mm-187 === Subject: Re: Silly question on limits, tensor 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 Euclidean ring] >To be Euclidean, there must be some 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). Even in the integers, e.g., when using continued fractions (or equivalently--and the reason I was once sort of up on this kind 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, in the latter case, you could say there's just one remainder, and it's that one , but you lose nothing that I can see by saying there are various pairs (q,r), for each of which r is of smaller size than b ; and I can imagine cases in which you might want actually to take a continued fraction expansion with some remainders positive (though not necessarily of smallest absolute value) and others negative (ditto). Certainly for the application to g.c.d.'s, uniqueness isn't used, is it? All that's used is that the measure of size doesn't admit any in'nite descending chains. Lee Rudolph === Subject: Re: Need help with proving that Z[i] is Euclidean Adjunct Assistant Professor at the University of Montana. Lee there must be some 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 drops in 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 ) = === 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? This isn't so hard since I've already proved that there exists a divisions algorithm for Gaussin integers. Am I right? Once again, thank you for all the help. Yours Pierre ps. Sorry if there are any mistakes in my spelling; Enlish isn't my native language. ds === Subject: Re: Need help with proving that Z[i] is Euclidean Adjunct Assistant Professor at the University of Montana. 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 a gcd, but as I noted in my previous reply, the exact same argument as you 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, the remainder is smaller than the immediate previous one (that is, show (1) above) then that is enough, by induction, to show that the remainder in step n is strictly smaller than the remainders in steps 1 through 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 ) = === 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? 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 Stephen J. 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: 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 Stephen J. 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 balls 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 Townsend I 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 ïunboundedness' When traversing the surface of a sphere, one can go in any direction forever, even a straight line, and never encounter an ïedge' of the surface. What is the topological invariant that represents this property? l8r, Mike N. Christoff === Subject: Re: Topological property that represents ïunboundedness' Christoff in any direction > forever, even a straight line, and never encounter an ïedge' 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 ïunboundedness' > When traversing the surface of a sphere, one can go in any direction > forever, even a straight line, and never encounter an ïedge' 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. l8r, Mike N. Christoff === Subject: Re: Topological property that represents ïunboundedness' Michael N. Christoff go in any direction >> forever, even a straight line, and never encounter an ïedge' 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 latter one falls off the edge pretty rapidly. So what is the topological distinction between these two surfaces? What in the topology makes one 'nite and not the other? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: irrationality of sqrt(2): easy >Dirk Van de moortel 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.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: irrationality of sqrt(2): easy question Adjunct Assistant Professor at the University of Montana. Robin Chapman >> Doesn't it follow easily from unique factorization into primes? > >Sledgehammer -> nut. In Mexico, the expression is swatting §ies by 'ring cannons at them... -- = It's not denial. I'm just very selective about what I accept as reality. --- Calvin ( Calvin and Hobbes ) = === 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 === 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 ? > > 'eld of rational numbers. In this case I doubt that there is a nice formula. In the case of nxn matrices the problem would be equivalent to counting the number of nonsingular (regular) 0-1 nxn matrices. Below is the entry in the OEIS which gives 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 are not 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 according to this entry equal to 270345669985440 ID Number: A055165 URL: http://www.research.att.com/projects/OEIS?Anum=A055165 Sequence: 1,6,174,22560,12514320,28836612000,270345669985440 Name: Number of regular n X n matrices with rational entries equal to 0 or 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 matrices Example: 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_sequence A071095 A024277 A012177 Keywords: nonn,nice,hard Offset: 1 Author(s): Ulrich Hermisson (uhermiss(AT)server1.rz.uni-leipzig.de), Jun 18 2000 If 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 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), 2000 === Subject: Re: rank of a random {0,1} matrix (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 There must be a typo here because the formula as stated does not make sense. On the left you have k to infty so the right hand side should not involve k. -- Rouben Rostamian === 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 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 $m times n$ 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 rank test.) George Marsaglia === 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 found 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 $m times n$ 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 Here 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, and p1+p2+,..pt=1. How to measure the probability of the rank of M === Subject: Re: rank of a random {0,1} matrix 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), 2000 If 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: 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 you that it is greater than the probability that you will get others 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 at http://www.dcproof.com/Gallery.htm Keep those submissions coming, folks! Dan Visit DC Proof Online at http://www.dcproof.com > Calling all aspiring mathematicians! > > Have 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 in your > proofs. (See Documenting and Viewing your Proof in the User Reference > Guide.) > > Suggested topics: Introductory theorems in logic, set theory, number theory > and group theory. Or introduce any axioms as you see 't -- it's YOUR proof! > > 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: re:R-Integration A bounded function on [a, b] is Riemann integrable iff the set of points of discontinuity 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 any 0 < x < 1 (with integration 0) So if talking about improper Riemann integration, i.e. int_0^1 g = lim_{x->0^+} int_x^1 g. Then g is still integrable. therfore g is bounded and integrable on interval [a, 1] for any a with 0 < a < 1. for the given g , it is not R-integrable is my view correct? 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 === $$f_w91tn_g 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 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: hyperbolas Quantumcat > 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 advance. I'm not sure what 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 apex is 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 i1CLmLB16112; >On 11 Feb 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 not himself he`s refering to, but to his own uncle, Al ? Then, we might at least 'nd some sympathy (or at least) some explanations. === Subject: Re: prime relative to involution In 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, then n 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 of exponents. There doesn't seem to be any reason to do such a factorization, 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 are hyperprime. If you require the base to be hyperprime as well it's true; 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 constructed quite easily out of primes, they're probably not a much richer concept. Further, these properties are rather trivial, so if someone needs to use them they would quickly develop them on their own. Looking that over, it sounds kinda harsh. I don't mean to burst your bubble here; when I 'rst heard about the Laplace transformation in high school I remember I tried to develop something similar using limits. After working on it on and off for a week I or so eventually found out that it reduced to something silly like T(f(x)) = f'(x)/f(x) or something like that. === Subject: Re: 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 ï4' on the > denominator changes to a ï2.') > > J So sorry for the bleeps in my non-solution . Yes, the original question did involve (sqrt(5)-1)/(2 *p)i not sqrt(5)+1 and Lindemann's lemma and my 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 can be 'xed. Sorry for the whole === Subject: Re: 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. === Subject: Re: need help!! > > > On 11 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: 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 . > > J dear 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 this answer,and if you plot it ,please show me thank you ,hupo === Subject: Re: need help!! 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 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 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 > 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. You started 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 variable 5 , 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 algebraic number theory. Both of those arguments are evidently not suf'cient for you, so you claim that the answer is no. You say essentially, suppose c is an algebraic integer which 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, it turns out that one can easily show: b = c + 7/c, which is an algebraic integer because it is assumed that c 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 = 0 which is a non-monic polynomial in z. There is a theorem which says that a non-monic irreducible 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. Your desired conclusion, that z cannot be an algebraic integer, does not follow. As you know, your conclusion, that there can be no algebraic integer z satisfying [*], con§icts, as I noted 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 different and 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 cannot be an algebraic integer, you are not really arguing with us mere mortals in sci.math. You are arguing with Richard Dedekind. 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. Nora Baron (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 >irreducible 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. 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. 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 ) = === Re: JSH: Math is hard > Nora Baron z^2 - 6bz + 252 = 0 > > > >which is a non-monic polynomial in z. > > > > There is a theorem which says that a non-monic > >irreducible > > 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 not an integer. Anyway, deal with that 'rst. If it does turn out to be an integer, then start worrying about primitivity and 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 ) > = === Subject: Re: JSH: Math is hard Arturo Magidin 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. -Bill Dubuque === Subject: Re: JSH: Math is hard Adjunct Assistant Professor at the University of Montana. Bill Dubuque 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 ) = === Re: JSH: Math is hard Apparently to hard for JSH. === Subject: 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. -- 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: Re: JSH: Math 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, from http://www.mentalhealth.com/book/p45-eat1.html#Head_1 == Anorexia Nervosa People who intentionally starve themselves suffer from an eating disorder called anorexia nervosa. The disorder, which usually begins in young people around the time of puberty, involves extreme weight loss--at least 15 percent below the individual's normal body weight. Many people with the disorder look emaciated but are convinced they are overweight. Sometimes they must be hospitalized to prevent starvation. # Deborah developed anorexia nervosa when she was 16. A rather shy, studious teenager, she tried hard to please everyone. She had an attractive appearance, but was slightly overweight. Like many teenager girls, she was interested in boys but concerned that she wasn't pretty enough to get their attention. When her father jokingly remarked that she would never get a date if she didn't take off some weight, she took him seriously and began to diet relentlessly--never believing she was thin enough even when she became extremely underweight. # Soon after the pounds started dropping off, Deborah's menstrual periods stopped. As anorexia tightened its grip, she became obsessed with dieting and food and developed strange eating rituals. Every day she weighed all the food she would eat on a kitchen scale, cutting solids into minuscule pieces and precisely measuring liquids. She would then put her daily ration in small containers, lining them up in neat rows. She also exercised compulsively, even after she weakened and became faint. She never took an elevator if she could walk up steps. # No one was able to convince Deborah that she was in danger. Finally, her doctor insisted that she be hospitalized and carefully monitored for treatment of her illness. While in the hospital, she secretly continued her exercise regimen in the bathroom, doing strenuous routines of sit-ups and knee-bends. It took several hospitalizations and a good deal of individual and family outpatient therapy for Deborah to face and solve her problems. Deborah's case is not unusual. People with anorexia typically starve themselves, even though they suffer terribly from hunger pains. One of the most frightening aspects of the disorder is that people with anorexia continue to think they are overweight even when they are bone-thin. For reasons not yet understood, they become terri'ed of gaining any weight. Food and weight become obsessions. For some, the compulsiveness shows up in strange eating rituals or the refusal to eat in front of others. It is not uncommon for people with anorexia to collect recipes and prepare gourmet feasts for family and friends, but not partake in the meals themselves. Like Deborah, they may adhere to strict exercise routines to keep off weight. Loss of monthly menstrual periods is typical in women with the disorder. Men with anorexia often become impotent. == Your post is rather like sending an email to Deborah, pointing out that she is 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 deal of individual and family outpatient therapy for Deborah to face and solve her problems. My feeling is that James may need a similar level of care and treatment if he is ever going to recover from his condition. I doubt that any communication with him, via usenet, will change his behavior materially. So, I guess we just continue making fun of him, pointing out errors in his arguments 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 the poster. So... James, you ing moron, why don't you [etc, etc, etc...] -- Clive Tooth http://www.clivetooth.dk === Subject: Re: JSH: Math is 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 hard If 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 improvise fwbrown@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 hard >[...] > >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 seemed at 'rst to be nicer than the people here), every journal editor on the planet... Guffaw. >James Harris ************************ David C. Ullrich === Subject: Re: 3-D analogue of pythagorean theorem Continued... The question raised by Ausurosh is a very important one for geometry and applied sciences. Perhaps ought to be brought to the notice of all young senior school/college students. I myself had (and still have with respect to visualization) such questions in my mind for forty years plus and without fully satisfying answers. The lengths,areas, volumes and hypervolumes in N dimensional space are componented along mutually perpendicular / orthogonal directions, the sum 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/ vectors Area , 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 right triangle/tritetrahedron are known. We take sections parallel to the planes (x=constant etc.) to come to lower order space. But I can't visualize the sections of HyperVolume into Volume by any means of representation. Hoping someone helps towards the literature. === Subject: A tricky integral by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1CMBLk18462; 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). === Subject: Re: A tricky integral howardg@aol.com ( 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 some values of A, B, or Y. David === Subject: Re: Please help prove! ~~~~~~~~~~>_<~~~~~~~~~ A N Niel 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-Time The nature of space-time? The nature of cubic feet-seconds? The nature of cubic meter-days? space-time Still 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 Systems Unusual Numbering Systems Most people are familiar with 'xed radix numbering systems 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 also product based numbering system. For base 10: S=(1,10,10,10,...) P=(1,10,100,1000,...) The allowable coef'cients for position i are 0 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 position are 0 or 1. The coef'cient for the second position are 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 sums of f() = 1 and every real number in the range (0,1) is the sum of 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 the allowable coef'cients for the PREVIOUS position are 0 through the number of times x appears in the series. To represent integers we 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 given above. 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 = 0 1 = 1*1 = 1 (base 10) 10 = 1*4 + 0*1 = 4 11 = 1*4 + 1*1 = 5 20 = 2*4 + 0*1 = 8 21 = 2*4 + 1*1 = 9 100 = 1*6 + 0*4 + 0*1 = 6 101 = 1*6 + 0*4 + 1*1 = 7 110 = 1*6 + 1*4 + 0*1 = 10 111 = 1*6 + 1*4 + 1*1 = 11 120 = 1*6 + 2*4 + 0*1 = 14 121 = 1*6 + 2*4 + 1*1 = 15 1000 = 1*8 + 0*6 + 0*4 + 0*1 = 8 1001 = 1*8 + 0*6 + 0*4 + 1*1 = 9 1010 = 1*8 + 0*6 + 1*4 + 0*1 = 12 1011 = 1*8 + 0*6 + 1*4 + 1*1 = 13 1020 = 1*8 + 0*6 + 2*4 + 0*1 = 16 Every product base has a unique representation for each integer. 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 special rule for single digit representations. 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? Russell - 2 many 2 count === Subject: 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. J === Subject: Re: Unusual Numbering 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, and > you might already be familiar with them, but if you are > not you would be interested in Fibonacci bases. No, I had never seen 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 On Thu, 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 mean the sum of 1/f(n) then, yes, it does converge and that's a simple consequence of the fact that the 'bonacci sequence grows exponentially. (Converges to about 3.35988566) J === Subject: Re: Unusual Numbering Systems > > > 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.html LH === Subject: Re: Unusual Numbering Systems > > > > > > On Thu, 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.html Is there a series that sums to the golden ratio? Russell - 2 many 2 count === Subject: Re: Unusual Numbering Systems golden ratio? Just make one. You should know about geometric series, so 1 + r + r^2 + ... = 1/(1-r) = phi 1-r = 1/phi r = 1 - 1/phi = 1 - (phi - 1) = 2 - phi. Or, of course, you can start with any initial term different from 1 you like. J === Subject: Re: Unusual Numbering Systems Russell Easterly 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: Unusual Numbering Systems > 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 Russell 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 Systems Russell 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 the sequence to a series, using f_n squared = f_{n+1} f _{n-1} +- 1 LH === Subject: 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. === 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; use the MVT to get f((a,b)) bounded by some multiple of |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: 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 Lebesgue null 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' is continuous, 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) (all subsets of [a,b] WLOG) with sum_i (bi - ai) < e/M. The image f((ai,bi)) is an interval (not necessarily open) whose length is <= M*(bi-ai); therefore the union of the f(ai,bi) is Lebesgue measurable with Lebesgue measure <= sum_i M(bi-ai) < e. Since f(X) is a subset of 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's Lebesgue measurable and has Lebesgue measure zero. --Ron Bruck === Subject: Re: urgent analysis question 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 Cantelli >> >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 = 0 and (k - X_n)(k+1 - X_n)/k(k+1) >= 0 elsewhere. Both are very easy and have nothing to do with probability. (The second looked wrong at 'rst, but it's true because X_n is an integer, so k - 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 inequality you asked about: If m(X) < 1 then there exists c < 1 such that ||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: Analysis 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 === Subject: Re: Analysis > 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 ] ï > n for all X >0 and since (1 + nX) ï = n for all X we know that (1+X)^n > (1+nX) ( but we already knew this from the binomial theorem). The real question is whether or not (1 + X) ^n and (1 + nX ) intersect below X =0 ? === Subject: Re: Analysis > 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: Analysis > > 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? 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 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 used long before any discussion of linear approximations or concavity. For example, it can serve in an elementary proof of the Arithmetic-Geometric Mean Inequality, and in an equally elementary proof that {(1+1/n)^n} increases, {(1+1/n)^(n+1)} decreases, hence bracketing the 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 way is 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) du so that F(x) > 0, and 1 + n*x < (1+x)^n . The same formula shows that for n<0, you have the same ZVK(Slavek). === Subject: Re: Analysis Doug Norris approximation to (1+x)^n when x>-1? 0. === Subject: Re: 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? One question: what's wrong with induction on n? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Analysis I think you mean X>-1 John -- John T Lowry 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 jlowry100@earthlink.net > 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 > > === Subject: Re: Analysis > I think you mean X>-1 > > John > > -- > John T Lowry > 5217 Old Spicewood Springs Rd, #312 > Austin, Texas 78731 > (512) 231-9391 > jlowry100@earthlink.net > > 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 > > > > With x > -1; Induction works: The case for n=1 clearly holds, and (1+x)^{n+1} = (1+x)^n(1+x) >= (1+nx)(1+x) = 1+nx+x+nx^2 = 1+(n+1)x+nx^2 >= 1+(n+1)x. === Subject: Re: What is the Origin of Space and Time? > > And here I was, pondering the collapse of a 10- or 26-dimensional universe > in a false vacuum state into 4-dimensional, more stable universe with the > remaining dimensions curled upon themselves... Silly me. ;) Yo, dude, those extra spaces do curl upon themselves ... just like my kitty ... === Subject: Re: What is I was, pondering the collapse of a 10- or 26-dimensional >> universe in a false vacuum state into 4-dimensional, more stable universe >> with the remaining dimensions curled upon themselves... Silly me. ;) > > Yo, dude, those extra spaces do curl upon themselves ... just like my > kitty ... And, y'know, those extra dimensions have a natural harmonic that makes them purr just like a kitten as well...could that be the answer to Schroedinger's cat mystery? -- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other === people think, Mr. Feynman? Subject: re:What is the Origin of Wow, is this forum a serious one or not? o_o We can't understand what's not in 3D, so we cannot know. We see 3D, but we can't see it all at the same time, like a person drawn on a piece of paper can't see all the piece of paper. He can see all of a 1D piece of paper at the same time, though. We are in 3D, we can see all of a 2D sheet of paper at the same time. We're immersed in 4D but we can't see even a small part of it at a time. Only a 4D person can, or a 5D person but he can see the 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: What is the Origin of Space and Time? > > Uh, I think it is you who needs to do the homeowork. Everyone knows there > are exactly FIVE dimensions. Every hear about a little something called > ïTwilight Zone' (aka the 'fth dimension) !?!?! I bet you feel pretty dumb > right about now. > > You are only partially correct, sir. Our time space fabric is a combination of 5-dimensional Twilight Zone and 3-dimensional Cartesian Space, which gives EXACTLY 8 dimensions. Buckaroo Banzai was right. Who's feeling pretty dumb right now? === Subject: Re: (Reference) Book for learning mathematics from the ground up James R Newman's World of Mathematics should be available in learning > mathematics, but since I have completed high-school and > won't be going into college for another few years, I think > the best way to learn would be to do it by myself. I don't > like pressure either, and I'm not hoping to major in > mathematics in the future. I started thinking about getting > myself a book, but since I've completed the most basic > high-school maths, it would be unnecessary to buy a whole > series of books starting with explaining addition > completely. A reference book or some sort of dictionary > with examples and complete de'nitions would be the ideal. > Is mathematics a far too great 'eld to summarize into one > reference book? It would be great to have a book that I can > look things up in when I get to an advanced level. Of > course, the best way to learn math is to solve problems and > exercises, but that might be out of scope for a reference > book. I might add that I wish to learn mathematics up to, > let's say, a university level. > > So my question is: given my need, could any of you please > recommend some good reading for me? My wallet-size isn't > in'nite, so one or two major works === Subject: Got a speeding ticket and need to 'ght back Hello everyone, the cops got me speeding 10-15 mph above the speed-limit. My arguement is as follows. Given that the radar has a +/-5 mph standard deviation error, I claim that the cops did not get me beyound resonable doubt. Z = -10/5 = -2 P( Z < -2) = 0.03 3% chance that i didnt speed and the radar was at fault. My arguement to the judge is that 3 in 100 cars will recieve a false speeding ticket because the radar isnt accurate. And i am one of those 3. Even still, the chances that i went just anywhere from below ï5 mph above the speed limit' Z = -5/5 = -1 P(Z < -1) = 0.15% There is a 15% chance(huge chance) that the radar gave a wrong reading. Do u think the judge will buy the arguement? = Btw The best way to embarrass them is to teach High school students statistics and statiscal reasoning and embarrass them, by making them appear smarter than the cops -suresh === Subject: Re: Got a speeding ticket Duncan are not engineers.You should never build a computer. Dont ever send your children to college for engineering. They are most likely to drive the companies, where they work, bankrupt. If 15 out of every 100 instructions, a computer executes is in error, you know something, the computer is no more than random number generator. That's the restatement of your idiocy. You do not understand statistics. Go away . You do not understand pecentages. You know nothing about them. You do not even know how to use them. If you ever got a Ph.D in psychology, remember that all you did was bull your way thru college. How about if i told u, that there is 15% chance you are going to die tomorrow? Let's see: 85% chance that you wont die. Let's see the power of the 15% chance. You know 3/20 people hate you, and think you are ugly. Out of 20 times you drive, 3 times, the chances are you are going to hit something. chance are that our of 3 of 20 people, in public, are going to mug you. -suresh > Devanathan ) > mph above the speed-limit. My arguement is > >as follows. > > > > Given that the radar has a +/-5 mph standard deviation error, I claim that > >the cops did not get me beyound resonable doubt. > > > >Z = -10/5 = -2 > > > >P( Z < -2) = 0.03 > > > >3% chance that i didnt speed and the radar was at fault. My arguement to the > >judge is that 3 in 100 cars will recieve a false speeding ticket because > >the radar isnt accurate. And i am one of those 3. > > > >Even still, the chances that i went just anywhere from below ï5 mph above > >the speed limit' > >Z = -5/5 = -1 > > > >P(Z < -1) = 0.15% > > > >There is a 15% chance(huge chance) that the radar gave a wrong reading. > > > >Do u think the judge will buy the arguement? > > = > >Btw > >The best way to embarrass them is to teach High school students statistics > >and statiscal reasoning and embarrass them, by making them appear smarter > >than the cops > > > >-suresh > > > > Hi -suresh > > Give it up. You don't stand a chance. > > First you're applying statistical techniques to a laboratory process, > meaning doppler shift measurement. For doppler techniques to > be accepted under laboratory conditions the measurement > repeatability must be quite high if not in'nite. I don't know where > you got a any statistic which would show a +- 5mph (say +- 10% > if you were doing near 50 mph.)error, but you would have to be > ready to prove this point to the judge. > > Second; You _may_ attempt to claim that the measuring accuracy > of the cop is not 100%. Assuming you will attempt this and you > are able to prove that an ordinary cop might have an innaccuracy > rate of say 15%, well the judge is going to say to himself OK he's > probably 85% guilty and because I know already that most of the > speedometers on consumer automobiles might be anywhere up > to +-15% innacurate, if I add it all up there's a 70% to 100% chance > he was really speeding. > > GUILTY !, , > > Pay the 'ne! ........Next case....... > > Hmmmm.... > I wonder if this could become a pilot for a new TV show, > something like Americas Dumbest Court Cases . > > methinks, perhaps, methinks too much.......... Oh well...... > > Later, > Pepe le Pew aka Pat Sullivan === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My > arguement is > as Do you think the judge will even be listening? Real petit court is nothing like TV court. The judge is so bored he's completely zoned out and the only thing that's going to impinge on his awareness is that a slightly louder, longer and more annoying blip is on his radar. The only thing you will awaken in him is his desparate desire to make things go faster so he can get back to his crossword. Let's see, do I want to set a precedent here that potentially lets zillions of people legitimately speed 15 mph over the limit in, say, 15 mph School zones and causes state legislatures to rewrite a bunch of laws?......or should I just pound this big wooden hammer and make this noise go away? They don't select judges from the set of really good lawyers, they select them from the lawyers who can't stay awake during trials. Bart === Subject: Re: Got a speeding ticket and need to 'ght back If the cop really wants to get the speeder, he should speed clock him, with his own car. That's the only way out. Instead of hiding on some obscure corner and pulling a radar gun, isnt just going to cut it. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > If the cop really wants to get the speeder, he should speed clock him, > with his own car. That's the only way out. Instead of hiding on some > obscure corner and pulling a radar gun, isnt just going to cut it. The cop doesn't need a way out. In some areas, speeding ticket money goes directly to the local school system and there's considerable pressure to put 35 mph signs right behind a bush about 100 yds from a 55 mph sign and have the local cop hide nearby. Locals know about it, so outsiders are helping fund the music program. Darned nice of ïem. There's should and there's revenue . Go ahead and try to explain to the judge how all his friends and relatives are stupid, inbred and evil. The contempt charge will help buy that new playground equipment. Enjoy your stay in Hogwaller, and do come back soon. Bart === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > If the cop really wants to get the speeder, he should speed clock him, > with his own car. That's the only way out. Instead of hiding on some > obscure corner and pulling a radar gun, isnt just going to cut it. > > -suresh You can deny till your blue in the face... youre still going to be convicted. Just plead guilty and pay the 'ne. Stop wasting you own time trying to 'gure ways out of this. When is your court date? Rob === Subject: Re: Got a speeding ticket and need to 'ght back Everybody has to learn to play by my rules. Nobody has other options. Otherwise, you will goto . Here's a secret 1) I make the rules 2) I make the rules what is do u think the 3rd rule is? I make the rules That's all everyone, has to know. I always have it my way. Otherwise, i am going to torture you alive, just to have fun and make a buffon out of you. -suresh > > The Lord of Chaos (Suresh Devanathan) message > > If the cop really wants to get the speeder, he should speed clock him, > > with his own car. That's the only way out. Instead of hiding on some > > obscure corner and pulling a radar gun, isnt just going to cut it. > > > > -suresh > > You can deny till your blue in the face... youre still going to be > convicted. Just plead guilty and pay the 'ne. Stop wasting you own time > trying to 'gure ways out of this. When is your court date? > > > Rob > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Everybody has to learn to play by my rules. Nobody has other options. > Otherwise, you will goto . > > Here's a secret > 1) I make the rules > 2) I make the rules > > what is do u think the 3rd rule is? > I make the rules And yet, you still have to ask for help to get out of a speeding ticket. What's the matter? Can't tell the judge your rules ? Doug === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Everybody has to learn to play by my rules. Nobody has other options. > Otherwise, you will goto . > > Here's a secret > 1) I make the rules > 2) I make the rules > > what is do u think the 3rd rule is? > I make the rules > > That's all everyone, has to know. I always have it my way. Otherwise, i am > going to torture you alive, just to have fun and make a buffon out of you. > > > -suresh I was willing to help you. I had an idea and asked you for your court date. Instead you go off on a temper tantrum. No help for you. So sorry. There is a con'dent ignorance that comes with youth. The sooner you recognise that, the better. Rob === Subject: Re: Got a speeding ticket and need to 'ght back you are going to be burned alive. -suresh > > The Lord of Chaos (Suresh Devanathan) message > > Everybody has to learn to play by my rules. Nobody has other options. > > Otherwise, you will goto . > > > > Here's a secret > > 1) I make the rules > > 2) I make the rules > > > > what is do u think the 3rd rule is? > > I make the rules > > > > That's all everyone, has to know. I always have it my way. Otherwise, i am > > going to torture you alive, just to have fun and make a buffon out of you. > > > > > > -suresh > > I was willing to help you. I had an idea and asked you for your court date. > Instead you go off on a temper tantrum. No help for you. So sorry. There > is a con'dent ignorance that comes with youth. The sooner you recognise > that, the better. > > > Rob > > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > you are going to be burned alive. > -suresh Thats mature. Rob > > > > The Lord of Chaos (Suresh Devanathan) > message > > > Everybody has to learn to play by my rules. Nobody has other options. > > > Otherwise, you will goto . > > > > > > Here's a secret > > > 1) I make the rules > > > 2) I make the rules > > > > > > what is do u think the 3rd rule is? > > > I make the rules > > > > > > That's all everyone, has to know. I always have it my way. Otherwise, i > am > > > going to torture you alive, just to have fun and make a buffon out of > you. > > > > > > > > > -suresh > > > > I was willing to help you. I had an idea and asked you for your court > date. > > Instead you go off on a temper tantrum. No help for you. So sorry. > There > > is a con'dent ignorance that comes with youth. The sooner you recognise > > that, the better. > > > > > > Rob > > > > > > > > === Subject: Re: Got a speeding ticket and need to 'ght back above the speed-limit. My arguement is >as follows. You can claim anything that you want, however, each community has its standards at what level a ticket can be given with radar. In my state that is 10 mph or perhaps less. Arguing won't change that. Also, if the of'cer than followed you in his car -- then he also clocked your with his speedometer. In my State a State patrolman can give you a ticket at 1 mph over on the highway. Other of'cials have to catch you at 10 mph over on an Interstate or State highway. Now, understanding that the speed limit is set for the safety of others on the road in addition to you, why were you risking others life and limbs? 10 to 15 mph more could mean the difference between life and death of someone you hit when you loose control. That someone might be you. === Subject: Re: Got a speeding ticket and need to 'ght back > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My arguement is > as follows. > > Given that the radar has a +/-5 mph standard deviation error, I claim that > the cops did not get me beyound resonable doubt. [Stuff snipped] You could try arguing that the posted speed limit, as well as (probably) the speed limit laws on the books are arbitrarily vague since there is no preferred reference frame with respect to constant velocity. Thus a sign that indicates a speed limit of 55 mph technically provides no information, since it doesn't specify any reference frame. Tom Larson === Subject: Re: Got a speeding ticket and need to 'ght back > You could try arguing that the posted speed limit, as well as > (probably) the speed limit laws on the books are arbitrarily vague > since there is no preferred reference frame with respect to constant > velocity. Thus a sign that indicates a speed limit of 55 mph > technically provides no information, since it doesn't specify any > reference frame. I could. But the earth is the assumed reference frame. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) ews:c0imue$16o3tr$1@ID-182852.news.uni-berlin.de: > >> You could try arguing that the posted speed limit, as well as >> (probably) the speed limit laws on the books are arbitrarily vague >> since there is no preferred reference frame with respect to constant >> velocity. Thus a sign that indicates a speed limit of 55 mph >> technically provides no information, since it doesn't specify any >> reference frame. > I could. But the earth is the assumed reference frame. > > -suresh You could argue that just by standing still in the court room, you are travelling at about 790 mph for New Jersey, based on the rotational speed of the Earth. And, if you were driving East, you can claim that you were actually slowing down from your normal speed. - Tim Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > > > You could try arguing that the posted speed limit, as well as > > (probably) the speed limit laws on the books are arbitrarily vague > > since there is no preferred reference frame with respect to constant > > velocity. Thus a sign that indicates a speed limit of 55 mph > > technically provides no information, since it doesn't specify any > > reference frame. > I could. But the earth is the assumed reference frame. This becomes more entertaining than JSH threads. Is there a single crank on earth that didn't lost sence of humour? === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My arguement is > as follows. > > Given that the radar has a +/-5 mph standard deviation error, I claim that > the cops did not get me beyound resonable doubt. > > Z = -10/5 = -2 > > P( Z < -2) = 0.03 > > 3% chance that i didnt speed and the radar was at fault. My arguement to the > judge is that 3 in 100 cars will recieve a false speeding ticket because > the radar isnt accurate. And i am one of those 3. > > Even still, the chances that i went just anywhere from below ï5 mph above > the speed limit' > Z = -5/5 = -1 > > P(Z < -1) = 0.15% > > There is a 15% chance(huge chance) that the radar gave a wrong reading. > > Do u think the judge will buy the arguement? > = > Btw > The best way to embarrass them is to teach High school students statistics > and statiscal reasoning and embarrass them, by making them appear smarter > than the cops > > -suresh LOL, ya, that will sure get you out of a speeding ticket. heh heh. You were speeding, grow up and pay the 'ne for heavens sake. Why even waste the energy thinking about 'ghting it? Rob === Subject: Re: Got a speeding ticket and need to 'ght back > LOL, ya, that will sure get you out of a speeding ticket. heh heh. You > were speeding, grow up and pay the 'ne for heavens sake. Why even waste > the energy thinking about 'ghting it? u arent a lawyer are you? because you sound like one. If did goto law school, ask yourself, why you wasted all those years there. Apparently you missed the point of the constitution and the US judical system. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > > LOL, ya, that will sure get you out of a speeding ticket. heh heh. You > > were speeding, grow up and pay the 'ne for heavens sake. Why even waste > > the energy thinking about 'ghting it? > u arent a lawyer are you? because you sound like one. If did goto law > school, ask yourself, why you wasted all those years there. Apparently you > missed the point of the constitution and the US judical system. > > -suresh Suresh, I realize youre young and it seems fun to think about beating the system, but really, unless you have a leg to stand on, which you dont sinse you *were speeding, just get it over with and pay the 'ne. When the 2000 Mustang came out, a woderful car btw, I bought one. One bright and shiny day in central Oregon, on a deserted freeway, I decided to test her out. I hit its speed limiter, 116mph, in no time §at. It handled like it was on rails. Very nice car. Well I got a ticket near the end of my joy ride. 79mph. I argued that back in 96 the legislature passed a provision that allowed ALL drivers to drive whatever speed they chose as long as it was safe and reasonable for the conditions. I was right. They had... But, it didnt apply to Interstate freeways. I was lucky though, through proper respect I displayed for the court I was allowed a sentence that put me on 6 months of driving probation and a $98.00 'ne. I was overjoyed at the 'ne. It could have been hundreds... especially if I had been caught at top speed. So you see if you take responcibility for your actions, and not try to avoid them, the court is much more amenable to lowered 'nes, etc. Heck, my younger brother got out of a ticket for not wearing his seatbelt by offering to take a safe drivers course. There are ways around things like this... but you 'rst have to take responcibility... not avoid it. Btw, I still have no tickets on my driving record sinse 1986. Dont try ot *beat the system with arguments that have no chance of succeeding, rather, aproach the process with humility. Explore other avenues. One might be to take responsibility for your crime. Rob === Subject: Re: Got a speeding ticket and need to 'ght back > The speeding 10-15 mph above the speed-limit. > > My argument is as follows. > > > > Given that the radar has a +/-5 mph standard deviation error, > > I claim that the cops did not get me beyond reasonable doubt. > > They don't need to. > Driving is a privilege and not a right. > All the cops need to show is that it was more likely, > given the evidence, that you were speeding > anyone can he likes. Why should there be laws against racism? Hey, why even have a court? ïDriving is a privilege. The cops can make any rule they want. Why even goto court to 'ght a ticket?' Wait a second, by your reasoning ïNational security is a privilege. Or is it a right?' spelling has nothing to do with math. If you didnt learn that in high school, you are never going to learn it, in the rest of your life! Embrassing!!! completely! high school students are a lot smarter than you! it's a fact. Did u pass math/physics by a C+ ? -suresh === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > spelling has nothing to do with math. If you didnt learn that in high > school, you are never going to learn it, in the rest of your life! > Embrassing!!! completely! high school students are a lot smarter than you! > it's a fact. Ah, you're trolling! Why didn't you just say so in the 'rst place? Doug === Subject: Re: Got a speeding ticket and need to 'ght back hey buddy, here's some advice: avoid my theard. I am looking for intelligent people to talk to and debate with. Not some loser who has a lot of time, to waste mine. Here's my problem with you. A) you are dumb B) you do not understand mathematics So, go away. -suresh > The Lord of Chaos (Suresh Devanathan) > > > spelling has nothing to do with math. If you didnt learn that in high > > school, you are never going to learn it, in the rest of your life! > > Embrassing!!! completely! high school students are a lot smarter than > you! > > it's a fact. > > Ah, you're trolling! Why didn't you just say so in the 'rst place? > > Doug > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) am looking for > intelligent people to talk to and debate with. Not some loser who has a lot > of time, to waste mine. > Here's my problem with you. > A) you are dumb > B) you do not understand mathematics === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > hey buddy, here's some advice: avoid my theard. Hell, no. I won't avoid your thread, either. Anyone whose best argument in a mathematical discussion is you don't understand mathematics and you are dumb deserves what he gets. I hope they get you for contempt of court. Doug === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos math. If you didnt learn that in high > school, you are never going to learn it, in the rest of your life! > Embrassing!!! completely! high school students are a lot smarter than > you! it's a fact. > > Did u pass math/physics by a C+ ? > > -suresh Spelling has a lot to do with math. For example, I spell the word postulate as p-o-s-T-u-l-a-t-e, not p-o-s-u-l-a-t-e, as you did in a previous post. Also, perhaps it is one of the failings of math student or perhaps math teachers should stress it more, but math is so much more than 2 + 2 = 4 and other symbols. Try to prove something using symbols only. Hell, even your ïproof' about the standard deviation of the radar gun used words galore. Before you go making statements such as the standard deviation of the radar gun is +/- 5mph, you should actually learn what the standard deviation is. I mean, you said yourself that Any high school student can understand what I am saying. If I recall correctly from my last three years of teaching statistics, the standard deviation cannot be negative. Maybe you should actually pay attention in class. Oh, and by the way, the speed of light is not constant [http://www.telegraph.co.uk/news/main.jhtml? xml=/news/2002/02/17/waa117.xml&sSheet=/portal/2002/02/17/por_ right.html ] (http://tinyurl.com/2nrdb) or [http://www.news.harvard.edu/gazette/2001/01.24/01- stoplight.html] (http://tinyurl.com/3cnzk) - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > > They don't need to. > Driving is a privilege and not a right. > All the cops need to show is that it was more likely, > given the evidence, that you were speeding > anyone can he likes. Why should there be laws against racism? Hey, why even have a court? ïDriving is a privilege. The cops can make any rule they want. Why even goto court to 'ght a ticket?' Wait a second, by your reasoning ïNational security is a privilege. Or is it a right?' spelling has nothing to do with math. If you didnt learn that in high school, you are never going to learn it, in the rest of your life! Embrassing!!! completely! high school students are a lot smarter than you! it's a fact. > Did u pass math/physics by a C+ ? > -suresh > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > > The Lord of Chaos (Suresh Devanathan) message > > > > > They don't need to. > > Driving is a privilege and not a right. > > All the cops need to show is that it was more likely, > > given the evidence, that you were speeding > > > Well, it is a privilege is some circumstances. For example, you need a permit if you want to work in this country if you are not a citizen. Owning a business is a privilege. I could make a lot of money selling opium or child pornography, but I can't get a license to do so. Anytime you need a license to do something, that something is a priviliege. Otherwise, you wouldn't need a license to do it. Russell - 2 many 2 count === Subject: Re: Got a speeding ticket and need to 'ght back ... > Anytime you need a license to do something, > that something is a priviliege. Otherwise, > you wouldn't need a license to do it. could you elaborate a bit? > Russell > - 2 many 2 count === Subject: Re: Got a speeding ticket and need to 'ght back You are spewing this thread, into nonsense. Look, here's my solid agruement, if you missed my post. If you cannot talk the law of physics, or law of averages or statistics, i am not here to argue with you. Either you have no focus,or just plain dumb. I will post arguement again,if you missed it. If you cannot do math or statistics or physics, dont post, at all. Your posts are pointless, completely to me and just to everybody else. If you are one of those mensa folks, you are an embarrassment to their community. I was here to ask help from a statistical and scienti'c viewpoint. If you cannot answer that, you stand testimonial to , question of ïprobability of 'nding an idiot in mensa?'. I already have talked to 3 idiots, out of 5 or 6, ones, i have met there. So, there is about 1/2 probabilty that a member of mensa is a complete idiot. -suresh > > The Lord of Chaos (Suresh Devanathan) message > > > > The Lord of Chaos (Suresh Devanathan) > message > > > > > > > > They don't need to. > > > Driving is a privilege and not a right. > > > All the cops need to show is that it was more likely, > > > given the evidence, that you were speeding > > > > > > > Well, it is a privilege is some circumstances. > For example, you need a permit if you want > to work in this country if you are not a citizen. > > Owning a business is a privilege. > I could make a lot of money selling opium or > child pornography, but I can't get a license to > do so. > > Anytime you need a license to do something, > that something is a priviliege. Otherwise, > you wouldn't need a license to do it. > > > Russell > - 2 many 2 count > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Either you have no focus,or just plain dumb. Yeah, he's dumb, and I'm dumb, and everyone else is just plain dumb. Look, I can see how embarrassing this must be for you - you came in here to have a nice troll, and you're getting whacked around more than Paul Reubens' jewels at an all-night theater. Really, though, you brought it upon yourself. Doug === Subject: Re: Got a speeding ticket and need to 'ght back > The Lord of Chaos (Suresh Devanathan) >> Either you have no focus,or just plain dumb. > > Yeah, he's dumb, and I'm dumb, and everyone else > is just plain dumb. > > Look, I can see how embarrassing this must be for > you - you came in here to have a nice troll, and > you're getting whacked around more than Paul Reubens' > jewels at an all-night theater. Really, though, you > brought it upon yourself. sounds like damage control to me. and what do you know about the experince of paul reubens' jewels at an all-night theater? never mind... forget i asked. :) rgrds, === Subject: Re: Got a speeding ticket and need to 'ght back > and what do you know about the experince of > paul reubens' jewels at an all-night theater? I can read a newspaper. How about yourself? Doug === Subject: Re: Got a speeding ticket and need to 'ght back >> and what do you know about the experince of >> paul reubens' jewels at an all-night theater? > > I can read a newspaper. How about yourself? like everybody else who reads newspapers, i 'lter for interest. :) rgrds, === Subject: Re: Got a speeding ticket and need to 'ght back > > >> and what do you know about the experince of > >> paul reubens' jewels at an all-night theater? > > > > I can read a newspaper. How about yourself? > > like everybody else who reads newspapers, > i 'lter for interest. Good one, chief. I hear the WB network is hiring comedy writers. Doug === Subject: Re: Got a speeding ticket and need to 'ght back >> >> and what do you know about the experince of >> >> paul reubens' jewels at an all-night theater? >> > >> > I can read a newspaper. How about yourself? >> >> like everybody else who reads newspapers, >> i 'lter for interest. > > Good one, chief. I hear the WB network is > hiring comedy writers. send them your r .8esum .8e, you might serve as inspiration. :) rgrds, === Subject: Re: Got a speeding ticket and need to 'ght back @nwrdny02.gnilink.net: >> >> Good one, chief. I hear the WB network is >> hiring comedy writers. > > > send them your r .8esum .8e, you > might serve as inspiration. > > > >:) > > > rgrds, This thread has been more entertaining than anything I've seen on the WB. - Tim Timothy M. Brauch Graduate Student Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back I told you to ïgo away'. Do u speak english? Can you read or write in english? -suresh > The Lord of Chaos (Suresh Devanathan) > > Either you have no focus,or just plain dumb. > > Yeah, he's dumb, and I'm dumb, and everyone else is just plain dumb. > > Look, I can see how embarrassing this must be for you - you came in here > to have a nice troll, and you're getting whacked around more than Paul > Reubens' jewels at an all-night theater. Really, though, you brought it > upon yourself. > > Doug > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > I told you to ïgo away'. Do u speak english? Can you read or write in > english? Do u speak English? What - are you too much in a hurry to actually write out the word you ? Doug === Subject: Re: Got a speeding ticket and need to 'ght back may be, you missed the title of my theard. Read it: Got a speeding ticket and neet to 'ght back Ofcourse, i am in a hurry to 'nd an excellent arguement. -suresh > The Lord of Chaos (Suresh Devanathan) > > I told you to ïgo away'. Do u speak english? Can you read or write in > > english? > > Do u speak English? What - are you too much in a hurry to actually > write out the word you ? > > Doug > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > may be, you missed the title of my theard. Read it: Got a speeding ticket > and neet to 'ght back Ofcourse, i am in a hurry to 'nd an excellent > arguement. > Why? You know you were speeding. The honest thing is to admit it, pay the 'ne and learn to drive within the limit. > -suresh > > > The Lord of Chaos (Suresh Devanathan) > > > I told you to ïgo away'. Do u speak english? Can you read or write in > > > english? > > > > Do u speak English? What - are you too much in a hurry to actually > > write out the word you ? > > > > Doug > > > > > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > I told you to ïgo away'. Do u speak english? > Can you read or write in english? never mind the ticket thing, how did you come by the lord of chaos thing? rgrds, === Subject: Re: Got a speeding ticket and need to 'ght back > The Lord of Chaos (Suresh Devanathan ) > > > I told you to ïgo away'. Do u speak english? > > Can you read or write in english? > > > never mind the ticket thing, > > how did you come by > the lord of chaos thing? > That's translation of my name. Suresh = lord of the rain , lord of chaos ,... Kumar = Prince , handsome ,.... Devanathan = King of the Gods > > > rgrds, > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > > > The Lord of Chaos (Suresh Devanathan ) > > > > > I told you to ïgo away'. Do u speak english? > > > Can you read or write in english? > > > > > > never mind the ticket thing, > > > > how did you come by > > the lord of chaos thing? > > > That's translation of my name. > > Suresh = lord of the rain , lord of chaos ,... > Kumar = Prince , handsome ,.... > Devanathan = King of the Gods > > rgrds, Nothing in there about inteligence, huh? Rob === Subject: Re: Got a speeding ticket and need to 'ght back > > That's translation of my name. > > > > Suresh = lord of the rain , lord of chaos ,... > > Kumar = Prince , handsome ,.... > > Devanathan = King of the Gods > > > rgrds, > > Nothing in there about inteligence, huh? That shows the level of your intelligence. You are de'nitely not the deep thinker type. A Prince, is supposed to be intelligent. duh!!! > > > Rob > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > A Prince, is supposed to be intelligent. duh!!! I guess so, what with them being smart enough to be born into wealth. You kinda have to wonder what went wrong in your case? Doug === Subject: Re: Got a speeding ticket and need to 'ght back Go away. Your mom thinks you are the ugliest one in the family. -suresh > The Lord of Chaos (Suresh Devanathan) > > > A Prince, is supposed to be intelligent. duh!!! > > I guess so, what with them being smart enough to be born into wealth. > > You kinda have to wonder what went wrong in your case? > > Doug > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Go away. Your mom thinks you are the ugliest one in the family. C'mon. You can troll better than this. Go back to misspelling the words, or something - at least that exotic and interesting. Doug === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > A Prince, is supposed to be intelligent. duh!!! I thought that when Euler said There is no prince way to mathematics he was politely meaning the opposite? Or by Prince with capital P you do refer to some unfamous singer. That one looks intelligent for sure! === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > >> The Lord of Chaos (Suresh Devanathan ) >> >> > I told you to ïgo away'. Do u speak english? >> > Can you read or write in english? >> >> >> never mind the ticket thing, >> >> how did you come by >> the lord of chaos thing? >> > That's translation of my name. > > Suresh = lord of the rain , lord of chaos ,... > Kumar = Prince , handsome ,.... > Devanathan = King of the Gods your given name, or an assumed name? rgrds, === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > I told you to ïgo away'. Do u speak english? Can you read or write in > english? > > -suresh Hmm, I think the real question is, can *you* write English? - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back If you dont understand ïgo away', here's a link to dictionary.com http://www.dictionary.com Look up go and away . And put it together and you will understand what i m saying. In case, you do not understand, what together means, you can still look it up at http://www.dictionary.com -suresh The Lord of Chaos (Suresh Devanathan) > I told you to ïgo away'. Do u speak english? Can you read or write in > english? > > -suresh > > The Lord of Chaos (Suresh Devanathan) > > > Either you have no focus,or just plain dumb. > > > > Yeah, he's dumb, and I'm dumb, and everyone else is just plain dumb. > > > > Look, I can see how embarrassing this must be for you - you came in here > > to have a nice troll, and you're getting whacked around more than Paul > > Reubens' jewels at an all-night theater. Really, though, you brought it > > upon yourself. > > > > Doug > > > > > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > You are spewing this thread, into nonsense. Look, here's my solid > agruement, if you missed my post. If you cannot talk the law of > physics, or law of averages or statistics, i am not here to argue > with you. Either you have no focus,or just plain dumb. I will post > arguement again,if you missed it. If you cannot do math or statistics > or physics, dont post, at all. Your posts are pointless, completely to > me and just to everybody else. If you are one of those mensa folks, > you are an embarrassment to their community. I was here to ask help > from a statistical and scienti'c viewpoint. If you cannot answer > that, you stand testimonial to , question of ïprobability of 'nding > an idiot in mensa?'. I already have talked to 3 idiots, out of 5 or 6, > ones, i have met there. So, there is about 1/2 probabilty that a > member of mensa is a complete idiot. > > -suresh > By that reasoning, I §ipped a coin once and got heads all three times. I guess the probability of getting heads must be 100%. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My arguement is > as follows. If you really want to beat the ticket do what a lawyer friend of mine did. As part of your discovery demand that the police provide all of the maintenance history of the radar gun. Radar guns have to be regularly calibrated. The judge will (sometimes!) dismiss the case if the police don't provide you with the maintenance records, and the police are usually too lazy to do so. Russell - 2 many 2 count === Subject: Re: Got a speeding ticket and need to 'ght back Russell Easterly everyone, > > the cops got me speeding 10-15 mph above the speed-limit. My arguement > is > > as follows. > > If you really want to beat the ticket do what a lawyer friend of mine did. > As part of your discovery demand that the police provide all of the > maintenance history of the radar gun. Radar guns have to be > regularly calibrated. The judge will (sometimes!) dismiss the case if > the police don't provide you with the maintenance records, > and the police are usually too lazy to do so. > > > Russell > - 2 many 2 count > > In Colorado, radar guns must be calibrated with a tuning fork that has been validated by a particular state laboratory within 36 months of the date of the ticket, and many police organizations on Colorado have not bothered to get the validation. === Subject: Re: Got a speeding ticket and need to 'ght back > Russell (Suresh Devanathan) > > > Hello everyone, > > > the cops got me speeding 10-15 mph above the speed-limit. My arguement > is > > > as follows. > > > > If you really want to beat the ticket do what a lawyer friend of mine did. > > As part of your discovery demand that the police provide all of the > > maintenance history of the radar gun. Radar guns have to be > > regularly calibrated. The judge will (sometimes!) dismiss the case if > > the police don't provide you with the maintenance records, > > and the police are usually too lazy to do so. > > > > > > Russell > > - 2 many 2 count > > > > > > In Colorado, radar guns must be calibrated with a tuning fork that has > been validated by a particular state laboratory within 36 months of the > date of the ticket, and many police organizations on Colorado have not > bothered to get the validation. That is a VERY good point to bring up to the judge. That is, if this guy lives in Colorado. (...Starblade Riven Darksquall...) === Subject: Re: Got a speeding ticket Chaos (Suresh Devanathan) >> > > Hello everyone, >> > > the cops got me speeding 10-15 mph above the speed-limit. My >> > > arguement >> is >> > > as follows. >> > >> > If you really want to beat the ticket do what a lawyer friend of mine >> > did. As part of your discovery demand that the police provide all of >> > the maintenance history of the radar gun. Radar guns have to be >> > regularly calibrated. The judge will (sometimes!) dismiss the case if >> > the police don't provide you with the maintenance records, >> > and the police are usually too lazy to do so. >> > >> > >> > Russell >> > - 2 many 2 count >> > >> > >> >> In Colorado, radar guns must be calibrated with a tuning fork that has >> been validated by a particular state laboratory within 36 months of the >> date of the ticket, and many police organizations on Colorado have not >> bothered to get the validation. > > That is a VERY good point to bring up to the judge. That is, if this > guy lives in Colorado. > > (...Starblade Riven Darksquall...) I once heard that these tuning forks are hit against something to calibrate the radar guns, but cops sometimes hit them against things that weren't intended to be hit by a tuning fork. Perhaps you could show reasonable doubt because any small imperfections in the tuning fork can alter the frequency of the tuning fork. === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My > arguement is > as follows. > > Given that the radar has a +/-5 mph standard deviation error, I claim > that > the cops did not get me beyound resonable doubt. If I were the judge in this case, I would remind you that perjury, that is, lying under oath is illegal. Then I would ask you where you got this +/- 5mph fact > > Z = -10/5 = -2 > > P( Z < -2) = 0.03 > > 3% chance that i didnt speed and the radar was at fault. My arguement > to the judge is that 3 in 100 cars will recieve a false speeding > ticket because the radar isnt accurate. And i am one of those 3. Assuming the judge understands the math, which is highly unlikely since you say no cops would and it justs embarrasses them. In fact, maybe you will embarrass the judge by showing you are smarter than he is. I'm sure that will sit well with him. Besides, if I were the judge, I would say, Fine, then you only have to 97% of the ticket, plus court costs. > Even still, the chances that i went just anywhere from below ï5 mph > above the speed limit' > Z = -5/5 = -1 > > P(Z < -1) = 0.15% > > There is a 15% chance(huge chance) that the radar gave a wrong > reading. If I go up to the average person on the street and say there is an 85% chance of rain tomorrow at 5:00pm, do you think he would take an umbrella to work with him? If I said you have an 85% chance of winning the lottery, would you buy a ticket? If you tell me you have an 85% chance of having been actually speeding, even just just one mile over, I would say guilty then add a stupidity 'ne for trying to embarrass me and to reimburse the city for the time the of'cer has wasted being in court to challenge you. > Do u think the judge will buy the arguement? > = > Btw > The best way to embarrass them is to teach High school students > statistics and statiscal reasoning and embarrass them, by making them > appear smarter than the cops > > -suresh I would think the best way to embarrass them would to be 'nd some pictures of the male of'cers in women's underwear all huddled around livestock of some sort. But then, maybe I'm just sick. Besides, whether you know it or not, most police of'cers went to high school. So, in say 10 years, all of the of'cers would know statistics (since you are requiring it be taught to high schoolers) and you plan would 're. In fact, police of'cers would then know enough statistics to pull you over going 5mph *under* the speed limit and say there is an 85% chance you were speeding. - Tim I hate when people lie with math Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket judge understands the math, which is highly unlikely since >you say no cops would and it justs embarrasses them. In fact, maybe you >will embarrass the judge by showing you are smarter than he is. ... or full of (insert whatever) here. Attempts at embarrassing the judge may lead to contempt charges. Go to the Laws and see if you have missed anything there. If not, your out of luck in a US Court. Arguments may work if you snow a Jury -- but you are going before just a Judge who has heard every excuse in the book before. Mary === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Given that the radar has a +/-5 mph standard deviation error, I claim that > the cops did not get me beyound resonable doubt. In addition, where'd you get this +/-5mph standard deviation 'gure? It appears highly arbitrary. Doug === Subject: Re: Got a speeding ticket and need to 'ght back Approximately f lamda = c f lamda = 186 000 mps f -> frequency in hertz h = 60min = 60 * 60 sec f lamda = 186 000 = 186 000 mps f -> 1000,000,000 s^-1 (gigahertz) lamda = .000186 miles That's the radar's wavelength approximately. Take the radar's time window, it internally uses to compute the speed: 1/10 second ( the human eye can see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 sec approx 0.000186/ .1s = .00186 mps = 6.6 mph > The Lord of Chaos (Suresh Devanathan) > > > Given that the radar has a +/-5 mph standard deviation error, I claim > that > > the cops did not get me beyound resonable doubt. > > In addition, where'd you get this +/-5mph standard deviation 'gure? It > appears highly arbitrary. > > Doug > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) mps > > f -> frequency in hertz > > h = 60min > = 60 * 60 sec > > f lamda = 186 000 > = 186 000 mps > > f -> 1000,000,000 s^-1 (gigahertz) > > lamda = .000186 miles > > That's the radar's wavelength approximately. Take the radar's time window, > it internally uses to compute the speed: 1/10 second ( the human eye can > see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 sec > approx > > 0.000186/ .1s = .00186 mps = 6.6 mph Numeric aperture and aiming accuracy are better arguments if you were in traf'c. Was it a laser gun? Either way, it won't be admissable. You aren't an acknowledged expert or professional witness and your testimony will not be admitted as evidence. If you can haul in an Assistant Prof. in physics from a community college, that will probably §y. The persecution (no typo) will then seek to discredit your witness. Example, Persecutor Dr. Whore, are you being rembursed for appearing here. Dr. Whore, Yes, of course. I had expenses. Persecution, Your testimony is then purchased, Dr. Whore, is it not? Persecutor Dr. Whore, are you being rembursed for appearing here. Dr. Whore, No, of course not. My testimony carries with it no taint of bribery. Perscutor, Dr. Whore, what competent professional works for free? (Dr. Whore, A pro bono lawyer. ) You have to coach your witness and not be accused of coaching your witness. Your best insurance is to bring in a reporter with a videocam (and spare batteries and tape). Citation hearings have no court reporter and no documentation. There is nothing to go appelate. Make a certi'ed recording of the procedures and the heat is on. Cops and prosecutors are stupid. It works to their advantage. They have the Book, they are trained like seals to go exactly by the Book, and everything is scored only by the Book. What do you know? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Got a speeding the testing specs for the speci'c equipment to 'nd the error of the equipment. Mary Devanathan ) > >f -> frequency in hertz > >h = 60min > = 60 * 60 sec > >f lamda = 186 000 > = 186 000 mps > >f -> 1000,000,000 s^-1 (gigahertz) > >lamda = .000186 miles > >That's the radar's wavelength approximately. Take the radar's time window, >it internally uses to compute the speed: 1/10 second ( the human eye can >see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 sec >approx > >0.000186/ .1s = .00186 mps = 6.6 mph > > > > > > > > > > > > > > > >> The Lord of Chaos (Suresh Devanathan) >> >> > Given that the radar has a +/-5 mph standard deviation error, I claim >> that >> > the cops did not get me beyound resonable doubt. >> >> In addition, where'd you get this +/-5mph standard deviation 'gure? It >> appears highly arbitrary. >> >> Doug >> >> > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > Approximately > > f lamda = c > > f lamda = 186 000 mps > > f -> frequency in hertz > > h = 60min > = 60 * 60 sec > > f lamda = 186 000 > = 186 000 mps > > f -> 1000,000,000 s^-1 (gigahertz) > > lamda = .000186 miles > > That's the radar's wavelength approximately. Take the radar's time window, > it internally uses to compute the speed: 1/10 second ( the human eye can > see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 sec > approx > > 0.000186/ .1s = .00186 mps = 6.6 mph Your arguments garbage. Radar guns work, saddly, so pay your 'ne and dont waste the courts time. Rob === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) > Approximately > > f lamda = c > > f lamda = 186 000 mps > > f -> frequency in hertz > > h = 60min > = 60 * 60 sec > > f lamda = 186 000 > = 186 000 mps > > f -> 1000,000,000 s^-1 (gigahertz) > > lamda = .000186 miles > > That's the radar's wavelength approximately. Take the radar's time > window, it internally uses to compute the speed: 1/10 second ( the > human eye can see about 30 frames / sec ). Assuming Reaction time of a > driver = 1/10 sec approx > > 0.000186/ .1s = .00186 mps = 6.6 mph > [http://www.bushnell.com/productinfo/speedgun/speedster.html] The Speedster uses digital technology and DSP (Digital Signal Processing) to provide accurate real-time measurements to +/- 1.0 MPH. Commercial radar guns are accurate to about +/- 1mph. I would assume police of'cers use something just as good, if not better. But that is just an assumption. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back > The Lord of Chaos (Suresh Devanathan ) > The point of the math was to estimate the order of the error, ie 6.6 mph, because for one, i am not aware of the internals of the radar device. I used the analysis to give a good reason, as to where the +/-5 mph came from. And as you can see, from your own 'ndings of +/- 1 mph , the math is self-consistant. Unless the company that makes the radar, makes their internal patents public, i have a solid case. Btw, i am an electrical engineer. I can de'nitely look into the radar and i can see for myself, if it has no obvious §aws. I am sure, it has tons of §aws in the design. > > > [http://www.bushnell.com/productinfo/speedgun/speedster.html] > The Speedster uses digital technology and DSP (Digital Signal > Processing) to provide accurate real-time measurements to +/- 1.0 MPH. > > Commercial radar guns are accurate to about +/- 1mph. I would assume > police of'cers use something just as good, if not better. But that is > just an assumption. > > - Tim > > > Timothy M. Brauch > Graduate Student > Department of Mathematics > Wake Forest University > === Subject: Re: Got a speeding ticket and need to 'ght back Proofs are based on assumptions, unfortunately. That's how things are proved, based on axioms or things taken for granted. If you did not understand that, it's not my fault. Assumptions does not imply invalidity of reasoning. They imply that there are based on solid posulates. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back Are you a professional lawyer? because you seem to taking dumb. -suresh The Lord of Chaos (Suresh Devanathan) > Proofs are based on assumptions, unfortunately. That's how things are > proved, based on axioms or things taken for granted. If you did not > understand that, it's not my fault. Assumptions does not imply invalidity of > reasoning. They imply that there are based on solid posulates. > > > -suresh > > === Subject: Re: Got a speeding ticket and need to 'ght back high school students who take statistics are smarter than you. Besides, everything i am talking is at the high school level. Any high school can understand what i am saying. You have got to be stupid. You dont understand ïposulates'! you think they are signs of weakness. The Radar works based on posulates. One of them is that speed of light is constant and there is such a thing as wavelength, such a thing as velocity, etc. -suresh The Lord of Chaos (Suresh Devanathan) > Are you a professional lawyer? because you seem to taking dumb. > > -suresh > > The Lord of Chaos (Suresh Devanathan) message > > Proofs are based on assumptions, unfortunately. That's how things are > > proved, based on axioms or things taken for granted. If you did not > > understand that, it's not my fault. Assumptions does not imply invalidity > of > > reasoning. They imply that there are based on solid posulates. > > > > > > -suresh > > > > > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) than you. Besides, > everything i am talking is at the high school level. Any high school can > understand what i am saying. You have got to be stupid. You dont understand > ïposulates'! you think they are signs of weakness. The Radar works based on > posulates. One of them is that speed of light is constant and there is such > a thing as wavelength, such a thing as velocity, etc. You are posting to sci.math, sci.math.num-analysis and sci.physics. If you are a high school student who takes statistics then there is a high propability that the reader is better educated, more experienced and more intelligent than you are. In your case, that propability is very close to 1. === Subject: Re: Got a speeding ticket and need to 'ght back > The Lord of Chaos (Suresh Devanathan ) > and sci.physics. If > you are a high school student who takes statistics then there is a high > propability that the reader is better educated, more experienced and > more intelligent than you are. wait a minute, you used a conditional If , I am so not! Your agruement falls apart, by a false proposition. I guess, i do not even have to read the rest, of your arguement, since the condtional If is not even, a bit true to begin with. It's like this If assume that the other guy in the arguement is an idoit // (comment block) by virtue of the condition being true return I am always right end // comment : the end > > In your case, that propability is very close to 1. === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > 0.000186/ .1s = .00186 mps = 6.6 mph That's a lot of assumptions. You could have saved the time, and just slowed the hell down. Doug === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan) > 3% chance that i didnt speed and the radar was at fault. My arguement to the > judge is that 3 in 100 cars will recieve a false speeding ticket because > the radar isnt accurate. And i am one of those 3. That doesn't prove that you were one of those three. Doug === Subject: Re: Got a speeding ticket and need to 'ght back Assuming a US judge, he will say shut up, guilty, pay at the window on your way out. Slainte, Fletch The Lord of Chaos (Suresh Devanathan) > Hello everyone, > the cops got me speeding 10-15 mph above the speed-limit. My arguement is > as follows. > > Given that the radar has a +/-5 mph standard deviation error, I claim that > the cops did not get me beyound resonable doubt. > > Z = -10/5 = -2 > > P( Z < -2) = 0.03 > > 3% chance that i didnt speed and the radar was at fault. My arguement to the > judge is that 3 in 100 cars will recieve a false speeding ticket because > the radar isnt accurate. And i am one of those 3. > > Even still, the chances that i went just anywhere from below ï5 mph above > the speed limit' > Z = -5/5 = -1 > > P(Z < -1) = 0.15% > > There is a 15% chance(huge chance) that the radar gave a wrong reading. > > Do u think the judge will buy the arguement? > = > Btw > The best way to embarrass them is to teach High school students statistics > and statiscal reasoning and embarrass them, by making them appear smarter > than the cops > > -suresh > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) above the speed-limit. My arguement is > as follows. > > Given that the radar has a +/-5 mph standard deviation error, I claim that > the cops did not get me beyound resonable doubt. > > Z = -10/5 = -2 > > P( Z < -2) = 0.03 > > 3% chance that i didnt speed and the radar was at fault. My arguement to the > judge is that 3 in 100 cars will recieve a false speeding ticket because > the radar isnt accurate. And i am one of those 3. > > Even still, the chances that i went just anywhere from below ï5 mph above > the speed limit' > Z = -5/5 = -1 > > P(Z < -1) = 0.15% > > There is a 15% chance(huge chance) that the radar gave a wrong reading. > > Do u think the judge will buy the arguement? = Refusing to use proper spelling for English words identi'es you as a smartass, and you de'nitely don't want any judge to think that you are a smartass. I think a German judge would ask you if you are serious, and if you say yes, then he would get an expert witness to examine the camera very carefully. If it turns out that the camera couldn't have been so wrong so that you de'nitely have been speeding (even one mile above the limit), then you will be convicted and the cost for the expert witness will be added to your bill. And it won't be cheap. That is just common sense, as it prevents people who have no respect for other people's lives to use lame excuses like yours. And what makes you think that the error of a radar camera would have normal distribution anyway? === Subject: Re: proper spelling for English words identi'es you as a >smartass, and you de'nitely don't want any judge to think that you are >a smartass. Posts like this identify you as a jerk. === Subject: Re: Got a speeding ticket and need to 'ght back > And what makes you > think that the error of a radar camera would have normal distribution > anyway? It's upto the court to prove it's not. Normal Distribution wins by popularity vote. SAT scores are normal, error in experiments are normal, grades in a class are normal,etc. If the cops really want to prove it's not normal, they have go get the guy who invented the radar or the radar gun, or release the schematic of the radar or obscure patented techology. I am going to get a lot of high school students to use this argument. It's so simple that even a ninth grader can understand and argue against the prosecutor. And it's so embarrassing for the police if they do not understand what ninth graders understand. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back >> And what makes you >> think that the error of a radar camera would have normal distribution >> anyway? >It's upto the court to prove it's not. Normal Distribution wins by >popularity vote. SAT scores are normal, error in experiments are normal, >grades in a class are normal,etc. If the cops really want to prove it's not >normal, they have go get the guy who invented the radar or the radar gun, or >release the schematic of the radar or obscure patented techology. > Wrong. If the distribution is bounded, it can't be normal, full stop. All those normal-looking distributions divert usually strongly from normal as soon as one is out of the +/- 1.5 standard deviation interval. === Subject: Re: Got a speeding ticket and need to 'ght back > > Wrong. If the distribution is bounded, it can't be normal, full stop. > All those normal-looking distributions divert usually strongly from > normal as soon as one is out of the +/- 1.5 standard deviation interval. Earth to whoever you are. Welcome to the real world. Translation: almost all real world error is never bounded. > > === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) can't be normal, full stop. >> All those normal-looking distributions divert usually strongly from >> normal as soon as one is out of the +/- 1.5 standard deviation interval. > >Earth to whoever you are. Welcome to the real world. Translation: almost all >real world error is never bounded. Right. I'm often clocked going at negative speeds. (Also occasionally found to be exceeding the speed of light.) dave and that's on a bicyle! === Subject: Re: Got a speeding ticket and need to 'ght back rusin@vesuvius.math.niu.edu (Dave Rusin) bounded, it can't be normal, full stop. > >> All those normal-looking distributions divert usually strongly from > >> normal as soon as one is out of the +/- 1.5 standard deviation interval. > > > >Earth to whoever you are. Welcome to the real world. Translation: almost all > >real world error is never bounded. There's only one error that's unbounded in this thread. Make that two - my failure to kill'le you sooner. > Right. I'm often clocked going at negative speeds. (Also occasionally > found to be exceeding the speed of light.) > > dave and that's on a bicyle! I must have been going close to the speed of light a few times, as I certainly have been, if brie§y, shorter than expected. But maybe that was instead a function of hitting something? Phil -- Unpatched IE vulnerability: NavigateAndFind protocol history Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-Content.HTM Exploit: http://safecenter.net/liudieyu/NAFjpuInHistory/ NAFjpuInHistory-MyPage.HTM === Subject: Re: Got a speeding ticket and need to 'ght back > rusin@vesuvius.math.niu.edu speed of light a few times, as > I certainly have been, if brie§y, shorter than expected. > But maybe that was instead a function of hitting something? Ever heard of tunneling? If you werent that big, you would have tunneled to the other side of the galaxy. GED/QED > There's only one error that's unbounded in this thread. > Make that two - my failure to kill'le you sooner. Earth to dirt bag. We just received your transmission and we dont think, you are that important to care about your kill'le index. -suresh === Subject: Re: Got a speeding ticket and need to 'ght back Dave Rusin > can't be normal, full stop. > All those normal-looking distributions divert usually strongly from > normal as soon as one is out of the +/- 1.5 standard deviation interval. >> >>Earth to whoever you are. Welcome to the real world. Translation: almost all >>real world error is never bounded. > >Right. I'm often clocked going at negative speeds. (Also occasionally >found to be exceeding the speed of light.) > >dave and that's on a bicyle! I once bought a box of Captain Crunch with 20 pounds of cereal inside. -- Don't try to teach a pig how to sing. You'll waste your time and annoy the pig. === Subject: Re: Got a speeding ticket and need to 'ght back > I once bought a box of Captain Crunch with 20 pounds of cereal > inside. > That explains the box I picked up one day... When I took it home, not only was it empty, but also the half-full box that I had left sitting on top of the refrigerator was also now empty. I guess my box had -0.5 pounds of cereal in it. Seeing as error is unbounded, now things make sense. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Re: Got a speeding ticket and need to 'ght back The Lord of Chaos (Suresh Devanathan ) radar camera would have normal distribution > > anyway? > It's upto the court to prove it's not. Normal Distribution wins by > popularity vote. SAT scores are normal, error in experiments are normal, > grades in a class are normal,etc. If the cops really want to prove it's not > normal, they have go get the guy who invented the radar or the radar gun, or > release the schematic of the radar or obscure patented techology. > > I am going to get a lot of high school students to use this argument. It's > so simple that even a ninth grader can understand and argue against the > prosecutor. And it's so embarrassing for the police if they do not > understand what ninth graders understand. As I said, in Germany people have tried to use lame excuses before, and they come unstuck. Your assumption that a policeman wouldn't understand something that a ninth grader understands is completely unwarranted and irrelevant. Your assumption that a judge wouldn't understand the argument is also completely unwarranted and irrelevant. Your problem is that there will be expert witnesses who will testify that your argument is wrong, and you will pay. By the way, here is an argument that even an eighth grader will understand that shows how stupid you are: If the speed measurement of a radar camera were off by ten miles an hour in one in 'fteen cases, then one in 'fteen people who pass a speeding camera while driving exactly at the speed limit would get a speeding ticket. Compare your assumptionn with reality, and reality wins. === Subject: Re: Got a speeding ticket and need to 'ght back > The Lord of Chaos (Suresh Devanathan ) > of a radar camera would have normal distribution > > > anyway? > > It's upto the court to prove it's not. Normal Distribution wins by > > popularity vote. SAT scores are normal, error in experiments are normal, > > grades in a class are normal,etc. If the cops really want to prove it's not > > normal, they have go get the guy who invented the radar or the radar gun, or > > release the schematic of the radar or obscure patented techology. > > > > I am going to get a lot of high school students to use this argument. It's > > so simple that even a ninth grader can understand and argue against the > > prosecutor. And it's so embarrassing for the police if they do not > > understand what ninth graders understand. > > As I said, in Germany people have tried to use lame excuses before, and > they come unstuck. Your assumption that a policeman wouldn't understand > something that a ninth grader understands is completely unwarranted and > irrelevant. Your assumption that a judge wouldn't understand the > argument is also completely unwarranted and irrelevant. Your problem is > that there will be expert witnesses who will testify that your argument > is wrong, and you will pay. on the other hand, if it is in the good old US of A, not only will both the police of'cer and the judge completely fail to understand your defence (which is valid,) but will also refuse it on that basis (their failure to understand it, that is...) > By the way, here is an argument that even an eighth grader will > understand that shows how stupid you are: If the speed measurement of a > radar camera were off by ten miles an hour in one in 'fteen cases, then > one in 'fteen people who pass a speeding camera while driving exactly > at the speed limit would get a speeding ticket. Compare your assumptionn > with reality, and reality wins. === Subject: Re: pi + e is irrational? > Ok, complete disregard my last two posts. Here is what I meant to say: > Are you sure that it is unknown if pi + e is unknown? Quite apart from all the valid objections everyone else is making, I will note this: > let pi + e = q taken seriously by anyone. > and since e is a strictly increasing fuction we have How can a function be strictly increasing over the complex numbers? There's no ordering relation among complex numbers. Besides, e^2*pi*i = e^0. eyelessgame === Subject: Re: Is it known whether pi + e is an unsolved problem archive. Why isn't this > > >>problem solved? I know it must be deeper than I think, so could > > >>someone explain why this simple proof doesn't work: > > >> > > >>Assume pi + e is irrational > > >>Then let pi + e = q, where q is a rational number. Let i = sqrt(-1). > > >>Then > > >>i*(pi + e) = i*q > > >>then > > >>e^(i*(pi + e)) = e^(i*q) > > >>then > > >>e^(i*pi)*e^(i*e) = e^(i*q) > > >>then > > >>-e^(i*e) = e^(i*q) > > >>then squaring both sides: > > >>e^(2*(i*e)) = e^(2*(i*q)) > > >>and since e is a strictly increasing fuction we have > > >>(2*i)*e = (2*i)*q > > >>meaning > > >>e = q. > > > > If this result would be right, it actually proved something more > > surprising: > > > > pi+e = q (from the assumption) > > e = q (from the argumentation) > > > > coming to... > > > > pi = 0 (as a new ISO-norm for the exhausted student.... ) > > from which we can easily deduce that _all_ numbers are zero. > Therefore I suggest that, instead of calling it a new ISO-norm , > we should call it the new is0-norm. ;-) > > David Ahh oh thank you. yea ehh it's like one of those 1=0 proofs except instead of dividing by 0 you ignore the fact that i*pi is imaginary, heheh. Speaking of which, there is another proof like that and I was wondering if anyone would be interested in addressing it (for amusement, if anything) i = e^(pi*i/2) = e^(pi*(e^(pi*i/2))/2) = .. = e^(pi*(e^(pi*e^(pi*(...)/2)/2))/2) If we index the 'rst expression 1 , the second expression 2 , and the n th expression n , then the expression you get as (lim n -> in'nite) is composed of nothing but e raised to real powers (since only the highest exponent is imaginary, but there is no highest exponent at in'nite). But wouldn't this imply that i is real? Also, if you look at it a different way, it is imaginary: i = e^(pi*i/2) = e^(pi*(e^(pi*i/2))/2) = .. = ....^(pi*e^(pi*(e^(pi*e^(pi*i/2)/2))/2)) If you look at it this way, there is a highest exponent, and it contains i . It would seem that these are two irreconcilably different ways of looking at the same expression? Or are both of these expressions poorly constructed and impossible to de'ne? Please explain in epsilon-delta terms if necessary. Thought this was interesting. Joey === Subject: Re: Study groups in science > please visit > site > http://de.geocities.com/scienceworkshops/ > Its goal is to organize study groups on scienti'c topics like quantum > 'eld theory, > probabilistic inference or neural nets, to name a few topics I am > personally interested in > (of course, arbitray topics may be suggested). > The idea is to study some text (which is freely available over the > internet, by setting up > a study plan for 12 weeks and open a discussion groups where > participants can post questions, > or solutions to exercises. > The text must be of academic level, ranging from introductory texts to > post-graduate level. > Participation is free in these groups. > I would like to start the 'rst workshop in fall, so if you are > interested you might either > subscribe to our mailing list or propose your own workshop. useful online books. I don't think you should limit yourself to online books, but should also include online papers such as those of Einstein and classics such as The Principle Of Relativity from Dover books and The Principle Of Quantum Mechanics by Dirac which people are prepared to buy. It's also important that there are a few academics that are prepared to kindly give their time for free. It'll end up the blind leading the blind otherwise. Are you an academic? Funky === : What positive precautions are you taking to prevent the idiots morons and > : kooks from taking it over, as has happened in sci.physics? > > Perhaps the textbook and the equations will keep people like you away? You evidently did not realise that the morons and kooks I referred to included you. Franz === Subject: Re: Study groups in science Franz and kooks I referred to : included you. Of course I understood your pretext. You post it all over the group. I was merely pointing out that your obsession with these categorisations without recourse to understanding or, indeed, textual analysis only serves to decrease the quality of posts in this newsgroup. It is antirationalistic and the antithesis of science. So, these posts of yours are amusingly self-referential of you. -- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Limit Of # Of Coprime Integers > Leroy, > > For the original n=2 case, the multiplier out front should be > 1/m^3 rather than 1/m^2, of course. > > But also the numerical value seems wrong; > for the limit I get > product ( (1-1/p) + (1/p)*(1-1/p)^2 ) = 0.428... > (the product being taken over all primes) > while for the expression with zeta(3) I get > 12/pi^2 - 1/zeta(3) = 0.383... > > The factor in the in'nite product simpli'es to > 1 - 2/p^2 + 1/p^3 > which is unfortunate. If it could be expressed as products of > factors of the form > 1 - p^k > then we could write the in'nite product in terms of > 1/zeta(k) > but it can't be so expressed, so I'm at a loss. > > For general n, the in'nite product would be instead > product ( (1-1/p) + (1/p)*(1-1/p)^n ) > > Don Coppersmith You are right, I was *WRONG*!... I discovered I was wrong this morning when I noted that n*6/pi^2 - sum{k=3 to n+1} 1/zeta(j) is negative for all high enough n's, ...obviously wrong! And I did not even notice myself that I had an incorrect 1/m^2 , where I should have had a 1/m^3 . But, in any case, I did 'nally get the n=2 case: limit{m-> oo} (1/m^3) *sum{k=1 to m} sum{j=1 to m} H(m;k,j) = (6/pi^2) *product{p=primes} (1 -1/(p(p+1))), which is what you got. (I seem to remember seeing somewhere that the product in my representation is equal to some famous constant,...or perhaps that was with (p(p-1)) in the denominator of the fraction instead...) As for n > 2, I have not 'gured that out, nor will I investigate this positive integers <= m which are coprime > >to *both* k and j. > > > >(So, for instance, H(m;k,j) = H(m;kj,1).) > > > >(And H(kj;j,k) = phi(jk), the Euler phi function.) > > > > > >I believe that > > > >limit{m-> oo} > > > > m m > > --- --- > > 1 --- > > H(m;k,j) > >m^2 / / > > --- --- > > k=1 j=1 > > > > > > 12 1 > >= ---- - ------- , > > pi^2 zeta(3) > > > > > >which is, in linear-mode, > > > > > >limit{m-> oo} > > > >(1/m^2) *sum{k=1 to m} sum{j=1 to m} H(m;k,j) > > > > > >= 12/pi^2 - 1/zeta(3). > > > >(zeta(r) is, generally, sum{j=1 to oo} 1/k^r.) > > > >And, more generally, > > > > > >limit{m-> oo} > > > > m m m > > --- --- --- > > 1 ------- > > ... > H(m;k_1,k_2,..,k_n) > >m^(n+1) / / / > > --- --- --- > > k_1=1 k_2=1 k_n=1 > > > > n+1 > > --- > > 6*n 1 > >= ---- - / ------- , > > pi^2 --- zeta(j) > > j=3 > > > > > >which is, in linear-mode: > > > >limit{m-> oo} > > > >(1/m^(1+n))* > > sum{k_1=1 to m} sum{k_2=1 to m}...sum{k_n=1 to m} H(m;k_1,k_2,..,k_n) > > > > > >= 6*n/pi^2 - sum{j=3 to n+1} 1/zeta(j) . > > > > > >And H(m;k_1,k_2,..,k_n) = > > > >the number of integers, j, coprime with (k_1)*(k_2)*(k_3)*...*(k_n) > >and such that 1 <= j <= m. > > > > > > > > > >Am I right? At least, is my n=2 case right? > > === Subject: Re: Limit Of # Of Coprime Integers >.... > You are right, I was *WRONG*!... > > I discovered I was wrong this morning when I noted that > > n*6/pi^2 - sum{k=3 to n+1} 1/zeta(j) > > is negative for all high enough n's, > > ...obviously wrong! > > And I did not even notice myself that I had an incorrect > 1/m^2 , where I should have had a 1/m^3 . > > But, in any case, I did 'nally get the n=2 case: > > limit{m-> oo} > (1/m^3) *sum{k=1 to m} sum{j=1 to m} H(m;k,j) > > = (6/pi^2) *product{p=primes} (1 -1/(p(p+1))), > > which is what you got. > > (I seem to remember seeing somewhere that the product in my > representation is equal to some famous constant,...or perhaps that > was with (p(p-1)) in the denominator of the fraction instead...) With a (p(p-1)), the product would have been Artin's constant: http://www.worldwideschool.org/library/books/sci/math/ MiscellaneousMathemati calConstants/chap8.html Leroy > > As for n > 2, I have not 'gured that out, nor will I investigate > >... === Subject: the meaning of Jacobian? Can someone please explain to me what a Jacobian is about? I know how to use one, but I don't know its signi'cance. Someone may answer: A zero jacobian makes a singularity in a manifold because some fractions have a zero denominator. That's not my point. I want to know what the jacobian is in the 'rst place. For example, the schwarzian derivative measures the difference , if you will, between a given function and a mobius transformation. It's that kind of info that I'm looking for. Please respond to responses! Ted Shoemaker === Subject: Re: the meaning of someone please explain to me what a Jacobian is about? I know how > to use one, but I don't know its signi'cance. > > Someone may answer: A zero jacobian makes a singularity in a manifold > because some fractions have a zero denominator. That's not my point. > I want to know what the jacobian is in the 'rst place. > > For example, the schwarzian derivative measures the difference , if > you will, between a given function and a mobius transformation. It's > that kind of info that I'm looking for. > > Please respond to the newsgroup and > Ted Shoemaker Take a tiny box B around a point (u,v,w) in the domain; it will transform into some set F(B); the absolute value of the Jacobian is the limit of the ratio volume(F(B)) / volume(B) as the box shrinks to zero diameter. (Isn't this obvious from the change-of-variable formula for multiple integrals?) The sign of the Jacobian indicates is the transformation was right-handed (+) or left-handed (-). Now you can see that if the Jacobian is zero then F(B) collapses in'nitely faster than B (volumewise), indicating the presence === Subject: Re: the meaning of Jacobian? > > Isn't this obvious Now that you say === Subject: Re: Resolution out there's another approach to prove a problem with the old > > concepts about the ring of algebraic integers. > > > > Decker put forward the quadratic > > > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > > > where his a's are roots of > > > > a^2 - (x - 1)a + 7(x^2 + x). > > > > Letting x=2, you have > > > > a_1(2)^2 - a_1(2) + 42 = 0, which gives > > > > a_1(2) = (1 + sqrt(-167))/2 as one of two solutions. > > > > Now consider the quadratic > > > > y^2 - by - 7 = 0, which has as one of its roots > > > > (b + sqrt(b^2 + 28))/2. > > > > Note that root is an algebraic integer factor of 7 for all algebraic > > integers y, and b. > > > > Now consider > > > > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 > > > > which is > > > > (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z > > > > > > > and I can divide both sides by 4 to 'nally get > > > > 49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0. > > > > Not quite. Not that it affects your argument, but you actually get > > 7z^4 + bz^3 +(6b^2 + 83)z^2 - 6bz + 252 = 0 > Yeah Jones already noted that mistake in his reply in this thread. > > Importantly, for any integer b, such that it is irreducible over Q, > > *none* of the solutions for z can be an algebraic integer! > > > > Okay. So for the b's that make the polynomial above irreducible, > you've shown that > > (b + sqrt(b^2 + 28))/2 > > doesn't divide > > (1 + sqrt(-167))/2 > > in the algebraic integers. Not much of a surprise, in spite of > the fact that the former divides 7 and the latter divides 42. It is a surprise as I can let b be any algebraic integer, and thus check for every possible algebraic integer factor of 7. Doing so reveals that for algebraic integer b (b + sqrt(b^2 + 28))/2 is never a factor of (1+sqrt(-167))/2 in the ring of algebraic integers. It's a fascinatingly simple way to end the arguing. > > > Now then, imagine that there exists some algebraic integer b for which > > it is reducible over Q, then the root will be a fraction with a 7 or > > 49 in the denominator. > > > > > > > So consider the root c/7, which gives > > > > (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7, > > > > which would force (1 + sqrt(-167)) to be coprime to 7. > > > > Why would that be? The problem for your position is that 7 is prime, so if the polynomial with integer coef'cients de'ning z had a rational solution i.e. was reducible over Q, it'd have to have at least one fraction as a root with a denominator of 7. However (b + sqrt(b^2 + 28))/2 is itself a factor of 7, as (b + sqrt(b^2 + 28))/2[(b - sqrt(b^2 + 28))/2] = -7, so for (b + sqrt(b^2 + 28))c/7 would have to be coprime to 7, which would force (1 + sqrt(-167)) to be coprime to 7. So no matter how you look at it, (1 + sqrt(-167))/2 can't have a non-unit factor in common with 7 in the ring of algebraic integers. > > > Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the > > ring of algebraic integers!!! > > > But that's simply not true, as Dik reminded us in an earlier response. Such a simple denial is not the mark of a highly intelligent person. You need to face the mathematics, as I've done when I've been wrong. After all, isn't it the truth which is important? > > > > > > > Rick Decker, and think carefully before you post again, as not only are you on the line here, but so is Hamilton College. You misstep here, and you may let down an awful lot of people for no good reason. Fighting mathematics just isn't the way to go. It's just not logical. James Harris === Subject: Re: Resolution it, (1 + sqrt(-167))/2 can't have a > non-unit factor in common with 7 in the ring of algebraic integers. > > >> >> >Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the >ring of algebraic integers!!! > >> >> But that's simply not true, as Dik reminded us in an earlier response. > > > Such a simple denial is not the mark of a highly intelligent person. > > You need to face the mathematics, as I've done when I've been wrong. > > After all, isn't it the >>Rick > > > You need to look carefully at the argument Rick Decker, and think > carefully before you post again, as not only are you on the line here, > but so is Hamilton College. > > You misstep here, and you may let down an awful lot of people for no > good reason. > > Fighting mathematics just isn't the way to go. It's just not logical. > > > James Harris Still with the implied threats? Why not grow a pair, and get on with making good on the threats you are continually implying? Are you *that* much of a girly-boy? I thought as much. You have charged Rick Decker with fraud. You have suggested that you may instigate civil proceedings. You have stated that there will be unpleasant consequences from the actions that people on this newsgroup are taking, speci'cally in their denial of your ascendancy to the Throne of Mathematics. Yet you do nothing other than to prance around wearing that silly out't. What out't, you ask? Why, it's the poor me, no one treats me fairly, I'm the misunderstood genius, but only don't ever ask me to prove anything because I can't be bothered, plus I couldn't write a proof if my life depended on it satin baby-doll PJs, complete with pink fuzzy bunny-rabbit slippers. You may as well wear a rhinestone tiara and ballerina slippers. Problem is, you don't think anyone else can see through the bull. Your out't is as plain as the nose on your face, to coin a phrase, and everyone who knows a lick of mathematics can see it a mile away. Make threats, or don't. I don't care. Do you know what they call a person who makes threats and doesn't follow through? A coward. A bully. A sissy. A man doesn't threaten. That's the job of a person who doesn't have the courage to be a man. Here's the fact: you're wrong. You've always been wrong, and until you actually *learn* some mathematics, you'll *continue* to be wrong. The arguments that prove you wrong are beyond your grasp, for the simple reason that the *ONLY* mode of argument you comprehend is the elementary algebraic manipulation. You are outgunned and outclassed in every meaningful sense of the word, and your continuing attempts at using high-school algebra to confound the Evil Mathematics Cabal are as pathetic as they are amusing. If you spent half the time and effort as you're currently doing, but applied it towards *learning* some algebra, you would stand a much better chance at doing something worthwhile. Instead, your romantic notion of being the tortured genius is only serving to make you the complete model of the Usenet crackpot. Perhaps you should take Jim Ferry's recent note to heart. Dale. === Subject: Re: integer, and thus > check for every possible algebraic integer factor of 7. No, you can't 'nd them all that way. If you form a quadratic equation, such as x^2 + bx +7 = 0 and let ïb' change, you will *not* 'nd all possible algebraic integer factors of 7. You still have to deal with cubic equations, such as x^3 + px^2 + qx + 7 = 0 which will reveal other algebraic integer factors of 7 which did not appear in the quadratic. [snip typical triumphant leap to false conclusion by JSH] -- 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: Re: JSH: Open letter to Jim Ferry > > I'm intrigued by the questions raised by your recent posts, as for > > years you were this guy who came up with rather creative ways to > > insult me, and now I 'nd it hard not to 'gure you're just doing so > > again. > > Is it okay to ask you questions, talk in any familiar way, or in any way > act as if I wish you to reply or am addressing you? > > Those proscriptions apply only to David Ullrich, Virgil, and anyone who > posts under a palindromic pseudonym, right? > > Okay then. You ask a good question. Just what is my intent? I'm not > sure precisely why I'm doing what I'm doing, but I'll try to answer you > earnestly and respectfully. > > > Now I'll embarrass you a bit as from what I've read on the web you > > have one of the highest IQ's out there, so it seems to me there's > > probably some reason to what you're doing, and possibly I'm wrong > > about what it is. > > This does embarrass me because I'm certainly not one of the big 'sh on > sci.math. You must be basing this statement on the fact that I once > joined something called Mega, which purports to be a high-IQ society for > those of 1-in-a-million intelligence. I now realize that by joining, I > was implicitly making this arrogant claim about myself, but I reject that > claim. The fact that I was able to ace the math part of their test just > indicates that it wasn't hard enough, because it's easy to 'nd people > better at math than I. Indeed, you can 'nd lots of them on sci.math. > And some of them even take the time to analyze your work, James. You're not one of the big 'sh on sci.math, and I'm less curious about your past experiences with high IQ societies than you probably think. However, I had a theory, and testing it involved mentioning that facet of your public persona. > > Therefore, I'm going to give you the opportunity that I feel *I* don't > > get, which is the bene't of the doubt. > > You're asking me to clarify me position, which I am about to do. I think > it's fair to say, however, that others have given you the same opportunity, > i.e., that they've asked you to clarify your position. > > > Tell me succinctly and in a way that will minimize potential > > embarrassment for both of us, what it is that your up to, and no, none > > of this wild stuff about how great I supposedly am, or how I've proven > > FLT or any of that, as I just want you to say something that 'ts into > > a worldview that makes sense. > > > > What's your intent? > > You think that I'm mocking you, but you're not entirely sure. David > Ullrich thinks it's incredible that it's not obvious to you that I'm > mocking you. > > Well let me clear things up. Yes, I'm mocking you. I've made a series > of posts over the last six months in which I've appeared to be converted > to a religion in which you are the Messiah of Mathematical Truth being > cruci'ed my the benighted masses. Most people consider that absurd and > therefore conclude that I must have been being sarcastic. You, however, > do not consider the idea that you are the Messiah of Mathematical Truth > to be absurd. You consider it to be essentially correct -- a little off > somehow: a little over the top, or emotionally overblown, but basically > the correct attitude to take. > So you were lying. I just needed to make sure. The problem is that I'd concluded that very intelligent people see a much higher value in telling the truth than others. That lead me to consider the possibility that you were in fact sincere, but deluded and confused, possibly dealing with a lot of emotional pain from a dif'cult position--considering that I might be right--against tremendous social forces. But you weren't being brave. You were just being a smart-ass. James Harris === Subject: Re: JSH: Open letter to Jim Ferry > That lead me to consider the possibility that you were in fact > sincere, but deluded and confused, possibly dealing with a lot of > emotional pain from a dif'cult position--considering that I might be > right--against tremendous social forces. I was promoting the position that you are what you claim to be -- a great mathematical revolutionary misunderstood by the ignorant, corrupt mathematical establishment. So I suppose it was a natural reaction for you to read this and think, Wow! That guy's off his rocker! What an ill-founded, absurd, untenable position *that* is! Evidently I must have appeared to be suffering from TNPD: Transferred Narcissistic Personality Disorder. This is simply a belief system about someone else isomorphic to the belief system that person would have about him- or herself if he or she suffered from NPD. Astute of you to recognize this. Have you considered the possibility, however, that I really do suffer from this disorder, but am just *claiming* to be lying because I'd rather be thought a liar than a crazy person? As David Ullrich pointed out, what I've done can't even be called a lie, not even in a court of law. Thus, by claiming to be lying, I get off scot free. But if I really admitted to adhering to a belief system which is, as you point out, delusional (i.e., TNPD), then the rami'cations would be much more serious. I'd have to think about going to a mental health professional. In fact, the nagging doubt that perhaps the world was right would spur me to go because then I could settle the issue, like in the episode of The Simpsons where Homer's hand is stamped Not Insane . But the nagging doubt itself, and the fear it engendered would probably win out, so I wouldn't go. Much easier to claim to have been lying. And in this case, it's the truth, although I know that my claim is now suspect because I've admitted to lying (although, again, it was just sarcasm, and not lying) and, more importantly, because it is so outrageously convenient for me to make that claim rather than its alternative. -Jim Ferry === Subject: Re: JSH: Open letter were lying. I just needed to make sure. It wasn't a lie; it was a parody. As usual, you were the only one here who couldn't see the obvious truth. We've all been having a good laugh at your expense this whole time. Congratulations on 'nally catching up with the rest of the world (in this one small area), even if you *did* have to have it spelled out for you before you could see it. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@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: Open letter to Jim Ferry >> [...] >> >> You think that I'm mocking you, but you're not entirely sure. David >> Ullrich thinks it's incredible that it's not obvious to you that I'm >> mocking you. >> >> Well let me clear things up. Yes, I'm mocking you. I've made a series >> of posts over the last six months in which I've appeared to be converted >> to a religion in which you are the Messiah of Mathematical Truth being >> cruci'ed my the benighted masses. Most people consider that absurd and >> therefore conclude that I must have been being sarcastic. You, however, >> do not consider the idea that you are the Messiah of Mathematical Truth >> to be absurd. You consider it to be essentially correct -- a little off >> somehow: a little over the top, or emotionally overblown, but basically >> the correct attitude to take. >> > >So you were lying. I just needed to make sure. Lying is not really the right word, because it was _so_ obvious that he was being facetious. Except to you of course - he's exactly right when he says that the fact that it wasn't obvious to you that he was mocking you indicates some truly _awesome_ megalomania. The idea that sarcasm is not lying is even recognized by the law - people have been sued for libel over obvious parodies, and the court decided that no it wasn't libel _because_ it was obviously a parody. (Like if I said that George W. Bush was a draft dodger I _might_ get in trouble for that if I couldn't prove it was so, but if I claim that he's being paid by Martians to destroy the economy to make it easier for them to take over when they arrive nobody's going to call that a lie .) >The problem is that I'd concluded that very intelligent people see a >much higher value in telling the truth than others. > >That lead me to consider the possibility that you were in fact >sincere, but deluded and confused, possibly dealing with a lot of >emotional pain from a dif'cult position--considering that I might be >right--against tremendous social forces. > >But you weren't being brave. You were just being a smart-ass. > > >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Open letter to Jim Ferry > > >> [...] > > (Like if I said that George W. Bush was a draft dodger I > _might_ get in trouble for that if I couldn't prove it was so, Uh-oh. I think someone's going to dash off an e-mail to the White House. > but if I claim that he's being paid by Martians to destroy > the economy to make it easier for them to take over when > they arrive nobody's going to call that a lie .) > > > > ************************ > > David C. Ullrich === Subject: Re: JSH: Open letter to Jim Ferry