mm-383 === Subject: : Re: P vs NP and the analog machine> err> x - nmu = +/- sqrt(2n) sigma sqrt(N) sqrt( log N -1)> x = n mu +/- n sqrt(2) sigma sqrt( log N -1) x/(n mu) -> 1 +/- frac{sqrt(2) sigma sqrt( log N - 1)}{mu}> I will leave the math here.The Lord of Chaos (Suresh Devanathan)> One more thing although the % deviation = (standard deviation)/(average) goes to 0, the> deviation between the path with the least cost and the average cost> doesnt. It can be seen from the following analysis. The notation is sloppy Let P(a) be probability of getting a value anything under a, given that a> is> normally distributed. P(a) = 1/N! P ( (x - n mu)/( sqrt {2 n} sigma) ) = 1/N! For asymptocially small values P(a) would be about exp(-a^2)/ a exp(-a^2)/ a = 1/N! -a^2 = -log(N!)> a^2 = N (log N - 1)> a = sqrt(N) sqrt( log N - 1) x - nmu = sqrt(2n) sqrt(N) sqrt( log N -1)> x = n mu + n sqrt(2) sqrt( log N -1) x/(n mu) -> 1 + frac{sqrt(2) sqrt( log N - 1)}{mu} It does us no good to find just the average one.> The Lord of Chaos (Suresh Devanathan)> message> This is one thing i can prove.> Take an element a_ij. Let x = a_ij Assume that x has the pdf p(x)> Now, the total cost along certain tour can written as> cost = sum a_mn> Now, each a_mn also has the distribution p(x), by definition of p(x) .> For a tour of length n,> The distribution of costs would have the mean (n) mu and the standard> deviation sqrt(n) sigma> % deviation from the mean is frac{sqrt(n) sigma}{ n mu} => frac{1}{sqrt{n}} frac{sigma}{mu}> As n->infy, % deviation goes to 0. In a way, all paths relatively> have> about the same n mu> Of course, the proof makes implicit assumption that p(x) is not changing> with n.> Either, i have gone completely crazy, or have discovered something> completely interesting. It looks so deviously simple.> -suresh> As n->infy, % deviation goes to 0. In a way, all paths relatively> have> about the same n muThis works for large N. However, let me back up a few steps. You areassuming N! for all possible routes between some i and j (or k). Thisassumption may bewrong. To compute the possible paths of any length binomially weshould useN!/((N-2)!) since we require just 2 vertices for each path. Goingfurther still, and assuming random distribution of path choiceprobabilities, we could multiply our binomial by p^n * (p-1)^n-2 wherep should be 1/N.I think your math, above, for 'normalization' was accurate, but let merephrase it for clarity (if not redundancy). normalized = (random variable - finite mean) / standard deviationwhere standard deviation can be calculated by: sd = sqrt[ (1/n)*((n0^2)+(n1^2)+(n2^2...) - (averageexpectation)^2 ]In general, we could apply transitive closure (~composition) to findthe longest path lengths by assuming a fully converged (or total mesh)and removing increasing path lengths by squaring over all therelations (R^n). Where n=sqrt(nodes) should be a weak upper bound onthe total of squarings needed. In this way, the step which loses themost paths is found to be contain the average path length.=== === Subject: : Re: P vs NP and the analog machineto it, i say blah.I am not trying to insult you or anything. I am not a master incombinatorics. The terms you are using, i have bare conception of.References to definitions would help-suresh> The Lord of Chaos (Suresh Devanathan)> err> x - nmu = +/- sqrt(2n) sigma sqrt(N) sqrt( log N -1)> x = n mu +/- n sqrt(2) sigma sqrt( log N -1) x/(n mu) -> 1 +/- frac{sqrt(2) sigma sqrt( log N - 1)}{mu}> I will leave the math here. The Lord of Chaos (Suresh Devanathan)message> One more thing> although the % deviation = (standard deviation)/(average) goes to 0,the> deviation between the path with the least cost and the average cost> doesnt.> It can be seen from the following analysis. The notation is sloppy> Let P(a) be probability of getting a value anything under a, giventhat a> is> normally distributed. P(a) = 1/N!> P ( (x - n mu)/( sqrt {2 n} sigma) ) = 1/N!> For asymptocially small values P(a) would be about exp(-a^2)/ a> exp(-a^2)/ a = 1/N!> -a^2 = -log(N!)> a^2 = N (log N - 1)> a = sqrt(N) sqrt( log N - 1)> x - nmu = sqrt(2n) sqrt(N) sqrt( log N -1)> x = n mu + n sqrt(2) sqrt( log N -1)> x/(n mu) -> 1 + frac{sqrt(2) sqrt( log N - 1)}{mu}> It does us no good to find just the average one.> The Lord of Chaos (Suresh Devanathan)> message> This is one thing i can prove.> Take an element a_ij. Let x = a_ij Assume that x has the pdf p(x)> Now, the total cost along certain tour can written as> cost = sum a_mn> Now, each a_mn also has the distribution p(x), by definition ofp(x) .> For a tour of length n,> The distribution of costs would have the mean (n) mu and thestandard> deviation sqrt(n) sigma> % deviation from the mean is frac{sqrt(n) sigma}{ n mu} => frac{1}{sqrt{n}} frac{sigma}{mu}> As n->infy, % deviation goes to 0. In a way, all paths relatively> have> about the same n mu> Of course, the proof makes implicit assumption that p(x) is notchanging> with n.> Either, i have gone completely crazy, or have discovered something> completely interesting. It looks so deviously simple.> -suresh> As n->infy, % deviation goes to 0. In a way, all paths relatively> have> about the same n mu> This works for large N. However, let me back up a few steps. You are> assuming N! for all possible routes between some i and j (or k). This> assumption may be> wrong. To compute the possible paths of any length binomially we> should use> N!/((N-2)!) since we require just 2 vertices for each path. Going> further still, and assuming random distribution of path choice> probabilities, we could multiply our binomial by p^n * (p-1)^n-2 where> p should be 1/N.> I think your math, above, for 'normalization' was accurate, but let me> rephrase it for clarity (if not redundancy).> normalized = (random variable - finite mean) / standard deviation> where standard deviation can be calculated by:> sd = sqrt[ (1/n)*((n0^2)+(n1^2)+(n2^2...) - (average> expectation)^2 ]> In general, we could apply transitive closure (~composition) to find> the longest path lengths by assuming a fully converged (or total mesh)> and removing increasing path lengths by squaring over all the> relations (R^n). Where n=sqrt(nodes) should be a weak upper bound on> the total of squarings needed. In this way, the step which loses the> most paths is found to be contain the average path length.=== === Subject: : Re: Public Notice: Copyright Violations>> I'm not a copyright lawyer, but I know one. Here's some info you>> mgiht find enlightening...>> ...[F]air use is a highly fact-specific determination. Copyright>> Office document FL102 puts it this way: The distinction between>> 'fair use' and infringement may be unclear and not easily defined. >> There is no specific number of words, lines, or notes that may safely>> be taken without permission. Acknowledging the source of the>> copyrighted material does not substitute for obtaining permission.>> The document then quotes from the 1961 Report of the Register of>> Copyrights on the General Revision of the U.S. Copyright Law.,>> providing the following examples of activities that courts have held>> to be fair use:>> - Quotation of excerpts in a review or criticism for purposes of>> illustration or comment;>> - Quotation of short passages in a scholarly or technical work>> for illustration or clarification of the author's observations;>> - Use in a parody of some of the content of the work parodied;>> news report;>> Document FL102 is included in Copyright Office information kit 102>> (Fair Use), which can be ordered from the Copyright Office.>> Proponents of the implied license idea point out that Usenet postings>> are routinely copied and quoted, and anyone posting to Usenet is>> granting an implied license for others to similarly copy or quote>> that posting, too. It's not clear whether such implied license>> extends beyond Usenet, or indeed, what Usenet really means (does it>> include, for example, Internet mailing lists? Does it include>> netnews on CD- ROM?). If a posting includes an express limitation on>> the right to copy or quote, it's not at all certain whether the>> express limitation or the implied license will control. No doubt it>> depends on the specific facts. For example, was the limitation>> clearly visible to the person who did the copying? Was the>> limitation placed such that it would be visible only after the person>> who did the copying invested time and money to get the posting,>> believing it to be without any limitation? >> These theories are largely speculative, because there has been little>> litigation to test them in the courts. As a practical matter, most>> postings, with a small number of notable exceptions, are not>> registered with the Copyright Office. As such, to prevail in court,>> the copyright holder would need to show actual damages (see section>> 2.5). Since most of these cases will result in little or no actual>> damage, no cases have been be brought; it's simply too expensive to>> sue for negligible damages.>> Just some food for thought.>>> permission of the author.Why not?> Hey, that's cheating. -- === === Subject: : Re: Public Notice: Copyright ViolationsUh, you do know that quoting passages is fair use, presumably. And> that there are thousands of news servers copying your post without> expressed written permission?> Nooooooooooooooooooo! -- === === Subject: : Re: Public Notice: Copyright Violations Discussion, linux)> How can you tell whether something is going to be emvarassing> years later? Well in _your_ case it's very simple: don't make any> posts on mathematical topics!This is hardly sufficient.Consider his recent foray into international copyright law.-- A set having three members is a single thing wholly constituted byits members but distinct from them. After this, the theologicaldoctrine of the Trinity as 'three in one' should be child's play. --Max Black, _Caveats and Critiques_=== === Subject: : Re: Public Notice: Copyright Violations Discussion, linux)> Google has a _web page_ where one can find _complete_ quotes of> _all_ your posts.Holy cow! I just checked! Google *does* have pages and pages devotedto old and discarded arguments of James going back *years*. They*maintain* these pages, even though James has moved on.Man, are they in for a thrashing.-- This page contains information of a type (text/html) that can only be viewed with the appropriate Plug-in. Click OK to download Plugin. --- Netscape 4.7 error message=== === Subject: : Re: Squares Of Certain Areas: puzzleAnswer below:> I must admit that my formula for part-2 is simply a partial sum,> although an interesting partial sum, interesting since it is an> unexpected alternate representation of the straight-forward> partial-sum which gives the area as the sums of squares' areas (taken> over the k's from part-1).> If anyone can find a more simple representation of part-2's answer,> this would be interesting.But I will wait a couple days to give answer to this puzzle.Leroy QuetYou have a FIXED sequence of distinct positive integers, {k}, which is> to be determined.For all k's <= EACH positive integer m, we can take a number n (n <= m) of (geometric) squares,> each of *integer* side-length,where the j_th square is of the largest square> with an integer area that is <= m/(k_j).> But {k} is such that, if we lined the squares up vertically so that> their horizontally-running edges coincide and they do not overlap,> the length from the upper side of the top square to the lower side of> the bottom square will ALWAYS be m.2 questions: 1) Which integers make up {k_j}?2) Give a formula for the total area of the (n number of) squares.Example:> For m = 4,We have 1, 2, 3 in {k}, but not 4.So, we have the squares, as lined up: ------> ! ! !> ------ k=1, area = 4 <= 4/1> ! ! !> ==----> ! ! k=2, area = 1 <= 4/2> ==> ! ! k=3, area = 1 <= 4/3> --Height of the stack is m = 4. Area is 6.Leroy QuetMy solution: ||V||V||V||V||Vpart 1:{k} is the sequence of squarefree integers (including 1).In other words, m--- | ----- |/ | / m/ |--- |_V / k _|k=1k=squarefreealways equals m.(In linear-mode:sum{1<=k<=m,k=squarefree} floor(sqrt(m/k))= m, for all positive integers m.)Part 2:Straightforwardly, we have:area = m--- 2 | ----- |/ | / m/ |--- |_V / k _|k=1k=squarefree= sum{1<=k<=m,k=squarefree} floor(sqrt(m/k))^2,which also is:-m + 2 *sum{k=1 to m} a(k),where a(k) = squareroot of largest square number dividing k.So, even though this is not a closed-form in the strictest sense, S(m) - S(m-1) =-1 +2*a(m), is,where S(m) = the sum of the squares' areas for m. Leroy Quet=== === Subject: : Implied Recursion, Harmonic Numbers: puzzleLet {a(m,k)} be a sequence such that, for all positive integers n: n--- / n +2m | | / | k | H(m,n+1-k) a(m,k)--- /k=0= H(m,n+2) (n+1)/(n+1+2m),where the sum is, in linear-mode,sum{k=0 to n} binomial(n+2m, k) H(m,n+1-k) a(m,k).H(m,n) is a generalization of the harmonic numbers,H(0,n) = 1/n,H(m,n) =sum{k=1 to n} H(m-1,k),for all positive integers m and n.Puzzle:So, given the sequence of a's, what is, in terms of H(m,k)'s:--- / n +2m | | / | k2 | H(m,n+1-k2) a(m,k2),--- /k=0k<=n/2which is, in linear-mode,sum{0<=k<=n/2} binomial(n+2m, k2) H(m,n+1-k2) a(m,k2)???Leroy Quet=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?>I, gave a reference:>Complex Notation, chapter 12 ,Electrical Technology 3RD ED.>Edward Hughes Longmans ,page 338:>States:> OA*=OB+jOC=OA(COStheta+jSINtheta)>* Symbols representing phasors are printed in bold face italics, while those representing only magnitudes are printed in ordinary italics,..> What are these magnitudes then ?>>OB (which can obviously be either positive, zero or negative) is the real >>part of the complex number. OC (which can obviously be either positive, >>zero or negative) is the imaginary part of the complex number. OA (which >>is presumably positive or zero) is the magnitude of the complex number.>>theta is the argument of the complex number.>>Note that neither OB nor OC is the magnitude of the complex number.I would have thought that since you asked about OB and OC, you might have actually been interested in an answer to that question. Or has being told that ::Re(e^[iPi]) = Re(-1+i[0]) = -1 AND::Im(e^[iPi]) = Im(-1+i[0]) = 0made you lose interest in all other questions that you asked? Have youlost all interest in complex numbers as a consequence of finding out that::I,have a reason , with my due respect.::Panagiotis Stefanideswhen I saId that you had no reason to conclude that exp(i pi) = 0, and::I, have made myself very explicit.::My original question of the implication::of the imaginary component: e^[ipi]=j*0 (to the related proof)where you treated e^[i pi] as the imaginary part of itself?David McAnally -------=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?Of course, all the recent discussion has obscured my original comment on the generalized Lindemann theorem being stated and proven in Chapter 1 of Ivan Niven's Irrational Numbers, a topic of greater intrinsic interest than subsequent discussions. As a reminder, the theorem states that if a_1, ..., a_n are distinct algebraic numbers, then exp(a_1), ...,exp(a_n) are linearly independent over the algebraic numbers. The transcendence of e follows by taking a_1 = 0, a_2 = 1. The transcendenceof pi follows from taking a_1 = 0, a_2 = i pi. It is also possible to use the theorem to prove that: exp(b), sinh(b), cosh(b), tanh(b), coth(b),sech(b), csch(b), sin(b), cos(b), tan(b), cot(b), sec(b), and cosec(b), are all transcendental if b is a nonzero algebraic number, and also thatall nonzero values of log(b), ar sinh(b), ar cosh(b), ar tanh(b), ar coth(b), ar sech(b), ar csch(b), arc sin(b), arc cos(b), arc tan(b),arc cot(b), arc sec(b), and arc cosec(b), for algebraic numbers b, are transcendental.Other chapters of the book are also of interest. The second chapterdiscusses the (generalized?) Gelfond-Schneider Theorem: if a_1, a_2, ...,a_n, are nonzero algebraic numbers, and if values of log(a_1), ...,log(a_n), are such that log(a_1), ..., log(a_n), are linearly independentover Q, then 1, log(a_1), ..., log(a_n), are linearly independent over thealgebraic numbers. Taking a_1 = 1 and log(a_1) = 2 i pi, it immediatelyfollows that pi is transcendental. The famous result that if a is analgebraic number distinct from 0 and 1, and b is an irrational algebraicnumber, then all values of a^b are transcendental, also follows from thetheorem. Specifically, taking a = i and b = i, it follows that exp(pi)is transcendental.Later chapters discuss the constants associated with elliptic and related functions, giving further proofs that some minimum number of such constants from a given set are transcendental.David McAnally -------=== === Subject: : Re: e is transcendental (was: classes of transcendental numbers ?>Of course, all the recent discussion has obscured my original comment on >the generalized Lindemann theorem being stated and proven in Chapter 1 >of Ivan Niven's Irrational Numbers, a topic of greater intrinsic >interest than subsequent discussions. As a reminder, the theorem states >that if a_1, ..., a_n are distinct algebraic numbers, then exp(a_1), ...,>exp(a_n) are linearly independent over the algebraic numbers. The >transcendence of e follows by taking a_1 = 0, a_2 = 1. The transcendence>of pi follows from taking a_1 = 0, a_2 = i pi. It is also possible to >use the theorem to prove that: exp(b), sinh(b), cosh(b), tanh(b), coth(b),>sech(b), csch(b), sin(b), cos(b), tan(b), cot(b), sec(b), and cosec(b), >are all transcendental if b is a nonzero algebraic number, and also that>all nonzero values of log(b), ar sinh(b), ar cosh(b), ar tanh(b), >ar coth(b), ar sech(b), ar csch(b), arc sin(b), arc cos(b), arc tan(b),>arc cot(b), arc sec(b), and arc cosec(b), for algebraic numbers b, are >transcendental.Where b is nonzero in the cases of log(b), ar sech(b), ar csch(b), arc sec(b) and arc cosec(b), b is distinct from 1 and -1 in the case of ar tanh(b) and ar coth)b), and b is distinct from i and -i in the case of arc tan(b) or arc cot(b).>Other chapters of the book are also of interest. The second chapter>discusses the (generalized?) Gelfond-Schneider Theorem: if a_1, a_2, ...,>a_n, are nonzero algebraic numbers, and if values of log(a_1), ...,>log(a_n), are such that log(a_1), ..., log(a_n), are linearly independent>over Q, then 1, log(a_1), ..., log(a_n), are linearly independent over the>algebraic numbers. Taking a_1 = 1 and log(a_1) = 2 i pi, it immediately>follows that pi is transcendental. The famous result that if a is an>algebraic number distinct from 0 and 1, and b is an irrational algebraic>number, then all values of a^b are transcendental, also follows from the>theorem. Specifically, taking a = i and b = i, it follows that exp(pi)>is transcendental.The transendence of e is proven by taking a_1 = e and log(a_1) = 1.>Later chapters discuss the constants associated with elliptic and related >functions, giving further proofs that some minimum number of such >constants from a given set are transcendental.David McAnally -------=== === Subject: : Ratio / Proportion Question I would like to be able to solve the following problem.I would like to change the starting point for a unit circle. For example: 90=0 Degrees | | /45=800 Degrees |/ ----------------- 0=1600 Degrees | | 315= 2400 Degrees | | 270=3200 DegreesCould anyone direct me to how to solve this problem?Sean=== === Subject: : Re: Ratio / Proportion Question> I would like to be able to solve the following problem.>I would like to change the starting point for a unit circle. For example:> 90=0 Degrees> |> | /45=800 Degrees> |/> ----------------- 0=1600 Degrees> |> | 315= 2400 Degrees> |> |> 270=3200 Degrees>Could anyone direct me to how to solve this problem?>SeanOK, let's call your new units Seans instead of degrees. If you measurethe angle A clockwise measured in Seans from the positive y axis andare a distance r from the origin, thenx = r sin (9A/160) and y = r cos (9A/160), where units in the cos andsin functions are degrees.For example if you are 10 units at 1600 Seans then:x = 10 sin (9*1600/160) = 10 sin(90) = 10y = 10 sin (9*1600/160) = 10 cos(90) = 0,so you are at (10,0). Is that what you are looking for?=== === Subject: : Re: Ratio / Proportion QuestionNaturally I find a typo just after I post>x = 10 sin (9*1600/160) = 10 sin(90) = 10>y = 10 sin (9*1600/160) = 10 cos(90) = 0,>so you are at (10,0). Is that what you are looking for?The second line above is obviously y = 10 cos (...)=== === Subject: : Re: Ratio / Proportion QuestionNaturally I find a typo just after I post>x = 10 sin (9*1600/160) = 10 sin(90) = 10>y = 10 sin (9*1600/160) = 10 cos(90) = 0,so you are at (10,0). Is that what you are looking for?The second line above is obviously y = 10 cos (...)Yes, I can just subsitute r are 1 (to use the unit circle). I wouldlike to know if you could direct me to the topic that describes thisin more detail. Is this polar coordinate transformation? Thank you foryou help.Sean=== === Subject: : Re: Ratio / Proportion Question>Yes, I can just subsitute r are 1 (to use the unit circle). I would>like to know if you could direct me to the topic that describes this>in more detail. Is this polar coordinate transformation? Thank you for>you help.>SeanYes, it comes from polar coordinates. Polar coordinates give you:x = r cos(theta) y = r sin(theta)where theta is measured counterclockwise from the positive x axis. Inyour case if alpha is measured clockwise from the positive y axis,then theta is 90 - alpha. This gives youx = r cos(90 - alpha) = r sin(alpha)y = r sin(90 - alpha) = r cos(alpha)These would work except for the change of scale where you want tomeasure alpha in Seans which is why the 9/160 scaling factor isneeded.You can read about polar coordinates in most any trigonometry book.Also there is lots of stuff on the web. Type polar coordinates inGoogle and you will find lots of references like:http://www.ping.be/math/polar.htmhttp:// mathworld.wolfram.com/PolarCoordinates.html=== === Subject: : Re: Ratio / Proportion Question I would like to be able to solve the following problem.> I would like to change the starting point for a unit circle. For example:> 90=0 Degrees> |> | /45=800 Degrees> |/> ----------------- 0=1600 Degrees> |> | 315= 2400 Degrees> |> |> 270=3200 Degrees> Could anyone direct me to how to solve this problem?> SeanIf I understand your wishes, you wish to transform the usual degree measurements as marked above into non-degree measurements also as marked above, so that 90 degrees becomes 0 new units, 0 degrees become 1600 new units, and so on.This requires you to find the equation of a linear function for which(x,y) = (90,0) and (x,y) = (0, 1600) are two points.If x represents any measurement in degrees and y the corresponding measurement in the new units, according to your diagram, then y = (90 - x)*1600/180 = (90 - x)*80/9=== === Subject: : Re: e is transcendental>My,question has been resolved neatly and Laconicaly,>within two lines:>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>Im(e^[iPi]) = Im(-1+i[0]) = 0>Since there is acceptance to this,>any more saying has no value.Except to point out that it means that your original conclusion that exp(i pi) = 0 was not validly drawn from what preceded it.David McAnally -------=== === Subject: : Re: partitions of an integer n into k parts> I'd be grateful for suggestions where I could find on the web an > algorithm for the partitions of an integer into k parts, such as the 7 > partitions of 9 into 3 positive integers:> (1 7 1), (2 6 1), (3 5 1), (4 4 1), (2 5 2), (3 4 2), (3 3 3)> I do have an algorithm for compositions and I could remove the > permutations from its output, but that would be rather inefficient.PierreOn-line, try:http://mathworld.wolfram.com/Partition.htmlBut I will add some wisdom to sci.math myself:First, the number of partitions of n into k parts is the same as thenumber of partitions of n which have their largest part = to k.We can see this by rotating each partition's Ferrers diagram90-degrees:example:..... .. . .(6,2,1) <> (3,2,1,1,1,1)If P(k,n) = the number of partitions of n into partswhere the largest is <= k, not necessarily = k, then:1 + sum{n=1 to oo} P(k,n) x^n= product{n=1 to k} 1/(1 -x^n).So, we want, if we let p(k,n) be the number of partitions of n intoparts whose largest IS k,p(k,n) = P(k,n) - P(k-1,n).So, concluding, and if I did not err,if the number of partitions of n into exactly k parts is:p(k,n),thensum{n=1 to oo} p(k,n) x^n =product{n=1 to k} 1/(1 -x^n)-product{n=1 to k-1} 1/(1 -x^n)= x^k product{n=1 to k} 1/(1 -x^n),which means that:p(k,n) = P(k,n-k),for n>= k,and where P(k,0) is defined as 1.Right? (I am uncertain.) Leroy Quet=== === Subject: : Re: Permutation(?): + Integers Increasingly Arranged> Define an integer sequence as follows:a(1) = 1; And place the positive integers, once per integer, into the sequence> so that:(m+1) is positioned in the sequence such that there are exactly (m-1)> higher-valued terms between it and the term equal to m.> And the (m+1) is always less that the m whenever this is possible.As to hopefully eliminate ambiguity, here is the arrangement of the> first> 16 integers: (figured by-hand)1, 2, 13, 3, 6, *, 4, 11, *, 9, 5, * 7, *, *, *, *, *, *, 8,> *, 10, 16, 12, *, 14, *, *, *, *, *, *, *, *, *, *, *, *, *, 15,...> Of course, the *'s are yet to be determined and are each > 16.> Because of the higher-valued part of the definition, the sequence> will never get 'stuck', ie. unlike if we simply counted the number of> terms in-between, where the sequence might get into a region where the> nearest unused position is too far away.But I must ask, is this sequence a permutation of the positive> integers?> In other words, will every position eventually contain an integer?If so, here is the inverse permutation's first few terms:1, 2, 4, 7, 11, 5, 13, 20, 10,...(Maybe the inverse sequence would be easier to study, actually.)Leroy QuetSome kind contributers to the Encyclopedia Of Integer Sequences haveadded the above sequence for me at:http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A058747Leroy Quet=== === Subject: : Having trouble with some math that deals with x,y coordinatesBasicly, I have a point x,yAlso, I have an angle from 0 to 360 aAnd I have a distance dWhat I want to do is to start at point x,y.Face the angle aMove distance dand compute the new pointAnyone know the formula/math for this?=== === Subject: : Re: Having trouble with some math that deals with x,y coordinates> Basicly, I have a point x,y> Also, I have an angle from 0 to 360 a> And I have a distance d> What I want to do is to start at point x,y.> Face the angle a> Move distance d> and compute the new pointAnyone know the formula/math for this?Suppose (x,y) is the origin (0,0). The ray starting at the origin and making angle a with the x-axis consists of the points (d * cos a, d * sin a) where d is the distance from the origin. If you start instead at arbitrary (x,y) you translate everything accordingly. If (x',y') is the final point, thenx' = x + d * cos ay'= y + d * sin a=== === Subject: : Re: Having trouble with some math that deals with x,y coordinates> x' = x + d * cos a> y'= y + d * sin aThankyou, thats just the formula I need :)=== === Subject: : Generating random numbers with specific propertiesIs there any book, website, or whatever that has a good, detaileddescription of the algorithms needed to generate sequences of randomnumbers with a given distribution, mean, variance, higher-ordermoments (kurtosis, skew, etc.), autocorrelation, cross-correlationwith other such series, etc? In other words, you figure out theproperties and then you can either look up or figure out the algorithmto generate it.I'm finding it one sliver at a time all over the Web tailored for somevery specific uses plus documentation for commercial libraries thatmay generate such data given the parameters, but I'd like to know howto generate my own for all sorts of different cases/applications. Itjust comes up again and again each time I consider modeling something,so it's time to ask the experts. ;-)Any suggestions?=== === Subject: : Re: Generating random numbers with specific properties>Is there any book, website, or whatever that has a good, detailed>description of the algorithms needed to generate sequences of random>numbers with a given distribution, mean, variance, higher-order>moments (kurtosis, skew, etc.), autocorrelation, cross-correlation>with other such series, etc? In other words, you figure out the>properties and then you can either look up or figure out the algorithm>to generate it.>I'm finding it one sliver at a time all over the Web tailored for some>very specific uses plus documentation for commercial libraries that>may generate such data given the parameters, but I'd like to know how>to generate my own for all sorts of different cases/applications. It>just comes up again and again each time I consider modeling something,>so it's time to ask the experts. ;-)>Any suggestions?Indeed, there is much literature on the subject. Look up simulaiton in your library. One particularly comprehensive reference is the book by Devroye, Non-uniform Random Variate Generation. Also, take a look at Knuth, vol. 2.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: Generating random numbers with specific propertiesGenerate a series of pseudo-random numbers. Then use a filter to throw outthose that do not satisfy your condition.Or make the filter part of the generating loop.Roman=== === Subject: : Re: Generating random numbers with specific properties> Generate a series of pseudo-random numbers. Then use a filter to throw out> those that do not satisfy your condition.Is this considered the standard general approach to creating data witharbitrary parameters -- the one to fall back on when there isn't amore direct generator algorithm?Are there other such general approaches, or is this one by far thebest (again, in cases where there isn't a more direct approach)?=== === Subject: : Sum{k=2 to n} 1/ln(k) and prime-countingI was wondering:How good an approximation to pi(n), the number of primes <= n, issum{k=2 to n} 1/ln(k) ?It overestimates integral{2 to n} dx/ln(x),and so I would bet that it is not very accurate, relatively.But...For which n's, if any,is sum{k=2 to n} 1/ln(k) duality:: Doesn't tell much about that. Duality is negation in boolean algebras,isn't it ?The definitions of T and _|_ are dual in this regard, so there is some sensewhere boolean negation is represented in dualities. Paticularly with / and/ sharing a dual nature, in a boolean framework with excluded middle muchof the structural theory of boolean algebras can be understood through thestudy of dualities. However in the general case, duality can be used inmany circumstances to extend theorems on semilattices up to a completelattice, which saves some work and repetition, but does not confine us yetto any boolean model.-- ===-=-=-=-=-=== === Subject: : Re: 3 Squares Covering 1 Circle I was asked (via e-mail) to explain this here in more detail. If a sense, this is what I did when I found that Jake's> configuration yielded a saddle point. Three of the eigenlines> were hot -- 2 slightly, and 1 much more so. So I looked at> that hot line, and it suggested the twisted configuration that> produced the bona fide local maximum.Hi Jim,thanks a lot for your explanation, which rings a lot of bells.> But there is still a large gap between the general explanation,> which you gave in a vivid manner, and the (I think) awkward> looking matrix itself, full of lengthy terms.It would be a great pleasure for me, if you showed this beast,> which eventually gave you the best solution so far. Maybe I or> some of the interested lurkers feels so strong now as to try> by myself/herself/himself to play with the matrix and get an> improvement. Or else get a better feeling for why this should> be the best solution indeed.I like these geometric puzzles very much and I want to say> thanksto Leroy for this nice puzzle. The Hessian seems> to be a strong tool to achieve insight as well as results.You are welcome.:)But I should thank, in turn, 3 square napkins and a paper plate forthe inspiration...In any case, this is most likely a well-known problem.I know that, generally, such coverings are well-studied.(Although I know not the history of my particular problem.)Leroy Quetconfiguration so far,Rainer Rosenthal> r.rosenthal@web.de=== === Subject: : Mean distance to an ellipse from points within itThe mean distance to a circle from points within it can be shown to be 1/3of its radius, an easy result which is fairly well known. Last month, ingeometry.research, someone asked about the mean distance to an ellipse frompoints within it. That mean distance, M, can be expressed as an integral interms of the lengths, a and b, of the semiaxes. Details of the messycalculation are omitted here. The result expresses the mean distance inclosed form in terms of K and E, the complete elliptic integrals of thefirst and second kinds, resp. With e^2 = 1 - (b/a)^2, we get* M = 4 b/(3 pi) ( E(e^2) - 1/2 (1-e^2) K(e^2) )[Lest there be any notational confusion, I'm using essentially the sameconventions used in .]The mean distance can be approximated nicely by an expression in the form# 4 b/(3 pi) ( 1 - (1-pi/4) (b/a)^p )where p is a suitable positive constant. Note that, regardless of p,# gives M precisely at both extremes of eccentricity, e = 0 and e = 1.If we wish # to approximate the mean distance as accurately as possiblefor small eccentricities, it can be shown that p = pi/(8 - 2 pi), which isabout 1.83 . Using that value for p, # provides a lower bound for the meandistance, with its worst relative error being roughly -0.0021 .If we wish # to approximate the mean distance as accurately as possiblefor large eccentricities, it can be shown that p = 2 . Using that value forp, # provides an upper bound for the mean distance, with its worst relativeerror being roughly 0.0064 .If we wish # to approximate the mean distance as accurately as possibleover all eccentricities, then the value of p, between pi/(8 - 2 pi) and 2,which minimizes max|relative error| can be determined numerically. Thisgives p = 1.86906..., and # then provides |relative error| < 0.00087 .Example: For an ellipse having semiaxes lengths 10 and 5, find the meandistance to it from points within it.Using *, we get M = 20/(3 pi) ( E(3/4) - 1/8 K(3/4)) = 1.9979071667...Using # with p = 1.86906, we get 1.9974 as an approximation for M, and sorelative error is about -0.00025 .Has anyone seen a closed form, an approximation, or bounds for the meandistance previously?David W. Cantrell=== === Subject: : Re: RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem> Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem [Revised]> For every a, there is always at least one b such that a^n+b is always> composite, for every n> ??Try b = ma for any positive integer m. If you want to include n=0,you can choose q so gcd(q,a) = 1, and take m so ma = -1 mod q. In a slightly different direction:For every pair of integers a, b with a > 1 and ab <> 2 (thus leavingout the case of the Mersenne numbers), there are infinitely many primes p such that a^p + b is composite.Department of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2=== === Subject: : Re: RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem>Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem [Revised]>>For every a, there is always at least one b such that a^n+b is always>>composite, for every n>>??> Try b = ma for any positive integer m. If you want to include n=0,> you can choose q so gcd(q,a) = 1, and take m so ma = -1 mod q. In a slightly different direction:For every pair of integers a, b with a > 1 and ab <> 2 (thus leaving> out the case of the Mersenne numbers), there are infinitely many > primes p such that a^p + b is composite.> The case a+b <> +-1 is easy, because if q is a prime factor of a+b then q will divide a^p+b whenever p=1 mod(q-1). The case -b = a+-1 seems a bit trickier, a similar sort of thing seems to be going on, but I placethe old nose to the grindstone.> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem> Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem [Revised]> For every a, there is always at least one b such that a^n+b is always> composite, for every n> ??> JAs stated, it is trivially true, choose b a multiple of a, such that b+1 is composite. Adding the unstated restriction that a and b are relatively prime, it is still true: let b = -1 mod (a-1) if a>2, the case a=2 having already been dealt with. Adding the further, likewise unstated, restriction that a-1 and b+1 are relatively prime, it is still true, but the proof is somewhat more subtle. (but not a lot.)=== === Subject: : Re: the anticlassicalist }{ v: universal truths:: > And on these lattices we can make more explicit our theories ofnegation.: >: > On sci.logic, recently gave a very interesting post on Stonealgebras.: > When we want to model uses of various definitions, we need to show themas: > an axiom set. Stone algebras study the axiom set:: >: > ~(a) / ~(~(a)) <--> T::: Btw, can complex numbers model it ? Let's see:: ~ : r exp(i s) --> r/2 exp(i s/2):: a = 1:: ~(a) = -1/2:: ~(~(a)) = i/4:: let's say T is 0...:: Well, then I am sure a Lorentz-Moebius transform can save the situation, I: mean, provide a natural interpretation to the same formula graphics. The: Moebius group is strictly 3-transitive, and:: z / w <---> u:: can pass as denoting -the- moebius transform that swaps z and w whilekeeping: u invariant. Now idea if the thing can mimick a lattice any further,though.I think I would need a little more exposition. In particular, would such alattice define T and _|_. Also, there would be two fixed points in anymoebius transformation, so I'm wondering if an identification is assumed tobe made. Or maybe my confusion lies at what exactly the objects of thelattice are. In other words, are they transforms (the lattice being allMoebius transforms) or points under a given transform. Although thenegation definition does seem to be on points inside the transformationgiven, I have difficulty reading the / on points unambiguously.I can see how the symbols z / w <-> u can be used to single out atransformation, but wouldn't z / w <-> v (where v was the other fixedpoint of the transformation) denote the same transformation, implying thateither / was not unique or that some kind of collapse was in order?There is, however, a natural way to identify certain lattice substructuresto operators obeying Hecke algebras, if that is what you are looking at.Certainly, the study of modular forms and lattice structures is intricatelyconnected.: Isn't the symmetric group attached to the boolean algebra in such a manner: that extremely transitive (but non-symmetric) groups shine as interesting: analogues ?I think these kinds of questions are best analysed in the context of topoi,where we can generalise the relationships. At least, when you mentionsymmetric group attached to the boolean algebra, I think of the naturalpower objects in a topos defined in terms of a subobject classifier.Symmetric groups can be found in extensions of power objects in the categoryof sets, and we can lift the natural boolean structure of the subobjectclassifier in a standard way. In a general topos, however, all we have is aHeyting algebra defined on the subobject classifier, and the lifting impliesa certain, more general structure.Now I am aware of relationships that identify the permutation groups tocertain commutative dynamical flows (not quite in relation to groupcommutativity and k-commutativity), with other groups denoting flows ofdeeper structure (braidings and knots embedded, for example, with nocrossing type theorems), and I am also aware of the deep relationshipbetween such groups and Hopf algebras. But I have not gone so far as tomake the full connection except in the study of quantum groups and braidingcategories.So, although I cannot answer your questions, maybe what I have provided is astart for discussion?-- ===-=-=-=-=-=== === Subject: : series clarificationMy book shows how to derive the Maclaurin __polynomial__, and leaves it asan exercise to do the taylor polynomial.So from my understanding, the taylor polynomial will approximate somefunction f(x) at x0. In contrast, the Maclaurin polynomial approximatesf(x) at x0 =0.The text also says its convenient to write p(x) in the form:* p(x) = c0 + c1(x-x0) + c2(x-x0)^2+...+ cn(x-x0)^nversus p(x) = c0 + c1x + c2x^2 +...+cnx^nIs the only reason for writing the approx. polynomial p(x) in the form (*)so that we get zeros when we plug in x0 into p(x) and its derivatives tofind the constants c0...cn?=== === Subject: : Re: series clarification> The text also says its convenient to write p(x) in the form:> * p(x) = c0 + c1(x-x0) + c2(x-x0)^2+...+ cn(x-x0)^n> versus p(x) = c0 + c1x + c2x^2 +...+cnx^n> Is the only reason for writing the approx. polynomial p(x) in the form (*)> so that we get zeros when we plug in x0 into p(x) and its derivatives to> find the constants c0...cn?No, consider log x.You can use *, but not with x0 = 0.=== === Subject: : Re: a fibred link question>Well, I can't quite see why the automorphism should be expected >to come in a reasonable way from the embedding. If the automorphism doesn't come in a reasonable way from theembedding, then there's no reason to expect that the space youconstruct from the automorphism (that is, the mapping cylinder completed by the binding) should be the 3-sphere which was thetarget of the embedding. And, in fact, it isn't in your case.Yes, you started with a link L in S^3, and a Seifert surface A forthat link, but you didn't actually *use* the embedding of A in S^3 to construct the automorphism. You'd have had a similar problemif you'd started with a link that *is* fibered and a Seifert surface for it that *isn't* a fiber surface; you might find it illuminating to think about the simplest example of that, namely, the unknot Oand a maximally-uncomplicated Seifert surface F for O of genus 1(in other words, a Heegard torus in S^3 with an open disk removed).Why can't you construct g:F->F in this case?>I'm not so clear >on your alternate explanation either. I mean, the construction of >my first paragraph should give me some sort of 3-manifold for any >link L and g in the mapping class group of A, should it not? For >g=Id, this is obviously S^1 x S^2, so for g a dehn twist, i feel >(or at least i interpret Lickorish to the effect that) the result >should be cutting S^1 x S^2 along some surface and gluing back >with a dehn twist. Actually, now that I put it that way, it >seems that all I am doing is a dehn twist around one of the >S^2's, i.e. nothing at all, so I still get S^1 x S^2. Sound >good? But I still would like to understand what you meant. >link to Agora.Lee Rudolph=== === Subject: : Re: 6 Universali Domande Tranello Per Nascondere la Propria Ignoranza in MatematicaROTFL!!!!keroppi ha scritto nel messaggio> SIX ALL-PURPOSE TRICK QUESTIONS THAT WILL KEEP YOUR IGNORANCE FROM> BEING EXPOSED IN TREACHEROUS MATHEMATICS DISCUSSION GROUPS> 1. Can that result be restated in terms of category theory?> 2. Isn't the constant in that equation suspiciously close to the> square root of pi divided by e cubed?> 3. Wasn't your theorem prefigured in the work of Euler?> 4. Same as above, but replace Euler with Poincar.8e.> 5. Is any of this applicable to linear p-adic groups, p-adic Galois> groups, or p-affine Schur algebras?> 6. Yes, but can you prove that lemma for the case of n=3?> --> keroppi> www.geocities.com/rosacrux=== === Subject: : Re: No Set Contains Every Computable Natural charset=Windows-1252[earlier in the thread:]> The number of operations in Cantor's proof is not finite.[and now:]> The diagonal construction is not an algorithm?> The construction of a set in P(N) not in N is not an algorithm?> What is it? Wishful thinking?I think you might benefit by reading (or re-reading) the post by Dave Seaman 2-22 1:24Pm in this same thread, concerning both direct and indirect proofs. The diagonal set can be regarded as generated, fully and completely in a single step.--r.e.s.=== === Subject: : Re: No Set Contains Every Computable Natural> [earlier in the thread:]> The number of operations in Cantor's proof is not finite.> [and now:]> The diagonal construction is not an algorithm?> The construction of a set in P(N) not in N is not an algorithm?> What is it? Wishful thinking?> I think you might benefit by reading (or re-reading) the post> by Dave Seaman 2-22 1:24Pm in this same thread, concerning> both direct and indirect proofs. The diagonal set can be> regarded as generated, fully and completely in a single step.An infinite set in a single step!That is an impressive TM. Can I get one?=== === Subject: : Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vflykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <14-dnWhDt8HHB6fdRVn-uQ@comcast.com> Discussion, linux)>> Cantor didn't give an algorithm. Aside from that, um, yeah, good>> point.> The diagonal construction is not an algorithm?> The construction of a set in P(N) not in N is not an algorithm?> What is it? Wishful thinking?An existence proof.-- And a journal can beg me for the right to publish it [...] becauseI'd rather see it in People magazine [...] --James Harris on his simple proof of Fermat's last theorem=== === Subject: : Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vflykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> Discussion, linux)>> This is lame and sloppy thinking. I don't really care what you>> accept, but I won't get into silly arguments that Cantor's proof>> doesn't count if TMs have to halt within a finite number of steps.>> Try to be precise and correct in your reasoning. Don't use mushy and>> ignorant analogies.> There is nothing mushy about pointing out that Cantor's proof> requires an infinite number of operations.There is plenty mushy, beginning with the fact that the number ofoperations is utterly without meaning in Cantor's proof. There are nooperations that I can see. There is a proof of the existence of areal number not on the list.>> Whether the output was written before step omega is inconsequential.>> The machine didn't converge until step omega.>> No, you can't solve the halting problem for machines with infinite>> computations with our convention. Let m be any machine which, on>> input x, converges in the omega'th step and not before.> There is no such machine.> We are assuming that a TM can perform an infinite number of> operations in a finite amount of time. We specifically are NOT> assuming that a TM can perform an infinite number of operations> and then halt.When I say that the TM converged in the omega'th step, I mean that (1) he didn't halt in a finite step and(2) he produced a tape after an infinite number of steps.What do you want the halting problem decider to say in that case?Seems sensible to me to say that our decider ought to output 1 (thehalt output), but if you think he ought to output 0, then that'sfine. But in that case, my comment (2) doesn't apply.> Halt is an instruction. Any TM that performs a halt instruction must> do so in a finite number of steps.OK. (Though, in most conventions, halt is not an instruction.)The rest of your post is uncontroversial, as long as we adopt youridea of the halting problem for this setting.-- And I'm one of my own biggest skeptics as I had *YEARS* of wrongideas, and attempts that failed. Worse, for some of them it took*MONTHS* before I figured out where I screwed up. -- James Harris=== === Subject: : Re: No Set Contains Every Computable NaturalThis is lame and sloppy thinking. I don't really care what you>> accept, but I won't get into silly arguments that Cantor's proof>> doesn't count if TMs have to halt within a finite number of steps.Try to be precise and correct in your reasoning. Don't use mushy and>> ignorant analogies. There is nothing mushy about pointing out that Cantor's proof> requires an infinite number of operations.> There is plenty mushy, beginning with the fact that the number of> operations is utterly without meaning in Cantor's proof.Not finite has a meaning.The number of operations in Cantor's proof is not finite.(Except in the trivial case where the original list is finite.)> There are no> operations that I can see. There is a proof of the existence of a> real number not on the list.Then I prove there exists a natural number not on the input tape.>> Whether the output was written before step omega is inconsequential.>> The machine didn't converge until step omega.No, you can't solve the halting problem for machines with infinite>> computations with our convention. Let m be any machine which, on>> input x, converges in the omega'th step and not before. There is no such machine.> We are assuming that a TM can perform an infinite number of> operations in a finite amount of time. We specifically are NOT> assuming that a TM can perform an infinite number of operations> and then halt.> When I say that the TM converged in the omega'th step, I mean that> (1) he didn't halt in a finite step and> (2) he produced a tape after an infinite number of steps.The output on the tape converges after a finite number of steps.We might have to wait for an infinite number of steps to besure the output has converged.> What do you want the halting problem decider to say in that case?> Seems sensible to me to say that our decider ought to output 1 (the> halt output), but if you think he ought to output 0, then that's> fine. But in that case, my comment (2) doesn't apply.I think we agree that the definition of halt means the TM haltsafter a finite number of steps. Even if we allow a TM to haltafter an infinite number of steps, I don't see how we can saythe TM halts if it does so on the omega'th step.> Halt is an instruction. Any TM that performs a halt instruction must> do so in a finite number of steps.> OK. (Though, in most conventions, halt is not an instruction.)> The rest of your post is uncontroversial, as long as we adopt your> idea of the halting problem for this setting.OKLet's agree that our hyper-TM can perform an infinite number ofoperations in finite time, but it can NOT halt on step omega.In all other respects, the hyper-TM behaves like a normal TM.A hyper-TM can solve the halting problem, even for hyper-TM's.=== === Subject: : Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vflykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <87n079e239.fsf@phiwumbda.org> Discussion, linux)>> There is plenty mushy, beginning with the fact that the number of>> operations is utterly without meaning in Cantor's proof.> Not finite has a meaning.> The number of operations in Cantor's proof is not finite.> (Except in the trivial case where the original list is finite.)There are no discernible operations in Cantor's proof. It is not analgorithm. Maybe one of the local constructivists will disagree, Idon't know. But as far as the classical interpretation of Cantor'sproof goes, there is not a procedure, but the specification of a realnumber. And even if we *were* to view it algorithmically, we have noimposed convergence conditions (like completes in finitely many steps).>> There are no>> operations that I can see. There is a proof of the existence of a>> real number not on the list.> Then I prove there exists a natural number not on the input tape.No, you don't. , I'm not arguing that your proof fails becauseit converges only in omega and no fewer steps. Your proof failsbecause after omega steps, there is not a single natural number onyour tape. I've told you this. I've given the proof that your tapecontains no blanks (right of the start state). By the n+1st iterationof your loop, the nth state is not blank. Therefore, after omegasteps there are no blanks.Want to take omega steps to construct your tape? Fine. Go ahead.Take your time, we have all day here. You still utterly fail tocreate a natural number not on the list.Why are you confused on this point? Because, after I don't know howmany years, you are still incapable of going five minutes withoutcommuting quantifiers. You have a proof that, for each step n, thereis an x which is blank at n and you believe this is sufficient to show thatthere is an x such that for all n, x is blank at n. This is just thesame error over and over and over.>> When I say that the TM converged in the omega'th step, I mean that>> (1) he didn't halt in a finite step and>> (2) he produced a tape after an infinite number of steps.> The output on the tape converges after a finite number of steps.> We might have to wait for an infinite number of steps to be> sure the output has converged.This point is irrelevant but correct, more or less. However, I'vechosen to define the term convergence here as above, so yourobservation is a bit confusing with the definition I'm using.>> Halt is an instruction. Any TM that performs a halt instruction must>> do so in a finite number of steps.>> OK. (Though, in most conventions, halt is not an instruction.)>> The rest of your post is uncontroversial, as long as we adopt your>> idea of the halting problem for this setting.> OK> Let's agree that our hyper-TM can perform an infinite number of> operations in finite time, but it can NOT halt on step omega.> In all other respects, the hyper-TM behaves like a normal TM.> A hyper-TM can solve the halting problem, even for hyper-TM's.No, it can't, but I'm not going down this road with you. There is nosense in making the picture more complicated at this point. Let'swait until you grasp basic Turing machines, or perhaps those whichconverge in omega steps, before we go galloping through the hierarchyof transfinite ordinals.-- Just because you're ... in a Ph.d program it does not mean thatyou're up to the challenge of being a real mathematician. Only thosewho have a purity of mind and dedication to the truth as the highestideal have a chance. --James Harris, as Sir Galahad the Pure.=== === Subject: : Re: No Set Contains Every Computable Natural> There are no discernible operations in Cantor's proof. It is not an> algorithm.How can it construct a counterexample without an algorithm?It's not a proof unless it does.> Maybe one of the local constructivists will disagree, I> don't know. But as far as the classical interpretation of Cantor's> proof goes, there is not a procedure, but the specification of a real> number.An uncomputable real number.How do I know this number is not in my list if I can't compute it?> And even if we *were* to view it algorithmically, we have no> imposed convergence conditions (like completes in finitely many steps).Or any proof it exists.>> There are no>> operations that I can see. There is a proof of the existence of a>> real number not on the list.But we can't compute this real number.Very convenient.> Then I prove there exists a natural number not on the input tape.> No, you don't. , I'm not arguing that your proof fails because> it converges only in omega and no fewer steps.> Your proof fails> because after omega steps, there is not a single natural number on> your tape.We know there was a non-zero number of blanks on the initial tape.My TM can't overwrite the last blank on the tape.It is incapable of doing so.There will be a blank on the tape.> I've told you this. I've given the proof that your tape> contains no blanks (right of the start state). By the n+1st iteration> of your loop, the nth state is not blank. Therefore, after omega> steps there are no blanks.Then there is a blank on my tape at position omega.You keep talking about omega steps like it really means something.I have shown why a TM always performs a finite number ofoperations. There is no such thing as omega steps.There will still be blanks on the tape after any number of steps.> Want to take omega steps to construct your tape? Fine. Go ahead.Cantor uses omega steps to construct his missing real.Unless you are still claiming this real can't be constructed.> Take your time, we have all day here. You still utterly fail to> create a natural number not on the list.Cantor utterly fails to create a real number not on the list.You have already admitted that a hyper-TM can solvethe halting problem. Cantor's algorithm requires a hyper-TM.Why do you accept some hyper-TM proofs and not others?=== === Subject: : Re: No Set Contains Every Computable Natural> How can it construct a counterexample without an algorithm?How can Detroit construct a car without a buggy whip holder?> It's not a proof unless it does.It's not a vehicle unless it does.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W=== === Subject: : Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vflykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <87n079e239.fsf@phiwumbda.org> <878yitdkf4.fsf@phiwumbda.org> Discussion, linux)>> There are no discernible operations in Cantor's proof. It is not an>> algorithm.> How can it construct a counterexample without an algorithm?> It's not a proof unless it does., go look up the word proof. Come back when you find where itsays that algorithms are a necessary feature of existence proofs.>> Maybe one of the local constructivists will disagree, I>> don't know. But as far as the classical interpretation of Cantor's>> proof goes, there is not a procedure, but the specification of a real>> number.> An uncomputable real number.> How do I know this number is not in my list if I can't compute it?I never said that the real number was uncomputable. It is relativelycomputable, namely relative to the function f:N -> R given in thehypothesis. Note, however, that computability of reals meanssomething different than computability of naturals.No sense in discussing computability of reals if you don't understandcomputability of naturals. And there is no need for a number to becomputable in order to show that it is different than every number onthe list.Cantor is not restricted to computable methods. He's not doingcomputability theory.>> And even if we *were* to view it algorithmically, we have no>> imposed convergence conditions (like completes in finitely many steps).> Or any proof it exists.Any sequence of digits defines a real number. Cantor's proof yields asequence of digits. Therefore, it defines a real number.> There are no> operations that I can see. There is a proof of the existence of a> real number not on the list.> But we can't compute this real number.> Very convenient.Who said we can't? We can, *given* the function f:N -> R. If thefunction f is uncomputable, then of course, we can't really computethe number, but none of this is relevant.>> Then I prove there exists a natural number not on the input tape.>> No, you don't. , I'm not arguing that your proof fails because>> it converges only in omega and no fewer steps.>> Your proof fails>> because after omega steps, there is not a single natural number on>> your tape.> We know there was a non-zero number of blanks on the initial tape.> My TM can't overwrite the last blank on the tape.> It is incapable of doing so.> There will be a blank on the tape.Wrong, wrong, wrong. At any time n, there is a blank on the tape. At time omega, there areno blanks.>> I've told you this. I've given the proof that your tape>> contains no blanks (right of the start state). By the n+1st iteration>> of your loop, the nth state is not blank. Therefore, after omega>> steps there are no blanks.> Then there is a blank on my tape at position omega.There *is* no ing position omega on the tape, you twit.There is no position omega, you silly boy. > You keep talking about omega steps like it really means something.> I have shown why a TM always performs a finite number of> operations. There is no such thing as omega steps.> There will still be blanks on the tape after any number of steps.But, after he completes *every* step for n < omega, there are noblanks. I know you don't believe me. You're incapable ofunderstanding this point, evidently. But we started with a tape thathas squares enumerated 0,1,2,3,.... It does not have a square labeledomega. It simply doesn't. So you can't have a blank in the squaremarked omega.I'm done for now. You're just an idiot and the conversation goesround and round. Listen, you have fun with your delusions, 'kay?-- I've been thinking about my problems with getting any kind ofadmission that my math arguments showing the core error in mathematicsare correct, so I've gone to marketing books. -- James S. Harris, on when mathematics isn't enough=== === Subject: : Re: No Set Contains Every Computable Natural>> There are no discernible operations in Cantor's proof. It is not an>> algorithm.> Cantor is not restricted to computable methods. He's not doing> computability theory.>> And even if we *were* to view it algorithmically, we have no>> imposed convergence conditions (like completes in finitely many steps).> At any time n, there is a blank on the tape. At time omega, there are> no blanks.> There is no position omega, you silly boy.There is no time omega, either.> But, after he completes *every* step for n < omega, there are no> blanks. I know you don't believe me.You just said there is no position omega.How can a TM complete *every* step for n < omega if we bothagree there is no position omega?> You're incapable of> understanding this point, evidently. But we started with a tape that> has squares enumerated 0,1,2,3,.... It does not have a square labeled> omega. It simply doesn't. So you can't have a blank in the square> marked omega.And a TM can't perform omega steps.There will always be a blank on the tape.That is the point of the proof.It proves the non-existence of omega.=== === Subject: : integration with respect to a function?In reading a probability text I encountered this formula:INT(-oo,oo) F_Y (t-x) dF_X (x) = 1 if t >= 1, 0 otherwise. I need toconclude something about F_Y, but I was wondering what exactly is meant bythedF_X (x)in the integral. How should I interpret this?=== === Subject: : Re: integration with respect to a function?integral f(x) dg(x) ... this is a Stieltjes integral.=== === Subject: : Re: integration with respect to a function?>In reading a probability text I encountered this formula:>INT(-oo,oo) F_Y (t-x) dF_X (x) = 1 if t >= 1, 0 otherwise. I need to>conclude something about F_Y, but I was wondering what exactly is meant by>the>dF_X (x)>in the integral. How should I interpret this?F_X is almost surely a cumulative distribution function of a random variable X. This is a Stieltjes integral. It represents the same thing as E[F_Y(t - X)]. (This notation is a little bit ambiguous, since the subscript Y in the expectand is not a random variable.)By the way, this is the formula for the convolution of the distributions F_X and F_Y.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Vector Fields on non-compact manifoldsX-ID: SPhkMYZb8efQJMeRyMA7StPO6K5BlQNs19SrMjeOIU8YU97CkZgaZ1As a hobby, I_m studying the book Geometric Theory of DynamicalSystems by Palis / de Melo again. Almost from the onset they restrictthemselves to vector fields (resp. diffeomorphisms) on compactmanifolds. So, I'm trying to fill the gap in the case of vectorfieldson R^n:Let V be a smooth (i.e. C^r-) vector field on |R^n and p be a positivefunction on |R^n, then V^ = p*V has the same orbit structure as V (Iconvinced myself of this). Suppose h: |R^n -> S^(n) - {north pole} isa map of the n-sphere. I claim:There exists p as above, such that V^ lifts to a vector field on S^n -{north pole} that can be extended to a smooth vector field W on S^nsuch that W(north pole) = 0, a not necessarily hyperbolic singularity.Am I correct so far? If so, it would really suffice to studyvectorfields on compact manifolds, from a qualitative point of view.Another question is: What about diffeomorphisms, how would one arguehere?=== === Subject: : Re: Vector Fields on non-compact manifolds> Let V be a smooth (i.e. C^r-) vector field on |R^n and p be a positive> function on |R^n, then V^ = p*V has the same orbit structure as V (I> convinced myself of this). Are you sure about this? I don't really know what you mean by thesame orbit structure, but if you take V to be the unit field in R^2in the y-direction, and p(x.y)=1+y^2, then p*V isn't even globallyintegrable (y'=1+y^2 is solved by tan). I think that somehow qualifiesfor different orbit structures...> Suppose h: |R^n -> S^(n) - {north pole} is> a map of the n-sphere. I claim:There exists p as above, such that V^ lifts to a vector field on S^n -> {north pole} that can be extended to a smooth vector field W on S^n> such that W(north pole) = 0, a not necessarily hyperbolic singularity.What do you mean by that lift? You want to lift via the differentialof h, so that Dh(V^)=W^? You run into trouble if h is not anembedding, then. Or do you want something like W^(h(x))=V^(x)? Thenyou'd have to identify the tangential spaces of S^(n) and |R^nsomehow.. > Another question is: What about diffeomorphisms, how would one argue> here?Hmm, I don't really understand that question. You mean if h is adiffeomorphism? Cheersipp=== === Subject: : Re: Vector Fields on non-compact manifoldsX-ID: ZdW4vYZDoePlSfUXx16-8CZyFh4J0UUBKaVksxevLguhbi3ZrEglQO ipp Frank schrieb:>> Let V be a smooth (i.e. C^r-) vector field on |R^n and p be a positive>> function on |R^n, then V^ = p*V has the same orbit structure as V (I>> convinced myself of this). >Are you sure about this? I don't really know what you mean by the>same orbit structure, but if you take V to be the unit field in R^2>in the y-direction, and p(x.y)=1+y^2, then p*V isn't even globally>integrable (y'=1+y^2 is solved by tan). I think that somehow qualifies>for different orbit structures...No, what I meant is, they have the same orbits up tore-parametrization, the same *orbit-sets*. That is, if x(t) is asolution of x' = V(x), then I can find a function t -> gamma(t) withgamma'(t) > 0 for all t, such that t -> x^(t) = x(gamma(t)) is asolution of x^' = V^(x^). gamma can be found by solving the d.e.gamma' = p(x(gamma)).BTW. By making p decay fast enough as ||x|| -> infty, I can get thesolutions to be defined for all t, that is I can find gamma: |R -> I,where the open interval I is the domain of the solution x to theoriginal equation.>> Suppose h: |R^n -> S^(n) - {north pole} is>> a map of the n-sphere. I claim:>>> There exists p as above, such that V^ lifts to a vector field on S^n ->> {north pole} that can be extended to a smooth vector field W on S^n>> such that W(north pole) = 0, a not necessarily hyperbolic singularity.>What do you mean by that lift? You want to lift via the differential>of h, so that Dh(V^)=W^? You run into trouble if h is not an>embedding, then.Yes. But that's why I want to multiply V by a *decaying* function p tobegin with. Example for what I mean by lift ( on |R^1 ):V(x) = x^2. h(x): (-Pi,Pi) -> |R, y = h(x) = 2*arctan(x). If x is asolution of x' = V(x), then y = hox satisfies y' = 2*sin^2(y/2) = 1 -cos(y)= V^(x) on the interval (-Pi,Pi). Since V^is 2-Pi periodic Ihave a nice vector field on S^1, which has no singularity at thenorth pole y =Pi. Furthermore, I now have globally defined solutionson S^1. The solutions of x' = V(x), x(0) > 0 reach oo in finite time.But on S^1 they just travel through the north pole and approach x=0from the other side ;-)If I would now multply V by a function p that behaves like 1/x^2 for|x| -> oo p*V would lift to a vector field that has a singularity atthe north pole since the equation x' = 1 lifts to y' = 2*cos^2(y/2) =1 + cos(y) on (-Pi, Pi). Since cos'(Pi) =0, that singularity is nothyperbolic though (a saddle-node). That's what I meant by notnecessarily hyperbolic singularity.So - what I claim is, still, that I can make any vector field on |R^ninto a vector field on S^n with a singularity at the north pole.=== === Subject: : Re: Vector Fields on non-compact manifolds> Hmm, I don't really understand that question. You mean if h is a> diffeomorphism? Ok, sorry, I should have the the beginning of you post more careful ;)But I don't know what to say here, I think the restriction to compactmanifolds is due to the fact that vector fields on non-compactmanifolds may not be globally integrable, and has nothing to do withthe diffeomorphisms... I don't know that book, though...Cheersipp=== === Subject: : Re: Trying to unify axioms.> Hi Gregory L. Hansen, You said,> I think a lot of people invoke Godel's theorem > as a vague analogy for a osophical principle > that's been known for a long time > and that is more easily understood without the analogy.> Any logical system has a set of postulates > that cannot be explained, or derived > from more fundamental assumptions.> If they could be derived,> they would be conclusions, not postulates,> and not the axiomatic basis of a theory. .Lately, I've been failing to mention that > Godel's incompleteness theorem> is a merely strong argument against > trying to unify axioms.Number one, it can't be done.> Number two, it gets too convoluted. You're right, but Goedel's Theorem has nothing to do with it. Since Goedel's Theorem concerns set theory and mathematical axioms, not postulates. Postulates are like propositions, which are known to fit tidly into first order logic. Even more tidly than Goedel's Completeness Theorem concern Logic. Since even Goedel claimed they didn't.=== === Subject: : Re: Trying to unify axioms.> Nothing.You forever spewing ing imbecile, an axiom by definition is> irreducible and unprovable.False.First off, an axiom isn't necessarily irreducable, unless you want toclaim that reducing it makes the axiom not reducable and therefore notan axiom. Sometimes we have an axiom that we find out can be reducedinto other axioms. Does that therefore disprove the axiom, or destroythe usefulness of the axiom, or of taking it as such? No it doesn't.Second, axioms are not unprovable. They CAN be proven. For an exampleof this, consider the law of identity. Can you prove it? If not, thenhow do we even know it's true? We do know it's true, and it IS anaxiom, so that just proves that axioms are not unprovable.(...Starblade Riven Darksquall...)=== === Subject: : Re: Trying to unify axioms.> Nothing.You forever spewing ing imbecile, an axiom by definition is> irreducible and unprovable.False.First off, an axiom isn't necessarily irreducable, unless you want to> claim that reducing it makes the axiom not reducable and therefore not> an axiom. Sometimes we have an axiom that we find out can be reduced> into other axioms. Does that therefore disprove the axiom, or destroy> the usefulness of the axiom, or of taking it as such? No it doesn't.Second, axioms are not unprovable. They CAN be proven. For an example> of this, consider the law of identity. Can you prove it? If not, then> how do we even know it's true? We do know it's true, and it IS an> axiom, so that just proves that axioms are not unprovable.(...Starblade Riven Darksquall...)An axiom is unprovable by definition. Usually, if an axiom can beshown to be provable based on other axioms or postulates, it no longerremains an axiom and is technically a theorem, which by definition isprovable. And of course, nothing is ever provable in and of itself,which is why a single isolated statement can only be true bydefinition.=== === Subject: : Re: Trying to unify axioms.>> Nothing.>>> You forever spewing ing imbecile, an axiom by definition is>> irreducible and unprovable.>False.>First off, an axiom isn't necessarily irreducable, unless you want to>claim that reducing it makes the axiom not reducable and therefore not>an axiom. Sometimes we have an axiom that we find out can be reduced>into other axioms. Does that therefore disprove the axiom, or destroy>the usefulness of the axiom, or of taking it as such? No it doesn't.>Second, axioms are not unprovable. They CAN be proven. For an example>of this, consider the law of identity. Can you prove it? If not, then>how do we even know it's true? We do know it's true, and it IS an>axiom, so that just proves that axioms are not unprovable.>(...Starblade Riven Darksquall...)If an axiom were reducible or proveable, it would be a conclusion, not an axiom. The axioms would become the axioms used to reach that conclusion.-- Usenet is like a herd of performing elephants with diarrhea -- massive,difficult to redirect, awe-inspiring, entertaining, and a source of mind-boggling amounts of excrement when you least expect it. -- Gene Spafford, 1992=== === Subject: : Re: Trying to unify axioms.Nothing. You forever spewing ing imbecile, an axiom by definition is> irreducible and unprovable.False.> Second, axioms are not unprovable. They CAN be proven. For an example> of this, consider the law of identity. Can you prove it? If not, then> how do we even know it's true? We do know it's true, and it IS an> axiom, so that just proves that axioms are not unprovable.Idiot.> (...Starblade Riven Darksquall...) (...Bull Spewing Horse...)-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== === Subject: : Re: Trying to unify axioms.Hi Uncle Al, You mentioned,> an axiom by definition is irreducible and unprovable. .I was talking about Godel's incompleteness theorem,> which is about unifying axioms into a consistent set.You forever spewing ing imbecile, Godel's incompleteness theorem> is the Liar's Paradox writ large. If you have nothing to say, don't.BTW, the barber shaves herself.Sounds as if someone has been applying Occam's razor a bit tooliberally. An axiom is a self-evident truth. An example would beWhere there is life, there is death. This example is irreducibleand it is provable. Without the use of axioms, we would be unable tomake decisions. Sine qua non.As applies to Godel, an axiom need not be transitive or commutative,only reflexive.=== === Subject: : Re: Trying to unify axioms. > Hi Uncle Al, You mentioned,> an axiom by definition is irreducible and unprovable. .> I was talking about Godel's incompleteness theorem,> which is about unifying axioms into a consistent set. You forever spewing ing imbecile, Godel's incompleteness theorem> is the Liar's Paradox writ large. If you have nothing to say, don't. BTW, the barber shaves herself.Sounds as if someone has been applying Occam's razor a bit too> liberally. An axiom is a self-evident truth. An example would be> Where there is life, there is death. This example is irreducible> and it is provable. Without the use of axioms, we would be unable to> make decisions. Sine qua non.As applies to Godel, an axiom need not be transitive or commutative,> only reflexive.Euclid's Fifth Postulate (i.e., Playfair's Axiom) is so obvious it isplain wrong right here on Earth. A gravitation theorem can beconstructed with (metric, Einstein) or without (affine,Weitzenboek)the Equivalence Principle. All derived predictions are identical. Doa pair of local test masses in vacuum fall along parallel andnon-parallel trajectories simultaneously? osophers don't needwaste crocks; scientists do.> Where there is life, there is death.Spermatogonia are immortal. Hitler's showers opened their doors toshow death without life.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net!=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues: | > I know many models whose Heyting structure is far more simplistic: | > than the corresponding Boolean embedding.: |: | Can you name them? Heyting algebras are always infinite, afaik.:: Note that simplistic means excessively simplified.:: Boolean algebras are a special case of Heyting algebras, and there are: plenty of finite Heyting algebras even excluding finite boolean algebras.Yes, additional structure can make these potential infinities collapse. Iwanted to stress that both algebras are finitely generated, however, and itis only through the potential application of axioms over an infinity thatany infinities arise. This, I feel, is why the notion of potentialinfinity is so stressed in concstructive circles, because much of thedistinction in concepts only occurs over this realm, the finite sharing mostproperties with the boolean.: | > And since Heyting algebras have a potential universality hinted by the: | > Curry-Howard isomorphism, why is it so necessary to fall back on the: | > classical approach. I still do not see from where this desire arisesat: | > forcing the ontology of a model...: |: | Perhaps the following may help (and perhaps not :-) ).: |: | A Heyting algebra is a mathematical structure of some kind.: | It's defined as an infinite set, together with some operators and: | relations, that satisfies certain (first-order logic) conditions.: | So even to understand the notion of a Heyting algebra, most: | people require a good intuitive picture of the Tarski semantics: | of first order (classical) predicate logic.: |: | Constructivism has so many variants that, in order to study: | them well, most systems are defined and studied using classical: | means and classical thinking habits. In more vague terms:: | 'reasoning on the meta level is still classical'.:: I don't think pluralism is much of a reason, here. People talk about: constructivism as though there were a lot of constructivists around,: but really it's quite a small enterprise. Nearly all constructive: mathematics (by which I mean, mathematics that's done intentionally: constructively, not merely mathematics that happens to be constructive): that's done is done in Bishop's constructivism plus perhaps a few added: axioms. Markov's school is alleged to have used the assumption that all: functions from N to N are computable, for example.My first intention has not been to get mathematicians to workconstructively, only to teach the logical structure in an order ofincreasing specialisation. Heyting algebras are much more general thanmerely models for constructive mathematics. They also underly a huge numberof models in the sciences, which has been why I have listed so many of them.In fact, with the theory of causal sets, we can attach a natural Heytingstructure to many theories that share the underlying causal structure.: Part of what makes the situation confusing is that the ratio between: ordinary mathematics done constructively and metamathematics about: it is much lower than the ratio between ordinary mathematics done not: bothering with constructivity and the metamathematics of that. It's to: the point that Mathematical Reviews places constructive mathematics under: the 03 (logic) category. You might, for example, wonder whether such: things as Martin-Lof's type theories count as counterexamples to my claim: above. It's possible that somebody out there has been actually doing: mathematics in them, but not as far as I know.The Heyting structure extends to untyped lambda calculi as well, throughCurry-Howard, and CS students regularly study computability and mathematicalframeworks. Obviously, turing-completeness is often a requirement for anynew language proposed, and discovering turing-completeness in c++'s templatemetaprogramming mechanism was a crucial step to modern generativeprogramming paradigms, so much of the focus is on the cut-eliminationoperations and similar reduction theorems. However, illustrating reasoningin terms of the logical structure is not as well taught. I liken this tothe fact that the logical reasoning in quantum mechanics is rarely taught interms of orthomodular lattices, though doing so obviously prevents a lot ofthe conceptual difficulties associated with quantum mechanics.: If you were to count as schools of classical mathematics all the: different nonconstructive formal systems in which one could do: mathematics, the number would hugely exceed the number of constructive: formal systems. Even if you were to restrict yourself just to classical: theories in which _some_ mathematics has actually been done, you can find: set theorists who've taken as their starting points initial assumptions: of varying strengths.I believe this narrows the applicability of the logical structure far toomuch. There is some type of universality in Heyting structures not sharedby the Boolean that allows it to model propositions of a huge variety. Manyof the early foundationalists saw this and did much research in the area(Kleene, Tarski, etc.). With the semantical identification with S4, onefinds a deep identification with notions of possible worlds, descriptions ofnecessity, and the basic modality of science, computability, and proof.Certainly, as well, proof theory today is highly influenced by its Heytingstructure, and I am looking for why this is not so in more fields thatimplicitly have the structure hiding away in their analyses.: The kicker here is that my statement about Bishop constructivism applies: to classical mathematics as well. Most classical mathematics also starts: from Bishop's constructivism plus a few extra axioms. I would argue that: this is evidence that it would be a good choice of metatheory. There's a: basic asymmetry between a theory having extra axioms and one which simply: leaves them out. Having a metatheory which makes more assumptions than: the theory being considered sometimes leads to confusion. There's an odd: little theorem in topos theory whose proof starts something like this:: either every null object is an initial object, or there exists a null: object that isn't an initial object.... This is an application of the: law of excluded middle. It's sort of like allowing the classicality of: the metatheory to bleed into the object theory. And then this is overcome: by considering topos objects in toposes!There is a lot of research in topoi that uses the metatheory of classicalfirst-order predication, just as in many areas of math. But again, that isa choice for metatheory which does not in any way determine the model logicdetermined by the language inside the topoi. For example, numerous theorieswork inside the topos of directed graphs (automata, phylogenetics, etc.),and although a particular analyses may use classical logic to analyses themodel, propositions inside the model still must be evaluated in the logic ofthe topos (which is Heyting).I would like to hear more about your mentioned theorem, though, as itcertainly does sound like an odd beginning.: Many of the problems people believe they have with constructivism also: are less liable to apply while doing metamathematics. The theorems people: try to prove tend to be of low logical complexity, things like If X is: a theorem of S, then Y is a theorem of T. Theorems of that form are: guaranteed to be constructively valid if they are classically valid.: There's a tradition in logic of trying to pare down the principles needed: to prove metamathematical results to the most elementary kind. Making them: also constructive is just a natural next step in this direction.Yes, and of course you have Kleene truth and Goedel decidability and thestructure of metamathematics requiring evaluations of certain kinds ofpropositions in alternate logics for the sake of consistency...: | You may find that awful, but that's the way it is. One reason: | for this is that there is no reason to prefer one form of: | constructivism over another.:: But there is. All else being equal, one should be sparing with one's: assumptions. If I'm a Bishop constructivist, and you are a fan of the: Markov school's work, I can accomodate your results in my scheme of: things by considering them proofs of consequences of Markov's rule and: the recursiveness of all functions from N to N. One should only regard: assumptions as permanent axioms once it's clear that they've become: thoroughly intertwined in your work, and that your work really doesn't: go through without them.This kind of minimality is also why I believe that mathematicians often addvalue to constructive proofs, even when they may be less committal to theentire constructivist ansatz.: | You like Heyting algebras,: | someone else likes some other system. And unlike: | the several formalizations of classical logic, these variants: | are far from being mutually equivalent or translatable.:: Heyting algebras are quite standard. Heyting algebras bear the same: relationship to constructive reasoning as boolean algebras bear to: classical reasoning. They reflect the propositional calculus.:: One *could* consider (as logicians have, being very thorough) logics: intermediate between the usual propositional calculus of constructive: and classical mathematics, but I would be very surprised if anybody: were actually doing mathematics that way. Even on the level of predicate: calculus, is anybody seriously using something in between intuitionist: logic and classical logic?I want to point out that many regularly used models have a model logic withmore structure than just the definition of a Heyting algebra. For example,a topologist will work in a given topology, and this defines a logic withHeyting structure plus the additional structure defining the particulartopology.: | Using classical definitions of the lot allows us at least: | to understand all variants at the same time, and to make: | mutual comparisons.:: Having to use a classical metatheory for constructive reasoning creates: certain confusions. People wind up using techniques for circumventing: the assumption of excluded middle, such as topos theory, Kripke or Beth: models. Crudely speaking, the problem is that one has to keep going around: thinking, The law of excluded middle is actually true, but we're going to: pretend like it isn't. I don't know what problems the reverse is: supposed to create; one is just considering some extra assumptionI think this highlights the differnce between thinking _inside_ the model'slogic, and thinking _about_ the model's logic.: | There is plenty of room for more: | research and other points of view, but as something that: | is supposed to give students a good basis for further: | research, classical logic is still essential, constructivism: | an extra.:: Classical logic is overwhelmingly the popular choice, whether it deserves: it or not. Sort of like Microsoft Windows. It may make pragmatic sense to: defer to it, given the place it has in the world, but it's not a really: persuasive argument for its being inherently better.As a metatheory, yes. But models abound that do not have classical logicinherent to them. Often, like quantum mechanics, this is avoided by notreasoning inside the models logic directly but instead working with thealgebraic translation of the logic. But I believe this is merely becausepeople are not used to thinking of their model's structure in a logical wayand feel much more comfortable in algebraic or other formulations. TheHeyting algebraic structure, though, is itself quite common (though oftenunacknowledged).: | That may serve as an explanation why not many: | have responded very enthousiastic to your pamflet. :-):: I'm sorry to say that there are other reasons why some of us haven't: responded enthusiastically to Galathaea's pamphlet.:: It's probably rather rare for someone to be more sympathetic to the: constructivists than I am. I still feel like I don't know enough about it: to make a really fair evaluation of it, but as you can see I'm more: optimistic about it than people usually are. It seems to me that we need: both to know more constructive mathematics, and to have better reasons for: what we think it's pros and cons are. The lack of experience leaves our: general idea of the relationship between constructive and classical: mathematics resting on too little. On the other hand, weaknesses in: people's general ideas and justifications leads them to chalk certain: things up as advantages of classical reasoning, that certain specific: theorems can be proven, for instance, not taking other subtleties properly: into account, like the fact that they are taken to mean something: different constructively, and that alternative paths to the same practical: end result exist.What I have tried to show is that this is only one of many, many reasons forstudying the underlying logic. The bibliography I gave was only a smallfraction of the work I have seen, but it touches on foundational models innumerous fields.: Nevertheless, I'd just as soon not have someone trying to get people: interested in it in the manner Galathaea has been trying to.:: The excessive cross-posting is a bad sign. The fact alone that one has: had to try to justify it almost always means that one has gone too far.: And excessive cross-posting usually means that someone feels entitled to: grab attention at others' expense.Do you disagree with any of the points I have made in the towards aconstructive education or more focus... posts? Or do you believe that Ihave in some other way violated the constraints of the groups' topicalities?What I have tried to find is people in all of these groups from severaldirections (which I have worked hard to detail), to see what thatcommunities ideas are concerning the education proposal, because it is afractured and disparate community which I felt might share a common goal.I have seen many pleas against the cross posting. None of them have beenvery convincing in my opinion, since I have made it quite clear the pointsof topicality I want to discuss. Often these have been from people whoadmitted they were unfamiliar with the actual work in the topic they wereattempting to defend, and usually they were uniterested in making any effortto learn about it. All of my main posts have worked to make this absolutelyclear.: I would generally advise against being a self-proclaimed liar, even if: this is meant in a humorous way (which I don't know).Its just a fact that many psychologists have verified that most people(percentages close to unity) lie in their life. I've done it. I've writtenstories. Fiction. I like the idea of a fiction lying itself into reality,much like the mythology of galathaea. But also like a scientific model,which never knows itself to be true but seeks justification.I write it in my signature to annoy those who cannot get over it. Its anannoyance they will have to carry with them until they forget my signature(rather transitory unless they keep reading my threads), or until theyaccept at a much more fundamental level the metaphor and fiction thatunderlies their entire perception of the world and methods of modeling it.: Galathaea seems to me to be one of the people Barabara Sher, the career: counsellor, calls a skimmer, as opposed to a diver. A skimmer deals: with more things in less depth; a diver deals in fewer at greater depth.: There's nothing necessarily wrong with being a skimmer, but it seems to me: that there's a kind of effort required to be a skimmer who makes an actual: contribution, rather than just being the dilettant and tossing around: stuff you've heard about. As someone who's more of a diver, I'm not all: that good at advising someone how to be a good skimmer. The advice I'm: tempted to give is basically to be more like a diver: stick to specific: topics long enough to be sure you actually have something in your hands!Be careful here. This is not a very good distinction for how I exploretopics. I am an obsessive reader, going through anywhere from 800 to 1500pages a week, with copious notes, etc. A lot of my research has focused onstructural analyses of topics and foundationalist approaches, and so I havehad to skim huge pantheons of objects to become familiar with the variousterritories. However, my learning model includes going deeper and deeperinto the topics I feel need most exploring to understand the structuralquestions I want to answer. For example, I bohminised Witten's cubicbosonic string model during my analysis of extensions of realist ontologiesof quantum mechanics in order to demonstrate to myself that some of mynotions concerning the isomorphism of Bohm and its relation to quantisationcould carry over to some modern theories. I have done original research inthe study of functors from the category of Poisson manifolds to the categoryof Hilbert spaces. I have derived results on the combinatorial enumerationof certain magma types. Many would not consider these types of calculationsto be that of the skimmer type, and the classification is often used in aderogatory way.: It seemed to me that a lot of the examples, and maybe all of them, of: things whose logic is constructive (whatever that is supposed to mean,: specifically) that we've seen here, are just special cases of the: topological interpretation. This business about perception of the letter: W, for example; you can dress it up in the language of basins of: attraction for the dynamics of your visual cortex or whatever, but it: still boils down to talking about open sets of stimuli that get perceived: as W. Idealizing things a bit, one could say that the complementary: perception, that something is not a W, also corresponds to an open set.: Then since there are borderline cases, perception either as W or not W: doesn't cover all possible cases. It seems to me that Galathaea's: description made it sound rather more mysterious than it is.The region-connection calculus is more developed than that, as are analysesof pattern recognition and the classification problem, so I don't quiteagree here.: But that's a poor argument for dropping the law of excluded middle. It's: more realistic to say that the concept of being a W has a grey area on: its borders. So really one has an argument in favor of fuzzy logic. As far: as I can tell, the jury is still out on fuzzy logic. Such an argument in: favor of fuzzy logic is surely only a motivational argument. We can't say: so easily whether your logic is the right place to introduce awareness of: fuzziness.It is funny then that I've also pointed to the natural Heyting structure offuzzy logic in the literature!=): A lot of the discussion I've stayed out of just because there doesn't seem: to be all that much content in it. Let's please knock it off with the: massive cross-posting and deal more patiently with the various topics one: at a time.Most of the lack of content has been from those spamming their ownnewsgroups, not asking for intelligent discussion, just spamming withinsults and the like. I am always eager to go into more depth as timepermits me, and I have been struggling to give myself more and more time asthe questions turn more and more to a technical nature.: If someone wants to chip in on the mathematical side of constructivism,: try helping me satisfy some of my curiosity. I've had the question of the: degree to which the Jordan-Holder theorem is constructive on the back: burner for a long time. It's easy to see that the fact that any two: decomposition series have a common refinement is constructive. But then: given two decompositions with simple quotients, it's not clear to me that: we should be able to get isomorphisms between them in some order. We can: get a common refinement where not all the quotients are nontrivial, but: we have no way in general to determine whether a quotient group is: trivial. On the other hand, I haven't thought of a good counterexample,: either.So are you questioning the second isomorphism theorem of groups as not beingconstructive?Although I haven't explored this before, I never noticed anything hiding inthere that wasn't extendible to constructive definitions of groups or madeuse of bivalence. I thought it was a simple application of intersectionsand joins, but now you have me intrigued. Would you like to expand on this?: I'm interested in a strong version of the concept of simple group (with: apartness). I'm willing to assume that simplicity holds for the quotient: groups in the following form. If x and y are elements of the group, and: x<>1, then y can be expressed as a product of conjugates of x and x^{-1}.: I think that's a pretty strong assumption, but not crazy. I'm not sure: for instance whether it holds (constructively, of course) for the: classical simple Lie groups.Usually, I find it is more natural to approach problems like this in theopposite direction when looking for constructive deductions. In otherwords, I would define those groups first with a distinguished element x andall elements that can be constructed as products of conjugates (throughother constructed or defined elements) of x and its inverse. This class ofgroups is quite large. Then look at the structure required toconstructively prove elements apart from x that have this conjugationconstruction equal to the entire group as well. For finite groups, ofcourse, this can be carried out to completion constructively. For infinitegroups, of course, its much more difficult, although the relationship beingimplied is finite between all elements. Constructing the elements toconjugate through would get you there, though.I think this approach is very basic to the logic that I would desire beingtaught more. This is the computational approach so inherent toconstruction, that you build the structures you desire to study throughfinitely axiomatising the definitions and deduce constructive consequences,which does oppose the infinite axiomatics underlying certain classicalconstructions. It is very much the difference between bottom-up andtop-down approaches.I know this answer is kind of vague, but I would need to figure out betterwhat the obstructions are to such constructions, and I have not beenthinking of Jordan-Holder and similar theorems in a constructive light yet,so I will need to revisit some of my materials.Thank you for your interest, by the way!-- ===-=-=-=-=-=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues[...]>: One *could* consider (as logicians have, being very thorough) logics>: intermediate between the usual propositional calculus of constructive>: and classical mathematics, but I would be very surprised if anybody>: were actually doing mathematics that way. Even on the level of predicate>: calculus, is anybody seriously using something in between intuitionist>: logic and classical logic?> I want to point out that many regularly used models have a model logic with> more structure than just the definition of a Heyting algebra. For example,> a topologist will work in a given topology, and this defines a logic with> Heyting structure plus the additional structure defining the particular> topology.Which is generally of no interest to this topologist.[...]>: Nevertheless, I'd just as soon not have someone trying to get people>: interested in it in the manner Galathaea has been trying to.>: The excessive cross-posting is a bad sign. The fact alone that one has>: had to try to justify it almost always means that one has gone too far.>: And excessive cross-posting usually means that someone feels entitled to>: grab attention at others' expense.> Do you disagree with any of the points I have made in the towards a> constructive education or more focus... posts? Or do you believe that I> have in some other way violated the constraints of the groups' topicalities?You obviously have in respect of sci.lang. You alsoobviously either violated one of the first principles ofnetiquette, namely, that one should become familiar with theactual content and customary practices of a newsgroup beforeposting to it, or deliberately posted to sci.lang somethingthat you should have known was inappropriate. Quite clearlyyou *do* feel entitled to grab attention at others' expense.[...]> I have seen many pleas against the cross posting. Which is prima facie evidence that it was inappropriate. Itis ultimately the members of a newsgroup who determine whatis appropriate, not some document. If I want to readphysics, I'll go to sci.physics; I don't want it clutteringup sci.lang. If I want to read mathematics, I'll go tosci.math. If I want to read logic, I'll go to sci.logic.If I want to read osophy, I'm ill.> None of them have been> very convincing in my opinion, Which is largely irrelevant.[...]>: A lot of the discussion I've stayed out of just because there doesn't seem>: to be all that much content in it. Let's please knock it off with the>: massive cross-posting and deal more patiently with the various topics one>: at a time.> Most of the lack of content has been from those spamming their own> newsgroups, not asking for intelligent discussion, just spamming with> insults and the like.Which again is a very good indication that your content waswidely considered inappropriate. Like it or not, manynewsgroups are communities. Outsiders are not necessarilyunwelcome, but outsiders who barge in and presume to lecturefrom a pedestal are likely to get the rough reception thatthey've earned.Bluntly, you're a rude, arrogant bastard with the socialintelligence of a pet rock. On top of that you write someof the flabbiest, most turgid prose that it's been mymisfortune to read anywhere, let alone on Usenet, andexhibit several of the familiar stigmata of the Usenet crankor monomaniac. If you don't like your reception, mend yourmanners.[...]=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues: > I want to point out that many regularly used models have a model logicwith: > more structure than just the definition of a Heyting algebra. Forexample,: > a topologist will work in a given topology, and this defines a logicwith: > Heyting structure plus the additional structure defining the particular: > topology.:: Which is generally of no interest to this topologist.Whereas this topologist finds that much of the work by other topologists,much like work of the quantum physicists, merely re-expresses the logicalstructure in other constructs. Often, I find that it is not a question ofinterest, but more of just being unaware that the manipulations being donefit any kind of logical formalism. My findings are, of course, not final!: > Do you disagree with any of the points I have made in the towards a: > constructive education or more focus... posts? Or do you believethat I: > have in some other way violated the constraints of the groups'topicalities?:: You obviously have in respect of sci.lang. You also: obviously either violated one of the first principles of: netiquette, namely, that one should become familiar with the: actual content and customary practices of a newsgroup before: posting to it, or deliberately posted to sci.lang something: that you should have known was inappropriate. Quite clearly: you *do* feel entitled to grab attention at others' expense.Expense? I took the time to write several long posts concerning a topicrelevant to all newsgroups posted to. I am earnestly interested in aneducational deficiency which has been quite well demonstrated by many postsof others in this thread. I came out attacking no one. I was upfront aboutmy agenda. And I quarantined my ramblings to 2 intended (and 2 incidentaldue to newsserver problems) threads, all of which are not forced upon anynewsgroup reader and may easily be avoided. The netiquette problems arefrom those who come attacking, giving no constructive material in which todiscuss. That now includes you, mister Scott.As for sci.lang, obviously there are problems in understanding reasoning ontopological trees (and linguistic phylogeny) that have been evidencedseveral times recently in that forum. Plus the whole cognitive origins oflanguage, semiotics and natural language models, etc. features of myexposition that I've been willing to discuss in depth make your statementpatently false.Oh, did you miss the thread on modality in language as well?Yeah, I'm reading...Are you?: [...]:: > I have seen many pleas against the cross posting.:: Which is prima facie evidence that it was inappropriate. It: is ultimately the members of a newsgroup who determine what: is appropriate, not some document. If I want to read: physics, I'll go to sci.physics; I don't want it cluttering: up sci.lang. If I want to read mathematics, I'll go to: sci.math. If I want to read logic, I'll go to sci.logic.: If I want to read osophy, I'm ill.No, it's evidence of nothing of the kind! None of those pleas has everdescribed in any way how I have violated the topicality of their newsgroup.Even you didn't make any such description above. You just accuse and fightyour alpha games like you are the arbiter of truth and justice.Your desire to want to avoid certain discussions of an interdisciplinarynature is easily avoided by ignoring threads you find distasteful. I canwalk you through that procedure if you are having any difficulties.: > None of them have been: > very convincing in my opinion,:: Which is largely irrelevant.Unfortunately, that is quite relevant. Convincing me is the only waysomeone is going to get me to stop posting.: > Most of the lack of content has been from those spamming their own: > newsgroups, not asking for intelligent discussion, just spamming with: > insults and the like.:: Which again is a very good indication that your content was: widely considered inappropriate. Like it or not, many: newsgroups are communities. Outsiders are not necessarily: unwelcome, but outsiders who barge in and presume to lecture: from a pedestal are likely to get the rough reception that: they've earned.No, its an indication that there are quite a lot of jerks out there who,when faced with a topic they do not understand and do not want tounderstand, find solace in insults. I do like the fact that newsgroups arecommunities, particularly that they are communities of wide ranges of viewsabout the topics they discuss. There are certainly members, such asyourself mister Scott, who dislike the fact that others may begin adiscussion confident of the knowledge that they have such a right, butunfortunately you are in the wrong and I am in the right. And playing yourpower games, with their complete absence of any rational points, justillustrates to me that you recognise your complete lack of power in thiscircumstance.: Bluntly, you're a rude, arrogant bastard with the social: intelligence of a pet rock. On top of that you write some: of the flabbiest, most turgid prose that it's been my: misfortune to read anywhere, let alone on Usenet, and: exhibit several of the familiar stigmata of the Usenet crank: or monomaniac. If you don't like your reception, mend your: manners.I have never been rude to anyone who was not first rude to me, and then onlyenough to play the alpha game they initiated to its proper closure. I don'tseek contentless arguments; it is others who feel inclined to provide mewith such. I am a bastard; that is true. I was born with an unmarriedmother. I can be quite humble when speaking to others who engage inrational critique or otherwise educate me of my errors. I see you borrowedthe use of turgidity from the other newsgroup spammer. Feeling uncreativetoday?the writing process. I have tried to be exciting yet remain technicallycorrect. Sure, I've failed in places, but that is part of the process of myfinding out how to express these ideas.What I see is that there is a group of sterile, uncreative, alpha posturingjerks who enjoy attempting to crush any form of expression that does notconform to their galatea of elegance, and they joyfully make no contributionto the technical discussion to keep their spamming focused.Only part of my reception I do not like.Mend your manners, mister Scott.-- ===-=-=-=-=-=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continues > I came out attacking no one. I was upfront about> my agenda. And I quarantined my ramblings to 2 intended (and 2 incidental> due to newsserver problems) threads, all of which are not forced upon any> newsgroup reader and may easily be avoided. The netiquette problems are> from those who come attacking, giving no constructive material in which to> discuss. That now includes you, mister Scott.All right, gaga-la-t.8et.8ee, here, fished out of google=== === Subject: : Re: New human migration map based upon mtDNAs :: > Groups L1, L2, M & N can be likened to language groups.::: Since languages have been demonstrated to evolve at wildly: different rates, that can only be::: (1) a chance coincidence (helped by cherry picking the: language groups):: or:: (2) a self-delusion, like Ruhlen and Cavalli-Sforza's two: trees, linguistic vs genetic:: or:: (3) evidence that the mtDNA of speakers of any given language: evolves at the same rate as that language.Jacques, you are using these wildly different rates as your sword tocutand slash what you do not understand. ^^^^^^^^^^^^^^^^^^^^^^^^^^ [1] They in no way imply these threechoices of yours, and I _do_ accept that rates are quite far fromconstantacross linguistic evolution. Different rates can easily point to thesameor similar tree topologies. This is phylogenetic bioinformatics 101. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [2]And there is a hypothesis underlying this. That languages used bycommunities as humans migrate tend to maintain similarities frompreviouslanguages used by the groups, and that genetic isolation can often implysome amount of linguistic isolation. It's not pseudoscience (like Iwould ^^^^^^^^^^^^characterise your theory on the three possibilities), it is the samesort of^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [3]hypothesis that working archaeologists use to understand a language usedina find and hypothesise on where the people came from.: > Therefore, Jacques, there can be such a map. If sides 1 - 12contain: > too much theory for you, then ignore them.:: If the three points above contain too much logic for: you, then ignore them and hop onto the bandwagon, which: looks about to join a nice gravy train.I can offer some books on phylogenetic reconstruction if the ideas areso ^^^^^^^^^^^^^^^^^^^difficult you need to resort to name calling.^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [4]: I myself favour (3) above, as it is sufficiently ridiculous: to get published in Nature. And it goes a long way: towards proving the innateness of grammar and whatnot.: Can't lose with such a hand.I figure you'd like 3. Its the culmination of your pseudoscience. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ [5]The five segments highlighted above are five mealy-mouthedrepetitions of the same plain English words you are full of. None of them follows a demonstration, not even anargument, nor a piece of evidence, and none leads to what follows.And you, miserable mealy-mouthed two-faced drivel-dribblinglying snake, have the cheek of complaining that> netiquette problems are> from those who come attacking, giving no constructive material in which to> discuss. That now includes you, mister Scott.=== === Subject: : Re: the anticlassicalist }{ ii: the spectre continuesJacques, your first response to my first post (which was purely a scientificdescription of phylogenetic techniques with no attacks to anyone) in thatthread was:===-=-=-=-=-Yes, I know that. Like the drunk on all fours.Comes a policeman.Hello there, Sir, in a bit of trouble are we?I'm jusht looking for me housh keys.Oh dear. And whereabouts have you lost them?On t'other shide of the shtreetEr... and shouldn't you be looking there?It'sh dark there. There'sh light here.===-=-=-=-=-(...and you _were_ wrong in the places I pointed out.)Your response to my post in Out of Anatolia contained the lovely little:===-=-=-=-=-What does Heyting algebra have to contribute tounmasking such hocus pocus merchants? Let me guess:that is not what it is for. Perhaps it even helpscamouflage the snake oil as pure olive oil?===-=-=-=-=-Again you demonstrate my point about who is picking the fights, in brilliantclarity, for the small little section of usenet unfortunate enough to haveto partake in your ugliness.Littering the world with your hatred does not make you look strong.-- ===-=-=-=-=-=== === Subject: : Re: Analysis help>if f is differentiable on an interval containing zero and if>lim(x->0)f'(x) = L>then L = f'(0).>>Hint: What does the Mean Value Theorem tell you about >>(f(x) - f(0))/x when x is near 0?there exists a c in (x,0) or (0,x) such that> f'(c) = (f(x) - f(0))/x so there is a sequence of c_n's in the interval (and not equal to 0)> that approach 0, so (f(c_n) - f(0))/c_n -> f'(0)> and thus f'(c_n) -> f'(0) > so, since lim(x->0) f'(x) = L, f'(c_n) -> L, therefore L = f'(0).is that correct?Yes, although I would rather put it as a delta-epsilon argument.> Based on the exercise's location in the book (understanding analysis> by abbott) i don't think he intends you to use the MVT. so do you see> another way to do it?No, and I would be surprised if it turned out that there is anotherapproach which can be left as an exercise.Best regards,Jose Carlos Santos=== === Subject: : Re: Analysis help if f is differentiable on an interval containing zero and if>lim(x->0)f'(x) = L>then L = f'(0). Hint: What does the Mean Value Theorem tell you about> (f(x) - f(0))/x when x is near 0?>>That f'(0) exists and equals L.>>So the premise can be reduced to>>f differentiable on (a,b)0, for some a < 0 < bReally? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.>Yes, the (top line) of the premise can be reduced to ...> if f differentiable on (a,b)0, for some a < 0 < b and if> lim(x->0)f'(x) = L> then L = f'(0).> is true? No it's not - I _gave_ you a counterexample.> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you> claiming that f is not differentiable on (a,b){0}> for some a < 0 < b, that lim(x->0)f'(x) does not> equal 0, or that f'(0) = 0?Ok, it's a counterexample. f'(0) may not exist.Mean value theorem requires f' over all of (a,b).=== === Subject: : Re: Analysis help>>if f is differentiable on an interval containing zero and if>>lim(x->0)f'(x) = L>>then L = f'(0).>> Hint: What does the Mean Value Theorem tell you about>> (f(x) - f(0))/x when x is near 0?>That f'(0) exists and equals L.So the premise can be reduced to>f differentiable on (a,b)0, for some a < 0 < b Really? Let f(x) = -1 for x < 0, f(x) = 1 for x > 0.>>Yes, the (top line) of the premise can be reduced to ...>> if f differentiable on (a,b)0, for some a < 0 < b and if>> lim(x->0)f'(x) = L>> then L = f'(0).>> is true? No it's not - I _gave_ you a counterexample.>> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you>> claiming that f is not differentiable on (a,b){0}>> for some a < 0 < b, that lim(x->0)f'(x) does not>> equal 0, or that f'(0) = 0?>Ok, it's a counterexample. f'(0) may not exist.>Mean value theorem requires f' over all of (a,b).Actually there _is_ a very interesting/usefulversion of the result that does not require thatyou assume that f'(0) exists. But it does requirea hypothesis that does not appear in your version...=== === Subject: : Re: Analysis help if f is differentiable on an interval containing zero and if>>lim(x->0)f'(x) = L>>then L = f'(0).>> Hint: What does the Mean Value Theorem tell you about>> (f(x) - f(0))/x when x is near 0?>That f'(0) exists and equals L.if f differentiable on (a,b)0, for some a < 0 < b and if>> lim(x->0)f'(x) = L>> then L = f'(0).is true? No it's not - I _gave_ you a counterexample.Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you>> claiming that f is not differentiable on (a,b){0}>> for some a < 0 < b, that lim(x->0)f'(x) does not>> equal 0, or that f'(0) = 0?>Ok, it's a counterexample. f'(0) may not exist.>Mean value theorem requires f' over all of (a,b).> Actually there _is_ a very interesting/useful> version of the result that does not require that> you assume that f'(0) exists. But it does require> a hypothesis that does not appear in your version...Continuity of f over (a,b)=== === Subject: : Re: Analysis help>> if f differentiable on (a,b)0, for some a < 0 < b and if>> lim(x->0)f'(x) = L>> then L = f'(0).>> is true? No it's not - I _gave_ you a counterexample.>> Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you>> claiming that f is not differentiable on (a,b){0}>> for some a < 0 < b, that lim(x->0)f'(x) does not>> equal 0, or that f'(0) = 0?>Ok, it's a counterexample. f'(0) may not exist.>Mean value theorem requires f' over all of (a,b).The mean value theorem was used in this case on [0,b] and [a,0],and those don't require f'(0) to exist. But what else does themean value theorem require?Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: Analysis help=== === Subject: : Re: Analysis help > if f differentiable on (a,b)0, for some a < 0 < b and if > lim(x->0)f'(x) = L > then L = f'(0). > is true? No it's not - I _gave_ you a counterexample. > Let f(x) = -1 for x < 0, f(x) = 1 for x > 0. Are you > claiming that f is not differentiable on (a,b){0} > for some a < 0 < b, that lim(x->0)f'(x) does not > equal 0, or that f'(0) = 0? >>Ok, it's a counterexample. f'(0) may not exist. >>Mean value theorem requires f' over all of (a,b). >The mean value theorem was used in this case on [0,b] and [a,0], >and those don't require f'(0) to exist. But what else does the >mean value theorem require?f continuous on [a,b]----=== === Subject: : Re: Goading Godel> Ok ... we are not obliged to fall in obeisance to Hawking anymore than>> we are obliged to uncritically respect everything Maxwell or Einstein>> might have said: doubly so because this paper is in fact a>> transcribed informal lecture, where the freedom to vaporize is much>> higher, and triply so because this is Usenet. :-) Hawking is just invoking Godel's theorem as a vague analogy or>metaphor, the way thousands of others have done, in physics, theology,>psychology, and so on.That's how it looks to me. Godel's Theorem expressly applies to > infinite systems, and this does not necessarily include the universe. > There are generalisations that apply to finite cases, though, and he may > have been using the term loosely - depends on the audience.I don't think he is right anyway; in mathematics we would say > arithmetica is undecideable if we prove that we can't show that a > certain very complex statement is true or false. On the other hand, > physicists would not feel the project had failed if almost everything > flipped neutrino spins occasionally did not exist...That sounds about right ... except the undecidable question might notoccasionally did not exist. Or not even _that_ concrete ... maybeform sometimes do not exist can itself be proven...The infinite system argument seems weaker to me though ... the logicalsystem induced by the a set of physical laws is infinite even if theuniverse described by them is not: and even in a finite universe wealways find one potentially infinite direction -- that of forwardevolution in time.At heart there only seems to be one complexity argument: that theinfinite logical evolution of systems growing from finite seeds cannot in general be further circumscribed by finite means (that's thekind of sentence which either will bring a nod of recognition, or anuncharitable feeling I'm talking gibberish). The question is whetherin physics we even care about the pathological behavior of infinitesystems: we are content with the finite seeds.=== === Subject: : Re: Goading Godel>> That's how it looks to me. Godel's Theorem expressly applies to >> infinite systems, and this does not necessarily include the universe. >> There are generalisations that apply to finite cases, though, and he may >> have been using the term loosely - depends on the audience.>>> I don't think he is right anyway; in mathematics we would say >> arithmetica is undecideable if we prove that we can't show that a >> certain very complex statement is true or false. On the other hand, >> physicists would not feel the project had failed if almost everything >> flipped neutrino spins occasionally did not exist...>That sounds about right ... except the undecidable question might not>occasionally did not exist. Or not even _that_ concrete ... maybe>form sometimes do not exist can itself be proven...>The infinite system argument seems weaker to me though ... the logical>system induced by the a set of physical laws is infinite even if the>universe described by them is not: and even in a finite universe we>always find one potentially infinite direction -- that of forward>evolution in time.While getting to the 'end' of time may be difficult, time itself could be finite in a finite universe. There may be a configuration of maximum entropy, with no states having an image of it in their history. >At heart there only seems to be one complexity argument: that the>infinite logical evolution of systems growing from finite seeds can>not in general be further circumscribed by finite means (that's the>kind of sentence which either will bring a nod of recognition, or an>uncharitable feeling I'm talking gibberish). The question is whetherI'm not sure yet ;-)- Gerry Quinn>in physics we even care about the pathological behavior of infinite>systems: we are content with the finite seeds.=== === Subject: : Re: Goading GodelOk ... we are not obliged to fall in obeisance to Hawking anymore than> we are obliged to uncritically respect everything Maxwell or Einstein> might have said: doubly so because this paper is in fact a> transcribed informal lecture, where the freedom to vaporize is much> higher, and triply so because this is Usenet. :-) Hawking is just invoking Godel's theorem as a vague analogy or> metaphor, the way thousands of others have done, in physics, theology,> psychology, and so on.This may be true, but there are non-vague uses of Godel-like theoremsthat could apply to physical systems. If one assumes that any of thelogical theories of causation (like J.L. Mackie's) reflect physicalreality, then one is commited to the fact that the behaviour of aphysical system is not only unpredictable, but undecidable (in theTuring machine sense).So consider J.L. Mackie's INUS conditional analysis. C is an INUScondition for E if and only if there are sentences X and Y such that((C&X) v Y ) <-> E, but not C -> E or X -> E, where Y is a possiblyempty disjunction of the form (C1&X1)v. . . v(Cn&Xn).Suppose we take a list of INUS conditionals, and copy each of the C's,X's, and E's to the columns of a Turing table, the rows of a Turingtable, and the E's to the relevant positions. The operation of thisTuring machine will represent the behaviour of the correspondingcausal system. The well-known correspondence between Turing machinesand recursive functions, together with Rice's theorem, take care ofthe rest.'cid 'ooh=== === Subject: : Re: Goading GodelFor somebody so huffy you are incredibly careless about attribution(you snipped all of it, as if everybody out here were one amorphousrespondent).> You incorrectly assumed that > Godel's incompleteness theorem,> only applies to systems that implicitly or explicitly> contain within them arithmetic .Godel's theorem does exactly that.> Godel's incompleteness theorem can be applied to physics:No he didn't. I saw Professor Hawking give this talk live. What he> said was that he was beginning to think that there might not be a> complete physical model of reality with a finite number of axioms. > Godel's theorem, or any other mathematical theorem, would be of little> use at the present time; we don't have any certain hard-and-set rules> about the behavior of the universe, nor will we ever. That's the> nature of physics. There are no nonexistance theorems in real life.[snip a bunch of pop-science drivel]Funny ... that's about the way I thought of Hawking's analogy: perhapshis reputation argues otherwise -- but your take on it certainlydoesn't contradict this idea. Except, I suppose, that because of hisheroic reputation he is incapable of uttering pop-science dri....er... vague generalities? > Most of the above passage is utter nonsense. Your concept of> infinities needs some *serious* refining as it's so vague that it's> basically meaningless. For example, if you want to show that an> arbitrary infinite set can't be mapped to a finite set, just consider> the set of all real numbers, which cannot be put into correspondence> with the set of integers. Since every finite set can be considered as> a subset of the integers (there exists a bijective mapping), this> shows that you can't map ALL infinite sets to finite representations.You missed what I was talking about. I said that sometimes aninfinite number of cases could be circumscribed by a finite statement-- or words to that effect; which is by no means the same as saying aninfinite number of cases could be put in correspondence with a finitenumber of cases.I suggested that if we pose the test question: structure of the universe is to structure of the integers as a complete physical theory is to , that most nearly correct answer might be (b) a small set of axiomsspecifying and generating the structure of the integers, rather than(c) a complete listing of true theorems about the integers or (d) amachine capable of generating the list of all true theorems about theintegers or (a) the law of indecipherable operations (the choiceput in for people who choose the most academic sounding answer).In other words, taking electromagnetism as a test case, we take ourbasic theory to be complete when we have a small set of equationsgoverning the behavior of any possible solution, not when we havelisted all possible theorems about electromagetic fields. Godel'swork would seem to have more to say about the problems inherent in thelatter project than in the former, which was finished when Maxwell'sink dried.> I don't much like Hawking.This is an outright ridiculous statement from someone who's clearly> not intelligent enough to even consider such matters. I'm nowhere> NEAR competent to talk about M-theory, and from the nonsense you're> spouting it looks like you'll never even be near my modest amount of> intelligence. Aside from being a heroic figure for overcoming his> disabilities, Hawking is without a doubt one of the most important> physicists working today, and you are in no place to criticise him. > And furthermore, young man, if you haven't got anything nice (or> meaningful) to say, don't say anything.That indignant young man, together with the your utterly missing myarguments, pulls the fangs out of your acerbity, old fellow. Isuppose injecting an irrelevant and naked personal-opinion-singularityinto my post was not my finest moment, but I don't suppose Hawkingcares much what I think. What I'm not fond of is his publicpop-science guru persona: I think in this role his physicaldisabilities are an asset rather than otherwise.=== === Subject: : Re: Motivation for e; was : Re: Transcendental numbers other than pi and e?In sci.math, Stephen J. Herschkorn<40399C5B.7030104@ rutcor.rutgers.edu>:>> >[Btw, e arises when solving dy/dx = y, and in other ways.]>>Yes, I knew that, but that's still not a geometrically tangible idea,>>the way pi is. > Here are three ways to motivate the concept of e.> 1) Compound interest: The effective annual yield for a nominal annual > rate r compounded n times annually is (1 + r/n)^n. As n grows > arbitrariliy large, the effective yield approaches e^r.> 2) With some effort, for positive a, one can show that if f(x) = > a^x, then f is differentiable everywhere and f'(x) is proportional > to f(x). e is the unique value of a which renders as unity the > constant of proportionality. Geometircally tangible? Try drawing the > graph of such a function.> 3) Letting f(x) = log_a (x) (logariithm to the base a) for a > 0, > f'(x) is proportional to 1/x. e is the unique value of a which > renders as unity the constant of proportionality.4) The area under 1/t from 1 to x turns out to have many odd properties; in fact it looks a lot like a logarithm. Turns out it is a logarithm, with base e. Obviously this is 3), but coming from the reverse direction. If one takes L(a) = integral (t = 1, a) dt/t, then L(ab) = L(a) + integral(t = a, ab) dt/t = L(a) + integral(u = 1, b) (dt/du) (du/(au)) (where t = au) = L(a) + integral(u = 1, b) (adu/au) = L(a) + integral(u = 1, b) (du/u) = L(a) + L(b). Similar methods show L(1/a) = -L(a). Of course L(1) = 0.5) If we hypothesize a function f(x) = a0 + a1*x + a2*x^2 + ... + an*x^n + ..., the usual Maclaurin series, and f'(x) = f(x) and f(0) = 1, then we get the infinite series of identities a0 = 1 (the initial condition) a0 = a1 a1 = 2a2 a2 = 3a3 ... a{n-1} = n(an) ... and ultimately the absolutely convergent series we all know and love, exp(x) = 1 + x + x^2/2! + x^3/3!+ x^4/4! + ... + x^n/n! + ... exp(a+b) can probably be shown, with a *lot* of schlogging, to be exp(a)exp(b), and ultimately e = exp(1). (One of the tools that might be of assistance is the binomial expansion.)Still not horribly geometric, I know, but somehow it's a very natural(ahem) result.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Motivation for e; was : Re: Transcendental numbers other than pi and e?> In sci.math, Stephen J. Herschkorn> :>>[Btw, e arises when solving dy/dx = y, and in other ways.]>Yes, I knew that, but that's still not a geometrically tangible idea,>the way pi is. >>Here are three ways to motivate the concept of e.> 4) The area under 1/t from 1 to x turns out to have many odd properties;> in fact it looks a lot like a logarithm. Turns out it is a> logarithm, with base e.[ as another poster said, the x value for which the area is 1 is x = e]> Still not horribly geometric, I know, but somehow it's a very natural> (ahem) result.My contribution to some geometric meaning of e :Let a right triangle A(0),B(0),C(0), A(0)B(0) = 1 horizontal,B(0)C(0) = 1 vertical and a distance d = A(0)B(0)/n = 1/n C1 /| / | / | / | C0 + | /| | // | | / / | | / / | | / / | | / / | | /_____/______|____|A0 A1 B0 B1Translate segment A(0)B(0) by distance d to segment A(1)B(1)A(1)C(0) intersects the vertical from B(1) at C(1)This gives a new triangle A(1),B(1),C(1) with heighth(1) = B(1)C(1)Repeat the same step to get A(i+1),B(i+1),C(i+1) from A(i),B(i),C(i):Translate segment A(i)B(i) by d to A(i+1)B(i+1).A(i+1)C(i) intersects vertical from B(i+1) at C(i+1).After n steps, A(n)=B(0) because d=1/n and h(n)=B(n)C(n).When n increases, h(n) -> eProof :Thales gives h(i+1)/h(i)=1/(1-1/n)that is h(n) = h(n)/h(0) = ( 1/(1-1/n) )^nLn(h(n)) = n*Ln( 1/(1-1/n) ) ~ n*Ln(1+1/n) ~ n*1/n -> 1and h(n) -> e when n -> oo-- ippe(chephip at free dot fr)=== === Subject: : Re: how to finding eigenvalues and eigen vectors>hi >its hasber>i belong to the field of computer science.>i am working on face recognition systems.>my problem is that i have a matrix of size>4096*4096 and i have to find its eigenvalues and eigen vetors.[...]As others have said, that's a fairly big calculation. But I very muchdoubt you need to do it. Probably, you have n greyscale images, each 64by 64 pixels, and n is a lot less than 4096. Probably, you only need tofind the eigenvalues and eigenvectors of a n by n matrix which is real,symmetric, and has all eigenvalues real and non-negative. Probably, youare only interested in the largest 10 to 100 eigenvalues and theirassociated eigenvectors. All this makes your real problem much easierthan the one you asked about, but it is still non-trivial.I suggest you Google on eigenfaces and pca, or post tosci.stat.math.-- Graham Joneshttp://www.visiv.co.ukEmails to graham@visiv.co.uk may be deleted as spamPlease add a j just before the @ to ensure delivery=== === Subject: : Re: Question about Incompleteness (Godel)(Note: I'm replying after 5 posts in this thread even though GoogleGroups is only showing 2 at the moment)that answers things pretty clearly. Although to be absolutely sure Iunderstand the situation I need clarification of these two points:1. Is it a trivial corollary that the set of undecidable statements,is not only non-empty, but infinite (or even uncountable) (In some'interesting' way - obviously 'G & n=n' for all n is an infiniteundecidable set)2. What results are there about minimum complexity requirements fora statement to be potentially undecidable in a consistent formalsystem. (For instance, trivial statements like x+y > x for all x>0are certainly not undecidable in PA, and this is because (as Iunderstand) such a statement is expressable in a much weaker formalsystem which is consistent and complete.)I was really looking for. It led to me to the incredible 'Goodsteintheorem' and its proof (which is impossible in PA)=== === Subject: : Re: Question about Incompleteness (Godel)> There are plenty of natural/ordinary statements that are known to> be formally undecidable in given systems. Most famously, perhaps, the> Axiom of Choice is undecidable from the usual axioms of> Zermelo-Fraenkel, as are the Continuum Hypothesis (from both ZF and> ZF+Choice) and the Generalized Continuum Hypothesis (also from both ZF> and ZF+Choice). Yes. It's also known that these imply no new Pi_1-truths, i.e. truths of form Ax(P(x)=0), where P(x) is prim. rec. I don't know if this is best we can do, i.e. whether GCH and AC are actually conservative over ZFC for more than just Pi_1-sentences. Although both AC and GCH are interesting examples of mathematically undecidable sentences, they are not undecidable because of G.9adel's theorem - in any sense I'm aware of. > I believe certain results from Ramsey Theory are> formally undecidable in Peano Arithmetic.Yes. There is the Paris-Harrington theorem, which is a slightly tweaked finite form of the infinite Ramsey's Theorem (about existence of homogenous colourings). There are various variations, such as the Kirby-Paris theorem and what you have. There is also Friedman's Finite Form of the Kruskal Tree theorem, which is unprovable in theories much stronger than PA - in fact, it's provably beyond means of predicative mathematics.> But the important point is that while the Goedel statement G may seem> contrived, it is a perfectly valid 'ordinary' mathematical statement> about numbers. Is it? There seems to be a very clear sense in which it is not: it would not have arisen in the course of the study of number theory or related fields.If you just mean that there's nothing magically metamathematical about the statement, you're right, of course.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-osophicus=== === Subject: : Re: Question about Incompleteness (Godel)> Yes. It's also known that these imply no new Pi_1-truths, i.e. truths of > form Ax(P(x)=0), where P(x) is prim. rec. I don't know if this is best > we can do, i.e. whether GCH and AC are actually conservative over ZFC > for more than just Pi_1-sentences. G.9adel's interpretation of GCH and ZFC in L tells you (as firstpointed out, I believe, by Kreisel) that they have no new arithmeticalconsequences. Much stronger (optimal) results along these lines wereproved in the sixties - unfortunately I forget the reference.=== === Subject: : Re: Question about Incompleteness (Godel)> G.9adel's interpretation of GCH and ZFC in L tells you (as first> pointed out, I believe, by Kreisel) that they have no new arithmetical> consequences. Ah. I was thinking of something like that. Would you happen to have a refernce (or if it's just a simple illumination insight, sketch a proof or hint at the right direction)?-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-osophicus=== === Subject: : Re: Question about Incompleteness (Godel)> Ah. I was thinking of something like that. Would you happen to have a > refernce (or if it's just a simple illumination insight, sketch a proof > or hint at the right direction)? The arithmetic part is simple: for any arithmetical theorem A ofZF+V=L, the relativization of A to L is provable in ZF, but thisrelativization is equivalent in ZF to A itself, since the hereditarilyfinite sets in L are the hereditarily finite sets. Indeed, Godel'srelative consistency result concerning GCH and AC is much betterformulated as a proof that GCH and AC are conservative over ZF withrespect to theorems in finite mathematics. In an old FOM posting, Steve Simpson lists various results aboutconservative extensions, including ZFC + V=L is conservative over ZF for Pi^1_2 sentences. ZFC + GCH (actually ZFC + V=L(r) for some real r) is conservative over ZF for Pi^1_3 sentences. ZFC + GCH is conservative over ZFC (actually over ZF + a well ordering of the reals) for Pi^2_1 sentences.=== === Subject: : Re: Question about Incompleteness (Godel)>Goedel's proof shows that there are statements that are formally>undecidable. It says nothing about what form they will take, onlythat>they exist. If the set of formally undecidable statements isnonempty,>that means that statements may be there.Ok, great, so a minimal existence result. How about a response to mymore interesting supplementary questions:1. Is it a trivial corollary that this set has infinite or evenuncountable cardinality?2. What are the known results regarding minimal complexity for asentence to be undecidable? (If there are none, then I have a newtheorem, an undecidable sentence requires >2 symbols to state it,since the minimal meaningful sentence using 2 symbols is ~A, where Ais an axiom of the system,and ~A is false, hence decidable)And btw, since this is a fairly informal newsgroup, you should all tryto post more interesting replies than I've seen to my question. To behonest, I provided the best post by mentioning the Goodstein theorem,which admittedly followed from researching A. Magidin's initial reply.But as far as I can tell, Goodstein's theorem is the only reallycompelling demonstration of Godel as applied to Peano arithmetic.Restating results about AC and CH is lazy, boring, uninteresting and apretty good demonstration of why mathematics is losing its appeal indemocracies worldwide (big drop in quality of university applicants inthe UK anyway)JG=== === Subject: : Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana.>>Goedel's proof shows that there are statements that are formally>>undecidable. It says nothing about what form they will take, only>that>>they exist. If the set of formally undecidable statements is>nonempty,>>that means that statements may be there.>Ok, great, so a minimal existence result. How about a response to my>more interesting supplementary questions:My post was not a reply to that post, was it? So why should my replycontain answers to questions that you had not asked yet?>1. Is it a trivial corollary that this set has infinite or even>uncountable cardinality?It is trivial that it cannot be uncountable unless the (formal)language itself is uncountable: every formula is a finite sequence ofsymbols from the alphabet, hence there are at most countably manyformulas, well formed or not. I don't see why it would be a trivial corollary (unless you accepttrivial conjunctions of undecidable statements and other statements,which you explicitly excluded in your previous post). [.snip.]>And btw, since this is a fairly informal newsgroup, you should all try>to post more interesting replies than I've seen to my question. Limosnero, y con garrote...[1]>To be>honest, I provided the best post by mentioning the Goodstein theorem,Oh, well, then you should not have posted; you could have accomplishedeverything just by talking to yourself in the mirror, no? You gave thebest reply, after all.>which admittedly followed from researching A. Magidin's initial reply.>But as far as I can tell, Goodstein's theorem is the only really>compelling demonstration of Godel as applied to Peano arithmetic.>Restating results about AC and CH is lazy, boring, uninteresting and a>pretty good demonstration of why mathematics is losing its appeal in>democracies worldwide (big drop in quality of university applicants in>the UK anyway).Sigh. You must be a real hit at office hours...You asked a question. The level of your prior knowledge was not givenby the question, and could only be guessed. You asked explicitly ifthere were any 'ordinary' mathematical statements that wereundecidable. If you were so familiar with the independence resultsabout AC and CH, then your question was, to paraphrase you, lazy, boring,uninteresting, and a pretty good demonstration of why the big drop inquality of applicants: they don't think before asking, or they do notknow how to ask the question they really mean. You were already awareof statements that were independent, if you wanted to exclude those asanswers, you did a poor job of phrasing your question. To now complainbecause people answered the question you asked rather than thequestion you ->meant<- is rather annoying.As for restating results, unless you were expecting someone topresent you with a brand new proof of a brand new statement'sindependence, that was all you were goin to get in a fairly informalnewsgroup.-- ============================================================== ========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============[1] Beggar with a bat; asking for favors, and then complaining because those who freely give their time to help do not do so to the degree you would wish them to.=== === Subject: : Re: Question about Incompleteness (Godel)>>Something that has confused me for a while and I would appreciate any>>clarification:>>Godel's incompleteness proof essentially demonstrates that if you have>>a consistent formal system defining at least simple arithmetic then>>some statements will be undecidable in the axioms of that system. Now,>>in his proof, the undecidable statement constructed is the>>meta-statement G = 'G is not demonstrable'. The undecidable statement constructed is a statement about numbers; it> states there there does not exist a number which stands in a> well-defined relation with a given number y. This statement, which is> a statement purely about numbers, ->may be interpreted<- in the> metguage as stating that the formal statement with Goedel number y> has no proof in the system. But the statement itself is a statement in> the object language.>>How do we then conclude>>that 'standard' mathematical statements, like 'Every even number is>>prime or the sum of two primes' might be undecidable? Doesn't Godel>>only show that 'meta'-statements like G exist whch are undecidable?Goedel's proof shows that there are statements that are formally> undecidable. It says nothing about what form they will take, only that> they exist. If the set of formally undecidable statements is nonempty,> that means that statements may be there. >>Are there are proofs of incompleteness which construct (or exhibit>>directly) an 'ordinary' undecidable mathematical statement, rather>>than one like G, which seems like a meta-statement, of little>>'practical' interest.There are plenty of natural/ordinary statements that are known to> be formally undecidable in given systems. Most famously, perhaps, the> Axiom of Choice is undecidable from the usual axioms of> Zermelo-Fraenkel, as are the Continuum Hypothesis (from both ZF and> ZF+Choice) and the Generalized Continuum Hypothesis (also from both ZF> and ZF+Choice). I believe certain results from Ramsey Theory are> formally undecidable in Peano Arithmetic.But the important point is that while the Goedel statement G may seem> contrived, it is a perfectly valid 'ordinary' mathematical statement> about numbers.In particular, it asserts that a certain Diaphontine system has nosolutions. So, all true ...... but there is still a _big_ difference between the situationswrt AC, CH, GCH (and, on more elementary ground, the Euclidean parallel postulate) and the Goedel statement G in some theory T: The well-known undecidable (relative to [case-specific mumble])statements of set theory and geometry are undecidable becausethe base theories have models in which these statemenets are true, and (other) models in which they are false. Goedel-typestatements are true but not provable in _every_ model of the theory in which they are expressed.=== === Subject: : Re: Question about Incompleteness (Godel) Adjunct Assistant Professor at the University of Montana. [.snip.]>So, all true ...>... but there is still a _big_ difference between the situations>wrt AC, CH, GCH (and, on more elementary ground, the Euclidean >parallel postulate) and the Goedel statement G in some theory T: >The well-known undecidable (relative to [case-specific mumble])>statements of set theory and geometry are undecidable because>the base theories have models in which these statemenets are >true, and (other) models in which they are false. Goedel-type>statements are true but not provable in _every_ model of the >theory in which they are expressed.Huh? No, there are models in which the given Goedel statement is true,and models in which it is false; one may adjoin it or its negation tothe axioms (provided certain conditions are met) and obtain aconsistent system. Yes, the new system has a ->new<- Goedel statementwhich is formally undecidable in that system, but the old one is nottrue but not provable. -- ============================================================== ========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ================== === Subject: : Re: Question about Incompleteness (Godel)> Yes, the new system has a ->new<- Goedel statement> which is formally undecidable in that system, but the old one is not> true but not provable. For the usual cases, the Godel sentence for T is true, given that Tis consistent, but unprovable in T. Or in other words, given theequivalence in T of G and (a formalization of) T is consistent,that T is consistent is true, if T is consistent, but unprovable in T.=== === Subject: : Re: Question about Incompleteness (Godel)>How do we then conclude>that 'standard' mathematical statements, like 'Every even number is>prime or the sum of two primes' might be undecidable? Doesn't Godel>only show that 'meta'-statements like G exist whch are undecidable?As Goedel did it, the final undecidable statement is not a'meta-statement'. We now know it can have the form: Here is an explicit polynomial in many variables with integer coefficients: [long-ish polynomial written here] When natural numbers are plugged in for the variables, the result is never zero.Concrete but (as stated) uninteresting.=== === Subject: : Re: Question about Incompleteness (Godel) Goedel-type statements are true but not provable in _every_ model of the > theory in which they are expressed.What does it mean for a statement to be unprovable in every model of the theory in which they are expressed?By completeness theorem, if a sentence G is not provable from theory a (consistent) T there is a model in which T holds but G does not and if G is independent, there is a model in which both T and G hold as well as a model in which T holds but G does not. The G.9adel statement says that there is no natural number which codes the proof of the statement itself, and since it's independent of the theory T (provided T satisfies certain conditions, which you can slightly weaken by using the Rosser sentence) there is a model of T in which there is a non-standard natural number coding a proof of G and in addition there are models in which no such standard or non-standard number exists.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-osophicus=== === Subject: : Re: How big can a manifold be?Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a> neighborhood homeomorphic to R^n?and later:I'm mainly interested in 1-dimentional manifolds anyway.For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. I reduced this claim to the following set-theoretic statement, butcannot prove it from the first glance:Consider a linearly ordered by inclusion family of sets, all of themhaving the cardinality of the continuum. Then their union has thecardinality of the continuum as well.Simeon=== === Subject: : Re: How big can a manifold be?> Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a> neighborhood homeomorphic to R^n?and later:> I'm mainly interested in 1-dimentional manifolds anyway.For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. I reduced this claim to the following set-theoretic statement, but> cannot prove it from the first glance:Consider a linearly ordered by inclusion family of sets, all of them> having the cardinality of the continuum. Then their union has the> cardinality of the continuum as well. Because you can't prove it at first glance. You have to take a second glance and assume the Axiom of Choice is true. Then the proof is more trivial than the equivalent intersection of sets of the power of the continuum. Since in order to prove that it is true you have to *assume* that there is another set, A, which also has the power of the continuum, and for every set T in the ordered sets: T is a subset of A.Simeon=== === Subject: : Re: How big can a manifold be?>> I reduced this claim to the following set-theoretic statement, but>>cannot prove it from the first glance:>>Consider a linearly ordered by inclusion family of sets, all of them>>having the cardinality of the continuum. Then their union has the>>cardinality of the continuum as well.> Because you can't prove it at first glance. You have to > take a second glance and assume the Axiom of Choice is true. > Then the proof is more trivial than the equivalent intersection> of sets of the power of the continuum. Since in order to > prove that it is true you have to *assume* that there is another set, A,> which also has the power of the continuum, > and for every set T in the ordered sets: T is a subset of A.I don't follow:The original question may be interpreted as follows:Given an ordinal lambda, and a functional set f with domainlambda, alpha < beta < lambda => f(alpha) subset f(beta)and f(alpha) is of the cardinality of the continuum for eachalpha, the union of all f(alpha) is of the cardinality of thecontinuum.wlog, we can assume that if alpha < beta, then f(alpha) is notequal to f(beta). But that still allows lambda to be as big as2^c.=== === Subject: : Re: How big can a manifold be?> Is there any limit to the cardinality of a connected n-manifold, > that is a connected Hausdorff space every point of which has a> neighborhood homeomorphic to R^n?and later:> I'm mainly interested in 1-dimentional manifolds anyway.For 1-manifolds, the answer to your question appears to be that the > cardinality of the continuum is the limit. I reduced this claim to the following set-theoretic statement, but> cannot prove it from the first glance:Consider a linearly ordered by inclusion family of sets, all of them> having the cardinality of the continuum. Then their union has the> cardinality of the continuum as well.Unfortunately, this set-theoretic statement is false. Let k be any infinite cardinal. The ordinals of cardinality k are linearly ordered by inclusion, but their union has cardinality greater than k.=== === Subject: : Re: How big can a manifold be?>>> Is there any limit to the cardinality of a connected n-manifold, >> that is a connected Hausdorff space every point of which has a>> neighborhood homeomorphic to R^n?>>> and later:>>> I'm mainly interested in 1-dimentional manifolds anyway.>>> For 1-manifolds, the answer to your question appears to be that the >> cardinality of the continuum is the limit. >>> I reduced this claim to the following set-theoretic statement, but>> cannot prove it from the first glance:>>> Consider a linearly ordered by inclusion family of sets, all of them>> having the cardinality of the continuum. Then their union has the>> cardinality of the continuum as well.> Unfortunately, this set-theoretic statement is false. Let k be any > infinite cardinal. The ordinals of cardinality k are linearly ordered > by inclusion, but their union has cardinality greater than k.That works if c is an ordinal (for example, if we assume either AC or CH).What if it isn't?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : Re: How big can a manifold be?>> Consider a linearly ordered by inclusion family of sets, all of them>> having the cardinality of the continuum. Then their union has the>> cardinality of the continuum as well.Unfortunately, this set-theoretic statement is false. Let k be any > infinite cardinal. The ordinals of cardinality k are linearly ordered > by inclusion, but their union has cardinality greater than k.That works if c is an ordinal (for example, if we assume either AC or CH).> What if it isn't?Of course, I was assuming AC (and various other axioms of ZFC too).But I think you can do it in ZF, if you really want to.Let b be the least ordinal of cardinality not less than or equal to c. Then b is a cardinal. Let R be the reals. Consider the set X = { R cup a | a in b }. The elements of X are linearly ordered by inclusion, and all have cardinality c. The union of X is a superset of b. So cardinality of the union of X is greater than or equal to b, and therefore not equal to c.I don't know what I'm doing without AC, so I could easily have made a mistake, but it looks OK. It was much easier in ZFC.=== === Subject: : Re: How big can a manifold be?> Consider a linearly ordered by inclusion family of sets, all of them> having the cardinality of the continuum. Then their union has the> cardinality of the continuum as well.>>> Unfortunately, this set-theoretic statement is false. Let k be any >> infinite cardinal. The ordinals of cardinality k are linearly ordered >> by inclusion, but their union has cardinality greater than k.>> That works if c is an ordinal (for example, if we assume either AC or CH).>> What if it isn't?> Of course, I was assuming AC (and various other axioms of ZFC too).> But I think you can do it in ZF, if you really want to.> Let b be the least ordinal of cardinality not less than or equal to c. > Then b is a cardinal. Let R be the reals. Consider the set > X = { R cup a | a in b }. The elements of X are linearly ordered by > inclusion, and all have cardinality c. The union of X is a superset of > b. So cardinality of the union of X is greater than or equal to b, and > therefore not equal to c.> I don't know what I'm doing without AC, so I could easily have made a > mistake, but it looks OK. It was much easier in ZFC.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.=== === Subject: : simple surfaces and tangentsappreciate any help.1. Let U={(u1,u2) in R^2| -pi appreciate any help.> 1. Let U={(u1,u2) in R^2| -pi and define:> x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))> Prove that x is a simple surface.> Suppose:> x(u1,u2)=((2+cos(u1))*cos(u2),(2+cos(u1))*sin(u2),sin(u1))> =y((v1,v2))=((2+cos(v1))*cos(v2),(2+cos(v1))*sin(v2),sin(v1)) that should be x(u1,u2) = x(v1,v2)and for injectivity this equality should impy (u1,u2) = (v1,v2)> then (2+cos(u1))*cos(u2)=(2+cos(v1))*cos(v2)> 2cos(u2)+cos(u1)cos(u2)=2cos(v2)+cos(v1)cos(v2)> 2[cos(u2)-cos(v2)]+cos(u1)cos(u2)-cos(v1)cos(v2)=0> Stuck here..how do you show's its injective?Show that u1 = v1 and u2 = v2For example: from x(u1,u2) = x(v1,v2)you get 3 equations: ( (2+cos(u1))*cos(u2) , (2+cos(u1))*sin(u2) , sin(u1) ) = ( (2+cos(v1))*cos(v2) , (2+cos(v1))*sin(v2) , sin(v1) )or[1] { (2+cos(u1))*cos(u2) = (2+cos(v1))*cos(v2)[2] { (2+cos(u1))*sin(u2) = (2+cos(v1))*sin(v2)[3] { sin(u1) = sin(v1)square [1] and [2] and add to eliminate u2 and v2 anduse [3] to find that cos(u1) = cos(v1).Together with [3] this already implies u1 = v1.Put this information in [1] and [2] to find cos(u2) = cos(v2) sin(u2) = sin(v2)implying u2 = v2.Carefully write down each step.> Also when i compute the cross product of the derivative> I get a messy expression, what's the usual argument or> techinque in proving such expressions are 1-1?> For the cross product I get> (-cos^2(u1)*cos(u2),-cos^2(u1)sin(u2),-sin(u1)cos(u1)cos^2(u2) > -sin(u1)cos(u1)*sin(u2)*cos(u2))If you do it a bit more carefully, you get this: (2+cos(u1)) * ( cos(u1)cos(u2) , -cos(u1)sin(u2), -sin(u2) )Verify this, and then calculate the length of this vector for someu1 and u2.Can it be zero? Do you now understand why they tookthe number 2? What would happen if they had taken 1/2?> how do you compute x1,x2 and n as functions of u1 and u2???> 2. Prove that a(s) is a straight line iff all its tangent> lines are parallel.> Stuck on this one.=== === Subject: : Re: Transcendental numbers other than pi and e?>Yes, I knew that, but that's still not a geometrically tangible idea,>the way pi is. e is the continous 2. Justification follows.Take the difference equation: x_(k+1) - x_k = x_k<=> x_(k+1) = 2 x_k<=> x_k = x_0 * 2^kNow consider the equivalent differential equation: f'(x) = f(x)<=> f(x) = C * e^xThe number e arrises when we take a discrete form and use a limit toextend it to the continuum. In here the process takes us from 2 -> e.-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: Transcendental numbers other than pi and e?>>>[Btw, e arises when solving dy/dx = y, and in other ways.]>>Yes, I knew that, but that's still not a geometrically tangible idea,>the way pi is.Still not geometrically, but does it mean more to you when we express this by saying that f(x) = e^x is an eigenfunction of the differential operator corresponding to the eigenvalue 1? That is, there are infinite-dimensional vector spaces; the infinitely differentiable functions on R is one such space. Even in infinite-dimesional spaces, linear transformations (differentiation here) have eigenvalues and eigenvectors.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: Transcendental numbers other than pi and e? charset=Windows-1252Are there other known, established transcendental numbers besides pi> and e? If so, I'd be curious to know what they are called, and if> they can be characterized in a way that would be clear to my> mathematically simple brain. (For example, I've never found it easy> to understand were e comes from, but the idea that pi is the ratio> between a circle's circumfirence and diameter is clear enough.)It seems that those which arise naturally are few, or at least only a> few are _known_ to be transcendental.Liouville constructed an infinity of transcendental numbers using> continued fractions. Cantor proved that almost all real numbers are> transcendental. I just happen to be studying parts of the book by Allouche & Shallit(Automatic Sequences), who prove a very interesting theorem attributed to Bohmer. Paraphrasing ... Let t be an irrational in the unit interval, and let t_n = floor((n+1)t) - floor(nt) for n = 1,2,3,... Let b be an integer >= 2. Then sum(t_n / b**n, n>=1) is transcendental.That is, the characteristic word of an irrational in (0,1) (which always consists of 0's and 1's only) can be interpreted as the base-b representation of a transcendental number. So for each baseb >= 2, the theorem neatly gives uncountably many transcendentals, one for each irrational in (0,1).(The Fibonacci word 1011010110110..., or Rabbit Sequence, which is the characteristic word of the golden mean's fractional part, is the best-known example. Just this single word represents a countable infinity of transcendentals, corresponding to b = 2,3,4,...)--r.e.s.=== === Subject: : Re: Transcendental numbers other than pi and e? charset=Windows-1252> That is, the characteristic word of an irrational in (0,1) (which > always consists of 0's and 1's only) can be interpreted as the > base-b representation of a transcendental number. So for each base> b >= 2, the theorem neatly gives uncountably many transcendentals, > one for each irrational in (0,1).(The Fibonacci word 1011010110110..., or Rabbit Sequence, which is > the characteristic word of the golden mean's fractional part, is the > best-known example. Just this single word represents a countable > infinity of transcendentals, corresponding to b = 2,3,4,...)I forgot to add something for Steven O., who might like thisgeometrical interpretation of these transcendental numbers ...Take a piece of graph paper ruled into squares, and define x- and y-axes. Through the origin draw a staight line whose slopes is irrational, with 0 < s < 1. Now for the part of the line in the upper righthand quadrant, note the points where the line intersects the graph paper rulings, as you move away from the origin. These occur on a sequence of vertical and horizontal rulings, and the 0/1 sequence you get by coding vertical = 0 and horizontal = 1 is the characteristic word of t = s/(s+1).So pick any irrational slope in (0,1) and find the corresponding 0/1 sequence, say a,b,c,d,... Then for any integer base b > =2, say binary, the number represented by 0.abcd... is transcendental.--r.e.s.=== === Subject: : Re: Pythagoreian triangles and gaussian integersThis time I got really stuck on a certain problem. It goes like this:A natural number H can be the hypothenus? in a pythagoreian triangle> if it contains any of the prime factors of the form 4n+1.> If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the product of all> primes on the form 4n+1 then the number N of triangles with the> hypothenus h os given by the formula N=(P-1)/2 where> P=(2k_1+1)(2k_2+1)...(2k_n+1)> Now here's the actual assignment:> Calculate all natural numbers h<100 so that there are more than one> pythagoreian triangle with the hypthenus h and calculate all> pythagoreian triangles that correspond to these numbers hWhere are you getting stuck? Do you not know what a prime factor is? Do you know what a prime factor is, but not know what it means > to be of the form 4n + 1? Would it help if I ran through an example? Let H = 975. > Then H = Kh where K = 3 and h = (5^2)(13) > so n = 2, k_1 = 2, and k_2 = 1 > so P = (5)(3) = 15 > so N = (15 - 1) / 2 = 7 > so there should be 7 pythagorian triangles with hypotenuse h = (5^2)(13).Or maybe you are getting stuck on finding the actual triangles? > The key here is that surely somewhere you've been told that > if h = (m^2 + n^2)r with m > n then it is the hypotenuse of a triangle > whose sides are 2mnr and (m^2 - n^2)r. So all you have to do is find > all the ways of writing (5^2)(13) as (m^2 + n^2)r. > It helps to note that 5 = 2^2 + 1^2 and 13 = 3^2 + 2^2. > With some algebraic trickery (actually, this is where the Gaussian > integers of your subject header come in handy!) this leads to > 5^2 = 4^2 + 3^2 and (5)(13) = 8^2 + 1^2 = 7^2 + 4^2 and similarly > for (5^2)(13). Then don't forget that r can be 1, 5, 13, 25, or 65, > and you should be getting there.Actually there are a few things that I would really appreciate ifsomeone could help me understand.For a start, I'm really confused on how to find the triangles thatcorrespond to a certain hypothenuse. For e.g. a problem is to find thenumber h which corresponds to more than 4 triangles. Finding thisnumber isn't a problem (just put in N=5,6 etc until you find a numberof P which isn't a prime (N=7)). But then how do I find the triangles.I know you went on to explain that but I just can't relate that to thefacts that I'm given in the assignement. It says thatA gaussian integer z is pythagoreian if it's of the form N*w^2 whereN is a natural number and w^2 is neither all real or all imaginary.It then goes on to say that this gives a perfect way of findingtriangles. But how does it? I mean the number that I find isn't agaussian integer so how should I do? I realy need help on this oneThnxPierre=== === Subject: : Re: Pythagoreian triangles and gaussian integers>Calculate all natural numbers h<100 so that there are more than one>pythagoreian triangle with the hypthenus h and calculate all>pythagoreian triangles that correspond to these numbers h[...]> of P which isn't a prime (N=7)). But then how do I find the triangles.> I know you went on to explain that but I just can't relate that to the> facts that I'm given in the assignement. It says that> A gaussian integer z is pythagoreian if it's of the form N*w^2 where> N is a natural number and w^2 is neither all real or all imaginary.> It then goes on to say that this gives a perfect way of finding> triangles. But how does it? I mean the number that I find isn't a> gaussian integer so how should I do? I realy need help on this oneDepending on your skill on Gaussian integers (Z[i])2 methods for finding all triangles with a given hypotenusThat is all ways of x^2 + y^2 = h^2 for a given h.Both start from prime decomposition of h^21) elementary method :Use the Fibonacci relation(a^2 + b^2)*(c^2 + d^2) = (a*c +/- b*d)^2 + (a*d -/+ b*c)^2example with 5=1^2+2^2 and 13=2^2+3^2we derive5*13 = (1*2+2*3)^2 + (1*3-2*2)^2 = 8^2 + 1^2 and (1*2-2*3)^2 + (1*3+2*2)^2 = 4^2 + 7^2add each prime factor one at a time in the product to getthe list of all sums of squares = h^2.2) using Z[i]Use 5=(2+i)*(2-i), 13=(3+2*i)*(3-2*i) and s.o.So you have the decomposition in prime factors in Z[i] of h^2h^2 = x^2 + y^2 = (x+i*y)*(x-i*y)Hence x+i*y is a selected subset of prime factors from h^2Just choose any possible subset and calculate x+i*y for every subsetIn the example of h=5*13h^2 = (2+i)^2 * (2-i)^2 * (3+2*i)^2 * (3-2*i)^2x+i*y = e * (2+i)^r * (2-i)^(2-r) * (3+2*i)^s * (3-2*i)^(2-s)You let vary e (e=1,-1,i,-i is a unit), r=0,1,2, s=0,1,2.(the exponents r, 2-r because the two factors x+i*y and x-i*y have to be conjugate)In both methods, without special care, you get every solution several times.Choose carefully the subsets so you get the solutions only once.In our case r=0,s=0 is the same as r=2,s=2 for example (conjugate)You have to choose only half the values for r and s (and fix e) :r=0, s=(0,1,2)r=1, s=(0,1)That is 5 decompositions of 65^2=422563^2+16^239^2+52^233^2+56^225^2+60^265^2+0^2Note that this method gives the trivial decomposition too h^2 = h^2 + 0^2I have a java script using Z[i] to list all decompositions ofz = x^2 + y^2 (french, but doesn't matter for numbers)http://chephip.free.fr/exe001.html-- ippe(chephip at free dot fr)=== === Subject: : Re: Pythagoreian triangles and gaussian integersThis time I got really stuck on a certain problem. It goes like this:A natural number H can be the hypothenus? in a pythagoreian triangle> if it contains any of the prime factors of the form 4n+1.> If H=K*h where h=(p_1^k_1)(p_2^k_2)...p_n^k_n) is the product of all> primes on the form 4n+1 then the number N of triangles with the> hypothenus h os given by the formula N=(P-1)/2 where> P=(2k_1+1)(2k_2+1)...(2k_n+1)> Now here's the actual assignment:> Calculate all natural numbers h<100 so that there are more than one> pythagoreian triangle with the hypthenus h and calculate all> pythagoreian triangles that correspond to these numbers hWhere are you getting stuck? Do you not know what a prime factor is? Do you know what a prime factor is, but not know what it means > to be of the form 4n + 1? Would it help if I ran through an example? Let H = 975. > Then H = Kh where K = 3 and h = (5^2)(13) > so n = 2, k_1 = 2, and k_2 = 1 > so P = (5)(3) = 15 > so N = (15 - 1) / 2 = 7 > so there should be 7 pythagorian triangles with hypotenuse h = (5^2)(13).Or maybe you are getting stuck on finding the actual triangles? > The key here is that surely somewhere you've been told that > if h = (m^2 + n^2)r with m > n then it is the hypotenuse of a triangle > whose sides are 2mnr and (m^2 - n^2)r. So all you have to do is find > all the ways of writing (5^2)(13) as (m^2 + n^2)r. > It helps to note that 5 = 2^2 + 1^2 and 13 = 3^2 + 2^2. > With some algebraic trickery (actually, this is where the Gaussian > integers of your subject header come in handy!) this leads to > 5^2 = 4^2 + 3^2 and (5)(13) = 8^2 + 1^2 = 7^2 + 4^2 and similarly > for (5^2)(13). Then don't forget that r can be 1, 5, 13, 25, or 65, > and you should be getting there.thnx for the help, cleared some things out. Actually now my problem isto find those h<100. I got 5^2=25, 5*13=65 and 5*17=85. But thereshould be 2 more h<100.=== === Subject: : Re: Simple idea, mathematics and common-sense> ab = 1, > where again 'a' and 'b' are *irrational* the mathematicians have a> label for 'b' which is algebraic integer, but 'a' cannot be an> algebraic integer.> So they don't have a label for it!> You know how important naming is with human beings, and it makes it> that much harder for me to explain these ideas without a label.> I've named numbers like 'a' objectsTake b = sqrt(2). 'b' is *irrational*. Thus a=1/sqrt(2) is an object. - William HughesI meant with with 8a + b odd.Why not try to consider the idea versus just being a smart-k?If you were concerned with the truth, you might have noted that with> 8a+b even, it's clear that 'a' is a fraction like 1/2, like with your> example where it is 1/sqrt(2), versus being a smart-k by tossing> out that special case as if it were definitive.Now then, do you understand why your example doesn't apply for 8a+b> odd?Do you understand how your post is a smart-ky one, which invites> the kind of response that I just gave it?> James HarrisThe current conditions arei) ab=1ii) 8a + b is an odd integeriii) a and b are irrationalthen a is an object.Take 8a + b = 5thus a1= (5 + sqrt(-7))/16 and a2= (5 - sqrt(-7))/16 are objects.Thus a1*a2=1/8 is an object. - William Hughes=== === Subject: : Re: Simple idea, mathematics and common-senseIn sci.logic, James Harris<3c65f87.0402231427.392918ac@ posting.google.com>:>>>> ab = 1, >>> where again 'a' and 'b' are *irrational* the mathematicians have a>> label for 'b' which is algebraic integer, but 'a' cannot be an>> algebraic integer.>>> So they don't have a label for it!>>> You know how important naming is with human beings, and it makes it>> that much harder for me to explain these ideas without a label.>>> I've named numbers like 'a' objects>>>>> Take b = sqrt(2). 'b' is *irrational*. Thus a=1/sqrt(2) is an object.>>> - William Hughes> I meant with with 8a + b odd.[A1] You now might have the requirement that 8a + b = 2n + 1, n an integer. This, together with ab = 1, results in the quadratic8a + 1/a = 2n + 1or8a^2 - (2n + 1)a + 1 = 0which has as rootsa = ( (2n + 1) +/- sqrt(4n^2 + 4n + 1 - 32) ) / 16If n = 4 or n = -5, a is rational. No other rational aare possible. (The proof is left as an exercise.)[A2] It is also possible you meant 8a + b is an algebraic integer not divisible by 2, i.e., the quantity V = 4a + b/2 is not an algebraic integer. The problem is, if b is an algebraic integer, then 8a is an algebraic integer, and the problem divides into two subparts. [i] 4a is an algebraic integer. Therefore b/2 is not an algebraic integer. [ii] 4a is not an algebraic integer. I'm not sure what conclusions can be drawn here although I strongly suspect b/2 = V - 4a is an algebraic integer. However, I can't prove it without a bit of schlogging.> Why not try to consider the idea versus just being a smart-k?> If you were concerned with the truth, you might have noted that with> 8a+b even, it's clear that 'a' is a fraction like 1/2,[B] a = ( (2n) +/- sqrt(4n^2 - 32) ) / 16 = (n +/- sqrt(n^2 - 8)) / 8For n = 3 or -3, a is rational.> like with your> example where it is 1/sqrt(2), versus being a smart-k by tossing> out that special case as if it were definitive.> Now then, do you understand why your example doesn't apply for 8a+b> odd?> Do you understand how your post is a smart-ky one, which invites> the kind of response that I just gave it?> James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: I got low score on math test, please advise me and take a look> don't you think that such rule should have a> provision stateing that students need some> performance feedback prior to the drop deadline?> How do you know that he didn't get any performance> feedback before the deadline?> i believe he said that somewhere. now, can you answer> my question - perhaps in a direct manner this time?> Fine. RULES ARE RULES.> do you always follow rules blind-folded, or are there ocassions> where you wonder if some rules are well-founded?> Non-sequitur, since I agree completely with this rule. i am not particularly questioning whether you agree with that one> rule. my specific question to you, is whether you agree with the idea> that a withdrawal deadline should have a provision stating that> students need some performance feedback prior to the deadline. a simple 'yes' or 'no' will suffice.I'm not playing your lawyer games. I'm not on trial, and you're not> cross-examining me. Good night.i am definitely NOT playing any games. i am simply trying to determinewhy you objected to my post - that's all.> Doug=== === Subject: : divergent seriesS_1^inf [1-(ln(n))/n]^n = infhow to prove this?=== === Subject: : Re: divergent series>S_1^inf [1-(ln(n))/n]^n = inf>how to prove this?I assume you mean oo --- ln(n) n > ( 1 - ----- ) [1] --- n n=1When a < 1/2, log(1-a) > -a-a^2. Therefore, letting a = ln(n)/n,multiplying by n, and exponentiating, we get ln(n) n ( 1 - ----- ) > exp(-ln(n)-ln(n)^2/n) n 1 1 > - - [2] n 2For all n >= 1, ln(n)/n < 1/2, therefore, we have [2] for all n >= 1.Since each term of [1] is greater than 1/(2n), compare with the harmonicseries.Rob Johnson take out the trash before replying=== === Subject: : Re: divergent seriesea194576.0402240531.76b49fd1@posting.google.com...> S_1^inf [1-(ln(n))/n]^n = inf> how to prove this?ln(1-ln(n)/n) ~ -ln(n)/n , etc ... the general term of your series isequivalent to 1/n (and sum(1/n) diverges)=== === Subject: : Re: divergent series> S_1^inf [1-(ln(n))/n]^n = inf> how to prove this?S = sum, I guess.Limit comparison with 1/n=== === Subject: : Re: JSH> There is an internet and use-net phenomenon known as a troll. JSH is> a troll. A troll makes deliberately provocative posts to get people> to reply so he/she can make lots more posts, etc., wasting everyone's> time. This is what JSH does and he even gets angry when people don't> reply to his posts, further evidence of his troll status. There is> also a name for people who compulsively reply to trolls: troll fodder.Not to sound rude, but do you think you're the first person to piece thistogether?And, after countless other similar attempts have failed over the years, doyou think that *you* are the one who will succeed in ending the madness?=== === Subject: : Re: Cyclotomic fields by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OEHG812529;z=exp(2 pi i / n), where n is a power of an odd prime p. Then there is a rational prime, namely p, which has a unique prime ideal of Q(z) containing it. Is this possible in any other case? That is, does there exist w = exp(2 pi i / m) where m is an odd integer which is not a power of a prime and such that there is a prime q which is contained in a unique prime ideal of Q(w)?Just wondering,Mark>> Does anyone know of an example of a cyclotomic extension of the >> rationals which is cyclic, but not obtained by adjoining a primitive >> p^m root of unity for some prime p?>By cyclotomic extension, do you mean Q(exp(2 pi i / n)) for some n? >Then each s in the Galois group takes z to z^j, where z = exp(2 pi i / n)>and j is relatively prime to n. So we can identify the Galois group >with the multiplicative group of the integers (mod n). It's known >that this group is cyclic if and only if n is a power of an odd prime, >twice a power of an odd prime, 2, or 4. So the only hope is to take >n twice a power of an odd prime. But this gives the same extension as >taking n to be the power of the odd prime.>-- >Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)=== === Subject: : Re: Cyclotomic fields> z=exp(2 pi i / n), where n is a power of an odd prime p. Then there is> a rational prime, namely p, which has a unique prime ideal of Q(z)> containing it. That's not true. If q is a prime which is a primitive root modulo p^k,then q is inert in K = Q(exp(2 pi i/p^k)), that is q generates a primeideal in the ring of integers of K. That's infinitely many q for each p.> Is this possible in any other case? That is, does there> exist w = exp(2 pi i / m) where m is an odd integer which is not a power> of a prime and such that there is a prime q which is contained in a unique> prime ideal of Q(w)? Just wondering,I suppose what you really want is for the ideal generated by p to be a non-trivial power of a prime ideal. For this it is necessary that p ramifies, which is the case iff p | m. Also you need p to be inert inQ(exp(2 pi i / m')) where m = p^k m' and m' is prime to p. Now one casewhere this happens is where q = m' is prime and p is a primitive rootmodulo q. In particular if m = pq where p is a primitive root modulo qbut q is not a primitive root modulo p, then p is the only rational primewhich is a nontrivial power of a prime ideal. For instance, take p = 3 andq = 7. Then 3 is a square of a prime ideal in the integers of Q(exp(2pi i/21)) but 7 is the sixth power of a product of two distinctprime ideals.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OEFA111128;>>My,question has been resolved neatly and Laconicaly,>>within two lines:>>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>>Im(e^[iPi]) = Im(-1+i[0]) = 0>>Since there is acceptance to this,>>any more saying has no value.>Except to point out that it means that your original conclusion that >exp(i pi) = 0 was not validly drawn from what preceded it.>David McAnallyYes because I was not either happy with:e^[ipi]+1=0 as given in my referenced book:.....However ,since the transcendental numbers formulated by Liouville and G. Cantor,were somehow TECHNICAL,Charles Hermite proved that number e ,the basis of the natural logarithms ,is not algebraic That was important since it was being proved that it wastranscendental ,and related to pi via Euler's relationshipe^[iPi]+1 =0[ Eyler gave the general formula : e^[ix]=cosx+isinx in 1748 , gave e and its value 2.718 ,and new since 1728 the relationship: e^[iPi]+1 =0 ].......Panagiotis Stefanides=== === Subject: : Re: e is transcendental>My,question has been resolved neatly and Laconicaly,>within two lines:>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>Im(e^[iPi]) = Im(-1+i[0]) = 0>Since there is acceptance to this,>any more saying has no value.>>Except to point out that it means that your original conclusion that >>exp(i pi) = 0 was not validly drawn from what preceded it.>>David McAnally>Yes because I was not either happy with:>e^[ipi]+1=0 as given in my referenced book:Well, that is a matter of taste, rather than an objective observation of whether the result is true or false. In the field of complex numbers, exp(i pi)+1 = 0 is true, whether one likes the equality or not.>.....However ,since the transcendental numbers formulated by Liouville>and G. Cantor,were somehow TECHNICAL, Charles Hermite proved that>number e ,the basis of the natural logarithms ,is not algebraic That was>important since it was being proved that it was transcendental ,and>related to pi via Euler's relationship>e^[iPi]+1 =0Euler's identity is distinct from any consideration of whether or not e istranscendental. Spivak gives in his book, Calculus, a proof of thetranscendence of e which is completely unrelated to Euler's identity. Onthe other hand, the proof that pi is transcendental is very closelytrelated to the fact that e^[i pi]+1 = 0.>[ Eyler gave the general formula :> e^[ix]=cosx+isinx in 1748 ,> gave e and its value 2.718 ,and new since 1728> the relationship:> e^[iPi]+1 =0 ].......>Panagiotis StefanidesDavid McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: Having trouble with some math that deals with x,y coordinates by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OEF7r11027;>Basicly, I have a point x,y>Also, I have an angle from 0 to 360 a>And I have a distance d>What I want to do is to start at point x,y.>Face the angle a>Move distance d>and compute the new point>Anyone know the formula/math for this?How's your trig? Do sine and cosine ring any bells?The increment in the x-direction is d cos aThe increment in the y-direction is d sin aNow add the increments to the original positionto find the new positionx_new = x_orig + d cos ay_new = y_orig + d sin aOf course, you should do a bit of checking first.E.g., make sure that d is non-negativeGood luck=== === Subject: : re:Axioms defining a finite fieldYou're using the - operation, which is not well defined apriori andits properties yet to be shown, so this argument is incomplete. Andof course there's a lot more to be shown before it's a field. === === Subject: : re:Axioms defining a finite fieldI didn't realize that this was supposed to be a quiz. As far as -c isconcerned, you can define it as +(-c), where (-c) is the elementwhich added to c gives 0. I am aware that to complete this stage,uniqueness of the additive inverse has to be shown. === === Subject: : re:Axioms defining a finite fieldThe first part is a quiz. The second is more of an open challenge,as I don't know how much you can cut down the axioms so that theystill define a finite field. === === Subject: : re:Axioms defining a finite fieldFiniteness is also important for showing that a*b = b*a.I suspect that the a*1 = a axiom isn't essential. I might be wrong. === === Subject: : re:Axioms defining a finite fieldTo the best of recollection (this goes back over 50 years) (1) - (7)with a multiplicative inverse give a field (without any finiteness). Replacing the multiplicative inverse by your (9) and finiteness makesit work. I don't believe you can get away with any less. === === Subject: : re:Axioms defining a finite fieldBy the classical definition a field must be commutative, otherwiseit's called a skew field or something like that. The guys whodeveloped the concept originally (Galois, Dirichlet, Caushy) had onlycommutative fields in mind and I stick to their definition.As for a*1 = a, possibly it can be omitted and it's still would bepossible to prove that there is a unity, perhaps different from 1.Alternatively, maybe we can replace this axiom with just 1*1 = 1. === === Subject: : re:Axioms defining a finite fielda*b=b*a is not necessary for something to be a field. a*1=a is part ofthe definition of 1, i.e. there exists an element (called 1) sothat a*1=a. === === Subject: : Re: How to study for math?> I'm trying to see if my study habits are part of the reason why I can be> hot/cold with math. So how exactly do you study for proof-based math classes? I'm assuming you are trying to pass a test where you will be asked toprove things.1. Re-read and *understand* all proofs you have seen in course: duringclass, text book, homework, etc.You should be able to repeat the proofs without looking at thesolution. But not repeat them from pattern recognition, but becauseyou understand the logical argument as to how what you have justwritten is proof.2. Now re-read and memorize (unless they are accessable during thetest--IMO they always should be) all axioms, theorems, corollaries,etc. from course (presented in class, text book, homework, etc.)Understand these as well. One way I learn to understandtheorems/corollaries is to study the proof of the theorem/corollary.You should be able to write down a proof of the theorem/corollarywithout looking at the one provided. Again do not memorize andregurgitate the steps, understand the proof.3. The best thing to do at this point is to find related proofs topractice on. The best is unassigned homework problems. You need thesolutions available to compare your results to. If solutions are notavailable you can always as the TA/Prof to look at a few you havedone. Other texts or internet perhaps if you can't find sampleproblems.Now you are as ready for the exam as you can be. You will be asked toprovide really one of two types of proofs. A proof similar to what youhave already seen, or a new proof that lends itself to be proved withone or more of the the axioms/theorems/corollaries you have seen. Ineither case you have the knowledge to complete the problem.Now sometimes you have all the knowledge but don't see the how. Ialways start with a sketch. And then work with the given. If somethinglike prove every X that has a Y also has a Z, then start with an Xthat has a Y. Now condiser an X with a Y with and without a Z. Why iswithout a Z not possible? As you play with this kind of thinking,you will usually see a connection to some of the stuff you have beenstudying, i.e. you have seen this before in theroem 12.4.5.=== === Subject: : Re: How to study for math?>>Read the chapter laying in bed (hey, I was originally a history major). Hens lay, people lie. (Some more than others.)>>Lying in bed refers to being supported by the bed, but laying in bed would>>be OK for referring to being spread over the surface of the bed. >Nope. Saying you're laying in bed is not right, regardless of whether>you're spread all over the surface. (There _is_ a sense of the word>lay under which it might be regarded as correct, but that's not>what we're talking about here.) Lay is transitive, lie is>intransitive.Those garlands of flowers they use in Hawai'i are called leis. I thinkit's correct form to say to lei someone when you mean to put thegarland on someone. When some of the other students came back fromHawai'i, they told me they'd gotten lei'd. It took me the longesttime to figure this out. Still not sure what the bed has to do with it.dave=== === Subject: : Re: How to study for math?>Read the chapter laying in bed (hey, I was originally a history major). >>> Hens lay, people lie. (Some more than others.)>Lying in bed refers to being supported by the bed, but laying in bed would>be OK for referring to being spread over the surface of the bed. >>Nope. Saying you're laying in bed is not right, regardless of whether>>you're spread all over the surface. (There _is_ a sense of the word>>lay under which it might be regarded as correct, but that's not>>what we're talking about here.) Lay is transitive, lie is>>intransitive.>Those garlands of flowers they use in Hawai'i are called leis. I think>it's correct form to say to lei someone when you mean to put the>garland on someone. When some of the other students came back from>Hawai'i, they told me they'd gotten lei'd. It took me the longest>time to figure this out. I suspect they were just playing with your head. Kids these days, gotno respect...>Still not sure what the bed has to do with it.>dave=== === Subject: : Symmetric relation cannot be antisymmetric; Antisymmetric relation cannot be symmetric?Dear all,I've these thorny proofs which I've no idea how to start off with.I'm given 2 statements, which I'm told to decide if they're true orfalse and to prove or disprove them.The 2 statements are:(1) A symmetric relation cannot be antisymmetric.(2) An antisymmetric relation cannot be symmetric.Based on an example I thought about, {(1,1), (2,2), (3,3)}, a relationcan be both symmetric and antisymmetric at the same time and bothstatements above are _wrong_.But comes to the disproving, I'm stuck at which method to go aboutdisproving them.If we consider statement 1, and it essentially says:symmetric -> ~antisymmetricIf I go about using the direct proof method, then I will have toassumesymmetric and then somehow work forward until I get antisymmetric (toget a contradiction), but this need not necessarily be the case.Hence I'm thinking of a proof by contradiction where I should assume~symmetric and work forward to get ~antisymmetric. But this seems togive me a proof for another scenario, that is if a relation is notsymmetric then it is also antisymmetric (~symmetric ->~antisymmetric). So is it a proof by contradiction for symmetric ->~antisymmetric?I don't know if I'm on the right track. Anyone can enlighten me on theway to approach this proof?---Goh, Yong Kwanggohyongkwang@hotmail.comSingapore=== === Subject: : Re: Symmetric relation cannot be antisymmetric; Antisymmetric relation cannot be symmetric?> .... > I'm given 2 statements, which I'm told to decide if they're true or> false and to prove or disprove them.The 2 statements are:> (1) A symmetric relation cannot be antisymmetric.> (2) An antisymmetric relation cannot be symmetric.Based on an example I thought about, {(1,1), (2,2), (3,3)}, a relation> can be both symmetric and antisymmetric at the same time and both> statements above are _wrong_.But comes to the disproving, I'm stuck.... No, you're not. Your example shows that you understand the issue perfectly, and the existence of that counter-example is all you need to prove that in statements (1) and (2) the cannot should be replaced by can. === === Subject: : antisymmetric relation = ~symmetric?I was reading through course notes when this sentence befuddles me,Recall that antisymmetric mean aRb ^ bRa => a = b, or for all a not = b, aRb => ~bRa. In other words this relation is never symmetric!Well, when it says that this relation is never symmetric, does thismean that any antisymmetric relation is not symmetric?And I've always believed that an antisymmetric relation does notnecessarily implies that it is _not_ symmetric and vice-versa. So Ihit a contradiction. ;)What's the note trying to say or is there any way of interpreting thesentence?---Goh, Yong Kwanggohyongkwang@hotmail.comSingapore=== === Subject: : Re: antisymmetric relation = ~symmetric?> I was reading through course notes when this sentence befuddles me,Recall that antisymmetric mean aRb ^ bRa => a = b, > or for all a not = b, aRb => ~bRa. > In other words this relation is never symmetric!Well, when it says that this relation is never symmetric, does this> mean that any antisymmetric relation is not symmetric?And I've always believed that an antisymmetric relation does not> necessarily implies that it is _not_ symmetric and vice-versa. So I> hit a contradiction. ;)What's the note trying to say or is there any way of interpreting the> sentence?> .... You're right, and the word never is wrong. But don't think too even be that thinking about this has made your own understanding clearer. === === Subject: : Re: antisymmetric relation = ~symmetric?> Recall that antisymmetric mean aRb ^ bRa => a = b,> or for all a not = b, aRb => ~bRa.> In other words this relation is never symmetric!> Well, when it says that this relation is never symmetric, does this> mean that any antisymmetric relation is not symmetric?It is not symmetric provided that there exist a<>b with aRb. Do 2 suchelements always exist ?=== === Subject: : Inscribed circle in Pythagoreian trianglesHi everyone, once again.I'm having difficulties trying to prove a thing and hoped that maybesome of you felt likte helping me.If you have a Pythagoreian triangle, then the radius of the inscribedcircle is r=(a+b+c)/2. Now for those triangles that can be written asa square of a gaussian integer (e.g the triangle (3,4,5) can berepresented by the gaussian integer z=3+4i which is the square of(2+i)) the middlepoint? of the triangle is a multiple of the squaredgaussian integer. E.g. the middlepoint of the triangle 3+4i is 2+iwhich is obviously 1(2+i). Prove that so is the case for all trianglesof the form A+Bi for which A+Bi=(a+bi)^2 is trueif you draw an image you can easily see that the middlepoint will berepresented by m=(a-r)+ri if the triangle is on the form a+bu and theradius of the circle is r. Then I proceeded with inserting r=(a+b+c)/2into the equation which, after some basic algebra, givesm=((a-b+c)/2)+(a+b+c)i/2. Now am I on the right track? I then used thefact that a+bi is the square of another gaussian integer which I herecalled c+di. a+bi=(c+di)^2 gives that a=c^2-d^2 and that b=2cd. I putthis into the equation and got m=(c^2-d^2-c)/2+i(c^2-d^2+2cd-c)/2.Then I got stuck. Am I completely wrong or a bit on the way??Thnx for all helpPierre=== === Subject: : Re: Inscribed circle in Pythagoreian triangles> Hi everyone, once again.> I'm having difficulties trying to prove a thing and hoped that maybe> some of you felt likte helping me.> If you have a Pythagoreian triangle, then the radius of the inscribed> circle is r=(a+b+c)/2. Now for those triangles that can be written as> a square of a gaussian integer (e.g the triangle (3,4,5) can be> represented by the gaussian integer z=3+4i which is the square of> (2+i)) the middlepoint? of the triangle is a multiple of the squared> gaussian integer. E.g. the middlepoint of the triangle 3+4i is 2+i> which is obviously 1(2+i). Prove that so is the case for all triangles> of the form A+Bi for which A+Bi=(a+bi)^2 is true> if you draw an image you can easily see that the middlepoint will be> represented by m=(a-r)+ri if the triangle is on the form a+bu and the> radius of the circle is r. Then I proceeded with inserting r=(a+b+c)/2> into the equation which, after some basic algebra, gives> m=((a-b+c)/2)+(a+b+c)i/2. Now am I on the right track? I then used the> fact that a+bi is the square of another gaussian integer which I here> called c+di. a+bi=(c+di)^2 gives that a=c^2-d^2 and that b=2cd. I put> this into the equation and got m=(c^2-d^2-c)/2+i(c^2-d^2+2cd-c)/2.> Then I got stuck. Am I completely wrong or a bit on the way??> Thnx for all help> PierreI mean that the radius of the inscribed circle is r=(a+b-c)/2.Further, I used c in (a+bi)=(c+di)^2 when I meant (e+di)^2, to avoidconfusion since the hypothenuse is labeled as c. So the equation wasactually m=(e^2-d^2-c)/2+i(e^2-d^2+2ed-c)/2. However I solved thisquite soon after I posted.sorry for the trivial mistakes before, hope no one got too mad overthem :-)thnxPierre=== === Subject: : Re: Inscribed circle in Pythagoreian triangles> Hi everyone, once again.> I'm having difficulties trying to prove a thing and hoped that maybe> some of you felt likte helping me.> If you have a Pythagoreian triangle, then the radius of the inscribed> circle is r=(a+b+c)/2.... No it isn't. Draw a (3,4,5) triangle and see whether an inradius of 6 = (3 + 4 + 5)/2 looks likely. If c is the hypotenuse, then the incircle has diametera + b - c, so radius half that. This formula seems not to be well-known nowadays, although it occurred in classical Chinese mathematics. === === Subject: : Re: Inscribed circle in Pythagoreian triangles> Hi everyone, once again.> I'm having difficulties trying to prove a thing and hoped that maybe> some of you felt likte helping me.> If you have a Pythagoreian triangle, then the radius of the inscribed> circle is r=(a+b+c)/2. Now for those triangles that can be written asI have r=(a+b-c)/2> a square of a gaussian integer (e.g the triangle (3,4,5) can be> represented by the gaussian integer z=3+4i which is the square of> (2+i)) the middlepoint? of the triangle is a multiple of the squared> gaussian integer. E.g. the middlepoint of the triangle 3+4i is 2+i> which is obviously 1(2+i). Prove that so is the case for all triangles> of the form A+Bi for which A+Bi=(a+bi)^2 is true> if you draw an image you can easily see that the middlepoint will be> represented by m=(a-r)+ri if the triangle is on the form a+bu and theI understand that i is near u on the keyboard... so a+bi> radius of the circle is r. Then I proceeded with inserting r=(a+b+c)/2> into the equation which, after some basic algebra, gives> m=((a-b+c)/2)+(a+b+c)i/2. Now am I on the right track? I then used them=((a-b+c)/2)+(a+b-c)i/2strange that you got the right a-b+c with a bad r value...> fact that a+bi is the square of another gaussian integer which I here> called c+di. a+bi=(c+di)^2 gives that a=c^2-d^2 and that b=2cd. I put> this into the equation and got m=(c^2-d^2-c)/2+i(c^2-d^2+2cd-c)/2.> Then I got stuck. Am I completely wrong or a bit on the way??Care with your notations, c is already in use for c = |a+i*b| insidem=((a-b+c)/2)+(a+b-c)i/2 !I didn't go furtherMay be more work on this equation like :m=(a+bi)*(1+i)/2+...before using a+bi=(u+vi)^2 (avoid the c !)and also c^2 = (a+bi)*(a-bi), a-bi=(u-vi)^2, and so on-- ippe(chephip at free dot fr)=== === Subject: : Re: JSH: Behave Rick Decker>>You don't have to remove the webpage if you wish to be obsessive about>>the issue, but at least update it to note that I've dropped that math>>argument.You are all misusing the math newsgroups!Please put this discussion offline!At least remove sci.math and sci.math.num-analysis from the list ofnewsgroups to which you post any reply which is not focussed onmathematical contents!These are newsgroup about serious mathematics and numerical analysis,not about arguing mutual insults!Arnold Neumaier=== === Subject: : Re: JSH: Behave Rick Decker <403B730A.8030204@univie.ac.at> Discussion, linux)> These are newsgroup about serious mathematics and numerical analysis,> not about arguing mutual insults!I don't know about sci.math.num-analysis, but the mutual insults havebeen here in sci.math longer than you have. They have seniority.-- Even I, who know beyond doubt that my death will be caused by a sillygirl, will not hesitate when that girl passes by. -- Merlin, asreported by John Steinbeck.=== === Subject: : Re: JSH: Behave Rick Decker>That's how I found out the problem with the freaking ring of algebraic>integers!!!What problem?=== === Subject: : Re: JSH: Behave Rick Decker> Now so far Hamilton College is defending Decker's right to put up the> webpage, and I'm not saying he doesn't have that right! But he needs to at least reflect basic facts, like that I've admitted> the argument is flawed. If he does not, I'll push the case that he's deliberately misleading> readers, with malice aforethought, and depending on the Hamilton> College name and his position there so he's *making* the school> That's hilarious! This is the same threat you made to those who identified> your argument is flawed in the first place. First you threaten anyone who> claims you are wrong, then when you finally admit you are wrong, you> threaten everyone who posts your argument.Remember this one?http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ArmyMath.htmlwas also ripped off by:) LH=== === Subject: : Re: JSH: Behave Rick Decker Discussion, linux)> Remember this one?> http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ ArmyMath.html> was also ripped off byI prefer the phrase, lovingly archived by.-- At the Microsoft-sponsored cocktail reception in the Galaxy Ballroomthat evening, Robert Dees urges us 'to network on behalf of the peopleof Iraq,' -- Naomi Klein reports on Microsoft's efforts to further democracy. === === Subject: : Re: JSH: Behave Rick Decker> What is all this about Hamilton College?I live about 20 miles north of Hamilton College. As Elmo once said to Martha Stewart, Don't joke me.Nobody lives 20 miles north of here; from Hamiltonto the North Pole there's nothing but snow, snow,snow, ice, snow, slush, snow, snow, and more snow (toparaphrase Monty Python).I agree with the rest, though--the part I snipped.Rick=== === Subject: : The Dot and Line--A MovieThe animator Chuck Jones directed a animated feature in 1965 called The Dot and Line, which was nominated for an Academy Award. I happened to see part of it on the TMC channel this morning. Does anyone know if it's on video, VHS or DVD? I went to Amazon and looked and found mention of it on one Chuck Jones feature, but a reviewer noted that it was incomplete. I'm told it sometimes appears on The Cartoon Channel.-- Wayne T. Watson (121.015 Deg. W, 39.262 Deg. N, 2,701 feet, Nevada City, CA) Don't hassle me with your signs, Chuck -- Peppermint Patty, Peanuts Web Page: sierra_mtnview -at- earthlink -dot- net Imaginarium Museum: === === Subject: : Re: The Dot and Line--A Movie> The animator Chuck Jones directed a animated feature in 1965 called The Dot > and Line, > which was nominated for an Academy Award. I happened to see part of it on the > TMC > channel this morning. Does anyone know if it's on video, VHS or DVD? I went > to Amazon > and looked and found mention of it on one Chuck Jones feature, but a reviewer > noted > that it was incomplete. I'm told it sometimes appears on The Cartoon Channel.There is a book, from which I think the movie was made, called The Dot and the Line, A Romance in Lower Mathematics by Norton Juster See for exaxmplehttp://www.amazon.com/exec/obidos/tg/detail/-/ 1587170663/103-8735420-9207022?v=glance=== === Subject: : Re: Ideas for course on great ideas in (theoretical) CS?> I was looking at Amazon for the book you have mentained, but didn't find it.> Could you please tell me who the writer is?Here it is, along with a few related others. All can be foundat Amazon:Great Ideas in Computer Science - 2nd Edition: A Gentle Introduction by W. Bierman Publisher: MIT Press; 2nd edition (March 1, 1997) ISBN: 0262522233 ACM Turing Award Lectures: The First Twenty Years 1966-1985 Publisher: Addison-Wesley Pub Co; 1st edition (July 1, 1991) ISBN:0201548852by Jean-Luc Chabert (Editor), E. Barbin (Editor), R. Aasnogorodski, V.MalyshevPublisher: Springer Verlag; (September 1999) ISBN: 3540633693 People & Ideas in Theoretical Computer Science (Springer Series inDiscrete Mathematics and Theoretical Computer Science)by Cristian CaludePublisher: Springer Verlag; (October 1999) ISBN: 981402113X=== === Subject: : Divisions algorithm for Z[i]HiI posted here concerning this subject a few weeks ago and now I justwant to see that I got things right. I'm handing in a project done inmathematics but my teacher can't help me let alone say if I'm on theright track with things so I kind of depend on you guys.How would you prove that the remainder gets smaller every time whenthe divisions algorithm is applied to two gaussian integers and thenshow that the remainder is zero after a finite number of steps?I really appreciate all the help that the people here have been givingme during this projectThnxPiere=== === Subject: : Divisions algorithm for Z[i]HiI posted here concerning this subject a few weeks ago and now I justwant to see that I got things right. I'm handing in a project done inmathematics but my teacher can't help me let alone say if I'm on theright track with things so I kind of depend on you guys.How would you prove that the remainder gets smaller every time whenthe divisions algorithm is applied to two gaussian integers and thenshow that the remainder is zero after a finite number of steps?I really appreciate all the help that the people here have been givingme during this projectThnxPiere=== === Subject: : Re: Divisions algorithm for Z[i]> Hi> I posted here concerning this subject a few weeks ago and now I just> want to see that I got things right. I'm handing in a project done in> mathematics but my teacher can't help me let alone say if I'm on the> right track with things so I kind of depend on you guys.> How would you prove that the remainder gets smaller every time when> the divisions algorithm is applied to two gaussian integers and then> show that the remainder is zero after a finite number of steps?> I really appreciate all the help that the people here have been giving> me during this projectI believe it can be done with a purely geometric approach. View the points in a a 2D plane, create a lattice from the one you're dividing by. Norms correspond to distances in this setup.I can't remember what you do from there, but I do remember that it did resolve to something sensible.-- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp2nd and 3rd bug found after 10 more minutes: gethost.cBoth non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL)=== === Subject: : Re: Divisions algorithm for Z[i]>> Hi>> I posted here concerning this subject a few weeks ago and now I just>> want to see that I got things right. I'm handing in a project done in>> mathematics but my teacher can't help me let alone say if I'm on the>> right track with things so I kind of depend on you guys.>> How would you prove that the remainder gets smaller every time when>> the divisions algorithm is applied to two gaussian integers and then>> show that the remainder is zero after a finite number of steps?>> I really appreciate all the help that the people here have been giving>> me during this project>I believe it can be done with a purely geometric approach. >View the points in a a 2D plane, create a lattice from the one >you're dividing by. Norms correspond to distances in this setup.>I can't remember what you do from there, but I do remember that >it did resolve to something sensible.Yes, the geometric approach works perfectly well for Z[i],Z[sqrt(-2)]; Felix Klein had a lecture on it (Lecture VIII in theChelsea Edition of Lectures on Mathematics); very pretty...Say you want to divide a by b, b nonzero. The multiples of b form alattice on the complex plane, generated by the vector b and the vectori*b, which is b rotated 90 degrees counterclockwise. This tiles the plane with parallelograms; a lies in one suchparallelogram (and unless it lies on an edge, on one and only one suchparallelogram). This parallelogram will have corners given by(r+i*s)b, [(r+1) + i*s]b, [r+i*(s+1)]b, and [(r+1) + i*(s+1)]b. [*]The claim is that the distance from a to at least one of the cornersis strictly smaller than the size of b.Assume first that this claim is correct, and let [u+i*v]*b be thecorner in question. Then a - [u+i*v]b = x+i*y has size strictlysmaller than b, and we havea = q*b + p, quotient q and remainder p,where q = u+i*v, p=x+i*y, and the remainder has size strictly smallerthan the size of b.To establish the claim, consider the worst possible case for the pointa, which is when it lies in the intersection of the two diagonals ofthe parallelogram determined by the points [*]. In this case, sincewe have a square the two diagonals are of equal length. The length ofthe diagonals is sqrt(2)*length(b), so the midpoint of the square is(sqrt(2)/2)length(b) from each corner. Since sqrt(2)/2 < 1, this means that any point in the square will beless than length(b) away from at least one corner, and this gives thatthe remainder gets smaller.When we work in Z[sqrt(-2)] the situation is a bit more complicated,since now the parallelograms are not squares, but the argument canstill be pushed through in essentially the same way. The argument ofcourse breaks down for larger values (although I do not rememberexactly when it breaks down; I seem to recall that you get rhombiiinstead of parallelograms when we try sqrt(-3) or (1+sqrt(-3))/2).-- ============================================================== ========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ================== === Subject: : Re: Bayesian Class and Math/Stat Teaching Techniques>>department was just different. They liked theory and they liked formulas.>>They liked elegant solutions and proofs, even if they were irrelevant to>>application. I sensed a certain disdain for word problems and real world>>analogies and explanations to help the students conceptualize the theory>>because real math students don't need those crutches.>My experience, and that of other math/stat instructors whom I've>talked to, is quite the opposite. It's the STUDENTS who don't like>word problems, and resist applications (eg, to physics), because to do>them they have to actually understand the mathematical material (and>even some physics!), rather than just applying formulas without really>knowing what they're doing. This may not be true of real math>students, however, who ought to be able to do the word problems (but>who may find the standard ones to be too easy to be interesting).> Radford NealThe students want to know how to plug things into formulas and get answers. This is the LAST step in applying statistics,mathematics, or whatever. One can study proofs from axiomatic approaches bythemselves. But when one has a real-world problem, themost important, and often hardest, part is to translatethat problem into a pure mathematics or statistics problem,so what is known from those fields can be applied. Thisrequires knowing the CONCEPTS, and being able to formulatethe word problems, with the solution often having to bedone by computers or often by those who can use the powerof the subject, and possibly even extend it.The physicist applying mathematics, or the economist orbiologist applying statistics, have to state their formalassumptions, after which the full power can be used,sometimes showing that the assumptions are not what theuser thought they were.One needs a little more care with word problems than isoften the case. The economist may not be able to formulatephysics word problems, and vice versa. But when a wordproblem is properly formulated, solving it does not requireknowing physics or economics. If this is not the case, atleast the formulation is incomplete.Teaching statistical methods without concepts only gets themused as religion. In engineering, errors usually show themselvesquickly, but in statistics, this is not the case, and I know muchharm which has been done.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558=== === Subject: : Re: Bayesian Class and Math/Stat Teaching TechniquesI wonder if the course becomes more applied in future semesters? The twosubjects you've mentioned I don't think anyone would expect to be light on thetheory - especially anything called 'Mathematical Statistics'.Based on my experience (which is based on the exposure to statistics teaching ofonly one university mind you) I would expect a Masters of Applied Statistics toperhaps start off with some statistical theory courses and progress into what Iwould call 'applied statistics' topics like generalised linear models, survivalanalysis techniques, time series techniques etc. Now whether these would be what/you/ would call applied (case-study driven) I don't know - probably dependson the faculty culture, the teaching staff preferences etc. But a course thatcovers theory of GLMs, theory of survival analysis, theory of time seriesanalysis I would still look at as a valid Applied Statistics course. Ofcourse it would be /nice/ if each of those topics included real-lifecase-studies as well, but I don't think it would be false advertising if itdidn't. Basically, I think the applied in the degree title probably refers tothe topics covered - not the teaching style.Getting back to the Swiss Army Knife analogy, I think it would be valid to havethe theory of Swiss Army Knives topic in an Applied Knifework course, alongwith theory of cleavers, theory of bayonets etc.In my experience it is these 'applied' topics where the examples andcase-studies come into it more. There isn't an awful lot of 'real-world'examples that would be applicable to a Mathematical Statistics topic, but whendealing with survival analysis (as an example) there are lots of data sets,the theory and application.It might be worth checking out future topics and their syllabus before decidingwhether to withdraw.Kylie.> A few weeks ago I posted a message asking about books on Bayesian> Unfortunately I have since dropped the class and am wondering about whether> I should continue the degree (Masters in Applied Statistics) and would like> some thoughts from the thoughtful people here.> I have a BS in Electrical Engineering and an MBA, both from the University> of Michigan. What I really liked about the MBA program is that it was almost> 100% applied. Probably 50-70%+ of the classes was case study classes, a> trend mostly propagated and refined by Harvard Business School where> supposedly 100% of the classes are case study classes. Apparently more and> more law programs have more and more case study classes as well. Case study> classes are really as applied as you get because concepts and theories are> learned in the context of real world situations and circumstances. I am also> the type of person who is a very intuitive learner and has a much easier> time learning when I see how what I am learning relates to challenges in> real life (eg: business, which is what I do).> So given that the degree I am pursuing is called Masters in *APPLIED*> Statistics, I thought the the courses would be heavily applied and taught in> the context of solving real world problems. No dice. Both courses I took in> the first semester (part-time evening program) had heavy theory. The> Bayesian class was not even as bad as the other one (Mathematical> Statistics). There was essentially no attempt on the part of the professor> to relate the theory to real world programs or to even give real world> examples to illustrate the concepts. It was formula, theory, formula,> theory, theory, formula, etc. I asked him about that and he said there's no> way around the theory. I'm not trying to get around the theory but> theories and formulas mean nothing to me without real world context. I'm not> stupid either -- I score in the 99+ percentile on quantitative SAT/GMAT/GRE> and 98+ percentile overall.> My thinking right now is that my expectations were just off and disciplines> like Math/Statistics are just not as, ummm, progressive as Business/Law when> it comes to teaching (please -- no hate mail). Those teaching Math/Stat may> also be too smart and are not interested in mundane day-to-day> business/industry problems (hopefully that will stop the hate mail!). So> what's up with that? Why is a degree called Masters in Applied Statistics> so heavy in theory? I'm not interested in theory in the absence of> application. I enrolled in Masters in Applied Statistics to learn how to> use statistical techniques to solve real world problems, how to use> statistical software to solve real world problems, etc. and not to learn> esoteric statistical theory in the absence of application that I will surely> forget an hour after the final exam.> I am not trying to slam the field. I am interested in some opinions from> those in the field, especially those teaching it, to help me determine if I=== === Subject: : Re: Bayesian Class and Math/Stat Teaching Techniques> Just a comment on a detail here -> [ delete, some other stuff] In fact, I specifically remember a class in business school illustratinga> statistical problem and how/why the scientists at Morton Thiokol did not> catch the potential problem with the o-ring on the space shuttleChallenger> that caused the tragedy -- they didn't do tests nor have data points forthe> o-ring at low temperatures, and the temperature at launch was very coldfor> central Florida.> [ ... ]> As I recall reading about it, it eventually came out that there> were scientists/ engineers around who had wanted to scrub> the cold-weather launch. *Since* they did not have data,> they were wise enough to have serious doubts -- but they> were unable to convince the administrators, who were not> 'technical people.'> I wonder what the point was, in a business school class?> - Keep channels open to your technical people?> - or, S**t happens?> --> Rich Ulrich, wpilib@pitt.edu> http://www.pitt.edu/~wpilib/index.html> Taxes are the price we pay for civilization.just because there are many data points, one may still not have enoughinformation with which to proceed. All the data points they had indicatedthat the o-ring was fine but the had no data points for how the o-ring mightbehave in cold weather.=== === Subject: : Re: Bayesian Class and Math/Stat Teaching... (Space shuttle o-rings)> ... If I recall, the point was that>just because there are many data points, one may still not have enough>information with which to proceed. All the data points they had indicated>that the o-ring was fine but the had no data points for how the o-ring might>behave in cold weather.No. My fairly clear recollection is that they had enough data to bepositively suspicious that disaster might strike if they launched at atemperature lower than any previous launch. This WASN'T just ageneral conservative view that one shouldn't push into unexploredterritory. There were two problems. One was that the managers didn'tpay enough attention to the technical people's worries. The other wasthat the technical people didn't communicate well. In particular, Irecall that they produced a scatterplot of temperature versus amountof o-ring erosion FOR LAUNCHES WHERE THE O-RINGS WERE ERODED, whichshowed no clear correlation. Not shown on the plot were the launcheswith NO EROSION, which were all at warm temperatures. A more suitableplot would have made clear that there was plenty of evidence thaterosion was greater at low temperatures than at high temperatures, andthat launching at a temperature substantially lower than any previouslaunch was madness. Radford Neal=== === Subject: : Re: Bayesian Class and Math/Stat Teaching TechniquesA few weeks ago I posted a message asking about books on BayesianUnfortunately I have since dropped the class and am wondering about whether> I should continue the degree (Masters in Applied Statistics) and would like> some thoughts from the thoughtful people here.I have a BS in Electrical Engineering and an MBA, both from the University> of Michigan. What I really liked about the MBA program is that it was almost> 100% applied. Probably 50-70%+ of the classes was case study classes, a> trend mostly propagated and refined by Harvard Business School where> supposedly 100% of the classes are case study classes. Apparently more and> more law programs have more and more case study classes as well. Case study> classes are really as applied as you get because concepts and theories are> learned in the context of real world situations and circumstances. I am also> the type of person who is a very intuitive learner and has a much easier> time learning when I see how what I am learning relates to challenges in> real life (eg: business, which is what I do).So given that the degree I am pursuing is called Masters in *APPLIED*> Statistics, I thought the the courses would be heavily applied and taught in> the context of solving real world problems. No dice. Both courses I took in> the first semester (part-time evening program) had heavy theory. The> Bayesian class was not even as bad as the other one (Mathematical> Statistics). There was essentially no attempt on the part of the professor> to relate the theory to real world programs or to even give real world> examples to illustrate the concepts. It was formula, theory, formula,> theory, theory, formula, etc. I asked him about that and he said there's no> way around the theory. I'm not trying to get around the theory but> theories and formulas mean nothing to me without real world context. I'm not> stupid either -- I score in the 99+ percentile on quantitative SAT/GMAT/GRE> and 98+ percentile overall.My thinking right now is that my expectations were just off and disciplines> like Math/Statistics are just not as, ummm, progressive as Business/Law when> it comes to teaching (please -- no hate mail). Those teaching Math/Stat may> also be too smart and are not interested in mundane day-to-day> business/industry problems (hopefully that will stop the hate mail!). So> what's up with that? Why is a degree called Masters in Applied Statistics> so heavy in theory? I'm not interested in theory in the absence of> application. I enrolled in Masters in Applied Statistics to learn how to> use statistical techniques to solve real world problems, how to use> statistical software to solve real world problems, etc. and not to learn> esoteric statistical theory in the absence of application that I will surely> forget an hour after the final exam.I am not trying to slam the field. I am interested in some opinions from> those in the field, especially those teaching it, to help me determine if II can relate to some extent. While in theory I love theory, inpractice I often wonder how what I'm studying relates to what I wantThe best math course I ever had was diff eq using _DifferentialEquations with Applications and Historical Notes_ by George Simmonsas the text, in part because the applications gave me somethingconcrete with which to relate. I'm not surprised that the MathematicalStatistics course was heavily theory, and in fact many of the statbooks I've looked at appeared tilted that way. I've probably learnedmore statistics from books on signal processing than stat texts.Someone might argue that's why I know so little about statistics. :-)Perhaps it is just the particular school/program you're in, but I don'tknow. Maybe you should consider economics and econometrics, whichcan be heavy with stats and purport, at least, to deal with real worldproblems. Good luck.=== === Subject: : Re: Bayesian Class and Math/Stat Teaching Techniques boundary=------------070201070602030005080405----------------- ----------------------------------------------------I am a geologist whose career has been based on statistics. I suffered through the theory classes and the theorems. It did not help that many of my classmates were physics majors who were actually fairly narrow probabalists. The background math / theory has turned out to be critical. Far too often the applied instructors (in the business college, in chemometrics, etc.) don't remember (or care) that all procedures have limited validity (robustness). A prime advantage is to understand where procedures make no sense because the system of interest does not cooperate. Hence the mean has limited practical usefulness, as does the standard deviation, regression statistics, factors and the like. Using the procedures without understanding ( and investigating) the underlying assumptions can be professionally suicidal for an applied statistician. One of my avocations is evaluating the work of bozos like you in lawsuits. It's like hitting herring in a barrel.>>A few weeks ago I posted a message asking about books on Bayesian>>Unfortunately I have since dropped the class and am wondering about whether>>I should continue the degree (Masters in Applied Statistics) and would like>>some thoughts from the thoughtful people here.>>I have a BS in Electrical Engineering and an MBA, both from the University>>of Michigan. What I really liked about the MBA program is that it was almost>>100% applied. Probably 50-70%+ of the classes was case study classes, a>>trend mostly propagated and refined by Harvard Business School where>>supposedly 100% of the classes are case study classes. Apparently more and>>more law programs have more and more case study classes as well. Case study>>classes are really as applied as you get because concepts and theories are>>learned in the context of real world situations and circumstances. I am also>>the type of person who is a very intuitive learner and has a much easier>>time learning when I see how what I am learning relates to challenges in>>real life (eg: business, which is what I do).>>So given that the degree I am pursuing is called Masters in *APPLIED*>>Statistics, I thought the the courses would be heavily applied and taught in>>the context of solving real world problems. No dice. Both courses I took in>>the first semester (part-time evening program) had heavy theory. The>>Bayesian class was not even as bad as the other one (Mathematical>>Statistics). There was essentially no attempt on the part of the professor>>to relate the theory to real world programs or to even give real world>>examples to illustrate the concepts. It was formula, theory, formula,>>theory, theory, formula, etc. I asked him about that and he said there's no>>way around the theory. I'm not trying to get around the theory but>>theories and formulas mean nothing to me without real world context. I'm not>>stupid either -- I score in the 99+ percentile on quantitative SAT/GMAT/GRE>>and 98+ percentile overall.>>My thinking right now is that my expectations were just off and disciplines>>like Math/Statistics are just not as, ummm, progressive as Business/Law when>>it comes to teaching (please -- no hate mail). Those teaching Math/Stat may>>also be too smart and are not interested in mundane day-to-day>>business/industry problems (hopefully that will stop the hate mail!). So>>what's up with that? Why is a degree called Masters in Applied Statistics>>so heavy in theory? I'm not interested in theory in the absence of>>application. I enrolled in Masters in Applied Statistics to learn how to>>use statistical techniques to solve real world problems, how to use>>statistical software to solve real world problems, etc. and not to learn>>esoteric statistical theory in the absence of application that I will surely>>forget an hour after the final exam.>>I am not trying to slam the field. I am interested in some opinions from>>those in the field, especially those teaching it, to help me determine if I>>I can relate to some extent. While in theory I love theory, in>practice I often wonder how what I'm studying relates to what I want>The best math course I ever had was diff eq using _Differential>Equations with Applications and Historical Notes_ by George Simmons>as the text, in part because the applications gave me something>concrete with which to relate. I'm not surprised that the Mathematical>Statistics course was heavily theory, and in fact many of the stat>books I've looked at appeared tilted that way. I've probably learned>more statistics from books on signal processing than stat texts.>Someone might argue that's why I know so little about statistics. :-)>Perhaps it is just the particular school/program you're in, but I don't>know. Maybe you should consider economics and econometrics, which>can be heavy with stats and purport, at least, to deal with real world>problems. Good luck.>> === === Subject: : Re: Bayesian Class and Math/Stat Teaching Techniques> I am a geologist whose career has been based on statistics. I've worked in lots of fields myself. Right now my work is in meteorologyand climatatology. Lots of applied statistics in all the things I've done.I enjoy theory, but it is difficult to find teachers or books that relateit well to practice, but they are out there and can be found with someeffort.> I suffered through the theory classes and the theorems. It did not help > that many of my classmates were physics majors who were actually fairly > narrow probabalists. The background math / theory has turned out to be > critical. Far too often the applied instructors (in the business > college, in chemometrics, etc.) don't remember (or care) that all > procedures have limited validity (robustness). A prime advantage is to > understand where procedures make no sense because the system of interest > does not cooperate. Hence the mean has limited practical usefulness, as > does the standard deviation, regression statistics, factors and the > like. Using the procedures without understanding ( and investigating) > the underlying assumptions can be professionally suicidal for an > applied statistician. One of my avocations is evaluating the work of > bozos like you in lawsuits.Why do think I'm a bozo?> It's like hitting herring in a barrel.First, you have to get a barrel of herring. ;-) >A few weeks ago I posted a message asking about books on Bayesian>>Unfortunately I have since dropped the class and am wondering about whether>>I should continue the degree (Masters in Applied Statistics) and would like>>some thoughts from the thoughtful people here.>>I have a BS in Electrical Engineering and an MBA, both from the University>>of Michigan. What I really liked about the MBA program is that it was almost>>100% applied. Probably 50-70%+ of the classes was case study classes, a>>trend mostly propagated and refined by Harvard Business School where>>supposedly 100% of the classes are case study classes. Apparently more and>>more law programs have more and more case study classes as well. Case study>>classes are really as applied as you get because concepts and theories are>>learned in the context of real world situations and circumstances. I am also>>the type of person who is a very intuitive learner and has a much easier>>time learning when I see how what I am learning relates to challenges in>>real life (eg: business, which is what I do).>>So given that the degree I am pursuing is called Masters in *APPLIED*>>Statistics, I thought the the courses would be heavily applied and taught in>>the context of solving real world problems. No dice. Both courses I took in>>the first semester (part-time evening program) had heavy theory. The>>Bayesian class was not even as bad as the other one (Mathematical>>Statistics). There was essentially no attempt on the part of the professor>>to relate the theory to real world programs or to even give real world>>examples to illustrate the concepts. It was formula, theory, formula,>>theory, theory, formula, etc. I asked him about that and he said there's no>>way around the theory. I'm not trying to get around the theory but>>theories and formulas mean nothing to me without real world context. I'm not>>stupid either -- I score in the 99+ percentile on quantitative SAT/GMAT/GRE>>and 98+ percentile overall.>>My thinking right now is that my expectations were just off and disciplines>>like Math/Statistics are just not as, ummm, progressive as Business/Law when>>it comes to teaching (please -- no hate mail). Those teaching Math/Stat may>>also be too smart and are not interested in mundane day-to-day>>business/industry problems (hopefully that will stop the hate mail!). So>>what's up with that? Why is a degree called Masters in Applied Statistics>>so heavy in theory? I'm not interested in theory in the absence of>>application. I enrolled in Masters in Applied Statistics to learn how to>>use statistical techniques to solve real world problems, how to use>>statistical software to solve real world problems, etc. and not to learn>>esoteric statistical theory in the absence of application that I will surely>>forget an hour after the final exam.>>I am not trying to slam the field. I am interested in some opinions from>>those in the field, especially those teaching it, to help me determine if I>> I can relate to some extent. While in theory I love theory, in>practice I often wonder how what I'm studying relates to what I want>The best math course I ever had was diff eq using _Differential>Equations with Applications and Historical Notes_ by George Simmons>as the text, in part because the applications gave me something>concrete with which to relate. I'm not surprised that the Mathematical>Statistics course was heavily theory, and in fact many of the stat>books I've looked at appeared tilted that way. I've probably learned>more statistics from books on signal processing than stat texts.>Someone might argue that's why I know so little about statistics. :-)>Perhaps it is just the particular school/program you're in, but I don't>know. Maybe you should consider economics and econometrics, which>can be heavy with stats and purport, at least, to deal with real world>problems. Good luck.>--=== === Subject: : Re: Bayesian Class and Math/Stat Teaching Techniques boundary=------------080505090508040604090003----------------- ----------------------------------------------------Sorry . you are not a bozo. I was referring to JW. Maybe we should establish a business selling standard barrels of herring for the national shoot-the-herring-in-the-barrel tournament? Can we come up with a test to see whether or not the differences due to Matjes vs Schmaltz herring are signficant w/r to herring marknanship? .>>I am a geologist whose career has been based on statistics. >>I've worked in lots of fields myself. Right now my work is in meteorology>and climatatology. Lots of applied statistics in all the things I've done.>I enjoy theory, but it is difficult to find teachers or books that relate>it well to practice, but they are out there and can be found with some>effort.>> I suffered through the theory classes and the theorems. It did not help >>that many of my classmates were physics majors who were actually fairly >>narrow probabalists. The background math / theory has turned out to be >>critical. Far too often the applied instructors (in the business >>college, in chemometrics, etc.) don't remember (or care) that all >>procedures have limited validity (robustness). A prime advantage is to >>understand where procedures make no sense because the system of interest >>does not cooperate. Hence the mean has limited practical usefulness, as >>does the standard deviation, regression statistics, factors and the >>like. Using the procedures without understanding ( and investigating) >>the underlying assumptions can be professionally suicidal for an >>applied statistician. One of my avocations is evaluating the work of >>bozos like you in lawsuits.>>Why do think I'm a bozo?>> It's like hitting herring in a barrel.>>First, you have to get a barrel of herring. ;-)>>>>A few weeks ago I posted a message asking about books on Bayesian>>Unfortunately I have since dropped the class and am wondering about whether>>I should continue the degree (Masters in Applied Statistics) and would like>>some thoughts from the thoughtful people here.>>I have a BS in Electrical Engineering and an MBA, both from the University>>of Michigan. What I really liked about the MBA program is that it was almost>>100% applied. Probably 50-70%+ of the classes was case study classes, a>>trend mostly propagated and refined by Harvard Business School where>>supposedly 100% of the classes are case study classes. Apparently more and>>more law programs have more and more case study classes as well. Case study>>classes are really as applied as you get because concepts and theories are>>learned in the context of real world situations and circumstances. I am also>>the type of person who is a very intuitive learner and has a much easier>>time learning when I see how what I am learning relates to challenges in>>real life (eg: business, which is what I do).>>So given that the degree I am pursuing is called Masters in *APPLIED*>>Statistics, I thought the the courses would be heavily applied and taught in>>the context of solving real world problems. No dice. Both courses I took in>>the first semester (part-time evening program) had heavy theory. The>>Bayesian class was not even as bad as the other one (Mathematical>>Statistics). There was essentially no attempt on the part of the professor>>to relate the theory to real world programs or to even give real world>>examples to illustrate the concepts. It was formula, theory, formula,>>theory, theory, formula, etc. I asked him about that and he said there's no>>way around the theory. I'm not trying to get around the theory but>>theories and formulas mean nothing to me without real world context. I'm not>>stupid either -- I score in the 99+ percentile on quantitative SAT/GMAT/GRE>>and 98+ percentile overall.>>My thinking right now is that my expectations were just off and disciplines>>like Math/Statistics are just not as, ummm, progressive as Business/Law when>>it comes to teaching (please -- no hate mail). Those teaching Math/Stat may>>also be too smart and are not interested in mundane day-to-day>>business/industry problems (hopefully that will stop the hate mail!). So>>what's up with that? Why is a degree called Masters in Applied Statistics>>so heavy in theory? I'm not interested in theory in the absence of>>application. I enrolled in Masters in Applied Statistics to learn how to>>use statistical techniques to solve real world problems, how to use>>statistical software to solve real world problems, etc. and not to learn>>esoteric statistical theory in the absence of application that I will surely>>forget an hour after the final exam.>>I am not trying to slam the field. I am interested in some opinions from>>those in the field, especially those teaching it, to help me determine if I>>I can relate to some extent. While in theory I love theory, in>practice I often wonder how what I'm studying relates to what I want>The best math course I ever had was diff eq using _Differential>Equations with Applications and Historical Notes_ by George Simmons>as the text, in part because the applications gave me something>concrete with which to relate. I'm not surprised that the Mathematical>Statistics course was heavily theory, and in fact many of the stat>books I've looked at appeared tilted that way. I've probably learned>more statistics from books on signal processing than stat texts.>Someone might argue that's why I know so little about statistics. :-)>Perhaps it is just the particular school/program you're in, but I don't>know. Maybe you should consider economics and econometrics, which>can be heavy with stats and purport, at least, to deal with real world>problems. Good luck.>>-->> === === Subject: : Re: Stepwise regressionGup is right, I was thinking of recursive estimation (which you dostepwise), sorry for the misinformation.> [... moved top post to bottom...]> I'd like to know the advantages of a stepwise linear regression. Inwhat> way> does it differentiate itself from a normal linear regression?> for one, if the parameters (a&b in y= ax+b) are not constant then this> technique would stand a good chance of revealing this. may also be quicker if you have to do it in real time, not sure aboutthat> though.> Rob's reply is pretty nonsensical.> Stepwise regression is a way to put> predictor variables into a regression model one at a time. This is> especially useful if you have lots of variables and want to screen them,> as only significant predictors are added to the model. (In most> computer implementations you get to decide what significant means.)> The typical model you strive for is parsimonious, that is, it contains> only few predictors, and every predictor contributes significantly to> explaining the variation of the response.> The downside to stepwise regression is that you leave the selection> of important variables to the computer, and all it has to go by is> the data. If you do not check that the resulting model is plausible> based on your own understanding of the problem, you can end up> with pretty wacky models.> Consult an introductory stats text for further information --- this is> standard stuff, and every text should cover it.=== === Subject: : Re: Factoring RSA And the Reversible Multiplier(comments appreciated)>I have been looking into Edward Fredkin's Fredkin Gate built using ard>balls.Here's something that came to my mind.The RSA number to be factored m = p * qp and q are primes.[ stuff deleted ]The procedure would work best, if p != q. For this case notice that one of>the number in (p,q) have to be greater than other. So, the MSB of one, is at>a higher index than the other. By forcing the MSBs of A and B to take values>1 or 0, we will force just one ordering of (p,q) to appear at the input.[will explain later]. How does the idea sound?Not good. Fredkin gates have extra outputs beyond the values of the Boolean> functions they're implementing. To factor N-bit integers by reversing> a multiplier composed of Fredkin gates, you'd have to know the values of> (I believe) O(N^2) extra outputs. Since you only have 2N outputs (i.e.,> the product), you would have to do some kind of search on the extra outputs> that is probably no easier (probably harder) than the search you have to> do to factor large integers.The above comments are incorrect.As was explained in the original Conservative Logic paper, it isalways possible to build a circuit of Fredkin gates or iard BallModel gates that has only the informationally required number of extraoutputs (or inputs). In the case of a circuit designed to factor theproduct of 2 primes, the only informational uncertainty is whichoutput gets the larger factor. The design of the circuit can takeinto account such cases as pq=1 pq, pq=-p -q, pq=-1 -pq etc. Ofcourse, if the circuit cannot take advantage of the fact that bothfactors are primes, then the informational uncertainty very is muchgreater. My comments are not merely theoretical as I long agoprogrammed a large Mathematica system that implemented factoring bymeans of a reversible multiplier. Because it actually simulated thereversible digital logic, it was very slow. My conclusion was that,for the algorithm I implemented, a hardware version was not quitecompetitive with the best existing factoring algorithms. However, Istill think that the general approach could be successful.The circuit had the property that if you put 35 into the outputs and 5into one of the inputs, the other input became 7. (A Fredkin gatereally has no inputs or outputs, it is perfectly symmetrical and worksin either direction; the same is true of any well formed circuit ofFredkin gates.)Kumar's idea is not quite right either. If p>q, a number of the mostsignifigant bits of both can still be the same. However, as Kumarsuggests, one can take advantage of what knowledge one has about thepossibilities related to MSB's and also LSB's.The LSB's of the product also enforce constraints on the LSB's of thefactors. The problem is that such ideas are only useful for a few ofthe leading or trailing bits. It's all the bits in the middle thatpose the big problem!Ed F=== === Subject: : Joint Probability QuestionIf I have two independant rv x and y, both drawn from different normaldistributions; how can I determine the probability that x will be lessthan y?I'm leaving out the specifics of the distributions because this ispart of a homework question and I truely want to understand theproblem, not just get an answer.Any help would be appreciated.=== === Subject: : Re: Joint Probability Question>If I have two independant rv x and y, both drawn from different normal>distributions; how can I determine the probability that x will be less>than y?>I'm leaving out the specifics of the distributions because this is>part of a homework question and I truely want to understand the>problem, not just get an answer.More generally, if X and Y have joint density f, thenP{X < Y} = integral (y= -infty .. infty, x= -infty .. y, f(x,y) dx dy).If X and Y are independent, the f is the product of the marginal densties.If X and Y are normals (independent or not) with different means, I doubt you will get a closed form solution.Thank you *very* much for explicitly identifying this as homework!-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: Joint Probability Question> If I have two independant rv x and y, both drawn from different normal> distributions; how can I determine the probability that x will be less> than y?> I'm leaving out the specifics of the distributions because this is> part of a homework question and I truely want to understand the> problem, not just get an answer.> Any help would be appreciated.The typical approach to such problems uses a conditioning argument.That is, find P(x < y | y = b) and integrate over all possible values ofb.You have the probability density for y, so the integral can be found,by numerical integration if necessary.Since this is homework, I am leaving out the details, as well.=== === Subject: : convolution of measuresSuppose M and N are distributions (measures), and it's given thatM * N = d_1, where * indicates convolution and d_1 indicates the delta measureconcentrated at 1. How do you show that either M = d_c or N = d_c forsome constant c? This should probably follow from some really simplemanipulations with convolutions, but I lamentably don't see how to doit.=== === Subject: : Re: convolution of measures>Suppose M and N are distributions (measures), and it's given that>M * N = d_1, >where * indicates convolution and d_1 indicates the delta measure>concentrated at 1. How do you show that either M = d_c or N = d_c for>some constant c? This should probably follow from some really simple>manipulations with convolutions, but I lamentably don't see how to do>it.Here is a solution which works for finite measures; perhaps it can be modified by approximation to all measures?The problem is equivalent to showing that if X and Y are independent random variables and X + Y is almost surely 1, then either X or Y is almost surely constant. Actually, they both are, sinceVar(X + Y) = Var(X) + Var(Y) = 0.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: convolution of measuresOriginator: grubb@lola>Suppose M and N are distributions (measures), and it's given that>M * N = d_1, >where * indicates convolution and d_1 indicates the delta measure>concentrated at 1. How do you show that either M = d_c or N = d_c for>some constant c? This should probably follow from some really simple>manipulations with convolutions, but I lamentably don't see how to do>it.I am assuming that you are dealing with measures on the real lineand that convolution for you is defined (or equivalent to)M*N(E)=int N(E-x) dM(x).If you allow only positive measures, you can look at supportsas Edgar suggested. However, if you are allowing measures thatare not positive, the result is false. For example, letM=d_1 + (1/2)d_2N=sum_{n=0}^infty (-1/2)^n d_nThe point is that any measure close enough to the identity (i.e. d_0) will be invertible in the Banach algebra of measuresunder convolution with inverse given by the power seriesexpansion of (1-x)^(-1). Then convolve one of the measuresby d_1 to get an example.--Dan Grubb=== === Subject: : Re: convolution of measures> Suppose M and N are distributions (measures), and it's given thatM * N = d_1, where * indicates convolution and d_1 indicates the delta measure> concentrated at 1. How do you show that either M = d_c or N = d_c for> some constant c? This should probably follow from some really simple> manipulations with convolutions, but I lamentably don't see how to do> it.distributions where, in the line?Show that if x is in the support of M and y is in the support of N,then x+y is in the support of M*N. Then if either M or N (or both)have support with more than one point, then so does M*N.Definition. x is in the support of M if every neighborhoodof x has positive measure.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: Regression/prediction> I have a set of (x,y) points (units completed, time).> Using linear regression I have the equation y=a*x+b, and> the standard deviations of a and b.> Now, the problem is estimating time remaining and a confidence> interval.> The way I see it I have two choices,> where n=units completed, i.e. x in the last (x,y),> N=total units> 1. Time remaining=a*N+b-elapsed time> 2. Time remaining=(N-n)*aProbably neither, but it's hard to tell without actually lookingat your data set. If I read this correctly you have data of the form1st completed 9:202nd completed 9:303rd completed 9:45...You are almost surely dealing with autocorrelated data, that is,if the kth unit is delayed, then the k+1st will be delayed as well(because it started late). Differencing would probably help.Organize your data as1st completion took 20 minutes2nd completion took 10 minutes3rd completion took 15 minutes...Then you can compute the average time to complete one unit.It's not a regression problem at all, but the average time tocomplete one unit is your best estimate of what will happen tothe remaining (N-n).=== === Subject: : Re: Cubic FormulaIs there any reason why this formula, which is crucial to finding> solutions to most cubic equations, is not included in most high school> algebra textbooks?> Michael (1) Not enough time.> (2) Not enough patience.> (3) The formula is messy and hard to memorize.> (4) It is the method, rather than the final formula, that is applied; this> goes against the spirit of high-school math teaching.> (5) The famous obstacle: real cubics with real distinct roots lead to cube> roots of complex (non-real) numbers (casus irreducibilis). What is the method of finding the cube roots of any complex number? Michael=== === Subject: : Re: Cubic Formula> What is the method of finding the cube roots of any complex number?> MichaelWrite the complex number in polar form: z = r * exp(i*theta), where r and theta are real and r >= 0.Let s be the real cube root of the real number rthen the cube roots of z are are s*exp(i*theta/3)s*exp(i*(theta+2*pi)/3)s*exp(i*(theta+4*pi)/3) === === Subject: : Re: Cubic Formula <6930a3c6.0402240930.4b00367d@posting.google.com> Is there any reason why this formula, which is crucial to finding> solutions to most cubic equations, is not included in most high school> algebra textbooks?> Michael> (1) Not enough time.> (2) Not enough patience.> (3) The formula is messy and hard to memorize.> (4) It is the method, rather than the final formula, that is applied; this> goes against the spirit of high-school math teaching.> (5) The famous obstacle: real cubics with real distinct roots lead to cube> roots of complex (non-real) numbers (casus irreducibilis).> What is the method of finding the cube roots of any complex number?> MichaelStepping out of elementary algebra and trisecting an angle:if w = r*(cos(t) + i*sin(t) with r>0 then the cube roots of w arer^(1/3) * (cos((t+2*k*pi)/3) + i*sin((t+2*k*pi)/3), k=0, 1, -1}To improve an approximate cube root z of w, one can use a cubicallyconvergent version of Newton's Method: z_new = z - (z^3-w) * z/(2*z^3+w)Cheers, ZVK(Slavek).=== === Subject: : Re: Cubic Formula>Is there any reason why this formula, which is crucial to finding>solutions to most cubic equations, is not included in most high school>algebra textbooks?> MichaelYes, first of all it is *NOT* crucial to finding such roots, more> often than not you can easyly solve the equation by other means. Refer> to the same textbook which does not include the formula for more> explanation on this . So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It isnot like you can factor it into two binomials.Also, it is messy, hard or imposible to memorize, etc etc. I have no problem memorizing it, once I found a pattern. Michael=== === Subject: : Re: Cubic Formula> So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0?Long ago from self-study of an old College Algebra book (by C.H Lehmann, published in 1962) I found, I learned the following method.By Descartes' Rule of Signs, the given equation has no positive root and either three or one negative root.If it has a rational root, then this is of the form p/q where p is a divisor of the constant term and q is a divisor of the leading coefficient. Thus, in this case, the only possible rational roots are -1, 2, -4, and -8.Now check (e.g., by synthetic division) each of these values to find that -8 is a root and the remainder on division by x + 8 is x^2 + x + 1.More generally, one rarely has need for precise closed-form solutions of such equations and one uses approximation algorithms. Some, like Newton's method, use calculus (which I assume you have not studied yet). Yet there are others, e.g., Horner's method, which do not.Also, one can often make qualitative statements about the roots of a polynomial without explicitly finding them. I use calculus in the present example: Let f(x) be the given polynomial. Since f(x) approaches +/- infinity as x does, f has at least one real root (as do all real polynomials of odd degree). Since f maps nonnegative numbers to positive numbers, all its real roots are negative. It must have either one or three real roots, since if non-real z is a root, then so is the conjugate of z. (and f has exactly three complex roots, counting multiplicities).Now consider the derivative f'(x) = 3x^2 + 18x + 9 = 3(x^2 + 6x + 3). Since 6^2 - 4*3 is positive, this has two real roots, whence f has two relative extrema. The larger x value, -3 + sqrt(6), must be a local minimum. f(-3+sqrt(6)) > 0, so f has exactly one real root, and this is less than -3 - sqrt(6).I think you might enjoy looking at such an old-style College Algebra book. The discussion here comes under the topic, Theory of Equations.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : Re: Cubic FormulaEn el mensaje:6930a3c6.0402240927.66edac9d@posting.google.com, Michael Ejercito escribi.97:> Is there any reason why this formula, which is crucial to finding> solutions to most cubic equations, is not included in most high> school algebra textbooks?> Michael>> Yes, first of all it is *NOT* crucial to finding such roots, more>> often than not you can easyly solve the equation by other means.>> Refer to the same textbook which does not include the formula for>> more explanation on this .> So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It is> not like you can factor it into two binomials.What about synthetic division of x^3 + 9x^2 + 9x + 8 by (x - a), where a isa integr divisor of 8?-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: Cubic Formula> So how would you go about solving x^3 + 9x^2 + 9x + 8 = 0? It is>not like you can factor it into two binomials. x^3 + 9x^2 + 9x + 8 = 0<=> x^3 + x^2 + x + 8x^2 + 8x + 8 = 0<=> x(x^2 + x + 1) + 8(x^2 + x + 1) = 0<=> (8 + x)(x^2 + x + 1) = 0-- I'm not interested in mathematics that might have anythingto do with reality. -- Easterly, in sci.math=== === Subject: : Re: adjoining elements to a ring>I am having a very hard time understanding a section of my algebra>book about adjoining elements to a ring. my book (artin, algebra)>discusses it breifly, and I check out herstein but that didnt help>much either. In particular Im trying to how that Z[x]/(2x-1) is>isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2>for its denominator).I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and>phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have>the isomorphism. This hasn't worked so far.>I'm not sure why you want to do that. The natural map from Z to Q_2> would be an injection in my book, and have trivial kernel. (So does> the other one, I don't see how that gives the isomorphism you want.> You would just have two injections from Z to two other rings.) Consider instead the map from Z[x]/(2x-1) sending x to 1/2, or perhaps> the map from Z[x] sending x to 1/2 and finding its kernel.>However, I feel that my real problem is that I have only seen very>elementary examples of this construction (for example R[x]/(x^2+1)=C).>Are there books or online references that discuss this more clearly?Thank you...> Try Stewart's Galois Theory for some nice stuff about field> extensions.Is that what this subject is called? Field extensions? I tried lookingup information about this myself, but ring adjunction element didn'tturn up anything useful.=== === Subject: : Re: adjoining elements to a ring>>>I am having a very hard time understanding a section of my algebra>>book about adjoining elements to a ring. my book (artin, algebra)>>discusses it breifly, and I check out herstein but that didnt help>>much either. In particular Im trying to how that Z[x]/(2x-1) is>>isomorphic to Q_2 (the dyadic rations i.e. rations with a power of 2>>for its denominator).>>I have made the identifications j:Z-->Z[x], pi:Z[x]-->Z[x]/(2x-1), and>>phi:Z-->Q_2, so that if I could show ker(pi o j)=ker(phi) I would have>>the isomorphism. This hasn't worked so far.>> I'm not sure why you want to do that. The natural map from Z to Q_2>> would be an injection in my book, and have trivial kernel. (So does>> the other one, I don't see how that gives the isomorphism you want.>> You would just have two injections from Z to two other rings.) >>> Consider instead the map from Z[x]/(2x-1) sending x to 1/2, or perhaps>> the map from Z[x] sending x to 1/2 and finding its kernel.>>>>However, I feel that my real problem is that I have only seen very>>elementary examples of this construction (for example R[x]/(x^2+1)=C).>>Are there books or online references that discuss this more clearly?>>Thank you...>>>> Try Stewart's Galois Theory for some nice stuff about field>> extensions.>Is that what this subject is called? Field extensions? I tried looking>up information about this myself, but ring adjunction element didn't>turn up anything useful.Strictly speaking no, because you're adjoining things to ring, nonecessarily a field, but if you study field extensions of Q say you'llhave a good idea of what's going on. C is a degree two extension of Rfor instance.=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OHdp131321; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OHIQi29476 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OHIQb12185;>equation except factoring, quadratic formula, completing the>square,graphing or the square root method.>I went to http:// www.geocities.com/dirkie6/page4.html but those>methods are too complex.How about Newton's method? Guess, iterate to a better guess,etc. Do you consider this a solution?=== === Subject: : Re: Alternative ways to solve a quadratic equation>equation except factoring, quadratic formula, completing the>square,graphing or the square root method.I went to http:// www.geocities.com/d>irkie6/page4.html but those>methods are too complex.>How about Newton's method? Guess, iterate to a better guess,> etc. Do you consider this a solution?> One problem with it is that when there is a rational root and you do not guess it exactly right at first, you never get it exactly right.=== === Subject: : Re: Joint Probability Question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OHt7Z00427; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OHowi32671 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OHow114337;>> If I have two independant rv x and y, both drawn from different normal>> distributions; how can I determine the probability that x will be less>> than y?>> I'm leaving out the specifics of the distributions because this is>> part of a homework question and I truely want to understand the>> problem, not just get an answer.>> Any help would be appreciated.>The typical approach to such problems uses a conditioning argument.>That is, find P(x < y | y = b) and integrate over all possible values of>b.>You have the probability density for y, so the integral can be found,>by numerical integration if necessary.>Since this is homework, I am leaving out the details, as well.There's a much easier way to handle it in this special case. P[X < Y] = P[X-Y < 0]Now think about what you know (or should know) about the distribution of X-Y. This should lead you to an easy evaluation.Steve OakleyRemove iron for email=== === Subject: : Re: How far can one go with self-study nowadays?> I do not have a clue how to start. I was good at Maths and> should have done it at A-Level but never did. Huge regret.I've definitely got the time, but, dunno where to start.My personal opinion: I think that, for many learning universitymathematics for the first time, the biggest hurdle is mastering therigorous and formal style of mathematical language, notation, andthought.I recommend Elementary Number Theory by David Burton as a good placeto start. Try reading it and doing the exercises. If you post hereif you get stuck for a long time, I'm sure you can get help.Paul Epstein=== === Subject: : Re: How far can one go with self-study nowadays? <345l30tj09o9etg58pbci1g0jh611h8q2s@4ax.com> <9b562672.0402241013.8f60e9a@posting.google.com My personal opinion: I think that, for many learning university> mathematics for the first time, the biggest hurdle is mastering the> rigorous and formal style of mathematical language, notation, and> thought. I'd agree with this... when someone has been self-teaching mathematics to themselves for 10 years, they don't really need to be proving things to other people, and so sufficiently proving theorems to themselves usually suffices. Once they get to an education level that challenges their knowledge level, they have to start getting formal all of a sudden. For instance, I was differentiating functions while I was in grade 9 and integrating functions in grade 10, and felt pretty confident about solving calculus problems because I never came across any I couldn't solve (as a kid) and ended up with something stupid like 98% in high school calculus. Once I hit university and slept through most classes, the midterm was a shocker: Epsilons? Prove the least-upper-bound property of reals? (why prove that? seemed obvious enough) Prove the sequential characterization of limits? Prove Bolazano-Weierstrauss? (or, after that midterm, Bozo-WiseAss)J=== === Subject: : inequalityI trying to establish the inequality: (x^a - 1)/(x^b - 1) <= a/b I'm not sure if we need to restrict x, a, or b; but in the context of my problem a1.=== === Subject: : Re: inequality> I trying to establish the inequality: (x^a - 1)/(x^b - 1) <= a/b I'm not sure if we need to restrict x, a, or b; but in > the context of my problem a1.Supposed O < a < b and x > 1. Multiply through by (x^b - 1) and move everything to the right. The inequality becomes 0 <= (a/b)*(x^b - 1) - (x^a - 1). Call the last exression f(x), shich is defined for x >= 1. Note f(1) = 0. If you can show f'(x) >= 0 for x >= 1, you'll be done.=== === Subject: : Re: inequality> I trying to establish the inequality: (x^a - 1)/(x^b - 1) <= a/b I'm not sure if we need to restrict x, a, or b; but in> the context of my problem a1.> Supposed O < a < b and x > 1. Multiply through by (x^b - 1) and move> everything to the right. The inequality becomes 0 <= (a/b)*(x^b - 1) -> (x^a - 1). Call the last exression f(x), shich is defined for x >= 1. Note> f(1) = 0. If you can show f'(x) >= 0 for x >= 1, you'll be done. Calculus applied to another variable: first, for a>0 and x>0,elementary integration givesF(a) = (x^a-1)/a = (ln(x)) * integral[0 to 1] exp(a*u*ln(x)) duhenceF'(a) = (ln(x))^2 * integral[0 to 1] u * x^(a*u) duSo, F' is positive and F increases (unless x=1, of course).The original inequality is equivalent (with 00)to F(a) < F(b).Finito.Cheers, ZVK(Slavek).=== === Subject: : Re: inequalityEn el mensaje:ead84c41.0402241040.15164c9f@posting.google.com,Jerome Davies escribi.97:> I trying to establish the inequality:> (x^a - 1)/(x^b - 1) <= a/b> I'm not sure if we need to restrict x, a, or b; but in> the context of my problem a1.If 0 < a < b,Lim((x^a - 1)/(x^b - 1), x, 1) = Lim((ax^(a-1)/(bx^(b-1), x, 1) = a/b(By L'H.99pital rule)If 0 < a < b, Lim((x^a - 1)/(x^b - 1), x, +inf) = 0.Let f(x) = (x^a - 1)/(x^b - 1). Thenf'(x) = (-x^(a + b)(b - a) - ax^a + bx^b)/(x(x^b - 1)^2)The denominator is positive, and the numerator is negative if-x^(a + b)(b - a) - ax^a + bx^b < 0 <===>bx^b - ax^a < (b - a)x^(a+b)But bx^b - ax^a < bx^b - ax^b = (b - a)x^b < (b-a)x^(a+b)Then f'(x) < 0 for x > 1 anf f is decreasing from a/b to 0 when x go from 1to +inf, being a/b a uper bound.I think that if a < b < 0 also it is true, but if a < 0 < b that is false.-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: inequality> I trying to establish the inequality:> (x^a - 1)/(x^b - 1) <= a/b> I'm not sure if we need to restrict x, a, or b; but in> the context of my problem a1.Hint: Integrate exp(a*t) and exp(b*t) from 0 to ln(x), and compare.Cheers, ZVK(Slavek).=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OIiVi05040equation except factoring, quadratic formula, completing the> square,graphing or the square root method.I went to http://www.geocities.com/dirkie6/page4.html but those> methods are too complex.Factoring, quadratic formula, completing the square are all one method:the quadratic formula is proved by completing the square; if you can'tsee how to factor a quadratic, complete the square.Since you mention graphing, I take it that approximate methods areacceptable to you, so consider Newton's method (aka the Newton-Raphsonmethod): http://mathworld.wolfram.com/NewtonsMethod.html.What's the square root method?-- G.C.=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OH9Mi28590> equation except factoring, quadratic formula, completing the> square,graphing or the square root method.I went to http://www.geocities.com/dirkie6/page4.html but those> methods are too complex.> Well, any methods more advanced than the ones you mentionedwill be more complex. It will be hard to find a simplermethod than the ones you mentioned. If such a method existed,it would be presented more widely.But other methods do exist.I will give a method based on Galois theory.I wish to find the roots of the equation x^2 + bx + c = 0,expressed in terms of b and c, using radicals.Let r be a root.Note r^2 + b*r + c = 0.Using long division, I find that x^2 + bx + c = (x - r) * (x + (b + r)).Thus, the other root is -(b + r).I will assume that the polynomial x^2 + bx + c is irreducible.You can get around this assumption by assuming that b andc are variable parameters rather than complex numbers.It is easy to see that by adjoining r to the base fieldthat we get a splitting field and that the Galois groupis cyclic of order two.As prescribed by Galois theory, I find the two Lagrangeresolvents v_1 and v_2 given by: v_1 = r - (-b-r) = 2r + b v_2 = r + (-b-r) = -bSince v_1 + v_2 = 2r and r is not in the base field,then either v_1 or v_2 is not in the base field.Here, v_1 must not be in the base field, since -b is.Note that v_1 * v_1 = 4*r^2 + 4*b*r + b^2 = 4*(r^2 + b*r) + b^2 = 4*(-c) + b^2 = b^2 - 4cThus, v_1 * v_1 is in the base field since b and c are.It is easy to see that 1 and r is a basis for the splitting field.It is easy to see that 1 and v_1 is a basis also.But, 1 = 1 v_1 = b + 2rInverting, I get 1 = 1 r = (-b + v_1) / 2Since v_1 = sqrt(b^2 - 4c), I have r = (-b + sqrt(b^2 - 4c)) / 2This is the quadratic formula for one of the roots.The formula for the other root -(b + r) can be obtainedby eliminating the r.-- Hale=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OIuii06204X-Orig-Trace: DXC=kIXg>bnAYaPlmRcVd::IAPc[3Bf`1S5QPF0iV`L?jh__>nYnb3]1j]P< CUP2SXDnOFkB5QgBnAWKiXW<6[I>JenabflWQj] HL2J4S7LQ> equation except factoring, quadratic formula, completing the> square,graphing or the square root method.I went to http://www.geocities.com/dirkie6/page4.html but those> methods are too complex.Well, any methods more advanced than the ones you mentioned> will be more complex. It will be hard to find a simpler> method than the ones you mentioned. If such a method existed,> it would be presented more widely.But other methods do exist.I will give a method based on Galois theory.I wish to find the roots of the equation x^2 + bx + c = 0,> expressed in terms of b and c, using radicals.Let r be a root.Note r^2 + b*r + c = 0.Using long division, I find that x^2 + bx + c = (x - r) * (x + (b + r)).Thus, the other root is -(b + r).I will assume that the polynomial x^2 + bx + c is irreducible.> You can get around this assumption by assuming that b and> c are variable parameters rather than complex numbers.It is easy to see that by adjoining r to the base field> that we get a splitting field and that the Galois group> is cyclic of order two.As prescribed by Galois theory, I find the two Lagrange> resolvents v_1 and v_2 given by: v_1 = r - (-b-r) = 2r + b v_2 = r + (-b-r) = -bSince v_1 + v_2 = 2r and r is not in the base field,> then either v_1 or v_2 is not in the base field.> Here, v_1 must not be in the base field, since -b is.Note that v_1 * v_1 = 4*r^2 + 4*b*r + b^2> = 4*(r^2 + b*r) + b^2> = 4*(-c) + b^2> = b^2 - 4cThus, v_1 * v_1 is in the base field since b and c are.It is easy to see that 1 and r is a basis for the splitting field.> It is easy to see that 1 and v_1 is a basis also.But, 1 = 1 v_1 = b + 2rInverting, I get 1 = 1 r = (-b + v_1) / 2Since v_1 = sqrt(b^2 - 4c), I have r = (-b + sqrt(b^2 - 4c)) / 2This is the quadratic formula for one of the roots.The formula for the other root -(b + r) can be obtained> by eliminating the r.-- HaleAre you sure you couldn't find a heavier hammer to crack that egg?=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OCNHi01492> equation except factoring, quadratic formula, completing the> square,graphing or the square root method.I went to http://www.geocities.com/dirkie6/page4.html but those> methods are too complex.> You can do the following:ax^2 + bx + c = 0 is your equation. suppose m and n are roots of thisequation.a(x-m)(x-n) = 0 ax^2 - (m + n)x + mn = 0then you have the system : (A) m + n = -b (B) mn = c(A)^2-4(B) gives m-n, and you can solve the system.Actually if you go through symbolically with this, you'll end up withthe quadratic formula.=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OLdki21021required us to find a totally new method to solve a quadratic equationwith. It can involve the quadratic formula but has to be different.=== === Subject: : Re: the anticlassicalist }{ : ism for Daleks: OK. So if I'm already unknowingly using this logic, what is the point: you're trying to make? Believe me, it hasn't yet emerged from all the: verbiage.Well, from a practical point of view, understanding the logic oforthomodular lattices that arises in propositions on a Hilbert space canprevent a lot of early conceptual difficulties in the use of quantummechanics. Students first approaching the topic can understand that much ofthe strangeness of quantum mechanics, many quotes by various famousphysicists from early quantum foundations about the difficulty (or evenimpossibility) to understand the subject, can be dismissed as merelyproblems in understanding the logic of quantum mechanics. Because quantummechanics really is a well established model in physics which I believe canbe understood completely by mere humans, and I think that some people havefallen into certain antisymbolic modes of thought by a myriad of pop-physicsbook which agonise over this strangeness.As I keep trying to point out, my focus is on education with these series ofposts. The theory of models and their logics is crucial to all science.Indeed, it rigorously formalises the notion of a theory. Many models acrossdisciplines have a Heyting structure, and these are often unknown topractitioners in the various fields, so I use that as my archetype foreducation reform. There are many cases where the unfamiliarity with thelogical structure of the models used presents confusion to the studentsearly on.But understanding the logic of the models can also have importanttheoretical components as well. For quantum logic, as the case in question,understanding the structural differences in the logics of classical andquantum mechanics gives us information (of a cohomological nature)concerning quantisation programs. Although canonical quantisation is fineto mislead first year students about our understanding of quantisation, itignores the actual difficulties and ambiguities that arise. Instead oneneeds to go from prequantisations to metaplectic corrections and all sortsof fun stuff along the way. Avoiding obstructions to quantisation and(somewhat oppositely) understanding the ambiguities that arise in theprocess requires one be comfortable with logical structures of classical andquantum models.Extend to quantum field theories, strings, and such, and maybe it getsclearer why I see the necessity to alter education of logic in physicscurricula.And this does not even touch upon the theory of causal sets and otherdirections in physics where such an education would be beneficial.-- ===-=-=-=-=-=== === Subject: : Re: 'erf' function in CA possible improvement is to note that the even part of R(x),> (R(x)+R(-x))/2, is equal to 1/phi(x), thus only the odd part of R(x),> (R(x)-R(-x))/2, needs to be computed:double Phi(double x)> {long double s,t=0,b=1,pwr=x;> int i;> s=x;> for(i=2;s!=t;i+=2)> { b/=(i+1);> pwr=pwr*x*x;> t=s;> s+=pwr*b;> }> return .5+s*exp(-.5*x*x-.91893853320467274178L);> }Very nice, and a definite improvement*! It speeds up the convergence> considerably which, quite possibly, accounts for more accurate results.> i.e. w/CVF on Wintel, x phi(x)> 0.123 0.5489464510164368> 1.200 0.8849303297782917> 2.400 0.9918024640754040> 6.100 0.9999999994696567> -6.100 0.0000000005303433> -1.100 0.1356660609463827> 7.200 0.9999999999996979* Actually, it's an understatement considering the Syziphian effort at> LANL some odd thirty years ago. See, netlibfn lib.It is possible to compress the code further at the expense of clarity(warning: this is not for the faint of heart):double Phi(double x){long double s=x,t=0,b=x,x2=x*x,i=1; while(s!=t) s=(t=s)+(b*=x2/(i+=2)); return .5+s*exp(-.5*x2-.91893853320467274178L);}This avoids the unnecessary computation of i+1 followed by theunnecessary conversion of the result from int to long double,eliminates the unnecessary variable pwr, and precomputes x*x.=== === Subject: : Re: 'erf' function in C> It is possible to compress the code further at the expense of> clarity (warning: this is not for the faint of heart):> double Phi(double x)> {long double s=x,t=0,b=x,x2=x*x,i=1;> while(s!=t) s=(t=s)+(b*=x2/(i+=2));> return .5+s*exp(-.5*x2-.91893853320467274178L);> }> This avoids the unnecessary computation of i+1 followed by the> unnecessary conversion of the result from int to long double,> eliminates the unnecessary variable pwr, and precomputes x*x.Have you checked this algorithm in the tail? For, say, x = -50, I eitherget inf/nan or an incorrect answer (0.5), depending on which compiler Iuse. The compiler in my tests is gcc, specifically DJGPP or MingW. Theproblem is overflow in s.Furthermore, Phi(x) is not increasing in x for values between -13 and -7,see below (DJGPP using long double). x Phi(x)-13 -2.94464e-15-12 -2.89807e-15-11 6.27672e-16-10 6.49085e-16-9 6.65294e-16-8 1.28093e-15-7 1.28042e-12Best regards,Jesper Lund=== === Subject: : Re: Give me that old time ontology: (was: the anticlassicalist }{ i: linguistic negation): Well then, he was tersely codifying the mathematical structure as a: given, which is the easier hard direction. Going in the less easy: hard direction requires talking about something very much like: ontology, whether or not we use the word ... like speaking prose, and: all that. It's our old school chums deduction and induction.:: Basically, professor, it goes kind of like this (music please ...)::: Models as a Confidence Game:: In the beginning we are confronted with a blooming buzzing surface of: sensory confusion. Being cerebral beings, we create a partially: ordered heirarchy of model things, purported to be generative of this: surface which we've been presented. The models are essentially,: roughly and enoughly, our working ontology.:: There: I've used ontology in a sentence, and it wasn't so bad, was: it? ;-):: (Richard Herring vigorously spits and loudly calls for a pint).:: Some of these models; in a model itself received and conceived -- it's: all heirarchical, you see -- are for us evolutionarily preconceived:: in our lizard brains, ancient and tough, a world populated with: persistent three-dimensional objects is enough.:: Onto these models we multiply and add; mathematical annexes, now more: than a fad, allow us to predict with numeric decision the results of: some experiments with admirable precision. This predictive concision: increases our confidence that we're on to somethin', as we laboriously: peel back the skins of the onion -- and well it should: a rigorous: predictive structure is quite good.:: Confidence, confidence ... did I not add in my vision that each model: is invested with Bayesian strength; and though we sometimes forget we: should add at greater length they are all always subject to emprical: revision?:: Hssst! The doggerels out. But I think you may get the idea::: In the great chain of model being we may focus on some precise: mathematical bits, the triumphs of a natural science known to be: correct in a range of observation with a degree confidence only: predicated on a few assumptions; like sanity and the persistence of: order. Once given these bits, reasoning from their structure is: primarily downward -- deductive -- which is something which can: similarly be carried out with great rigor and confidence. But we may: tend to forget that these models were not themselves handed to us on a: platter -- or rather they were, if we adopt the stance that they are: known givens -- but each the product of protracted groping by groups: of the most talented people of their time, there being no mechanistic: crank we can turn to integrate as readily as we differentiate. And in: this groping, like it or not, something with a strong isometry to the: ologies will out; a discussion of models and our confidence in models,: hic nomen rosa.That was really beautiful. Thank you for that.I especially liked the punch line. Umberto Eco would be proud...-- ===-=-=-=-=-=== === Subject: : Saxon Math critique by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OJL5d08719; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OJJ1i08191 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OJJ1520298;I would like information regarding the Saxon Math program. How well does it meet national math standards, etc.?=== === Subject: : Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OJL5K08715; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OJBji07909 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OJBj919796;>>My,question has been resolved neatly and Laconicaly,>>within two lines:>>Re(e^[iPi]) = Re(-1+i[0]) = -1 AND>>Im(e^[iPi]) = Im(-1+i[0]) = 0>>Since there is acceptance to this,>>any more saying has no value.>Except to point out that it means that your original conclusion that >exp(i pi) = 0 was not validly drawn from what preceded it.>David McAnally>>Yes because I was not either happy with:>>e^[ipi]+1=0 as given in my referenced book:>Well, that is a matter of taste,O.k. then , as long there is no conflict with :Re(e^[iPi]) = Re(-1+i[0]) = -1 Panagiotis Stefanides. rather than an objective observation of >whether the result is true or false. In the field of complex numbers, >exp(i pi)+1 = 0 is true, whether one likes the equality or not.>>.....However ,since the transcendental numbers formulated by Liouville>>and G. Cantor,were somehow TECHNICAL, Charles Hermite proved that>>number e ,the basis of the natural logarithms ,is not algebraic That was>>important since it was being proved that it was transcendental ,and>>related to pi via Euler's relationship>>e^[iPi]+1 =0>Euler's identity is distinct from any consideration of whether or not e is>transcendental. Spivak gives in his book, Calculus, a proof of the>transcendence of e which is completely unrelated to Euler's identity. On>the other hand, the proof that pi is transcendental is very closely>trelated to the fact that e^[i pi]+1 = 0.>>[ Eyler gave the general formula :>> e^[ix]=cosx+isinx in 1748 ,>> gave e and its value 2.718 ,and new since 1728>> the relationship:>> e^[iPi]+1 =0 ].......>>Panagiotis Stefanides>David McAnally> Despite anything you may have heard to the contrary,> the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: OT observation: Sometime it seems that the exact mening of a statement is: not clear and that it is easier to rephrase the general meaning: in one's own terms with a negative attached.:: It's nice to see that John is happy?: Well, he certaintly isn't unhappy.:: (I don't know what you mean by happiness but: I agree that he seemed to be in better shape than: the last time we saw him.)Yeah, there seems to be some kind of internal order relationship implied ina lot of the use of the negation of contraries (I think my use of the termdouble-negative was likely improper, and accept the analyses of severalothers on this point).So when we say John is happy, it establishes a point happiness as being apart of the state of John. When we say that John is unhappy, it establishesa different point unhappiness as not being a part of the state of John.Something likehappiness ------------------ unhappinessand we seem to internally identify gradations between the states. Thedifference between the statements then seems to be that the negation ofcontraries does not assert the positive condition, it merely asserts thatthe contrary position is not the case and that anywhere elsewhere on thecontinuum is possible. Not being certain of the affirmative position forwhatever reason is one exemplar of this usage.This kind of negation can be modelled. One particularly nice thing I likein Heyting models is that one already gets relationships such as~(~(a)) -> aas an order relationship.-- ===-=-=-=-=-=== === Subject: : Q: Galois field represented by vector of subfieldsI have a question: I have a galois field in GF(2^n). Now, normalpolynomial bases is a_0 + a_1x + x_2x^2 + ... a_{n-1}x^{n-1} (usingpseudo-latex here), where a_i is in GF(2), and the representation ofthe member is the (bit) vector (a_0, a_1, ... a_{n-1}).Now, a more efficient representation would be instead of using avector of GF(2) components, we use a vector of GF(2^g), where g|n (gdivides n, n = 0 mod g). The member would then be a_0 + a_1x^g +a_2x^{g^2} + ..., and would be represented by the vector (a_0, a_1,...) where a_i is a member of GF(2^g). Instead of bitwise operations,we would now be doing word operations.Basically, I'm reading this paper:http://citeseer.nj.nec.com/dewin96fast.htmlThe question I have is this: is there a computationally feasible(polynomial time) conversion algorithm from the GF(2) polynomial basisrepresentation to the GF(2^g) polynomial basis representation?I want to be able to compute logarithms for some base in the GF(2^g)representation- obviously, I'm not moduloing by x^n, as convient asthat might be. This means I'm going to need some polynomial p whereinoperations in GF(2^g) are done mod p. Can p be any (primitive)polynomial? Or does it have to have special characteristics inrelationship to the full number field I'm computing in?Pointers to papers to read would be greatly appreciated.Brian=== === Subject: : Re: Q: Galois field represented by vector of subfieldsBrian Hurt a .8ecrit :I have a question: I have a galois field in GF(2^n). Now, normal> polynomial bases is a_0 + a_1x + x_2x^2 + ... a_{n-1}x^{n-1} (using> pseudo-latex here), where a_i is in GF(2), and the representation of> the member is the (bit) vector (a_0, a_1, ... a_{n-1}).The basis you are describing is a standard (polynomial) basis, not a normal basis. Moreover, conventionally, the order is reversed:* with a standard basis (a,b,c,...,z) means a x^(n-1) + b x^(n-2) + c x^(n-3) + ... + z,* with a normal basis (a,b,c,...,z) means a x + b x^2 + c x^4 + ... + z x^(2^(n-1)).Now, a more efficient representation would be instead of using a> vector of GF(2) components, we use a vector of GF(2^g), where g|n (g> divides n, n = 0 mod g). The member would then be a_0 + a_1x^g +> a_2x^{g^2} + ..., and would be represented by the vector (a_0, a_1,> ...) where a_i is a member of GF(2^g). Instead of bitwise operations,> we would now be doing word operations.Basically, I'm reading this paper:> http://citeseer.nj.nec.com/dewin96fast.htmlThe question I have is this: is there a computationally feasible> (polynomial time) conversion algorithm from the GF(2) polynomial basis> representation to the GF(2^g) polynomial basis representation?Yes. See P1363-A, Draft 8, 7,http://www.informatik.tu-darmstadt.de/TI/Veroeffentlichung/ Artikel/Kryptographie/PublicKeyAllgemein/ -- mmhttp://www.ellipsa.net/mm@ellipsa.no.sp.am.net ( suppress no.sp.am. )=== === Subject: : Re: Galois field represented by vector of subfields> I have a question: I have a galois field in GF(2^n). Now, normal> polynomial bases is a_0 + a_1x + x_2x^2 + ... a_{n-1}x^{n-1} (using> pseudo-latex here), where a_i is in GF(2), and the representation of> the member is the (bit) vector (a_0, a_1, ... a_{n-1}).> Now, a more efficient representation would be instead of using a> vector of GF(2) components, we use a vector of GF(2^g), where g|n (g> divides n, n = 0 mod g). The member would then be a_0 + a_1x^g +> a_2x^{g^2} + ..., and would be represented by the vector (a_0, a_1,> ...) where a_i is a member of GF(2^g). Instead of bitwise operations,> we would now be doing word operations.> Basically, I'm reading this paper:> http://citeseer.nj.nec.com/dewin96fast.html> The question I have is this: is there a computationally feasible> (polynomial time) conversion algorithm from the GF(2) polynomial basis> representation to the GF(2^g) polynomial basis representation?> I want to be able to compute logarithms for some base in the GF(2^g)> representation- obviously, I'm not moduloing by x^n, as convient as> that might be. This means I'm going to need some polynomial p wherein> operations in GF(2^g) are done mod p. Can p be any (primitive)> polynomial? Or does it have to have special characteristics in> relationship to the full number field I'm computing in?> Pointers to papers to read would be greatly appreciated.The polynomial you need for this is not arbitary since it must be anirreducible polynomial with coefficients in GF(2^g) and powers up to n/g ifit is to represent GF(2^n) (for g|n). There will generally be a number ofsuitable polynomials depending on the field sizes involved.The most efficient technique for moving between different fieldrepresentations is rather dependent on the field sizes and the specificrepresentations being used. For small fields (such as that used in AES) youmight find my note at:http://fp.gladman.plus.com/AES/gfmath.pdfuseful. However this is oriented towards small field sizes since it aims toanswer the most frequent questions I get about GF fields for efficient AESimplementation. Brian Gladman===XLSolutions Corporation (www.xlsolutions-corp.com) is proudTechniques.****San Francisco, CA --------------> March, 22-23****Boston, MA ---------------------> March, 25-26Reserve your seat now at the early bird rates! Payment due AFTER theclass. Course Description:This two-day beginner to intermediate R/S-plus course focuses on a broad spectrum of topics,from reading raw data to a comparison of R and S. We will learnthe essentials of data manipulation, graphical visualizationand R/S-plus programming. We will explore statistical data analysistools,including graphics with data sets. How to enhance your plots.We will perform basic statistics and fit linear regression models.Participants are encouraged to bring data for interactive sessions With the following outline:- An Overview of R- Data Manipulation and Graphics- Using Lattice Graphics- A Comparison of R and S-Plus- How can R Complement SAS?- Writing Functions- Avoiding Loops- Vectorization- Statistical Modeling- Project Management- Techniques for Effective use of R and S- Enhancing Plots- Using High-level Plotting Functions- Building and Distributing Packages (libraries) Early Bird Research: $995 (Includes course materials and snacks); Email us for group discounts.Email Sue Turner: sue@xlsolutions-corp.comPhone: 206-686-1578Visit us: www.xlsolutions-corp.com/training.htmPlease let us know if you and your colleagues are interested in thisclassto take advantage of group discount. Register now to secure your seat!Interested in R/Splus Advanced course? email us. Cheers,Elvis Miller, PhDManager Training.XLSolutions Corporation206 686 1578www.xlsolutions-corp.comelvis@xlsolutions-corp.com=== === Subject: : Re: Cyclotomic fields by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OKAxx13208; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OJqqi11347 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OJqq423059;>> z=exp(2 pi i / n), where n is a power of an odd prime p. Then there is>> a rational prime, namely p, which has a unique prime ideal of Q(z)>> containing it. >That's not true. If q is a prime which is a primitive root modulo p^k,>then q is inert in K = Q(exp(2 pi i/p^k)), that is q generates a prime>ideal in the ring of integers of K. That's infinitely many q for each p.I think maybe I wasn't explaining myself properly. I meant that the ideal generated by p is a power of a prime ideal in Q(z), not that this is the only prime with that property. Sorry.>> Is this possible in any other case? That is, does there>> exist w = exp(2 pi i / m) where m is an odd integer which is not a power>> of a prime and such that there is a prime q which is contained in a unique>> prime ideal of Q(w)? Just wondering,>I suppose what you really want is for the ideal generated by p to be a >non-trivial power of a prime ideal. For this it is necessary that p >ramifies, which is the case iff p | m. Also you need p to be inert in>Q(exp(2 pi i / m')) where m = p^k m' and m' is prime to p. Now one case>where this happens is where q = m' is prime and p is a primitive root>modulo q. In particular if m = pq where p is a primitive root modulo q>but q is not a primitive root modulo p, then p is the only rational prime>which is a nontrivial power of a prime ideal. For instance, take p = 3 and>q = 7. Then 3 is a square of a prime ideal in the integers of >Q(exp(2pi i/21)) but 7 is the sixth power of a product of two distinct>prime ideals.>-- >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_Mark Motley=== === Subject: : More transcendentals charset=Windows-1252The other transcendentals than e and pi got my quriosity perched.So besides the obvious generalization to Lousville's construction, where onehas all the digits after the decimal point equal to k except the digits atpositions n! which might equal m=/=k, are there any famous decimals that areconstructed by using digits taken from a particular sequence a_n?For example, what about:0.11235813.... where the digits are the consequtive elements of theFibonnaci sequence?Better yet, what about the decimal where the digits are the elements of thesequence a_n=[e^n]?what about a_n=2^n?(For Champeronne's contant, a_n=n).Without knowing anything about this, my mind goes to a general theorem thatwould take care of many sequences en-masse. Any ideas or are such resultsnon-trivial?(I am interested of course only in sequences a_n:N->N, withlim{n->+oo}a_n=+oo).--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable=== === Subject: : Re: More transcendentals> The other transcendentals than e and pi got my quriosity perched.So besides the obvious generalization to Lousville's construction, where one> has all the digits after the decimal point equal to k except the digits at> positions n! which might equal m=/=k, are there any famous decimals that are> constructed by using digits taken from a particular sequence a_n?You might be interested in one or more of the following. MR1834705 (2002b:11096)Martinez, PatrickSome new irrational decimal fractions.Amer. Math. Monthly 108 (2001), no. 3, 250--253.11J72If $a_1,a_2,a_3,dots$ is a strictly increasing sequence of positive integers, denote by Dec${a_k}$ the decimal fraction $0.(a_1)(a_2)(a_3)dots$. The author shows that if Dec${a_k}$ is rational then there are a positive constant $C$ and a real number $x>1$ such that $a_k geq Cx^k$ for all $kgeq 1$. This in particular gives the well-known result that Dec${p_k}=0.23571113171923dots$ is irrational. The author shows that the result is optimal by showing that for each real $z>1$ there is a rational decimal fraction Dec${a_k}$ and a positive constant $C$ such that $a_k leq Cz^k$ for all $kgeq 1$.Reviewed by W. W. AdamsMR1274981 (95d:11088)Mercer, A. McD.(3-GLPH)A note on some irrational decimal fractions.Amer. Math. Monthly 101 (1994), no. 6, 567--568.11J72For a sequence of positive integers $a_1,a_2,cdots$ denote by Dec${a_k}$ the decimal fraction $0.(a_1)(a_2)cdots$. So if $p_k$ denotes the $k$th prime then Dec${p_k}=0.2357111317cdots$. The result of this paper is that if the integers $a_k$ are strictly increasing and if there is an integer $rgeq 0$ such that $sum_{k=1}^infty k^r/a_k = infty$ then Dec${a_k}$ is irrational. So, for example, $0.14916cdots$ is irrational. The result was known previously in the case $r=0$ ref[N. Hegyv.87ri, Amer. Math. Monthly 100 (1993), no. 8, 779--780; MR 94e:11080].Reviewed by W. W. AdamsMR0965057 (90a:11082)Loxton, J. H.(5-MCQR); van der Poorten, A. J.(5-MCQR)Arithmetic properties of automata: regular sequences.J. Reine Angew. Math. 392 (1988), 57--69.11J85 (11K16 11R04 68Q70)Not very much is known about the decimal expansions of algebraic irrational numbers. One might conjecture that these expansions are random, or even normal. In this paper it is shown that they cannot be generated by a finite automaton. The result appears as a corollary of a theorem on algebraic independence. Therefore, heavy Mahler-type machinery is used.Reviewed by F. SchweigerMR0873878 (88d:11065)Shan, Zun(PRC-HEF)A note on irrationality of some numbers.J. Number Theory 25 (1987), no. 2, 211--212.11J72 (11A63)For positive integers $g,h$, both $ge 2$, and a nonnegative integer $n$, define $(g^n)_h$ to be the number $g^n$ written in base $h$. Set $a_h(g)coloneq 0.(g^0)_h(g^1)_h(g^2)_hcdots$. n K. Mahleren ref[same journal 13 (1981), no. 2,break 268--269; MR 82j:10061] showed by a $p$-adic method that $a_h(g)$ is irrational when $h=10$ and n P. Bundschuhen ref[ibid. 19 (1984), no. 2, 248--253; MR 86a:11028] generalized this to any $h$. The present author gives the following elementary proof of Bundschuh's result. Assume $a_h(g)$ is rational. Then its decimal must be periodic with length $L$, say. Consider two possible cases. (1) If $log_h g$ is rational, it is easily seen that in the expansion of some $(g^n)_h$ the digit 0 occurs consecutively with length longer than $2L$, yielding a contradiction. (2) If $log_h g$ is irrational, using the well-known Diophantine approximation theorem of Kronecker, it is shown that some $(g^n)_h$ contains consecutive 0's of length greater than $2L$, again yielding a contradiction.Reviewed by Vichian LaohakosolMR0713564 (85b:11059)Zhu, Yao Chen(PRC-ASBJ); Ren, Jian HuaThe transcendence of the values of certain gap series at rational points. (Chinese. English summary)J. Northwest Univ. 12 (1982), no. 4, 1--8.11J81Let $sigma(z)=sumsb {n=1}sp infty alphasb nzsp {csb n}$, where ${csb n}$ is a monotone increasing sequence of natural numbers and the $alphasb n$ $(n=1, 2,cdots)$ are nonzero rational numbers. Let $Msb n$ be the least common denominator of $alphasb 1,cdots, alphasb n$. In the present paper, the following theorem is proved. Theorem: Suppose that $vert alphasb nvert levert betasb nvert $ for all sufficiently large $n$, where the $betasb n (n=1, 2,cdots)$ are real numbers satisfying $vert betasb {m+n}vert le Kvert betasb mvert ,vert betasb nvert (m, n=1, 2,cdots)$ for some constant $K$. Suppose further that $limsupsb {ntoinfty}csb n/ csb {n+1}=M<1$, $limsb {ntoinfty}(log Msb n)/csb n=0$. Then $sigma(p/q)$ is transcendental for any integers $qge 2$ and $p$ satisfying $(p, q)=1$, $vert pvert le qsp delta$, where $delta$ is a given positive number with $delta<1-M$. Some consequences of the theorem are also given in the paper, i.e., Theorems 7 and 8 in the book by Th. Schneider [Introduction to the transcendental numbers (German), see p. 35, Springer, Berlin, 1957; MR 19, 252], and the transcendence of K. Mahler's decimal fractions [Comm. Pure Appl. Math. 29 (1976), no. 6, 717--725; MR 54 #12675]. The proof of the theorem uses a consequence of D. Ridout's result [Mathematika 4 (1957), 125--131; MR 20 #32], i.e., Lemma 1 in the paper. We would like to point out that the inequality $vert alpha-p/qvert 1$, then $sigma(a)$ is a transcendental number, where $E_{n_nu}^{(k_nu)}$ is defined explicitly in terms of $a_nu,n_nu,k_nu,g$. The author also proves the transcendence of $sigma(a)$ for several special cases.Reviewed by Ming-Chit LiuMR0480372 (58 #539)Mahler, K.On some special decimal fractions.Number theory and algebra, pp. 209--214.Academic Press, New York, 1977.10F35 (10A30)Let $f(x)$ be a polynomial of degree $mgeq 1$ which assumes positive integer values for $x=1,2,3,cdots$. Let $sigma(f)$ be the decimal fraction obtained by writing the decimal forms of $f(1),f(2),cdots$ successively after the decimal point. Many years ago the author proved that $sigma(f)$ is always transcendental [Nederl. Akad. Wetensch. Proc. 40 (1937), 421--428; Zbl 17, 56]. Now he proves that, for every sufficiently large positive integer $N$, among the first $10^{N/m}$ digits after the decimal point of $$ left((10^1-1)(10^2-1)cdots(10^N-1)right)^{m+1}sigma, $$ there are at most $(m+1)N^3$ digits distinct from 9.{For the entire collection see MR 56 #11675.}Reviewed by W. J. LeVequeMR0424716 (54 #12675)Mahler, KurtOn a class of transcendental decimal fractions.Comm. Pure Appl. Math. 29 (1976), no. 6, 717--725.10F35 (10A30)Some forty years ago the author [Mathematica B (Zutphen) 6 (1937), 22--36; Jbuch 63, 155; Akad. Wetensch. Amsterdam Proc. 40 (1937), 421--428; Zbl 17, 56] proved that the decimal $0123456789101112cdots$ is transcendental. In the present paper he generalizes his earlier result as follows: Let $n$ be a positive integer variable and let $alpha(n)$ be a positive integer-valued function. Construct a nonterminating decimal by placing after the decimal point each of the one-digit numbers $1,2,cdots,9$, each repeated $alpha(1)$ times; then each of the 2-digit numbers $10,11,cdots,99$, each repeated $alpha(2)$ times; then each of the three-digit numbers $100,101,cdots,999$, each repeated $alpha(3)$ times, etc. The resulting number is proved to be trancsendental. In addition, exact conditions for this decimal to be a Liouville number are given.Reviewed by David Lee HillikerAmerican Mathematical Society American Mathematical Society201 Charles StreetProvidence, RI 02904-6248 USA-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)=== === Subject: : Re: For what purpose ? The Michelson-Morley experiment was good enough> to suggest that the speed of light in a vacuum> is the same for all observers.> Bull Mr. Relf.> The MMX showed that the velocity of the light was relative> to the air it was passing through, as predicted by Maxwell eh!> keith steinWhat about ring lasers, Keith?Do they work? :-)Paul=== === Subject: : Re: Binary To Ternary>Given a binary number>$$(a_{n-1} a_{n-2} ldots a_2 a_1 a_0)_2,$$>is there an efficient way to convert it to the ternary number>$$(a_{n-1} a_{n-2} ldots a_2 a_1 a_0)_3$$>without explicitly using the normal or binary splitting radix conversion>techniques?>>No. What could possibly be faster?> I believe the answer is yes rather than no, since radix conversion> techniques usually involve division or subtraction, and have nothing > to do with the requested conversion, which only requires n multiplies> and adds. Perhaps an example will help. Given a value such as 101101_2 > (or 45_10) he wants the value of 101101_3, that is, wants > 1+3(0+3(1+3(1+3(0+3(1))))) or 280_10.-jiwThank you James Waldby. You gave me an insight. Can anyone point me to information regarding the evaluation of polynomials with constrained coefficients in either of the sets {0,1} OR {0,1,2} at a fixed point, more specifically, at the point 3?=== === Subject: : Re: Alternative ways to solve a quadratic equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OKY0115033; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OKVRi14873 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OKVRb26229;>> equation except factoring, quadratic formula, completing the>> square,graphing or the square root method.>>> I went to http:// www.geocities.com/dirkie6/page4.html but those>> methods are too complex.>>>>> Well, any methods more advanced than the ones you mentioned>> will be more complex. It will be hard to find a simpler>> method than the ones you mentioned. If such a method existed,>> it would be presented more widely.>> But other methods do exist.>>> I will give a method based on Galois theory.>>> I wish to find the roots of the equation x^2 + bx + c = 0,>> expressed in terms of b and c, using radicals.>>> Let r be a root.>>> Note r^2 + b*r + c = 0.>> Using long division, I find that>>> x^2 + bx + c = (x - r) * (x + (b + r)).>>> Thus, the other root is -(b + r).>>> I will assume that the polynomial x^2 + bx + c is irreducible.>> You can get around this assumption by assuming that b and>> c are variable parameters rather than complex numbers.>>> It is easy to see that by adjoining r to the base field>> that we get a splitting field and that the Galois group>> is cyclic of order two.>>> As prescribed by Galois theory, I find the two Lagrange>> resolvents v_1 and v_2 given by:>>> v_1 = r - (-b-r) = 2r + b>>> v_2 = r + (-b-r) = -b>>> Since v_1 + v_2 = 2r and r is not in the base field,>> then either v_1 or v_2 is not in the base field.>> Here, v_1 must not be in the base field, since -b is.>>> Note that>>> v_1 * v_1 = 4*r^2 + 4*b*r + b^2>> = 4*(r^2 + b*r) + b^2>> = 4*(-c) + b^2>> = b^2 - 4c>>> Thus, v_1 * v_1 is in the base field since b and c are.>>> It is easy to see that 1 and r is a basis for the splitting field.>> It is easy to see that 1 and v_1 is a basis also.>>> But,>>> 1 = 1>>> v_1 = b + 2r>>> Inverting, I get>>> 1 = 1>>> r = (-b + v_1) / 2>>> Since v_1 = sqrt(b^2 - 4c), I have>>> r = (-b + sqrt(b^2 - 4c)) / 2>>> This is the quadratic formula for one of the roots.>>> The formula for the other root -(b + r) can be obtained>> by eliminating the r.>>> -- Hale>Are you sure you couldn't find a heavier hammer to crack that egg?Yes. He probably could have. Albeit, it was a pretty elegant hammer was it not? MM=== === Subject: : Documents about multiplicative order in general and mersenne numberscould anybody of you tell me where to find material about thefollowing topics: - multiplicative order in general - divisors of mersenne numbersUnfortunately I couldn't find anything except for some smallerBecause divisors of Mersenne-numbers with an exponent which is primecan only be numbers with the same multiplicative order 2.Any and all help is appreciated :o)Thank you in advance.Juergen Bullinger=== === Subject: : Re: Documents about multiplicative order in general and mersenne numbers> could anybody of you tell me where to find material about the> following topics: - multiplicative order in general> - divisors of mersenne numbershttp://www.cerias.purdue.edu/homes/ssw/cun/ is the homepage for the Cunningham project with is concerned with factoring numbers of the form b^n + 1 and b^n - 1. Material on multiplicative order can be found in pretty much any book with a title like Introduction to Number Theory, Elementary Number Theory, etc.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)=== === Subject: : Bounds on MomentsSuppose x is a non-negative random variable. Given that the firstmoment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What arethe bounds upon m3=E(x^3)?Is there a general relationship for such bounds?--Douglas=== === Subject: : Re: Bounds on Moments>Suppose x is a non-negative random variable. Given that the first>moment of x, E(x^1) and E(x^2) are m1 and m2, respectively. What are>the bounds upon m3=E(x^3)?>Is there a general relationship for such bounds?> See the thread I started with the subject, Sufficient condition for particularly informative.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu=== === Subject: : cute li'l' graph theory question?Can somebody help me?Let (X,Y,E) be a bipartite graph, such that every node in X has degreeequal to n, and every node in Y has degree less than or equal to 2n. show that there exists a matching of at least order |X|/2.G=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?> I'm taking a math history class and the assignment is to write a paper> on whether Cantor was just misinformed and ultimately innocent despite> the harm he did mathematics... or whether he intentionally strove to> murder mathematics.As I understand it, one of the main ideas of the Jewish mysticalteachings known as kaballah is that the world of the infiniteis a higher, more perfect world, and that by seeking knowledge ofthat world, we are learning about god. It appears that Cantor wasstrongly influenced by those teachings. So the conclusion would bethat he innocently tried to merge religion with mathematics. > when he was young he was sexually abused by a math teacher. Is that a fact?=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?I have to confess, this is homework, so I am not looking for a full> answer, just some hints where to look.> I'm taking a math history class and the assignment is to write a paper> on whether Cantor was just misinformed and ultimately innocent despite> the harm he did mathematics... or whether he intentionally strove to> murder mathematics.> So far I've read into his biography and I am leaning toward the 2nd> case, since when he was young he was sexually abused by a math> teacher. That leads me to believe he had a big grudge against> mathematics in general.> If I adopt that route, one of my main arguments in this paper will be> that it would be impossible for anyone to be SO wrong about infinite> numbers purely be coincidence. I mean, he'd have to be really tripping> on some hard stuff if we're to believe he REALLY believed the nonsense> he published about infinite numbers.I'm sure most of you have written a paper to this effect some time or> other in one of your mathematics history classes :-)Nathaniel Deeth> Age 11> Math GeniousNathaniel,This is supposed to be done on your own. You are not allowed to askanyone else for help, as this is cheating.Your teacher,Dr. Ben Zona=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?I have to confess, this is homework, so I am not looking for a full> answer, just some hints where to look.> I'm taking a math history class and the assignment is to write a paper> on whether Cantor was just misinformed and ultimately innocent despite> the harm he did mathematics... or whether he intentionally strove to> murder mathematics.> So far I've read into his biography and I am leaning toward the 2nd> case, since when he was young he was sexually abused by a math> teacher. That leads me to believe he had a big grudge against> mathematics in general.> If I adopt that route, one of my main arguments in this paper will be> that it would be impossible for anyone to be SO wrong about infinite> numbers purely be coincidence. I mean, he'd have to be really tripping> on some hard stuff if we're to believe he REALLY believed the nonsense> he published about infinite numbers.I'm sure most of you have written a paper to this effect some time or> other in one of your mathematics history classes :-)Nathaniel Deeth> Age 11> Math Genious***********************Oh Dear! Does J. Harris have any kids? Or even worse, is he cloning? Besides the fact that the message of N. Deeth shows clear signs of mature stupidity andthus he's suspicious of being quite older than he say he is, he ALREADY knowswhat the harm Cantor did to mathematics is, and he's only interested infinding out why (so much for intellectual honestity...). Would he REALLY be that young, Nathaniel is one of the youngest trolls I've ever met, though notquite the dumbest: for this you need to work harder, my boy!Tonio=== === Subject: : Re: Cantor: ignorant, harmless fool or intentional liar?>>Nathaniel Deeth>>Age 11>>Math GeniousI will not comment on you being a genious (such a sad destiny...), but> so young and already a crank and a troll?But he's the oldest age 11 troll on usenet :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_=== === Subject: : Re: Simple problem>I have a tank containing 200ml of a Solution A. I have a mechanism>which removes 0.1ml from the tank, and feeds in a 0.1ml drop of>Solution B. The tank is stirred continuously, and so there is no>problem mixing A and B between drops. My questions are:>1) How do I determine the proportions of A and B in the tank after a>given number of drops (n) - I am assuming it is a simple case of>(0.995) to the power of n - am I right?Almost, .1/200 = .0005, so the proportion of A in the tank is .9995^n.>2) More interestingly, how do I determine how many drops I need in>order to achive a certain proportion? For instance, how many drops>will give me 10%, 1%, 0.1%, 0.01% A?Solve the equation (.9995)^n = .1, .01, .001, or .0001 using logarithms.Rob Johnson take out the trash before replying=== === Subject: : Interesting problemI have the following problem:A and B are two disjoint sets whose union is |R+. Both A and B are closedunder sum and multiplication. Is it possible that neither A nor B is theVOID set?Thank u for your attention.{V}=== === Subject: : Square factors of the odd values of shape 3M + 2 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1OLYJQ20652; by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i1OLMKi19590 by proapp.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 $, proapp) id i1OLMKP30738; Does somebody knows such question ? I've start to develop such possibilities:1) Allready it is known, that 3M + 2 could not to be some square as it has 0 or 1 mod 3 ; there should be considered only some factor of 3M + 2 as sqare value: let 3M + 2 = (3m + 2)(3n + 1) = 9mn + 3(m + 2n) + 2 = = 3(3mn + m + 2n) + 2 where 3n+1 will create such squares of 1 mod 3 , but once 3M + 2 should be odd so also 3n + 1 ; our square factors should be of shape 3*2p + 1 ; 6p + 12) How much of such factors could we achieve ? lets see: 6*1 + 1 = 7 6*2 + 1 = 13 6*3 + 1 = 19 6*4 + 1 = 25 = 5^2 6*5 + 1 = 31 6*6 + 1 = 37 6*7 + 1 = 43 6*8 + 1 = 49 = 7^2 6*9 + 1 = 55 and so on some therms of arithmetic progression: 61; 67; 73; 79; 85; 91; 97; 103; 109; 115; 121 = 11^2 127; 133; 139; 145; 151; 157; 163; 169 = 13^2 ... and so on it looks, that only some squares of bigger prime numbers could be such square factors ... Analysing this progression, somebody could also noticed, that if some therm will be square so we'd find next prime number !? Is there some mistake or certain limit ? Thank You for Your attention Ro === === Subject: : Re: Minimally simple finite groups?>> For L_2(p^e) to be minimal simple, its order must not be divisible by 60>> (otherwise it contains A_5).> Ah. Why must it contain A_5 if its order is divisible by 60?Hi Jim,Maybe the following idea can help to answer your question.If in a field F the equationsa^2 + a - 1 = 0b^2 + 1 = 0have a solution, then the matrices 0 1A = -1 a -1 bB = b 0( which are in SL(2,F) )satisfy the relations A^5 = B^3 = (AB)^4 = [A, (AB)^2] = 1.However these are defining relations of the group SL(2,5) (with generators A,B).Hence, SL(2,F) contains a homomorphic image of SL(2,5). -1 0 Since (AB)^2 = 0 -1we have a proper embedding unless F has characteristic 2,in which case there is an embedding A_5 -> SL(2,F) = PSL(2,F).It remains to see for which finite F=GF(q) the above two equationscan be solved. Maybe the condition that 60 divides |L_2(q)| is sufficient for this. Anvita=== === Subject: : Re: Minimally simple finite groups?[...]> The subgroups of PSL(2,q) were all classified by L.E. Dickson in about> 1900. I am not sure what the best reference for that is. It is in> Huppert's book Endliche Gruppen but that is in German of course!Is it covered in Carter's Simple Groups of Lie Type? How aboutthe Atlas? (Not that a copy of that can be had for love normoney, as far as I've been able to discover. :-(> Anyway, the subgroups are roughly cyclic groups, dihedral groups> of order dividing q-1 or q+1 (q odd) or 2(q-1) or 2(q+1) (q even),> semidirect products PD for a p-group P of order dividing q and cyclic> group D of order dividing q-1,Isn't P always a Sylow p-subgroup, and elementary abelian? And|D| = q-1 for q even, |D| = (q-1)/2 for q odd, D a Cartansubgroup and so PD a Borel subgroup? Are the Borel groups themaximal parabolics? Is the normalizer N(D) of D always dihedral?> A_4, S_4 whenever 24 divides order,? PSL(2,8) has order 504 = 2^3 * 3^2 * 7, but no subgroupisomorphic to S_4 or A_4.> A_5 whenever> 60 divides order, and PSL(2,r) and sometimes PGL(2,r) where q is a power> of r.Under what circumstances PGL(2,r)? Guessing from PSL(2,9) =~A_6, maybe when the order-r subgroups split into 2 conjugacyclasses? (Which seems like it should be easy to see by lookingat the Borel groups?)> In other words, A_5 is the only `sporadic' simple subgroup of> PSL(2,q). The fact that it occurs whenever 60 divides the order probably> follows from the fact that SL(2,5) has a 2-dimensional complex> representation, but don't push me for details there!-- === === Subject: : Re: Minimally simple finite groups?>[...]>> The subgroups of PSL(2,q) were all classified by L.E. Dickson in about>> 1900. I am not sure what the best reference for that is. It is in>> Huppert's book Endliche Gruppen but that is in German of course!>Is it covered in Carter's Simple Groups of Lie Type? How about>the Atlas? (Not that a copy of that can be had for love nor>money, as far as I've been able to discover. :-( John Roberts-Jones=== === Subject: : Re: Minimally simple finite groups?>[...]>> The subgroups of PSL(2,q) were all classified by L.E. Dickson in about>> 1900. I am not sure what the best reference for that is. It is in>> Huppert's book Endliche Gruppen but that is in German of course!>Is it covered in Carter's Simple Groups of Lie Type? How about>the Atlas? (Not that a copy of that can be had for love nor>money, as far as I've been able to discover. :-(I don't believe that the full description of the subgroups is in eitherof those. You will find maximal subgroups of individual groups in theAtlas.>> Anyway, the subgroups are roughly cyclic groups, dihedral groups>> of order dividing q-1 or q+1 (q odd) or 2(q-1) or 2(q+1) (q even),>> semidirect products PD for a p-group P of order dividing q and cyclic>> group D of order dividing q-1,>Isn't P always a Sylow p-subgroup, and elementary abelian? And>|D| = q-1 for q even, |D| = (q-1)/2 for q odd, D a Cartan>subgroup and so PD a Borel subgroup? Are the Borel groups the>maximal parabolics? Is the normalizer N(D) of D always dihedral?The largest such P is a Sylow p-subgroup and is elementary abelian.But I was describing an arbitrary subgroup of PSL(2,q).Again, the maximal D has the order you say, and the maximal PD is aBorel subgroup. Yes, the normalizer of D is dihedral provided D isnontrivial.>> A_4, S_4 whenever 24 divides order,>? PSL(2,8) has order 504 = 2^3 * 3^2 * 7, but no subgroup>isomorphic to S_4 or A_4.Yes you are right. The S_4 occur only for odd q.>> A_5 whenever>> 60 divides order, and PSL(2,r) and sometimes PGL(2,r) where q is a power>> of r.>Under what circumstances PGL(2,r)? Guessing from PSL(2,9) =~>A_6, maybe when the order-r subgroups split into 2 conjugacy>classes? (Which seems like it should be easy to see by looking>at the Borel groups?)I think, for odd q, PSL(2,q) contains PGL(2,r) iff q is an even power of r.For example PSL(2,p^2) contains PGL(2,p), but PSL(2,p^3) does not.Of course, for even q, PSL(2,q) = PGL(2,q).Derek Holt.=== === Subject: : Re: Clean coordinates on a unit sphere> Hi - I'm a computer science student working in the field of vision and > graphics. I'm looking for a clean way to represent a _direction_ in 3d > space (this can also be imagined as a position on a unit sphere).I don't like the traditional methods of using a 3d vector or spherical > coordinates for the following reasons:- 3d Vector e.g. (x, y, z): This is of course very common - can be > rotated with matrices, quaternions... but it also describes length, > which I don't need.- Spherical coordinates e.g. (theta, phi): This is more appropriate as > it does not describe length. But it still appears ugly to me since > multiple coordinate pairs can describe the same point e.g. (x, pi/2).Is there a better way? I'm sure I'm not the first who asked this > question. It seems similar to the problem of creating a 2d map of earth, > so it's probably been thought about before... And since there doesn't > seem to be a _popular_ solution, it's probably hard - but that's what > makes it FUN !!!My ideal solution would map a unit sphere with a uniform density (if > that makes any sense to any one). Could a person use some type of > tesselation of a sphere with an arbitrary density???Any ideas are good ideas! Please keep the language simple - I'm not as > smart as you.> NathanHow about using unit vectors (x,y,z), i.e., satisfying x^2+y^2+z^2=1?This seems clean, unless by clean you really mean efficient, since(x,y,z) does contain redundant information. For efficiency, you couldjust store (x,y,sign(z)), since then z is uniquely determined by z =sign(z)*sqrt(1-x^2-y^2). But then you're back some reduncency, sinceon the equator x^2+y^2=1, the sign of z is automatically zero.You say My ideal solution would map a unit sphere with a uniformdensity. Do you mean that you want a bijective map f:U-->S from asubset of the Euclidean plane to the unit sphere that preserves area,or at least so that for (nice) subsets V of U, f satisfiesArea(f(V))=c*Area(V) for a constant c that does not depend on V? I'mnot quite sure how otherwise to interpret your word density.Joe Silverman=== === Subject: : Re: Clean coordinates on a unit sphere> Hi - I'm a computer science student working in the field of vision and> graphics. I'm looking for a clean way to represent a _direction_ in 3d> space (this can also be imagined as a position on a unit sphere).> I don't like the traditional methods of using a 3d vector or spherical> coordinates for the following reasons:> - 3d Vector e.g. (x, y, z): This is of course very common - can be> rotated with matrices, quaternions... but it also describes length,> which I don't need.> - Spherical coordinates e.g. (theta, phi): This is more appropriate as> it does not describe length. But it still appears ugly to me since> multiple coordinate pairs can describe the same point e.g. (x, pi/2).> Is there a better way? I'm sure I'm not the first who asked this> question. It seems similar to the problem of creating a 2d map of earth,> so it's probably been thought about before... And since there doesn't> seem to be a _popular_ solution, it's probably hard - but that's what> makes it FUN !!!> My ideal solution would map a unit sphere with a uniform density (if> that makes any sense to any one). Could a person use some type of> tesselation of a sphere with an arbitrary density???> Any ideas are good ideas! Please keep the language simple - I'm not as> smart as you.You might try the following to map the surface of a unit sphere:Represent each point by two coordinates...# its longitude, from -pi to pi# its straight-line distance from the plane of the equator, from -1 to 1You will notice that this maps the surface of the sphere (area 4*pi) onto arectangle (area 4*pi).Also, any shape that you draw on the rectangle is mapped to a shape of thesame area on the sphere. This may be your uniform density criterion.I know that we still have the problem of the longitude of the poles beingundefined.-- Clive Toothhttp://www.clivetooth.dk=== === Subject: : Re: Clean coordinates on a unit sphere> Represent each point by two coordinates...# its longitude, from -pi to pi> # its straight-line distance from the plane of the equator, from -1 to 1You will notice that this maps the surface of the sphere (area 4*pi) onto a> rectangle (area 4*pi).> Also, any shape that you draw on the rectangle is mapped to a shape of the> same area on the sphere. This may be your uniform density criterion.Very interesting... I've never heard of that option before. Yes, you've suggestion!Any more suggestions out there?Nathan=== === Subject: : missing pages from Spanier's Algebraic TopologyI ordered a copy of Spanier's Algebraic Topology, and it arrivedtoday, minus page 13-14. This is mildly frustrating; although Idon't really need to see the definition of a category again, it feelsquite incomplete, and disrupts the flow of exposition something awful. I suppose I could complain to the bookseller, but I doubt it's hisfault, and I fear it will take some time to get a replacement fromSpringer. Instead, I was wondering if someone with access to ascanner would just scan these two pages (page 13 and 14 in the 1991printing--basically the first two pages of chapter 1) and email me thescans at npenton@hotmail.com. Anyone who would take the few minutesto do this for me will be rewarded with my undying gratitude andpossibly the soul of my first-born son.Nathan