mm-196 === Subject: Re: Silly question on limits, tensor products> In a nutshell, I see this line in my text : Let u(t) be a smooth curve in> the vector space U and v(t) be a smooth curve in the vector space V.> > lim [ u(t+h) tensor [v(t+h) - v(t)]/h ]> h--> 0> > => > u(t) tensor dv/dt> > Fine and dandy. But I am stepping back a bit and asking myself the> following: There was one step that was skipped. That is :> > lim [ u(t+h) tensor [v(t+h) - v(t)]/h ] => h--> 0> > lim u(t+h) tensor lim ( [v(t+h) - v(t)]/h )> h ----> 0 h --> 0> > Why is that true? (I know, it is a stupid stupid question) The way I> justify this is by basically hand-waving and saying it makes sense, but I> am not convinced that you can do this type of thing for every occasion in> which a limit breaks up into two limits. When can you do this sort of> thing with limits where one breaks up into two and when can't you? I know> I'm being a bit vague, but if someone has a quick counterexample or can> understand what I'm trying to say then I'd really appreciate the insight.If I have my de'nition right, a 2-tensor in this situation is a bilinear map T from U x V into R. But then there are numbers a(i,j) such that T((x1,...,xn),(y1,...,yn)) = sum a(i,j)*xi*yj,where we sum over all (i,j). Therefore T is continuous on U x V, which easily implies your result.As for a general discussion fof this sort of thing, I feel it's better to deal with these situations as they arise. In time you'll get a feel for it and the proofs should take a matter of seconds. === Subject: Re: Need help with proving that Z[i] is Euclideanmagidin@math.berkeley.edu (Arturo Magidin) of measuring the size of the>elements. What you want to show is that given two elements a and b of>Z[i], with b different from 0, then it is possible to divide a by b,>which means 'nding a ->unique<- quotient q and a ->unique<- remainder>r subject to the following two conditions:>> (1) a = b*q + r; and> (2) either r=0, or the size of r is strictly smaller than the> size of b....>The remainder/residue? Well, that's the de'nition of division>algorithm, so you must actually be trying to prove that you can>->de'ne<- a division algorithm for the Gaussian integers; that is,>that you can 'nd q and r as described above.Well, you're the trained algebraist, and I'm not, but I don'tsee why you're so insistent about the uniqueness of q and r,so long as (1) and (2) are satis'ed (and I sort of think I remember P. M. Cohn explicitly not requiring uniqueness, though of course in his case he's throwing away a lot of the other properties most people want in their rings, as well).Even in the integers, e.g., when using continued fractions(or equivalently--and the reason I was once sort of up on thiskind of stuff--when 'nding a reduction algorithm for SL(2,Z) modular symbols), it's sometimes handy to go for the smallest positive remainder and sometimes handy to go for the remainder of smallest absolute value. Sure, inthe latter case, you could say there's just one remainder,and it's that one, but you lose nothing that I can seeby saying there are various pairs (q,r), for each ofwhich r is of smaller size than b; and I can imaginecases in which you might want actually to take a continuedfraction expansion with some remainders positive (thoughnot necessarily of smallest absolute value) and othersnegative (ditto). Certainly for the application to g.c.d.'s, uniqueness isn't used, is it? All that'sused is that the measure of size doesn't admit anyin'nite descending chains.Lee Rudolph === Subject: Re: Need help with proving that Z[i] is Euclidean Adjunct Assistant Professor at the University of Montana.>magidin@math.berkeley.edu (Arturo Magidin) way of measuring the size of the>>elements. What you want to show is that given two elements a and b of>>Z[i], with b different from 0, then it is possible to divide a by b,>>which means 'nding a ->unique<- quotient q and a ->unique<- remainder>>r subject to the following two conditions:>> (1) a = b*q + r; and>> (2) either r=0, or the size of r is strictly smaller than the>> size of b.>...>>The remainder/residue? Well, that's the de'nition of division>>algorithm, so you must actually be trying to prove that you can>>->de'ne<- a division algorithm for the Gaussian integers; that is,>>that you can 'nd q and r as described above.>>Well, you're the trained algebraist, and I'm not, but I don't>see why you're so insistent about the uniqueness of q and r,>so long as (1) and (2) are satis'ed (and I sort of think I >remember P. M. Cohn explicitly not requiring uniqueness, >though of course in his case he's throwing away a lot of >the other properties most people want in their rings, as well).Fair enough. It does not really matter so long as the remainder dropsin size, for the purposes of the Euclidean algorithm. -- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Need help with really appreciate it.The case is that I've already got the proof that there is a divisionalgorithm de'ned for Gaussian integers. The way I 'gure it, I haveto prove that the remainder that shows up when I use the algorithmfor the second, third..n times is smaller than the one that showedup after n-1 steps, right?This isn't so hard since I've already proved that there exists adivisions algorithm for Gaussin integers. Am I right?Once again, thank you for all the help.YoursPierreps. Sorry if there are any mistakes in my spelling; Enlish isn't mynative language. ds === Subject: Re: Need help with proving that Z[i] is Euclidean Adjunct for all the help, really appreciate it.>The case is that I've already got the proof that there is a division>algorithm de'ned for Gaussian integers. The way I 'gure it, I have>to prove that the remainder that shows up when I use the algorithm>for the second, third..n times is smaller than the one that showed>up after n-1 steps, right?No need: all you have to show is that: (1) each time you apply the division algorithm, the remainder is smaller than the divisor; and (2) there cannot be an in'nite decreasing sequence of Gaussian integers; that is, these Gaussian integers cannot get smaller and smaller and smaller, inde'nitely, without reaching $0$.You would also have to explain why the Euclidean algorithm yields agcd, but as I noted in my previous reply, the exact same argument asyou have for the integers will work.>This isn't so hard since I've already proved that there exists a>divisions algorithm for Gaussian integers. Am I right?Yes, if somewhat convoluted. If you can show that at each step, theremainder is smaller than the immediate previous one (that is, show(1) above) then that is enough, by induction, to show that theremainder in step n is strictly smaller than the remainders in steps 1through n-1. You still need to show the process terminates, of course.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Rational sines and cosinesDoes there exist a rational number r such that 2r is not an integer and both the sine and cosine of pi r are rational?Put another way, does there exist a right triangle with integer sides and all angles of rational degree measure?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Rational sines and cosines> Does there exist a rational number r such that 2r is not an integer > and both the sine and cosine of pi r are rational?The only rational r between 0 and 1/2 such that sin pi r is rational are r = 0, 1/6, and 1/2. The only rational r between 0 and 1/2 such that cos pi r is rational are r = 0, 1/3, and 1/2. By the various symmetries inherent in the trig functions you can now work out all the rational r, etc.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Rational sines and cosines>> >Does there exist a rational number r such that 2r is not an integer >>and both the sine and cosine of pi r are rational?>> >>The only rational r between 0 and 1/2 such that sin pi r is rational >are r = 0, 1/6, and 1/2. >>The only rational r between 0 and 1/2 such that cos pi r is rational >are r = 0, 1/3, and 1/2. >>By the various symmetries inherent in the trig functions you can now >work out all the rational r, etc.>So the answer to my question is no. How do we know these facts?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Rational sines and cosines> > >> > > >> >>Does there exist a rational number r such that 2r is not an integer > >>and both the sine and cosine of pi r are rational?> >> > > >The only rational r between 0 and 1/2 such that sin pi r is rational > >are r = 0, 1/6, and 1/2. > >> >The only rational r between 0 and 1/2 such that cos pi r is rational > >are r = 0, 1/3, and 1/2. > >> >By the various symmetries inherent in the trig functions you can now > >work out all the rational r, etc.> >> So the answer to my question is no. How do we know these facts?2 cos pi r = exp(ir) + exp(-ir) is an algebraic integer (because exp(ir) and exp(-ir) are algebraic integers) and a rational number (because, by hypothesis, cos pi r is rational) so it's an ordinary integer (because the only rationals that are algebraic integers are the ordinary integers) so it's -2, -1, 0, 1, or 2 (since absolute value of cos pi r is at most 1) so we know what r can be. Similarly for sin pi r. There may be a way to do it without invoking algebraic number theory, but then again the algebraic number theory I've invoked is 'rst-week-of-class stuff.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Dogs, ¤eas, and hairy ballsGib Bogle (or somebody else of thusly:> I have made an interesting (to me) observation about my dog (a bitch, by> the way).The third element of the subject line is therefore a bit puzzling. Now I've got ... ah, I see what you mean.-- Paul TownsendI put it down there, and when I went back to it, there it was GONE!Interchange the alphabetic elements to reply === Subject: Topological property that represents Ounboundedness'When traversing the surface of a sphere, one can go in any directionforever, even a straight line, and never encounter an Oedge' of the surface.What is the topological invariant that represents this property?l8r, Mike N. Christoff === Subject: Re: Topological property that represents Ounboundedness'> When traversing the surface of a sphere, one can go in any direction> forever, even a straight line, and never encounter an Oedge' of the surface.> What is the topological invariant that represents this property?> > l8r, Mike N. Christoff> > > I believe that is called a surface without boundary.> Let me also qualify that I'm interested in objects with 'nite surfaces.perhaps compact surface without boundary... === Subject: Re: Topological property that represents Ounboundedness'Michael N. Christoff traversing the surface of a sphere, one can go in any direction> forever, even a straight line, and never encounter an Oedge' of thesurface.> What is the topological invariant that represents this property?>Let me also qualify that I'm interested in objects with 'nite surfaces.l8r, Mike N. Christoff === Subject: Re: Topological property that represents Ounboundedness'> > Michael N. Christoff the surface of a sphere, one can go in any direction>> forever, even a straight line, and never encounter an Oedge' of the> surface.>> What is the topological invariant that represents this property?> > Let me also qualify that I'm interested in objects with 'nite surfaces.Hmmm. Let's compare the real plane R^2 with the unit open disc.In the 'rst you can potter along, and never reach an edge. In the latterone falls off the edge pretty rapidly. So what is the topologicaldistinction between these two surfaces? What in the topology makes one'nite and not the other?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: irrationality of sqrt(2): easy question> >> >Dirk Van de moortel > > How to prove a^2 even => a even without using irrationality of>> > sqrt(2)?> > Doesn't it follow easily from unique factorization into primes?Sledgehammer -> nut.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: irrationality of sqrt(2): easy question Adjunct Assistant Professor at the University of Montana.>> Doesn't it follow easily from unique factorization into primes?>>Sledgehammer -> nut.In Mexico, the expression is swatting ¤ies by 'ring cannons atthem...-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: rank of a random {0,1} matrixGiven a nxk random matrix R, k>n, each entry of the matrix is either 1or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,what is the probability that === Subject: Re: rank message> Given a nxk random matrix R, k>n, each entry of the matrix is either 1> or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,> what is the probability that that you mean 0-1 matrices in the 'eld of rational numbers.In this case I doubt that there is a nice formula. In the case of nxnmatrices the problem would be equivalent to counting the number ofnonsingular (regular) 0-1 nxn matrices. Below is the entry in the OEIS whichgives the 'rst few terms. Note that the sequence is characterized as hard(as well as nice). This means it is hard to 'nd the next terms. (If you arenot familiar with the OEIS you can get to it using the URL below.)Since it is nice, people would have tried to 'nd a formula. :-)For example the number of 7x7 nonsingular matrices (full rank) is accordingto this entry equal to 270345669985440ID Number: A055165URL: http://www.research.att.com/projects/OEIS?Anum=A055165Sequence : 1,6,174,22560,12514320,28836612000,270345669985440Name: Number of regular n X n matrices with rational entries equal to 0or 1.Comments: All eigenvalues are nonzero.Links: E. W. Weisstein, Link to a section of The World of Mathematics. Index entries for sequences related to binary matricesExample: For n=2 the 6 matrices are {{{0, 1}, {1, 0}}, {{0, 1}, {1, 1}},{{1, 0}, {0, 1}}, {{1, 0}, {1, 1}}, {{1, 1}, {0, 1}}, {{1, 1}, {1, 0}}}.See also: Cf. A056990, A056989, A046747, A055165, A002416, A003024(positive de'nite matrices). Also A046747(n) + a(n) = 2^(n^2) = total number of n X n (0, 1)matrices, sequence A002416. Adjacent sequences: A055162 A055163 A055164 this_sequence A055166 A055167 A055168 Sequence in context: A078535 A003720 A002884 this_sequenceA071095 A024277 A012177Keywords: nonn,nice,hardOffset: 1Author(s): Ulrich Hermisson (uhermiss(AT)server1.rz.uni-leipzig.de), Jun 18 2000If you browse the OEIS you may 'nd sequences corresponding to 2xn and 3xn(0-1) matrices of full rank. I haven't checkedl--Edwin Clark === Subject: Re: rank of a random {0,1} matrix> Given a nxk random matrix R, k>n, each entry of the matrix is either 1> or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,> what is the probability that the answer.Cooper explained this in (Theorem 1 in [1]) that: If $V$ is a $n times k$ binary random matrix with entries independently and identically distributed as $Pr{v_{ij} = 1} =0.5 $ and $Pr{v_{ij}=0}=0.5$, then [ lim_{k to infty} Pr{ rank(V) = n } = prod_{j=k-n+1}^{infty} left( 1- 0.5^{j} right ) ][1] C. Cooper. textit{On The Distribution of Rank of a Random Matrix Over a Finite Field}, Random Structures and Algorithms 17(3:4), 2000 === Subject: Re: rank of a random {0,1} matrix>Cooper explained this in (Theorem 1 in [1]) that: >If $V$ is a $n times k$ binary random matrix with entries independently >and identically distributed as $Pr{v_{ij} = 1} =0.5 $ and >$Pr{v_{ij}=0}=0.5$, >then >[ lim_{k to infty} Pr{ rank(V) = n } = prod_{j=k-n+1}^{infty} >left( 1- 0.5^{j} right ) ]>>[1] C. Cooper. textit{On The Distribution of Rank of a Random Matrix Over >a Finite Field}, Random Structures and Algorithms 17(3:4), 2000There must be a typo here because the formula as stated does notmake sense. On the left you have k to infty so the right handside should not involve k.-- Rouben Rostamian === Subject: Re: rank of a random {0,1} matrixroy Given a nxk random matrix R, k>n, each entry of the matrix is either 1> > or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,> > what is the probability that the rank of the Cooper explained this in (Theorem 1 in [1]) that:> If $V$ is a $n times k$ binary random matrix with entries independently> and identically distributed as $Pr{v_{ij} = 1} =0.5 $ and> $Pr{v_{ij}=0}=0.5$,> then> [ lim_{k to infty} Pr{ rank(V) = n } = prod_{j=k-n+1}^{infty}> left( 1- 0.5^{j} right ) ]>> [1] C. Cooper. textit{On The Distribution of Rank of a Random Matrix Over> a Finite Field}, Random Structures and Algorithms 17(3:4), 2000mentions a result I derived years ago for use in a testsof randomness, in which various bits from a sequence ofrandom integers are used to form mxn matrices, then theranks determined over the 'eld mod 2.Using TeX notation: The rank of a random $mtimes n$ binary matrix takesthe value $r=1,2,ldots,min(m,n)$ with probability$$2^{r(n+m-r)-mn}prod_{i=0}^{r-1}frac{(1-2^{i-n })(1-2^{i-m})}{(1-2^{i-r})}.$$I pointed out in a sci.math query a few years agothat the result could extended to 'elds mod p, a prime.But it will not apply to 0-1 matrices over the real (orrational) 'eld, where counting the number of ways to geta 0-1 mxn matrix of rank r is much more dif'cult.(A current version of the Diehard battery of tests of randomness isavailable athttp://www.csis.hku.hk/~diehard/ ,including the binary rank test.)George Marsaglia === Subject: Re: rank of a a nxk random matrix R, k>n, each entry of the matrix is either 1> > > or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,> > > what is the probability that the rank of the answer.> > Cooper explained this in (Theorem 1 in [1]) that:> > If $V$ is a $n times k$ binary random matrix with entries independently> > and identically distributed as $Pr{v_{ij} = 1} =0.5 $ and> > $Pr{v_{ij}=0}=0.5$,> > then> > [ lim_{k to infty} Pr{ rank(V) = n } = prod_{j=k-n+1}^{infty}> > left( 1- 0.5^{j} right ) ]> >> > [1] C. Cooper. textit{On The Distribution of Rank of a Random Matrix Over> > a Finite Field}, Random Structures and Algorithms 17(3:4), 2000> > mentions a result I derived years ago for use in a tests> of randomness, in which various bits from a sequence of> random integers are used to form mxn matrices, then the> ranks determined over the 'eld mod 2.> Using TeX notation:> > The rank of a random $mtimes n$ binary matrix takes> the value $r=1,2,ldots,min(m,n)$ with probability> > $$2^{r(n+m-r)-mn}prod_{i=0}^{r-1}> frac{(1-2^{i-n})(1-2^{i-m})}{(1-2^{i-r})}.$$> > I pointed out in a sci.math query a few years ago> that the result could extended to 'elds mod p, a prime.> > > But it will not apply to 0-1 matrices over the real (or> rational) 'eld, where counting the number of ways to get> a 0-1 mxn matrix of rank r is much more dif'cult.> > > (A current version of the Diehard battery of tests of randomness is> available at> > http://www.csis.hku.hk/~diehard/ ,> > including the binary I got a new question.Let M= (m_ij) be a (mxn) matrix with entries of a 'nite 'eld{a1,a2,a3..at},Pr(m_ij = a_1) = p1, Pr(m_ij = a_2) = p2, ...,Pr(m_ij = a_t) = pt, andp1+p2+,..pt=1. How to measure the probability of the rank === Subject: Re: rank of a random {0,1} matrix> matrix R, k>n, each entry of the matrix is either 1> > or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5, what is the probability that the rank of the matrix is n ?> > explained this in (Theorem 1 in [1]) that: > If $V$ is a $n times k$ binary random matrix with entries independently > and identically distributed as $Pr{v_{ij} = 1} =0.5 $ and > $Pr{v_{ij}=0}=0.5$, > then > [ lim_{k to infty} Pr{ rank(V) = n } = prod_{j=k-n+1}^{infty} > left( 1- 0.5^{j} right ) ]> > [1] C. Cooper. textit{On The Distribution of Rank of a Random Matrix Over > a Finite Field}, Random Structures and Algorithms 17(3:4), 2000If Cooper is really talking about matrices over a 'nite 'eld, then his answer is not to your question. E.g., the matrix 1 1 0 0 1 1 1 0 1 has rank 3 as a real matrix but rank 2 over the 'eld of two elements.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: rank of a random {0,1} matrix> Given a nxk random matrix R, k>n, each entry of the matrix is either 1> or 0 with equal probability, i.e., Pr{r_ij=1}=0.5, Pr{r_ij=0}=0.5,> what is the probability that the rank of the matrix is n ?I can't tell you what the exact value is, but I can tell youthat it is greater than the probability that you will getothers to do your homework for you.Carlos--PS: Interesting problem, though! === Subject: Re: Aspiring mathematicians, send me your proofs!First off the mark, from Australia, Mark Hurd!See the DC Proof User's Gallery athttp://www.dcproof.com/Gallery.htmKeep those submissions coming, folks!DanVisit DC Proof Online at http://www.dcproof.comDan Christensen your proofs published at my website. Use my DC Proof software (FREE)to> generate your proofs in HTML format (see File / Make HTML File option).> Don't worry about making mistakes. They are impossible in DC Proof!>> Send your proofs as HTML attachments to me at: dc@dcproof.com>> Be sure to include a caption , an introduction and lots of comments inyour> proofs. (See Documenting and Viewing your Proof in the User Reference> Guide.)>> Suggested topics: Introductory theorems in logic, set theory, numbertheory> and group theory. Or introduce any axioms as you see 't -- it's YOURproof!>> Suggested limit: 100 lines (¤exible).>> Optional: Include your name. If a student, your college, university or> school, and year.>> Download my free DC Proof software at>> http://www.dcproof.com>> Includes self-study tutorial (see excerpts at my homepage).>> Dan === Christensen> Toronto, Canada>>Subject: A bounded function on [a, b] is Riemann integrable iff the set ofpoints ofdiscontinuity has Lebesgue measure 0.Now for f in your example, clearly it's R-integrable with value 0.For g, it's not bounded when x -> 0. But g is R-integrable on [x,1] for any0 < x < 1 (with integration 0) So if talking about improperRiemann integration,i.e. int_0^1 g = lim_{x->0^+} int_x^1 g. Then g is stillintegrable.therfore g is boundedand integrable on interval [a, 1] for any a with 0 < a < 1.for the given g , it is not R-integrableis my view Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption I would like to know how to 'nd the gradient of the tangent at theturning point of a hyperbola that doesn't have asymptotes goingstraight up and straight across. Finding the turning point of a nicesimple hyperbola is easy since the gradient is 1 or -1, that'sobvious, but I'm not quite sure how to 'nd them Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: hyperbolasQuantumcat> I would like to know how to 'nd the gradient of the tangent at the> turning point of a hyperbola that doesn't have asymptotes going> straight up and straight across. Finding the turning point of a nice> simple hyperbola is easy since the gradient is 1 or -1, that's> obvious, but I'm not quite sure how to 'nd them gradient means in this context. But guessing a little,if then gradient is a rate of change of an angle, then it's value at an apexis probably a simple function of the angle between the asymptotes.LH === Subject: Re: A 'nite set that actually has more elements than an in'nite one. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id Exploring the Universe by Seeds gives lower-dimensional>> projections as examples of different types of constructions of the>> unverse (closed, ¤at and open.)>>OK, git, how do you reconcile WMAPS data plus Sloan Digital Sky Survey>data plus Jeffryy Weeks connected dodecahderal universe 't (and its>impled intrinsic chirality) with a 2-D projection?>>Google>Jeffrey Weeks dodecahedron 153 hits>Jeffrey Weeks dodecahedral 82 hits>>Why don't you tell us how to ¤atten a Seifert-Weber dodecahedral>space?> >> A plane is given as a ¤at universe, the surface of a sphere as a>> closed universe, and the surface of a saddle as an open universe.>>You don't know about the subject. There are lots of compact>minimal surfaces with that are Euclidean without being ¤at planes. >There are fully *eight* simply-connected geometric 3-manifolds with>compact quotients,Why doesn`t Uncle Al just admit right here and now that it's nothimself he`s refering to, but to his own uncle, Al? Then, we mightat least 'nd some sympathy (or at least) some explanations. === Subject: Re: prime relative to involutionIn the following I write pn to mean p_n (p sub n) to save space/time.> De'ne a natural number n to be hypercomposite if there exist two natural numbers> a and b such that a^b=n (b is not 1),and hyperprime if no such numbers exist.If n has the prime factorization n = p1^a1 p2^a2 p3^a3 ... pn^am, thenn is hyperprime iff d = gcd(a1, a2, a3, ... am) = 1. If we let c =p1^(a1/d) p2^(a2/d) ... pn^(am/d) then n = c^d with c hyperprime.> Then,> Every natural number is either a hyperprime or can be represented uniquely > as the result of a left to right involution with a hyperprime base and normal> prime exponents;e.g:81=(3^2)^2,256=((2^2)^2)^2.Trivial -- using the above def'n of c and d, factor d into q1 q2 q3... qm (with the possibility that qi = qj) and n = ((((c ^ q1) ^ q2) ^q3) ... ^ qm). Of course this is unique only up to the order ofexponents. There doesn't seem to be any reason to do such afactorization, though.> Every natural number is either a hyperprime or can be represented uniquely> as the result of a right to left involution with hyperprime exponents;e.g:> 81=3^(2^2),256=2^(2^3).This is clearly not true; i.e. 4^3 = 8^2, and both 3 and 2 arehyperprime. If you require the base to be hyperprime as well it'strue; you can just write n = c^d as above, then let d = (c')^(d'), d'= c'' ^ d'' etc. until d(n) is hyperprime and then n = c ^ (c' ^ (c''^ c''' ...)) ^ d(n).> Does any one 'nd this representation theorem important?Probably not. First of all, since hyperprimes can be constructedquite easily out of primes, they're probably not a much richerconcept. Further, these properties are rather trivial, so if someoneneeds to use them they would quickly develop them on their own.Looking that over, it sounds kinda harsh. I don't mean to burst yourbubble here; when I 'rst heard about the Laplace transformation inhigh school I remember I tried to develop something similar usinglimits. After working on it on and off for a week I or so eventuallyfound out that it reduced to something silly like T(f(x)) = f'(x)/f(x)or something like that. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> > > > I will use this to show that > > > if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is > > > irrational.)> > > > But that wasn't the original question. The original question > > was whether x / pi is irrational. A whole nother kettle of 'sh.> > To make this question actually look interesting, note that if > cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5.> > (Also, the original question has sqrt(5)-1 instead of sqrt(5)+1 .> Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The question is > to show that this 2/5 value turns transcendental when that O4' on the > denominator changes to a O2.')> > JSo sorry for the bleeps in my non-solution.Yes, the original question did involve (sqrt(5)-1)/(2 *p)i notsqrt(5)+1 and Lindemann's lemma andmy proof are only true if x ne 0. Of course e^0 = 1.I will go back to the drawing board and see if my original proof canbe 'xed.Sorry for the whole nother kettle of Steiner === Subject: Re: need help!!> > > > > > > > > > > > i can not integrate from arcsin[2/(3+cosx)]> > > > > > if u can ,please say me by hupo19@yahoo.com> > > > > > i know answer by i dont know how it solve> > > thank you > > > hupo> > > > What answer do you have?> > J> > dear jim> my answer is =x*arcsin[2/(3+cosx)]This answer is incorrect, and looks like it came from a computer. Thecomputer misunderstood cosx to be a constant or something, not thecosine of x. A general inde'nite solution for this integral is notknown; you can, however, use a computer to estimate the de'niteintegral between two points. === Subject: Re: need help!!> > > > > > > > > > > > > i can not integrate from arcsin[2/(3+cosx)]> > > > > if u can ,please say me by hupo19@yahoo.com> > > > > > i know answer by i dont know how it solve> > > > thank you > > > > hupo> > > > > > What answer do you have?> > > J> > > > dear jim> > my answer is =x*arcsin[2/(3+cosx)]> > This answer is incorrect, and looks like it came from a computer. The> computer misunderstood cosx to be a constant or something, not the> cosine of x. A general inde'nite solution for this integral is not> known; you can, however, use a computer to estimate the de'nite> integral between two points.hi 1.if you say that my answer is wrong then what is it?2.if my qustion has not answer from anyway theni think that if we CAN plot it , and we CAN calculate area [for example from -1 to +1] we recived answer thank you for yuor help hupo === Subject: Re: need help!! 2.if my qustion has not answer from anyway theni think that if we CAN plot it ,> and we CAN calculate area [for example from -1 to +1] we recived answer> thank you for yuor help> hupo For your example of -1 to 1, the integral of arcsin(2/(3+cos(x))) is about 1.097132985 .J === Subject: Re: need help!!> > > hi > > 1.if you say that my answer is wrong then what is it?> > 2.if my qustion has not answer from anyway theni think that if we CAN plot it ,> > and we CAN calculate area [for example from -1 to +1] we recived answer> > thank you for yuor help> > hupo> > For your example of -1 to 1, the integral of arcsin(2/(3+cos(x))) is > about 1.097132985 .> > Jdear jim how you 'nd this answer[1.097132985]for [-1,1]? do you plot it? if you dont plot it please say me that ,how do you 'nd thisanswer,and if you plot it ,please show methank you ,hupo === Subject: Re: need help!! example of -1 to 1, the integral of arcsin(2/(3+cos(x))) is about 1.097132985 .> > > > J> > how you 'nd this answer[1.097132985]for [-1,1]? do you plot it? > if you dont plot it please say me that ,how do you 'nd this> answer,and if you plot it ,please show me There are computer programs that use numerical methods to evaluate areas. I used Maple, some people use Mathematica, there are probably many more. Perhaps there are even java applets available on the web to do these. Or, if you only have that one function to work on, you can write your own program to implement one of the many integration approximation routines given in a standard calculus book. Or, you could write out a taylor series for your function, and then integrate them term-by-term, and keep as many terms as you want (more terms -> more accuracy.) I did not plot the graph... but I could.J === Subject: Re: need help!! 2.if my qustion has not answer from anyway theni think that if we CAN plot it ,> and we CAN calculate area [for example from -1 to +1] we recived answer> thank you for yuor help 1. Yes, your answer is incorrect. The is no answer to what is it because there probably is no closed form for an inde'nite integral for the function you gave. 2. Yes, of course, the function is integrable (i.e. the upper and lower Riemann sums will converge to one another) but this does not mean that there is a function expressible with elementary functions whose derivative will be the function you initially gave. If you need to be 'nding the area under your function, there are numerical methods to 'nd that area within any precision you need (provided you specify the bounds of integration.)J === Subject: JSH: Math is hardI did a neat trick yesterday by 'nally realizing that I coulddirectly challenge posters making various claims about factors in thering of algebraic integers by using my own quadratic:y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2With what I've been calling the Decker quadratic various posters haveasserted that you can have some algebraic integer functions w_1(x) andw_2(x), wherew_1(x) w_2(x) = 7which vary as x varies, so I simply used a method to directly check byusing a quadratic with a variable b, so that I can let b be *any*algebraic integer.So, in fact, b = w_1(x) - w_2(x), is a possibility, which directlychallenges those people.What I end up with, at x=2, from the Decker quadratic is7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0which is important because it's non-monic, but *all* of its roots areactually checks against various possibilities for w_1(x) and w_2(x).It's easy to see that it takes away an integer possibility for b, butthe method reveals that no algebraic integer can possibly work, atall.Why does it have to work?Oddly enough, you can 'gure that out from what happens at b=0, asthen you havey^2 - 7 = 0.ANY polynomial that you end up with must always allow for thatpossibility if you just let b=0.I've already noticed attempts by Dik Winter and Rick Decker to dodgethe result, which doesn't surprise me, though it's ironic that I coulduse a non-monic primitive with integer coef'cients to prove my case;however, it's not hard to see why their position can't be true.You see, no matter how high you go (like Winter posted a now clearlyfalse polynomial from Keith Ramsay of degree 22) your polynomial would*have* to allow for b=0, which means that it is non-monic with aleading coef'cient with a factor that is 7.You may see posters using various techniques, and making claims orotherwise running from the issue as a lot of people have invested timeand energy on a false position:David Ullrich, Arturo Magidin, Dik Winter, C. Bond, Nora Baron,Keith Ramsay, Rick Decker, among othersEven just a couple of people would be enough for arguments to go onand on as people with egos refused to deal with theirs being bruised.However, the fact remains that what I've shown is a clearcounterexample to their claims, and you know, and I know that math ishard.These people may not be capable of accepting the mathematical truth.Maybe you're not either, but I hope that some of you will come over tothe side of mathematics, to believing things that are actuallymathematically correct, no matter how much it hurts, and no matter howmany other people refuse to do so.James Harris === Subject: Re: JSH: Math is hard> I did a neat trick yesterday by 'nally realizing that I could> directly challenge posters making various claims about factors in the> ring of algebraic integers by using my own quadratic: y^2 - by - 7 = 0, which has as one of its roots > > (b + sqrt(b^2 + 28))/2> > With what I've been calling the Decker quadratic various posters have> asserted that you can have some algebraic integer functions w_1(x) and> w_2(x), where> > w_1(x) w_2(x) = 7> > which vary as x varies, so I simply used a method to directly check by> using a quadratic with a variable b, so that I can let b be *any*> algebraic integer. So, in fact, b = w_1(x) - w_2(x), is a possibility, which directly> challenges those people.> > What I end up with, at x=2, from the Decker quadratic is> > 7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0> I must say this is an interesting approach. I am going to 'll in a little background here. Youstarted with the Decker polynomial P(x) = 7*(25^x^2 + 30*x + 2)and considered a factorization of the form P(x) = (5 a_1(x) + 7)*(5 a_2(x) + 7),and then, viewing this as a factorization in the variable5, require that a_1(x) and a_2(x) are roots of a^2 - (x - 1)*a + 7*(x^2 + x).Then you let x = 2. This means that a_1(x), for example, is a_1(2) = (1 + sqrt(-167))/2.Now you say, can this number have an algebraic integer factor in common with 7 which is not equal to 7 ? The answer is yes. This can be shown by elementary Galois theory or by an elementary argument in algebraicnumber theory. Both of those arguments are evidently notsuf'cient for you, so you claim that the answer is no. You say essentially, suppose c is an algebraic integerwhich divides 7 and which also divides a_1(2) as above.Thus[*] (1 + sqrt(-167))/2 = c * z,where z is another algebraic integer. Now, you note that c can be written in the form c = (b + sqrt(b^2 - 28))/2 for some other algebraic integer b. Speci'cally, itturns out that one can easily show: b = c + 7/c,which is an algebraic integer because it is assumed thatc is a divisor of 7. This means that (1 + sqrt(-167)) = (b + sqrt(b^2 - 28)) * z. 7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0which is a non-monic polynomial in z. There is a theorem which says that a non-monicirreducible primitive polynomial with integer coef'cients cannot have an algebraic integeras a root. You (and we) have applied that theoremmany times in these discussions. Does that theorem apply here ? If it does, you are right. The theorem does not apply. The polynomial in z above does not have *integer* coef'cients. Ingeneral, b is an algebraic integer, but it isnot an ordinary integer. Your desired conclusion,that z cannot be an algebraic integer, does not follow. As you know, your conclusion, that there canbe no algebraic integer z satisfying [*], con¤icts, as Inoted above, with other basic results in (1) Galois theory, and (2) algebraic number theory. Also Keith Ramsay has shown that (1 + sqrt(-167))/2 has an algebraic integer factor in common with a root of an 11th degree monic polynomial with integer coef'cients and constant term equal to a power of 7. You have thus been proved wrong about this in 3 differentand largely independent ways. There is another way to view what is going on here. Recall that we started with[*] (1 + sqrt(-167))/2 = c * z. The general de'nition of c is c = GCD(a_1(2), 7) = GCD((1 + sqrt(-167))/2, 7). This means that z = (1 + sqrt(-167))/(2*c),which is an algebraic integer by de'nition of GCD. The existence of the GCD function in algebraic integers was proved by Dedekind. In continuing to insist that, e.g., z cannotbe an algebraic integer, you are not really arguing withus mere mortals in sci.math. You are arguing with RichardDedekind. Nora B. > which is important because it's non-monic, but *all* of its roots are> actually checks against various possibilities for w_1(x) and w_2(x).> > It's easy to see that it takes away an integer possibility for b, but> the method reveals that no algebraic integer can possibly work, at> all.> > Why does it have to work?> > Oddly enough, you can 'gure that out from what happens at b=0, as> then you have> > y^2 - 7 = 0.> > ANY polynomial that you end up with must always allow for that> possibility if you just let b=0.> > I've already noticed attempts by Dik Winter and Rick Decker to dodge> the result, which doesn't surprise me, though it's ironic that I could> use a non-monic primitive with integer coef'cients to prove my case;> however, it's not hard to see why their position can't be true.> > You see, no matter how high you go (like Winter posted a now clearly> false polynomial from Keith Ramsay of degree 22) your polynomial would> *have* to allow for b=0, which means that it is non-monic with a> leading coef'cient with a factor that is 7.> > You may see posters using various techniques, and making claims or> otherwise running from the issue as a lot of people have invested time> and energy on a false position:> > David Ullrich, Arturo Magidin, Dik Winter, C. Bond, Nora Baron,> Keith Ramsay, Rick Decker, among others> > Even just a couple of people would be enough for arguments to go on> and on as people with egos refused to deal with theirs being bruised.> > However, the fact remains that what I've shown is a clear> counterexample to their claims, and you know, and I know that math is> hard.> > These people may not be capable of accepting the mathematical truth.> > Maybe you're not either, but I hope that some of you will come over to> the side of mathematics, to believing things that are actually> mathematically correct, no matter how much it hurts, and no matter how> many other people refuse to do so.> > > James Harris === Subject: Re: JSH: Math is hard Adjunct Assistant Professor at the University of Montana. [.snip.]>> 7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0>>which is a non-monic polynomial in z. >> There is a theorem which says that a non-monic>irreducibleover Q> primitive polynomial with integer >coef'cients cannot have an algebraic integer>as a root. You (and we) have applied that theorem>many times in these discussions.>> Does that theorem apply here ? >> If it does, you are right.>> The theorem does not apply. The polynomial in z >above does not have *integer* coef'cients. In>general, b is an algebraic integer, but it is>not an ordinary integer. Even if b is an integer, one would need to show that this polynomialis both primitive and irreducible. Primitive is easy: if b is aninteger, then either it is a multiple of 7 or not; if it is not amultiple of 7, the polynomial is primitive by looking at thecoef'cients of z^4 and z^3; if it is a multiple of 7, then 6b^2+83 iscoprime to 7, so the polynomial is primitive.But one would also have to show it is irreducible...Even if we extend the result to the 'eld of de'nition, and replacethe notion of primitivity with the corresponding one using ideals(so it does not require gcd's over the ring of integers), it wouldhave to be shown that this polynomial is irreducible over the 'eld ofde'nition, as well as primitive; and here we will run into biggerproblems with primitivity, since it could be, at least in principle,true that gcd(7, b, 6b^2+83) is not a unit. Not to mention irreducibility, of course...-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: JSH: Math is hard> > [.snip.]> > > >> > 7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0> >> >which is a non-monic polynomial in z. > > There is a theorem which says that a non-monicirreducible> > over Q> > > primitive polynomial with integer > >coef'cients cannot have an algebraic integer> >as a root. You (and we) have applied that theorem> >many times in these discussions.> >> > Does that theorem apply here ? > >> > If it does, you are right.> >> > The theorem does not apply. The polynomial in z > >above does not have *integer* coef'cients. In> >general, b is an algebraic integer, but it is> >not an ordinary integer. > > Even if b is an integer, one would need to show that this polynomial> is both primitive and irreducible. Yes, of course. My thought process was: maybe the toughest criterion is integrality of b: it is almost certainly notan integer. Anyway, deal with that 'rst. If it does turnout to be an integer, then start worrying about primitivityand irreducibilility. (Later) As I noted, the choices for b are restricted: b = c + 7/c, where c = GCD(((1 + sqrt(-162))/2, 7). If b were an integer, then since c^2 - b*c + 7 = 0,I would then conclude that c is a root of a degree 2 polynomial. with integer coef'cients. However I believe Keith Ramsay has shown that c is the the root of an 11th degree irreducible polynomial. So it seems to me we can forget about b being an integer.> Primitive is easy: if b is an> integer, then either it is a multiple of 7 or not; if it is not a> multiple of 7, the polynomial is primitive by looking at the> coef'cients of z^4 and z^3; if it is a multiple of 7, then 6b^2+83 is> coprime to 7, so the polynomial is primitive.> > But one would also have to show it is irreducible...> > Even if we extend the result to the 'eld of de'nition, and replace> the notion of primitivity with the corresponding one using ideals> (so it does not require gcd's over the ring of integers), it would> have to be shown that this polynomial is irreducible over the 'eld of> de'nition, as well as primitive; and here we will run into bigger> problems with primitivity, since it could be, at least in principle,> true that gcd(7, b, 6b^2+83) is not a unit. > I think Bill Dubuque answered that part satisfactorily. Nora B.> Not to mention irreducibility, of course...> > -- > => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> => > Arturo Magidin> magidin@math.berkeley.edu === Subject: Re: JSH: Math is hard> > ... since it could be, at least in principle,>> true that gcd(7, b, 6bb+83) is not a unit.Since 1 = 12(7)+6b(b)-(6bb+83) any commondivisor of 7, b, 6bb+83 must divide 1in any ring.-Bill Dubuque === Subject: Re: JSH: Math is hard Adjunct Assistant Professor at the University of Montana.>> >> ... since it could be, at least in principle,>> true that gcd(7, b, 6bb+83) is not a unit.>>Since 1 = 12(7)+6b(b)-(6bb+83) any common>>divisor of 7, b, 6bb+83 must divide 1>>in any ring.Silly me, of course.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: JSH: Math is hardApparently to hard for JSH. === Subject: Re: JSH: Math is hard[snip handwaving claptrap intended to resuscitate a dead argument]Why don't you post an argument which begins by stating what you areattempting to prove (instead of a tirade against your critics) and followit with a step by step proof (instead of clanking your sword)? Youconstantly post these things which have some vague purpose, prompted bysome vague statement, made in some unidenti'ed post, and then proceed to¤ing expressions and equations with poorly motivated substitutions, badmath and worse logic, and 'nally reach some kind of unsupportedconclusion which has no identi'able bearing on any previous problem. Whatwas the purpose of this post? You did not refute anything. What do youthink you proved, or disproved?Your original line of reasoning has been so conclusively refuted so manytimes that it really isn't worth anyone's time to step through your newarguments. They're already known to be patently false. If you want to getserious attention to any new post, state the purpose clearly and show alogical connection between your starting position and your conclusion. Youhave failed to do that -- again. It appears that you are somehow incapableof constructing a valid proof.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: message>> [snip handwaving claptrap intended to resuscitate a dead argument]>> Why don't you post an argument which begins by stating what you are> attempting to prove (instead of a tirade against your critics) and follow> it with a step by step proof (instead of clanking your sword)? You> constantly post these things which have some vague purpose, prompted by> some vague statement, made in some unidenti'ed post, and then proceed to> ¤ing expressions and equations with poorly motivated substitutions, bad> math and worse logic, and 'nally reach some kind of unsupported> conclusion which has no identi'able bearing on any previous problem. What> was the purpose of this post? You did not refute anything. What do you> think you proved, or disproved?>> Your original line of reasoning has been so conclusively refuted so many> times that it really isn't worth anyone's time to step through your new> arguments. They're already known to be patently false. If you want to get> serious attention to any new post, state the purpose clearly and show a> logical connection between your starting position and your conclusion. You> have failed to do that -- again. It appears that you are somehow incapable> of constructing a valid proof.I agree with all of what you say.However...I quote, at length, fromhttp://www.mentalhealth.com/book/p45-eat1.html#Head_1 == NervosaPeople who intentionally starve themselves suffer from an eating disordercalled anorexia nervosa. The disorder, which usually begins in young peoplearound the time of puberty, involves extreme weight loss--at least 15percent below the individual's normal body weight. Many people with thedisorder look emaciated but are convinced they are overweight. Sometimesthey must be hospitalized to prevent starvation.# Deborah developed anorexia nervosa when she was 16. A rather shy, studiousteenager, she tried hard to please everyone. She had an attractiveappearance, but was slightly overweight. Like many teenager girls, she wasinterested in boys but concerned that she wasn't pretty enough to get theirattention. When her father jokingly remarked that she would never get a dateif she didn't take off some weight, she took him seriously and began to dietrelentlessly--never believing she was thin enough even when she becameextremely underweight.# Soon after the pounds started dropping off, Deborah's menstrual periodsstopped. As anorexia tightened its grip, she became obsessed with dietingand food and developed strange eating rituals. Every day she weighed all thefood she would eat on a kitchen scale, cutting solids into minuscule piecesand precisely measuring liquids. She would then put her daily ration insmall containers, lining them up in neat rows. She also exercisedcompulsively, even after she weakened and became faint. She never took anelevator if she could walk up steps.# No one was able to convince Deborah that she was in danger. Finally, herdoctor insisted that she be hospitalized and carefully monitored fortreatment of her illness. While in the hospital, she secretly continued herexercise regimen in the bathroom, doing strenuous routines of sit-ups andknee-bends. It took several hospitalizations and a good deal of individualand family outpatient therapy for Deborah to face and solve her problems.Deborah's case is not unusual. People with anorexia typically starvethemselves, even though they suffer terribly from hunger pains. One of themost frightening aspects of the disorder is that people with anorexiacontinue to think they are overweight even when they are bone-thin. Forreasons not yet understood, they become terri'ed of gaining any weight.Food and weight become obsessions. For some, the compulsiveness shows up instrange eating rituals or the refusal to eat in front of others. It is notuncommon for people with anorexia to collect recipes and prepare gourmetfeasts for family and friends, but not partake in the meals themselves. LikeDeborah, they may adhere to strict exercise routines to keep off weight.Loss of monthly menstrual periods is typical in women with the disorder. Menwith anorexia often become impotent. == pointing out that sheis not overweight and should start eating properly.That approach will not work. You yourself must _know_ that it won't work.To quote from the above: It took several hospitalizations and a good dealof individual and family outpatient therapy for Deborah to face and solveher problems. My feeling is that James may need a similar level of care andtreatment if he is ever going to recover from his condition. I doubt thatany communication with him, via usenet, will change his behavior materially.So, I guess we just continue making fun of him, pointing out errors in hisarguments and pleading with him to get help.Does he need to recover? I don't know. His condition is not, presumably,life-threatening.Would he be happier if he could be cured? Who knows ...~~~~Anyway, in my estimation the main reason behind most posts to usenet[including this one] is not to help others but to boost the ego of theposter. So...James, you ing moron, why don't you [etc, etc, etc...]-- Clive Toothhttp://www.clivetooth.dk === Subject: Re: JSH: Math is hard> > [snip handwaving claptrap intended to resuscitate a dead argument]> > Why don't you post an argument which begins by stating what you are> attempting to prove (instead of a tirade against your critics) and follow> it with a step by step proof (instead of clanking your sword)? You> constantly post these things which have some vague purpose, prompted by> some vague statement, made in some unidenti'ed post, and then proceed to> ¤ing expressions and equations with poorly motivated substitutions, bad> math and worse logic, and 'nally reach some kind of unsupported> conclusion which has no identi'able bearing on any previous problem. What> was the purpose of this post? You did not refute anything. What do you> think you proved, or disproved?> > Your original line of reasoning has been so conclusively refuted so many> times that it really isn't worth anyone's time to step through your new> arguments. They're already known to be patently false. If you want to get> serious attention to any new post, state the purpose clearly and show a> logical connection between your starting position and your conclusion. You> have failed to do that -- again. It appears that you are somehow incapable> of constructing a valid proof.I don't think you're being fair. As the subject of his post states, JSH 'nds math hard. His post proves it.Gib === Subject: Re: JSH: Math is hardIf it's so hard, why not just go shopping?Does this mean they're coming out with a chubby Doofus Barbie now?-- Wayne Brown (HPCC #1104) | 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 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Math is hardX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>[...]>>You may see posters using various techniques, and making claims or>otherwise running from the issue as a lot of people have invested time>and energy on a false position:>>David Ullrich, Arturo Magidin, Dik Winter, C. Bond, Nora Baron,>Keith Ramsay, Rick Decker, among others>>Even just a couple of people would be enough for arguments to go on>and on as people with egos refused to deal with theirs being bruised.>>However, the fact remains that what I've shown is a clear>counterexample to their claims, and you know, and I know that math is>hard.>>These people may not be capable of accepting the mathematical truth.>>Maybe you're not either, but I hope that some of you will come over to>the side of mathematics, to believing things that are actually>mathematically correct, no matter how much it hurts, and no matter how>many other people refuse to do so.Right. Even if _everyone_ else refuses to do so. Everyone on sci.math,every mathematician we harass via email (even the ones who seemedat 'rst to be nicer than the people here), every journal editor onthe planet...Guffaw.>James Harris************************David C. Ullrich === Subject: Re: 3-D analogue of pythagorean theoremContinued...The question raised by Ausurosh is a very important one for geometryand applied sciences. Perhaps ought to be brought to the notice of allyoung senior school/college students. I myself had (and still havewith respect to visualization) such questions in my mind for fortyyears plus and without fully satisfying answers.The lengths,areas, volumes and hypervolumes in N dimensional space arecomponented along mutually perpendicular / orthogonal directions, thesum of the squares conserving Pythogorean Rule is valid.Let [u,v,w, ..] be a short notation/operator for u^2+v^2+w^2+ ..Length, 2Dimns l=[lx,ly] Theorem of Pythogorus,e.g.,forces/ vectorsArea , 3Dimns A=[Ax,Ay,Az] Elasticity stress componeting,e.g., vectors/tensors for stress/moment of inertia/curvature Volume, 4Dimns V=[Vx,Vy,Vz,V4] What is this? HyperVolume, 5Dimns H=[Hx,Hy,Hz,H4,H5] and this? Geometrical imagination of length and area in 2D and 3D as righttriangle/tritetrahedron are known. We take sections parallel to theplanes (x=constant etc.) to come to lower order space. But I can'tvisualize the sections of HyperVolume into Volume by any means ofrepresentation. Hoping someone helps towards the literature. === Subject: A tricky integral by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 17:11:21 -0500Does anyone know if there is a function for the following integration with respect to X please :SQRT[1 + A*(SIN(X))^2 + B*(SIN(X))^4]where A and B are positive constants and the limits are from zero to Y (pi/2 < Y < pi). === Subject: Re: A tricky integral> Does anyone know if there is a function for the following integration> with respect to X please : SQRT[1 + A*(SIN(X))^2 + B*(SIN(X))^4]>> where A and B are positive constants and the limits are from zero to Y> (pi/2 < Y < pi).Mathematica seems to give an answer. It is very lengthy, quite horri'c,in terms of incomplete elliptic integrals of all three kinds.BTW, I'd almost be willing to bet that it's incorrect for at least somevalues of A, B, or Y.David === Subject: Re: Please help prove! ~~~~~~~~~~>_<~~~~~~~~~> > 2. A system of linear equations over a 'eld F has 5 solutions => F is> > isomorphic to Z_5.> > Fill in the blank:> The set of solutions of a system of linear equations is a ...... .> The number of elements of a ..... is a power of the characteristic.> 5 is prime.But keep in mind when you 'll in the blanks that the system was not said to be homogeneous.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The Nature of Space-TimeThe nature of space-time?The nature of cubic feet-seconds?The nature of cubic meter-days?space-timeStill funny, still stupid,Still worshipped like the false god it is.Space is not part of time,time is not part of space.They are 2 different measurement factors.They should remain seperated,not joined as if they are one.Got a cubic inch-hour?How many cubic mm-seconds are in it? === Subject: Unusual Numbering SystemsUnusual Numbering SystemsMost people are familiar with 'xed radix numberingsystems like base ten and base two.There are also product based numbering systems.A product base uses two series: a base series, B,and the product of the base series, P.All 'xed radix numbering systems are alsoproduct based numbering system.For base 10:S=(1,10,10,10,...)P=(1,10,100,1000,...)The allowable coef'cients for position i are0 through S_(i+1)-1.The factorial base is an example of a product base.S=(1,2,3,4,...)P=(1,2,6,24,...)The allowable coef'cients for the lowest order positionare 0 or 1. The coef'cient for the second positionare 0 through (3-1) or 0,1, and 2.321 (base !) = 3*6 + 2*2 + 1*1 = 23 (base 10)Another intersting product base is the root of prime powers:S=(1,2,3,2,5,7,2,3,...)P=(1,2,6,12,60,...)4121 (base RPP) = 4*12 + 1*6 + 2*2 + 1*1 = 59 (base 10)We can combine two 'xed radix bases into one product base:S=(1,2,3,2,3,...)P=(1,2,6,12,36,...)2121 (base 2&3) = 2*12 + 1*6 + 2*2 + 1*1 = 35 (base 10)Recently, I have become interested in what I call series bases.Let f() be a series such that the limit of partial sumsof f() = 1 and every real number in the range (0,1) is the sumof a subset of f(). For example:f() = (1/2, 1/6, 1/6, 1/12, 1/36, 1/36, ...)Every distinct value, x, in f() represents a position and theallowable coef'cients for the PREVIOUS position are 0 through thenumber of times x appears in the series. To represent integerswe take the inverse of each term in f().(We need to add 1 as the 'rst term in the inverse series.)Inverse of f() = (1, 2, 6, 6, 12, 36, 36, ...)We see that this f() represents the product base 2&3 givenabove. Are there series bases that are not product bases?Combine the series' for base 2 and base 3,(1/2,1/4,1/8,...) and (1/3, 1/3, 1/9, 1/9, ...),to get f() = (1/2, 1/3, 1/3, 1/4, 1/8, 1/9, 1/9, ...)The limit of this f() is 2. Divide each term by 2.f() = (1/4, 1/6, 1/6, 1/8, 1/18, 1/18, ...)Inverse f() = (1, 4, 6, 6, 8, 18, 18, ...)0 = 01 = 1*1 = 1 (base 10)10 = 1*4 + 0*1 = 411 = 1*4 + 1*1 = 520 = 2*4 + 0*1 = 821 = 2*4 + 1*1 = 9100 = 1*6 + 0*4 + 0*1 = 6101 = 1*6 + 0*4 + 1*1 = 7110 = 1*6 + 1*4 + 0*1 = 10111 = 1*6 + 1*4 + 1*1 = 11120 = 1*6 + 2*4 + 0*1 = 14121 = 1*6 + 2*4 + 1*1 = 151000 = 1*8 + 0*6 + 0*4 + 0*1 = 81001 = 1*8 + 0*6 + 0*4 + 1*1 = 91010 = 1*8 + 0*6 + 1*4 + 0*1 = 121011 = 1*8 + 0*6 + 1*4 + 1*1 = 131020 = 1*8 + 0*6 + 2*4 + 0*1 = 16Every product base has a unique representation for eachinteger. This series base does not have that property.8 and 9 have more than one representation.This series base has no representation for 2 or 3.We can get around this problem by having a specialrule for single digit representations.Is there a series base that is not a product baseand every integer greater than some n has an uniquerepresentation?Is there a series base that has a 'nite representationfor every real algebraic number?Russell- 2 many 2 count === Subject: Re: Unusual Numbering Systems> Is there a series base that is not a product base> and every integer greater than some n has an unique> representation?> > Is there a series base that has a 'nite representation> for every real algebraic number? Um, I'm not sure if this satis'es your criteria, andyou might already be familiar with them, but if you arenot you would be interested in Fibonacci bases.J === Subject: Re: Unusual Numbering SystemsJim Nastos base that is not a product base> > and every integer greater than some n has an unique> > representation?> >> > Is there a series base that has a 'nite representation> > for every real algebraic number?>> Um, I'm not sure if this satis'es your criteria, and> you might already be familiar with them, but if you are> not you would be interested in Fibonacci bases.No, I reference.Too bad the inverse Fibonacci series doesn't converge.I could make an interesting series base if it did.Russell- 2 many 2 count === Subject: Re: Unusual Numbering Systems > No, I reference.> Too bad the inverse Fibonacci series doesn't converge.> I could make an interesting series base if it did. What do you mean by inverse 'bonacci series ? If you meanthe sum of 1/f(n) then, yes, it does converge and that's a simpleconsequence of the fact that the 'bonacci sequence growsexponentially. (Converges to about 3.35988566)J === Subject: Re: Unusual Numbering SystemsRussell there a series base that is not a product base> > > and every integer greater than some n has an unique> > > representation?> > >> > > Is there a series base that has a 'nite representation> > > for every real algebraic number? Um, I'm not sure if this satis'es your criteria, and> > you might already be familiar with them, but if you are> > not you would be interested in Fibonacci bases....> Too bad the inverse Fibonacci series doesn't converge.> I could make an interesting series base if it did.My own, very brief, spiel about Fibo base:http://www3.telus.net/ldh/math/'bo.htmlLH === Subject: Re: Unusual Numbering SystemsLarry Hammick a series base that is not a product base> > > > and every integer greater than some n has an unique> > > > representation?> > > >> > > > Is there a series base that has a 'nite representation> > > > for every real algebraic number?> > >> > > Um, I'm not sure if this satis'es your criteria, and> > > you might already be familiar with them, but if you are> > > not you would be interested in Fibonacci bases.> ...> > Too bad the inverse Fibonacci series doesn't converge.> > I could make an interesting series base if it did.> My own, very brief, spiel about Fibo base:> http://www3.telus.net/ldh/math/'bo.htmlIs there a series that sums to the golden ratio?Russell- 2 many 2 count === Subject: Re: Unusual Numbering Systems > Is there a series that sums to the golden ratio? Just make one. You should know about geometric series,so 1 + r + r^2 + ... = 1/(1-r) = phi1-r = 1/phir = 1 - 1/phi = 1 - (phi - 1) = 2 - phi.Or, of course, you can start with any initial termdifferent from 1 you like.J === Subject: Re: Unusual Numbering Systems> Is there a series that sums to the golden ratio?There are series that sum to any given real number, namely the 'rst differences of any sequence converging to that number form such a series, and there are in'nitely many such sequences. === Subject: Re: in message>> > Is there a series that sums to the golden ratio?>> There are series that sum to any given real number, namely the 'rst> differences of any sequence converging to that number form such a> series, and there are in'nitely many such sequences.Do you happen to know one?I am too lazy to compute one by hand.I did a search and found numerous methods using continued fractions,but no simple series.Russell- 2 many 2 count === Subject: Re: Unusual Numbering Systems> Virgil series that sums to the golden ratio?> >> > There are series that sum to any given real number, namely the 'rst> > differences of any sequence converging to that number form such a> > series, and there are in'nitely many such sequences.> > Do you happen to know one?> I am too lazy to compute one by hand.> I did a search and found numerous methods using continued fractions,> but no simple series.> > > Russell> - 2 many 2 count> > Work out the decimal representation for the golden ratio, then the series that adds on one more digit of that expansion with each term will do the trick nicely. === Subject: Re: Unusual Numbering SystemsRussell Easterly> > > Is there a series that sums to the golden ratio?> >> > There are series that sum to any given real number, namely the 'rst> > differences of any sequence converging to that number form such a> > series, and there are in'nitely many such sequences.>> Do you happen to know one?> I am too lazy to compute one by hand.2 - 1/2 + 1/6 - 1/15 + 1/40 - 1/104 + etc = phi.The denominators are the products of consecutive Fibonacci numbers.I got this just by writing lim (f_{n+1} / f_n} = phi and converting thesequence to a series, usingf_n squared = f_{n+1} f _{n-1} +- 1LH === Subject: urgent analysis questionI'm having dif'culty with a problem and was hoping someone might beable to help. The problem is the following:Let X be a Lebesgue null subset of R (the reals) and let f: R -> R bea continuously differentiable function. Prove === Subject: Re: urgent analysis question> I'm having dif'culty with a problem and was hoping someone might be> able to help. The problem is the following: Let X be a Lebesgue null subset of R (the reals) and let f: R -> R be> a continuously differentiable function. Prove f(X) is a null set.If (a,b) is an interval, how large can f((a,b)) be? (f' is bounded in[-N,N] so work with the intersection of the image in [-N,N] tostart with; use the MVT to get f((a,b)) bounded by some multipleof |b-a| ...) === Subject: Re: urgent analysis question>> I'm having dif'culty with a problem and was hoping someone might be>> able to help. The problem is the following:>> >> Let X be a Lebesgue null subset of R (the reals) and let f: R -> R be>> a continuously differentiable function. Prove f(X) is a null set.> > If (a,b) is an interval, how large can f((a,b)) be? (f' is bounded in> [-N,N] so work with the intersection of the image in [-N,N] to> start with;Ooops ... no .. work on the intersection of the original with [-N,N],of course ...> use the MVT to get f((a,b)) bounded by some multiple> of |b-a| ...) === Subject: Re: urgent analysis dif'culty with a problem and was hoping someone might be> able to help. The problem is the following:> > Let X be a Lebesgue null subset of R (the reals) and let f: R -> R be> a continuously differentiable function. Prove f(X) is a null set.Look up Sard's Lemma for the general case.For this easy case, note that it's enough to prove for BOUNDED Lebesguenull subsets of R. (Write X as the countable union of [-n,n] cap X.)So WLOG we may assume X is contained in an interval [a,b]. f' iscontinuous, therefore bounded on [a,b], say |f'(x)| <= M for all x,where M > 0.Given e > 0 we can cover X by countably many open sets (ai,bi) (allsubsets of [a,b] WLOG) with sum_i (bi - ai) < e/M. The imagef((ai,bi)) is an interval (not necessarily open) whose length is <=M*(bi-ai); therefore the union of the f(ai,bi) is Lebesgue measurablewith Lebesgue measure <= sum_i M(bi-ai) < e. Since f(X) is a subsetof this union, therefore the Lebesgue outer measure of f(X) is < e. Therefore the Lebesgue outer measure of f(X) is 0, which means it'sLebesgue measurable and has Lebesgue measure zero.--Ron Bruck === Subject: Re: urgent analysis question> I'm having dif'culty with a problem and was hoping someone might be> able to help. The problem is the following:> > Let X be a Lebesgue null subset of R (the reals) and let f: R -> R be> a continuously differentiable function. Prove f(X) is a null set.Hint: Cover X by countably many intervals whose lengths add to something small. Now, how large can the length of f(I) be if f' is bounded on the interval I? This should lead to a proof in the case X is bounded, and if you can handle that case, well ... === Subject: Re: [Help] Borel CantelliX-DMCA-Noti'cations: idea of showing next problem;>> > >Let A_1, A_2, ... be events in probability space. De'ne X_n=A_1+A_2+...+A_n>> >and s_n = E(X_n). Suppose s_n ->inf and ||X_n/s_n||_2 ->1.>> >> ??? You say X_n is the sum of some _events_, ie _sets_, and then>> X_n appears to be a random variable, ie a _function_. This>> makes no sense. Was X_n actually supposed to be the sum of>> the indicator functions of those sets?>>Yes, X_n is sum of indicator function of those sets.> >> >Show that>> > {X_n=0} <= (k - X_n)(k+1 - X_n)/k(k+1) >> >for each positive integer k.>> >> Again, you ask us to show that a set is <= a function; I>> don't know what this means (do you really want that>> the indicator function of that set is <= the right side>> or what?)>>This is the same, it is the indicator function.Ok. Then you need to show that (k - X_n)(k+1 - X_n)/k(k+1) >= 1 where X_n = 0and (k - X_n)(k+1 - X_n)/k(k+1) >= 0 elsewhere.Both are very easy and have nothing to do withprobability. (The second looked wrong at 'rst,but it's true because X_n is an integer, sok - X_n < 0 < k+1 - X_n is impossible.)>> >> Also I wonder if the above is _really_ what you want>> to prove. The reason I wonder is that it has nothing>> whatever to do with the hypotheses s_n ->inf >> and ||X_n/s_n||_2 ->1...>> >>I would also like to show following;>>2)By appropriate choice of k, deduce that Sum_0^inf A_i >= 1 a.s.>(again it is indicator function)This just says that the union of the sets A_i has measure 1.Which is clear from the hypotheses, without the inequalityyou asked about: If m(X) < 1 then there exists c < 1 suchthat ||f||_1 <= c||f||_2 for all f supported on X, by...>3)Prove that Sum_m^inf A_i >=1 a.s. for 'xed m. (again indicator>function)>>4)Deduce that P{omega in A_i i.o.} = 1>>To summarize I used the same symbol for a set and its indicator>function.Why?************************David C. Ullrich === Subject: AnalysisI saw this one before but I forgot the proof. Any hints will help. Provethat (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1.Steven === Subject: Re: message> I saw this one before but I forgot the proof. Any hints will help. Prove> that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1.> Steven>>Here is what I have (and don't have). I know that (1+X)^n and (1+nX)intersect at (0,1). Since [ (1+X)^n ] O > n for all X >0 and since (1 + nX)' = n for all X we know that (1+X)^n > (1+nX) ( but we already knew thisfrom the binomial theorem).The real question is whether or not (1 + X) ^n and (1 + nX ) intersect belowX =0 ? === Subject: message> I saw this one before but I forgot the proof. Any hints will help. Prove> that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1.Two questions:(1) What's the linear approximation to (1+x)^n when x>-1?(2) What's the concavity of the function (1+x)^2?Doug === Subject: Re: AnalysisDoug Norris before but I forgot the proof. Any hints will help. Prove> > that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1.>> Two questions:>> (1) What's the linear approximation to (1+x)^n when x>-1?Yes I realize that the linear approximation to (1+x)^n when x>-1 IS 1 + nx.I'm just lost after this.>> (2) What's the concavity of the function (1+x)^2?Concave upward>> Doug>> === Subject: Re: Analysis Doug Norris one before but I forgot the proof. Any hints will help. Prove> > > that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1.> >> > Two questions:> >> > (1) What's the linear approximation to (1+x)^n when x>-1?>> Yes I realize that the linear approximation to (1+x)^n when x>-1 IS 1 + nx.> I'm just lost after this.>> >> > (2) What's the concavity of the function (1+x)^2?> Concave upward>> >> > Doug The pedagogical dif'culty is as follows: the above Bernoulli Inequality(proved easily by induction, as already noted) can be covered and usedlong before any discussion of linear approximations or concavity. For example, it can serve in an elementary proof of theArithmetic-Geometric Mean Inequality, and in an equally elementary proofthat {(1+1/n)^n} increases, {(1+1/n)^(n+1)} decreases, hence bracketingthe number e. Look Ma, no Calculus.Since the proof is Public Domain, here it is, in a sharper form:(Statement) If x>-1, x non-zero, and n>=2 integer then (1+x)^n > 1+n*x.(Start) For n=2; right side is 1 + 2*x, left side is (1+x)^2 = 1 + 2*x + x^2 > right side.(Progress) Let (1+x)^n > 1 + n*x under the original assumptions. Then for n replaced by (n+1), right side is 1+(n+1)*x ; left side: (1+x)^(n+1) = (1+x) * (1+x)^n > (1+x) * (1+n*x) = 1 + (n+1)*x + n*x^2 > 1 + (n+1)*x.Finito.To prove the same with n allowed to be real and >1, the easy wayis to use Calculus: verify by usual methods that (1+n*x)/(1+x)^n = 1 - n*(n-1)*x^2*F(x) where F(x) = integral[0 to 1] u/(1+x*u)^(n+1) duso that F(x) > 0, and 1 + n*x < (1+x)^n .The same formula shows that for n<0, you have the same inequality, but ZVK(Slavek). === Subject: Re: Analysis> What's the linear approximation to (1+x)^n when x>-1?0.