mm-243 The number of replies is the thread's rating. Apparently, post's quality has> nothing to do with the rating. Now, face it: the JSH show has the highest> rating in sci.math.> That's what makes it so sad - that the rantings of that pathetic loony>> seem to be stuff that so many contributors to sci.math are so interested>> in. >I'm puzzled why this bothers you so much. The huge majority of posts on >sci.math are about mathematics. > I'm glad you said that, because I wanted to but hesitated to do so > because after all I'm one of the guys who's part of the problem> according to various posts here.Why not tell it like it is, Ullrich? You're the epitome of a mathiesadist!--John> Seems to me that an even huger majority of the _threads_ are> about mathematics, making it even easier to filter out the crap> if you want.>It's easy to skip the other threads, >assuming a moderately decent newsreader.> ************************> David C. === UllrichSubject: Re: consecutive composite integers factors and primes, I found that for any sequence>of consecutive composite numbers there is always one integer that has>a prime factor larger than any other prime factor of any of the other>integers. Further, this prime is not raised to any power.>> How does the Further,... apply to this example: 8, 9?>Perhaps he intended maximal sequence, 8,9,10 has 5 as max.However the maximal sequences 4 and 12 have 2 squared andthe maximal sequence 18 has 3 squared.Thus the conjecture needs be limited tomaximal sequences of composite numbers === single equation, such as f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly a plane, but not a line.> I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line,> last time I looked.> Lee RudolphNote that z is unrestricted. It === Infinity>> Between any two odd integers is an even integer, between any two even> integers is an odd integer. The density in their union of either is> one half.What does that last sentence mean?> Here I equate density with measure in the unit interval.Then the integers have density zero. > I don't care if you ignore gravity, it won't do you much good, I'm> here only concerned with considering a model where the rationals and> irrationals alternate in the reals.Which is still as stupid as trying to ignore the law of gravity.> If there are more irrationals than rationals and rationals and> irrationals are disjoint and distinct, then, where they are each> totally ordered, then there necessarily would be irrationals with no> rationals between them. Yet, there are not.In the set of rationals, there are no irrationals.In the set of irrationals, there are no rationals.In the set of reals, there are countably many rationasl ans uncontably many reals arranged so that between any two distinct reals there are countably many rationals and uncountably many reals.In fact, there is a order preserving bijection from any open interval, (a,b), with a < b, to the set of all reals, namely f:(a,b) -> R: x |-> (a-b)*x/[(x-a)*(x-b)], and if a and b are rational, the mapping f carries rationals to rationals and irrationals to irrationals. > I'm trying to think of a function between the unit interval's reals> and irrationals. The claim is that one exists because the rationals> map onto the integers and the integers don't map to the reals, thus> that the irrationals map onto the reals else the reals would be a> union of two sets that don't map onto the reals. Yet, a construction> explicitly mapping each element of the irrationals to each element of> the reals is not given. I'm also still looking for a mapping between> R[0,1)^N and R[0,1).A reverse bijection, from the reals to the irrationals, was given in at least 2 versions in a prior posting in this thread, so just take the inverse bijection of either of them.> I like to think that the rationals and irrationals alternate and that> the function f(x)=x+iota maps Q[0,1) onto P(0,1), and f(x)=x-iota maps> Q(0,1] to P(0,1).You may like to think a lot of things, but that does not make them true.> Then again I think the impulse function evaluates to half infinity> at zero, and consider the Gamma function on negative integers to have> values of various finite multiples of a scalar infinity.Again you reveal that your wiring is short circuited.> Now I'm looking at the post about mapping R <-> P. I don't> immediately grasp vector space over a field and linearly> independent. You have the sequence b being a sequence of reals each> linearly independent over Q, and a set C of reals of {b_0, b_1, ...}> linearly independent over Q, with the initial sequence element b_0> being a rational. RQ=P, you claim that R injects into P by> f(b_n)=b_{n+1} and f(c)=c. Why do you have braces around n+1 instead> of parentheses? Standard newsnet notation for a compound subscript.Then you have F(c)=c, for c in C. I think you mean> that c in C is not an element of the sequence b. Then you say to> extend that to all of R by linearity over Q. So you claim a function> f(r)=p for r in R and p in P to be defined for all reals. What's r> for f(r)=pi? What's p for f(2)? Why f and F, presumably a shift-key> error?> http://mathworld.wolfram.com/LinearlyIndependent.html> http://mathworld.wolfram.com/VectorSpace.html> http://www.wikipedia.org/wiki/Vector_space:A set V is a vector space over a field F, if given an operation> vector addition defined in V, denoted v+w for all v, w in v, and an> operation scalar multiplication in V, denoted a*v for all v in V and a> in F, the following 10 properties hold for all a, b, in F and u, v,> and w in V:> 1. v+w E V> 2. u+(v+w) = (u+v)+w> 3. v+0 = v> 4. v-v = 0> 5. v+w = w+v> 6. a*v E V> 7. a*(b*v) = (a*b)*v> 8. 1*v=v> 9. a*(v+w) = a*v + a*w> 10. (a+b)*v = a*v + b*v> Those each hold for V = R and F = Q. Properties 1 through 5 indicate that V is an abelian group under> vector addition. The intersection of all subspaces containing a given set of vectors> is called their span; if no vector can be removed without diminishing> the span, the set is called linearly independent.> So you say each element of the sequence represents a set of vectors or> a set of a vector, I'm not sure which, and that it is linearly> independent over Q because removing that vector from the set of> vectors would diminish the span of the intersection of the subspaces> of the vector space.I have this set of reals B = {b_0,b_1,b_2, ....} whose members are are, as vectors over Q, linearly independent and so that b_0 is a non-zero rational. Every real in span(B) is a linear combination of finitely many of the members of B. There are other reals which are linearly independent of the span of B, the set of which I called C, and which is a (vector) subspace of R. Given any real r, then there are unique rationals p,q, and a unique real b in span(B) and a unique real c in C such that r = p*b+q*c. Thus R is the direct sum of subspaces span(B) and C.It may be proven that there is only one Q-linear function, say f, from R to RQ and such that f(b_0) = b_1, f(b_1) = b_2 etc., and f(c) = c for every c in C. This function is a bijection.> Please neaten that up provide a more self-contained explanation. Also> explain. While you're at it show a bijection between R^N and R.The explanation is sufficient for those who know a little math. Learn a bit about vector spaces and it may become clear to you, or don't and it won't. > Some talk here is about the nosntandard treatment of the reals: the> hyperreals. One thing to note is that *R, the hyperreals, as a set> contains the same elements as R, the reals. It's just a different way> to consider them.The Robinson formulation of *R has more than a single member corresponding to a member of R, it has uncountably many coresponding to eanc member of R, which are infinitesimally close to each other, as well as some which do not correspond to any member of R.> About the uniform probability distributions over intervals of reals,> that's not about making some new definition of what a probability> distribution is.It is well known, to those who understand what pdf's (probability density functions) are that there cannot be a uniform pdf on R. Uniform pdf's require finite real intervals or finite sets to operate on. You cannot make one on a countably infinite set nor on an unbounded real interval.> It's about applying the characteristics of a> probability distribution to an infinite population. Since a uniform pdf must have certain properties to be a pdf, there are limits on what sets they can exist on, and N, Q and R are outside those limits.[garbage deleted]> Of course that's ludicrous but at the same time it allows us to> consider the realm of thought in concern of this issue and to then> talk about the probability of selecting a given element of the natural> integers assuming a uniform probability distribution over the> integers. At least we seem to have some agreement that a uniform> probability distribution over an interval of the reals exists, and a> simple method to sample an element of an interval of the reals exists.We have no agreement that such a distribution exists, because it cannot.> infinitesimals, it talks about 1-infinitesimals, 2-infinitesimals,> etcetera, n-infinitesimals, with the oo-infinitesimal being zero.Your knowledge of infinitesimals is nil. Until you get a reasonable understanding of the standard reals, your hopes of understanding anything about non-standard reals is === please explain these integrators differece?> 1. 1/(1-z^(-1))> lot!Write the difference equations. You can see that 1. uses the currentinput in the sum, 2. is one input behind and 3. is the same as 2. butscaled by T.Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth === Chuck Simmons chrlsim@earthlink.netSubject: Re: limit points open interval about any r in R contains numerous rationalsother than r, r is a limit point of Q.If you want to construct a sequence in Q with limit r.For each integer n > 0, let I_n = (r - 1/n, r + 1/n)and pick a rational q_n in I_n(If r is a rational, pick q_n so it isn't r.)To conclude, show lim(n->oo) q_n = r> well, I know this is true, and I can prove it. the reason i ask though> is because i doubt myself on this.>> let (xn) be a sequence in the set of all rationals 0<=p<=1. Further,> suppose xn is an enumeration of the rationals between 0 and 1. i think> that all rationals between 0 and 1 are cluster points of xn, mainly> because all are limit points. i could also work it from the definition> since a rational lies between any two rationals, but i started> doubting myself for some reason.>That is correct, every rational is a cluster point of an enumeration ofthe rationals. How is that used to show the limit points of Q are R?> can someone tell me if I am wrong about anything in the === f,g continuous, then so are max(f,g) and min(f,g)> After drawing some graphs, this seems pretty obvious for the single> point a0 -- max(f,g) has 2 cases: it equals to f or g. Either is> continuous. However, this question implies continuous on R, not just> at a single point. Any ideas how to approach this?There is a slight complexity. The point may lie on both f and g. You should try working directly from the definition for continuity andplaying around with the inequalities associated === Single FractionX-NFilter: 1.2.0>How do I do this?>>Express the following as a single fraction:>>4/3ab - 5/6bcMultiply the first term by 2c/2c and the second by a/a.>>and>>(m^2 + 2)/(m^2 + m) - (m - 2)/mNote that the denominator of the first term is m*(m+1). Use the sameapproach described above.<>a! - b! = n^2>>Of course, a, b and n are positive integers.>>I think the only two solutions are:>>2! - 1! = 1^2>>3! - 2! = 2^2[snip my proof outline]>I'm not sure how easy it is to get an effective upper bound using this>line of argument. We actually don't need anything quite as strong as the>n^0.6 result; I think any prime gap smaller than, say, 0.3 n/log n will>probably do. The primes in (b/2,b] are handled quite well by the Rosser->Schoenfeld inequalities.As luck would have it, I recently learned of Pierre Dusart's improvementsto Rosser-Schoenfeld, proving that for x >= 3275 there is always a primein the interval (x, x + x/(2 ln^2 x)]. So it looks like a full solutionis easily possible (indeed, to the following more general question: whenis the difference of two factorials a powerful number?). === too long ago, someone here asked for more rigorous references for issue of The American Mathematical Monthly, there is a glowing review of the bookHubbard JH, Hubbard BB. Vector Calculus, Linear Algebra, and Differential Forms.Might be worth looking at.-- Stephen J. Herschkorn === composite integers> Examining a table of factors and primes, I found that for any sequence> of consecutive composite numbers there is always one integer that has> a prime factor larger than any other prime factor of any of the othersNice conjecture!It is equivalent to saying (more neatly perhaps) that in any adjacentsequence of numbers, the largest prime factor occurs only once.It seems obviously true, though I can't see a slick proof off-hand.I'm sure there must be one. It is possible to produce a heuristicdemo that can doubtless be turned into a formal proof with effort.Suppose your two largest equal prime factors are P, and we may as wellput them at the ends of the interval WLOG. Then we have an interval oflength P which must be filled up by factors less than P. This would meanthe P interior numbers have to be removed one-by-one by ditching the evens,then the 3-multiples, then the 5-multiples, and so on up to P.But we can remove at most (1/2)*(2/3)*(4/5)*(6/7)*...*(1 - 1/P) this way;and a quick heuristic integral integral shows this to be 1/(log P).So the most numbers we can reduce to by removing those with (even at least one!) small prime factor is one log-Pth of the lot, and the lot is of size P. That's P/logP. Nowhere near zero or one.So it can't be done... not by a huge amount!Nice question though.------------------------------------------------------- ----------------------- Bill Taylor W.Taylor@math.canterbury.ac.nz-------------------------------- ---------------------------------------------- Chebychev said it -- I'll say it again; There is always a prime between n and 2n.------------------------------------------------------------ === Removed by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id h97NALo30990 id 1A70y9-0003Ht-2rX-AntiAbuse: This header was added to track abuse, please include it with any abuse reportX-AntiAbuse: Primary Hostname - host.safe-hosting.comX-AntiAbuse: Original Domain - mathforum.orgX-AntiAbuse: Originator/Caller UID/GID - [32148 32148] / [47 12]X-AntiAbuse: Sender Address Domain - host.safe-hosting.comYou (post-sci-math@mathforum.org) have been removed from the IGDNYC mailing list. We apologize for === by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h980fWW04789;>Hi Everyone!>>I have an odd request, here goes...>>I purchased an Expedition with the keypad entry - no one knows what>the combination is and I don't have time or money to take it to the>dealer...>>I'd like to try combinations at my leisure until I find the one that>works. What I need is to know every possible combination using the>numbers 1,3,5,7,9. Numbers may repeat, so that needs advance for your help. Please email me the results.>My email for sites on how to reset the code. You may not livelong enough to try all the codes yourself. For example, a six-digitcode may require up to 1 million attempts (half a === function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h980fWX04793;Depends how much security you are looking to achieve. You must assume,for example, that the algorithm you are using (e.g., DES) will eventually become known. Then the only security left is the key.Many other considerations enter. Do you need to protect === Re: looking for smooth function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h983NPO15303;>Hello sci.math,>>I've got an array size 100 of integer values (range: 0-5). Each plot in >the array represents a 1ms time window. I currently have these graphed >as stairs in MatLab (which looks like a bar graph). I want to make >this function smooooooth, but i'm not sure of a method to apply to it.>>What steps should I take to complete interpolation between points on a uniform grid:Target coordinate: . . . . . . . | . . . . . . . .Binomial coefficients: 1 5 10 10 5 1Reciprocal of offset: -1/7 -1/4 -1/1 1/2 1/5 1/8Product times 56: -8 -70 -560 280 56 7Alternate signs: 8 -70 560 280 -56 7Divided by total which is 729.Thus producing coefficients to use for the === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h983NPO15307;I don't get how to use the hint. Now quite sure how to attack this problem either. I dont think I understand. Any help please?Let {f_n} denote the characteristic function of the rationals in the interval [0, 1]. Show that there is no sequence of continous functions {f_n} such that lim f_n(x) = f(x) for all x in [0, 1].HINT: If I_n is a nested sequence of closed intervals such that the length of I_n tends to 0 as n tends to infinity then === Re: The Octic x^8-x^7+29x^2+29 Revisited by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h984e7P20172;>> z^7-7z^6-2763z^5-19523z^4+1946979z^3+34928043z^2+119557031z- 3247^2=0>>MAPLE tells me this has cyclic Galois group of order 7. Thus>its roots may be ordered z_0, ..., z_6 in such a way that>a^7 = b in Z[zeta]>where zeta = exp(2 pi I/7) and>a = sum_{j=0}^6 z_j zeta^j>is a Galois resolvent.>>I calculated the roots numerically to 50 dp and tried various orderings.>I came across one which gave>a^7 = -12392836399104 + 20856562728960 zeta + 26834412666880 zeta^2> -13437774020608 zeta^3 + 2576491954176 zeta^4 + 25860547313664 zeta^5> (snip)I came across this reply after I made my second post. Turns out the solution to the septic in question (a variant of it, really) wasalready in cyberspace, waiting to be plucked.With === help me? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h985IgK22585;>
>I am
stuck on a proof. I have gotten a series down to where I need
to>prove the following:>> lim c^1/n = 1 where c>0>n ->
===
Calculus Question>A single equation, such as f2(x,y,z)=c2, can
describe in a 3d space >a surface, possibly a plane, but not a
line.> I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a
line,> last time I looked.> Lee RudolphI stand corrected, but
it is not a very efficient way of doing lines, and certainly
===
there a Big-O listing for know problems somewhere on the
===
===
>> 1)prove that if f,g continuous, then so are max(f,g) and
min(f,g) >> Any ideas how to approach this? >Proving a
function continuous at any point >proves it continuous at
every point.f(x) = 0 when x <= 0 = 1 when 0 < xis continuous
===
at 1. Thus it's continuous at 0?----Subject: Re: hw help --
===
f,g continuous, then so are max(f,g) and min(f,g) >After
drawing some graphs, this seems pretty obvious for the single
>point a0 -- max(f,g) has 2 cases: it equals to f or g. Either
is >continuous. However, this question implies continuous on R,
not >just at a single point. Any ideas how to approach
this?Show min(x,y), max(x,y), RxR -> R are continuous.Then as
f,g are continuous, so are compositions min(f,g),
===
max(f,g)----Subject: Re: non-Hausdorff homeomorphism to
===
non-Hausdorff space locally homeomorphic to R^n?No. I use
theorem locally closed Hausdorff S ==> S HausdorffS locally
closed Hausdorff when for all p in S, some open U nhood p for
which cl U is a Hausdorff subspaceLet the non-Hausdorff space
be S. As I don't know definition of locallyhomeomorphic I'll
do my best to intuit what it means.If p in S, then there's
some open U nhood p homeomorphic to an a nhood V of R^n about
q, the image of pAs R^n is normal, we can find a closed nhood
K of q contained in V.Now the homeomorphic image of K is a
closed Hausdorff nhood of p.Thus I hope I've shown S is
===
Exponentative closure>A further question along those lines is
what is is the set, E, of>numbers generated by repeated sums
of products of rational powers of>rationals? It is a subset of
the algebraics, but does it form a>field? I believe so: if x_1
is a member of E, then so are its conjugates x_2, x_3, ...,
x_n.Note that x_1 x_2 ... x_n is rational (being the constant
term of a monic polynomial over the rationals whose roots are
x_1, x_2, ..., x_n).And then 1/x_1 = x_2 ... x_n / (x_1 x_2
... x_n) is in E.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
===
quantum echo> --I could have baked both cakes yesterday.> --I
could have baked both cakes today.> --I could have baked this
one yesterday> and that one today.> --I could have baked that
one yesterday and> this one today.>> Your example doesn't
work!>> How am I going to distinguish 'one cake' from 'the
other cake'!>Since they now are identical-with-somethings, if
they shared>all the same properties they would be one cake
rather than>two. THerefore, since they are two cakes, they
don't share>all their properties, and I can distinguish them
by means>of any property that one has and the other doesn't.>
>> you mean argument by GPS or by just determining their
global positioning> with sattellites we have a difference
even if there are no other clear> attributes of different
classes besides location?> That is a call which no logician
qua logician has any particular> competence> to make.> Are you
saying it is possible for something to exist at location A and>
location B at the same time? I thought it was inference by the
rule of> non-contradiction.I have no idea whether this is
possible or not. My suggestion that quanta may satisfy
'~(x=x)' is meant to translateinto logical terms the claim by
some physicists that quanta'lack individuality' (Heller
suggests that quanta 'lack haecceities').The scenario with
cakes-to-be suggests that the behaviour of otherentities
without individuality resembles that of quanta in therelevant
respect. Heller's description of that behaviour followsmy
sig.> --John> JJ>> 1 I. The Problem, and the Problem with
the Problem,> 2 of Identical Articles and Quantum Statistics>
3> 4 Suppose we have a box with two qualitatively> 6 bouncing
around inside. We think of the box> 7 as having a left (*l*)
and a right (*r*) side.> 9 at random without interacting, so
that their> 10 motions are independent; in particular we> 12
that we may neglect collisions. What are the> 13 chances for
finding one or both on one side> 14 or the other?> 15> 16 Many
find the following reasoning persuasive.> 19 in *r*, and 2 in
*l* and 1 in *r*. These should> 20 be equally likely, so that
each has a probability> 21 of 1/4, or a probability of 1/4 for
two in *l*,> 22 1/4 for two in *r*, and 1/2 for one on each
side.> 23> 24 This stylized example is a simple mock-up for a>
25 kind of situation that can occur with quantum entities> 26
and properties. For many of these situations the> 27
probabilities are in fact found to be 1/3 for each> 28 of the
three cases: two in *l*, two in *r* and one> 29 on each side.
Many interpreters have found this> 30 fact utterly
astonishing.> 31> 32 But on the face of it, there is a very
simple> 33 resolution of the puzzle: give up supposing> 34
that there are two qualitatively identical but> 36 there are
two *quanta*, as I'll put it, to which> 37 the notion of being
numerically distinct does not> 38 apply. . . (p.
===
is that given any>rational that the value greater than it
and less than any other>greater is irrational,>> There
is no such number, as several different people have shown you.>
In non-standard analysis, there might be, however.> See Alain
Robert's book about NSA. Rather than being> irrational, it
would be non-standard, though.>> I have yet to see any
standard or non-standard model of the reals in> which there is
a smallest positive number.Who says that that is meant?In nsa,
there can be a nonstandard number that can be said to
be'greater than (a given standard rational) and less than any
other (standard) greater'That it is not unique, who cares?In
nsa, though, there are both rational and irrational
non-standard numberswith this property. So, the statement that
such values are necessarily irrationalis not true, there.But
who knows what this poster's intuition may eventually lead
===
qn>I don't get how to use the hint. Now quite sure how to
attack this>problem either. I dont think I understand. Any
help please?>Let {f_n} denote the characteristic function of
the rationals in the>interval [0, 1]. Show that there is no
sequence of continous functions>{f_n} such that lim f_n(x) =
f(x) for all x in [0, 1].I think you meant f is the
characteristic function, not {f_n}.>HINT: If I_n is a nested
sequence of closed intervals such that the>length of I_n tends
to 0 as n tends to infinity then (intersection n=1>to
n=infinity)I_n is not empty.Further hints: I_n should be a
closed interval on which f_n > 1/2.What do you know about any
point in the intersection of these?Now how can you arrange it
so that the fact that the intersection is nonempty produces a
contradiction?Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
===
consecutive composite integers>Examining a table of factors
and primes, I found that for any sequence>of consecutive
composite numbers there is always one integer that has>a prime
factor larger than any other prime factor of any of the
other>integers. Further, this prime is not raised to any
power.>> How does the Further,... apply to this example: 8,
9?>Perhaps he intended maximal sequence, 8,9,10 has 5 as
max.But the example he gave was 1500, 1501, ... 1509 which is
not maximal.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
===
How to calculate the total coverage area of a few
number of circles isabout 10. I think that the calculation
time of Monte Carlo integration mightbe too long. Is it
possible to use numerical intergration, i.e calculatingthe
area enclosed by the envelope of those circles? But how can I
get theexpression of this envelop easily in my simulation? I
also need to considerthe sunk parts while integrating the
area, right? The calculation is relatedto computer graphics.
Supeng>> In my simulation, N circles with the same radius r
are randomly> placed. Let P_i denote the center of circle i.
For any i, p_i lies> within the coverage range of at least
one other clicle, i.e. at least> one other circle contains
p_i. How to calculate the total coverage> area of the N
overlaped circle? The method should be easy to be>
implemented by programming for simulation.> You can calculate
the intersection of any two circles> analytically:>
http://mathworld.wolfram.com/Circle-CircleIntersection.html>
However, your problem might involve a large number of these,>
and in addition you need to calculate overlaps of 3 circles,>
4 circles, etc.>> Edelsbrunner has inclusion-exclusion
formulas that depend only on> overlaps of at most three
circles: The union of balls and its dual> shape,
http://portal.acm.org/citation.cfm?id=161139>> -- > David
Eppstein http://www.ics.uci.edu/~eppstein/> Univ. of
California, Irvine, School of Information & Computer
===
couldn't send email to Jim Heckman, so I post my message
doesn't answer all your questions, just ask. I've> dug out> my
own decades-old notes on the groups of order 16 and 32,> and>
now have presentations, permutation reps, conjugacy> classes,>
subgroup structure, automorphisms, etc., for all of them.>
and 32,> if they aren't too hard to find. Email them to me if
they> are too much to post.I can neither e-mail nor post them,
since at this late date Ihave only hard copies. And we're
talking dozens of pages; thereare 51 groups of order 32, after
all.[...]> I just got GAP, and don't know how to use it yet.>
If anyone could tell me some of the commands I should use to>
get info on groups with 16,24,32 elements, (as well as what>
the commands mean), I would be grateful.Sorry, I don't use GAP
myself. I should probably learn it one ofthese years.> BTW,
could you define a perfect group for me.A perfect group G is
one that's equal to its own commutatorsubgroup G' = [G,G].
Note that this is equivalent to G having nonon-trivial
homomorphism to an abelian group.> Also, GAP speaks of
polycyclic and finitely presented> groups.I forget what
polycyclic means, and somewhat surprisingly can'tfind it in
the index of any of the books I have readily at hand.I assume
finitely presented means presented with a finite numberof
generators and relations.> I hope to find an intro to groups
on the web (with some definitions), and> download it.If you've
already worked through Hungerford, you're probablyready for
something a little beyond intro level. Two books thatI find
myself referring to constantly, and will probably nevergrok
completely, are Aschbacher's _Finite Group Theory_ andDixon
and Mortimer's _Permutation Groups_. Beware that theformer is
intended for specialists, and is written in a veryterse
===
Interestingly, there are no fewer than four recent pop books>
on the Riemann Hypothesis. (Probably because of its>
appearance on the Clay Mathematics Institute list of
problems.)> One sees:> Derbyshire, Prime Obsession> du Sautoy,
The Music of the Primes> Sabbagh, The Riemann Hypothesis>
Edwards, Riemann's Zeta FunctionWhatever your concept of pop
book is, it certainly does not applyto Edwards' book. Besides,
it is not recent, since it was originallypublished in 1974.Best
===
Infinity(On RAF)> But who knows what this poster's intuition
may eventually lead to?We should all worry :-(-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
===
===
Subject: non-Hausdorff homeomorphism to R^n>is there a
non-Hausdorff space locally homeomorphic to R^n?> No. I use
theorem> locally closed Hausdorff S ==> S HausdorffSo what do
you do about the real line with double origin?(Quotient of R x
{0,1} by the smallest equivalence ~with (x,0) ~ (x,1) for x =/=
0 ).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless
to say, I had the last laugh. Alan Partridge, _Bouncing Back_
===
there any algorithm to calculate the deep holes of leech
lattice?> Yes, it's not hard to find one that calculates all
holes of any lattice.> Could you please tell me the method? I
probably don't need all the deep> holes but I wanna find some
hole is the circumcentre of some n-simplex with vertices inthe
lattice L (of rank n). To find all holes then find all
curcumcentres of n-simplices with verticesin L.Problem:
infinitely many of them!Resolution: really only want to find
those in a fundamentalparallelotope. If this has diameter D,
then only need to considervertices distance <= 2D from origin.
Now finite problem.Simple, but hopeless in practice.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
===
Salut!>> j'ai qq problemes a resoudre le probleme suivant:>>
> (c'est de l'electronique a la base je vous l'accorde mais les
outils> de resolution sont des maths alors ...)> Supposons un
circuit RC banal aliment.8e par une tension U0 continue.>
Que'ce que c'est, un circuit RC banal? Est-il comme ca:>> On a
en s.8erie un g.8en.8erateur + une Capa + une r.8esistance.>>
(U0) ---R---C--- (0)> La tension U0, est-elle continue ou
constante?>> la tension est constante et non continue comme je
l'ai .8ecrit ci dessus:U(t)=U0> C'est comme ca alors:> (U0)
---C---R--- (0)> La tension aux bornes de la resistance,
ur(t), verifie l'equa-diff> R (C(t) (U0 - ur(t)))' = ur> ou>
ur'(t) + (1+R C'(t))/(r C(t)) ur(t) = C'(t)/C(t) U0.> On peut
bien trouve la solution generale de cet equation, mais c'est>
tres complique. Alors, il faut simplifier.> C(t) = C0 + DC
sin(w t)> avec C0 >> DC. Soit d=DC/C0, alors,> ur'(t) + (1/(R
C0) + (w cos(w t) - sin(w t)/(R C0)) d ) ur(t) > = d w V0
cos(w t) + reste> ou la reste est de l'ordre d^2. Si> ur(t) =
ur0(t) + d ur1(t) + reste> l'equa diff sera> ur0'(t) + 1/(R
C0) ur0(t) > + d ( ur1'(t) + ur0(t)/(R C0) + (w cos(w t) -
sin(w t)/(R C0)) ur1(t))> = d w V0 cos(w t) + reste.> Ca donne
deux equaions approximantes pour ur0(t) et ur1(t):> ur0'(t) +
1/(R C0) ur0(t) = 0> ur1'(t) + ur1(t)/(R C0) + (w cos(w t) -
sin(w t)/(R C0)) ur0(t))> = w V0 cos(w t)> Si ur(0)=0 on a
aussi ur0(0)=0 et ur1(0)=0 (je croix que tu peux> trouver la
solution pour le cas ur(0) > 0 toi-meme).> Ca donne ur0(t) = 0
et> ur1'(t) + ur1(t)/(R C0) = w V0 cos(w t)> avec la solution>
ur1(t) = U0 R^2 C0^2 w^2/(1 + R^2 C0^2 w^2) sin (w t)> + U0 R
C0 w/(1 + R^2 C0^2 w^2) cos(w t)> - U0 R C0 w/(1 + R^2 C0^2
w^2) exp(- t /(R C0))> Finalement, si R C0 w est grand, alors
> R^2 C0^2 w^2/(1 + R^2 C0^2 w^2) ~ 1> et> R C0 w/(1 + R^2
C0^2 w^2) ~ 0> ca donne> ur1(t) ~ U0 sin(w t)> et > ur(t) ~ U0
d sin(w t) = U0 DC/C0 sin(w t).> HTH,> Michael.Merci pour ton
===
RelationshipTo all,Forgive me if I sound naive, I have an
appreciative but highlyabstracted perception of mathematics.
My Question: I have been told that most of mathematics
(geometry, algebra, etc.)is derived, or at least reducible, to
arithmetic. Can arithmetic befurther simplified into boolean
logic?? Which I understand isequivalent or corresponds highly
to propositional logic (correct me ifI'm wrong here).Is this
what Bertand Russell was attempting to accomplish in
PrincipiaMathematica?? Or did he mean logic in a different
===
Find IntegralHI all,Dont know how to start with calculation of
integralTriple Integral from 0 to Inf .Integrant:
1/(1+x^a+y^b+z^c)dxdydzCalculate the integral and find a,b,c
for which integral convergent.(Should be solved using a Gamma
===
functions I think)IhorSubject: Re: dissolving Russel's
dissolves Russel's Paradox.> In this theory, it is possible to
get good theorems of ordinal number.> More description about
the set theory is available on my home page>
http://boat.zero.ad.jp/~zbi74583/another02.htm>> I appreciate
any comment about it.>> .81@ 1.Axiom of Free Class.> .81@>Not
using plain text in newsgroup makes reading difficult for
somebrousers. For example what does ^A@ mean?> A. ma
[A]A[A]B[A]x[A]y A ma x &.81@B ma y & A=B -> x=yWhat does that
mean? A makes for all A, for all B, for all x, for all y A
makes x and ?? B makes y and A = B implies x = yDon't
understand. What do you mean by 'A makes' ?> This axiom means
that any Atsumari makes only one class.> A. el x[A]A[A]B ([A]a
a el A <-> a el B) -> A=B> This axiom means that an Atsumari is
decided by its members.> A.F [E]x{E]B ([A}a a el B <-> F(a)) &
B ma x> This is the Schema of comprehention axioms.>> Where
A,B,C,...are variables for Atsumari, which means collection or
set or pile in Japanese.> a,b,c,...are variables for Class,>
[A] means all, [E] means exist, [E]! means exist only one, ma
means makes,> el means element, -> means then, <-> means
equivalent,> V means or, & means and, Atsu is an abbreviation
of Atsumari> ~ means negation, {a,b,c} means the Atsu which
===
just got GAP, and don't know how to use it yet.>> If anyone
could tell me some of the commands I should use to>> get info
on groups with 16,24,32 elements, (as well as what>> the
commands mean), I would be grateful.>>Sorry, I don't use GAP
myself. I should probably learn it one of>these years.To use
GAP effectively, you will need to devote several hours to
readingthe documentation (on-line) and playing with it. Here
is sample sessionconcerning groups of order 16. Note that the
generators of a group G aretyped in as G.1, G.2,..., but
(confusingly) they are printed by GAP asf1, f2,... gap>
NumberSmallGroups(16);14gap> G:=SmallGroup(16,11);gap> IsAbelian(G);falsegap>
Size(Centre(G));4gap> IsCyclic(Centre(G));falsegap>
NilpotencyClassOfGroup(G);2gap>
Size(CommutatorSubgroup(G,G));2gap> Order(G.3);2gap> G.1 in
Centre(G);falsegap> G.4*G.3*G.2*G.1;f1*f2*f3gap>
Size(AutomorphismGroup(G));64gap>
IsomorphismGroups(G,DihedralGroup(16));failfor i in [1..14] do
> Print(i, ,IsomorphismGroups(SmallGroup(16,i),
DihedralGroup(16)),n);> od;1 fail2 fail3 fail4 fail5 fail6
fail7 Pcgs([ f1, f2, f3, f4 ]) -> [ f1*f2, f1*f4, f3, f4 ]8
fail9 fail10 fail11 fail12 fail13 fail14 fail> Also, GAP
speaks of polycyclic and finitely presented>> groups.>>I
forget what polycyclic means, and somewhat surprisingly
can't>find it in the index of any of the books I have readily
at hand.A polycyclic group is one with a finite series of
normal subgroups in whichthe factor groups are cyclic. An
equivalent definition is a solvable groupin which all
===
Mass is a quantity of matter>> I'm sure that people would
stand in line for blocks to get a signed>copy!!>As a matter of
fact RJ, I have already written a couple, and can't even
give>'em away! As long as the gravy train keeps running,
nobody wants to rock the>gravy boat.Your metaphors are as
===
mixed as your understanding of elementaryphysics.Subject: Re:
to decide the satisfiability of those>formulas (since Primes
is in P) but it should be hard to find a>satisfying
assignment, since the satisfying assignment yields a
factor>(and that is supposed to be hard). Since all SAT solver
I know, yield a>satisfying assignment if the formula is
satisfiable they are not>optimal.> But the general problem of
producing satisfying assignments is at most> linearly harder
than deciding satisfiability, since you can use O(n)>
applications of a SAT decider to generate a satisfying
assignment, if> one exists.Sure. Such a statement holds for
the worstcase complexity of SAT. Butworstcase is not relevant
when we consider those special (and easy todecide) formulas
one gets by transforming the factoring problem.Jochen
===
Linear Algebra proof> I would like to ask your help with
theorem:Given an order n real matrix A, the function which
maps the real> variable lambda into the real variable
det(A-lambda*I) is a polynomial> of degree n> Hi Andrea, Below
are two proofs that came to my mind. In fact the> proofs below
work on any general field (the first proof works on any>
integral domain because any integral domain is a subring of
its field> of fractions).> Sketch of the 1st proof: Every
matrix is upper triangularizable over> its algebraic closure.
With that in mind, with no loss of generality> one may assume
that A is an upper triangular matrix. Now the rest> easily
follows. Done!> Sketch of the 2nd proof: Let A in M_n( F[x])
where F is a field whose> characteristic does not divide n!.
Think of Det(A) as a function of> the rows of A, say a_1, ...,
a_n. We can write Det(A)= det(a_1, ...,> a_n). Now using the
definition of determinant, one can easily verify> that>
det(A)' = det((a_1)', ..., a_n)+ ... + det(a_1, ..., (a_n)').
(*) > Now back to your problem; set f(lambda):= det(A-
lambda*I). In light> of (*), it is not difficult to see that
f^{(n)}(lambda) = (-1)^n*n!> and f^{(k)}(lambda) = 0 for all
lambda in F and k >= n+1 where> f^{(k)} denotes the k-th
derivative of f w.r.t. lambda. Now we are> done in view of the
Taylor formula. qed.> Hope that helps.> Best wishes, Bamdad. >
PS I do not check this place, nor my yahoo account on a
regular basis;> so sorry if I am unable to get back to your
possible reply.I post this reply anyway, in order to thank you
and all the otherusers of the ng: maybe you'll read it later
on. The posts in thisthread were very useful to me. I am now
sure that my proof is ok, butI like Bamdad's second proof
more: the first one he proposes is IMHOincorrect because it
relies on Schur's Decompostion, but the proof ofSchur's
Decomposition (at least, the one I know) uses the fact
thateigenvalues of an nxn real matrix are n, eventually
===
===
Re: 0! = 1> Did someone know a simple demonstration of>> 0! = 1
?>x! = (x+1)! / (x+1)3! = 4!/4= 24/4 = 62! = 3!/3 = 6/3 = 21! =
===
someone know a simple demonstration of>0! = 1 ?n! is equal to
the product of j as j varies from 1 to n by steps of 1.0! is
therefore an empty product, and the empty product evaluates to
BaB>>B -> BB>B -> aBb>B -> bBa>B -> lambda>I doubt there is a
way to prove it, or find a base caseIt's easy to see that B
gives any string with the same numberof a's and b's, so we
just need to show that if a string hasone more a's than b's it
can be written as BaB.Say s is such a string, and say s[n] is
the initial substringof s of length n. Now s = s[length(s)]
has more a's thanb's, so there exists an n such that s[n] has
this property.Let N be the _smallest_ n such that s[N] has
more a'sthan b's. Then s[N-1] must have the same number of a's
and b's and the N-th character must be a,
===
know a simple demonstration of>>0! = 1 ?Well this is actually
the _definition_ of 0!, so it doesn'tneed to be
demonstrated.But one can explain why this definition is a good
idea:For example with this definition the equation (n+1)! =
(n+1)*n!remains true for n = 0. With this definition the
formulafor binomial coefficients C(n,k) in terms of
factorialsgives the right answer for k = 0 and k = n. With
thisdefinition the traditional expression for the Taylorseries
of a function comes out right. Etc.>Tx>>PS Sorry about my
===
someone know a simple demonstration of> 0! = 1 ?n! is the
number of bijections from an n-element set to an n-element
set. Therefore 0! = 1.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
===
this?> Express the following as a single fraction:> 4/3ab -
5/6bc> (1) find a common denominator. The least common
denominator is a> good one to use.What, for example in the
first one, is the lowest common denominator? Is it(3ab)(6bc)?
If it is, do I then multiply that out? If not, what is
===
this?> Express the following as a single fraction:> 4/3ab -
5/6bc> (m^2 + 2)/(m^2 + m) - (m - 2)/m> You do it the same way
you do it for fractions in arithematic.> The general formula is
derived thus> a/b + r/s = as/bs + br/bs = (as + br)/bsYep, I
understand the basic priniciple, but I just don't know how to
===
Re: Express As Single Fraction>>How do I do this?>Express the
following as a single fraction:>4/3ab - 5/6bc>> Multiply the
first term by 2c/2c and the second by a/a.I'm obviously doing
it wrong but that appears to leave two differentdenominators.
===
Express As Single FractionJohn E 
following as a single fraction:>>4/3ab - 5/6bc> (1) find a
common denominator. The least common denominator is a>> good
one to use.>> What, for example in the first one, is the
lowest common denominator?> Is it (3ab)(6bc)? If it is, do I
then multiply that out? If not, what> is it?Factorize all
denominatars as much as possible, including numbers:3ab =
3*a*b6bc = 2*3*b*cThe lowest common denominator is the lowest
common multiple of thedenominators. To get it, take all
factors raised to the greatest exponentthat appears. In this
case,lcm(3ab, 6bc) = 2*3*a*b*c = 6abcThen multiply and divide
each fraction by the quotient between the commondenominator
and its denominadtor:4/(3ab) - 5/(6bc) = 4*(2c)/((3ab)*(2c) -
Larrosa Ca.96estroA Coru.96a
===
Properties of f(x) given the graph of f'(x)What conclusions
about function f(x) can be drawn given the graph ofits
derivate, function f'(x)?That is where the ordinate is f'(x)
===
=?ISO-8859-1?Q?Ces=E0ro-convergence_-_analysis_question?=Hi
all,Regular convergence implies Ces.88ro-convergence. The
inverse statement is not generally true. Can one give
conditions on a sequence to guarantee that when you have a
sequence that is Ces.88ro-converging it is also true that it
===
As Single Fraction> John E  escribi.97
a single fraction:>> 4/3ab - 5/6bc> (1) find a common
denominator. The least common denominator is a>> good one to
use.> What, for example in the first one, is the lowest common
denominator?> Is it (3ab)(6bc)? If it is, do I then multiply
that out? If not, what> is it?>> Factorize all denominatars as
much as possible, including numbers:>> 3ab = 3*a*b>> 6bc =
2*3*b*c>> The lowest common denominator is the lowest common
multiple of the> denominators. To get it, take all factors
raised to the greatest exponent> that appears. In this case,>>
lcm(3ab, 6bc) = 2*3*a*b*c = 6abc>> Then multiply and divide
each fraction by the quotient between the common> denominator
and its denominadtor:>> 4/(3ab) - 5/(6bc) = 4*(2c)/((3ab)*(2c)
- 5*a/((6bc)*a) = (8c - 5a)/(6abc)Ahhh.... I get it now.
===
that? Rudin's texts are known for their concinnity>and
concision, not for detailed proofs.concinnity. What a
the phrase concinnity and concision in an Australianmovie[*].
Good words should not be wasted.* That Eye, The Sky with Peter
Coyote.-- G. A. Edgar
===
joint density function problemQuestion 2.73 p.71:Let f(x,y) =
{1 0<=x<=1, 0<=y<=1; 0 otherwisebe the joint density function
of X and Y. Find the density functionof Z = XY.My work:Instead
of using U and V as in previous problems, I used U and Z.So Let
U=X. Z = XY given. Y=Z/X so Y=Z/U.Setting up the partials for
the Jacobian:dx/du=1 dx/dz=0dx/dz= - dy/dz=1/uJ=1/uSo I
get:g(u,z) = { 1/u 0<=u,z<=1; 0 otherwiseThey want the
marginal:g(z) = Integral(1/u) duI didn't know what parameters
to use. The answer in the book isg(z)={-ln z 0 Can you construct a set E in
[0, 1] s.t. for every open interval I in [0,1] m(I intersect E)
> 0 & m(I intersect E^c) > 0 > m is lebesgue measure > E^c is
the complement of E A doubt. You say the Lebesgue measure. Are
you implying that theconstructed set must be Lebesgue
measurable? Or the problem somehowgives us a notion that, even
if the wording Lebesgue measure isreplaced by Lebesgue outer
measure,it is impossible to find aLebesgue immeasurable set
satisfying the condition as the problemindicates?> This is so
tricky! I was thinking something with the generalized Cantor
set but everything I'm trying isn't working. Any suggestions?
===
 scribbled the following:>>Did
someone know a simple demonstration of>>0! = 1 ?> Well this is
actually the _definition_ of 0!, so it doesn't> need to be
demonstrated.> But one can explain why this definition is a
good idea:> For example with this definition the equation>
(n+1)! = (n+1)*n!> remains true for n = 0. With this
definition the formula> for binomial coefficients C(n,k) in
terms of factorials> gives the right answer for k = 0 and k =
n. With this> definition the traditional expression for the
Taylor> series of a function comes out right. Etc.David
Ullrich explained the mathematical side. I can show you
aphilosophical reason why it's a good idea.The number x! is
intuitively understood as the number of *different*orderings
you can put x different items into.Suppose x=0. Meaning you
don't have any items. For better visualisation,think of
someone holding an empty basket and asking: How
many*different* orderings can I put all the balls in this
basket in?Well, there aren't any balls in that basket, so you
could move ballsaround to your heart's content, and the
ordering of balls inside thebasket would never change.
Therefore there is *ONE* possible orderingof no balls: the one
that the basket already contains.Thus 0! = 1.-- /-- Joona
Palaste (palaste@cc.helsinki.fi) ---------------------------|
Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA
N+++|| http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/I am looking for myself. Have you seen me
===
exerciseHello good people,I am taking a course in probability
and for some reason found theBayes' Rule topic not so easy to
understand.I need your help with showing me the right way to
solve a problem. Idon't need only the solution as this is not
a test. I would reallyappreciate it if someone could show me
the different steps involved inthe solution with some
explanations.I have to solve the following problem:In a
certain region of the country it is known from past
experiencethat the probability of selecting an adult over 40
years of age withcancer is 0.05. If the probability of a
doctor correctly diagnosing aperson with cancer as having the
disease is 0.78 and the probabilityof incorrectly diagnosing a
person without cancer as having thedisease is 0.06, what is the
probability that a person is diagnosed ashaving cancer?Again,
I'd love to see the steps and way to the solution so I can
===
greatly appreciate some comments upon the correctness of the>
assertion about the following equation (1) under the given
conditions.> y = (a^m + b^m)(a^m - b^m) (1)> Conditions: (y,
a, b ) = 1; m is a non-integer > 0; prime p > 3.> Assertion:
If y is a p-th power then both a^m + b^m and a^m - b^m>
separately be p-th power.If you're happy with 1^p being a p-th
power, then the first half of your assertion is unnecessary. If
you want the unit to not bedescribed as a p-th power, then your
assertion is vacuously true.And what algebraic structure are
you working in? I see no reason to not assume you're working
in C, and therefore every number is a p-th
===
on the valuation you are using. If your valuationis the normal
absolute value, then you are correct. But know that foreach
prime there exists a valuation called the p-adic valuatian,
andin these cases your question can't even be
===
body or mass of material substance: It's a misnomer to callit
a weight; because it's really not:which is the mutual thrust
that they exert on each other when they obstructeach other
from simultaneously occupying, and/or passing through the
exactsame place; because they can't: This is the phenomena of
impenetrability.It's how matter interacts, and aggregates into
masses.So any object or body of material substance _is a mass_,
and will exertforce on any other object; body or mass of
material substance when any twoor more of them resist each
other's passage through the same place: Theforce they exert
will either reverse, stop, slow or change each
other'sdirection of motion.All masses of material substance
resting on Earth's surface are constantlybeing restrained from
free falling further toward Earth's center by a mutualforce
called weight; which we all feel as heaviness, and which can
bemeasured on weight-scales: Where weight-scales are
calibrated by standardmasses; whose weight, divided by the
acceleration at which they will freefall is a constant:Thus a
slug is a standard _unit_ of mass; as is a gram and a
kilogram. Theseunits of mass and subdivisions thereof, are
often simply called weights;which they aren't: They are
masses, whose inertia is a ratio of the thrustexerted on,
and/or by them, to the acceleration - and/or deceleration -
thatit causes:Now it's time for all of us to realize that a
customary pound is a unit offorce, and a slug or a subdivision
thereof is a mass: Just as a metricnewton is a unit of force,
===
Re: consecutive composite integers>>Examining a table of
factors and primes, I found that for any sequence>>of
consecutive composite numbers there is always one integer that
has>>a prime factor larger than any other prime factor of any
of the other>>integers. Further, this prime is not raised to
any power.>How does the Further,... apply to this example: 8,
9?>>Perhaps he intended maximal sequence, 8,9,10 has 5 as
max.> But the example he gave was 1500, 1501, ... 1509 which
is not maximal.Also, the largest prime factor of a number in
that range is 751, not503. The conjecture still holds, but not
for the reason given.-- Dave SeamanJudge Yohn's mistakes
revealed in Mumia Abu-Jamal
ruling.> message> 
Maybe the only point is that I fear James being overwhelmed
by> evil.> Hmmm.> I have to ask myself, Why should I care?
James may be the> reincarnation> of Gauss, but is it really
any skin off my nose if he goes> unrecognized?> I've been
worrying about the guy for months (ever since I> realized>
that> he> was not, in fact, a crank, but a genius) and
defending him on> this> newsgroup.> My reward? Laughter and
bile from the peanut gallery. And not> a> word> from>
James himself. A word of advice to Prof. Connes: Don't waste>
your> time> on> James Harris. If he loves being the
solitary genius so much,> let> him> fight> his own damn
battles. He's not worth losing sleep over.>> Well damn it, I'm
losing sleep at least partly because of your> scary> dream!!!>>
Good writing their Jim Ferry, and I have to give you credit
for> that,> but hey, how many years were you ridiculing me,
and now you expect> me> to just go, hey, pal?>> Show me you're
serious and wade in and respond to some of these>
ant-mathematicians in the current battle.>> Prove your
sincerity, and um, keep posting any interesting dreams> you>
have, as I found it interesting puzzling over that one.>>
Exactly! This Jim Ferry now thinks you are a genius, but has
he> defended>> you in your current battle? No!>> I suspect
that he is not *fully* committed to your view of>
mathematics.> That>> is just a feeling, and I suppose I could
be wrong.>The primary focus is the odd definition error in
core.>Once that's accepted for what it is, and most
importantly FIXED, then>it's not about committing to my view
of mathematics, but about showing>your commitment to
mathematics itself.>There is ONE mathematics.>> I, however,
have been trying to find out more about your Object> Ring,>
but I>> feel you are pushing me away.>I have spent some effort
guiding you along at the Mega Foundation>discussion area.>>
Yes, indeed.>> But, I expect you remember that I did ask
you a question that I was> unable> to answer.>> I was
using the notation [a,b,c] to represent an ordered triple of>
complex> numbers. And I was wondering if the ordered triple
[1,2,8] was an> element of> the Object Ring.>> You
lazy!!! You have the definition, figure it out for>
yourself!!!>What amazes me is how often people are willing to
ask someone else to> do> their work for them.>If you're smart
enough, answer your own question.>I'm curious to see if you
can.>I've given the definition for the object ring, so no
am ashamed to say I still cannot figure it out.> Sounds like a
ploy.> You have to understand that your record Clive Tooth is
rather long and> involves some rather...sleazy behavior...like
that attack webpage, and> quite a few negative postings over a
period of YEARS.> That was just... just... boyish high
spirits, James. And you joined in the> fun by threatening to
sue me for libel!! Ah... happy days...Well, it was effective
writing on your part as even when you took thatwebpage down
that other guy copied it for his own webpage, and thensome
other, um, person ran that robot program.And *someone* out
there was at one point trying a somewhat meek denialof service
attack on one of my old websites, or, weirdly enough, theywere
trying to convince me I had thousands of hits per day on
mywebpages!!!That's one of the reasons I was happy with MSN as
I never even knewhow many hits my pages were actually getting
with them, but there wasno worry about them charging me
extra!> You don't get the benefit of the doubt from me but
have to make the> extra effort yourself, so quit being
lazy!!!> Oh... James... You have to admit that I did help you
out, on the Mega board,> with sqrt(i) which you didn't realize
was a complex number. Can't you help> me out just a little with
this one?That was a silly error. Like I said there though I was
reaching as Ireally *wanted* to believe that the object ring
isn't a subset ofcomplex numbers, but didn't have the proof,
so I guess I did what I'vedone many times before and told
myself what I wanted to hear.As for your question, again, quit
being lazy!!!Figure it out yourself, or even go get help from
someone else besidesme, as I've spent enough time with you
already, and given yourhistory, it's not sensible to spend any
more.> And I noticed that you said that Jim Ferry is on the
team! How can I get> on the team if you won't help me when I
am struggling? By the way, is> anybody else on the team? I
think that all the team-members should have> well-defined
roles for the up-coming battle.So far the team as I call it
are people who recognize that my workis indeed correct, and so
far seem to only be very high IQ people.I guess that's not too
surprising. > Could you help me please?> It sounds to me
like you think you have some angle for even more> negativity
and I've given enough time explaining.>Remember mathematics is
a continuing process.>The object ring is fascinating in and of
itself, so you can't expect>me to know all the answers just
because I'm a discoverer.>After all, if it were that easy,
then math research would have ended>long ago with the first
mathematician explaining it all.>> Your friend.>> Clive>Time
will tell.> Yes, as I said, time will tell.> Very
===
interesting point about reduced form is that it so far has not
beenabsolutely necessary except for the dimensional analysis.
That is tosay that a number like: ( - 1 + 2 * 3 # 4 ) can be
worked with even though it is not in its reduced form. So
farnone of the math, including product and graphical analysis
fails towork with non-reduced numbers. The reduced form of the
above numberis: ( + 1 * 2 # 3 ).Is this is the reduction you
are speaking of? The neat thing is thatthe graphical analysis
coherent. I believe that the same will happen with
thetetrahedral. The origin is at the center of a tetrahedron.
the polesgo out from the origin through the points that form
the tetrahedron.This is the symmetrical mapping of four-signed
math in cartesianspace. Any point in RxRxR can be uniquely
defined in four-signed math.Since I don't have the math I
can't say to have proven this. I can seeit though. Making this
assumption and putting the #-pole in the idirection (as in
i,j,k) we get the following partial transform for afour-signed
value x: a = n(x) - ( 1/3 )( m(x) + p(x) + s(x) )where a is the
i component of the three dimensional vector ai + bj +ck.n(x) is
the # (number) component, m the - (minus), p the + (plus), ands
the * (star).Now putting the minus pole in he ij plane and
going in a left-handeddirection we see that the one third
component yields an angle of: pi - arccos( 1 / 3 )where arccos
is the inverse cosine.This should be the angle from the center
to any two corners of thetetrahedron.Please note that I am not
proving this. I'd like to find that angle ina book somewhere or
sharpen up my triginometry skills so that I couldverify this.If
this is true then there should be no problem with a clean
RxRxRtransformation for four-signed math. The same concept
should workupward beyond our sight to five
signed(pi-arccos(1/4)) and beyond.Please keep in mind the
simple sum rule: - x + x * x # x = 0.> I've done some work on
the four-signed math this weekend. It looks like > the key to
making it work is carefully defining a reduced form. Under >
certain definitions of reduced from, n-signed math exists in
R^(n-1), > Under other definitions, n-signed math is
isomorphic to C. I need to > finish some more details, then
I'll let you know what I come up with. I > think that thinking
in terms of n-tuples will make things easier when > dealing
with general n-signed math. Otherwise you may need to switch
to > subscripted signs.I will try to work with your
representation, but I am pretty happywith the symbolics that
are used here since it is more arithmetic. Atsome point I'd
like to go to code. Do you know C or C++?I think four signed
is as far as I'd like to go for now in symbolicformat.I've
written n-signed code for C++ and performed the
Mandelbrotmapping for four-signed in the simplest planes. Some
look like theusual Mandelbrot and some are simple solids. It's
quite likely thatthere are some bugs in my code so don't take
my results too seriously.I can't get too excited about the
Mandelbrot mapping anyhow. I wasvery disappointed when my
three-signed code spat out the usualmandelbrot shape. I was
really hoping for something new. What does theMandelbrot test
mean anyways?It has become apparent to me in writing code that
the symbols used forsign should really include a zero sign
below the minus sign. This zerosign is identical to the
highest sign ( # for four-signed, * forthree-signed, + for
two-signed). Since the reals are symbollicallyfaulty in this
way I will continue on with the symbols I have chosenfor
now.I'm looking forward to your results but I fear that you
===
exercise> In a certain region of the country it is known from
past experience> that the probability of selecting an adult
over 40 years of age with> cancer is 0.05. If the probability
of a doctor correctly diagnosing a> person with cancer as
having the disease is 0.78 and the probability> of incorrectly
diagnosing a person without cancer as having the> disease is
0.06, what is the probability that a person is diagnosed as>
having
cancer?Seehttp://www.acad.sunytccc.edu/instruct/sbrown/stat/
falsepos.htmfor a useful look at problems like this, although
===
spacetime; answer to critic.> I know it is a bit stupid, but>
1> how do you prove that a discrete topology is
metrizable?Write down a metric for it :-)The simplest possible
metric should give the discrete topology ...> 2> X is an set of
all positive integers and>
T={{},{1,2,3,4...},{2,3,4...},{3,4...},{4...},...} Why is
(X,T) not> metrizable?Is it Hausdorff?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
===
question>Hi all,>>Regular convergence implies
Ces.88ro-convergence. The inverse statement is >not generally
true. Can one give conditions on a sequence to guarantee >that
when you have a sequence that is Ces.88ro-converging it is also
true >that it is regularly converging?Yes. The fact that
convergence implies Cesaroo convergence is anabelian theorem,
and the fact that Cesaro convergence plus someextra condition
implies convergence is a tauberian theorem, wherethe extra
condition is a tauberian condition.It's fairly easy to show
that if the sequence s_n is Cesaro convergentand also (s_{n+1}
- s_n)*n -> 0 then the sequence s_n is convergent - if I recall
correctly this is the world's first tauberian theorem,proved by
Tauber. It's true but not nearly as easy to show that youonly
need to assume that (s_{n+1} - s_n)*n is bounded - this is
dueto Littlewood I think.(I'm assuming here you really want to
talk about convergenceof sequences - of course there are
corresponding results forsums of series...)At least I've read
that Littlewood's theorem is not so easy, andthe proofs I've
seen are not quite elementary. Thinking aboutit just now it
seems that one can give a very simple proof ofLittlewood's
theorem. Maybe there's a stupid error below, maybe the theorem
is really not supposed to be that hard,or maybe this is a huge
event in the history of mathematics:Assume that (i) (s_1 + ...
+ s_n) / n -> 0 and(ii) |s_{n+1} - s_n| <= c/nfor all n. We
want to show that s_n -> 0. Suppose not;then there exists d >
0 such that (iii) |s_N| >= d.for infinitely many N. So we may
as well assume(iii') s_N >= dfor infinitely many N. Suppose N
satisfies (iii'). Now forn > N we have s_n >= d - c sum_N^n
1/j ~ d - c log(n/N) >= d/2as long as c log(n/N) < d/2. This
condition is the sameas n < N * exp(d/(2c)) = N * K = M; note
that K > 1.Hence(iv) (s_N + ... + s_M) / M >= (d/2)(M - N)/M =
(d/2)(K - 1)/K = C > 0.But (iv) shows that |s_1 + ... + s_M|/M
does not tendto zero, since the difference between this and
theleft side of (iv) is smaller than |s_1 + ... + s_{N-1}| /
===
Micha************************Subject: Re: Chess/Go/etc:
these have even ever been played,> games such as Chess, Go,
etc, where the game-boards are without> any subdivision?> For
example, each piece's base's shape/size would be significant.
The> pieces would move (or be placed on the gameboard or
removed from> it) under certain rules, but the exact
distance/direction could be> within a continuous interval.>
For instance, in Chess, the pieces (except knight) would not
be> allowed to touch other pieces while moving. (So, either
this game> would be played on a computer so as to be precise
in moving, or the> game would take on aspects of Operation if
the pieces are moved by> hand {no touching pieces to each
other!}.)> I think, with Chess, what is considered a capture
might be ambiguous.> Go, on the other hand, might be very
interesting if played> continuously, especially if the pieces
are different-sized circles.> (Each player would have the same
number of pieces as their opponent> of any given radius, since
bigger pieces would be more valuable> than smaller pieces.)>
Any other games that might be interesting if played this way,>
either on a computer or by hand?Modifying the basic bridging
game (Twixt, Hex, etc.) to be played on acontinuous board
makes a fun game, but would require a computerimplementation
to play seriously (too much room for argument about sizeof
circles, thickness of lines, etc.). Somebody should write
one.The basic idea is that you have a square or rectangular
board. Playerstake turns drawing circles, and trying to make a
chain of their owncircles from one end to the other. If two
circles overlap, the firstone drawn gets the overlapping
space, but the other one can controlwhatever space it has that
is distinct from earlier circles. Player onegoes from top to
bottom and player two goes across. If the board issquare, each
player draws two circles per turn (except the very firstturn).
If rectangular, it is (ideally) carefully calibrated so that
theextra distance required of the first player neutralizes the
===
=?ISO-8859-1?Q?Ces=E0ro-convergence_-_analysis_question?=> Hi
all,> Regular convergence implies Ces.88ro-convergence. The
inverse statement is > not generally true. Can one give
conditions on a sequence to guarantee > that when you have a
sequence that is Ces.88ro-converging it is also true > that it
are called Tauberian conditions after the firsttheorem of this
===
simple demonstration of>>0! = 1 ?>> n! is equal to the product
of j as j varies from 1 to n by steps of 1.> 0! is therefore an
empty product, and the empty product evaluates to 1.>> David
McAnallyYes, and why does the empty product evaluate to 1?
Because that is themultiplicative identity.Likewise, the empty
sum evaluates to 0. You shouldn't change the value of anumber
===
Graph, Four Color Theorem Visiting Assistant Professor at the
University of Montana.>>The only minimal counter-example to
the FCT is K5!> No, K5 is NOT a counterexample to the Four
Color Theorem, because the>> 4 color theorem states that any
->planar<- graph can be colored with>> at most 4 colors in
such a way that no two adjacent vertices share the>> same
color.>>The conjecture that there exists a 5-chroma graph may
be recolored to>4-chroma is false.> There is no such
conjecture.>Let H be any subgraph of G, where G has n
vertices and H has n-1>vertices. Then, the description of H
seems to imply that the deletion>of 'any' vertex from G will
make chi(H)<=4.> This is true if G is a minimal
counterexample for the 4 Color Theorem.>>But this
interpretation is generally false and is valid only
for>n=5!!!> The triple exclamation points make you look like
a raving loon. So>> start by removing them.>>Point taken, thank
you. Could you explain why?Can I explain why the triple
exclamation points make you look like araving loon? Because
they do. It makes it seem like you are jumping upand down,
yelling, spitting, and foaming at the mouth. That's themental
image they conjure up.>> Then note that the original argument
started by ->assuming<- that the>> FCT is ->false<-, from
which we deduce that if this is the case, then>> among them,
there is one with the least number of vertices. Call n the>>
number of vertices of this HYPOTHETICAL counterexample. Then,
by the>> definition of n, any graph with fewer than n vertices
must be>> 4-colorable. In particular, if you took this
HYPOTHETICAL example G,>> and removed one vertex, then the
resulting graph would be 4-colorable.> What exactly are you
having trouble understanding about the above>> argument? Try
to answer without using a ->single<- exclamation point.>>I
understand the argument perfectly. Then why did you think
somebody was claiming the Four Color Theoremwas ->false<-?>I
have given the problem some>thought and I have concluded that
HYPOTHETICAL G is impossible.Good for you. > No>graph meets
all three criteria; ie, G is 5-chroma, G is planar, H
is>4-chroma.Good for you. But your argument seems to be no
such G can exist,because then G would be K5, and that does not
even begin to makesense. G cannot be K5 if it is assumed to be
planar.Indeed, the proof of the Four Color Theorem rests on
showing thatthere does not exist any graph G which requires 5
colors, is planar,and such that the removal of any vertex
results in a graph which canbe colored with only 4 colors. But
you have not given any coherentargument to establish this
proposition that I can see anywhere. Allyou have done is yell
like a loon that G would be K5!, which
about what I accept as reality. --- Calvin (Calvin and
===
Four Color Theorem|I understand the argument perfectly. I have
given the problem some|thought and I have concluded that
HYPOTHETICAL G is impossible. No|graph meets all three
criteria; ie, G is 5-chroma, G is planar, H is|4-chroma.I
assume by this last clause you mean that all the graphs one
obtains bydeleting a single vertex from G are 4-chromatic.This
conclusion is equivalent to the four color theorem. If the four
colortheorem is true, then no planar graph has chromatic number
5. On the otherhand, on the assumption that your conclusion
above is correct, the fourcolor theorem follows by induction
on the number of vertices. Once we'veshown it's true for
planar graphs of up to n vertices, then it also mustbe true
for planar graphs of n+1 vertices, since whatever graph H is,
it'salready been shown to have chromatic number <=4 (and its
chromatic numberdiffers from that of G by at most 1).So the
only way you can reach such a conclusion is by an argument
which isat most one short paragraph shorter than a proof of
the four color theorem.I would assume that you're just relying
upon the existing proof, except thatit wouldn't usually take
some thought to conclude that a 5-chromaticplanar graph having
a certain kind of subgraph doesn't exist, given thatno
5-chromatic planar graph exists at all.I just hate to see
someone go away still confused, so I hope your clarityon the
argument has reached the point of recognizing that this
conclusionyou state above is very far from trivial, without
taking the proof of thefour color theorem for granted. If
there's some simple way to showsuch a G doesn't exist, a
number of smart people have failed tosee it despite working
===
one-way hash function>> I need to find an algorithm that can
produce a unique non-predictable> 12>> digit (0-9) number for
any given 12 digit number. This is to be used to>> create a
unique barcode on a ticket that cannot be predicted. It is
not>> required that the original seed number be computed from
the resulting>> barcode, so some form of one-way hashing
function would be acceptable.>> Any help in this problem would
be appreciated.>> The simplest way is to encrypt the first
number using AES or> 3DES. You will have to convert the result
from binary to decimal.> Any extra digits can be thrown away.
Change the key variable> regularly and keep the old ones
secret.>> But shouldn't the bar codes be unique?>> Your
grade ciphers like AES there will be very few> duplicates. I
assume that you throw away extra digits bycomputing the block
cipher output modulo 10^12.When the block cipher is
cryptographically strongthis should give a random map, in
other wordsyou can regard the output as a random number
withrange [0..10^12).So when you take the entire set of
outputs for all10^12 inputs this set should behave like a set
of10^12 random numbers that are independently drawnfrom a
uniform distribution.Now let's compute the probability that
one specificnumber is not present in this output set,
thisprobability is (1 - 1/n)^n and when n is largethis is
approximately equal to 1/e = 0.368 .So about 36% of the
possible outputs never occur,this means that at least 36% of
the inputs willlead to a duplicate output value.> The simplest
way to ensure that the bar codes are> unique is to add a prime
number to the previous> value. Lap round when you get to the
top (or a prime> number near the top). Start at a weird
value.Could you be more precise, I am afraid that Idon't
understand your algorithm.Anyhow, I can't see that your
algorithm wouldbe unbiased, or how it would prevent
===
with a Bayes Rule exercise>In a certain region of the country
it is known from past experience>that the probability of
selecting an adult over 40 years of age with>cancer is 0.05.
If the probability of a doctor correctly diagnosing a>person
with cancer as having the disease is 0.78 and the
probability>of incorrectly diagnosing a person without cancer
as having the>disease is 0.06, what is the probability that a
person is diagnosed as>having cancer?A = person has cancer =>
P(A) = 0.05Ac = person is healthy => P(Ac) = 0.95B = diagnosis
is positiveBc = diagnosis is negativeP(B | A) = ???, P(B | Ac)
= ???Now, P(B | A) = P(A | B) * P(B) / P(A) => P(B) = P(A) *
P(A | B)Bayes' formula gives us:P(A | B) = P(A)*P(B | A) /
(P(A)*P(B | A) + P(Ac)*P(B | Ac))so the final result can be
calculated by plugging the second equationinto the first
===
Hello good people,>> I am taking a course in probability and
for some reason found the> Bayes' Rule topic not so easy to
understand.> I need your help with showing me the right way to
solve a problem. I> don't need only the solution as this is not
a test. I would really> appreciate it if someone could show me
the different steps involved in> the solution with some
explanations.> I have to solve the following problem:>> In a
certain region of the country it is known from past
experience> that the probability of selecting an adult over 40
years of age with> cancer is 0.05. If the probability of a
doctor correctly diagnosing a> person with cancer as having
the disease is 0.78 and the probability> of incorrectly
diagnosing a person without cancer as having the> disease is
0.06, what is the probability that a person is diagnosed as>
having cancer?>> Again, I'd love to see the steps and way to
the solution so I can do> similar exercises by myself.>>
===
Boolean Algebra - Arithmetic Relationship> ... Can arithmetic
be further simplified into boolean logic??Yes. Computers - or
rather those who design computers, are quite good at
===
 scribbled the following:>Did
someone know a simple demonstration of>0! = 1 ?>> n! is equal
to the product of j as j varies from 1 to n by steps of 1.>>
0! is therefore an empty product, and the empty product
evaluates to 1.> Yes, and why does the empty product evaluate
to 1? Because that is the> multiplicative identity.> Likewise,
the empty sum evaluates to 0. You shouldn't change the value of
a> number by not adding anything to it.If I know my algebra,
then an identity for an operation (let's callit #) is such an
i, that for all x, x#i=x. But there is another concepttoo:
such a j, that for all x, x#j=j. For multiplication, this is
0.For addition, there doesn't seem to be one. What is this
concept called?-- /-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------| Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++||
http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/As we all know, the hardware for the PC is great,
===
Properties of f(x) given the graph of f'(x)You can tell where
its max/min are. You can also tell where f(x) isincreasing and
decreasing. All functions f(x) will be the same with
theexception that they will differ by a constant.Lurch> What
conclusions about function f(x) can be drawn given the graph
of> its derivate, function f'(x)?>> That is where the ordinate
===
2*cos(2^k*arccos(x/2)) irreducible for each > positive integer
k? If so, how is it proved?I received email from Joe Silverman
answering this question and giving a good deal of additional
information on the factorization of Chebeshev polynomials.
--------------------------------------------------------------
-Comments from Joe Silverman:The n'th Chebeshev polynomial (up
to factors of 2 used by different authors)is the polynomial
F_n(x) with the property that F_n(2*cos(t)) = 2*cos(n*t).So if
we put t = arccos(x/2), then we get F_n(x) =
2*cos(n*arccos(x/2)).This is your formula.If we write cos(t) =
(e^{it} + e^{-it})/2, and for ease of notation,let z = e^{it},
then we get F_n(z+z^{-1}) = z^n + z^{-n}.This is another
characterization of F_n.The roots of F_n(x), from your formula
or from this alternative, are x = 2*cos(pi*(j+1/2)/n) for j =
0,1,2,...,n.Notice this can also be written as x = e^{it} +
e^{-it} with t = (pi*(j+1/2))/n.it's easy to see that the
splitting field of F_n(x) is the real subfieldof the 4n'th
cyclotomic field. In other words, the roots of F_n(x)generate
the field K_{4n} = Q(e^{pi*i/2*n}+e^{-pi*i/2*n}).The degree of
this field is [K_n:Q] = phi(4*n)/2, where phi(n) is Euler'sphi
function. On the other hand, the degree of F_n is n. So the
conclusionis the following:Proposition: F_n(x) is irreducible
over Q if and only if phi(4*n) = 2*n.The case you're asking
about is n = 2^k, and indeed phi(4*2^k) = phi(2^{k+2}) =
2^{k+1} = 2*2^k,so your polynomials are irreducible. Further,
I think this is probably theonly case that phi(4*n) = 2*n, so
the only case that F_n is irreducible.Hope this is of some
help. Feel free to post this, if you
want.------------------------------------------------------It
jbuddenh@REMOVEtexas.net to address bar and edit out
===
object; body or mass of material substance: It's a misnomer to
call>it a weight; because it's really not:>>which is the mutual
thrust that they exert on each other when they obstruct>each
other from simultaneously occupying, and/or passing through
the exact>same place; because they can't: This is the
phenomena of impenetrability.>>It's how matter interacts, and
aggregates into masses.>>So any object or body of material
substance _is a mass_, and will exert>force on any other
object; body or mass of material substance when any two>or
more of them resist each other's passage through the same
place: The>force they exert will either reverse, stop, slow or
change each other's>direction of motion.>>All masses of
material substance resting on Earth's surface are
constantly>being restrained from free falling further toward
Earth's center by a mutual>force called weight; which we all
feel as heaviness, and which can be>measured on weight-scales:
Where weight-scales are calibrated by standard>masses; whose
weight, divided by the acceleration at which they will
free>fall is a constant:>>Thus a slug is a standard _unit_ of
mass; A slug is a little used 20th century invention which
exists only inone specialized subsystem of mechanical
units--and a unit whichdoesn't even have an official
definition.>as is a gram and a kilogram. These>units of mass
and subdivisions thereof, are often simply called
weights;>which they aren't: They are masses, whose inertia is
a ratio of the thrust>exerted on, and/or by them, to the
acceleration - and/or deceleration - that>it causes:>>Now it's
time for all of us to realize that a customary pound is a unit
of>force,You know better, Donald. Pounds everywhere have
always been units ofmass. Pounds force are such a recent
bastardization that they areuniquely identified by that
name.Back in 1959, the national standards laboratories of the
United Statesof America, Canada, the United Kingdom of Great
Britain and NorthernIreland, the Union of South Africa,
Australia, and New Zealand gottogether and agreed on a common
definition of the most commonly usedpound, the avoirdupois
pound. They defined it as a unit of massexactly equal to
0.45359237 kg.Of course, you already knew that, Dishonest Don.
This is for thebenefit of anybody else who hadn't been paying
attention.> and a slug or a subdivision thereof is a
mass:Always. But they only exist in one specialized subsystem
ofmechanical units, a system which includes none of those
pints you areso fond of, not U.S. liquid pints, not U.S. dry
pints, not imperialpints, and not those ill-defined Dense
Donny Pints either.So is a hyl, another obsolete unit of mass
which like the slug wasonly used in calculations. Look it up;
it isn't so hard to find withinternet search engines. What is
the base unit of force in the systemin which the hyl (aka the
TME from a German acronym, or the mug, orthe metric slug) is
the derived unit of mass?> Just as a metric>newton is a unit
of force, And so is the poundal in English units.>and a
kilogram or a subdivision thereof is a mass!Kilograms force
were quite proper and legitimate units before theintroduction
of the International System of Units in 1960, and westill see
many vestiges of their use. For example, I have a torquewrench
in meter kilograms and they are still readily available inthose
units.Good grief, those kilograms force even have another name,
thekilopond, a name more common in Europe than it ever was in
America.Gene
Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/
Gentlemen of the jury, Chicolini here may look like an idiot,
and sound like an idiot, but don't let that fool you: He
===
know my algebra, then an identity for an operation (let's
call> it #) is such an i, that for all x, x#i=x. But there is
another concept> too: such a j, that for all x, x#j=j. For
multiplication, this is 0.> For addition, there doesn't seem
to be one. What is this concept called?It's called a
===
talk@vixen.cso.uiuc.edu:>> So if you think of the three hands
as rotating unit vectors,>> their sum will never be zero. So
here's another problem. Assign>> angular velocities to the
three unit vectors so that the mimimum>> length of their sum
is as large as possible.> There are various answers depending
on what additional restrictions> you apply. For j = 1,2,3, let
Case j allow up to j hands to have the> same angular velocity.
Define subcase (a) by requiring all hands to> meet at some
time, and subcase (b) by not requiring this.> Case 1a: maxmin
length is sqrt((47 - 14 sqrt(7))/27) ~= 0.607346.> The second
hand moves three times as fast as the minute hand in the>
frame of reference of the hour hand.Yeah, that's the
interesting case, and that's the right answer.We can restate
it as trying to evaluate:sup over {a_1 fun by threatening to sue me for libel!! Ah...
happy days...>> Well, it was effective writing on your part as
even when you took that> webpage down that other guy copied it
for his own webpage, and then> some other, um, person ran that
robot program.>> And *someone* out there was at one point
trying a somewhat meek denial> of service attack on one of my
old websites, or, weirdly enough, they> were trying to
convince me I had thousands of hits per day on my>
webpages!!!>> That's one of the reasons I was happy with MSN
as I never even knew> how many hits my pages were actually
getting with them, but there was> no worry about them charging
me extra!>> You don't get the benefit of the doubt from me
but have to make the> extra effort yourself, so quit being
lazy!!!> Oh... James... You have to admit that I did help you
out, on the Megaboard,> with sqrt(i) which you didn't realize
was a complex number. Can't youhelp> me out just a little with
this one?>> That was a silly error. Like I said there though I
was reaching as I> really *wanted* to believe that the object
ring isn't a subset of> complex numbers, but didn't have the
proof, so I guess I did what I've> done many times before and
told myself what I wanted to hear.>> As for your question,
again, quit being lazy!!!>> Figure it out yourself, or even go
get help from someone else besides> me, as I've spent enough
time with you already, and given your> history, it's not
sensible to spend any more.I am not sure who to turn to,
James.I would guess that neither Arturo nor Nora would be able
to help me withthis. They just don't seem to have grasped the
basic concepts of the ObjectRing.> And I noticed that you said
that Jim Ferry is on the team! How can Iget> on the team if you
won't help me when I am struggling? By the way, is> anybody
else on the team? I think that all the team-members should
have> well-defined roles for the up-coming battle.>> So far
the team as I call it are people who recognize that my work>
is indeed correct, and so far seem to only be very high IQ
people.James, I know you have been in the Army. Perhaps we
could organise the Teamalong Army lines... with ranks! You
could be the General, of course...and... um... I could be a
colonel, say. And that Jim Ferry could be aprivate.Talking
about Jim Ferry, I wonder if he could help me with my question
aboutthe ordered triple of complex numbers. I don't know where
===
Re: Can you help me?>
>I am stuck on a proof. I have
gotten a series down to where I need to>prove the following:>
lim c^1/n = 1 where c>0>n -> inf.>Do you have any
pointers?>
>Can't you take | nth-root(c) - 1 | and times
it by (nth-root(c) +1)/(nth-root(c) + 1) ? Work with that. I
===
Re: Properties of f(x) given the graph of f'(x)> What
conclusions about function f(x) can be drawn given the graph
of> its derivate, function f'(x)?>> That is where the ordinate
is f'(x) and abscissa is x.This question becomes reasonably
interesting if it is read asimplying simply that one is given
the graph of f' in the waythat graphs are usually given,
namely, as rather fuzzy drawings.(After all, if the real
intention of the question was to askwhat conclusions about
f(x) can be drawn given f'(x)?, whybring graphs in at all?)
When one is given a fuzzy graph, it might not be at allclear
where (if it all all) it touches, or crosses, the(equally
fuzzy) x-axis, nor where (if at all) the functionit represents
is increasing or decreasing.Here's a precise version of the
question: what conclusionsabout a continuously differentiable
function f on a closedinterval [a,b] can be drawn, given a
continuous functiong on [a,b], an epsilon > 0, and the
knowledge that |f'(x)-g(x)| is at most epsilon for x in
===
But my other interest is homebrewing, > What's your favorite
style? I like to brew IPA's. But, so far, my best> brew is a
lagered Bohemian Pilsner. Smooooth. I'm having a little>
problem with mashing ... poor yield. Math + Homebrew = FUN.I
probably to an IPA every other batch, because they're
readyearly and I seem to run out of beer so fast. I often
makethe Sister Star variety because I like heavily hopped
versions.But my favorites are double bocks, even though they
have toage forever. The thing that increased my mash
efficiency the most was doinga mash-out. (In a 10-gallon
batch) just before sparging I drainoff 3 gallons of wort,
bring to almost boil, and then add it backto the mash. This
brings the temp up into the 165 range and thesparging rinses
out much more fermentables. (But you gotta' becareful not to
go higher than 170.) But maybe you do that already.The other
thing that helped me a lot was paying attention to pHand
adding the right amount of calcium carbonate to the mash
water.Gotta keep those enzymes happy.rec.crafts.brewing is a
great group. I've nicknamed my liver Kenny,by the
===
for your question, again, quit being lazy!!!> Figure it out
yourself, or even go get help from someone else besides> me,
as I've spent enough time with you already, and given your>
history, it's not sensible to spend any more.Translation: You
haven't the slightest idea how to answer the question,so
you're stalling and hoping someone else will do it for you. Of
course,you'd still be quite happy to take the credit for the
answer.-- Wayne Brown | When your tail's in a crack, you
improvisefwbrown@bellsouth.net | if you're good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 --
===
extension proof> Well, it happens a lot that the _existence_
of the extension is > used to define measures, by first
defining them on a (non-sigma)> field... when you do that you
need the uniqueness to know that > you've defined _a_
measure.>In other words, is understanding of it>important in
terms of understanding new material down the road? > Other
people may have different opinions: If you're just learning>
measure theory my advice would be to skim through this part
as> quickly as possible and concentrate on the stuff coming up
that> gets used in applications of measures, as opposed to
constructions> of measures - if it turns out you get into
something where this is> important there will be plenty of
time to go back to the details.>Your>help is always
appreciated.>> ************************>To everyone who
===
 scribbled the following:>> If I
know my algebra, then an identity for an operation (let's
call>> it #) is such an i, that for all x, x#i=x. But there is
another concept>> too: such a j, that for all x, x#j=j. For
multiplication, this is 0.>> For addition, there doesn't seem
to be one. What is this concept called?> It's called a zero.So
intuitively, if #' is a repetitive #, which means that x#'i is
x#xperformed (i-1) times in succession, then the zero of #' is
theidentity of #? For example multiplication is repetitive
addition. 0 isthe additive identity and the multiplicative
zero. If we switch thedefinition of a zero around so that it
means that for all x, j#x=j, thenwe get that exponentation is
repetitive multiplication, and that 1 isthe multiplicative
identity and the exponentative zero.Is this making any
sense?-- /-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------| Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++||
http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/My absolute aspect is probably... - Mato
===
guys, is that all masses have inertia! That is it requiresa
net impulse - the product of a net force [f] and its duration
[t] - tochange a body's present velocity [vi], to some other
velocity [vt]; duringwhich period of time, the body is
displaced a distance [s]: The ratio ofthis impulse [ft], to
the time rate of displacement [s/t = (vt-vi)] that itcauses is
a constant: That is ft/(s/t) = ft/(vt-vi); which can be
writtenmore concisely as f/t/s = ft/ (vt-vi)! Isn't _that_
===
XMyVFeZFge7G89fUpJ2h01-WQJOrIUZ4WGuO+BOtk+4uSH-Iah7-U6James
2*cos(2^k*arccos(x/2)) irreducible for each> positive integer
k? If so, how is it proved?> I received email from Joe
Silverman answering this question> and giving a good deal of
additional information on the> factorization of Chebeshev
polynomials.>
--------------------------------------------------------------
-> Comments from Joe Silverman:> The n'th Chebeshev polynomial
(up to factors of 2 used by different authors)> is the
polynomial F_n(x) with the property that> F_n(2*cos(t)) =
2*cos(n*t).> So if we put t = arccos(x/2), then we get> F_n(x)
= 2*cos(n*arccos(x/2)).> This is your formula.> If we write
cos(t) = (e^{it} + e^{-it})/2, and for ease of notation,> let
z = e^{it}, then we get> F_n(z+z^{-1}) = z^n + z^{-n}.> This
is another characterization of F_n.> The roots of F_n(x), from
your formula or from this alternative, are> x =
2*cos(pi*(j+1/2)/n) for j = 0,1,2,...,n.> Notice this can also
be written as> x = e^{it} + e^{-it} with t = (pi*(j+1/2))/n.>
it's easy to see that the splitting field of F_n(x) is the
real subfield> of the 4n'th cyclotomic field. In other words,
the roots of F_n(x)> generate the field> K_{4n} =
Q(e^{pi*i/2*n}+e^{-pi*i/2*n}).> The degree of this field is
[K_n:Q] = phi(4*n)/2, where phi(n) is Euler's> phi function.
On the other hand, the degree of F_n is n. So the conclusion>
is the following:> Proposition: F_n(x) is irreducible over Q
if and only if phi(4*n) = 2*n.> The case you're asking about
is n = 2^k, and indeed> phi(4*2^k) = phi(2^{k+2}) = 2^{k+1} =
2*2^k,> so your polynomials are irreducible. Further, I think
this is probably the> only case that phi(4*n) = 2*n, so the
only case that F_n is irreducible.> Hope this is of some help.
Feel free to post this, if you want.by essentially the same
argument you see that the odd chebyshev polynomials ofprime
order are irreducible if you eliminate the trivial factor x
(zero x=0)sincerelyKlaus>
------------------------------------------------------> It
jbuddenh@REMOVEtexas.net to address bar and edit out
===
talk > sup over {a_1 Where the a_k and t are real. >> Let f(n)
be the maxmin length for n hands. Then f(1)=1, f(2)=0,> and
f(3)=sqrt((47 - 14 sqrt(7))/27).> Some questions:> 1. Is f(n)
always attainable by specific values of the a_i?> (And if so,
is it attainable by integer values as in the case> of f(3)?)>
2. Is f(n) monotonically increasing?> 3. What are some other
values for f(n)? (4- or 5-handed clock.)> 4. f(n) has a
trivial upper bound of sqrt(n+1), but this is pretty> gross.
How about a substantial improvement?> I have a particular
interest in question 1.Some comments:* It seems like a good
idea to normalize a_1 = 0 (i.e., replace a_k with a_k - a_1
for all k).* The hands all meet at t=0, z=1. Define T to be
the minimum positive value of t such that all values of
exp(a_k*t*i) are identical. With the normalization a_1 = 0,
this is equivalent to requiring that each a_k*t is an integer
multiple of 2 pi. This minimum can fail to exist in two ways:
(1) because no t exists to make the values identical, in which
case let T = infinity, and (2) because a_k = 0 for all k, in
which case let T = 0. Let's assume that n > 1 so that the
latter case never occurs (because the a_k are unequal).* If T
= infinity, the function |sum_{k=1}^{n} exp(ak*t*i)| attains
all values between 0 and n, so this case can be ignored.
(Exercise for the reader.)* Now we now apply a further
normalization (for finite, positive T): let T = 2 pi. I.e.,
replace each a_k by (2 pi/T) * a_k (making the complete
normalization to replace each a_k by (2 pi/T) * (a_k - a_1)).
This makes the a_k's co-prime integers (meaning that the GCD
of all of them together is 1, not that they're pairwise
co-prime).* So the candidate a_k's are n-tuples like
(0,3,7,8,11): i.e., n-tuples of increasing integers beginning
with 0. I would now apply one final normalization. Replacing
each a_k by a_n - a_k results in the same solution (i.e.,
frame of reference is now the fastest hand, and we look at the
clock in the mirror). In the example, this yields (0,3,4,8,11).
The final normalization is to take the smaller of the two (in
lexicographical order), which is (0,3,4,8,11) in this case.*
Now the key to proving things about this problem is to obtain
a bound on the minimal sum, showing that for big (a_k) it's
small, leaving only a few cases to check. In particular, the
answer to 1 should be yes, because the maximal minimum sum
will be obtained for small (a_k), not approached by a sequence
of large (a_k).* Okay, that's a lot of hot air, and no real
proofs, but I it's how I'd approach the problem if I wanted to
prove things.-- | Jim Ferry | Center for Simulation
|+------------------------------------+ of Advanced Rockets ||
http://www.uiuc.edu/ph/www/jferry/ +------------------------+|
jferry@[delete_this]uiuc.edu | University of Illinois
===
arcsin(x)=x+1/2/3*x^3+...and using an integration operation
show well-known sum:oo 1S ------------------- = (pi^2)/6 n=1
===
Visiting Assistant Professor at the University of Montana.>You
can tell where its max/min are. You can also tell where f(x)
is>increasing and decreasing. And also where f is concave up,
concave down, and where its points ofinflection are.f is
increasing where f' is positive; f is decreasing where f'
isnegative. f has critical points where f' does not exist or
is equal tozero, which can usually be figured out from the
graph of f'.Also, f is concave up where f' is ->increasing<-,
and f is concavedown where f' is ->decreasing<-. The
inflection points of f are wheref' has relative
about what I accept as reality. --- Calvin (Calvin and
===
calculate the total coverage area of a few circles?> ...Randy
Poe...:>In my simulation, N circles with the same radius r are
randomly>placed. Let P_i denote the center of circle i. For any
i, p_i lies>within the coverage range of at least one other
clicle, i.e. at least>one other circle contains p_i. How to
calculate the total coverage>area of the N overlaped circle?
The method should be easy to be>implemented by programming for
simulation....>
http://mathworld.wolfram.com/Circle-CircleIntersection.html>
However, your problem might involve a large number of these,>
and in addition you need to calculate overlaps of 3 circles,>
4 circles, etc.> Edelsbrunner has inclusion-exclusion formulas
that depend only on> overlaps of at most three circles: The
union of balls and its dual> shape,
http://portal.acm.org/citation.cfm?id=161139> [...] The number
of circles is about 10. I think that the calculation > time of
Monte Carlo integration might be too long. Is it possible to >
use numerical integration [...] need to consider the sunk parts
while> integrating the area [...] The sunk parts presumably are
accounted for by the inclusion-exclusionformulae of
Edelsbrunner. Here is an alternative plane-sweep method
thatwould be ok for ~ 10 circles (its complexity is probably
O(N^3) as stated):Make a list of the N*(N-1) circle
intersection points and the 2Npoints p_i with x +/- r that are
not in the interior of anothercircle, and sort into ascending
order by x coordinate, then sum theareas of vertical zones
bounded by these critical points. Thereare no arc
intersections within a zone. A zone may contain
disjointsegments but each segment is bounded above and below
by an arc of acircle, and left and right by straight lines.For
example, if we have 3 circles of radius 5 and centers at
(5,12),(8,9), and (9,5), the event-points list for the plane
sweep is: 0.0 12.0 3.3 7.3 4.0 5.0 4.1 5.9 9.7 13.7 12.9 8.1
13.0 9.0 14.0 5.0The following edges or intersections are
interior and not relevant: 3.0 9.0 4.4 7.0 9.6 10.0 10.0
===
sure that people would stand in line for blocks to get a
signed>copy!!>As a matter of fact RJ, I have already written a
couple, and can't evengive>'em away! As long as the gravy train
keeps running, nobody wants to rockthe>gravy boat.>> Your
metaphors are as mixed as your understanding of elementary>
physics.I think his metaphors are funny!But the idea is that
THE ESTABLISHMENT (shiver, shiver)controls the TROOTH (tm) and
suppresses publication of HERESY, especiallybecause they're
making lotsa money is certainly neither novel nor
humorous.You'd think that he could at least GIVE them away to
giving this some serious study.(Nudge,nudge, wink, wink) Bob
===
[snipped] ...>... [I think Ruby-Lackov can >tolerate a
small amount of bias in f. If not, I'm sure someone will post
>another suggestion.]> Depends what you mean by tolerate.
The security> proofs for Luby-Rackov certainly don't hold up
if> there's any knowledge at all of the f() functions. I was
thinking that, with sample size limited to 4, the slightly
biased set of functions from which the four particular
functions used are drawn could not be distinguished from the
unbiased set of functions from which the four function should
have been drawn. Given hash output words w distributed
uniformly in 0..2**32-1, one way the code could avoid bias is
by using w%1000000 when w>=967296 (which is 2**32%1000000) and
repeating with a new w value otherwise. The probability of
going on to a second w value is about 1/4440. The probability
of exhausting all four words available from an MD5 hash even
once in 10**12 barcodes 1/338. Code that just unconditionally
used the first w%1000000 would, on average, in 1000000 trials,
produce 999775 output values uniformly distributed in 0..999999
and 225 output values uniformly distributed in 0..967295. I
conjecture that the number of plaintext-ciphertext pairs that
an attacker would need to take advantage of that bias would
exceed the number of barcodes the OP intends to print. In any
event, I agree that qualms about the bias can be removed by
using the uniform-in-[0..N-1] algorithms you mention. I think
Benjamin posted good code for one.> While on that subject,
I'll also point out that> the proofs require four
*independent* f()> functions, not one that is reused. You can>
simulate this with f_n = hash(key, n, data) for n = 1..4. My
mistake.> That said, for such a small data input, you'd>
probably be safe ignoring those two nits. Problems> are more
likely to surface somewhere else.>>You would have to be
careful in the selection of your hash function.>>All standard
hash functions have 2**n different outputs, and>>I don't know
any hash function that produces 10**6 outputs. >>An example of
a bad hash function would be to take the first 20 bits>>mod
1000000 of a standard cryptographical hash, because this
hash>>function is extremely biased (some values occur twice as
often as the>>others). When you use a larger number of bits the
bias is reduced.> Yes, there's a regular subject here about how
to> get unbiased uniform random in [0..N-1] given a>
pseudorandom bitstream (such as generated by a> hash
function), avoiding this bias that> creeps in when you least
===
you. If we allow the same numbers such as {2,2} ?> Suppose we
take x numbers out of y numbers in a decreasing sequence.>
Say, take 2 numbers from {1,2,3} and arrange as {2,1}. What is
the> number of possible combinations ?> y choose x (assuming
===
come in!!!!>And I noticed that you said that Jim Ferry is on
the team! How can I get>on the team if you won't help me when
I am struggling? By the way, is>anybody else on the team? I
think that all the team-members should have>well-defined
roles for the up-coming battle.Sarcasm. Sigh.>>So far the team
as I call it are people who recognize that my work>>is indeed
correct, and so far seem to only be very high IQ people.I must
reiterate: I don't recognize that your work is indeed correct.
(Maybemy IQ is not high enough...) My position is that it's
not absurd to think itcould be correct rather than to assume,
arrogantly, that if I don't understandit, it must be wrong.
I'm sickened by this carnival of dogs driven wild by rawmeat,
and I'm outraged that James is being treated like raw meat.
James Harrisis a child of God. We are all children of God.
Where did it all go so wrong?I am supporting James because if
he is correct (and I'm entirely unable to assigna number to
the probability of that), it will have a profound impact on
society.For the better, despite short-term chaos.> James, I
know you have been in the Army. Perhaps we could organise the
Team> along Army lines... with ranks! You could be the
General, of course...> and... um... I could be a colonel, say.
And that Jim Ferry could be a> private.I'd prefer that when you
dream about James's privates, my name not spring tomind. James
would probably prefer people not to think about his privates
atall, especially people named Tooth. Or Ferry, for that
matter.But enough silliness. I don't take orders from James,
nor does James give themat all, as far as I know. At first I
didn't realize what my role in all thiswas to be, but it's
becoming clear to me now. James needs a plan, and forreasons
not clear to me, I have been receiving a plan. It keeps me
awake atnight. It distracts me from my work. It is vast and
beautiful, and I worrythat I'll never get it typed up, or
worded right. I also worry that Jameswill simply reject it out
of hand because I've been such an asshole all theseyears. In
fact, why would his plan even come to me? Why wouldn't it
justcome to him? That would seem more direct. Maybe it's too
much for one personto do the math and receive the plan. It is,
after all, a vast plan. Alreadyit's wearing me out, and I've
only typed up some notes, sketching the broadthemes. I want to
just chuck it all -- I don't owe James anything, not really--
but there's a reason I can't, a reason that has nothing to do
with James,or his math. I'd sort of like to say, but I can
imagine the kind of reactionI'd get on *this* newsgroup.
Anyway, better to save it for the final product.Here's to the
revolution!> Talking about Jim Ferry, I wonder if he could
help me with my question about> the ordered triple of complex
numbers. I don't know where to turn...Oh poor Clive. It's
really burning you up, isn't it? As James said, figure itout
yourself, or just forget about it. Stop pretending to be
James's buddy.Colonel Clive? Colonel of Sarcasm! Colonel of
Chuckles! Colonel of, I don'tknow, figure something else out,
wise guy.Sorry, that was uncalled for. Clive Tooth, too, is a
child of God. It's justthat I wish you wouldn't bother James
with such nonsense at a time like this.His day approaches
rapidly.-- | Jim Ferry | Center for Simulation
|+------------------------------------+ of Advanced Rockets ||
http://www.uiuc.edu/ph/www/jferry/ +------------------------+|
jferry@[delete_this]uiuc.edu | University of Illinois
===
guys, is that all masses have inertia! That is itrequires> a
net impulse - the product of a net force [f] and its duration
[t] -to> change a body's present velocity [vi], to some other
velocity [vt];during> which period of time, the body is
displaced a distance [s]: The ratioof> this impulse [ft], to
the time rate of displacement [s/t = (vt-vi)]that it> causes
is a constant: That is ft/(s/t) = ft/(vt-vi); which can
bewritten> more concisely as f/t/s = ft/ (vt-vi)! Isn't _that_
inertia?Why are you polluting this newsgroup with your topic.
This is the thirdthread you started with almost the same
topic, isn't it?-- AndreasFor replying, remove the fruit from
===
NYScientific Programmer - New York, NYThis client is a
technology leader specializing in software solutionsfor life
science research and development.They are seeking
computational scientists to create the nextgeneration of
computer-aided drug design methods, algorithms, andsoftware.
Two Ligand-based drug design positions are available
forindividuals with a background in 3D pharmacophore
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not strictmodeling) in the New York City
office.Responsibilities include: Participate in scientific and
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implementation.Extend and maintain large software packages
written by others, as wellas develop new applications.Required
qualifications include: Ph.D. in computational chemistry,
biochemistry, or biology. Post-doctoral research and/or
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experience in ligand-based drug design withspecialization in
development of 3D pharmacophore models. Experiencein
methodology development for conformational analysis,
pharmacophorefeature definition and 3D pharmacophore model
identification ispreferred. Experience in pharmacophore
modeling, 3D ligand-based drugdesign, and structure-based
design experience is a plus.Strong algorithm development and
methodology development skills. Strong programming experience.
Good knowledge of C/C++ andobject-oriented design is a plus.
Experience writing and maintaininglarge (100,000-line-plus)
software packages is preferred.Peer-reviewed publications in
computational chemistry, bioinformatics,cheminformatics, or
computational medicinal chemistry.For consideration, please
send CV (including publications, thesistopic, academic scores
and references) toresumes@workwondersstaffing.net.Benefits
include medical, dental, 401(k), flexible spending account,3+
===
Computational Chemist - NYComputational Chemist - NY A great
company in New York is seeking computational chemists(junior,
senior and management level). Here is more information:
Candidates should have world-class credentials in
computationalchemistry, biology, or physics, or in a relevant
area of computerscience or applied mathematics, and must have
unusually strongresearch and software engineering skills.
Relevant areas of experiencemight include the computation of
protein-ligand binding free energies,molecular dynamics and/or
Monte Carlo simulations of biomolecularsystems, application of
statistical mechanics to biomolecular systems,free energy
perturbation methods, and methods for speeding up --
butspecific knowledge of any of these areas is less critical
thanexceptional intellectual ability and a demonstrated track
record ofachievement. Current areas of activity for the firm
include structure-and ligand-based drug design, protein
structure determination throughhomology modeling, molecular
mechanics force field development, denovo drug design
algorithms, and the development of special-purposehardware to
accelerate computational chemistry simulations. This client
will offer above-market compensation to candidates oftruly
exceptional ability. Please send your resume (including list
ofpublications, thesis topic, and advisor, if applicable),
along with ahistory of academic performance (including GPAs as
well as SAT, GRE,and other standardized test scores)
===
single precision floating point. -- Example routine needed?I'm
trying to write ATAN2 function for a small basic language that
hasIEEE single precision math.. *,/,+.-, SQRT(), SIN(), COS(),
TAN() areavailible in the language. I've tried a few methods
I've found but the results are way off due tolow precision,
rounding, etc. Are there any repositorys of old fortran
routines or algorithms that Icould use to get a good accuracy
single precision routine. Speed orspace aren't as important as
===
Scientific Programmer - New York, NYWho does the composition
on these things?I feel overqualified.-- Ron
Hardinrhhardin@mindspring.comOn the internet, nobody knows
===
graph of f'(x)Arturo Magidin  a
.8ecrit dans le message de> And also where f is concave up,
concave down, and where its points of> inflection are.> [...]>
Also, f is concave up where f' is ->increasing<-, and f is
concave> down where f' is ->decreasing<-.Is a function with
f'' > 0 called concav up in English? Is convex asynonym for
that?I'm just wondering, because in German, such functions are
calledkonvex (f''>0) and konkav (f''<0).-- AndreasFor replying,
===
very much, Randy and Prof. Eppstein. The number of circles is>
about 10. I think that the calculation time of Monte Carlo
integration might> be too long. Is it possible to use
numerical intergration, i.e calculating> the area enclosed by
the envelope of those circles? But how can I get the>
expression of this envelop easily in my simulation? I also
need to consider> the sunk parts while integrating the area,
right? The calculation is related> to computer graphics. Could
Supengquoted, but it sounds like it contains an efficient
algorithmto do exactly what you want to do. Note that it's
from acomputer graphics conference.Monte Carlo integration is
a quick way of estimating theintegral you're talking about
doing explicitly. On my Solarismachine here's the result of a
quick run with 10 circlesand a half million points. That took
9 seconds of real time,6.6 seconds of CPU time, and converged
to 4 decimal places.By the way, another method that occurs to
me is to actuallyrender the circles in some pixelated medium
and then countcolored pixels. -
Randy------------------------------------------Circle centersC
= 0.1942 0.1138 0.0846 0.9897 0.9635 0.5098 0.4557 0.0639
0.6524 0.0272 0.0005 0.0413 0.3786 0.4947 0.0858 0.8082 0.5010
0.4129 0.3872 0.9048Circle radiians = Columns 1 through 8
0.9391 0.4621 0.9122 0.2243 0.6262 0.2088 0.4072 0.7326
Columns 9 through 10 0.3542 0.2420N A(est) clocktime
CPUtime20000 4.791736 0.3326 0.270040000 4.768794 0.6383
0.540060000 4.775305 0.9956 0.810080000 4.774684 1.3191
1.0800100000 4.772452 1.6082 1.3500120000 4.772566 2.0246
1.6300140000 4.770211 2.4158 1.9000160000 4.768678 2.9425
2.1700180000 4.769965 3.2977 2.4400200000 4.769445 3.5792
2.7000220000 4.769358 3.9876 2.9700240000 4.767554 4.4163
3.2400260000 4.769772 4.7735 3.4900280000 4.769901 5.1083
3.7600300000 4.769807 5.4628 4.0100320000 4.768561 5.8629
4.2600340000 4.766387 6.1869 4.5300360000 4.768725 6.5527
4.7900380000 4.769789 7.0027 5.0500400000 4.768623 7.3632
5.3100420000 4.767096 7.7723 5.5800440000 4.767117 8.0896
5.8400460000 4.765990 8.4463 6.1000480000 4.764828 8.8470
===
or is it weightCut<>> A slug is a little used 20th century
invention which exists only in> one specialized subsystem of
mechanical units--and a unit which> doesn't even have an
official definition.>Cut> Gentlemen of the jury, Chicolini
here may look like an idiot,> and sound like an idiot, but
don't let that fool you: He> really is an idiot.> Groucho
MarxJeepers Gene, I think all of your knowledge about _so
many_ differentsystems of weights and measures must have
driven you buggy as Chicolini:According to Webster's online,
definition #7 for Slug is : the gravitationalunit of mass in
the foot-pound-second system to which a pound force canimpart
an acceleration of one foot per second per second and which is
equalto the mass of an object weighing 32 pounds.There _is_
===
proof of the following;Let X(n) be a function defined for all
positive integers n,X(n) = 0, n evenX(n) = 0, if the prime
factors of n include repeated primesX(n) = 1, if the prime
factors of n do not include repeated primesexample; x(15) =
x(3*5) = 1, x(45) = x(3*3*5) = 0Prove that (1/n)*integral(
===
= 1>Did someone know a simple demonstration of>>0! = 1
?>>Tx>>PS Sorry about my englishRefer to the implication from
http://www.wikipedia.org/wiki/Factorialn!=n(n-1)!Let n=1; you
know that 1!=1n! becomes when n=1,1*(1-1)!=1Divide both sides
===
A N Niel  scribbled the
following:>> If I know my algebra, then an identity for an
operation (let's call>> it #) is such an i, that for all x,
x#i=x. But there is another>> concept too: such a j, that for
all x, x#j=j. For multiplication, this>> is 0.>> For addition,
there doesn't seem to be one. What is this concept>> called?>>
It's called a zero.As to what the _concept_ is called, my
answer is absorption. And in ageneral context, instead of
zero, one might use the term absorptiveelement.> So
intuitively, if #' is a repetitive #, which means that x#'i is
x#x> performed (i-1) times in succession, then the zero of #'
is the> identity of #? For example multiplication is
repetitive addition. 0 is> the additive identity and the
multiplicative zero. If we switch the> definition of a zero
around so that it means that for all x, j#x=j, then> we get
that exponentation is repetitive multiplication, and that 1
is> the multiplicative identity and the exponentative zero.>
Is this making any sense?Yes. But exponentiation isn't
commutative. Perhaps one might say that1 is left-absorptive,
and that exponentiation has no
===
proved?> Has anyone ever seen a proof of the following;> Let
X(n) be a function defined for all positive integers n,> X(n)
= 0, n even> X(n) = 0, if the prime factors of n include
repeated primes> X(n) = 1, if the prime factors of n do not
include repeated primes> example; x(15) = x(3*5) = 1, x(45) =
x(3*3*5) = 0> Prove that (1/n)*integral( X(n) ) converges with
increasing n.?integral( X(n) )??-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
===
homeomorphisms are open and closed bijections. Does thatmean
that if f:X->Y is a homeomorphism, then f(boundary(A))
=boundary(f(A)) for all sets A subset X?-- /-- Joona Palaste
(palaste@cc.helsinki.fi) ---------------------------|
Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA
N+++|| http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/I will never display my bum in public again. -
===
I see it guys, is that all masses have inertia! That is it>
requires> a net impulse - the product of a net force [f] and
its duration [t] -> to> change a body's present velocity [vi],
to some other velocity [vt];> during> which period of time, the
body is displaced a distance [s]: The ratio> of> this impulse
[ft], to the time rate of displacement [s/t = (vt-vi)]> that
it> causes is a constant: That is ft/(s/t) = ft/(vt-vi); which
can be> written> more concisely as f/t/s = ft/ (vt-vi)! Isn't
_that_ inertia?>> Why are you polluting this newsgroup with
your topic. This is the third> thread you started with almost
the same topic, isn't it?>> --> Andreas> For replying, remove
the fruit from my address.>>You've got to learn; like it or
===
Re: a baseball odds calculation problem>>>here's the
problem, specifically:>>there are three divisions in the
american league. every season, each one> of>these divisions
will be clinched by one team on a certain date. this is> a>
>mathematical certainty.>>past baseball history tells us
that dates before september are extremely>rare for teams
clinching. in fact, the later days of september, roughly> the>
>15th through the 28th, are where the likelihood of teams
clinching is the>greatest.>> You need to specify the
probability distribution of the date of> clinching for each
division. For example, assuming the the three> divisions are
independent and have the same distribution, completely>
supported (a model of your approximation) by {15, 16,..., 28},
then the> probability is sum(k=15..28, p(k)^3), where p(k) is
the probability of> clinching on the date k. So if you are
saying that each division is> equally likely to be clinched on
each day and is independ of the others,> the probability is
14*(1/14)^3 = 1/196 = .0051 (0.51%), approximately.> Or if you
assume identical triangular distributions, p(k) =>
min(k-14,29-k)/56, the probability (still assuming
independence) is> 2/56^3 sum(k=1..7, k^3) = 1/112 = .0089
(0.89%), approximately.> Another model would be a discretized
normal over the whole season.> Perhaps you want to make an
empirical estimate of the distribution; you> may then do the
clinching dates over the last 27 full> baseball seasons. in
those seasons, there have been 126 different clinches> (4
clinches per year from 1975-1993 excluding 1981 strike year
and then 6> and i was able to find 96 of those dates (they're
not very easy to find). i> did an estimate based on this
here's the distribution of clinches i found:[...data
omitted...]It seems you are not satisfied with the quality of
the answers you got so farand want a more accurate answer. But
I suspect you collected the wrong data.It would be far more
informative to know how many days from the endof the season
the clinch occurred. That is because the last (scheduled)
dayof a season is a Sunday, whose date obviously varies from
===
the idea is that THE ESTABLISHMENT (shiver, shiver)>controls
the TROOTH (tm) and suppresses publication of HERESY,
especially>because they're making lotsa money is certainly
neither novel nor humorous.Well that's the standard crackpot
conspiracy theory, but somehow it'seven more ludicrous when
adapted to the scientific community.>You'd think that he could
at least GIVE them away to social friends who>would say Gee,
study.>(Nudge,nudge, wink, wink) The trouble is, most people
have learned elementary physics in schoolso it narrows down
===
single precision floating point. -- Example routine needed?>
I'm trying to write ATAN2 function for a small basic language
that has> IEEE single precision math.. *,/,+.-, SQRT(), SIN(),
COS(), TAN() are> availible in the language.> I've tried a few
methods I've found but the results are way off due to> low
precision, rounding, etc.> Are there any repositorys of old
fortran routines or algorithms that I> could use to get a good
accuracy single precision routine. Speed or> space aren't as
suggestions:(i) See The Standard C Library by P J Plauger,
published by PrenticeHall. You will find two short pages in C.
Do you need to code inFortran?(ii) Google for CORDIC; with it
===
Principle of Finite InductionI'm hoping to get help in how to
approach the following problem on numbertheory:Use the 2nd
prinjicple of finite induction to establish thata^n - 1 = (a -
1)(a^(n - 1) + a^(n - 2) + a^(n - 3) + ... + a + 1) for alln
geq 1the trouble i'm having is there's a hint:a^(n + 1) - 1 =
(a + 1)(a^n - 1) -a(a^(n - 1) - 1)and i am clueless as to what
===
graph of f'(x) Visiting Assistant Professor at the University
of Montana.>Arturo Magidin  a
.8ecrit dans le message de>> And also where f is concave up,
concave down, and where its points of>> inflection are.>>
[...]>> Also, f is concave up where f' is ->increasing<-, and
f is concave>> down where f' is ->decreasing<-.>>Is a function
with f'' > 0 called concav up in English? Is convex a>synonym
for that?>I'm just wondering, because in German, such
functions are called>konvex (f''>0) and konkav (f''<0).Yes, it
is; in fact, I learned it convex and 'concave', but mostbooks
now seem to use concave up and concave down instead, asbeing,
I assume (a) easier to remember and (b) more evocative.
Ofcourse, you still have students who don't remember if
concave upmeans opens up or the round part points up, so I'm
not sure itaccomplished that much. (Except that f''>0 means f'
goes up meansconcave up may be
what I accept as reality. --- Calvin (Calvin and
===
have inertia> You've got to learn; like it or not:What do I
have to learn? I'm a happy SI user who never gets
confuseddistinguishing between mass and force, so I don't have
a problem withthat topic.No one asked you to start three
threads. It seems that you're the onlyone here having
problems.> Fruit for your address; very appropriate(:^)Yeah, I
like apples!-- AndreasFor replying, remove the fruit from my
===
Assistant Professor at the University of Montana.>I've read
that homeomorphisms are open and closed bijections. Does
that>mean that if f:X->Y is a homeomorphism, then
f(boundary(A)) =>boundary(f(A)) for all sets A subset X?I was
going to say depends on how you define boundary, but then
Iremembered we already had ->that<- conversation. Looking back
isboundary(A) = cl(A) - int(A).A homeomorphism is a continuous
bijection that sends open sets to opensets (equivalently, a
bijection which sends open to open and closed toclosed). Thus,
if f:X->Y is a homeomorphism, it follows that for all
A,f(int(A)) = f( union{B : B is open, B contained in A}) =
union{ f(B): B is open, B contained in A} (since f is
bijective) = union{ f(B): f(B) is open, f(B) contained in
f(A)} = int(f(A))and likewisef(clos(A)) = f( intersection{B: B
closed, B contains A}) = intersection {f(B): B closed, B
contains A} = intersection {f(B): f(B) closed, f(B) contains
f(A)} = clos(f(A)).Sof(boundary(A) ) = f(cl(A) - int(A)) =
f(cl(A)) - f(int(A)) = cl(f(A)) - int(f(A)) =
selective about what I accept as reality. --- Calvin (Calvin
and
===
of Finite Induction>I'm hoping to get help in how to approach
the following problem on number>theory:>Use the 2nd prinjicple
of finite induction to establish that>a^n - 1 = (a - 1)(a^(n -
1) + a^(n - 2) + a^(n - 3) + ... + a + 1) for all>n geq 1>>the
trouble i'm having is there's a hint:a^(n + 1) - 1 = (a +
1)(a^n - 1) ->a(a^(n - 1) - 1)>and i am clueless as to what to
do with it.The 2nd Principle of Induction (applied to the
natural numbers) is thestatement that if a property P of
natural numbers, if[for all n, (for all kp(n) ] then
for all n, P(n).That is, you assume the property is true for
all natural numbers lessthan n, and from that you prove it is
true for n. If you can do thatfor each n, then you have proven
it for all n.In this case, your induction hypothesis is that
for all k<(n+1), youhavea^k - 1 = (a-1) (a^{k-1} + a^{k-2} +
... + a + 1).and you want to prove thata^(n+1) - 1 = (a-1)(a^n
+ a^{n-1} + ... + a + 1).So, use the hint and the induction
hypothesis.Caveat: your hint presupposes that n>1; so you
would need to do thecase n=1 separately; this is called a
special case of your
about what I accept as reality. --- Calvin (Calvin and
===
satisfiabilitybut lets just restate this in a more
interestingway that immediately leads to an
interestingresearch direction, along the lines of
exploringstructural properties of SAT factoring instancesIve
been advocating: what is the smallest x-SATformula that can be
used to factor an arbitrary sized number?by electrical
engineering principles a half adderand full adder can be
implemented in 5-SAT, soas I recall (working from memory on my
old SAT generator) 5-SAT issufficient. but is it necessary?RE
has a reason that if 3-SAT is sufficient then we have some
reason to believe that factoring might be in P.if it is not,
then we have a demonstration of thepower of nonuniform versus
uniform computation. if factoring can be done with 3-SAT but
is not in P, then factoring is in nonuniform P, but not in
uniform P.actually, now that I think about it, this is
interestingbecause for the same reason5-SAT has C(n,5) term
possibilities.so we may already have some kind of implication
here fornonuniform vs uniform power of factoring. we can find
theC(n,5) clauses that create the factoring problem, butnot
solve them.I guess factoring is in the C(n,5) clause class
andIve refuted my own claim that the clauses go up
exponentially. hmmmmmmnote: uniform= computations done by
TMsuniform= computations done by arbitrary circuit families>
Why should all the clauses in this formula have at most three
literals?> You can't necessarily represent a function in N
variables using a 3-CNF> formula. The number of functions that
can be represented that way is very> small by a counting
argument: there are 2^(2^N) functions in N variables,> but
only 2^(O(N^3)) possible 3-CNF formulas. What are the odds
that> factoring is one of those lucky functions for
===
boundariesArturo Magidin  scribbled
the following:>>I've read that homeomorphisms are open and
closed bijections. Does that>>mean that if f:X->Y is a
homeomorphism, then f(boundary(A)) =>>boundary(f(A)) for all
sets A subset X?> I was going to say depends on how you define
boundary, but then I> remembered we already had ->that<-
definition you are using is> boundary(A) = cl(A) -
int(A).(snip)> So> f(boundary(A) ) = f(cl(A) - int(A)) =
f(cl(A)) - f(int(A))> = cl(f(A)) - int(f(A))> =
much easier.-- /-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------| Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++||
http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
------------/Roses are red, violets are blue, I'm a
===
factoring to satisfiability
3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21:> what is the
smallest x-SAT> formula that can be used to factor an
arbitrary sized number?> by electrical engineering principles
a half adder> and full adder can be implemented in 5-SAT, so>
as I recall (working from memory on my old SAT generator)
5-SAT is> sufficient. but is it necessary?5-SAT can be reduced
to 3-SAT with only a constant factor blowup in number of
clauses.More specifically, any 5-clause (t1 + t2 + t3 + t4 +
t5) can be replaced by three 3-clauses (t1 + t2 + x)(-x + t3 +
y)(-y + t4 + t5) where x and y are new variables.-- David
Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California,
===
Chess/Go/etc: Continuous Game-Boards?Nick Wedd
 scribbled the followingon sci.math:> For
Go, I assume:> The difference between Japanese (territory)
rules and Chinese (area) > rules would be very important in
this game, unlike in normal Go where it > rarely matters. For
one thing, under Chinese rules, you get captured > pieces back
and can use them again.What's the precise difference between
these rules?-- /-- Joona Palaste (palaste@cc.helsinki.fi)
---------------------------| Kingpriest of The Flying Lemon
Tree G++ FR FW+ M- #108 D+ ADA N+++||
http://www.helsinki.fi/~palaste W++ B OP+
|----------------------------------------- Finland rules!
===
all. conversion of factoring to SAT does NOT (*always*or
*usually*) give easyinstances as was stated in another msg on
this thread.it MAY or MAY NOT give easy instances. if you want
to factor a prime number then the resulting SAT clause mustbe
easy based on the new agrawal proof. and it wouldbe extremely
interesting to try to find a SAT algorithmthat runs in P time
on prime number factoring instances...I think it is virtually
guaranteed that the basic DPLLetc algorithms do NOT.. and so
why not??and in fact if you look at a SAT algorithm that does
giveP time performance for factoring prime numbers.. it
mightgive more insight into the ASK algorithm, and possiblynew
ways to optimize it.this all falls into a basic strategy &
generalresearch program of analyzing computationalproblems via
their boolean circuits, which imho is morefundamental and will
ultimately lead to more/most insightinto the inherent problem
complexities.I think this will soon be shown for some number
theory problem(s).soon gauged in research time which is
glacial prior to a tipping point. 5-10 years would be soon> As
primality testing has been shown to be in P would not a
reduction of> SAT to factoring be a proof that NSAT in in P?
===
explanations and links were of great value for
===
minute talk> * If T = infinity, the function |sum_{k=1}^{n}
exp(ak*t*i)| attains all> values between 0 and n, so this case
can be ignored. (Exercise for the> reader.)Oops. Not correct. I
was thinking of the case where each pair (a_j,a_k)has an
irrational ratio. The T = infinity case should never pertain,
butJustifying this is more complicated than I thought.-- | Jim
Ferry | Center for Simulation
|+------------------------------------+ of Advanced Rockets ||
http://www.uiuc.edu/ph/www/jferry/ +------------------------+|
jferry@[delete_this]uiuc.edu | University of Illinois
===
point. -- Example routine needed?> I'm trying to write ATAN2
function for a small basic language that has> IEEE single
precision math.. *,/,+.-, SQRT(), SIN(), COS(), TAN() are>
availible in the language.> I've tried a few methods I've
found but the results are way off due to> low precision,
rounding, etc.>A Basic language with Sin, Cos, and Tan, should
also have at least an ArcTan? I mention that because a division
result sent to an ArcTan and combinedwith some simple quadrant
===
satisfiabilityanother obvious point. a proofthat C(n,3) clauses
are not possible to specifyfactoring instances (or arbitrary
other problemsfor that matter) would be quite interesting.it
probably would not be too difficult to construct.just look for
some bit relationships in factoringthat are more complex than
the C(n,3) relationships;these could easily be identified
empirically and thenpossibly generalized into a proof without
too muchtrouble/hassle/effort.also, we are referring to a
natural complexity class thathas been studied in the
literature somewhat.bounded width circuits I believe they are
called, or bounded input circuits.note that bounded width
measures are apparentlymore fine-grainedthan P vs NP. the
various NP complete problems mayhave different bounded-width
complexities.not too much is known in this area so far.another
theory-edge.> Why should all the clauses in this formula have
at most three literals?> You can't necessarily represent a
function in N variables using a 3-CNF> formula. The number of
functions that can be represented that way is very> small by a
counting argument: there are 2^(2^N) functions in N variables,>
but only 2^(O(N^3)) possible 3-CNF formulas. What are the odds
that> factoring is one of those lucky functions for
===
Continuous Game-Boards? 
In message
, Joona I Palaste >Nick
Wedd  scribbled the following>on
sci.math:>> For Go, I assume:> The difference between
Japanese (territory) rules and Chinese (area)>> rules would be
very important in this game, unlike in normal Go where it>>
rarely matters. For one thing, under Chinese rules, you get
captured>> pieces back and can use them again.>>What's the
precise difference between these rules?The basic difference is
that with Japanese rules, you count points for territory
surrounded and lose points for pieces you have lost, while
with Chinese rules, you count points for territory surrounded
_or_ occupied, and ignore pieces you have lost. Usually this
makes no difference, there is a simple proof of this depending
on the fact that each player has made the same number of
moves.Other differences (which have little practical effect):
With Japanese rules you don't count points inside a seki; with
Chinese rules you do. With Japanese rules bent four in the
corner is rules to be dead; with Chinese rules, you play it
out, and it almost always turns out to be dead anyway.
Japanese put handicap stones on the marked hoshi points;
Chinese allow the recipient of the handicap to choose where to
put them.There are more details at
http://www.britgo.org/rules/compare.htmlNick-- Nick Wedd
===
Game-Boards? Visiting Assistant Professor at the University of
Montana.>Nick Wedd  scribbled the
following>on sci.math:>> For Go, I assume:> The difference
between Japanese (territory) rules and Chinese (area) >> rules
would be very important in this game, unlike in normal Go where
it >> rarely matters. For one thing, under Chinese rules, you
get captured >> pieces back and can use them again.>>What's
the precise difference between these rules?Don't know if they
are all of the differences, but:In Chinese rules, you get your
captured pieces back and you return toyour opponent his
captured pieces. At the end of the game, you totalthe number
of positions ->occupied<- by your pieces, and add them tothe
number of empty positions ->controlled<- by your pieces.
Youcompare your total to your opponents (plus or minus the
handicap), andhighest total wins.In Japanese rules, you keep
the pieces you capture from your opponent,he/she keeps the
pieces he captures from you. At the end of the game,you use
your prisoners to occupy territory he controls, and heuses the
prisoners to occupy territory you control. At the end, youcount
the number of empty positions controlled by your pieces,
hecounts the number of empty positions he controls, and after
adding theappropriate handicap the highest total wins. If you
have moreprisoners than your opponent has empty but controlled
territory, hehas a negative
what I accept as reality. --- Calvin (Calvin and
===
have inertia> You've got to learn; like it or not:Plonk
Him..I'm even wasting bandwith with this advice.Bob
===
f'(x)Arturo Magidin  a .8ecrit dans
le message de>Is a function with f'' > 0 called concav up in
English? Is convexa>synonym for that?>I'm just wondering,
because in German, such functions are called>konvex (f''>0)
and konkav (f''<0).>> Yes, it is; in fact, I learned it convex
and 'concave', but most> books now seem to use concave up and
concave down instead, as> being, I assume (a) easier to
remember and (b) more evocative. Of> course, you still have
students who don't remember if concave up> means opens up or
the round part points up, so I'm not sure it> accomplished
that much. (Except that f''>0 means f' goes up means> concave
AndreasFor replying, remove the fruit from my
===
Visiting Assistant Professor at the University of Montana.>In
message , Joona I
Palaste >>Nick Wedd  scribbled the
following>>on sci.math:>For Go, I assume:>The difference
between Japanese (territory) rules and Chinese (area)>rules
would be very important in this game, unlike in normal Go
where it>rarely matters. For one thing, under Chinese rules,
you get captured>pieces back and can use them again.>>What's
the precise difference between these rules?>>The basic
difference is that with Japanese rules, you count points for
>territory surrounded and lose points for pieces you have
lost, while >with Chinese rules, you count points for
territory surrounded _or_ >occupied, and ignore pieces you
have lost. Usually this makes no >difference, there is a
simple proof of this depending on the fact that >each player
has made the same number of moves.Which is certainly not the
case under Japanese rules. At least the wayI learned them: a
player is allowed to pass, and then the other playermay either
pass (terminating the game), or play.Is this different under
Chinese
what I accept as reality. --- Calvin (Calvin and
===
one-way hash function> I need to find an algorithm that can
produce a unique non-predictable 12> digit (0-9) number for
any given 12 digit number. This is to be used to> create a
unique barcode on a ticket that cannot be predicted. It is
not> required that the original seed number be computed from
the resulting> barcode, so some form of one-way hashing
function would be acceptable.> Any help in this problem would
be appreciated.> Mark.1. Create a 10 element array with digits
0-9 in linear order (0 in 0,1 in 1, etc.)2. For each digit of
the 12-digit number:2a. Shuffle the 10-element array.2b. Using
the original decimal digit as an offset into the array, lookup
===
artinian group cards are dealt to them.There is no asumption
>on the number of cards a player receive. In eachround, all
players with 2 or more cards pass >one card to the left andone
card to the right. Prove that eventually, all players but one
have>exactly one card.I do not have the solution, so any hint
would be highly appreciated.Please >restrain from posting
messages saying this is obvious, andthe like, because what is
n this is, of course, a cellular automaton. You can allow
anypositive integer states for each of the n cells. The
transition rule issuch that if a cell has state x >=2 then the
state never exceeds x and ifit has state x < 2 then it either
stays the same or increase by 1 or 2.If N is the maximum value
of a cell state, then the maximum value of thestates in the
next generation can be at most N, unless N = 2 and then
amaximum of 3 is possible in the next generation. It follows
that there arethere are only a finite number of possible
global states, so eventuallyit is periodic.One question might
be to figure out which of Wolfram's four classes thiscellular
automaton lies in.These remarks don't answer your question,
===
Chess/Go/etc: Continuous Game-Boards?

In message
, Arturo Magidin >>In
message , Joona I
Palaste>Nick Wedd  scribbled the
following>on sci.math:> For Go, I assume:>> The difference
between Japanese (territory) rules and Chinese (area)> rules
would be very important in this game, unlike in normal Go
where it> rarely matters. For one thing, under Chinese rules,
you get captured> pieces back and can use them
again.>>What's the precise difference between these
rules?>>The basic difference is that with Japanese rules, you
count points for>>territory surrounded and lose points for
pieces you have lost, while>>with Chinese rules, you count
points for territory surrounded _or_>>occupied, and ignore
pieces you have lost. Usually this makes no>>difference, there
is a simple proof of this depending on the fact that>>each
player has made the same number of moves.>>Which is certainly
not the case under Japanese rules. At least the way>I learned
them: a player is allowed to pass, and then the other
player>may either pass (terminating the game), or play.>>Is
this different under Chinese rules?It is the same under
Chinese rules. A player may always pass - two consecutive
passes end the game.So, unless there are passes before the
final two passes, each player has played the same number of
stones; except that Black may have played one more point than
White. So you are right - it can make one point of
difference.The North American rules are a compromise between
Chinese and Japanese rules, tending to be closer to the
Chinese rules. They deal with this one-point problem by
requiring White to make the last move.Nick-- Nick Wedd
===
unique factorizationI studying a different way of proving that
the sqrt(2) is irrational, or nth roots in general. It proceeds
as follows:Assume the sqrt(2) is rational. Thus, m/n = sqrt(2).
Which then implies2^1 * n^2 = 2^0 * m^2Now, by unique
factorization, a perfect square will have all even exponentsin
its prime factorization. Since, the LHS of the above equation
has an oddexponent, namely 2^1, can I then conclude that the
sqrt(2) is irrational?It seems to me that some step is
missing? Shouldn't there be somethingabout the RHS being even
and the LHS being odd ? How does one arrive at theconclusion
that if a square isn't perfect, then it is irrational?
Iunderstand intuitively that if a sqrt() is not perfect, then
it isirrational, but I don't see how one can logically
===
there a name for this operation?Let the members of set Ai be
ai1, ai2, ... aiNi.input, returns as an answer{ [a11, a21,
a31, ...], [a12, a21, a31, ...], [a13, a21, a31, ...], ...,
[a1N1, a21, a31,...], [a11, a22, a31, ...], [a11, a23, a31,
...], ..., [a11, a2N2, a31,...], ..., [a11, a21, ..., aM1],
[a11, a21, ..., aM2], ..., [a11, a21, ...,aMNM] }In other
words, the answer is a subset of the Cartesian
product,containing only those sets that differ from [a11, a21,
a31, ...] in atmost one place.I'm providing this operation as
one of a number of possibilities in atest-case generation
program. So far I'm just calling itperturbation which is
===
teaching a class this term in which students are wrestlingwith
projective planes. I gave them an exercise to identify
(i.e.,coordinatize) the plane over the field of 4 elements. It
can bedescribed as an assembly of 21 points, some subsets of 5
of whichare called lines. When I posed the problem, I just
assigned names A, B, C, ... to thepoints and gave the lines as
5-tuples. Later, it occurred to meI might have been able to
find a different assignment of thepoints into the alphabet in
such a way that each line was( a permutation of ) the letters
in a 5-letter word.So here is the challenge: how well can you
do ? Can you embedthese letters back into the alphabet so that
all, or most, ofthese 5-tuples become English words? (Interpret
word as you wish.)(I would be willing to take a solution which
works only foran affine plane inside here, if necessary.)By
the way, I hereby claim my title as inventor of a new
field.Google finds no match for projective linguistics except
forrandom-word generators.daveThe 21 lines are:{A, B, C, D,
Q}{A, F, K, P, R}{A, G, L, N, T}{A, H, J, O, S}{A, I, E, M,
U}{B, E, L, O, R}{B, F, J, N, U}{B, H, K, M, T}{B, I, P, G,
S}{C, E, J, P, T}{C, F, L, M, S}{C, G, K, O, U}{C, I, R, H,
N}{D, E, K, N, S}{D, F, I, O, T}{D, G, J, M, R}{D, H, L, P,
U}{E, F, G, H, Q}{I, J, K, L, Q}{M, N, O, P, Q}{Q, R, S, T,
===
this?>> Express the following as a single fraction:>>
4/3ab - 5/6bc>> (1) find a common denominator. The least
common denominator is a> good one to use.> What, for example
in the first one, is the lowest common denominator? Is it>
(3ab)(6bc)? If it is, do I then multiply that out? If not,
what is it?6abc = (3ab)(2c) = (6bc)(a), so 6abc is a common
multiple of 3ab and 6bc.Since (2c) and (a) have no common
factor greater than 1, nothing can be left out of 6abc and
still have a common multiple of 3ab and 6bc, so 6abc is the
===
sqrt(2) by unique factorization>I studying a different way of
proving that the sqrt(2) is irrational, or n>th roots in
general. It proceeds as follows:>>Assume the sqrt(2) is
rational. Thus, m/n = sqrt(2). Which then implies>>2^1 * n^2 =
2^0 * m^2>>Now, by unique factorization, a perfect square will
have all even exponents>in its prime factorization. Since, the
LHS of the above equation has an odd>exponent, namely 2^1, can
I then conclude that the sqrt(2) is irrational?Yes.>It seems
to me that some step is missing? Shouldn't there be
something>about the RHS being even and the LHS being odd
?Consider the factors of 2 that appear in either n or m or
both; thereis an even number of factors of 2 in n^2, and an
even number offactors of 2 in m^2. So the RHS has, in total,
an even number offactors of 2 (0 + the number of factors in
m^2), while the LHS has anodd number of factors of 2 (1 + the
number of factors in 2).This is impossible.This would be the
long-winded explanation. What you have noticed issimply that
all the primes on the right hand side appear raised to aneven
exponent, while all but ONE of the primes on the left hand
sideappear raised to an even exponent, the exception being
raised to anodd exponent. That's impossible by unique
factorization.> How does one arrive at the>conclusion that if
a square isn't perfect, then it is irrational?No such
conclusion is reached. Or do you mean, that an integer whichis
not a perfect square has an irrational square root? >
I>understand intuitively that if a sqrt() is not perfect, then
it is>irrational, but I don't see how one can logically
conclude that from the>above equation.The above equation only
proves it for 2. A similar argument yields theresult for
sqrt(p) for any prime p; then it is not hard to prove it bya
similar argument for any product of the form p_1*...*p_r
wherep_1,...,p_r are distinct primes.Then simply note that
positive integer greater than 1 can be writtenuniquely as
n=k*s, where s is a perfect square, and k is a product
ofdistinct primes. So then sqrt(n) = sqrt(k)*sqrt(s), sqrt(s)
is aninteger, and sqrt(k) is
about what I accept as reality. --- Calvin (Calvin and
===
of the normal distributionHi to everyone!!Some of you know
where I can find the formula of the k-th absolute momentfor
the normal distribution?My own calculations are different from
the Mathematica output and I am notsure about what to
===
===
continuity> Subject: Re: hw help -- continuity>> 1)prove
that if f,g continuous, then so are max(f,g) and min(f,g)>>
Any ideas how to approach this?>Proving a function
continuous at any point>proves it continuous at every
point.> f(x) = 0 when x <= 0> = 1 when 0 < x> is continuous at
1. Thus it's continuous at 0?But 1 is not any point, it is a
specific point.To restate less ambiguously:If you prove f
continuous at an arbitrary point, then it is continuous at
every pointAnd to prove it at an arbitrary point, you only
need show that for an arbitrary x, g(x) = lim_{y -> x}
===
that given any>>rational that the value greater than it and
less than any other>>greater is irrational,>There is no such
number, as several different people have shown you.>> In
non-standard analysis, there might be, however.> See Alain
Robert's book about NSA. Rather than being> irrational, it
would be non-standard, though.> I have yet to see any standard
or non-standard model of the reals in> which there is a
smallest positive number.> Who says that that is meant?> In
nsa, there can be a nonstandard number that can be said to be>
'greater than (a given standard rational) and less than any
other (standard) > greater'> That it is not unique, who
cares?Ross wants to use non-standard numbers to confirm his
hypothesis that there is a next real after any real, and that
the irrationals and rationals alternate on the real line or on
some non-standard real line. So he cares. On the other hand,
his hypothesis is way out in left field, where he has been
===
--I could have baked both cakes yesterday.> --I could have
baked both cakes today.> --I could have baked this one
yesterday> and that one today.> --I could have baked that one
yesterday and> this one today.>> Your example doesn't work!>
How am I going to distinguish 'one cake' from 'the other
cake'!>> Since they now are identical-with-somethings, if they
shared>> all the same properties they would be one cake rather
than>> two. THerefore, since they are two cakes, they don't
share>> all their properties, and I can distinguish them by
means>> of any property that one has and the other
doesn't.>you mean argument by GPS or by just determining their
globalpositioning>with sattellites we have a difference even if
there are no otherclear>attributes of different classes besides
location?>> That is a call which no logician qua logician
has any particular> competence> to make.>> Are you saying
it is possible for something to exist at location A and>
location B at the same time? I thought it was inference by the
rule of> non-contradiction.>> I have no idea whether this is
possible or not.>> My suggestion that quanta may satisfy
'~(x=x)' is meant to translate> into logical terms the claim
by some physicists that quanta> 'lack individuality' (Heller
suggests that quanta 'lack haecceities').> The scenario with
cakes-to-be suggests that the behaviour of other> entities
without individuality resembles that of quanta in the>
relevant respect. Heller's description of that behaviour
follows> my sig.>Metaphysical Background, Thomas McTighe
asserted that the quiddity of athing is nothing other than
unity itself. Hence, by virtue of its positivecontent, the sun
differs not at all from the moon or any other particularthing.
The diversity which is exhibited by the natural world is
merely theproduct of accidental differences; no object
possesses any specific formwhich interposes itself between a
particular existing thing and the sourceof their being e.g.
the Absolute.15 All individual entities are nothing morethan
differing contractions of the whole devoid of any being of
their own....because the restricted quiddity of a thing is the
thing
itself.http://www.crvp.org/book/Series01/I-10/chapter_ii.htm
Aristotle and Aquinas and Scotus and Bonaventura all believed
that humanminds can conceive and express the intelligibilities
or quiddities of thingsand their properties, intelligibilities
that are not simply mind-dependent.We can capture in thought
and language the actual natures of things,spelling out their
genera and specific differences. Definition brackets
ordelimits for us as knowers just what it is we attempt to
understand andnothing else. The mind-independent
thing-substance or the characteristicsthat we are attempting
to define measure the epistemic correctness of adefinition.
Such real (as opposed to nominal) definition relies on
theintelligible and perceptible characteristics
thing-substances exhibit toperception and thought for
understanding what they are and for picking outindividuals of
a type. In this way the epistemological realism of
thedefinition corresponds to an ontological realism of actual
formal featuresin mind-independent
entities.http://www.sunysb.edu/philosophy/faculty/lmiller/
Delinonaliud.htmAfter all, the three tenets that largely
define Nicholas's 'metaphysic ofcontraction' seem altogether
remote from Anselm's Scholasticism. For Anselmhas no use for
the triad of notions (1) that there is an
infinitedisproportion between the Creator and His creatures,
(2) that, therefore,finite minds can never positively know
what God is, given the alleged ground(3) that He is the
Coincidence of opposites, i. e., is undifferentiated'Being'
itself, which, with respect to its Quiddity, can never be
conceivedby anyone except
itself.http://www.cla.umn.edu/jhopkins/CusaAnselm.pdf>
--John> JJ>> 1 I. The Problem, and the Problem with the
Problem,> 2 of Identical Articles and Quantum Statistics> 3> 4
Suppose we have a box with two qualitatively> 6 bouncing around
inside. We think of the box> 7 as having a left (*l*) and a
right (*r*) side.> 9 at random without interacting, so that
their> 10 motions are independent; in particular we> 12 that
we may neglect collisions. What are the> 13 chances for
finding one or both on one side> 14 or the other?> 15> 16 Many
find the following reasoning persuasive.> 19 in *r*, and 2 in
*l* and 1 in *r*. These should> 20 be equally likely, so that
each has a probability> 21 of 1/4, or a probability of 1/4 for
two in *l*,> 22 1/4 for two in *r*, and 1/2 for one on each
side.> 23> 24 This stylized example is a simple mock-up for a>
25 kind of situation that can occur with quantum entities> 26
and properties. For many of these situations the> 27
probabilities are in fact found to be 1/3 for each> 28 of the
three cases: two in *l*, two in *r* and one> 29 on each side.
Many interpreters have found this> 30 fact utterly
astonishing.> 31> 32 But on the face of it, there is a very
simple> 33 resolution of the puzzle: give up supposing> 34
that there are two qualitatively identical but> 36 there are
two *quanta*, as I'll put it, to which> 37 the notion of being
numerically distinct does not> 38 apply. . . (p.
===
along those lines is what is is the set, E, of>numbers
generated by repeated sums of products of rational powers
of>rationals? It is a subset of the algebraics, but does it
form a>field? > I believe so: if x_1 is a member of E, then so
are its conjugates x_2, > x_3, ..., x_n.> Note that x_1 x_2 ...
x_n is rational (being the constant term of a > monic
polynomial over the rationals whose roots are x_1, x_2, ...,
x_n).> And then 1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in
E.> Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel > University of
British Columbia > Vancouver, BC, Canada V6T 1Z2If we say that
a set, P, is exponentially closed if for all p,q in P,p^q is P,
then can the complex numbers be defined as the field that
isexponentially closed? Or even more simply, the set that
isexponentially closed and contains -2 and 0? If not, what are
theproperties of the smallest field, S, that is exponentially
closed(cardinality, algebraic completeness, convergence
ofCauchy-Sequences)?Note: that I am only considering fields
===
do you propose to integrateX(n) since it is defined only over
theintegers? Do you mean integrate asa Stieltje's integral?You
can lead a horse's ass to knowledge, but you can't make him
===
Visiting Assistant Professor at the University of Montana.
argument perfectly. I have given the problem some>thought and
I have concluded that HYPOTHETICAL G is impossible. No>graph
meets all three criteria; ie, G is 5-chroma, G is planar, H
is>4-chroma.As has been noted, if by H you mean any graph
obtained by removinga single vertex from G, then your
statement is exactly equivalent tothe 4 Color Theorem.Reading
back through the thread, I ->think<- I'm beginning
tounderstand what exactly it is that you are finding
problematic. You are arguing that a G which requires 5 colors,
and with theproperty that removing any vertex results in a
planar graph thatrequires just 4 colors, and with a minimal
number of vertices amongall graphs that satisfy that
condition, must be equal to K5.This is true, but note that we
have dropped the key property ofplanar from the assumptions of
G. Rather, we are expected to take aplanar G which requires 5
colors, and with the property that removingany vertex results
in a graph that requires just 4 colors, and withthe minimal
number of vertices among all graphs that satisfy
thatcondition. Clearly, any graph G that satisfies the
conditions of (a) Being planar; (b) Requiring 5 colors; (c)
Removing any vertex results in a planar graph that requires
only 4 colors;also satisfies just (b) and (c); so a graph
which satisfies (a)-(c)and has a minimal number of vertices
among all graphs satisfying thiscondition would necessarily
have 5 or more vertices (since a graphthat satisfies just (b)
and (c) and has a minimal number of verticesamong all graphs
satisfying (b) and (c) has exactly 5 vertices). Butthe only
minimal graph that satisfies (b) & (c), namely K5, does
notsatisfy (a). So the number of vertices of this hypothetical
G will bestrictly greater than 5.But after that, you seem to be
arguing that in fact this hypotheticalG would necessarily have
no more than 5 vertices. And I think thatthe reason you are
making this argument is that you point out that inorder to
reduce G to a 4-colorable graph, you must remove all
verticesof some color, which means that the vertex you removed
was the onlyvertex of the given color; which in turn means that
each vertex is theonly vertex of its color, which means G has 5
vertices, which means Gis K5, which is a contradiction.Or
something like that.Now, if that is not what you are arguing,
then you may ignore thispost and everything that follows.So,
assuming I got the gist of your argument correct, the error is
inthe step that goes from the vertex you removed was the only
vertex ofthe given color to each vertex is the only vertex of
its color.The fact that G-{v} can be 4 colored but G cannot
means that for eachvertex v, there exists a coloring C(v),
which ->depends on v<-, withthe property that v is the only
vertex of its color. However, if v andw are two distinct
vertices, there is no guarantee that the coloringof G-{v} is
compatible with the coloring of G-{w}; that is, there is
acoloring C(v) which depends on v and in which v is the only
vertex ofits color, and there is a different coloring C(w)
which depends on win which w is the only vertex of its color,
but there is no reason toassume that w is the only vertex of
its color under the coloring C(v),and there is no reason to
assume that v is the only vertex of itscolor under the
coloring C(w). This was mentioned by Erick Wong
40morgoth.sfu.caAgain, consider the 5-cycle, that is the graph
consisting of 5vertices, {1,2,3,4,5}, with adjacencies
1-2-3-4-5-1 (so each n isadjacent to n+1 (mod 5) ).Removing
any vertex results in a 2-chromatic graph; the graph
itself,however, is not 2-chromatic, it requires 3 colors. For
->each<-vertex, there is a coloring of G in which that vertex
is the onlyvertex of its color. However, there is no 3
coloring of the graph inwhich ->each<- vertex is the only
vertex of its color, and there is noreason to assume that this
is the case from the fact that there is acoloring for each
vertex.That is, we are encountering a typical fallacious
exchange ofquantifiers. We have:(1) For every vertex v, there
exists a coloring C such that v is the only vertex of its
color;and you seem to be interpreting this as being equivalent
to(2) There exists a coloring C such that for every vertex v, v
is the only vertex of its color.The two statements are not
equivalent; (2) implies (1), but (1) doesnot imply
what I accept as reality. --- Calvin (Calvin and
===
closure>Can the reals be defined using repeated exponentative
closure?>By the exponentative closure F, I define F/x as the
set of all the>zeroes of all the polynomial functions with
coeffeicients AND>exponents in F. For example, the the
algebraics are the exponentative>closure of the integers.
Thus, it can be written A=Z/x. Does>C=A/x? If not what does
A/x equal? Can C be generated by>repeatedly exponentatively
closing the integers a finite number of>times? If so, how
many? A countable number of times? An uncountable>number of
times?The set of all that is obtained is countable. This is
becauseeach polynomial has only a finite number of
coefficients andexponents, and returns only a finite number of
results. This is assuming that a clear definition exists of x^y
if x is negative and takes the value |x|^y; more can be
done.Doing this more than omega (the order type of the
integers)times yields nothing new, because there are only a
finite numberof arguments for each extension.-- This address
is for information only. I do not claim that these viewsare
those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
===
start with calculation of integral>Triple Integral from 0 to
Inf .>Integrant: 1/(1+x^a+y^b+z^c)dxdydz>Calculate the
integral and find a,b,c for which integral convergent.>(Should
be solved using a Gamma functions I think)You might start by
showing thatint_0^infinity (A+x^a)^(-q) dx = A^(1/a-q) Gamma(q
- 1/a) Gamma(1/a)/(b Gamma(q))under suitable assumptions on A,
a, q. This is related to the Beta function integral.Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
===
Continuous Game-Boards?>I was just wondering today if there
have even ever been played, games such>as Chess, Go, etc,
where the game-boards are without any subdivision?On a
slightly related basis, I've been thinking about
somethingsimilar in the past, but I was rather day-dreaming
about quantum-likedelocalized pieces, or continuous
distribuitions of pieces. Needlessto say I had no actual idea
of one such game...Rather than chess something like draughts
would be more suitable IMHOfor such an anti-discretization,
but in any case it would be hard todefine what a move could
be, or how to eat an opponent's piece (ehm, acomponent of the
opponent's distribuition of pieces).Funny to think of, having
such a game to be played on a computer, atsome point a (much
finer) discretization should be introducedanyway...Just my 2
(Euro)cents,Michele-- > Comments should say _why_ something is
being done.Oh? My comments always say what _really_ should have
===
Re: polysigned numbersNote: I'm posting through Google since
my regular news-feed seems tobe screwed up.> An interesting
point about reduced form is that it so far has not been>
absolutely necessary except for the dimensional analysis. That
is to> say that a number like:> ( - 1 + 2 * 3 # 4 ) > can be
worked with even though it is not in its reduced form. So far>
none of the math, including product and graphical analysis
fails to> work with non-reduced numbers. The reduced form of
the above number> is:> ( + 1 * 2 # 3 ).> Is this is the
reduction you are speaking of? This is one of two possible
reductions. This one corresponds to thetetrahedral
interpretation, RxRxR. There is a second possiblereduction:(
-1 +2 *3 #4) = (*2 #2) which directly corresponds with ( i -2
-3i 4) = (-2i 2) so is isomorphic (again) with C. > The neat
thing is that> the graphical analysis performs the same
that the same will happen with the> tetrahedral. The origin is
at the center of a tetrahedron. the poles> go out from the
origin through the points that form the tetrahedron.> This is
the symmetrical mapping of four-signed math in cartesian>
space. Any point in RxRxR can be uniquely defined in
four-signed math.> Since I don't have the math I can't say to
have proven this. You can fairly easily use trig to compute
the (x,y,z) from the(-,+,*,#) coordinates.> I can see> it
though. Making this assumption and putting the #-pole in the
i> direction (as in i,j,k) we get the following partial
transform for a> four-signed value x:> a = n(x) - ( 1/3 )(
m(x) + p(x) + s(x) )> where a is the i component of the three
dimensional vector ai + bj +> ck.> n(x) is the # (number)
component, m the - (minus), p the + (plus), and> s the *
(star).> Now putting the minus pole in he ij plane and going
in a left-handed> direction we see that the one third
component yields an angle of:> pi - arccos( 1 / 3 )> where
arccos is the inverse cosine.> This should be the angle from
the center to any two corners of the> tetrahedron.> Please
note that I am not proving this. I'd like to find that angle
in> a book somewhere or sharpen up my triginometry skills so
that I could> verify this.> If this is true then there should
be no problem with a clean RxRxR> transformation for
four-signed math. The same concept should work> upward beyond
our sight to five signed(pi-arccos(1/4)) and beyond.> Please
keep in mind the simple sum rule:> - x + x * x # x = 0.Notice
that *both* simplifications obey the sum rule, but
representvery different structures. This will be important if
you developethis out to 5-signed math.> I've done some work on
the four-signed math this weekend. It looks like > the key to
making it work is carefully defining a reduced form. Under >
certain definitions of reduced from, n-signed math exists in
R^(n-1), > Under other definitions, n-signed math is
isomorphic to C. I need to > finish some more details, then
I'll let you know what I come up with. I > think that thinking
in terms of n-tuples will make things easier when > dealing
with general n-signed math. Otherwise you may need to switch
to > subscripted signs.> I will try to work with your
representation, but I am pretty happy> with the symbolics that
are used here since it is more arithmetic. At> some point I'd
like to go to code. Do you know C or C++?> I think four signed
is as far as I'd like to go for now in symbolic> format.I'm
reasonably comfortable with both. I'm working on C++ right
now.> I've written n-signed code for C++ and performed the
Mandelbrot> mapping for four-signed in the simplest planes.
Some look like the> usual Mandelbrot and some are simple
solids. It's quite likely that> there are some bugs in my code
so don't take my results too seriously.> I can't get too
excited about the Mandelbrot mapping anyhow. I was> very
disappointed when my three-signed code spat out the usual>
mandelbrot shape. I was really hoping for something new. What
does the> Mandelbrot test mean anyways?I believe it will
determine if your interpretation corresponds with Cor not. the
3-signed is equivalent to C, so Mandelbrot looked thesame.
Since 4-signed as you are doing it is *not* equivalent to
C,Mandelbrot calculations will produce different results.> It
has become apparent to me in writing code that the symbols
used for> sign should really include a zero sign below the
minus sign. This zero> sign is identical to the highest sign (
# for four-signed, * for> three-signed, + for two-signed).
Since the reals are symbollically> faulty in this way I will
continue on with the symbols I have chosen> for now.I wouldn't
worry about it.> I'm looking forward to your results but I fear
that you are straying> from the zero sum rule.I'm not, I just
want to make sure the tetrahedral model works in aconsistent
===
Visiting Assistant Professor at the University of Montana.>>A
further question along those lines is what is is the set, E,
of>numbers generated by repeated sums of products of rational
powers of>rationals? It is a subset of the algebraics, but
does it form a>field? > I believe so: if x_1 is a member of
E, then so are its conjugates x_2, >> x_3, ..., x_n.>> Note
that x_1 x_2 ... x_n is rational (being the constant term of a
>> monic polynomial over the rationals whose roots are x_1,
x_2, ..., x_n).>> And then 1/x_1 = x_2 ... x_n / (x_1 x_2 ...
x_n) is in E.> Robert Israel israel@math.ubc.ca>> Department
of Mathematics http://www.math.ubc.ca/~israel >> University of
British Columbia >> Vancouver, BC, Canada V6T 1Z2>>If we say
that a set, P, is exponentially closed if for all p,q in
P,>p^q is P, then can the complex numbers be defined as the
field that is>exponentially closed? Or even more simply, the
set that is>exponentially closed and contains -2 and 0? >Note:
that I am only considering fields with characteristic 0.No;
either set is too small.I assume that your operation ^:PxP ->
P is just a partial operation,not defined at (0,0); so just
redefine it to give ^(0,0) = 0.The smallest field of
characteristic zero that is exponentially closedwould be the
closure of Q under the operation ^. Since ^ is a
finitaryoperation (takes only a finite number of arguments),
the closure of Qis obtained as follows:S_0 = Q;T_{n} = S_n cup
^(S_n,S_n)where ^(S_n,S_n) is the image of (S_n,S_n) in C; that
is, all numbersof the form p^q with p,q in S_n, under that
definition; S_{n+1} = T_{n} cup +(T_{n},T_{n}) cup -(T_n) cup
*(T_n,T_n) cup ^{-1}(T_n-{0})where +(T_{n},T_{n}) is the sum
of any two elements of T_n, -(T_n) isthe additive inverse of
any element of T_n, *(T_n,T_n) is the productof any two
elements of T_n, and ^{-1}(T_n-{0}) is the
multiplicativeinverse of any nonzero element of T_nThen the
closure of Q isS_{omega} = Union_{nIf
not, what are the>properties of the smallest field, S, that is
exponentially closed>(cardinality, algebraic completeness,
convergence of>Cauchy-Sequences)?So S=S_{omega}.Cardinality is
clearly countable, by the argument above (a standardargument of
General Algebra). You would not get convergence of
Cauchysequences: since this is countable, there is a real
number which isnot in the set, and of course that real number
is the limit of acauchy sequence of rationals, which are all
in S. I suspect algebraic completeness will also fail: if r is
an algebraicnumber such that Q(r) is not solvable by radicals,
how would you get rin S? Of course, since you also have
non-algebraic numbers in S, thisis hardly a 'proof', more of a
'I would look at these kinds of numbersto try to settle the
question in the
about what I accept as reality. --- Calvin (Calvin and
===
me?>>
I am stuck on a proof. I have gotten a series down
to where I need to>>prove the following:>> lim c^1/n = 1 where
c>0>>n -> inf.>>Do you have any pointers?>>
There is a
proof of this in (among many others) What is Mathematicsby
Courant and Robbins.It starts with Bernoulii's
inequality(1+x)^n >= 1+n*x if n is a positive integerand x >=
0.Rewrite this as (1+y)^(1/n) <= 1 + y/nand choose an
===
Game-Boards? Visiting Assistant Professor at the University of
Montana.>In message , Arturo
Magidin [.snip.]>The basic difference is that with Japanese
rules, you count points for>territory surrounded and lose
points for pieces you have lost, while>with Chinese rules,
you count points for territory surrounded _or_>occupied, and
ignore pieces you have lost. Usually this makes
no>difference, there is a simple proof of this depending on
the fact that>each player has made the same number of
moves.>>Which is certainly not the case under Japanese rules.
At least the way>>I learned them: a player is allowed to pass,
and then the other player>>may either pass (terminating the
game), or play.>>Is this different under Chinese rules?>>It is
the same under Chinese rules. A player may always pass - two
>consecutive passes end the game.>>So, unless there are passes
before the final two passes, each player has >played the same
number of stones; except that Black may have played one >more
point than White. So you are right - it can make one point of
>difference.Well, no, it can make an arbitrarily large
difference (limited to thenumber of positions on the board, of
course). You play, I pass. Thenyou play, I pass. Then we
alternate plays until the final twopasses. Then it makes 2
points difference, does it not? And since Imay pass an
arbitrary number of times, would that not allow a
largedifference? The difference would equal the absolute value
of thedifference in the number of passes each player made
before the finaltwo
what I accept as reality. --- Calvin (Calvin and
===
me?>I am stuck on a proof. I have gotten a series down to
where I need to>prove the following:> lim c^1/n = 1 where
c>0>n -> inf.>Do you have any pointers?The limit of a discrete
series is the same as the limit of a smooth functionwith those
values. Why not say:lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] =
===
analysis: construct this set ...>Can you construct a set E in
[0, 1] s.t. for every open interval I in [0,1] m(I intersect
E) > 0 & m(I intersect E^c) > 0 >m is lebesgue measure >E^c is
the complement of E >This is so tricky! I was thinking
something with the generalized Cantor set but everything I'm
trying isn't working. Any suggestions? Ideas?I have previously
posted an example of such a set, which isBorel measurable. Use
the Cantor set idea, but with differentsized parts removed
from the intervals still in, and added tothe intervals already
removed. It can be done in terms of thedigits to any base.--
This address is for information only. I do not claim that
these viewsare those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
===
energy/matter equationsTechnical typo correction on the math
formula below for the exotic vacuum zero point stress-energy
density residual inflation field whose local control could be
applied to the Doomsday WMD described by Sir Martin Rees in
Chapter 9 of his book Our Final Hour.tuv(x)vac =
(c^4/8piG*)/(x)zpfguv(x)Lp*^2 = hG*/c^3 = Lp^4/3(c/Ho)^2/3
(t'Hooft-Susskind) ~ (1 fermi)^2 ~ (10^-13 cm)^2Ho is from
cosmology (Hubble parameter), c/Ho ~ 10^28 cmLp^2 =
hG(Newton)/c^3 ~ (10^-33 cm)^2/zpf(x) = Lp*^-2[1 - Lp*^3|Higgs
Field(x)|^2]is the unified dark energy/matter zero point vacuum
field.Giant vacuum superfluid local inflation wave from BCS
spontaneous U(1)em symmetry breaking pair instability in the
-mc^2 Fermi sphere band of the Dirac electron micro-quantum
vacuum isPSI(vac) = (0|e+(x)e-(x)|0) = |Higgs
Field(x)|e^i(Goldstone Phase(x) - Integral TaA^au(x)dx^u)Ta
are the generalized charges of standard model of lepto-quarks
and gauge forces.[D^uDu + V(|Higgs Field)] PSI(vac) =
0V(|Higgs Field|) in large scale FRW limit of wavelet
transform is the potential of inflationary cosmologyis the BIT
FROM IT Diff(4) covariant Landau-Ginzburg nonlinear local field
equationDu is the Diff(4) covariant derivativeguv(x) = nuv +
(1/2)[du(x),v + dv(x),u]is Einstein's gravity c-number local
geometrodynamic field where nuv is the flat constant Minkowski
metric, and the elastic world crystal lattice distortion field
isdu(x) = Lp*^2 (Goldstone Phase(x) - Integral
TaA^av(x)dx^v),uin 4D where ,u is the ordinary partial
derivativeis the IT FROM BIT Bohm guidance constraint
analogous tov = (h/m)GradPhase - (e/mc)Ain the micro-quantum
===
Re: Can you help me?>I am stuck on a proof. I have gotten a
series down to where I need to>prove the following:>>
lim c^1/n = 1 where c>0>n -> inf.>>Do you have any
pointers?>> The limit of a discrete series is the same as the
limit of a smoothfunction> with those values. Why not say:>>
lim[n->oo, c^(1/n)] = lim[x->oo,c^(1/x)] = lim[x->0,c^x] = c^0
= 1 (sincec> =/= 0).Well, at least if the function has a limit,
then the series has one too.Sometimes the function might not
even have a limit. In this case, f(x) =c^(1/x) does have a
===
this set ...>> Can you construct a set E in [0, 1] s.t. for
every open interval I in>[0,1] m(I intersect E) > 0 & m(I
intersect E^c) > 0 > m is lebesgue measure >> E^c is the
complement of E >A doubt. You say the Lebesgue measure. Are
you implying that the>constructed set must be Lebesgue
measurable? Or the problem somehow>gives us a notion that,
even if the wording Lebesgue measure is>replaced by Lebesgue
outer measure,it is impossible to find a>Lebesgue immeasurable
set satisfying the condition as the problem>indicates?By saying
m(...) where m is Lebesgue measure, it's clearly asking
formeasurable sets. As far as outer measure m^* is concerned,
you can do a lot better: there is a (nonmeasurable) set E such
that for every open interval I, m^*(E intersect I) = m(I) and
m^*(E^c intersect I) = m(I).Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
===
someone please explain what lopital's rule is? my teacher
likesto say things like, now, we could use lopital's rule, but
that'd betoo easy and you don't know it, so naturally it piqued
my interest.sorry to ask such a broad question, but i really
don't know anythingat all about it to possibly narrow the
===
... The example BBB in>> the TeXBook (my edition anyway) is
pretty obviously a CMR I and R>> run together. (I think it
looks better than any currently available>> doublestroke font
and have considered adapting cmr to produce>> something along
those lines.)>>You mean like the bbm.sty and fonts do?No, I
don't mean like that. I'm pretty sure I've checked all
currently available (free) doublestroke fonts. There's
ablackboard.ps by Olaf Kummer available on CTAN that lists
them (or at least those that were known to the author). And I
usually check out any new ones announced here.As to bbm:
compare the example R in the TeXBook (in index underblackboard
bold) to the bbm R. Not at all alike.Dan-- Dan Luecking
Department of Mathematical SciencesUniversity of Arkansas
Fayetteville, Arkansas 72701luecking at uark dot
===
explain what lopital's rule is? my teacher likes> to say
things like, now, we could use lopital's rule, but that'd be>
too easy and you don't know it, so naturally it piqued my
interest.It's L'Hopital's rule (with a circumflex over the
o).It's the principle that if f(x) -> 0 and g(x) -> 0as x -> a
but if f'(x)/g'(x) -> L as x -> a thenf(x)/g(x) -> L as x ->
polemics from me on how evil and useless this ruleis and how
it should never be mentioned to impressionable undergraduates
:-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless
to say, I had the last laugh. Alan Partridge, _Bouncing Back_
(14 times)