infinite series
1 + 2 * sum (r^k * cos(k*x) , k = 1..infinity)
equals (1 - r^2) / (1 + r^2 - 2 * r * cos(x))
(why? Express cos(n*x) as Real(exp(i*n*x)) and do the
geometric series)
then integrate the series, multiplied by cos(n*x), term by
term, and do
some algebra to relate r to q. Do distinguish cases n=1 and
n>0 .
(The series converges uniformly, so integration term by term is
legitimate.)
ZVK(Slavek).
===
Subject: Re: Trigonometric Integral
> Dear all,
> How to solve the following integral
> Int (cos nx /(a - b cos x), x = 0.. Pi ), n =
0,1,2,3,4,........,
> a, b are real numbers
> Thank you,
> N. Karjanto
Here's Maple...
> assume(a,real);assume(b>0);additionally(a>b);
> int(1/(a+b*cos(x)),x=0..Pi);
Pi/(a^2-b^2)^(1/2)
> int(cos(x)/(a+b*cos(x)),x=0..Pi);
Pi*(-a+(a^2-b^2)^(1/2))/(b*(a^2-b^2)^(1/2))
> int(cos(x)^2/(a+b*cos(x)),x=0..Pi);
-a*Pi*(-a+(a^2-b^2)^(1/2))/(b^2*(a^2-b^2)^(1/2))
> int(cos(x)^3/(a+b*cos(x)),x=0..Pi);
1/2*Pi*(b^2*(a^2-b^2)^(1/2)-2*a^3+2*a^2*(a^2-b^2)^(1/2))/(
b^3*(a^2-b^2)^(1/2))
> int(cos(x)^4/(a+b*cos(x)),x=0..Pi);
-1/2*a*Pi*(b^2*(a^2-b^2)^(1/2)-2*a^3+2*a^2*(a^2-b^2)^(1/
2))/(b^4*(a^2-b^2)^(1/2))
> int(cos(x)^5/(a+b*cos(x)),x=0..Pi);
1/8*Pi*(3*b^4*(a^2-b^2)^(1/2)+8*a^4*(a^2-b^2)^(1/2)-8*a^5+
4*a^2*b^2*(a^2-b^2)^(1/2))/(b^5*(a^2-b^2)^(1/2))
> int(cos(x)^6/(a+b*cos(x)),x=0..Pi);
-1/8*a*Pi*(3*b^4*(a^2-b^2)^(1/2)+8*a^4*(a^2-b^2)^(1/2)-8*
a^5+4*a^2*b^2*(a^2-b^2)^(1/2))/(b^6*(a^2-b^2)^(1/2))
> int(cos(x)^20/(a+b*cos(x)),x=0..Pi);
1/65536*a*Pi*(65536*a^19-20480*a^12*b^6*(a^2-b^2)^(1/2)-1215
5*b^18*(a^2-b^2)^(1/2)-32768*a^16*b^2*(a^2-b^2)^(1/2)-1372
8*a^4*b^14*(a^2-b^2)^(1/2)-24576*a^14*b^4*(a^2-b^2)^(1/2
)-17920*a^10*b^8*(a^2-b^2)^(1/2)-16128*a^8*b^10*(a^2-b^2
)^(1/2)-65536*a^18*(a^2-b^2)^(1/2)-14784*a^6*b^12*(a^2-b^2
)^(1/2)-12870*a^2*b^16*(a^2-b^2)^(1/2))/(b^20*(a^2-b^2)^(1
/2))
===
Subject: Re: Squaring A Circle...(almost)
> So, FOR FUN, what are some other more efficient
> approximate-circle-squaring algorithms (using only
> straight-edge and compass) you all can come up with??
approximations to pi, Quarterly Journal of Mathematics XLV,
1914,
pp. 350372; it's reproduced at Pi: a source book, edi
by
Lennart Berggren, Jonathan Borwein and Peter Borwein) which
contains
a remarkable approximate quadrature; it is based upon the fact
that
pi is very close to the fourth root of 81 + 361/22.
===
Subject: Re: Squaring A Circle...(almost)
>> We all (should) know it is impossible to square a circle
with only a
>> straight-edge and compass and pencil.
>I'm sorry, but you never make it clear what you mean by
square a
>circle. Please explain.
<OT>
It is a well known expression (as others have explained to
you). You
did no real harm asking, but you would have better wai a while
to
see if he really was missing some fundamental detail or if
others
could understand his question unambiguously, and if so, had
you still
problems, then you could ask: what is this 'squaring of the
circle'
you all seem to be so familiar with?
And since we're on sci.math, I'll add my informal answer too:
it's one
of the top-three crank attracting topics in mathematics ;-)
</OT>
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Squaring A Circle...(almost)
>I'm sorry, but you never make it clear what you mean by
square a
>circle. Please explain.
> It is a well known expression (as others have explained to
you). You
> did no real harm asking, but you would have better wai a
while
Sorry, I'm actually reading and posting on rec.puzzles, where
we're
less math savvy :)
Moshe
--
http://runslinux.net :: moshe at runslinux dot net :: AIM:
Jehsom
===
Subject: Re: Squaring A Circle...(almost)
>We all (should) know it is impossible to square a circle with
only a
>straight-edge and compass and pencil.
>>I'm sorry, but you never make it clear what you mean by
square a
>>circle. Please explain.
> <OTIt is a well known expression (as others have explained
to you). You
> did no real harm asking, but you would have better wai a
while to
> see if he really was missing some fundamental detail or if
others
> could understand his question unambiguously, and if so, had
you still
> problems, then you could ask: what is this 'squaring of the
circle'
> you all seem to be so familiar with?
> And since we're on sci.math, I'll add my informal answer
too: it's one
> of the top-three crank attracting topics in mathematics ;-)
> </OT>
Although that and duplicating the cube seem to have fallen out
of fashion with the crank fringe lately. BTW, what did you
have in mind for the third entry in your top-three list?
Rick
===
Subject: Re: Squaring A Circle...(almost)
>> And since we're on sci.math, I'll add my informal answer
too: it's one
>> of the top-three crank attracting topics in mathematics ;-)
>Although that and duplicating the cube seem to have fallen out
>of fashion with the crank fringe lately. BTW, what did you
>have in mind for the third entry in your top-three list?
Huh?!? To be honest I *didn't* really have something in mind
for the
third entry. But we all know which is the very first one,
don't we?
It's that F* (:-) thing...
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Squaring A Circle...(almost)
Michele Dondi a .8ecrit:
[...]
> Huh?!? To be honest I *didn't* really have something in mind
for the
> third entry. But we all know which is the very first one,
don't we?
> It's that F* (:-) thing...
Shhhtt ! you'll wake them up (there is one who is half
awakened a few
threads above)
--
===
Subject: Re: Squaring A Circle...(almost)
) Although that and duplicating the cube seem to have fallen
out
) of fashion with the crank fringe lately. BTW, what did you
) have in mind for the third entry in your top-three list?
The four-colour map theorem, of course. The really really cool
bit about
that one is that the theorem is actually true, so you can't
just brush them
off with 'that's proven to be impossible' or something.
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the
statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
===
Subject: Re: Squaring A Circle...(almost)
Arithmetic/geometric mean series construction maybe?
Should be easy to implement geometrically,
converges fast.
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
===
Subject: Re: Squaring A Circle...(almost)
Hauke Reddmann a .8ecrit:
> Arithmetic/geometric mean series construction maybe?
> Should be easy to implement geometrically,
> converges fast.
Problem of convergent constructions is the loss of accuracy
during the
repea process...
As far as we want a *practical* construction and not only
theory...
The same drawback is in the repea constructs needed by high
number of
sides polygons, or summing the epsilon length segments
resulting of
repea bissector constructs as in the original post.
regards
--
===
Subject: Re: Squaring A Circle...(almost)
Hello
Oscar Lanzi III a .8ecrit:
> If you inscribe an n-sided regular polygon in the circle, and
> circumscribe an n-sided regular polygon about the same
circle, then you
> can approximate the circumference by adding one-third the
circumscribed
> perimeter to two-thirds the inscribed perimeter. You may use
any n that
> gives a constructible polygon, and the larger the n you use
the more
> accurate you will be.
> Let's try this out with n = 4. Taking the unit length to be
the circle
> radius, we get a circumscribed perimeter of 4 and an
inscribed perimeter
> of 2*sqrt(2) = 2.283. Then the approxima circumference is
> 4/3*(1+sqrt(2)) = 4/3*2.414 = 3.219. Which is sort of OK.
> Increase n to 6 and we get:
> Circumscribed perimeter = 2*sqrt(3) = 3.464
> Incscribed perimeter = 3
> Approxima circumference = 3.155 (!)
[...]
There are two different problems with squaring the circle
1) make a square whose AREA equals the area of the given circle
2) make a segment whose LENGTH (or a square whose PERIMETER)
is the
perimeter of the given circle
These two are equivallent because it is possible to construct
A^2 from A
or sqrt(A) from A with straight-edge and compass.
So squaring the circle is equivallent to construct a segment
of length PI.
Of cause impossible, but any good aproximation of PI leeds to
approximate squaring.
* 22/7 = 3.142... (easy with Thales to construct x/R=22/7. To
avoid a 22
units segment, just construct 3+1/7)
* sqrt(2)+sqrt(3)=3.146... easy to construct sqrt(2) with
diagonal of a
square and sqrt(3) from height of an equilateral triangle.
* sqrt((4+(3-sqrt(3)/3)^2)=3.14153...
* 1/2*(2+3/5)*sqrt(1+(2+1/5)^2) = 3.1415919
On a tangent in A to the circle centered in O, construct
AB=2*R, BC=R/5,
CD=2*R/5.
The circle of centre A and radius=OC intersects OA in E, on
the O side
of OA.
Parallel to OD in E intersect the tangent in F. AF~2*PI
* 355/113=3.1415159292... (there is a shorter construction
than Thales
on a 355R and 113R segments !)
* 4root(9^2+(19/22)^2)=3.1415926525... with
4root()=sqrt(sqrt())
Note (PI=3,1415926535....)
IMHO the sqrt(2)+sqrt(3) is the best one (=easiest) here.
5*10^-3 is a
*good* practical accuracy.
Given diameter AB, center O.
the circle at C and D, (C between O and E). OE=R*sqrt(3)
Circles centered in D and B of radius=OB=R intersects at F.
OF=R*sqrt(2)
Circle of radius OF centered in O intersects CD in G (D
between O and G)
EG ~ PI*R is half the perimeter. Double it to get the
perimeter.
To get the square with same area :
Let H on CD with EH=R, C between E and H.
The perpendicular to CD in H intersects the circle of diameter
EG in K
EK^2 = EH*EG ~ R*R*PI. EK is the side of the (approximate)
equivallent
square.
Regards.
--
===
Subject: Re: Squaring A Circle...(almost)
ippe che a .8ecrit:
[...]
> * 4root(9^2+(19/22)^2)=3.1415926525... with
4root()=sqrt(sqrt())
as said :
--
===
Subject: Re: Squaring A Circle...(almost)
>> We all (should) know it is impossible to square a circle
with only a
>> straight-edge and compass and pencil.
>I'm sorry, but you never make it clear what you mean by
square a
>circle. Please explain.
To square a circle means to construct a square with area equal
to
that of a given circle.
--
Greg
phobos78-marslink-net
Replace dashes and move in by 1 planet to reply.
===
Subject: Re: Squaring A Circle...(almost)
>> We all (should) know it is impossible to square a circle
with only a
>> straight-edge and compass and pencil.
> I'm sorry, but you never make it clear what you mean by
square a
> circle. Please explain.
Squaring the circle means constructing (using only a straighge,
compass, and pencil) a square with the same area as the unit
circle.
--
<fedya at bestweb dot net>
The way I see it, you raised three children who could knock
out and hog-
tie a perfect stranger, you must be doing *something* right.
Marge Simpson, <http://www.snpp.com/episodes/7G01.html>
===
Subject: postage stamp problem
i've been reading the discussion about the (generalized)
postage stamp
problem. Is there a reference to this problem somewhere in
standard
texts, or can someone describe the basic problem and what that
proof
is?
===
Subject: Re: postage stamp problem
> i've been reading the discussion about the (generalized)
postage stamp
> problem. Is there a reference to this problem somewhere in
standard
> texts, or can someone describe the basic problem and what
that proof
> is?
See:
http://mathworld.wolfram.com/PostageStampProblem.html
===
Subject: Inverse Function Theorem--stronger result, better
proof
A Stronger Inverse Function Theorem
Stephen A. Meigs
Has anyone heard of this result or proof before? It's so
simple and
straightforward, I suspect it must be in literature. Makes one
wonder
why people bother with proving the inverse function theorem in
normal
way, though.
It is shown that if a map is differentiable on a neighborhood
of a
point at which its derivative is continuous and invertible,
the map is
invertible on an open neighborhood of the point. The usual
hypothesis
that the derivative be continuous throughout the neighborhood
is
unnecessary.
Strong Injective Function Theorem. Suppose that y = f(x) is a
differentiable function of x on some neighborhood of p (where
f: R^n
-->R^n), and that f'(x) = Dy/Dx is invertible and continuous
at p.
Then y is an injective function of x on some neighborhood of p.
Lemma. Let A be an invertible n x n matrix. Then there is M >
0 such
that ||AX|| >= M ||X|| for every column vector X.
Proof of lemma. Since the unit sphere is compact, we may
choose M > 0
such that ||AX|| >= M whenever ||X|| = 1. Thus, for non-zero
X, ||AX||
= ||A((X/ ||X||)|| ||X|| >= M ||X||. QED
Proof of theorem. Using Lemma, choose M > 0 so that ||Dy/Dx(p)
X|| >=M
||X|| for all X. Let U be a neighborhood of p that is convex
(with
respect to x) and such that ||Dy/Dx - Dy/Dx(p)|| < M on U.
Using a
one-dimensional Henstock integral, it is clear that delta_y
(q)(r) is
the integral of Dy/Dx delta_x along a line segment between
distinct
points q and r. Between distinct points of U, the difference
between
Dy/Dx delta_x and Dy/Dx (p) delta_x has norm < M ||delta_x||.
Thus,
the norm of the difference between their integrals on a line
segment
between distinct points q and r of U is less than M ||delta_x
(q)(r)||. I.e., || delta_y (q)(r) - Dy/Dx (p) delta_x (q)(r)||
< M
||delta_x (q)(r)||. Thus, delta_y(q)(r) = 0 would imply
||Dy/Dx(p)
delta_x (q)(r)|| < M ||delta_x (q)(r)||, which would
contradict our
choice of M. QED
Strong Inverse Function Theorem. Suppose that y is a
differentiable
function of x on some neighborhood of p and that Dy/Dx is
invertible
and continuous at p. Then on some neighborhood U of p, y is a
homeomorphism with respect to x.
Proof of theorem. By the preceding theorem, there is a
neighborhood U
of p on which y is an injection with respect to x. The result
follows
immediately from the invariance of domain theorem. QED
step314 @ aol.com
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
> Strong Inverse Function Theorem. Suppose that y is a
differentiable
> function of x on some neighborhood of p and that Dy/Dx is
invertible
> and continuous at p. Then on some neighborhood U of p, y is a
> homeomorphism with respect to x.
> Proof of theorem. By the preceding theorem, there is a
neighborhood U
> of p on which y is an injection with respect to x. The
result follows
> immediately from the invariance of domain theorem. QED
This result is not stronger than the standard theorem. It has
a weaker
hypothesis and a weaker conclusion. Also, appealing to
invariance of domain
brings in weapons of mass destruction that should be kept
underground for
the big wars.
Shouldn't this be the strong form? Theorem: Suppose y is a
differentiable
function in a neighborhood of p and Dy/Dx is invertible and
continuous at
p. Then y maps a neighborhood U of p bijectively onto a
neighborhood V of
y(p), with y^(-1) differentiable in V and continuous at y(p).
I believe this is true, and I think it follows from the
standard proofs
quite easily (although I haven't checked this super carefully).
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
> Shouldn't this be the strong form? Theorem: Suppose y is a
differentiable
> function in a neighborhood of p and Dy/Dx is invertible and
continuous at
> p. Then y maps a neighborhood U of p bijectively onto a
neighborhood V of
> y(p), with y^(-1) differentiable in V and continuous at y(p).
> I believe this is true, and I think it follows from the
standard proofs
> quite easily (although I haven't checked this super
carefully).
Yes, looking at the standard proofs it does seem to follow
quite easily.
Looking quickly at the standard proof it seems the standard
proof involving
the contraction lemma doesn't really seem meaningfully to use
Dy/Dx being
continuous at any place other than p. I don't know why they
didn't claim to
prove the stronger result, guess they weren't interes in it.
So I
suppose
my result is not as interesting as I first thought, and is
mainly just the
observation that local injectivity is a little easier to prove
than local
homeomorphism onto an open set.
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
> A Stronger Inverse Function Theorem
> Stephen A. Meigs
> Has anyone heard of this result or proof before? It's so
simple and
> straightforward, I suspect it must be in literature. Makes
one wonder
> why people bother with proving the inverse function theorem
in normal
> way, though.
> It is shown that if a map is differentiable on a
neighborhood of a
> point at which its derivative is continuous and invertible,
the map is
> invertible on an open neighborhood of the point. The usual
hypothesis
> that the derivative be continuous throughout the
neighborhood is
> unnecessary.
> Strong Injective Function Theorem. Suppose that y = f(x) is a
> differentiable function of x on some neighborhood of p
(where f: R^n
> -->R^n), and that f'(x) = Dy/Dx is invertible and continuous
at p.
> Then y is an injective function of x on some neighborhood of
p.
> Lemma. Let A be an invertible n x n matrix. Then there is M
> 0 such
> that ||AX|| >= M ||X|| for every column vector X.
> Proof of lemma. Since the unit sphere is compact, we may
choose M > 0
> such that ||AX|| >= M whenever ||X|| = 1. Thus, for non-zero
X, ||AX||
> = ||A((X/ ||X||)|| ||X|| >= M ||X||. QED
> Proof of theorem. Using Lemma, choose M > 0 so that
||Dy/Dx(p) X|| >=M
> ||X|| for all X. Let U be a neighborhood of p that is convex
(with
> respect to x) and such that ||Dy/Dx - Dy/Dx(p)|| < M on U.
Using a
> one-dimensional Henstock integral, it is clear that delta_y
(q)(r) is
> the integral of Dy/Dx delta_x along a line segment between
distinct
> points q and r. Between distinct points of U, the difference
between
> Dy/Dx delta_x and Dy/Dx (p) delta_x has norm < M
||delta_x||. Thus,
> the norm of the difference between their integrals on a line
segment
> between distinct points q and r of U is less than M ||delta_x
> (q)(r)||. I.e., || delta_y (q)(r) - Dy/Dx (p) delta_x
(q)(r)|| < M
> ||delta_x (q)(r)||. Thus, delta_y(q)(r) = 0 would imply
||Dy/Dx(p)
> delta_x (q)(r)|| < M ||delta_x (q)(r)||, which would
contradict our
> choice of M. QED
> Strong Inverse Function Theorem. Suppose that y is a
differentiable
> function of x on some neighborhood of p and that Dy/Dx is
invertible
> and continuous at p. Then on some neighborhood U of p, y is a
> homeomorphism with respect to x.
> Proof of theorem. By the preceding theorem, there is a
neighborhood U
> of p on which y is an injection with respect to x. The
result follows
> immediately from the invariance of domain theorem. QED
First, you should not call this strong inverse function
theorem because
it
is rather weak. The inverse function theorem I know also
states that the
inverse function is differentiable and even gives its
derivative at y(p),
which is of course (Dy/Dx)^(-1).
Second, I think that your proof is inappropiate for an
introductory
analysis
course because the proof of the invariance of domain theorem
uses algebraic
topology. Or do you know a proof without? I would be very
interes in
knowing one...
--
Just because you're paranoid
Don't mean they're not after you
reverse my forename for mail!
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
First, you should not call this strong inverse function theorem
because
it
> is rather weak. The inverse function theorem I know also
states that the
> inverse function is differentiable and even gives its
derivative at y(p),
> which is of course (Dy/Dx)^(-1).
This is fairly easy corollary. That y is differentiable with
respect to x
at
q means delta_y(q) - Dy/Dx(q) delta_x(q) = E(q)
||delta_x(q)||, where E(q)
is near zero near q. Well, since the determinant is a
continuous function,
it is clear that on a neighborhood near p, Dy/Dx is
invertible. Thus, for q
in this neighborhood, there is by the lemma an N > 0 such that
||Dy/Dx (q)
delta_x (q)|| >= 2N ||delta_x(q)|| . Since near q, ||Dy/Dx (q)
delta_x (q)
-
delta_y(q)|| is small compared with ||delta_x(q)|| , it
follows that there
is neighborhood of q on which ||delta_y(q)|| >=N
||delta_x(q)||. If
Dy/Dx(q)
has inverse T(q), then on this neighborhood, ||T(q) delta_y(q)
-
delta_x(q)|| = ||T(q) E(q)|| ||delta_x(q)||< = ||T(q)||
||E(q)||
(1/N)||delta_y(q)||. Moreover, near q, ||T(q)|| ||E(q)|| (1/N)
is near 0,
so
T(q) is by definition a derivative of x with respect to y at
q. (By ||A||,
I
mean the maximum value left-multiplication by A attains on the
unit
sphere.)
>Second, I think that your proof is inappropiate for an
introductory
analysis
> course because the proof of the invariance of domain theorem
uses
algebraic
> topology. Or do you know a proof without? I would be very
interes in
> knowing one...
This is a true drawback. However, an advanced mathematician
needs to learn
the invariance of domain theorerm anyway; thus insofar as
advanced
mathematicians are concerned, it is not much of a drawback.
> --
> Just because you're paranoid
> Don't mean they're not after you
> reverse my forename for mail!
I found this abstract about a strong Inverse Function Theorem,
http://zb.msri.org/cgi-bin/zmen/ZMATH/en/quick.html?first=1&
maxdocs=3&type=h
tml&an=0663.26008&format=complete ,
but I no longer remember enough German to make sense of it.
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
> A Stronger Inverse Function Theorem
> Stephen A. Meigs
> Has anyone heard of this result or proof before? It's so
simple and
> straightforward, I suspect it must be in literature. Makes
one wonder
> why people bother with proving the inverse function theorem
in normal
> way, though.
> It is shown that if a map is differentiable on a
neighborhood of a
> point at which its derivative is continuous and invertible,
the map is
> invertible on an open neighborhood of the point. The usual
hypothesis
> that the derivative be continuous throughout the
neighborhood is
> unnecessary.
[snip]
> Strong Inverse Function Theorem. Suppose that y is a
differentiable
> function of x on some neighborhood of p and that Dy/Dx is
invertible
> and continuous at p. Then on some neighborhood U of p, y is a
> homeomorphism with respect to x.
But if the map is not continuous throughout the neighbourhood,
how is it
going to be
a homeomorphism of that neighbourhood?
> Proof of theorem. By the preceding theorem, there is a
neighborhood U
> of p on which y is an injection with respect to x. The
result follows
> immediately from the invariance of domain theorem. QED
> step314 @ aol.com
===
Subject: Re: Inverse Function Theorem--stronger result, better
proof
> A Stronger Inverse Function Theorem
> Stephen A. Meigs
Has anyone heard of this result or proof before? It's so
simple and
> straightforward, I suspect it must be in literature. Makes
one wonder
> why people bother with proving the inverse function theorem
in normal
> way, though.
It is shown that if a map is differentiable on a neighborhood
of a
> point at which its derivative is continuous and invertible,
the map is
> invertible on an open neighborhood of the point. The usual
hypothesis
> that the derivative be continuous throughout the
neighborhood is
> unnecessary.
> [snip]
Strong Inverse Function Theorem. Suppose that y is a
differentiable
> function of x on some neighborhood of p and that Dy/Dx is
invertible
> and continuous at p. Then on some neighborhood U of p, y is a
> homeomorphism with respect to x.
> But if the map is not continuous throughout the
neighbourhood, how is it
going to be
> a homeomorphism of that neighbourhood?
Proof of theorem. By the preceding theorem, there is a
neighborhood U
> of p on which y is an injection with respect to x. The
result follows
> immediately from the invariance of domain theorem. QED
step314 @ aol.com
It's trivial that the map is continuous throughout the
neighborhood
(differentiability implies continuity). What I am not assuming
is that the
derivative is continuous throughout the neighborhood.
===
Subject: Isometries
(This is for an independent study so don't worry about the hw
aspects)
In the 1st chapter of Boothby's Intro to differentiable
manifolds he asks
us to prove that there exists an isometry from an abstract
euclidean
vector space into R^N. In this he says a space is euclidean if
is has a
positive definite inner product. Now if W is a subset of R^n I
can think
of all kins of isometries but having not dealt uch with
abstract vector
spaces I was wondering where to start. Any suggestions would
be helpful
(books, websites,etc...)
===
Subject: Re: Isometries
Content-transfer-encoding: 8bit
> (This is for an independent study so don't worry about the
hw aspects)
> In the 1st chapter of Boothby's Intro to differentiable
manifolds he asks
> us to prove that there exists an isometry from an abstract
euclidean
> vector space into R^N. In this he says a space is euclidean
if is has a
> positive definite inner product. Now if W is a subset of R^n
I can think
> of all kins of isometries but having not dealt uch with
abstract vector
> spaces I was wondering where to start. Any suggestions would
be helpful
> (books, websites,etc...)
Assuming the abstract euclidean space is of dimension N, as you
indicate in your followup post...
Find an orthonormal basis for your space, call it {f_1, ...,
f_N}.
Define T(sum_k a_k f_k) = sum_k a_k e_k, where e_k is the
usual k-th
basis element of R^N, i.e.
e_k = (0,0,...,1,0,0,...)
^
|
k-th position
--Ron Bruck
===
Subject: Re: Isometries
> In the 1st chapter of Boothby's Intro to differentiable
manifolds he asks
> us to prove that there exists an isometry from an abstract
euclidean
> vector space into R^N. In this he says a space is euclidean
if is has a
> positive definite inner product. Now if W is a subset of R^n
I can think
> of all kins of isometries but having not dealt uch with
abstract vector
> spaces I was wondering where to start. Any suggestions would
be helpful
> (books, websites,etc...)
What is this in the plane? Mapping an ellipse onto a circle.
===
Subject: Re: Isometries
>> In the 1st chapter of Boothby's Intro to differentiable
manifolds he
asks
>> us to prove that there exists an isometry from an abstract
euclidean
>> vector space into R^N. In this he says a space is euclidean
if is has a
>> positive definite inner product. Now if W is a subset of
R^n I can think
>> of all kins of isometries but having not dealt uch with
abstract vector
>> spaces I was wondering where to start. Any suggestions
would be helpful
>> (books, websites,etc...)
> What is this in the plane? Mapping an ellipse onto a circle.
It is kind of hard to tell exactly what this means. I assume
it is more
complica than just a mapping of a geometric object into
another one.
Cause can't a function space be considered a euclidean vector
space.
===
Subject: Re: Isometries
the abstract vector space has the same dimension as R^n
===
Subject: Re: Hilbert spaces in quantum mechanics
>I had a course on quantum mechanics, but I'm having trouble
>relating the maths to the physics.
...
>We say that the set of state vectors for a system is a
>Hilbert space. Is there a concrete realisation of this?
>My question is: what is state space? Does
>quantum mechanics actually have rigorous mathematical
>foundations? How come textbooks always gloss over these
>contradictions as if they all make sense?
If you want to known the QM postulates, you can look in the
books by
Cohen-Tannoudi et. al., Quantum Mechanics. Also, the book by
Streater &
Wightman, PCT, Spin & Statistics and all that, has an example
of
working
with a QFT in a purely mathematical setting.
QM was developed empirically from the de Broglie waves to the
Schrodinger
equation. But as one did not find a Schrodinger equation for
relativity or
the weak and strong forces, this paved the way for QFT and
gauge theories,
that in the bottom just has an abstract Hilbert space. One way
to make
such theories practically computable is via Feynman path
integrals, which
provides a Green's function without having to know it being
derived
from
a differential equation. It is not known, though, from a purely
mathematical point of view, why this method works.
Eventually, this lead to movements such as string theory,
which reasons
that the true foundation of QM should be some kind of
generalization of
Feynman path integrals, not originating from underlying
differential
equations. But such movements have so far been unsuccessful in
terms
creating of computable theories with the capacity to predict
observations
and experiments.
So the answer is that some parts of QM can be formalized into
pure math,
but QM as a whole is purely empirical, and there is no known
common
mathematical framework, except the abstract Hilbert space
formulation
then.
Hans Aberg
===
Subject: Re: Hilbert spaces in quantum mechanics
Originator: grubb@lola
>> How then can we possibly
>> write something like
>> <x,x'>=delta(x-x') ?
>> <x,x> is not in C.
>I don't know why you would write that. Perhaps you could
elaborate on
>the context in which you would see that written.
Actually, this type of thing is written fairly often. Usually,
sense
can be made out of it by taking and integral over some
variable (like x
above)
and interpreting the result usung integral transforms. For
example,
it is common to write something like
int_{-infty}^infty e^{ixy} dx = 2pi delta(y).
This is ultimately a way of writing the Fourier inversion
formula.
If you integrate over y and switch the integrals, you get and
expression
for the inversion formula at y=0. Of course, this is not
rigorous
as it stands. On the other hand, it seems clear to me that
this integral
can actually be interpre as a weak integral in the space of
distributions, but I have yet to see a good treatment of such
things.
--Dan Grubb
===
Subject: Re: Derivative of the complex variable function
exp(i*Arg z)
>Is it possible to say something about the derivative of such
functions
>of the complex variable as:
exp(i*Arg z)
or
z/|z|
where z belongs to C excepting (0,0).
It seems that according to the Cauchy-Riemann conditions these
>functions are not differentiable, isn't it?
> Correct.
>If so, how this corresponds with the fact that exp(z) is
analytic on
>C?
> So what? The problem isn't the exp, it's the Arg.
>Another question is: what is possible to say about the
>differentiability of the functions |z| and Arg(z), where z is
>complex?
> They aren't differentiable anywhere. A real-valued function
of a
> complex variable is never differentiable unless the
derivative is 0.
> This is easy to see from Cauchy-Riemann.
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
Robert,
thank you very much for your explanation.
Igor
===
Subject: Re: Automobile Speeds.
> If my car manual and some car reviews are a guide, in the
rest of
> Europe, as well as using kilometres instead of miles, and
litres
> instead of gallons (1, 2), you measure fuel economy as
l/100km rather
> than mpg as we do in the UK and the US. So low numbers are
good for
> you but high numbers are good for us.
>This is not entirely true. Both l/100km and km/l are used.
[...]
>The changeover is caused by an EC directive from 1993 which
standardises
>the usage figures. It standardises, amongst others, the way
to measure.
I always liked the comment that, dimensionally speaking,
mileage has
units of reciprocal area. Thus you can report your efficiency
in
cm^(-2) or something like that.
Exercise 1. Complete the following sentence: My car gets ___
to the
acre.
Exercise 2. Describe a physical meaning for the unit of area
associa
to your automobile's efficiency.
dave
===
Subject: Re: Automobile Speeds.
> If my car manual and some car reviews are a guide, in the
rest of
> Europe, as well as using kilometres instead of miles, and
litres
> instead of gallons (1, 2), you measure fuel economy as
l/100km rather
> than mpg as we do in the UK and the US. So low numbers are
good for
> you but high numbers are good for us.
This is not entirely true. Both l/100km and km/l are used.
> [...]
>The changeover is caused by an EC directive from 1993 which
standardises
>the usage figures. It standardises, amongst others, the way
to measure.
> I always liked the comment that, dimensionally speaking,
mileage has
> units of reciprocal area. Thus you can report your
efficiency in
> cm^(-2) or something like that.
> Exercise 1. Complete the following sentence: My car gets ___
to the
acre.
There are no units in the numerator, so the sentence is
complete as given.
> Exercise 2. Describe a physical meaning for the unit of area
associa
> to your automobile's efficiency.
What did the farmer get when he stepped on a rake? Two acres.
> dave
===
Subject: The Lost Proof of Fermat
p is a prime number
x,y, and z, are positive integers.
x^p + y^p = z^p
[x^p + y^p]^w = z^[p]^n
then
z^[wp] = z^[p]^n
w = p = n
z^[np] = z^[p]^n
np = p^n
n = p^[n-1]
n = n^[n-1]
n^[n-2] = 1
n^[0] = 1
n-2 = 0
w = p = n = 2
===
Subject: Re: The Lost Proof of Fermat
x-no-archive: yes
windows-nt)
> p is a prime number x,y, and z, are positive integers.
> x^p + y^p = z^p
> [x^p + y^p]^w = z^[p]^n
This equation implies that w = n. It tells us nothing about p.
> z^[np] = z^[p]^n
> np = p^n
This is nonsense. (z^p)^n = Z^(np); you seem to think that
z^(p^n) = z^(pn), which is false.
--Len.
===
Subject: Re: The Lost Proof of Fermat
> This is nonsense. (z^p)^n = Z^(np); you seem to think that
> z^(p^n) = z^(pn), which is false.
lol.
Everyman thinks he's the next Will Hunting or Srinivasa
Ramanujan.
--
Demetrius Merisi di Carravallo
===
Subject: Re: The Lost Proof of Fermat
Here's what I imagined Fermat probably had in mind. Just a
wild guess!!!
;-)
Let's see what he was aware of : binomial theorem, calculus.
Take the equation
X^n + Y^n = Z^n
There are 3 variables, so we need to introduce 2 other
equations.
Suppose, we differentiate both sides of the equation by some
variable t
we would have
X^{n-1} dX/dt + Y^{n-1} dY/dt = Z^{n-1} dZ/dt
Take dX/dt , dY/dt, dZ/dt , we can replace them with a',b',c'
because
dX/dt,
dY/dt, dZ/dt can take any value. We can regard them as unknown.
X^{n-1} a' + Y^{n-1} b' = Z^{n-1} c'
Now, suppose we normalize by c'. We can rewrite the above
equation as
a X^{n-1} + b Y{n-1} = Z^{n-1}
Now let's differentiate the system by t'. We would end with
d X^{n-1} + e X^{n-2} + f Y^{n-1} + g Y^{n-2} = m Z^{n-1} + n
Z^{n-2}
Notice {n-2} , this proof would only apply where n>2.
Now, replace X/Z = w_x , Y/Z = w_y
We would have
w_x^n + w_y^n = 1
a w_x^{n-1} + b w_y^{n-1} = 1
w_x^n = 1 - w_y^n
aw_x^{n-1} = 1 - b w_y^{n-1}
w_x = a ( 1- w_y^n)/(1 - b w_y^{n-1})
Substituting the above equation into
w_x^n + w_y^n = 1
We get
a^n( 1 - w_y^n)^n/(1- b w_y^{n-1})^n = 1 - w_y^n
which leads to
a^n( 1 - w_y^n)^{n-1} = (1 - b w_y^{n-1})^n
Similarly we can obtain an equation for w_x
b^n( 1- w_x^n)^{n-1} = ( 1- a w_x^{n-1})^n
We can get an expression for Z, by substituting X = w_x Z, and
Y = w_y Z,
into
d X^{n-1} + e X^{n-2} + f Y^{n-1} + g Y^{n-2} = m Z^{n-1} + n
Z^{n-2}
Proving Fermat's last theorem would would equal to proving that
expressions
obtained for X,Y,Z can not be distinct integers. Ofcourse,
this step could
be hard. Still thinking!!! May be just a wild wild guess.
> p is a prime number
> x,y, and z, are positive integers.
> x^p + y^p = z^p
> [x^p + y^p]^w = z^[p]^n
> then
> z^[wp] = z^[p]^n
> w = p = n
> z^[np] = z^[p]^n
> np = p^n
> n = p^[n-1]
> n = n^[n-1]
> n^[n-2] = 1
> n^[0] = 1
> n-2 = 0
> w = p = n = 2
===
Subject: Re: The Lost Proof of Fermat
> p is a prime number
> x,y, and z, are positive integers.
> x^p + y^p = z^p
> [x^p + y^p]^w = z^[p]^n
> then
> z^[wp] = z^[p]^n
> w = p = n
> z^[np] = z^[p]^n
> np = p^n
> n = p^[n-1]
> n = n^[n-1]
> n^[n-2] = 1
> n^[0] = 1
> n-2 = 0
> w = p = n = 2
Aha! so we can't have
(3^2 + 4^2)^3 = (5^2)^3
'cos then w = n = 3 and p = 2.
Brilliant!
--
===
Subject: Re: The Lost Proof of Fermat
Adjunct Assistant Professor at the University of Montana.
You are kidding, right?
>p is a prime number
>x,y, and z, are positive integers.
>x^p + y^p = z^p
>[x^p + y^p]^w = z^[p]^n
>then
>z^[wp] = z^[p]^n
>w = p = n
No. You conclude that z^[wp] = z^[pn], from which you get
(since z>1)
that wp = pn, hence w=n. You cannot conclude it for p here,
because it
would not follow.
>z^[np] = z^[p]^n
>np = p^n
The exponentiation rules state that (a^b)^c = a^{bc}. So the
right
hand side is z^[pn], just like the left hand side.
>n = p^[n-1]
>n = n^[n-1]
>n^[n-2] = 1
>n^[0] = 1
>n-2 = 0
>w = p = n = 2
HINT: The only thing you apparently used was that z was a
positive
real number greater than 1. Explain why your 'argument' does
not show
that 2^3 + 3^3 = z^3, where z is the real cube root of 35.
--
===
Subject: Possible Interpretation of the Wave Equation &
Non-linearity
Given a newtonian system
m d^2 x/dt^2 = - d/dx V(x)
Suppose, a discrete approximation of this system was simula
using a
finite state machine, just using a few number of varibale x_i
, x_{i,-1},
x_{i,-2} and t . Over time the error will compound and the
results will be
away from original as time t progress. On the otherhand,
suppose, one can
invent a system, that can introduce many many degrees of
freedom to the
above system, such that the average of that system remains
true to the
above equation....[i am thinking of a better way to explain it
].
That could be an interesting way to interpret the wave
equation, wouldn't
it?
-suresh
===
Subject: Re: A problem
I hope you can give me some advice in resolving the following
> question, which is a Number Theory problem. I've tried and
tried but
> can't figure out the right way to solve it.
> The problem is:
> Let q(x,y) be a primitive (that is, with (a,b,c)=1)
quadratic form
> ax^2+bxy+cy^2,
> and n a positive integer.
> Show the existence of two integers s,t , with (s,t)=1, such
that
> (q(s,t),n)=1.
> Use Chinese remainder theorem to reduce to case n = p^r, p
prime.
> Notice that is equivalent to n = p.
> Then one of (1,0), (0,1) and (1,1) works.
You mean, these are the values to assign to s and t?
I thought that the condition (a,b)=1 was defined just for a>1
and b>1,
not for a, b taking values 0 and-or 1.
And, the condition a,b,c coprime seems not necessary to me, if
I solve
the problem in this way you told me.
Anyway, thank you for your advice!
Francesco
P.S.: your site is very good, by the way. Your pdf's helped me
a lot
in last year's exams!
===
Subject: functions needed
I need to construct such a family of functions which
satisfy following constrains:
f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d.
for all positive integers a,b,c,d.
Thanks
===
Subject: Re: JSH: Questioning my conclusions
>>Here's some identifying to find Decker's post, which also
shows that
>>he is indeed at Hamilton College:
Good to have that confirmed, my paychecks say the same thing.
>>I just don't accept that mushing position as it's not
mathematics.
Regardless of whether you accept it or not, it's correct, as
a number of respondants have poin out.
> See my previous reply on this. The mushing position is
correct.
> I am glad you have chosen to focus on x = 3. When you do
that,
> we are down to the heart of the matter. Everything in the
equations
> is just numbers and we can think about factorizations of
numbers
> instead of the perhaps confusing background of functions and
> polynomials.
> Let A = (5 a_1(3) + 7),
> B = (5 a_2(3) + 7), and
> C = (25*3^2 + 30*3 + 2) = 257.
Nora, you left out a couple of 30's. Replace 257 with 317.
Rick
===
Subject: Re: JSH: Questioning my conclusions
>>Here's some identifying to find Decker's post, which also
shows that
>>he is indeed at Hamilton College:
> Good to have that confirmed, my paychecks say the same thing.
>I just don't accept that mushing position as it's not
mathematics.
> Regardless of whether you accept it or not, it's correct, as
> a number of respondants have poin out.
See my previous reply on this. The mushing position is correct.
I am glad you have chosen to focus on x = 3. When you do that,
> we are down to the heart of the matter. Everything in the
equations
> is just numbers and we can think about factorizations of
numbers
> instead of the perhaps confusing background of functions and
> polynomials.
Let A = (5 a_1(3) + 7),
B = (5 a_2(3) + 7), and
C = (25*3^2 + 30*3 + 2) = 257.
 Nora, you left out a couple of 30's. Replace 257 with 317.
you are right, I intended obviously to let x = 3 but
when I did the arithmetic I made an error. C should
indeed be 317. Fortunately 317 is coprime to 7, so everything
else works as I sta.
Curious side note: x = 2 is an interesting case also. Note that
a1 = (1 + sqrt(-167))/2, and that C = 162, which is coprime
to 7. However,
(5*7 + 7) * (5*5 + 2) = 1134 = 7*162 = 7*C.
In this case, exactly as claimed by Harris, you can divide
7 out of the first factor and 1 out of the second factor to
obtain
(5*1 + 1)*(5*5 + 2) = 162 = C,
which is certainly a factorization with all the coefficients
algebraic integers, etc..
A bit puzzling, till you realize that 7 is not equal
to a_1(2) as defined above, and 5 <> b_2(2).
Harris has conveniently ignored 's note that
a_1(2) = (1 + sqrt(-167))/2 has a factor of 7 that is the
root of a monic polynomial of degree 11.

> Rick
===
Subject: Re: JSH: Questioning my conclusions
>>Here's some identifying to find Decker's post, which also
shows that
>>he is indeed at Hamilton College:
> Good to have that confirmed, my paychecks say the same thing.
>I just don't accept that mushing position as it's not
mathematics.
> Regardless of whether you accept it or not, it's correct, as
> a number of respondants have poin out.
It's not algebraically correct.
Here is algebra, so that you can see what it looks like Decker.
Originally a slight modification to your example gives
(5a_1(x) + 7)(5b_2(x)) + 2) = 7(25x^2 + 30x + 2),
and the question is how to divide both sides by 7 and remain
in the
ring of algebraic integers. That's important. It's not just
about
dividing by 7, but can you do so and remain in the ring of
algebraic
integers?
the left have sqrt(7) as a factor, then you have
(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =
25x^2 + 30x + 2.
The problem though is that 2/sqrt(7) is not in the ring of
algebraic
integers.
However, algebra covers more inclusive rings, so I can display
that
and readers can see that now you're in a ring which has
2/sqrt(7).
The algebra doesn't mind.
Now then, even if you now say, hey, but wait, (5b_2(x) +
2)/sqrt(7)
will give a result--that is, if you do the operations, work it
out and
get a value--for *some* x, or even some family of x's that IS
in the
ring of algebraic integers, I say, ok, but here and now,
whether it
works or not for some x, you need 2/sqrt(7), which is not
available in
the ring of algebraic integers.
That's algebra Decker.
Mushing everything together in the hope that somehow, someway
it will
all work out isn't algebra Decker. Algebraically there's one
answer,
and it's not debatable, as algebra isn't a democratic process.
Algebraically, to try and remain in the ring of algebraic
integers,
your only option is to have
(5a_1(x)/7 + 1)(5b_2(x)/7 + 2) = 25x^2 + 30x + 2
and I know that it bothers you because it's possible to show
that
5a_1(x)/7 in general is not an algebraic integers, but sorry
Decker,
that's what logically follows from the algebra.
The simple fact is that you can't even present any other
outcome
without having members that aren't in the ring of algebraic
integers,
like I showed with 2/sqrt(7) and it's basic algebra.
The conclusion that follows algebraically is that you can't
divide 7
from both sides and guarantee that you're still in the ring of
algebraic integers.
You're forced out, as the algebra clearly shows.
===
Subject: Re: JSH: Questioning my conclusions
>It turns out, of course, that it's easy enough to calculate
b_2(x),
>from
>a_2(x) = b_2(x) - 1
>and
>a_2(x)^2 - (x - 1)a_2(x) + 7(x^2 + x) = 0,
Why not simply calculate b_2(x) by determining the roots of
your
equation?
In other words: Solve a^2 - (x - 1)a + 7(x^2 + x)=0
You'll find that:
a = (1/2)*(-1 + x +/- sqrt(1 - 30 x - 27 x^2))
Since you wan a1(0) to be 0:
a1(x) = (1/2)*(-1 + x + sqrt(1 - 30 x - 27 x^2))
This means:
a2(x) = (1/2)*(-1 + x - sqrt(1 - 30 x - 27 x^2))
So:
b2(x) = a2(x)+1= (1/2)*(1 + x - sqrt(1 - 30 x - 27 x^2))
>as substituting gives
>b_2(x)^2 - 2b_2(x) + 1 -x b_2(x) + x + b_2(x) -1 + 7(x^2 + x)
= 0
>which is
>b_2(x)^2 -(x+1) b_2(x) + 7x^2 + 8x = 0
>which gives
>b_2(x) = ((x+1)+/- sqrt(x^2 + 2x + 1 - 28 x^2 - 32 x))/2
>which is
>b_2(x) = ((x+1) +/- sqrt(-27x^2 - 30x + 1))/2.
No, the +/- should simply be a -.
===
Subject: bounded linear functionals in Lp
I'm looking for information on applications in pure math of the
theorem that states that the dual of Lp is canonically
isomorphic to
Lq (1<p<inf, 1/p+1/q=1). I'm searching for material in order to
motivate it to a group of undergrads...
Any help apprecia,
nojb.
===
Subject: Re: bounded linear functionals in Lp
>I'm looking for information on applications in pure math of
the
>theorem that states that the dual of Lp is canonically
isomorphic to
>Lq (1<p<inf, 1/p+1/q=1). I'm searching for material in order
to
>motivate it to a group of undergrads...
I suspect another topic would be a better idea.
>Any help apprecia,
>nojb.
************************
===
Subject: Re: bounded linear functionals in Lp
> I'm looking for information on applications in pure math of
the
> theorem that states that the dual of Lp is canonically
isomorphic to
> Lq (1<p<inf, 1/p+1/q=1). I'm searching for material in order
to
> motivate it to a group of undergrads...
Tell 'em their iPods depend on it.
===
Subject: Re: need help in understanding Torkel's ZFC comment
*** PART ONE  FOR THOSE INTERES IN PSYCHOLOGY ***
> You'd probably have more fruitful discussions with people if
you
> didn't argue just for the sake of arguing, asking questions
that
> you know the answer to very well.
Actually, I really don't know how you could possibly know that
the
context that I referred to is so different from the current
one, since
I never indica where I saw that.
The fruit borne of a conversation is a function of many
factors.
Maybe it would be better to address the technical merits of
what I say
rather than what you perceive to be my motives.
> it's clear that you know that you're asking silly questions
I don't know how you could know that either. My point (believe
it or
not) was that you are being very presumptuous.
> I wasn't actually claiming to know what some author thought,
just offering
a
> likely explanation for how it could happen that you saw
> those things lis as axioms
You sta it as fact, not speculation: The author didn't think
the
difference mattered, but in the present context it does. As a
mathematician, I would hope that you would be careful to
distinguish
between conjecture and theorem. You can always ask me for the
context
first, you know. Here, let me save you the trouble. The entire
theory crea by Cantor and Dedekind can be reduced to seven
axioms.
is from Investigations in the Foundations of Set Theory I by
Ernst
Zermelo, 1908, Mathematische Annalen 59, 261-281.
> Also it would work better if you didn't pull this sort of
swithcheroo
> I haven't said _anything_ here about the interpretation of
the
> intuitive concepts that ZFC is attempting to formalize.
It's not essential that you did, but for the record:
Your first post was: ZFC has infinitely many axioms. Those
infinitely
many
axioms can be described by finitely many axiom schemata'.
My response was: I agree with (this) interpretation (but) I
have
always felt that axiom schemas are just rules of inference.
> previous things you said indicate you haven't been paying
> attention to who said what in you replies
Since the premise is false, we need not be concerned with the
above
conclusion.
> What in the world is illformed here?
not conforming to the rules of grammar
(http://dictionary.reference.com/search?q=illformed) In this
case,
the rules are that axiomatic systems consist of definitions,
axioms,
rules of inferences and (consequently) theorems. Axiom schemas
unnecessarily (redundantly) go outside of this formalism. I
consider
it a violation of Occam's Razor: one should not increase,
beyond what
is necessary, the number of entities required to explain
anything (or
plurality should not be posi without necessity) We don't need
axiom schemas.
> I imagine you also agree that triangles have three sides?
Great.
> (That's an attempt at explaning what I find amusing about
what
> you say you agree to...)
Damned if I do and damned if I don't. If agreeing with a pos
statement is deserving of ridicule, what does that say about
the
author of the statement itself?
> This is just silly. _Any_ axiom or axiom scheme can be
formalized
> as a rule of inference.
Me: I have always felt that axiom schemas are just rules of
inference.
Your 1st. response: I've always felt that automobiles were
actually
pieces of fruit. (ridiculing the idea - speaking of silliness)
Your new response: 'Any' axiom or axiom scheme can be
formalized as a
rule of inference. (accepting the idea)
What was that about swithcheroo? (This incongruousness is
similar
to the idea that what you post is, presumably, of
significance, but
for me to agree to it is comparable to concluding that a
triangle has
3 sides.)
You seem to be saying (a) that my comment above is wrong, and
(b)
everyone already knows what I'm saying. While each is a way of
discrediting someone's work, unfortunately they are
incompatible. I
prefer the latter, and appreciate your affirming my point.
Yes, the method of translating an axiom schema into a rule of
inference that I described is quite general. The only question
is,
would it be better to apply this method to ZFC, i.e., to
express the
axiom schemas of ZFC as rules of inference instead? I say yes
and
cite Occam's Razor. I also think it is better since axioms and
rules
are used differently, and calling a rule of inference an axiom
schema only clouds that distinction. (I am also not aware of
anyone
having done this in ZFC before. Are you?)
> this has nothing to do with the
> distinction between axioms and axioms schemes
It shows that it is a syntactic distinction (and is thus
redundant, as
I discuss above.)
*** PART TWO  FOR THOSE INTERS IN MATHEMATICAL LOGIC ***
> You haven't given any evidence for the better part.
Thanks for asking. I explained my formalization of Separation
and
Replacement (the axiom schemas mentioned by others), and
compared them
to ZFC's formalisms:
Separation is: P[TW] , Q[SE] => P^Q[SE]
In ZFC: existsX allY ( Y e X = Y e S ^ A(y) )
Replacement is: P(x)[SE] , Q(I,x)[TW] => (eA)P(A)^Q(A,x)[SE]
In ZFC: existsX allY e S ( existsZ A(Y,Z) .88 existsZ e X
A(Y,Z))
The entire set of ZFC axioms (schemas) is a hodgepodge of
various
combinations of a small number of simpler (more primitive)
definitions, axioms and rules of inference that I introduced in
http://www.arxiv.org/html/cs.lo/0003071 . These rules of
inference
are also used to construct computer programs and derive
theorems from
the Theory of Computation (as I detail in this paper.) Do you
agree
with the principle that a formalism that has wide
applicability is
better than one that applies to only one domain?
I also maintain that it is easier to work with a set of
primitives
than it is to work with a set of combinations of those
primitives. It
is also more revealing as to the true nature of Set Theory to
develop
simpler primitives. After all, the goal of axiomizing is to
express a
theory in terms of a set of primitives. The more primitive the
better. Would you agree with that principle, at least?
I entered into this discussion with the express goal of
obtaining some
formal uses of ZFC so that I could compare how the same
results are
derived using my formalism. To date, nobody has come through
with the
formal proofs of consistency of subsets of ZFC that were
described.
> in ZFC when they're doing mathematics _anyway_.)
I think that shows that ZFC has failed in its goal of
formalizing Set
Theory. (There are plenty of rigorous proofs in Propositional
Calculus.)
>So what are the Rules of Inference of ZFC then?
> You're lecturing us on this stuff and you really don't know?
> Wow.
No, I am pointing out that ZFC is an axiomitization without any
(expressed) rules of inference! In reality, there are rules of
inference that are either expressed as axioms (schemas) or are
part
of ZFC axioms that are actually combinations of simpler
primitives.
> Hint: there's really no such thing as the rules of inference
of
> ZFC per se.
That's right. What kind of an axiomatic system is that?
> ************************
>
===
Subject: Re: need help in understanding Torkel's ZFC comment
>[...]
>> I wasn't actually claiming to know what some author
thought, just
offering a
>> likely explanation for how it could happen that you saw
>> those things lis as axioms
>You sta it as fact, not speculation: The author didn't think
the
>difference mattered, but in the present context it does. As a
>mathematician, I would hope that you would be careful to
distinguish
>between conjecture and theorem. You can always ask me for the
context
>first, you know. Here, let me save you the trouble. The entire
>theory crea by Cantor and Dedekind can be reduced to seven
axioms.
>is from Investigations in the Foundations of Set Theory I by
Ernst
>Zermelo, 1908, Mathematische Annalen 59, 261-281.
1908? For heaven's sake, it _really_ never occured to you that
the
precise meaning of a technical term might change a teensy bit
in a _century_? GIven that mathematical logic barely exis
at that time and was developed during the last century, when
someone tells you something about the distinction between
axiom and axiom scheme you're puzzled by the fact
that a 100 friggin years ago someone appeared to be using
the terms in a way that's not consistent with the distinction
being made at present?
Wow.
>> Also it would work better if you didn't pull this sort of
swithcheroo
>> I haven't said _anything_ here about the interpretation of
the
>> intuitive concepts that ZFC is attempting to formalize.
>It's not essential that you did, but for the record:
>Your first post was: ZFC has infinitely many axioms. Those
infinitely
>many
>axioms can be described by finitely many axiom schemata'.
Yes, I said that. And that says nothing whatever about the
interpretation of the intuitive concepts that ZFC is trying
to formalize.
>My response was: I agree with (this) interpretation (but) I
have
>always felt that axiom schemas are just rules of inference.
>> previous things you said indicate you haven't been paying
>> attention to who said what in you replies
>Since the premise is false, we need not be concerned with the
above
>conclusion.
>> What in the world is illformed here?
not conforming to the rules of grammar
>(http://dictionary.reference.com/search?q=illformed) In this
case,
>the rules are that axiomatic systems consist of definitions,
axioms,
>rules of inferences and (consequently) theorems. Axiom schemas
>unnecessarily (redundantly) go outside of this formalism. I
consider
>it a violation of Occam's Razor: one should not increase,
beyond what
>is necessary, the number of entities required to explain
anything (or
plurality should not be posi without necessity) We don't need
>axiom schemas.
Sorry to break this to you, but you're making a fool of
yourself.
Yes, not conforming to the rules of grammar is indeed what
ill-formed means. The fact that you feel something violates
Occam's razor does not show that it does not conform to the
rules of grammmar... (unless it should be just like Charlie
Boo wants thinks it should be is a rule of grammar. It's not.)
>> I imagine you also agree that triangles have three sides?
Great.
>> (That's an attempt at explaning what I find amusing about
what
>> you say you agree to...)
>Damned if I do and damned if I don't. If agreeing with a pos
>statement is deserving of ridicule, what does that say about
the
>author of the statement itself?
>> This is just silly. _Any_ axiom or axiom scheme can be
formalized
>> as a rule of inference.
>Me: I have always felt that axiom schemas are just rules of
>inference.
>Your 1st. response: I've always felt that automobiles were
actually
>pieces of fruit. (ridiculing the idea - speaking of silliness)
>Your new response: 'Any' axiom or axiom scheme can be
formalized as a
>rule of inference. (accepting the idea)
Here's a hint: Are just rules of inference is not the same
thing
as can be formalized as rules of inference.
>What was that about swithcheroo? [...]
If you want to get these things straight you have to read
and write much more carefully.
>> in ZFC when they're doing mathematics _anyway_.)
>I think that shows that ZFC has failed in its goal of
formalizing Set
>Theory. (There are plenty of rigorous proofs in Propositional
>Calculus.)
Uh, rigorous proof is not the same thing as formal proof.
>>So what are the Rules of Inference of ZFC then?
>> You're lecturing us on this stuff and you really don't know?
>> Wow.
>No, I am pointing out that ZFC is an axiomitization without
any
>(expressed) rules of inference! In reality, there are rules of
>inference that are either expressed as axioms (schemas) or
are part
>of ZFC axioms that are actually combinations of simpler
primitives.
>> Hint: there's really no such thing as the rules of
inference of
>> ZFC per se.
>That's right. What kind of an axiomatic system is that?
It's an (axiomatic) _mathematical theory_, _not_ a
system of formal logic.
Some of my comments that you've objec to have been
based on my assumption that you couldn't be as ignorant
as you're pretending to be. Maybe you are. I'll explain:
The reason there are no rules of inference in ZFC per se
is because ZFC is a mathematical theory. There are
various axioms, conveniently summarized in finitely
many schemes, and the theorems of ZFC are precisely
the logical consequences of those axioms.
There are many different but equivalent ways to
formalize the notion of logical consequence.
These formalizations are called formal systems.
They have, or can have, _differing_ rules of inference.
When we switch from one formal system to another
we _change_ what the rules of inference are. But
we don't change what ZFC is - it's still precisely
the set of logical consequences of those axioms.
Asking what the rules of inference of ZFC are makes
no sense, just as it makes no sense to ask what
the rules of inference of any other mathematical
theory, eg group theory, are. Mathematical theories
do not have rules of inference - it's formal systems
of logic that have rules of inference.
> ************************
>>
************************
===
Subject: Re: Questions about Hatcher's Algebraic Topology book
- the idea of n-cycles in simplicial/singular homology is to
replace
> the idea of loops in the context of the fundamental group,
right?
> However, it is no clear to me whether relevant topological
information
> is not lost when a free abelian group of n-cycles is formula.
> Of course information is lost when forming chain groups.
> The point is that even so, some useful information is
preserved
> in chain complexes (and can be calcula).
In fact, you could say the point is that information *is* lost!
Hopefully enough information is lost so that the relevant
information
for the situation at hand is sif out. This is what you're
trying to
do when you apply a functor -- looking at the simplified
image,free of
irrelevant details.
===
Subject: Re: Isoperimetric Zepp
> For example, the 3-d version of the Zepp would be an
axisymmetric
> shape whose profile obeys the equation
> x(s) |y(s)| = alpha kappa(s)
> for some alpha.
Oops: that's not a good 3-d generalization. This would maximize
the integral of x over the volume subject to the constraint
that
the length of the cross-section is 1, which is a strange thing
to
want. It would make more sense to constrain the surface area to
be 1.
Assuming that the force-balance formula remains valid in 3-d,
then the equation for the surface would be that F(x,y,z) is
proportional to the mean curvature. If we assume F(x,y,z) is
axisymmetric about x, then we can formulate this in terms of a
cross-section F(x,y). The mean curvature could then be
expressed as
(1/2) ( phi'(s) - cos(phi(s)) / y(s) ),
yielding the equation
F(x(s),y(s)) = alpha ( phi'(s) - cos(phi(s)) / y(s) ), or
F(x(s),y(s)) = alpha ( kappa(s) - x'(s)/y(s) ).
Example:
Let F(x,y) = 1. Then phi(s) = 2 pi (s-1/4) is a solution:
1 = alpha (2 pi - sin (2 pi s) / (-sin(2 pi s) / (2 pi))
= 4 pi alpha, so alpha = 1/(4 pi). Check.
I.e., a sphere has (locally) maximal volume (at least among
axisymmetric shapes) for unit surface area.
-Jim Ferry
===
Subject: Re: Isoperimetric Zepp
> One shape quite similiar to the Zepp is the lemniscate.
> I was very happy when I found out that the lemniscate is the
> solution of the given problem of maximizing the enclosed
> value, if the price is given by
> F(x,y) = sqrt(x^2 + y^2)
> i.e. where the price varies with the distance. The apex angle
> of the lemniscate is 90 degrees, the one of the Zepp is
slightly
> larger than 80 degrees.
know that F must be proportional to curvature, this means that
the lemniscate should have the equation |r| = alpha kappa, for
some alpha. And yes, this checks out: the curvature of the
lemniscate
r^2 = a^2 cos(2 theta)
is given by
kappa = 3|r|/a^2.
-Jim Ferry
===
Subject: Re: Isoperimetric Zepp
> I was very happy when I found out that the lemniscate is the
> solution of the given problem of maximizing the enclosed
> value, if the price is given by
F(x,y) = sqrt(x^2 + y^2) (1)
And yes, this checks out: the curvature of the
> lemniscate
> r^2 = a^2 cos(2 theta)
> is given by
> kappa = 3|r|/a^2.
As I said: I was very happy ...
And not to forget: F(x,y) = const.)
All in all we are a little wiser than on April 25 last year:
There was a thread star on April 21 then, which states
(1) as P(phi,rho) = rho, using polar coordinates:
I wrongly withdrew my statements about the maximizing
property of the lemniscate. Sorry. But this thread is still
interesting because of the nice shapes emerging from
P = rho^t for t other than t=1 (lemniscate).
There were nice links given by Hermann Kremer, which are
still functioning:
http://www-history.mcs.st-and.ac.uk/~history/Curves/
Rhodonea.html
http://xahlee.org/SpecialPlaneCurves_dir/Rose_dir/rose.html
http://xahlee.org/SpecialPlaneCurves_dir/
specialPlaneCurves.html
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: Isoperimetric Zepp
Jim Ferry <corklebath@hotmail.com> a .8ecrit dans le message de
> Hello sci.math,
> Consider the problem of maximizing the integral of some
function F(x,y)
> over a simply connec region whose boundary C is allowed to
vary
> subject to these two constraints:
> 1) C contains the origin (0,0), and
> 2) C has length 1.
> In general, the extremal curve C must satisfy
> F(x(s),y(s)) = alpha kappa(s),
> for some constant alpha.
> I derived this rather laboriously using the calculus of
variations
Can you give hints for the proof of result.
I am very interes by this proof. I have very searched but
without result
Many thanks.
Excuse my bad English
===
Subject: Re: Isoperimetric Zepp
Jim Ferry <corklebath@hotmail.com> a .8ecrit dans le message de
> Hello sci.math,
Consider the problem of maximizing the integral of some
function F(x,y)
> over a simply connec region whose boundary C is allowed to
vary
> subject to these two constraints:
1) C contains the origin (0,0), and
> 2) C has length 1.
> In general, the extremal curve C must satisfy
F(x(s),y(s)) = alpha kappa(s),
for some constant alpha.
I derived this rather laboriously using the calculus of
variations
> Can you give hints for the proof of result.
> I am very interes by this proof. I have very searched but
without
result
> Many thanks.
> Excuse my bad English
Well, I don't want to attempt a rigorous proof where I have be
careful
about
just how smooth everything is. The elegant, simple route to a
proof would
rely on this physical intuition (which I sta in my original
post):
Let F represent a scalar potential. It yields a normal force F
at the
boundary which is counterbalanced by the tension force
alpha*kappa.
My original calculus of variations argument is more convolu,
but it
does get to the same result. I don't know how interesting it
is, but I
have it more or less written up, so here it is:
Start by writing F(x,y) = g_x(x,y) - f_y(x,y), where the
subscripts x and y
denote partial derivatives. We assume f and g are C^1
functions, so that
we may apply Green's Theorem:
Intg F(x,y) dx dy = Intg f(x,y)(s=0,1) x'(s) + g(x,y) y'(s) ds.
We use the angle phi(s) as the fundamental description of the
curve, where
x'(s) = cos(phi(s)), or x(s) = Intg(t=0,s) cos(phi(t)) dt, and
y'(s) = sin(phi(s)), or y(s) = Intg(t=0,s) sin(phi(t)) dt.
Note that x(0) = y(0) = 0, so we automatically satisfy the
condition of
the origin being on the curve without having to incorporate a
Lagrange
multiplier to enforce this. We do have to enforce that the
curve is
closed, however:
Intg(s=0,1) cos(phi(s)) ds = Intg(s=0,1) sin(phi(s)) ds = 0.
The quantity we are trying to extremize, is (with Lagrange
multipliers
already thrown in):
J = Intg(s=0,1) f(x(s),y(s)) cos(phi(s)) + g(x(s),y(s))
sin(phi(s)) +
lam_1 cos(phi(s)) + lam_2 sin(phi(s)) ds
Substitute phi(s) + dphi(s) for phi(s). Then
dx(s) = -Intg(t=0,s) sin(phi(t)) dphi(t) dt, and
dy(s) = Intg(t=0,s) cos(phi(t)) dphi(t) dt, so
dJ = Intg(s=0,1) (
dx(s) ( f_x(x(s),y(s)) cos(phi(s)) + g_x(x(s),y(s))
sin(phi(s)) ) +
dy(s) ( f_y(x(s),y(s)) cos(phi(s)) + g_y(x(s),y(s))
sin(phi(s)) ) +
dphi(s) ( -(f(x(s),y(s)) + lam_1) sin(phi(s)) +
(g(x(s),y(s)) + lam_2) cos(phi(s)) ) ) ds
Now let p(s) = f_x(x(s),y(s)) cos(phi(s)) + g_x(x(s),y(s))
sin(phi(s)), and
q(s) = f_y(x(s),y(s)) cos(phi(s)) + g_y(x(s),y(s)) sin(phi(s))
We note that
Intg(s=0,1) dx(s) p(s) ds
= -Intg(s=0,1) Intg(t=0,s) p(s) sin(phi(t)) dphi(t) dt ds
= -Intg(t=0,1) Intg(s=t,1) p(s) sin(phi(t)) dphi(t) ds dt
= -Intg(s=0,1) Intg(t=s,1) p(t) sin(phi(s)) dphi(s) dt ds
= -Intg(s=0,1) sin(phi(s)) dphi(s) Intg(t=s,1) p(t) dt ds,
and similarly that
Intg(s=0,1) dy(s) q(s) ds =
Intg(s=0,1) cos(phi(s)) dphi(s) Intg(t=s,1) q(t) dt ds.
Hence,
dJ = Intg(s=0,1) dphi(s) (
-sin(phi(s)) Intg(t=s,1) p(t) dt +
cos(phi(s)) Intg(t=s,1) q(t) dt +
( -(f(x(s),y(s)) + lam_1) sin(phi(s)) +
(g(x(s),y(s)) + lam_2) cos(phi(s)) ) ) ds.
Therefore, an extremal solution must satisfy
-sin(phi(s)) (f(x(s),y(s)) + lam_1 + Intg(t=s,1) p(t) dt) +
cos(phi(s)) (g(x(s),y(s)) + lam_2 + Intg(t=s,1) q(t) dt) = 0.
Noting that
d/ds (f(x(s),y(s)) + lam_1 + Intg(t=s,1) p(t) dt) =
f_x(x(s),y(s)) cos(phi(s)) + f_y(x(s),y(s)) sin(phi(s)) +
-(f_x(x(s),y(s)) cos(phi(s)) + g_x(x(s),y(s)) sin(phi(s))) =
(f_y(x(s),y(s)) - g_x(x(s),y(s))) sin(phi(s)) =
-F(x(s),y(s)) sin(phi(s)),
and similarly that
d/ds ((g(x(s),y(s)) + lam_2 + Intg(t=s,1) q(t) dt) =
F(x(s),y(s)) cos(phi(s)),
allows us to express the necessary condition more simply:
sin(phi(s)) (mu_1 + Intg(t=0,s) F(x(t),y(t)) sin(phi(t)) dt) +
cos(phi(s)) (mu_2 + Intg(t=0,s) F(x(t),y(t)) cos(phi(t)) dt) =
0,
where mu_1 and mu_2 are constants. We now take the derivative
of this
equation with respect to s:
-cos(phi(s)) phi'(s) (mu_1 + Intg(t=0,s) F(x(t),y(t))
sin(phi(t)) dt) +
sin(phi(s)) phi'(s) (mu_2 + Intg(t=0,s) F(x(t),y(t))
cos(phi(t)) dt) =
F(x(s),y(s)).
We treat the previous two equations as a linear system of two
variables,
for
which we now solve:
mu_1 + Intg(t=0,s) F(x(t),y(t)) sin(phi(t)) dt =
-cos(phi(s)) F(x(s),y(s)) / phi'(s), and
mu_2 + Intg(t=0,s) F(x(t),y(t)) cos(phi(t)) dt =
sin(phi(s)) F(x(s),y(s)) / phi'(s).
We take the derivative of each of these:
F(x(s),y(s)) sin(phi(s)) = -(d/ds) (cos(phi(s)) F(x(s),y(s)) /
phi'(s)),
F(x(s),y(s)) cos(phi(s)) = (d/ds) (sin(phi(s)) F(x(s),y(s)) /
phi'(s)).
Finally, we rearrange each equation to get
cos(phi(s)) (d/ds) (F(x(s),y(s)) / phi'(s)) = 0, and
sin(phi(s)) (d/ds) (F(x(s),y(s)) / phi'(s)) = 0,
and we conclude that F(x(s),y(s)) / phi'(s) must be constant.
QED, but too much work for such a simple result! To my mind,
the simple
balance of forces argument is as compelling as the above long
one. Neither
is a real proof, but I'm not going to attempt a real proof
because I can't
believe that this result could be new. As Rainer poin out,
Euler worked
on this stuff, so it is material that has been in good hands.
Euler
wouldn't
haved proved this rigorously, however. Indeed, for the
simplest case of
F(x,y) = 1, the result reduces to the classic one that a
circle is the
shape
of maximal area for a given perimeter. But, according to
Mathworld, this
was
not demonstra rigorously until 1841 (by Steiner). See
http://mathworld.wolfram.com/IsoperimetricProblem.html
-Jim Ferry
===
Subject: Re: Rationals are Uncountable
The intersection may be empty, a closed interval [lo, hi],
or a semi-open interval [lo, r) or (r, hi],
where r must be rational, but lo and hi may be rational or
irrational.
Each intersec interval must have rational bounds
and be a proper subset of the previous interval.
Let the sequence of intervals be (a[j], b[j]).
Define lo=supremum(a_j) and hi=infinum(b_j).
If lo< hi the intersection is a subset of [lo, hi].
If there is an a_j=lo, then the interval is (lo, hi]. lo must
be rational.
If there is a b_j=hi, then the interval is [lo, hi). hi must
be rational.
If there is neither an a_j=lo or a b_j=hi, then the interval
is [lo, hi].
There may not be both an a_j=lo and a b_k=hi.
After max(j, k) intervals,
shrinking would require a_j's greater than their supremeum
or b_k's less than their infinum.
The same analysis applies if lo=hi.
[x, x] consists of a single point.
[x, x) and (x, x] are empty.
(-1/j, 1/j) produces [0, 0]={0}.
(0, 1/j) produces the empty set.
(-1/j, 1+1/j) produces [0, 1].
(0, 1+1/j) produces (0, 1].
(-1/j, 1) produces [0, 1).
(0, ceil(sqrt(2)*2**j)*2**-j) produces (0, sqrt(2)]
===
Subject: Re: Rationals are Uncountable
> The intersection may be empty, a closed interval [lo, hi],
> or a semi-open interval [lo, r) or (r, hi],
> where r must be rational, but lo and hi may be rational or
irrational.
That was very informative.
I hope you don't mind if I ask some stupid questions.
1) (-1/j, 1/j) produces [0, 0]={0}.
2) (0, 1/j) produces the empty set.
I want to understand why (1) is a singleton set and (2) is
empty.
I will use a shorthand notation for the intersection.
Let w mean omega and let (-1/w,1/w) represent the intersection
of the set of intervals (-1/j, 1/j) over all j.
Define 1/w = 0 and 1/-w = 0.
Rewrite the formulas above:
1) (1/-j, 1/j) produces (1/-w, 1/w)
2) (0, 1/j) produces (0, 1/w)
Both formulas convert to (0,0) = empty set.
How can I assume formula (1) is not empty?
Using your definitions, it makes sense that
formula (1) produces [0,0].
Let's assume 1/w =/= 0.
1) (1/-j, 1/j) produces (1/-w, 1/w)
2) (0, 1/j) produces (0, 1/w)
Clearly (1/-w, 1/w) is not empty and must contain [0,0].
But, (0, 1/w) is not empty because 1/w is not zero.
I don't really see how one proves (2) is empty
without also proving (1) is empty.
Russell
- Learning is easy. Unlearning is harder.
===
Subject: Re: Rationals are Uncountable
> I hope you don't mind if I ask some stupid questions.
> 1) (-1/j, 1/j) produces [0, 0]={0}.
> 2) (0, 1/j) produces the empty set.
> I want to understand why (1) is a singleton set and (2) is
empty.
Okay, let's use Archimedes' Axiom as described here:
http://mathworld.wolfram.com/ArchimedesAxiom.html
For any positive real or rational x, there is a j such that
j*x > 1.
Then x > 1/j. Since x > 1/j, x is in neither (-1/j, 1/j) nor
(0, 1/j),
and is therefore in neither infinite intersection.
If x is negative, it is automatically in none of the (0, 1/j)
intervals,
and the above demonstration applied to -x finds a (-1/j, 1/j)
interval
that does not include x. Again, x is in neither intersection.
If x is 0, we always have -1/j < x < 1/j, so it is in the first
intersection. On the other hand, we never have 0 < x < 1/j, so
it is
not in the second intersection.
> I will use a shorthand notation for the intersection.
> Let w mean omega and let (-1/w,1/w) represent the
intersection
> of the set of intervals (-1/j, 1/j) over all j.
> Define 1/w = 0 and 1/-w = 0.
You have to be careful with definitions like this. If you
don't know
what you are doing, it is easy to reach false conclusions.
Actual work
with infinities in the context of limits requires proof that
the usual
steps still work.
> Let's assume 1/w =/= 0.
First, you defined 1/w = 0. Now you assume 1/w =/= 0. Your
mistake
isn't even interesting. The inevitable bogosity of assuming a
contradiction is on your head.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: Rationals are Uncountable
> For each positive integer n, let a_n = 0 and let b_n = 1/n.
> Then for each n, (a_n, b_n) contains a positive number.
> For example (a_1000, b_1000) contains 1/1001. Furthermore,
> if k is a fixed positive integer, the intersection of
> (a_1, b_1), (a_2, b_2), ..., (a_k, b_k) also contains a
positive
> number. However, the intersection of all of the intervals in
the
> infinite sequence (a_1, b_1), (a_2, b_2), (a_3, b_3), ...
> does not contain a positive number (in fact it doesn't
contain
> anything at all). It is easy to see why. In fact to prove it
to
> you, name any positive number you like and post it. I will
then
> show you an interval in that infinite sequence that your
positive
number
> isn't in, and hence it can't be in the intersection.
> Congratulations.
> You have just shown Cantor's first uncountability proof to
be false.
> Cantor assumes such a number exists.
I have done no such thing, and Cantor assumed no such thing.
> This is a quote from wikipedia:
The two monotone sequences a and b move toward each other.
> By the gaplessness of R, some point c must lie between them.
> The claim is that c cannot be in the range of the sequence
x, and that is
> the contradiction.
Perhaps you need to learn how to read: when you don't
understand a word
look it up.
> You have demonstra that the reals are not gapless.
> You also prove they aren't dense.
> (You can try Virgal's tactic and claim it makes a difference
if a_n is
> constant.)
It's not a tactic. In my sequence of nes intervals, the
sequence of
left-hand endpoints is not a monotone sequence and it is not
moving
anywhere.
I officially give up reading and posting to this thread (my
patience is not
good). It is not good for my mental health to argue against
someone who
can't even read. By the way, perhaps you should stop learning
math from
an encyclopedia. Now, I can't stand it when someone runs away
in an
argument before I get to prove my point, so just in case you
can't stand
it either, feel free to do a google search for Leonard
Blackburn
and Knox to find my homepage where you can find my email
address (my
email address above is defunct) in case you want the last word.
Have a nice day.
-Leonard
> Russell
> - 2 many 2 count
===
Subject: Re: Rationals are Uncountable
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<GKmdnTfqKvsnMJDdRVn-jg@comcast.com>
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<8rudnRsOcq3o4I3dRVn-hQ@comcast.com>
<aa503d8.0401230623.7e24d086@posting.google.com>
Discussion, linux)
> It's not a tactic. In my sequence of nes intervals, the
> sequence of left-hand endpoints is not a monotone sequence
and it is
> not moving anywhere.
In many definitions of monotone, constant sequences are
monotone,
but not *strictly* monotone.
--
But he himself was not to blame for his vices. They grew out
of a
personal
defect in his mother. She did her best in the way of flogging
him while an
infant... but, poor woman! she had the misfortune to be
left-handed, and a
child flogged left-handedly had better be left unflogged. --
E.A. Poe
===
Subject: Re: Rationals are Uncountable
<bukhf0$2evn$1@msunews.cl.msu.edu>
<yKGdnTJavO2eS5Dd4p2dnA@comcast.com>
<bukt3q$l4n$1@mozo.cc.purdue.edu>
<nKydnYyvmbotY5DdRVn-jw@comcast.com>
<bul34g$mf5$2@mozo.cc.purdue.edu>
<tPudnYKsV6t4gpPd4p2dnA@comcast.com>
<bum5rh0pp8@drn.newsguy.com>
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<fab2a28f.0401220738.409c7044@posting.google.com>
<99a0b764.0401221214.6a666a05@posting.google.com>
Discussion, linux)
> The intersection may be empty, it may be a single point (not
> necessarily rational), it may be a closed or semi-closed
interval.
> The endpoints need not be rational.
> An open interval is not allowed because that would imply
that the
> lower bound reached its supremum and the upper bound reached
its
> infinum.
> After that, the sequence could not both nest and decrease.
Oops! My example (1/n, 1 - 1/n) wasn't nes. Silly me.
--
Jesse Hughes
We will run this with the same kind of openness that we've run
Windows. Steve Ballmer, speaking about MS's new .Net project.
===
Subject: Re: Rationals are Uncountable
What do you think about the monotone mapping?
In brief the proof would not generate infinite sequences a and
b if
the function X N->R is monotone, or piecewise monotone.
Please explain why any function from N to R would necessarily
be
non-monotone in the requisite way to generate two infinite
sequences a
and b.
Thank you,
Ross
===
Subject: Re: Rationals are Uncountable
> What do you think about the monotone mapping?
> In brief the proof would not generate infinite sequences a
and b if
> the function X N->R is monotone, or piecewise monotone.
> Please explain why any function from N to R would
necessarily be
> non-monotone in the requisite way to generate two infinite
sequences a
> and b.
> Thank you,
> Ross
Point 1: The X in this discussion is not a function, it is the
set we
are trying to show is uncountable.
Point 2: The question is not whether some one function must
necessarily
provide the two sequences but whether there are pairs of
sequences that
approach each other in the describe fashion.
One can show example of sequences that work the desired way,
and a
number of such pairs of sequences have been shown here.
===
Subject: Re: Rationals are Uncountable
> What do you think about the monotone mapping?
In brief the proof would not generate infinite sequences a and
b if
> the function X N->R is monotone, or piecewise monotone.
Please explain why any function from N to R would necessarily
be
> non-monotone in the requisite way to generate two infinite
sequences a
> and b.
Thank you,
Ross
> Point 1: The X in this discussion is not a function, it is
the set we
> are trying to show is uncountable.
> Point 2: The question is not whether some one function must
necessarily
> provide the two sequences but whether there are pairs of
sequences that
> approach each other in the describe fashion.
The proof (Wiki's Cantor's) assumes a function, arguably X(n)
as has
been used in this thread, from N onto R, the range of which
forms a
sequence x from which is formed two sequences a and b, to
derive
contradiction.
If the function's range does not form the sequence x as was
described
as specifically non-monotone, changing signs of consecutive
differences infinitely many times, generating infinite
sequences a and
b, and with a form of convergence of each of a and b to some
value,
then it does not fulfill the conditions (for any function
N<->R) and
thus imply the conclusions presen therein.
So the question is why or why not is a function from N onto R
necessarily satisfying those non-monotone conditions.
Consider it this way, in mapping the integers onto the
rationals: the
range contains only integers, and thus no non-integer values
are
mapped. That's so for only a very specific class of functions
from Z
onto Q, and is not true for all functions from Z to Q.
In this case, the function is assumed to generate two infinite
sequences a and b, yet there are classes of functions that
would not.
So actually a question is whether any function X would
necessarily
provide the infinite, monotone, convergent sequences a and b,
and why
or why not.
If you integrate x^2 and evaluate it over [0,1], you get an
exact
solution.
Ross
===
Subject: Re: Rationals are Uncountable
What do you think about the monotone mapping?
In brief the proof would not generate infinite sequences a and
b if
> the function X N->R is monotone, or piecewise monotone.
Please explain why any function from N to R would necessarily
be
> non-monotone in the requisite way to generate two infinite
sequences
a
> and b.
Thank you,
Ross
Point 1: The X in this discussion is not a function, it is the
set we
> are trying to show is uncountable.
Point 2: The question is not whether some one function must
necessarily
> provide the two sequences but whether there are pairs of
sequences that
> approach each other in the describe fashion.
 The proof (Wiki's Cantor's) assumes a function, arguably
X(n) as has
> been used in this thread, from N onto R, the range of which
forms a
> sequence x from which is formed two sequences a and b, to
derive
> contradiction.
The notation in the proof, see below, never uses X(n). It does
use x_n
as members of a sequence to show that such a sequence cannot
ennumerate
R.
The statements in this thread were about the countabluluty of
an
arbitrary set X.
> If the function's range does not form the sequence x as was
described
> as specifically non-monotone, changing signs of consecutive
> differences infinitely many times, generating infinite
sequences a and
> b, and with a form of convergence of each of a and b to some
value,
> then it does not fulfill the conditions (for any function
N<->R) and
> thus imply the conclusions presen therein.
It is the allegation that such a sequence can ennumerate R
that leads to
the properties that you describe here. So it is the
ennumeration that
one is trying to show impossible. The other stuff is just
details.
> So the question is why or why not is a function from N onto R
> necessarily satisfying those non-monotone conditions.
If a sequence of reals is to ennumerate all of R, Cantor shows
that it
must have those properties. then he shows that no sequence of
reals can
have those properties. The conclusion is that no such
ennumeration is
possible, so the reals are not ennumnerable.
> Consider it this way, in mapping the integers onto the
rationals: the
> range contains only integers, and thus no non-integer values
are
> mapped. That's so for only a very specific class of
functions from Z
> onto Q, and is not true for all functions from Z to Q.
> In this case, the function is assumed to generate two
infinite
> sequences a and b, yet there are classes of functions that
would not.
In fact, that is the whole point. There are no functions which
do what
they must do to ennumerate all the reals.
> So actually a question is whether any function X would
necessarily
> provide the infinite, monotone, convergent sequences a and
b, and why
> or why not.
The answer is that lots of them exist. Any injection of N into
Q or R
whose range is dense in its codomain will be such a function,
or even
one whose range is dense in some open subset of Q or R.
Since there are bijections from N to Q and Q is dense in R,
any such
bijection from N to Q would serve.
> If you integrate x^2 and evaluate it over [0,1], you get an
exact
> solution.
> Ross
**************************************************************
***********
**************************************************************
***********
Quo from Wikipeia, the theorem in question, and its proof.
<< begin quote>
The theorem
Suppose a set R is
linearly ordered, and
densely ordered, i.e., between any two members there is
another,
and
has no endpoints, i.e., smallest or largest members, and
has no gaps, i.e., if it is partitioned into two sets A and B
in
such a way that every member of Ais less than every member of
B, then
there is a boundary point c, so that every point less than cis
in A and
every point greater than c is in B.
Then R is not countable.
The proof
The proof begins by assuming some sequence x1, x2, x3, ... has
all of R
as its range. Define two other sequences as follows:
a1 = x1.
b1 = xi, where i is the smallest index such that xi is not
equal to a1.
an+1 = xi, where i is the smallest index greater than the one
considered
in the previous stepsuch that xi is between an and bn.
bn+1 = xi, where i is the smallest index greater than the one
considered
in the previous stepsuch that xi is between an+1 and bn.
The two monotone sequences a and b move toward each other. By
the
gaplessness of R, some point c must lie between them. The
claim is
that c cannot be in the range of the sequence x, and that is
the
contradiction. If c were in the range, then we would have c =
xi for
some index i. But then, when that index was reached in the
process of
defining a and b, then c would have been added as the next
member of one
or the other of those two sequences, contrary to the
assumption that it
lies between their ranges.
<< end quote >>
===
Subject: Re: Rationals are Uncountable
> If the function's range does not form the sequence x as was
described
> as specifically non-monotone, changing signs of consecutive
> differences infinitely many times, generating infinite
sequences a and
> b, and with a form of convergence of each of a and b to some
value,
> then it does not fulfill the conditions (for any function
N<->R) and
> thus imply the conclusions presen therein.
> So the question is why or why not is a function from N onto R
> necessarily satisfying those non-monotone conditions.
If there is no suitable entry for a_{n+1} (for example), then
the
function's range doesn't include any values in the interval
(a_n,b_n).
The density of the number system was part of the hypothesis,
so there
are values in that interval, and we have our contradiction.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: Rationals are Uncountable
> What do you think about the monotone mapping?
In brief the proof would not generate infinite sequences a and
b if
> the function X N->R is monotone, or piecewise monotone.
Please explain why any function from N to R would necessarily
be
> non-monotone in the requisite way to generate two infinite
sequences a
> and b.
Thank you,
Ross
> Point 1: The X in this discussion is not a function, it is
the set we
> are trying to show is uncountable.
> Point 2: The question is not whether some one function must
necessarily
> provide the two sequences but whether there are pairs of
sequences that
> approach each other in the describe fashion.
Yes it is.
Considering two approaching sequences is just a shortcut
we have been using. Since it is a shortcut, we need to
remember what it really represents.
You can't prove X is uncountable until you at least attempt to
prove it
might
be countable. Cantor's proof requires a sequence that is
countable.
He then constructs a number not in the sequence.
This isn't very interesting unless the sequence in question
covers an
interval, X.
It would be very difficult to construct some of the
intersections
we have been throwing around. Some might be impossible.
For example. Assume some function, f(), produces a sequence of
rational
x_i,
such that Cantor's proof constructs the following sets A and B:
a_n = 0-1/n
b_n = 1/n
If X contains every rational in the set [-1,1],
we have proven the rationals are uncountable.
> One can show example of sequences that work the desired way,
and a
> number of such pairs of sequences have been shown here.
You better hope this is not true.
I would really like an example of the sequence, x_i, I
describe above.
Russell
- 2 many 2 count
===
Subject: Re: Rationals are Uncountable
What do you think about the monotone mapping?
In brief the proof would not generate infinite sequences a and
b if
> the function X N->R is monotone, or piecewise monotone.
Please explain why any function from N to R would necessarily
be
> non-monotone in the requisite way to generate two infinite
sequences
a
> and b.
Thank you,
Ross
Point 1: The X in this discussion is not a function, it is the
set we
> are trying to show is uncountable.
Point 2: The question is not whether some one function must
necessarily
> provide the two sequences but whether there are pairs of
sequences that
> approach each other in the described fashion.
> Yes it is.
Since you have publicly declared yourself not to understand
the theorem
in question, your assurances of the truth of that falsehood do
not
convince.
> Considering two approaching sequences is just a shortcut
> we have been using. Since it is a shortcut, we need to
> remember what it really represents.
But you have declared that you do not know what it really
represents, so
how can you remember what you do not know?
> You can't prove X is uncountable until you at least attempt
to prove
> it might be countable.
Sort of. The proofs of uncountability are all based on showing
that
countability proofs must fail.
Cantor's proof requires a sequence that is countable.
> He then constructs a number not in the sequence.
> This isn't very interesting unless the sequence in question
covers an
> interval, X.
Since such a sequence never can cover an interval, as Cantor's
original diagonal proof shows, you are only saying that there
are no
interesting sequences.
> It would be very difficult to construct some of the
intersections
> we have been throwing around. Some might be impossible.
In a strict constructionist sense, possibly. But the strict
constructionists reject the standard set of real numbers, and
a lot
more, so the question of countability becomes moot.
> For example. Assume some function, f(), produces a sequence
of rational
> x_i,
> such that Cantor's proof constructs the following sets A and
B:
> a_n = 0-1/n
> b_n = 1/n
> If X contains every rational in the set [-1,1],
> we have proven the rationals are uncountable.
The above garbage is only proof of Russell's mental defects,
and of
nothing else.
> One can show example of sequences that work the desired way,
and a
> number of such pairs of sequences have been shown here.
> You better hope this is not true.
> I would really like an example of the sequence, x_i, I
describe above.
Not your desires, idiot, but the desires of the Cantor in the
theorem
you have so badly mangled.
Cantor redux:
Relevant definitions and notations:
Let (X,<) be a totally ordered infinite set.
For a and b in X, a < b, let (a,b) = {x in X: a < x and x < b}
be called
an open interval in X.
A sequence f:N -> P(X): n -> (a_n,b_n) of such open intervals
is
nes if a_n < a_{n+1} < b_{n+1} < b_n for each n in N.
Note that the existence of any nes sequence in X requires that
X be
infinite (at least countably infinite, and possibly uncountably
infinite).
I(f) or I({(a_n,b_n)}) shall denote the intersection of all
intervals
in a sequence of intervals, usually of nes intervals.
Theorem: Given X as above, if for every sequence of nes
intervals f,
I(f) contains at least one member of X, then X is uncountably
infinite,
but if ANY I(f) is empty, X may countable.
Cantor then gives a valid proof, which I shall not paraphrase
here.
One may show, at least to those with eyes to see it, that
taking X = R
allows us to conclude that R is uncountable. In fact, we have
in R that
both LUB({a_n}) and GLB({b_n}) must be in R and thus in I(f),
though
one may have LUB({a_n}) = GLB({b_n}).
One may also sho, at least to those who have the eyes for it,
that
taking X = Q, does NOT allow us to conclude that Q is
uncountable,
because LUB({a_n}) and GLB({b_n}) may be the same but not in
Q, so
I(f) is sometimes empty in Q. And the theorem does NOT then
say that Q
is uncountable.
Note that this is not a proof that Q is countable, it merely
fails to
prove that Q is uncountable.
There are also uncountable sets (e.g., the irrationals) which
this
theorem will fail to prove uncountable.
===
Subject: Re: Rationals are Uncountable
Russell Easterly says...
>For example. Assume some function, f(), produces a sequence
of rational
>x_i, such that Cantor's proof constructs the following sets A
and B:
>a_n = 0-1/n
>b_n = 1/n
>If X contains every rational in the set [-1,1],
>we have proven the rationals are uncountable.
Cantor's proof would never construct those particular
sets A and B, starting with the set X containing all
rationals in the set [-1,1].
The way that Cantor's proof goes is that you start with
an enumeration x_0, x_1, ... of all the elements in the
set X. Then at stage n, you pick a_n and b_n so that
x_n is *not* in the interval (a_n, b_n).
For your particular values of a_n, b_n, the interval
(a_n, b_n) always contains the number 0. That is not
following Cantor's procedure. If 0 appears in the
list X (and it does, in your case), then it must
be that there is a k such that 0 = x_k. Then at
stage k, if you're following Cantor's procedure,
you must choose a_k and b_k so that 0 is *not*
in the interval (a_k, b_k). The choice a_k = 0 - 1/k,
b_k = 1/k violates Cantor's requirement.
If instead you had star with the set X' = all nonzero
rationals in [-1,1], then your particular choice for a_n
and b_n would work fine. Cantor's procedure would give
a number c that is not in X', and that number would in
fact be 0.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Rationals are Uncountable
> For example. Assume some function, f(), produces a sequence
of rational
> x_i,
> such that Cantor's proof constructs the following sets A and
B:
> a_n = 0-1/n
> b_n = 1/n
Then 0 can't be in the sequence x_i. If it were, the
construction would
have made it either a_n or b_n when it reached that position.
> If X contains every rational in the set [-1,1],
Which it doesn't.
> I would really like an example of the sequence, x_i, I
describe above.
-1, 1, -1/2, 1/2, -1/3, -2/3, 1/3, 2/3, -1/4, -3/4, 1/4, 3/4,
-1/5,
-2/5, ...
This contains every NON-ZERO rational in the set [-1,1].
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: Rationals are Uncountable
> I am guessing that the intersection of something like:
(0-1/n, 1+1/n)
> would be written as: [0,1].
> Yes.
> I would write it as: (0-1/n, 1+1/n)
> The intersection of all the (0-1/n,1+1/n) inteverals
includes no
> negative numbers.
> The interval (0-1/n, 1+1/n) does include negative numbers,
for example,
> -1/(n+1) .
I agree.
That is why I hesitate to call it [0,1].
But I will if you do.
I would rather just say its an open set.
Its hard to come up with an n such that 1/n = 0.
What happens if Cantor's missing number is negative?
Russell
- 2 many 2 count
===
Subject: Re: Rationals are Uncountable
I am guessing that the intersection of something like: (0-1/n,
1+1/n)
> would be written as: [0,1].
Yes.
I would write it as: (0-1/n, 1+1/n)
The intersection of all the (0-1/n,1+1/n) inteverals includes
no
> negative numbers.
The interval (0-1/n, 1+1/n) does include negative numbers, for
example,
> -1/(n+1) .
> I agree.
> That is why I hesitate to call it [0,1].
> But I will if you do.
> I would rather just say its an open set.
No one is asking you to call (0-1/n, 1+1/n) anything but
itself.
What is being asked is that you call the _intersection_ of all
the
(0-1/n, 1+1/n) as [0,1], which it is.
The intersection of all of them is not an open set, it is a
closed set,
namely, [0,1] = {x in R: 0 <= x <= 1}
> Its hard to come up with an n such that 1/n = 0.
Nobody asked you to. Why do you set yourself such impossible
and
irrelevant tasks when no one wants you to do them?
> What happens if Cantor's missing number is negative?
Cantor isn't missing any numbers, but you seem to be.
Given any irrational number c, then a_n = floor(c*10^n)/10^n
is a
sequence of rationals monotonically increasing towards c, and
b_n = ceil(c*10^n)/10^n is a sequence of rationals decreasing
monotonically towards c, such that the intersection of all the
intervals
(a_n,b_n) contains only one real number, c itself, which is
irrational.
If we restrict the intervals (a_n,b_n) to intervals in the set
of
rationals, that intersection is empty, because c, its only
possible
member, is not rational.
Is this your, not Cantor's, missing c?
===
Subject: Re: Rationals are Uncountable
Given an infinite set of nes open intervals with rational
bounds
> where every interval is a proper subset of the previous
interval.
> Is the intersection of all these intervals:
1) empty
> 2) exactly one rational
> 3) exactly one irrational
> 4) (2) or (3) depending on the sequence
> 5) an open interval with rational bounds
> 6) none of the above
Actually the intersection of a nes decreasing sequence of open
> intervals with rational endpoints can be any one of 1)
through 5).
It
> could also be a closed interval with rational or irrational
bounds or
> an open interval with rational or irrational bounds.
The intersection may be empty, it may be a single point (not
> necessarily rational), it may be a closed or semi-closed
interval.
> The endpoints need not be rational.
> An open interval is not allowed because that would imply
that the
> lower bound reached its supremum and the upper bound reached
its
> infinum.
> After that, the sequence could not both nest and decrease.
> Thank god somebody knows this stuff.
> This is the most reasonable answer I've seen.
> So the answer is 1, 4, and 6.
> I am interes in why 5 is never possible.
How do you get 1,4, or 6 out of any one of 1) through 5)?
> Let A be an open interval and B an open subset of A.
> A n B = B
> Since B is an open interval, the intersection is also an
open interval.
> How can we get a closed interval from the intersection of
nes open
> intervals?
> I am guessing that the intersection of something like:
(0-1/n, 1+1/n)
> would be written as: [0,1].
> I would write it as: (0-1/n, 1+1/n)
The intersection of all of the (0-1/n, 1+1/n), over all
positive
natural number, n, is the closed interval [0,1].
Anyone who would write it as (0-1/n, 1+1/n) is being
unnecesssarily
stupid, Russal.
===
Subject: Re: Rationals are Uncountable
> Let A be a set and B a proper subset of A.
> Take the intersection of A and B.
> AxB = B
Is this a correct statement?
> Yes.
> Does it matter if A and/or B are infinite?
> No, but that's not the case in question. The issue is not
about the
> intersection of infinite sets, but about the intersection of
> infinitely many sets.
Please explain the issue about the intersection of infinitely
many sets.
Why would this be different from every finite number of
intersections?
Russell
- many 2 count
===
Subject: Re: Rationals are Uncountable
permission for an emailed response.
> Please explain the issue about the intersection of
infinitely many sets.
> Why would this be different from every finite number of
intersections?
Hrm, it just is. Not everything that is true of a finite
number of
intersections is true of an infinite number of intersections.
This is
one example; there are plenty more.
For example, finite intersections of open sets are always
open. But
infinite ones, often not.
===
Subject: Re: Rationals are Uncountable
Let A be a set and B a proper subset of A.
> Take the intersection of A and B.
> AxB = B
Is this a correct statement?
Yes.
Does it matter if A and/or B are infinite?
No, but that's not the case in question. The issue is not
about the
> intersection of infinite sets, but about the intersection of
> infinitely many sets.
> Please explain the issue about the intersection of
infinitely many sets.
> Why would this be different from every finite number of
intersections?
The intersection of a finite nes sequence of sets, each but
the first
being a proper subset of all previous sets, is always the last
set in
the sequence.
That cannot be true for any infinite nes sequence of sets,
since
there is no last one in the sequence.
What has to be in the finite case cannot be in the infinite
case.
===
Subject: Re: p-adic zero of a polynomial in extension Q_p(zeta)
> I wouldn't use Hensel. I'd note that as long as r > = 2,
> (1 + c lambda^r)^p = 1 + p c lambda^r (mod lambda^{p+r}).
> Take b and write it as 1 + p c_2 lambda^2
> for some integer c_2 in Q(zeta_p).
> Now b/(1 + c_2 lambda^2) = 1 (mod lambda^{p+2}) so
> write b/(1 + c_2 lambda^2)^p = 1 + p c_3 lambda^3.
> Keep going: for each j >= 2,
> b/(1 + c_j lambda^j)^p = 1 + p c_{j+1} lambda^{j+1}.
> Then b = a^p where a = product_{j=2}^infinity(1 + c_j
lambda^j).
Thanks a lot!
One typo puzzled me a bit, but now I understand your solution.
===
Subject: factoring
factoring
: what are the factors of 48?
24.01.04 00:04
I'll just briefly comment on the vague question (later), from
this
thread.
Would it be good if a scientific calculator had a function.
what is the least prime factor.
48
= 2*24.
shows 2 and then allows least prime factor (again) of 24, etc.
Hit F.
What is the next prime number after x.
Hit Z.
myprog SeriCalc4S. BASIC64.
> Messages 11-16 from thread factoring
> Message 14 in thread
> Subject: factoring
 Hi
> Is there an easier way to factor a number other than
remembering? For
> instance, what are the factors for 48?
> My memory is bad, and I was
just wondering if anyone had a different, easier
> method of finding all the digits?
Well, writing on a piece of paper (like this monitor) may be
ok.
However, we probably desire proof.
the Fundamental theorem of Arithmetic may say
the factorisation of every positive integer into primes
is unique up to the order of primes.
I will show.
there may also be a very simple formula for
the number of **divisors. ( including primes and non-prime
divisors.)
Hit D.
I have also studied what is the least positive integer
with 360 distinct positive integer divisors.
I will show.
myprog COMPOSE.
> Thanks.
> Message 15 in thread
> Subject: Re: factoring
 ...
> For small numbers, trial division is probably the best.
yes (Well, easiest.)
If you do not just want the (repea) prime
factors.
> For large numbers, it's unknown what's the best way. RSA
Laboratories is
> offering serious money (well, serious to me, since I'm not
one of those
> highly-paid computer geeks) for factorizations:
>
http://www.rsasecurity.com/rsalabs/challenges/factoring/
numbers.html
I have a notable achievement.
Factor 1580,187,223
is the least prime divisor of
(10^9)^(10^9)+3.
sci.math.
[raised to a power]
1 billion raised to the 1 billion, add 3.
10,ooo days plus a heartbeat [seconds]
= 864 million+1
divides billion raised to the billion +1.
search google, eis njas research lookup.
> If you want to factor for finding a common denominator (for
adding
> fractions), you're probably ahead to find the gcd of the two
numbers
using
> the Euclidean algorithm, and then use the formula that lcm =
product of
the
> two numbers/gcd. If that's gobbledygook to you, google for
euclidean
> algorithm.
my program also prints the convergents of a continued fraction
approximation. You can pick out one of desired precision or
has good factor.
Hit W.
some of my programs show sums of squares Hit K ,
multiples Hit J,
or powers and roots, Hit Y.
I will find the gcd greatest common divisor of bush and gore
votes.
Hit H.
(Over 40 maths and number theory/ statistics functions.)
> Jon Miller
> Message 16 in thread
> Subject: Re: factoring
 Thanks.
> Sure, try each positive integer up to the square root of the
number. For
48:
> 1 goes into 48 48 times. 2 goes into 48 24 times. 3 goes
into 48 16 times.
4
> goes into 48 12 times. 5 does not go into 48. 6 goes into 48
8 times. Stop
> here since the square root of 48 is less than 7(recall that
7 is the
square
> root of 49 and 49 is more than 48) So the (positive) factors
of 48 are
> 1,2,3,4,6,8,12,16,24, and 48. I hope this was helpful.
> Steven
>
I will show with my program. SeriCalc4S.
(THIS program found Mersenne prime exponent =
1*2*34*44432+1.= 302 3177.) Palindrome Table Mountain.
/ / /T TTT .
BEGIN, ENTER START NUMBER (Expression), Q. END
?48
RESULT (1) 48 ROUND TO NEAREST (+ve) INTEGER
DIVISORS OF INTEGER, R% = 48
1 * 48 DSUM= 49 DD%= 2
2 * 24 DSUM= 75 DD%= 4
3 * 16 DSUM= 94 DD%= 6
4 * 12 DSUM= 110 DD%= 8
6 * 8 DSUM= 124 DD%= 10
DSUM/R% = 2.58333333
NO. OF DIVISORS , DD% = 10
DSUM / number R% = 2.58333333
ABUNDANT.
I just play around with some add-on serial calculations.
RESULT ..(3) 48 *?100 multiply
(4) 4800
Z. FIND NEXT PRIME
SMALL INCREASE = 1 , PRIME = 4801
(P-1) / 2 = 2400= 2 X 1200
PRIME = (5) 4801
Z. FIND NEXT PRIME
(4802 = 2 * 2401
( 4803 = 3 * 1601 )
(4804 = 4 * 1201
( 4805 = 5 * 961 ) ** I saw square factor 31 here!
(4806 = 2 * 2403
( 4807 = 11 * 437 )
(4808 = 8 * 601
( 4809 = 3 * 1603 )
(4810 = 2 * 2405
( 4811 = 17 * 283 both prime )
(4812 = 4 * 1203
INCREASE = 12 , PRIME = 4813
(P-1) / 2 = 2406= 2 X 1203
PRIME = (6) 4813 -?8
so I went back.. subtract 8.
(7) 4805
then I genera another multiple of 31
so it should have a common factor.
HIT N.
BEGIN, ENTER START NUMBER (Expression), Q. END
?31*16 [ I liked this because I chose a Perfect Number.]
(8) 496 H.C.F.
(Highest Common Factor) ?RES(7) result above.
N% = 4805
(9) 31 answer greatest common divisor hcf.
compare the decomposition of each number into primes.
RECALL RES () ENTER SUBSCRIPT 0-999,
OR (ENTER) LIST ?7
(10) 4805 LEAST PRIME FACTOR
5 * (11) 961 LEAST PRIME FACTOR
31 * (12) 31 LEAST PRIME FACTOR
PRIME (13) 31
RECALL RES () ENTER SUBSCRIPT 0-999,
OR (ENTER) LIST ?8
(14) 496 LEAST PRIME FACTOR
2 * (15) 248 LEAST PRIME FACTOR
2 * (16) 124 LEAST PRIME FACTOR
2 * (17) 62 LEAST PRIME FACTOR
2 * (18) 31 LEAST PRIME FACTOR
PRIME (19) 31
and all divisors of result (7).
RECALL RES () ENTER SUBSCRIPT 0-999,
OR (ENTER) LIST ?7
(20) 4805 ROUND TO NEAREST (+ve) INTEGER
DIVISORS OF INTEGER, R% = 4805
1 * 4805 DSUM= 4806 DD%= 2
5 * 961 DSUM= 5772 DD%= 4
31 * 155 DSUM= 5958 DD%= 6
NO. OF DIVISORS , DD% = 6
DSUM / number R% = 1.23995838
DEFICIENT.
deflate by factor 5.
(21) 4805 /?5
(22) 961 SQUARE ROOT, v
(23) 31 ^?(1/3) cube root.
(24) 3.14138065 an approximation to pi.
Hit N.
BEGIN, ENTER START NUMBER (Expression), Q. END
?PI
(25) 3.14159265 Optn Y
: disp Roots x^(1/y),& powers x^y.
1 3.14159265 3.14159265
2 1.77245385 9.8696044
3 1.46459189 31.0062767 ** PI cubed.
index y | x^(1/y) | x^y
: x=3.14159265
4 1.33133536 97.409091
5 1.25727412 306.019685
6 1.21020324 961.389194
7 1.17766403 3020.29323
8 1.15383507 9488.53102
9 1.13563528 29809.0993
10 1.12128235 93648.0475
11 1.10967408 294204.018
index y | x^(1/y) | x^y
: x=3.14159265
Escape at line. 2680
15.11.00 18:21 nzdt*..
A calculation from year 2000 votes.
?2910492 Bush
(3) 2910492 CTD FRACTION,
RATIO TO DENOM 'A' (default 1)
(ENTER NO. OR EXPRESSION) ?2910192 Gore Florida. ********
2910192 = 2910192
CALCULATE B / A =2910492 / 2910192
= 1.00010309 recipr = 0.999896925
CTD-FRAC | NUMERATOR / DENOMIN |=DECIMAL
1 1 / 1 = 1
9700 9701 / 9700 = 1.00010309
1 9702 / 9701 = 1.00010308
1 19403 / 19401 = 1.00010309
1 29105 / 29102 = 1.00010309
3 106718 / 106707 = 1.00010309
2 242541 / 242516 = 1.00010309***** a very good approximation.
I don't remember if it is an exact reduced fraction - NO.
example of gcd HCF above.
(4) 2910492 rem.
?BUSH/GORE FLORIDA 15/11/00.
(5) 2910492 LEAST PRIME FACTOR
2 * (6) 1455246 LEAST PRIME FACTOR
2 * (7) 727623 LEAST PRIME FACTOR
...
13 * (12) 691 LEAST PRIME FACTOR
PRIME (13) 691 *?4*81*13
prime factorisation of number of
votes. (14) 2910492 -?300
subtract
(15) 2910192 Gore LEAST PRIME FACTOR
2 * (16) 1455096 LEAST PRIME FACTOR
2 * (17) 727548 LEAST PRIME FACTOR
2 * (18) 363774 LEAST PRIME FACTOR
2 * (19) 181887 LEAST PRIME FACTOR
3 * (20) 60629 LEAST PRIME FACTOR
19 * (21) 3191 LEAST PRIME FACTOR
PRIME (22) 3191 *?24*19
(23) 1455096 *?2
(24) 2910192 Optn Y
: disp Roots x^(1/y),& powers x^y.
1 2910192 2910192
2 1705.92848 8.46921748E12
3 142.77118 2.46470489E19
index y | x^(1/y) | x^y
: x=2910192
4 41.3028872 7.17276447E25
5 19.6238551 2.08741218E32
6 11.9486895
: x=2910192
(25) 2910192 +?300 difference in votes.
(26) 2910492 Optn
Y : disp Roots x^(1/y),& powers x^y.
1 2910492 2910492
2 1706.01641 ** 8.47096368E12
3 142.776085 2.4654672E19
index y | x^(1/y) | x^y
: x=2910492
4 41.3039515 7.17572257E25
5 19.6242597 2.08848831E32
6 11.9488947
: x=2910492
...
ENTER NO./EXPRESSION TO FACTORISE,
< 2^31, 0=END?24
= 24
ENTER STEP BETWEEN NUMBERS, 1 / return= EVERY NUMBER, default
step set to default = 1.
24 = 2 2 2 3 ******* prime divisors.
25 = 5 5
26 = 2 13
27 = 3 3 3
28 = 2 2 7
29 = 29
29 prime..1th
30 = 2 3 5
31 = 31
31 prime..2th
32 = 2 2 2 2 2
33 = 3 11
34 = 2 17
35 = 5 7
36 = 2 2 3 3
37 = 37
37 prime..3th
38 = 2 19
39 = 3 13
40 = 2 2 2 5
41 = 41
41 prime..4th
Escape AT LINE 210
RESIDUAL FACTOR X%=41. START NO. X$=24 = X1%= 24
NO. PRIMES=4 Y% =41 TIME = 7.799999999 SECS
PROFILES/factors OF INTEGERS 1-1,000,000,
BY DS don.MCDONALDxx, don.lotto@paradise.net.nz
WGTN, N.Z., (04) 389-6820
ENTER START NO. expression, 1 - N <= 2E9+, null = next,
Q.UIT? 24 alt.math *********
24, 576, 4.90 2.2.2.3 4.166666666
*******
NO. | SQUARE | SUM SQS | SQR | FACTORS | RECIP.|
<enter num, N% (input N$), I% (current), CR, Q.uit>?
25, 625, 5.00 5.5. 4
(1) 25 +0 = 5^2 + 0^2.
(2) 16 +9 = 4^2 + 3^2.
(3) 9 +16 = 3^2 + 4^2.
26, 676, 5.10 2.13 3.846153846
(1) 25 +1 = 5^2 + 1^2.
27, 729, 5.20 3.3.3. 3.703703703
28, 784, 5.29 2.2.7 3.571428571
29, 841, 5.39 29 3.448275861
(1) 25 +4 = 5^2 + 2^2.
NO. | SQUARE | SUM SQS | SQR | FACTORS | RECIP.|
<enter num, N% (input N$), I% (current), CR, Q.uit>?
30, 900, 5.48 2.3.5 3.333333333
31, 961, 5.57 31 3.225806451
32, 1024, 5.66 2.2.2.2.2. 3.125000001
(1) 16 +16 = 4^2 + 4^2.
33, 1089, 5.74 3.11 3.03030303
34, 1156, 5.83 2.17 2.941176471
(1) 25 +9 = 5^2 + 3^2.
(2) 9 +25 = 3^2 + 5^2.
ENTER START NO. expression, 1 - N <= 2E9+, null = next,
Q.UIT?Q
**** PGM PROFILE E N D. *sp. ram.profono phon # off
CALCULATE RECURSIVE FACTORISATIONS (OF 360),
CALCULATE least (SMALLEST) INTEGER WITH (360) positive
DivisORS,
BY Don S MCDONALD, WELLINGTON, NZ (published year 1976.)
2^10*3^6*5^4*7^4*11^2*13^2*17^2*19^2*23^2*29*31*37*41*43*47*53
*59
= 1.08961211E37 HAS 2^8*3^5*5^2*7*11
= 119750400 DivisORS.
=947,374+1 distinct factorisations.
SHORTEST PROGM THAT DOES STHG V INTG IN <26 STATMTS.
PRIMES.
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79
83 89 97
ENTER POSITIVE INTEGER (EXPRESSION) 0= END, RETN = REPEAT?24
24=24 hit spaces (better than return). min.= 8388608  =8388608
2 12 (1) min.= 6144  =6144
2 6 (2) min.= 480  =480
2 3 (3) min.= 420  =420
3 4 (4) min.= 360  =360 *******
3 8 (5)=1152
4 6 (6)=864
SMALLEST (4) =360 xx smallest pos integ with ,..
24 = 24 DivisORS.
**** PROGM COMPOSE E N D. *spool ram recurfact8 close
> Message 13 in thread
> Subject: Factoring
 I was wondering if somebody could show me an example formula
for
> producing a four gear ratio, starting with a decimal. I
would like to
> show all possible combinations between 20 and 100. I can
easily
> accomplish this using a two gear ratio formula. Any help
would be
> apprecia.
don: Is this rela to cycling and 21 speed mountain biking? It
may be
practical mathematics. ...
Is this rela to finding- for example- rational approximations
to
(decimal) 1/PI. I show the latter , Above.
> Rick
cheers
don.mcdonald
23.01.04 23:48
.MZD04.MZD04-Jan.BrainSleep
.MZD04.MZD04-Jan.michClark
.MZD04.MZD04-J an.watsfact48 > this file.
===
Subject: Re: factoring
> : what are the factors of 48?
4 and 8
Anything else is an exercise for a computer, or someone who
needs to get
more sunshine...
===
Subject: Re: factoring
Don McDonald <don.lotto@paradise.net.nz> schreef in bericht
> factoring
> : what are the factors of 48?
> 24.01.04 00:04
Those are some of its divisors, but not all of them!
===
Subject: Re: factoring
>factoring: what are the factors of 48?
<hundreds of lines>
Jeez, keep it simple. Just divide by the lowest primes to
exhaustion.
(1) divide by 2: once leaves 24. twice leaves 12. 3x leaves 6,
4x
leaves 3.
(2) divide by 3: once leaves 1. done.
Thus, prime factors of 48 are 2,2,2,2,3. Multiply these
together in
all combinations to get *all* factors.
willy
===
Subject: mathematical formulation of this number
representation problem?(a
strange binary number representation)
Dear all,
5=101, 6=110, 7=111, these are all binary representation of
decimal
numbers;
but I can represent 7=100(-1), which only needs two non-zero
digits to
represent it; by saving the number of non-zero digits, we can
reduce the
computation in arithmatic. this is called canonical signed
digit(CSD)
representation.
but this scheme has no effect on 5=101, 6=110, since they
don't have at
least three consequtive 1's in their binary representation.
now I want to ask if there are any mathematical treatment
behind this CSD
representation? given a decimal number, how to directly write
out its CSD
without even looking at its binary format first? and any
property, theories
about this CSD numbering system?
thanks a lot,
-walala
===
Subject: JSH: Problem with Decker, Madigin, Ullrich et al
Now I can show you over and over again *basic* algebra, but
for many
of you mathematics is the antithesis of what it was to those
who came
before you, as to them it was a refuge from the often crazy
world of
opinions, beliefs, religion, and just plain nuttiness which
most
people take for gran.
For them logic and preciseness in mathematics made it
beautiful.
Your generation clearly hates that in mathematics.
I have outlined a position that follows from algebra basic
enough to
explain to a kid in school, but most of you have refused to
accept it
because you don't *like* it!!!
What does liking have to do with it?
What in the hell is wrong with you people?
Do you despise the mathematicians who came before you? The
one's who
actually had principles?
Do you hate Gauss? Despise Dedekind? Spit on Archimedes? Feel
disgust for Newton? Blast Euler?
What is WRONG WITH YOUR PEOPLE?
Mathematics is about logic, and accepting what follows
mathematically
because it is true. Not because it makes you feel good. Not
because
it makes you feel smart, or because you can brag to people
about what
you know, but because it's TRUE.
If you have decided to continue to turn your back on the
spirit of
mathematics, to spit on the legacy of so many who came before
you, to
try and destroy the value in the search for truth, then you
are just
debased animals.
Pretenders to a grand tradition, who have no hope, no freedom,
no
place to turn to, as God in Heaven despises you for despising
the
greatest thing of all: Truth.
===
Subject: Re: Problem with Decker, Madigin, Ullrich et al
> What is WRONG WITH YOUR PEOPLE?
I dunno. What is wrong with YOUR PEOPLE? Because whomever the
hell
you've hired to do your BASIC ALGEBRA, they're not
representing you well.
===
Subject: Re: JSH: Problem with Decker, Madigin, Ullrich et al
> Do you hate Gauss? Despise Dedekind? Spit on Archimedes?
Feel disgust
> for Newton? Blast Euler?
> What is WRONG WITH YOUR PEOPLE?
Gauss, Dedekind, Archimedes, Newton, and Euler didn't
constantly introduce
new meanings, known only to themselves, for standard
terminology. For
example, if they said they'd proved something, that meant it
was true, and
they could demonstrate it, whereas for you, proof means have
not been
shown a counterexample yet.
--
===
Subject: Re: Problem with Decker, Madigin, Ullrich et al
> Now I can show you over and over again *basic*
dip
===
Subject: Re: JSH: Problem with Decker, Madigin, Ullrich et al
>Now I can show you over and over again *basic* algebra, but
for many
>of you mathematics is the antithesis of what it was to those
who came
>before you, as to them it was a refuge from the often crazy
world of
>opinions, beliefs, religion, and just plain nuttiness which
most
>people take for gran.
>For them logic and preciseness in mathematics made it
beautiful.
>Your generation clearly hates that in mathematics.
>I have outlined a position that follows from algebra basic
enough to
>explain to a kid in school, but most of you have refused to
accept it
>because you don't *like* it!!!
Uh, right. You should really cite a post where someone has said
they didn't accept your claims because they don't like them.
Cuz it looks to everyone else like people have much better
reasons - no, that's not the only way this factors, no, the
case x = 0 does not imply anything about x <> 0, etc.
>What does liking have to do with it?
>What in the hell is wrong with you people?
>Do you despise the mathematicians who came before you? The
one's who
>actually had principles?
>Do you hate Gauss? Despise Dedekind? Spit on Archimedes? Feel
>disgust for Newton? Blast Euler?
>What is WRONG WITH YOUR PEOPLE?
>Mathematics is about logic, and accepting what follows
mathematically
>because it is true. Not because it makes you feel good. Not
because
>it makes you feel smart, or because you can brag to people
about what
>you know, but because it's TRUE.
>If you have decided to continue to turn your back on the
spirit of
>mathematics, to spit on the legacy of so many who came before
you, to
>try and destroy the value in the search for truth, then you
are just
>debased animals.
>Pretenders to a grand tradition, who have no hope, no
freedom, no
>place to turn to, as God in Heaven despises you for despising
the
>greatest thing of all: Truth.
Bull. There are a lot of crackpots around, but you're the
_only_ one
I've _ever_ seen say If this is wrong don't bother telling me,
I
don't want to know.
Huh, I just realized how paradoxical that is: the day you
_told_
us you didn't care about the truth you were telling the truth.
>
**************************
As far as I'm concerend you're trying to wait until I die, so
I figure
maybe you should die instead. How about that, eh? Wouldn't
that be a
better twist?
You refuse to follow the math, so the great Powers that control
reality and *speak* in mathematics decide to kill you instead
of me.
So what do you think about that, eh? Oh, can't hear Them
talking?
Well, I guess that's because you don't really understand
Mathematics,
the true language, which is THE language.
They're talking about you now, and They agree with my
assessment, and
will not penalize me as They allowed the others like Galois
and Abel
to be penalized.
They will kill you instead.
speaking on Weird factorization, genius
===
Subject: Re: ugly sum
> sum_{i=1}^{infty}frac{mu^iln{i!}}{i!}
> What do you want to know about it? For large positive
> mu, it is approximately log( gamma(mu) ) * exp( mu )
Now that's something; i would of course prefer some closed
form but it does seem unlikely. I would be interes in
the analysis for your result. can you post it in latex?
This sum is the only component of the natural entropy of
a poisson process that i can't get a nice form for. Using
the coarse bounds i<i*ln(i)<i^2 gives u<H(p[u])<u(1+u) where
p[u](i)=exp(-u)u^i/i! is the poisson process and H is natural
entropy. This is not a very rewarding result.
p.s.
For the benefit of those not used to latex (at least one poster
has asked), i'll translate the summand written above as
frac{mu^iln{i!}}{i!} as
i
u log(i!)
----------
i!
steve
===
Subject: Re: Recommend a Book?
> Hi all,
> I am a graduate - well 4 years ago anyway. I havnt been
using maths to
any
> real standard and I don't want to forget the knowledge.
> I wonder if anyone could recommend a good reference book - I
would like
the
> book to contain details/tutorials etc on topics such as
calculus,
Fourier,
> ODE/PDE - matrix ops... I know these topics are wide ranging
- so I
wonder
> does a good book covering these topics decently exist?
Maybe something like a review book for the GRE Mathematics
Subject Exam?
===
Subject: Chaos Question re Strange Attractors
Will any given subset (other than the null set) of a strange
attractor
ultimately flow to equal and fill out the entire strange
attractor?
Thanks -
L
===
Subject: Re: Chicken Nugget Problem (Number Theory)
Mathematicians call this type of problem a Frobenius Problem,
though
it has other names. MathWorld calls it the Coin Problem, see
http://mathworld.wolfram.com/CoinProblem.html
It says there that there are efficient methods to solve the
problem if
you have three numbers a,b,c. That is, given three positive
integers
a,b,c with gcd(a,b,c)=1, it is possible to efficiently find the
smallest number N which cannot be written as a*x+b*y+c*z using
nonnegative integers x,y,z.
However, if you instead start with n numbers a1,a2,...,an and
ask for
the smallest number that cannot be written as a sum
a1*x1+a2*x2+...+an*xn, then the problem is NP-hard, which
means that
it is a very hard computational problem, and certainly that no
efficient algorithm is known.
The MathWorld site has a bunch of references for this much
studied
problem.
Joe Silverman
> Recently I came across this problem in my textbook and I've
been quite
> interes to know the answer to.
> Here it is,
> Chicken Nuggets used to be sold at a hamburger chain in
packages of 6, 9,
20
> pieces. What is the largest number of pieces you could not
order exactly?
> I'd appreciate any help on this problem.
> Gavin
===
Subject: Re: Chicken Nugget Problem (Number Theory)
> Recently I came across this problem in my textbook and I've
been quite
> interes to know the answer to.
Here it is,
Chicken Nuggets used to be sold at a hamburger chain in
packages of 6,
9, 20
> pieces. What is the largest number of pieces you could not
order
exactly?
I'd appreciate any help on this problem.
Gavin
> As I recall, the answer is 43. This is a modern example of
Diophantine
equations
> and yields to classic methods for that branch of the arts.
Well, I believe an upper bound is 6*9*20, so one could use
brute force. Are
there any computationally more efficient methods, or perhaps a
better upper
bound?
-Michael.
===
Subject: Re: Chicken Nugget Problem (Number Theory)
>> Recently I came across this problem in my textbook and I've
been quite
>> interes to know the answer to.
>> Here it is,
>> Chicken Nuggets used to be sold at a hamburger chain in
packages of 6,
>9, 20
>> pieces. What is the largest number of pieces you could not
order
>exactly?
>> I'd appreciate any help on this problem.
>> Gavin
>> As I recall, the answer is 43. This is a modern example of
Diophantine
>equations
>> and yields to classic methods for that branch of the arts.
>Well, I believe an upper bound is 6*9*20, so one could use
brute force.
Are
>there any computationally more efficient methods, or perhaps
a better
upper
>bound?
A better upper bound is 43.
Here is a straightforward solution:
Using 6's and 9's, you can get 3n for any for n>=2
Hence, using one 20 + 6's and 9's, you can get 3n+2 for n>=8
Using two 20's + 6's and 9's, you can get 3n+1 for any n>=15
You cannot get 43, because 43 = 1 (mod 3), and so you would
need at
least two 20's to get 43.
Derek Holt.
===
Subject: Re: Chicken Nugget Problem (Number Theory)
* mareg@mimosa.csv.warwick.ac.uk
> A better upper bound is 43.
> Here is a straightforward solution:
> Using 6's and 9's, you can get 3n for any for n>=2
> Hence, using one 20 + 6's and 9's, you can get 3n+2 for n>=8
> Using two 20's + 6's and 9's, you can get 3n+1 for any n>=15
> You cannot get 43, because 43 = 1 (mod 3), and so you would
need at
> least two 20's to get 43.
<Nitpicking>
To get any 3n+2 you actually _need_ one 20 (or 4, 7, ... 20s).
To get any 3n+1 you actuall _need_ two 20s (or 5, 8, ... 20s).
Therefore you cannot make 43, which is 3*14+1 so you need to
make a
3 which is clearly impossible.
So, it remains to show how you can make any number _higher_
than 43
which you indicate, but just indicate.
</Nitpicking>
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Chicken Nugget Problem (Number Theory)
>* mareg@mimosa.csv.warwick.ac.uk
>> A better upper bound is 43.
>> Here is a straightforward solution:
>> Using 6's and 9's, you can get 3n for any for n>=2
>> Hence, using one 20 + 6's and 9's, you can get 3n+2 for n>=8
>> Using two 20's + 6's and 9's, you can get 3n+1 for any n>=15
>> You cannot get 43, because 43 = 1 (mod 3), and so you would
need at
>> least two 20's to get 43.
><Nitpicking To get any 3n+2 you actually _need_ one 20 (or 4,
7, ... 20s).
True, but you do not need that fact.
> To get any 3n+1 you actuall _need_ two 20s (or 5, 8, ...
20s).
> Therefore you cannot make 43, which is 3*14+1 so you need to
make a
> 3 which is clearly impossible.
I said that!
> So, it remains to show how you can make any number _higher_
than 43
> which you indicate, but just indicate.
No, I gave a complete proof! Worthy of full marks, I would say!
></Nitpicking--
>Jon Haugsand
> Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
> http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Chicken Nugget Problem (Number Theory)
* mareg@mimosa.csv.warwick.ac.uk
> To get any 3n+2 you actually _need_ one 20 (or 4, 7, ...
20s).
> True, but you do not need that fact.
Well, how do you prove that 43 is not expressable then?
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Volume of simplex inside sphere (coding theory,
linear algebra)
I have a unit sphere in a vector space (with say 20
dimensions) and I
want to place some points on the surface of the sphere so that
their
convex hull includes most of the volume inside the sphere. How
many
points do I need to enclose most (say, half) of the volume?
Equivalently, I want to find a bunch of vectors with norm 1,
such that
I can multiply the implied matrix by a vector with every
element
0<=e<=1 to be able to get most vectors with norm less than 1.
I fear it requires 2^N points. For a hypercube, you need a
point at
each corner to generate the entire space. But the volume of a
hypersphere is much smaller than a hypercube, and my intuition
is
often faulty in high-dimensional space.
There's a obvious connection to Shannon's coding theory here
if the
points represent symbols with unit energy. I may try his trick
of
generating an average random code and see what fraction of the
area
it encloses.
===
Subject: Newbie Questions: Series
Dear all,
I have two newbie questions. First, if one sums the following:
1*k + 2*k^2 + 3*k^3 + ..... + (n-1)*k^(n-1) + n*k^n
with k = some constant number, what will the result be? Anyone
has any
idea? Does this have a special name?
Second question: Is there any place (websites, books, etc.)
where I
could find finite sums and infinite series - including this
one? I
have not been able to find one which, fo instance, include the
one in
my first question, thus far.
Subject: re:Newbie Questions: Series
===
By making k a variable, the first problem can be solved by
elementary
calculus.
mx^m=x*d(x^m)/dx
Therefore sum=x*dz/dx, where z=(1-x^(n+1))/(1-x).
I'll let you do the rest. When you're finished let k=x.
----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet
News==----
http://www.newsfeed.com The #1 Newsgroup Service in the World!
>100,000
Newsgroups
---= 19 East/West-Coast Specialized Servers - Total Privacy
via Encryption
=---
===
Subject: Re: Cyclotomic polynomials
Please provide the reference for the paper by Paul Garrett
that you
mentioned in your recent post. Thanks.
Robert
> I came across an interesting paper by Paul Garrett on
cyclotomic
> polynomials.
> This will be familiar to number theorists, but I will post
it anyway.
> I also recall doing several problems from Hungerford like the
> following;
> if p is prime, n > 1 and odd;
> F_(p^k)(x) = F_p(x^p^(k-1)) ; F_2n(x) = F_n(-x) when n is
odd.
> (I call F_n the cyclotomic polynomials).
> For those not used to them, they come up as irreducible
polynomials
> associa with the roots of unity, i.e., x^n - 1 = 0.
> There are 2 expressions for F_n that are very useful;
> F_n = (x - a_1)(x - a_2) ... (x - a_m) ; where m = phi(n) is
the
> Euler fcn. of n,
> and the a_i are primitive roots of unity. If we are looking
at
> x^n - 1 = 0, then the complex solns. are exp[i*2*pi*k/n],
which is
> primitive
> iff (k,n) = 1, i.e. if k is in Z*_n. In this case x^n = 1
but x^(n/d)
> != 1 (d|n ; d < n).
> The other useful formula is
> F_n(x) = [x^n - 1]/D ; where D = prod(d < n; d|n) F_d(x)
> This allows calculation of F_n from the F_d where d|n and d
< n.
> Its a good exercise to calculate all of them up to about 15
or 20.
> F_1 = x - 1 ; F_2 = x + 1 ; F_3 = x^2 + x + 1 =( x^3 - 1)/(x
- 1) ;
> F_4 = x^2 + 1, etc.
> Based on fairly extensive hand calculations, one might
suspect that
> all
> coefficients of all cyclotomic polynomials are either +1, 1,
or 0,
> but this is not true. It is true for n prime, and for n
having at most
> 2 distinct
> prime factors, but not generally.
> The smallest n where F_n has coefficients not +/- 1, 0, is
> n = 105 = 3 x 5 x 7 (there must be at least 3 odd primes).
> One gets F_105 = [x^105 - 1]/D ; D = F_1 * F_3 * F_5 * F_7 *
F_15 *
> F_21 * F_35
> = F_3 * F_15 * F_21 *(x^35 - 1) ; then
> (x^105 - 1)/(x^35 - 1) = x^70 + x^35 + 1 ;
> continuing like this, and multiplying
> top and bottom by appropriate F_d's and x^d - 1, noting that
> deg(F_105) = phi(105) = 2.4.6 = 48, and that the symmetry of
the
> binomial coefs.
> implies symmetry of the coeffs. of F_105 (coefs of
decreasing powers
> are the same
> as read for increasing powers of x), one gets a formula for
> F_105 = N/D ; where both N and D are products of several F_d.
> Garrett then assumes that assumes that |x| < 1, and expands
things
> like
> 1/(1 - x^21) = 1 + x^21 + x^42 + x^63 + ..., and gets all
the coefs
> up to the x^24, since the coefs of x^25 to x^48 will be the
same
> as x^23 down to x^0 = 1.
> Finally, he gets a term -2*x^7 (I have not checked all his
> calculations.)
> which means also a -2*x^41 term.
> He also has some interesting stuff on factoring polynomials.
> Van
===
Subject: Advances in Quantum Gravity
Announcement.
The Nova Science Publishers (NY, USA) continues the series of
the
booksRelativity, Gravitation, Cosmology. We have published one
with
the 14 paperswhich have been selec from about 40 submissions
last
year. In future it is assumed to launch a Journal with the
same title.
Gravitation, Cosmology:New Development. It will be dedica to
the
1. Dilaton gravity.
2. Quantum Mechanical Phases, Neutrino and Gravity. Photon and
Gravity.
3. Spin connection and 4-potential. Axion, Torsion and Notoph.
4. Curvature as a Scalar Field over the Minkowski Space.
5. Multidimensional Gravity. De Sitter Gravity. Weyl Approach.
6. Relativistic Quantum Mechanics Approach to Gravity (a la S.
Weinberg). Parity Violation.
7. Non-commutative Space-time.
We are ready to consider other candidates for the Editorial
Board of the
Book Series/Journal.
Jack Sarfatti replied:
I do have a pre-print at
http://qedcorp.com/APS/EmergentGravity.pdf
that may be suitable. It is also a CERN preprint.
The approach is novel and I invite feedback.
My idea is that Einstein's gravity including both dark energy
and dark
matter are emergent macro-quantum phenomena in the sense of
Oliver
Penrose's ODLRO (Off Diagonal Long Range Order) inside the
vacuum.
This is consistent with P.W. Anderson's More is different and
Andrei
Sakharov's metric elasticity. The result is a
background-independent
non-perturbative model consistent with Loop Quantum Gravity. I
give a
toy model for the formation of the vacuum condensate inflation
scalar
field domina by bound virtual electron-positron pairs that
make the
globally flat quantum electrodynamic vacuum unstable.
Einstein's
c-number space-time emerges as the more stable vacuum quite
similar to
the superconducting vacuum in the BCS model for a condensate
of real
electron pairs where the attractive virtual phonons overpower
the
repulsive virtual photons in the real electron pair coupling.
I use
general notions of the two fluid model where the residual
random zero
point vacuumfluctuations from all dynamical fields are the
normal
fluid. I also use Hagen Kleinert's 4D elastic world crystal
Planck
lattice model of Einstein's gravity that is consistent with,
i.e. a
special case of, spin-network -> spin foam loop quantum
gravity. In
addition I use Bohm's pilot wave interpretation of quantum
theory. The
key mathematical results are:
1. Hagen Kleinert's World Crystal Lattice Distortion Field =
(Area
Quantum of Loop Quantum Gravity)(Partial Derivative of the
Goldstone
Phase of the Vacuum Coherent Inflation Field)
i.e.
du(x) = (alpha')PSI(x),u
(alpha') = Ed Witten's string parameter in units of area that
is
proportional to (string tension)^-1
PSI is the complex scalar local macro-quantum coherent vacuum
order
parameter inflation field of zero thermodynamic entropy.
Eq (1) is analogous to Bohm's guidance constraint in a quantum
liquid
v = (h/m)Grad(Phase of quantum bit Pilot Wave)
That is, the quantum of circulation h/m for vortex line
defects in the
case of the single-valued superfluid macro wave function for a
quantum
liquid is replaced by Witten's alpha') for the elastic-plastic
world
crystal lattice.
The world crystal lattice is replaced in the final theory by a
spin
foam. All world crystal lattices are spin foams, but not vice
versa.
Einstein's IT FROM BIT (J.A. Wheeler) local metric tensor
field guv(x)
for curved c-number space-time is then
guv(x) = Globally Flat Metric + Strain Tensor of the Elastic
Distortion
Field
This is non-perturbative, the second term on RHS is not small
compared
to first term.
The Area dependence of Einstein's curved space-time metric
field guv
is consistent with the t'Hooft-Suskind world hologram
generalization
of the Bekenstein-Hawking quantized entropy of the area of the
event
horizon of a black hole that is also a very complex quantized
string state.
guv(x) is background-independent, i.e. a dynamical field of
spin 2 for
the normal fluid linear gravitons in the macro-coherent curved
condensate dynamical background.
Sakharov's metric elasticity is a special case of P.W.
Anderson's
generalized phase rigidity that keeps local classical physics
immune
from micro-quantum environmental decoherence. This explains
why the
macroworld is local without giant Schrodinger Cat states. The
macro-quantum coherence inflation field obeys a local covariant
Landau-Ginzburg BIT FROM IT equation with a renormalizable
cubic
nonlinearity.
Sakharov's metric elasticity ~ (QED alpha)(Witten's alpha') ~
(QED
alpha)/(String Tension) ~ G(Newton)/c^4 = (Planck Area)/hc
Note duality between Loop Gravity Quantum of Area and String
Tension
Quanta (i.e. links or bits of the quantized string).
This is not a two metric theory like the Yilmaz theory.
Einstein's local (passive) general coordinate transformation
emerge as
inhomogeneities in the local phase transformations of the
vacuum
coherence inflation field. Indeed, the world crystal
distortion field
is the compensating gauge force field needed to restore the
4-parameter
spontaneously broken translational symmetry group T4
responsible for
tidal curvature. Note that Hagen Kleinert has shown that tidal
curvature
comes from only the disclination defects in the world crystal
lattice
vacuum coherent local order parameter (inflation field). This
will
correspond to some topological property of the spin foam. Any
torsion
field from locally gauging the Lorentz group O(1,3) would
correspond to
dislocation topological defects in Kleinert's theory. In
addition,
locally gauging the 4 special conformal transformation (rela
to the
deSitter spaces) and the single dilation operator of the 15
parameter
Penrose twistor Conformal Group will lead to additional
dynamical
compensating gauge force fields.
Both dark energy and dark matter making ~ 96% of all the large
scale
stuff of the Universe are w = (pressure/energy density) = -1
exotic vacuum residual zero point stress-energy current
density fields
with the local equation
3. tuv(Vacuum) = (Dimensionless) QED electron-virtual
photon-electron
coupling)(String Tension)(Quantum of Area)^-1[(Quantum of
Area)^3/2|PSI(x)|^2 - l]guv(x)
4. Define the real local scalar world hologram zero point
energy exotic
vacuum field
/zpf(x) = (Loop Quantum of Area)^-1[(Loop Quantum of
Area)^3/2|Vacuum
Coherence|^2 - 1]
Einstein's exotic vacuum local field equation
Guv(x) + /zpf(x)guv(x) = 0
In the weak curvature field Newtonian gravity limit reduces to
the
Poisson equation
Laplacian of Exotic Vacuum Gravity Potential ~ - c^2/zpf
with the sign convention that /zpf > 0 is anti-gravity
universal
repulsion and /zpf < 0 is universal gravity attraction.
Neither of these non-classical quantum effects are limi by the
weakness of large scale G(Newton).
The large scale limit of /zpf is Einstein's cosmological
constant
where we use the FRW cosmological metric in the usual way.
On the micro-scale in stringy vortex core topological defects
of the
single-valued inflation field PSI(x) there is strong exotic
vacuum dark
matter attraction that prevents the electric charge of
spatially
extended lepto-quarks from exploding. This solves the old
Abraham-Lorentz-Poincare stress problem of the electron's
self-energy of
100 years ago. Large micro-scale warping of these geon strings
of
Wheeler's Mass without mass etc. make these spatially extended
momentum transfers in high energy scattering as shown in deep
inelastic
electron-hadron scattering for examples. This model is
consistent with
observed universal slope the hadronic resonances, which are
seen to be
simply the exotic vacuum dark matter universal gravity on the
short
scale. The universality of the Regge slope at (1Gev)^-2 shows
it is an
equivalence principle gravity phenomenon.
w = -1 dark matter macro-geons as in the spherical Galactic
halos will
look like w = 0 CDM to distant observers.
In this theory dark matter detectors will never click with the
right
stuff. They will only click with false positives because dark
matter is
space. Dark matter is simply exotic vacuum with positive
pressure, just
as dark energy is simply exotic vacuum with negative pressure.
Which is
which depends on the local intensity of the vacuum coherence
field. That
the universe as it expands is more domina by repulsive dark
energy
than it was in the past is a matter of dynamics of the local
/zpf(x)
field. It may be contingent in the sense of Lenny Susskind's
Landscape?
Comments, questions, refutations solici.
===
Subject: Convergent subsequence
charset=Windows-1252
===
Can there be a (non-random) non-convergent sequence {a[n], n
in N} for
which
NO convergent subsequence exists?
Thanks in advance,
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Convergent subsequence
> Can there be a (non-random) non-convergent sequence {a[n], n
in N} for
which
> NO convergent subsequence exists?
> Thanks in advance,
> --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
The sequence {n: n in N} fits your specifications, if you
insist on convergence to a real number. It is unbounded,
of course.
Consider this:
A theorem: Every sequence of reals has a monotone subsequence.
(We may not know in advance whether it would be a
(non-strictly)
increasing or decreasing one.)
Draw your own conclusion.
ZVK(Slavek).
===
Subject: Re: Convergent subsequence
charset=utf-8
===
Ioannis <morpheus@olympus.mons>
[CapitalEth][EDouble
Dot][Micro]
.b3
.b9[EDo
ubleDot].b9
> Can there be a (non-random) non-convergent sequence {a[n], n
in N} for
which
> NO convergent subsequence exists?
Duh,
take an eventually constant sequence.
Sorry, let me rephrase that:
Can there be a (non-random) non-convergent sequence {a[n], n
in N} for
which
NO non-eventually constant convergent subsequence exists?
> Thanks in advance,
--
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: JSH: Focusing on counting prime numbers
> My analysis continues to indicate that my research on
counting prime
> numbers *should* be more accessible than my other math
research which
> is more abstract, and clearly more difficult. Still I also
recognize
> that significant parts of my prime counting research are
beyond a lot
> of people simply because that research extends into partial
> differential equations.
[...big clip for brevity...]
> Yup, despite the research done on prime numbers through the
entire
> math history of the world, mathematicians as a group can
only produce
> *longer* expressions if they use their own research on
counting prime
> numbers!
 Want more on prime counting? Then see my blog archives:


*
Slightly off-topic, but if you want some interesting reading,
check
out Prime Obsession, by English writer John Derbyshire, which
starts
out with estimates of the number of primes less than n, then
goes on
to the most important remaining unsolved problem in
mathematics, the
conjecture of Riemann -- that the non-trivial zeros of the zeta
function have real part 1/2.
This is a very readable and well-written book -- no degree in
math is
required.
The odd-numbered chapters are full of mathematics -- well
documen
and explained from the ground up. The even-numbered chapters
are
historical and biographical in nature.
In the appendix, there is a ten-verse song:
Where are the zeros of zeta of s?
by Tom M. Apostol.
First verse:
Where are the zeros of zeta of s?
GBF Riemann has made a good guess:
They're all on the critical line, sta he,
And their density's one over two pi log T.
Tenth (last) verse:
There's a moral to draw from this long tale of woe
That every young genius among you must know:
If you tackle a problem and seem to get stuck,
Just take it mod p and you'll have better luck.
Happy reading!
earle
*
===
Subject: Re: GCD proof...
> 2. Show that (a, b, c) = ax + by + cz for some integers x,
y, z.
It's essentially the same proof that works for (a,b). So:
(1) To make notation easier, let g = (a,b,c).
(2) Let d be the smallest _positive_ integer that can be
written as
ax+by+cz for some integers x,y,z.
(3) Since g always divides ax+by+cz, it divides d.
(4) Now divide a by d and look at the remainder, say a = dq+r
with r
between 0 and d-1 inclusive.
(5) Substituting, we get
(a-r)/q = d = ax+by+cz,
so
a-r = a(qx)+b(qy)+c(qz),
so
r = a(1-qx)+b(qy)+c(qz).
(6) Thus r is a nonnegative linear combination of a,b,c. It's
strictly
smaller than d, which was the smallest positive linear
combination.
Therefore r=0.
(7) This proves that a=dq, so d divides a. A similar argument
shows
that d divides b and that d divides c. Therefore d less than
or equal
to g, since g is the greatest common divisor of a,b,c.
(8) This proves that g divides d and that d is less than or
equal to
g, and hence d = g.
Clearly the same proof works for any number of integers.
Joe Silverman
> This is in relation to the chicken nuggets that I 've pos,
which I
cannot
> obtain a proof to as well. Can someone outline what they
would do? a full
> proof is not neccessary.
> 1. Show that (a, (b,c)) = ((a,b), c) for any three numbers
a, b, c.
Define
> the greatest common disvisor of a, b, c, call it (a, b, c)
and show that
(a,
> b, c) = (a, (b, c)).
> 2. Show that (a, b, c) = ax + by + cz for some integers x,
y, z.
===
Subject: Re: GCD proof...
> This is in relation to the chicken nuggets that I 've pos,
which I
cannot
> obtain a proof to as well. Can someone outline what they
would do? a full
> proof is not neccessary.
> 1. Show that (a, (b,c)) = ((a,b), c) for any three numbers
a, b, c.
Define
> the greatest common disvisor of a, b, c, call it (a, b, c)
and show that
(a,
> b, c) = (a, (b, c)).
Use the theorem that says that if
a = P1^A1 P2^A2 ... Pn^An
b = P1^B1 P2^B2 ... Pn^Bn
are the prime factorizations of a and b, then
gcd(a,b) = P1^min(A1,B1) P2^min(A2,B2) ... Pn^min(An,Bn)
A huge fraction of elementary gcd problems reduce to very
similar min
problems when looked at from that direction. Your problem
essentially
becomes one of showing that min(a, min(b,c)) = min( min(a,b),
c) for the
first part, and defining min(a,b,c) for the second part.
> 2. Show that (a, b, c) = ax + by + cz for some integers x,
y, z.
And this is one of the problems that DOESN'T become a min
problem.
--
===
Subject: Re: JSH: Limi intellects
Harris
> says...
>
[snip]
> After all you don't have to believe that [people who
disagree with
you]
>> are correct to not find their disagreement strange do you?
>
>Yes I do. There are any number of people with wacky ideas
that I feel
>are just wrong. At times I may even try to communicate that
to such a
>person.
>
I think this marks a fundamental difference between us. If
someone
disagrees
> with me then I at least consider it possible that they are
doing so in
good
> faith - genuinely believing their position to be correct.
> What if they then demonstrate they aren't?
> Even if they are wacky ideas that [you] feel are just wrong
they do
not imply
> some dark ulterior agenda.
> I didn't say anything about a dark ulterior agenda as I
merely
> poin out a position which boils down to *most* people don't
bother
> bumping heads in endless arguments with people they think
are nutty,
> and wrong.
> It's just a waste of time.
> What's weird about my case is that there are these posters
like Dik
> Winter who has been diligiently replying to my posts for
YEARS. The
> guy has freaking web pages over old arguments of mine that
I've myself
> admit were wrong!
> It's like some weird obsessive cult of sci.math posters.
> Now if you say that's normal, then I suggest to you that
you're not
> willing to tell the truth.
That's what I was trying to get you to consider. But with this
fundamental a
> lack of common understanding I don't think it's possible.
 Well at least you're sounding like a rational person who
will probably
> wander off now.
> I on the other hand have to keep facing this weird crew who
WILL NOT
> GO AWAY!!!
> Like David Ullrich, the poster child for obsessive behavior,
who not
> only obsessively replies to my posts, he maintains that it's
*my*
> fault!!!
It IS your fault. If you wouldn't post, maybe you'd be happier
because
nobody would insult you. James, I think a 4 year old is more
mature than
you
are. GROW UP.

>Research is difficult. When I'm pushing the mathematical
envelope I
>get cranky.
I've decided I've discovered enough, and can settle down and
enjoy the
>fruits of my labors.
So I'm no longer into extreme math.
>
Well enjoy whatever it is you plan to do instead.
> Publish papers. It's time for me to get a few papers
published. So I
> still have some difficulties to face, but it'll be a lot
easier than
> actively researching.
> Mathematics is some discipline. No wonder mathematicians
find ways to
> react wackily to my proven results, like my prime counting
discovery,
> as possibly they're already quite gone anyway after a few
years of
> active research.
> Like it fries their brains.
>
===
Subject: Dilemma with undergraduate real analysis
Right now I'm taking the undergraduate year-long 'real
analysis' class. To
be
honest the professor is terrible and the way he goes about
things is to
make
the class so easy that even the worst students do great. The
main thing
that
has happened is that I'm not learning anything. The problem is
that I'm
going
to graduate school next year and I bet they will expect me to
be taking
their
graduate level analysis class with Lebesgue Integration and
the rest.
I have two questions. Do I need to know this before taking a
class at the
graduate level (I assume so but it could be radically
different..right?).
What
are books that are good for self-study that teach all the
basics about
metric
spaces, riemann/riemann-stieltjes integrals, and the rest?
Randy