8632 Subject: Re: Minimally simple finite groups? >Which of the finite simple groups are minimally simple, i.e., >have all of their proper subgroups solvable? Obviously the only >alternating group that qualifies is A_5 =~ L(2,4) =~ L(2,5), and >I know the list also includes L(2,7) =~ L(3,2), L(2,8), L(2,13), >... On the other hand, I also know it *doesn't* include L(2,9) =~ >A_6 or L(2,11), both of which contain subgroups isomorphic to >A_5. The classification of minimal simple groups was a consequence of John Thomspon's classification of nonsolvable groups in which nontrivial solvable subgroups have solvable normlaizers, which was one of the big classification theorems that came before CFSG. This is in Bull. Amer. Math. Soc. 74, 1968, 383-437. The simple groups coming out of Thompson's Theorem are L_2(q), Sz(q), L_3(3), M_11, A_7, U_3(3). Of course, these are not all minimal simple. L_3(3) is, but M_11, A_7, U_3(3) are not. Sz(2^e) is minimal simple whenever it does not contain a smaller Sz(2^f), which I guess is equivalent to e being prime. For L_2(p^e) to be minimal simple, its order must not be divisible by 60 (otherwise it contains A_5). Also it must not contain any smaller simple L_2(p^f), so if p>=3, then we must have e=1. But L_2(2^e) and L_2(3^e) will be minimal simple for e prime provided their order is no divisible by 60. Something like that, anyway! Derek Holt. Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K0Bfi06959; >> 2) [(A-->B)(A'-->B')]' = AB' U A'B >> 3) [(A'-->B)(A-->B')]' = A'B' U AB > >2) is just symmetric difference, assuming the existence of some universe The symmetric difference is indeed AB' U A'B. You can also find it in Hewitt and Stromberg (1965) p. 4. It obeys the associative and commutative laws and even a version of the distributive law under intersection (p. 6 ibid). However, Hewitt and Stromberg never got as far as A-->B and so on since they weren't looking for it. It definitely helps to not be looking for something else even when walking. See my replies to the other posts if I get that far today. Osher Subject: Genetics and Math-Ability Just curious: What is the current scientific opinion regarding how math-ability is inherited? Is it, in general, a recessive trait or is it dominate? (I say recessive.) I am only half-serious, since math ability is most probably a result of many traits and experiences and education. But I was wondering today because, even though there are some in my family who were math-teachers, almost everyone else had/has little ability with mathematics...even very little ability as relative to me... ('relative to me'...snicker, snicker...) Subject: Re: Genetics and Math-Ability There is an old saying about inheritance of mathematical ability. Unlike most interitance, this goes from a man to his son-in-law. The explanation being, I guess, that when the professor's daughter is of the right age, she meets the professor's current student, and marries him. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Subject: Re: Genetics and Math-Ability >There is an old saying about inheritance of mathematical ability. >Unlike most interitance, this goes from a man to his son-in-law. >The explanation being, I guess, that when the professor's daughter is >of the right age, she meets the professor's current student, and >marries him. My wife, her sister and her first cousin are also married to mathematicians. Obviously there's an inherited tendency to marry mathematicians. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 Subject: Re: Genetics and Math-Ability > There is an old saying about inheritance of mathematical ability. > Unlike most interitance, this goes from a man to his son-in-law. > The explanation being, I guess, that when the professor's daughter is > of the right age, she meets the professor's current student, and > marries him. I have known three cases of identical twins, one of whom was a mathematician and the other one wasn't. In one case, the other twin went into physics, but in the other two they did something unrelated to math. To me that says that while genetics may play a role, there is a lot more to it than that. In my case, I am the only one in a rather large extended family who went into math. I did have a distanct cousin who was a physicist, but that is the nearest, AFAIK. Even if it were genetic, it wouldn't make sense to ask if it was dominant or recessive unless it was a simple Mendelian trait (essentially, one gene). Even simple Mendelian traits aren't always. For example, the common blood groups are supposedly simple Mendelian and they mainly are in the sense that the blood type is mostly determined by one gene. But more than one gene is involved in producing the A and B alleles and it is entirely possible (though unlikely) for two type O parents to have an A or B offspring, contrary to what you read. Mathematical ability _and interest_ are clearly determined, insofar as they are genetic, by many, many genes and their interaction with the environment. Subject: Re: Genetics and Math-Ability > Just curious: > What is the current scientific opinion regarding how math-ability is > inherited? Math ability depends entirely on what is currently fashionable in math. If cloning is currently fashionable, and you can add, you are a genuis. If not, well there's always chemistry. > Is it, in general, a recessive trait or is it dominate? > (I say recessive.) > I am only half-serious, since math ability is most probably a result > of many traits and experiences and education. > But I was wondering today because, even though there are some in my > family who were math-teachers, almost everyone else had/has little > ability with mathematics...even very little ability as relative to > me... > ('relative to me'...snicker, snicker...) > Subject: Re: Genetics and Math-Ability > Just curious: > What is the current scientific opinion regarding how math-ability is > inherited? > Is it, in general, a recessive trait or is it dominate? > (I say recessive.) I don't have any formal stats about it, but it seems not to run in families very much. Among the famous names, there were several Bernoulli's and a couple of Jacobi's. Today there are the Chudnovsky's and the Voevodsky's, but that's all I can think of, not that I know much about it. But I'll bet a lot of mathematicians have near relatives in other sciences. I'd be interested in any stats about birth order, because temperament seems to matter in this game. Consider two brothers, one a doctor and the other a math professer. I'll bet the doctor is the older of the two. LH Subject: Re: Genetics and Math-Ability >I don't have any formal stats about it, but it seems not to run in families >very much. Among the famous names, there were several Bernoulli's and a >couple of Jacobi's. Today there are the Chudnovsky's and the Voevodsky's, >but that's all I can think of, not that I know much about it. Among those I know are the three Browder brothers (Felix, Andrew and William), the two Fefferman brothers (Charles and Robert), the two Hsiang brothers (Wu-chung and Wu-yi), and the two Borwein brothers (Peter and Jonathan). There are many parent-and-child pairs. >But I'll bet a lot of mathematicians have near relatives in other sciences. >I'd be interested in any stats about birth order, because temperament seems >to matter in this game. Consider two brothers, one a doctor and the other a >math professer. I'll bet the doctor is the older of the two. The way I heard it, mathematicians are most likely to be firstborn. I don't know about physicians. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 Subject: Re: Genetics and Math-Ability >>I don't have any formal stats about it, but it seems not to run in >>families very much. Among the famous names, there were several Bernoulli's >>and a couple of Jacobi's. Today there are the Chudnovsky's and the >>Voevodsky's, but that's all I can think of, not that I know much about it. > Among those I know are the three Browder brothers (Felix, Andrew and > William), the two Fefferman brothers (Charles and Robert), the two > Hsiang brothers (Wu-chung and Wu-yi), and the two Borwein brothers > (Peter and Jonathan). There are many parent-and-child pairs. In Britain we have Mary Rees (Liverpool) and Sarah Rees (Newcastle), Daughters of David Rees (Exeter emertius). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Subject: Re: Genetics and Math-Ability I don't have any formal stats about it, but it seems not to run in > families very much. Among the famous names, there were several > Bernoulli's and a couple of Jacobi's. Today there are the Chudnovsky's > and the Voevodsky's, but that's all I can think of, not that I know much > about it. And apparently Euler had something like 11 or 12 children, but we didn't even see a 2 or 3 generation reign of Eulers like the Bernoulli family had. (At least not to my knowledge.) J Subject: Re: Genetics and Math-Ability > Just curious: > What is the current scientific opinion regarding how math-ability is > inherited? > Is it, in general, a recessive trait or is it dominate? > (I say recessive.) > I am only half-serious, since math ability is most probably a result > of many traits and experiences and education. > But I was wondering today because, even though there are some in my > family who were math-teachers, almost everyone else had/has little > ability with mathematics...even very little ability as relative to > me... > ('relative to me'...snicker, snicker...) > Here's my experience. My father was really good at math, but grew up in a time when trivial pursuits like math for math's sake weren't as important as learning a trade. My sister (the oldest in the family) had to go to summer school in high school because of math, twice. My oldest brother is a chemist of some sort and did well enough in math, but nothing spectacular. My older brother has a master degree in math and teaches high school math. And there is me, the youngest, who is getting a masters degree in math and planning on startting a Ph.D. program in August for math. So it seems the math ability in my family increased with each child. Maybe it is just a product of my environment, but we all went to the same grade school and high school and even had the same teachers. I know that is not conclusive, or probably doesn't even have a point either way, but that's just how it is in my family. - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University email is: news (dot) post (at) tbrauch (dot) com Subject: mathcad 11 I have plugged the expression Sum(n=1->m)(10^n)/n, and asked for a symbolic solution. The answer that came back involved, among other things, the natural log of -9, a red m superimposed on a parenthesis, and something illegible (black)superimposed on a plus sign. Did I do something wrong? I have to say that I am completely at sea with computers. Any help will be appreciated, including the proper answer, if it exists, for my sum. Thanks, C. F. Connie Eaton Subject: Re: mathcad 11 > I have plugged the expression Sum(n=1->m)(10^n)/n, and asked for a > symbolic solution. The answer that came back involved, among other > things, the natural log of -9, a red m superimposed on a > parenthesis, and something illegible (black)superimposed on a plus > sign. Did I do something wrong? I have to say that I am completely > at sea with computers. Any help will be appreciated, including the > proper answer, if it exists, for my sum. Hmmm. The same bug (and it is a bug) occurs in Mathcad 2001. Subject: Notation Question I'm wondering if there is a standard name (i.e. something like the Kronecker Delta Function) for the following: (read d subscript i for any integer i) d_i = 1 ( i >= 0 ) = 0 ( i <= -1 ) I realize that this is basically a discrete form of the Heaviside function, but somehow I doubt that this function goes by that name. Is there a standard name for this thing in the math/physics world? Thanks, JB Subject: Re: Church-Turing compared to Zuse-Fredkin thesis (two new papers) Distribution: inet <4030e88c.32841454@netnews.att.net <40311933.36090852@netnews.att.net <40311cb1.36985181@netnews.att.net> <5jeYb.26213$Ov3.20124@newssvr25.news.prodigy.com <%etYb.24944$2c3.7452@newssvr27.news.prodigy.com The web page states, by implication, > * All algebraic numbers are roots of Diophantine equations of > degree 1 or 2. > Cube root of 2 is an algebraic number which is not the root of any > Diophantine equation of degree 1 or 2. Hence, the web page's statement > is disproved by this example. > As I understand it, we may say that all algebraic numbers are roots > of Diophantine equations; but that's all. > I understand this differently: A Diophantine equation is an equation in > which only integer solutions are allowed. > So I don't see how bringing up Diophantine equations is relevant. Whoops! I misspoke: read polynomial equation with integer coefficients, not Diophantine equation. Sorry about that. Still, the web page is wrong. > That website focuses on numbers: rational, irrational and constructible. > ---------|----------------Real Numbers-----------|------------ > Rational numbers Irrational numbers > are all algebraic are algebraic or > transcendental Correct, after re-formatting. :) > All algebraic numbers (including algebraic irrationals) > of the 1st or 2nd degree, or those having a power of two, > e.g., x2, x4, x8, x16, x32, x64, x128, ..., etc., if given a line > segment of unit length, are numbers that can be constructed by the > classical Greek geometric method of straightedge and compass. > SH: The author is pointing out that some algebraic irrationals > are constructible. For instance the square root of 2. Right. > The cube root of 2 is not constructible and it well known that > in general cube roots (trisecting etc.) are not constructible. Right. > His explanation disallows including cube roots, which is > the type of counterexample you provided and thought injured > the correctness of his description. Right. His explanation disallows cube roots from being called algebraic; however, cube roots *are* called algebraic; thus, his explanation is flawed. I don't know what that page means by algebraic equation. > You can find numerous definitions on Google. No, actually, I can't. If I had, you can be assured I wouldn't be posting otherwise! MathWorld doesn't know the term, and neither does anyone on the first page of Google results. Post a URL to the definition if you want to use it. However, I have made the assumption above that when you write algebraic equation, you mean zero = some polynomial with integer coefficients. Is that right? -Arthur Subject: Re: Church-Turing compared to Zuse-Fredkin thesis (two new papers) > [Followups set to sci.math, who may be able to explain what an > algebraic number is better than I can.] I and the website author http://members.ispwest.com/r-logan/narrative.html understand that an algebraic number is the solution to a algebraic equation. Your criticism is unfounded and I think is caused by not realizing that an algebraic equation and polynomial equation with integer coefficients, mean just about the same thing. Knowing that is not so uncommon for somebody who feels qualified to point out errors in math websites. > > The web page states, by implication, > > * All algebraic numbers are roots of Diophantine equations of > degree 1 or 2. > > Cube root of 2 is an algebraic number which is not the root of any > Diophantine equation of degree 1 or 2. Hence, the web page's statement > is disproved by this example. > > As I understand it, we may say that all algebraic numbers are roots > of Diophantine equations; but that's all. > I understand this differently: A Diophantine equation is an equation in > which only integer solutions are allowed. > So I don't see how bringing up Diophantine equations is relevant. > Whoops! I misspoke: read polynomial equation with integer > coefficients, not Diophantine equation. Sorry about that. Still, > the web page is wrong. polynomial equation with integer coefficients, That looks to me like the definition of algebraic equation: http://www.cut-the-knot.org/do_you_know/numbers.shtml# algebraic ... Real roots of such equations are said to be algebraic. In other words, a number a is called algebraic if it satisfies an algebraic equation: Pn(a)=0 for some polynomial Pn(x) with integer coefficients. Which is P_n(a)=0 for some polynomial P_n(x) with integer coefficients in case the formatting is lost. > That website focuses on numbers: rational, irrational and constructible. > ---------|----------------Real Numbers-----------|------------ > Rational numbers Irrational numbers > are all algebraic are algebraic or > transcendental > Correct, after re-formatting. :) > All algebraic numbers (including algebraic irrationals) > of the 1st or 2nd degree, or those having a power of two, > e.g., x2, x4, x8, x16, x32, x64, x128, ..., etc., if given a line > segment of unit length, are numbers that can be constructed by the > classical Greek geometric method of straightedge and compass. > SH: The author is pointing out that some algebraic irrationals > are constructible. For instance the square root of 2. > Right. > The cube root of 2 is not constructible and it well known that > in general cube roots (trisecting etc.) are not constructible. > Right. > His explanation disallows including cube roots, which is > the type of counterexample you provided and thought injured > the correctness of his description. > Right. His explanation disallows cube roots from being called algebraic; however, cube roots *are* called algebraic; thus, > his explanation is flawed. Wrong. Read this again. All algebraic numbers (including algebraic irrationals)...[SH: then follows a rule excluding cube roots as algebraic solutions to algebraic equations] can be *constructed* His rule is explaining that irrational cube roots are not solutions to algebraic equations which are at the same time constructible. He is defining _constuctible_ algebraic irrational solutions which of course are some but not all solutions to algebraic equations having irrational solutions. Of course he calls cube roots algebraic (or more precisely algebraic equations which have algebraic irrational solutions). being called algebraic; satisfy algebraic equations such as x^ 2 - 2 = 0 and x^3 - 11 = 0. I previously quoted this from his webpage. x^3 -11 = 0 will have the solution: x = the cube root of 11 The author is stating that this cube root solution is not constructible. Not, that the cube root is not an algebraic solution; because he gives an example of algebraic irrational cube root equation in his definition. I don't know what that page means by algebraic equation. > You can find numerous definitions on Google. > No, actually, I can't. If I had, you can be assured I wouldn't > be posting otherwise! MathWorld doesn't know the term, and neither > does anyone on the first page of Google results. Post a URL to > the definition if you want to use it. > However, I have made the assumption above that when you write algebraic equation, you mean zero = some polynomial with integer > coefficients. Is that right? > -Arthur http://members.ispwest.com/r-logan/glossary.html#glo same author Yes. The author of the website you find flawed also provides the polynomial with integer coefficients definition as a generalized equation. Algebraic number - A real number that is the root of an equation in the form ... [ see webpage for equation with correct formatting] http://mathforum.org/dr.math/faq/faq.impossible.construct.html Three geometric construction problems from antiquity puzzled mathematicians for centuries: the trisection of an angle, squaring the circle, and duplicating the cube. Are these constructions impossible? The impossibility proofs depend on the fact that the only quantities you can obtain by doing straightedge-and-compass constructions are those you can get from the given quantities by using addition, subtraction, multiplication, division, and by taking square roots. These numbers are called Euclidean numbers, and you can think of them as the numbers that can be obtained by repeatedly solving the quadratic equation. These three problems require either taking a cube root or constructing pi. A cube root is not a Euclidean number, and Lindemann showed that pi is a transcendental number, which means that it is not the root of an *algebraic equation with integer coefficients*, making it too non-Euclidean. The author of the website to which you object is not stating that cube root solutions are not the roots of algebraic equations. He is saying they are not constructible [see above] quote. Transcendental numbers like Pi, are the numbers which are neither constructible nor the root of an algebraic equation (polynomial equation with integer coefficients). Algebraic equations have integer coefficients. What Pi and cube roots have in common is their inability to be constructed-- not that they both are non-algebraic equations-- and the website author doesn't claim they are both non-algebraic as you impute, but that they are both non-constructible. SH: As to finding results on google it was easier for me since I knew that an algebraic number was a solution generated by an algebraic equation. Actually what I thought was strange about your post was that you had a cmu.edu email address and were imprecise/limited with your math definitions. This was a google result: http://www.mayer.dial.pipex.com/samples/pi/tex4ht/pisample.html A complex number is algebraic over Q if it is a root of a polynomial equation with rational coefficients. http://groups.google.com/groups?q=%22algebraic+equation%22+ root+definition&h l=en&lr=&ie=UTF-8&oe=UTF-8&selm=2540%40tecsun1.tec.army.mil& rnum=1 2. :TRANSCENDENT 1a 3. a. incapable of being the root of an algebraic equation with rational coefficients (Pi is transcendental number). An AltaVista search for algebraic equation returned AltaVista found 8,739 results Stephen Subject: Re: combitorics charset=Windows-1252 > Torrey loves Matt on some days, but Torrey hate Matt on other days. > Torrey also loves Paul on some days, but Torrey hates Paul on other > days. > Torrey also loves Sandy on some days, but she never hates Sandy. > Sandy hates Torrey on some days, but loves her on other days > Matt loves Sandy on some some days, but hates her on other other days > Paul Loves Matt on some days, but Paul hates other days. > paul hates Sandy on some days. > how can I figure out how many possible combinations there are that > desribes the love and hate relationship between Torrey, Matt, Paul, > and Sandy? > I have 32. but I'm not sure if i'm right. and i'm doing it brute force > way. A description of the love/hate relationships in the group would be given by a table as follows, with 24 True/False values, but the given information is only enough to fix the values shown: How A Feels About B Person A Person B Love Hate -------- -------- ------------------- Torrey Matt T T Torrey Paul T T Torrey Sandy T F Matt Torrey Matt Paul Matt Sandy T T Paul Torrey Paul Matt T T Paul Sandy ? T Sandy Torrey T T Sandy Matt Sandy Paul It's not quite clear how Paul feels about Sandy, but if the ? is known to be T, then that leaves 10 values undetermined; i.e, there are 2**10 = 1,024 different tables consistent with the given information, and each one corresponds to a different set of love/hate relationships in the group. --r.e.s. Subject: Re: combitorics sorry. I did leave out some information, about paul and sandy and I left out gary. here's more complete information. Torrey loves Matt on some days, but Torrey hates Matt on other days. Torrey loves Paul on some days, but Torrey hates Paul on other days. Torrey loves Sandy on some days, but Torrey never hates Sandy. Sandy hates Torrey on some days, but Sandy loves Torrey on other days. Sandy loves Gary on some days but Sandy never hates Gary. Matt loves Sandy on some days, but Matt hates Sandy on other days Paul Loves Matt on some days, but Paul hates other days. Paul hates Sandy on some days, but Paul never loves Sandy. Gary hates Sandy, but Gary never loves Sandy. Gary Loves Torrey on some days but Gary hates Torrey on other days. I have reduced the problem to just hate and love relationship. is there anything wrong with doing that? because I'm ignoring the time/day information. what about the time/love/hate relationships? is there a way to figure out what pattern they have? with the information above, taking only the love and hate relationships, there are 32 possibles. what happens if i take into consideration the days? or is there a way to do that? what other information will one need in order to do that? I think gary and paul should be nicer to sandy. thanks all in advance. sean > Torrey loves Matt on some days, but Torrey hate Matt on other days. > Torrey also loves Paul on some days, but Torrey hates Paul on other > days. > Torrey also loves Sandy on some days, but she never hates Sandy. > Sandy hates Torrey on some days, but loves her on other days > Matt loves Sandy on some some days, but hates her on other other days > Paul Loves Matt on some days, but Paul hates other days. > paul hates Sandy on some days. > > how can I figure out how many possible combinations there are that > desribes the love and hate relationship between Torrey, Matt, Paul, > and Sandy? > I have 32. but I'm not sure if i'm right. and i'm doing it brute force > way. > A description of the love/hate relationships in the > group would be given by a table as follows, with > 24 True/False values, but the given information is > only enough to fix the values shown: > How A Feels About B > Person A Person B Love Hate > -------- -------- ------------------- > Torrey Matt T T > Torrey Paul T T > Torrey Sandy T F > Matt Torrey > Matt Paul > Matt Sandy T T > Paul Torrey > Paul Matt T T > Paul Sandy ? T > Sandy Torrey T T > Sandy Matt > Sandy Paul > It's not quite clear how Paul feels about Sandy, but > if the ? is known to be T, then that leaves 10 values > undetermined; i.e, there are 2**10 = 1,024 different > tables consistent with the given information, and > each one corresponds to a different set of love/hate > relationships in the group. > --r.e.s. Subject: Re: combitorics > I'm not sure if this is a good group to post this. > I have a combinatorics problem. well, a puzzle really. I'm not sure > what kind of problem this is. so I would love to get some feedbacks > from the group. ( as to the nature of the problem as well as how to > solve it) > Let's say you have a little sister named Torrey. And she has quite a > quite a number of friends in kindergarten. let's say her friendship > goes as follows. > Torrey loves Matt on some days, but Torrey hate Matt on other days. > Torrey also loves Paul on some days, but Torrey hates Paul on other > days. > Torrey also loves Sandy on some days, but she never hates Sandy. > Sandy hates Torrey on some days, but loves her on other days > Matt loves Sandy on some some days, but hates her on other other days > Paul Loves Matt on some days, but Paul hates other days. > paul hates Sandy on some days. > to make it a bit simpler... > Torrey loves and hates Matt > Torrey loves and hates Paul > Torrey loves Sandy > Sandy loves and hates Torrey > Matt loves and hates Matt > Paul loves and hates Matt > Paul hates Sandy > how can I figure out how many possible combinations there are that > desribes the love and hate relationship between Torrey, Matt, Paul, > and Sandy? > I have 32. but I'm not sure if i'm right. and i'm doing it brute force > way. > any helpful thoughts and comments are appreciated. > thank you all. > sean you're right. 32 = 2^5. There are two possibilities: either torrey loves matt or hates him. In either case there are two more possibilities: either torrey loves Paul or hates him. After these cases we have 2 * 2 = 4 possibilities: (love, love), (love, hate), (hate, love), (hate, hate). Imagine a 2x2 square with each of these in them. Clearly if you multiply the number of possibilities for the first thing (2) by the number of possibilities for the second thing (2), you get the total number of possibilities. Now there are 5 choices, so the total is 2^5 = 2 * 2 * 2 * 2 * 2 = 32. Subject: Re: combitorics > Torrey loves and hates Matt > Torrey loves and hates Paul > Torrey loves Sandy > Sandy loves and hates Torrey > Matt loves and hates Matt > Paul loves and hates Matt > Paul hates Sandy Well, there are five loves and hates relationships in the above description. Assuming that they are independent, then the total number of possibilities are 2^5=32. -Michael. Subject: combitorics I'm not sure if this is a good group to post this. I have a combinatorics problem. well, a puzzle really. I'm not sure what kind of problem this is. so I would love to get some feedbacks from the group. ( as to the nature of the problem as well as how to solve it) Let's say you have a little sister named Torrey. And she has quite a quite a number of friends in kindergarten. let's say her friendship goes as follows. Torrey loves Matt on some days, but Torrey hate Matt on other days. Torrey also loves Paul on some days, but Torrey hates Paul on other days. Torrey also loves Sandy on some days, but she never hates Sandy. Sandy hates Torrey on some days, but loves her on other days Matt loves Sandy on some some days, but hates her on other other days Paul Loves Matt on some days, but Paul hates other days. paul hates Sandy on some days. to make it a bit simpler... Torrey loves and hates Matt Torrey loves and hates Paul Torrey loves Sandy Sandy loves and hates Torrey Matt loves and hates Matt Paul loves and hates Matt Paul hates Sandy how can I figure out how many possible combinations there are that desribes the love and hate relationship between Torrey, Matt, Paul, and Sandy? I have 32. but I'm not sure if i'm right. and i'm doing it brute force way. any helpful thoughts and comments are appreciated. thank you all. sean Subject: malta-new discovery Can you believe that the oldest stone building in the world has a mathematical perfection? Check: www.star-mysteries.com Subject: Re: malta-new discovery > Can you believe that the oldest stone building in the world has a > mathematical perfection? Check: > www.star-mysteries.com Doug Subject: Re: malta-new discovery X-No-Archive: yes While still snuggled in a 'spider hole', ivanmohoric@volja.net (temacnik) scribbled: >Can you believe that the oldest stone building in the world has a >mathematical perfection? I believe someone could PRETEND that it does. To reply by email, remove the XYZ. Lumber Cartel (tinlc) #2063. Spam this account at your own risk. This sig censored by the Office of Home and Land Insecurity.... Subject: malta-new discovery Can you believe that the oldest stone building in the world has a mathematical perfection? Check: www.star-mysteries.com Subject: Re: malta-new discovery > Can you believe that the oldest stone building in the world has a > mathematical perfection? Check: > www.star-mysteries.com Hey a real money making scam. They make stupid claims and you buy the ------------------------------------------- Quote Ancient Architects On Malta I - NEW Malta archipelago can be proud of the oldest stone-build structures on our planet supposed to be ancient temples build by ignorant farmers five millennia ago. Yet they are geometrically perfect. Who designed them any why? BUY THIS ARTICLE --------------------------------------------- Subject: 1=ma+nb m,n belongs to Z a and b is realtively prime. Is there any proof of this? or just an axiom? m,n belongs to Z a and b is realtively prime. ma+nb=1 does not hold in {Z+}? If so prove it. Subject: Re: 1=ma+nb > m,n belongs to Z > a and b is realtively prime. > Is there any proof of this? or just an axiom? > m,n belongs to Z > a and b is realtively prime. > ma+nb=1 does not hold in {Z+}? If so prove it. If a and b are members of Z and are relatively prime (have no non-unit common factors, then there are m and n in Z such that a M + b N = 1. If a and b are of the same sign, then m and n must be of opposite signs. If a and b are of opposite signs then m and n must be of the same sign. One method of finding suitable m and n for given a and b is through the extended Euclidean algorithm. See, for example: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm Subject: Re: 1=ma+nb Virgil > m,n belongs to Z > a and b is realtively prime. > Is there any proof of this? or just an axiom? ... > See, for example: > http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm A similar procedure, called Blankinship's algo, goes like this, for a=39 and b=23: 39 1 0 23 0 1 16 1 -1 7 -1 2 2 3 -5 1 -10 17 ;and now 1 = -10*39 + 17*23 The first two rows are a 1 0 b 0 1 and each subsequent row is a linear combination of the two previous rows, as in Euclid's algo. When the leftmost column reaches gcd(a,b), you can read off m and n. LH Subject: Re: 1=ma+nb >> snip > A similar procedure, called Blankinship's algo, goes like this, for a=39 and > b=23: >snip > and each subsequent row is a linear combination of the two previous rows, as > in Euclid's algo. When the leftmost column reaches gcd(a,b), you can read > off m and n. > LH This is great and much more suited to a spreadsheet than the algorithm I currently use. It is also obvious that it works, since lhs=a*rhs_1+b*rhs_2 is an invariant on each line, and is guaranteed to reach lhs=gcd(a,b) by Euclid. Who or why or when or what is Blankinship? Good on him/her. JJ Subject: Re: 1=ma+nb John Jones > This [Blankinship's algo] is great and much more suited to a > spreadsheet than the algorithm I currently use. ... > Who or why or when or what is Blankinship? Good on him/her. Nobody seems to have noticed this tidy routine before W. A. Blankinship in 1963. I learned it only recently, from this: http://mathworld.wolfram.com/BlankinshipAlgorithm.html LH Subject: Re: 1=ma+nb In-reply-to: Larry Hammick , Bill Dubuque Virgil >> m,n belongs to Z >> a and b is realtively prime. >> >> Is there any proof of this? or just an axiom? >... >> See, for example: >> http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm >A similar procedure, called Blankinship's algo, goes like this, for a=39 and >b=23: >39 1 0 >23 0 1 >16 1 -1 >7 -1 2 >2 3 -5 >1 -10 17 ;and now 1 = -10*39 + 17*23 >The first two rows are >a 1 0 >b 0 1 >and each subsequent row is a linear combination of the two previous rows, as >in Euclid's algo. When the leftmost column reaches gcd(a,b), you can read >off m and n. Last April, I developed this algorithm and called it the Euclid-Wallis Algorithm: Bill Dubuque mentioned that he had previously posted this algorithm, which he called the Extended Euclidean Algorithm: . I see that Blankinship published this algorithm in 1963 and it has an entry at MathWorld: m,n belongs to Z >a and b is realtively prime. >Is there any proof of this? or just an axiom? >m,n belongs to Z >a and b is realtively prime. >ma+nb=1 does not hold in {Z+}? If so prove it. This is called Bezout's Identity. I have a page about it at Let A be a subset of (X,T). Then A is dense in X iff for every non-empty > open subset U of X, A / U != {}. > I have seen two definitions of 'separable topological space': > a) (X,T) is separable if there exists A X, where A is countable and > dense in X > b) (X,T) is separable if there exists A X, where A is > countable and dense in X > Given def (a), its simple to prove that all countable spaces X are > automatically separable since the subset X is countable and non-trivially > intersects every non-empty open subset. However, this won't work given def > (b). I first became suspicious when I saw a proof that all countable spaces > were separable that seemed ridiculously complex in comparison to the > seemingly obvious 1 step proof above. I later found definitions of > separable like def (b). > l8r, Mike N. Christoff > I have never seen definition (b). I have seen the symbol $subset$ > used to mean subset, not proper subset, so maybe that is what you saw. > Certainly a one-point space with the discrete topology is to be > considered separable, even though it has no dense proper subsets. I've read the other follow-ups, and I think you've hit the nail on the head: the OP is confused about the meaning of subset, and is taking A subset B to mean A is a proper subset of B. I have never seen definition (b) either. It would be very weird for finite sets NOT to be separable. (And it would make the useful fact, A subspace of a separable metric space is also separable false. --Ron Bruck Subject: Re: Definition of Separable Space (basic topology question) > Let A be a subset of (X,T). Then A is dense in X iff for every non-empty > open subset U of X, A / U != {}. > > I have seen two definitions of 'separable topological space': > > a) (X,T) is separable if there exists A X, where A is countable and > dense in X > b) (X,T) is separable if there exists A X, where A is > countable and dense in X > > Given def (a), its simple to prove that all countable spaces X are > automatically separable since the subset X is countable and non-trivially > intersects every non-empty open subset. However, this won't work given def > (b). I first became suspicious when I saw a proof that all countable spaces > were separable that seemed ridiculously complex in comparison to the > seemingly obvious 1 step proof above. I later found definitions of > separable like def (b). > > > > l8r, Mike N. Christoff > I have never seen definition (b). I have seen the symbol $subset$ > used to mean subset, not proper subset, so maybe that is what you saw. > Certainly a one-point space with the discrete topology is to be > considered separable, even though it has no dense proper subsets. > I've read the other follow-ups, and I think you've hit the nail on the > head: the OP is confused about the meaning of subset, and is taking A subset B to mean A is a proper subset of B. > I have never seen definition (b) either. It would be very weird for > finite sets NOT to be separable. (And it would make the useful fact, A subspace of a separable metric space is also separable false. Ok. I'm tired and don't immediately see anything wrong with this: (X,T): X = {1,2,3,4} T = { X,{},{1,2},{3,4} } A = {1,3} Finite space with dense proper subset... l8r, Mike N. Christoff Subject: Re: Definition of Separable Space (basic topology question) Content-transfer-encoding: 8bit > > > Let A be a subset of (X,T). Then A is dense in X iff for every > non-empty > > open subset U of X, A / U != {}. > > > I have seen two definitions of 'separable topological space': > > > a) (X,T) is separable if there exists A X, where A is > countable and > > dense in X > > b) (X,T) is separable if there exists A X, where A is > > countable and dense in X > > > Given def (a), its simple to prove that all countable spaces X are > > automatically separable since the subset X is countable and > non-trivially > > intersects every non-empty open subset. However, this won't work > given def > > (b). I first became suspicious when I saw a proof that all countable > spaces > > were separable that seemed ridiculously complex in comparison to the > > seemingly obvious 1 step proof above. I later found definitions of > > separable like def (b). > > > > > l8r, Mike N. Christoff > > I have never seen definition (b). I have seen the symbol $subset$ > used to mean subset, not proper subset, so maybe that is what you saw. > > Certainly a one-point space with the discrete topology is to be > considered separable, even though it has no dense proper subsets. > I've read the other follow-ups, and I think you've hit the nail on the > head: the OP is confused about the meaning of subset, and is taking A subset B to mean A is a proper subset of B. > I have never seen definition (b) either. It would be very weird for > finite sets NOT to be separable. (And it would make the useful fact, A subspace of a separable metric space is also separable false. > Ok. I'm tired and don't immediately see anything wrong with this: > (X,T): > X = {1,2,3,4} > T = { X,{},{1,2},{3,4} } > A = {1,3} > Finite space with dense proper subset... Presumably re: my remark It would be very weird for finite sets NOT to be separable. I didn't mean that ALL finite sets wouldn't be separable. (Although, if the topology is Hausdorff, it means exactly that.) It suffices for a SINGLE finite set not to be separable to set off my weirdness alarm. But I've lost the thread of what I was thinking when I said it would falsify subspace of separable metric space is separable. I'm sure I had something in mind... --Ron Bruck Subject: Re: Definition of Separable Space (basic topology question) > N. > > > Let A be a subset of (X,T). Then A is dense in X iff for every > non-empty > > open subset U of X, A / U != {}. > > > > I have seen two definitions of 'separable topological space': > > > > a) (X,T) is separable if there exists A X, where A is > countable and > > dense in X > > b) (X,T) is separable if there exists A X, where A is > > countable and dense in X > > > > Given def (a), its simple to prove that all countable spaces X are > > automatically separable since the subset X is countable and > non-trivially > > intersects every non-empty open subset. However, this won't work > given def > > (b). I first became suspicious when I saw a proof that all countable > spaces > > were separable that seemed ridiculously complex in comparison to the > > seemingly obvious 1 step proof above. I later found definitions of > > separable like def (b). > > > > > > > > l8r, Mike N. Christoff > > > I have never seen definition (b). I have seen the symbol $subset$ > > used to mean subset, not proper subset, so maybe that is what you saw. > > > Certainly a one-point space with the discrete topology is to be > > considered separable, even though it has no dense proper subsets. > > I've read the other follow-ups, and I think you've hit the nail on the > head: the OP is confused about the meaning of subset, and is taking > A subset B to mean A is a proper subset of B. > > I have never seen definition (b) either. It would be very weird for > finite sets NOT to be separable. (And it would make the useful fact, > A subspace of a separable metric space is also separable false. > > Ok. I'm tired and don't immediately see anything wrong with this: > (X,T): > X = {1,2,3,4} > T = { X,{},{1,2},{3,4} } > A = {1,3} > Finite space with dense proper subset... > Presumably re: my remark It would be very weird for finite sets NOT to > be separable. I didn't mean that ALL finite sets wouldn't be > separable. (Although, if the topology is Hausdorff, it means exactly > that.) It suffices for a SINGLE finite set not to be separable to set > off my weirdness alarm. > But I've lost the thread of what I was thinking when I said it would > falsify subspace of separable metric space is separable. I'm sure I > had something in mind... The following assumes topological spaces, and subspaces must be non-empty. Let X be a separable metric space, and S a single element subspace of X. Assume A is a proper dense subset of S under the induced topology. Then A must equal {}, but it must also non-trivially intersect S, which is impossible. l8r, Mike N. Christoff Subject: Re: Definition of Separable Space (basic topology question) >Let A be a subset of (X,T). Then A is dense in X iff for every non-empty >open subset U of X, A / U != {}. >I have seen two definitions of 'separable topological space': >a) (X,T) is separable if there exists A X, where A is countable and >dense in X >b) (X,T) is separable if there exists A X, where A is >countable and dense in X >Given def (a), its simple to prove that all countable spaces X are >automatically separable since the subset X is countable and non-trivially >intersects every non-empty open subset. However, this won't work given def >(b). I first became suspicious when I saw a proof that all countable spaces >were separable that seemed ridiculously complex in comparison to the >seemingly obvious 1 step proof above. I later found definitions of >separable like def (b). The integers with the usual topology is separable. Definition (a) is the correct one. BTW, finite spaces are separable. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 Subject: Re: Definition of Separable Space (basic topology question) >Let A be a subset of (X,T). Then A is dense in X iff for every non-empty >open subset U of X, A / U != {}. >I have seen two definitions of 'separable topological space': >a) (X,T) is separable if there exists A X, where A is countable and >dense in X >b) (X,T) is separable if there exists A X, where A is >countable and dense in X >Given def (a), its simple to prove that all countable spaces X are >automatically separable since the subset X is countable and non-trivially >intersects every non-empty open subset. However, this won't work given def >(b). I first became suspicious when I saw a proof that all countable spaces >were separable that seemed ridiculously complex in comparison to the >seemingly obvious 1 step proof above. I later found definitions of >separable like def (b). > The integers with the usual topology is separable. Right. This would follow from the fact that any topology on a countable space is separable. > Definition (a) is the correct one. BTW, finite > spaces are separable. I consider finite spaces to be a subset of the class of countable spaces. However, from what I recall, this isn't a universal definition. http://en.wikipedia.org/wiki/Countable A countable (or denumerable) set is a set which is either finite or countably infinite. Is there something else I'm missing here? l8r, Mike N. Christoff Subject: Re: Definition of Separable Space (basic topology question) >Let A be a subset of (X,T). Then A is dense in X iff for every non-empty >open subset U of X, A / U != {}. >I have seen two definitions of 'separable topological space': >a) (X,T) is separable if there exists A X, where A is countable > and >dense in X >b) (X,T) is separable if there exists A X, where A is >countable and dense in X >Given def (a), its simple to prove that all countable spaces X are >automatically separable since the subset X is countable and non-trivially >intersects every non-empty open subset. However, this won't work given > def >(b). I first became suspicious when I saw a proof that all countable > spaces >were separable that seemed ridiculously complex in comparison to the >seemingly obvious 1 step proof above. I later found definitions of >separable like def (b). > The integers with the usual topology is separable. > Right. This would follow from the fact that any topology on a countable > space is separable. Oops! I get you. ie: no proper subset's closure equals the integers.... duh > Definition (a) is the correct one. BTW, finite > spaces are separable. I was thinking you may have meant a similar thing about finite sets (ie: no dense proper subset), but: (X,T): X = {1,2,3,4} T = { X,{},{1,2},{3,4} } A = {1,3} l8r, Mike N. Christoff Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFc18878; >>[Snip: in Doctorowese A' is the completement of A, A+B is the union >>of A and B and AB is the intersection of A and B] >> >>> 2) (A')' = A >> >>> This doesn't correspond to differentiation, >>Yet another thing that doesn't correspond to differentiation >>is that >>AB' + A'B =/= (AB)'. >>Indeed I am at a loss reading these posts to find any analogy between >>complementation and differentiation (save that Doctor Osherow uses >>the same notation for both). >Evidently you've forgotten that derivatives satisfy > (f + g)' = (f')(g'). Contrary to appearance, actually added the last two lines above as a joke expanding on Robin Chapman's claim that I used A + B to be the union of A and B, or else Ullrich himself wove this claim into Robin's post (it's hard to figure it out where this quote came from, so I'm inclined to believe that Ullrich invented this claim based on the nature of Ingenious Imitation as it pervades world colleges at the faculty level). He then asserts that (f + g)' = (f')(g'). This also is his invention, the purpose of which beyond emotionality escapes me. I think that he is asserting a contradiction and projecting it upon the external world in the manner of some with whom we are very familiar from recent experiences. Osher Doctorow Subject: Re: Sets That Resemble Derivatives Somewhat >>>[Snip: in Doctorowese A' is the completement of A, A+B is the union >>>of A and B and AB is the intersection of A and B] >>> >>>> 2) (A')' = A >>> >>>> This doesn't correspond to differentiation, >>>Yet another thing that doesn't correspond to differentiation >>>is that >>>AB' + A'B =/= (AB)'. >>>Indeed I am at a loss reading these posts to find any analogy between >>>complementation and differentiation (save that Doctor Osherow uses >>>the same notation for both). >>Evidently you've forgotten that derivatives satisfy >> (f + g)' = (f')(g'). > Contrary to appearance, actually added the last > two lines above as a joke expanding on Robin Chapman's claim that I used A > + B to be the union of A and B, or else Ullrich himself wove this claim > into Robin's post (it's hard to figure it out where this quote came from, > so I'm inclined to believe that Ullrich invented this claim based on the > nature of Ingenious Imitation as it pervades > world colleges at the faculty level). He then asserts that (f + g)' > = (f')(g'). This also is his invention, the purpose of which beyond > emotionality escapes me. I think that he is asserting a contradiction > and projecting it upon the external world in the manner of some > with whom we are very familiar from recent experiences. Why don't you talk proper? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFf18885; >there's a general theme that some people have explored that says that >the fact that certain sorts of differential operators obey >leibniz-like identities is conceptually subordinate to the fact that >certain sorts of boundary operators in geometry and topology obey >leibniz-like identities, roughly (or sometimes precisely, depending on >the particular concept of boundary being used) boundary(x X y) = >(boundary(x) X y) + (x X boundary(y)). This looks interesting, although I think that themes of conceptual subordination may be a bit premature here, but I am very encouraged by the lack of hostility in your message and by your ability to detect relationships. How did a nice guy like you get in the same thread as Robin and Ullrich? Osher Doctorow Subject: Re: Sets That Resemble Derivatives Somewhat >>there's a general theme that some people have explored that says that >>the fact that certain sorts of differential operators obey >>leibniz-like identities is conceptually subordinate to the fact that >>certain sorts of boundary operators in geometry and topology obey >>leibniz-like identities, roughly (or sometimes precisely, depending on >>the particular concept of boundary being used) boundary(x X y) = >>(boundary(x) X y) + (x X boundary(y)). > This looks interesting, although I think that themes of conceptual > subordination may be a bit premature here, but I am very encouraged > by the lack of hostility in your message and by your ability to > detect relationships. How did a nice guy like you get in the same > thread as Robin and Ullrich? Detect what relationship. As I pointed out there is no analogy between complementation and boundary/derivative. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Subject: Re: Sets That Resemble Derivatives Somewhat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1K2cFZ18867; >>in what? >No. The right side looks similar, 'cos of your denoting complement >by the same notation as derivative, but the right side is completely >different! >One idea? Not the idea of it being a limit of difference >quotients methinks. >> Derivations are a topic in the literature on Lie Algebras and other >> structures, but (2) and (3) have a rather different flavor. >I.e., by not resembling derivation at all? >applause! >Like er yeah! I mean yeah if x + h is different to x then er yeah >then x + h is outside x. Wow, groovy! Aha! The universe itslef. Next you'll be giving us the answer >to life the universe and everything. >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Francis Wheen, _How Mumbo-Jumbo Conquered the World I omit Robin Chapman's quotation of Lacan on penises, and I omit my comments to which these are replies (except that Lacan was Robin's own quotation) for brevity. We don't have to really worry about compliments in what (Did I write compliment? Perhaps a Freudian slip). Robin is in charge of that. The next comment concerns the right side looking similar because of my denoting complement by the same notation as derivative.... Let's stop there for a moment. So far, nothing's wrong beyond beyond being novel in use of symbols. I mean, if a student of yours claimed A is similar to B because the symbols used to express A and B, perhaps he hasn't given the answer to a pile of homework, but it might even be deeper! Here I use the ordinary notation for complements of sets, prime ('). I point out that something = AB' + A'B in equation (2), and I leave it for readers to notice that this is a similar form to the right side of (fg)' = fg' + f'g. (No, I'm not defining f and g here - take a guess!) Now, Matt Grime who I doubt wants to get involved in this argument did point out that AB' + A'B is the symmetric difference, which Robin didn't, so Matt wasn't as upset as Robin at that point. If a student in one's Math class said: Look at (fg)' = fg' + f'g. Isn't the right hand side similar in symbols to (AB)' = AB' + A'B, one might well think of giving that student an A for noticing unusual relationships, though admittedly nothing yet has been done to apply it to one's pet projects. However, Robin apparently took me to be a college teacher (right on!), and I can understand his disappointment that I only seemed to be at the level of an A student up to that point. Or perhaps Robin grades on the basis of homework exclusively, in which case slightly different problems arise so to speak. Next, Robin used the word 'cos to mean because. Robin, cos with an argument is an abbreviation for cosine, and we discourage students for using it without an argument unless explicitly stated, but it's O.K. this time. Perhaps you mean some relationship between 'cos and something else which escapes me, and if so, do explain. (Is that a current keyword, 'cos'?) But Robin continues: 'cos of your denoting complement by the same notation as derivative, but the right side is completely different! So Robin doesn't detect the relationship between AB' + A'B and fg' + f'g, so we know that this isn't an I.Q. test. I have also been chastised for using the same symbol for sets like A, B and functions like f, g, in order to point out a similarity in pattern - not bad for an I.Q. test, but this isn't an I.Q. test (well, not by definition, anyway). The next comment of Robin is: One idea? Not the idea of it being a limit of difference quotients methinks. Here Robin's Spirit of Science-Math as opposed to Letter of Science-Math involves the idea that the limit is found in derivatives but not in set comple- ments. If this were stated as a simple fact, it would be interest- ing but still as a non-similarity one suspects that life would be much more Creative looking for Similarities, and despite the warning Robin has not yet eliminated Similarities. I won't quibble about Letters of Science-Math such as for example methinks, which are presumably not intended to conjure up confusion between drama and Mathematics, unless of course it is in the same code as 'cos, one an expanding code and the other a contracting code (this might in fact be the relationship that we're looking for!). I could go on and on like this, but I think that most readers get the idea. Osher Doctorow Subject: Re: Sets That Resemble Derivatives Somewhat >>>in what? >>No. The right side looks similar, 'cos of your denoting complement >>by the same notation as derivative, but the right side is completely >>different! >>One idea? Not the idea of it being a limit of difference >>quotients methinks. >>> Derivations are a topic in the literature on Lie Algebras and other >>> structures, but (2) and (3) have a rather different flavor. >>I.e., by not resembling derivation at all? >>applause! >>Like er yeah! I mean yeah if x + h is different to x then er yeah >>then x + h is outside x. Wow, groovy! > Aha! The universe itslef. Next you'll be giving us the answer >>to life the universe and everything. >>Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >>Francis Wheen, _How Mumbo-Jumbo Conquered the World > I omit Robin Chapman's quotation of Lacan on penises, and I omit my > comments to which these are replies (except that Lacan was Robin's > own quotation) for brevity. For brevity, or for deliberate obscurity? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Subject: Fresnel integrals 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2) 2) int[-inf to inf] exp(-x^2) dx = sqrt(pi) Eqn (2) is easy to show by looking at the square of the left side (a double integral) and switching to polar coordinates. I tried the same approach with (1). (There are other ways to get (1), I know.) The double integrand will be (cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2 The first term on the right integrates to zero okay, but the second term, in polar coordinates, runs into a convergence problem. Moreover the conversion to polars requires some justification, because the Fresnel integrals are not absolutely convergent. Anybody know any quick fix for this one? I get the feeling I'm missing something obvious... TIA, Larry Subject: Re: Fresnel integrals In-reply-to: Larry Hammick 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2) >2) int[-inf to inf] exp(-x^2) dx = sqrt(pi) >Eqn (2) is easy to show by looking at the square of the left side >(a double integral) and switching to polar coordinates. I tried the >same approach with (1). (There are other ways to get (1), I know.) >The double integrand will be >(cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2 >The first term on the right integrates to zero okay, but the second >term, in polar coordinates, runs into a convergence problem. >Moreover the conversion to polars requires some justification, >because the Fresnel integrals are not absolutely convergent. >Anybody know any quick fix for this one? I get the feeling I'm >missing something obvious... This is usually done by contour integration, which we will do below. However, let us try fixing up your idea. Let |oo 2 2 A+iB = | (cos(x ) + i sin(x )) dx [1] |-oo Then we can try the same trick that worked for exp(-x^2): 2 (A+iB) |oo |oo 2 2 2 2 = | | (cos(x ) + i sin(x ))(cos(y ) + i sin(y )) dx dy |-oo |-oo |oo |oo 2 2 2 2 = | | (cos(x + y ) + i sin(x + y )) dx dy |-oo |-oo |2pi |oo 2 2 = | | (cos(r ) + i sin(r )) rdr d@ | 0 | 0 |oo = pi | (cos(r) + i sin(r)) dr [2] | 0 However, this integral does not converge, so I don't think this method will work. Let's try contour integration. Let w = (1-i)/sqrt(2), so that w^2 = -i. Let z = wx. Then |oo 2 2 | (cos(x ) + i sin(x )) dx |-oo |oo 2 = | exp(ix ) dx |-oo 1 |oo w 2 = - | exp(-z ) dz [3] w |-oo w The path of integration is a line through the origin with slope -1. For x > 0, along the vertical path from the path of integration in [3] to the x-axis, | |0 2 | | | exp(-(x+iy) ) i dy | | |-x | |0 2 2 <= | exp(-(x - y ) dy |-x |x 2 2 = | exp(-x + (y-x) ) dy | 0 |x <= | exp(-xy) dy | 0 <= 1/x A similar situation holds for x < 0. Since exp(-z^2) has no poles and the integrals along the paths connecting the path of integration and the x-axis vanish at oo, we have [3] 1 |oo 2 = - | exp(-z ) dz w |-oo 1+i = ------- sqrt(pi) sqrt(2) = (1+i)sqrt(pi/2) [4] Thus, [4] tells us that |oo 2 |oo 2 | cos(x ) dx = | sin(x ) dx = sqrt(pi/2) |-oo |-oo Rob Johnson 1) int[-inf to inf] cos x^2 dx = sqrt(pi/2) > 2) int[-inf to inf] exp(-x^2) dx = sqrt(pi) > Eqn (2) is easy to show by looking at the square of the left side > (a double integral) and switching to polar coordinates. I tried the > same approach with (1). (There are other ways to get (1), I know.) > The double integrand will be > (cos x^2)(cos y^2) = (cos(x^2 - y^2) + cos(x^2 + y^2)) /2 > The first term on the right integrates to zero okay, but the second > term, in polar coordinates, runs into a convergence problem. > Moreover the conversion to polars requires some justification, > because the Fresnel integrals are not absolutely convergent. > Anybody know any quick fix for this one? I get the feeling I'm > missing something obvious... Dunno, but (1) is the wrong thing to consider --- you should consider integral_{-infinity}^infinity exp(i x^2) dx of which (1) is the real part. Of course these are *improper* integrals). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ Subject: Re: x^2 + y^4 = z^4 >Would you explain why? A squarefree positive integer n is a congruent number if and only if n is the area of a right triangle with rational sides. As examples, 1, 2, 3 are not congruent numbers while 5, 6, 7 are congruent numbers. n - sides of rational Pythagorean triangle 5 - (3/2, 20/3, 41/6) 6 - (3, 4, 5) 7 - (35/12, 24/5, 337/60) A number n is a congruent number if and only if n(m^2) is the area of a rational right triangle. In particular, that 1 is not a congruent number is equivalent to there being no (rational) right triangle whose area is the square of an integer. See Koblitz's book Introduction to Elliptic Curves and Modular Forms for more info. John Robertson Subject: Re: x^2 + y^4 = z^4 > the equation x^2 + y^4 = z^4 has no positive integer solutions. >> Is the proof ... short enough for some kind soul to post it >An elementary proof is based on descent. That is, assume a >solution exists in positive x, y, z, and show that another one can be >found with smaller z. Are you sure? That's true of x^4 + y^4 = z^2 where infinite descent is used at http://www.math.toronto.edu/mathnet/plain/questionCorner/ fermat4.html However I see no way to modify the method used there to this problem. ---- Subject: Re: x^2 + y^4 = z^4 > Subject: Re: x^2 + y^4 = z^4 > >> the equation x^2 + y^4 = z^4 has no positive integer solutions. > >> Is the proof ... short enough for some kind soul to post it > > >An elementary proof is based on descent. That is, assume a > >solution exists in positive x, y, z, and show that another one can be > >found with smaller z. > Are you sure? > That's true of x^4 + y^4 = z^2 where infinite descent is used at > http://www.math.toronto.edu/mathnet/plain/questionCorner/ fermat4.html > However I see no way to modify the method used there to this problem. > ---- Here is an elementary proof by descent. We may assume WLOG that x, y and z are pairwise coprime, and that there is a minimal solution with xyz <> 0. There are two cases: I. x odd, y even, z odd Assume that y is minimal and nonzero. x = a^2 - b^2 y^2 = 2ab z^2 = a^2 + b^2 with a, b coprime and of opposite parity [known parametric solution] a = c^2 b = 2d^2 c, d coprime, c odd (alternatively a = 2c^2, b = d^2, which leads to the same result) [follows from y^2 = 2ab with gcd(a,b) = 1] a = e^2 - f^2 b = 2ef e, f coprime and of opposite parity [follows from z^2 = a^2 + b^2 with a odd, b even, a coprime to b] d^2 = ef c^2 = e^2 - f^2 [combining the above] e = g^2 f = h^2 [follows from d^2 = ef with gcd(e,f) = 1] c^2 = g^4 - h^4 c^2 + h^4 = g^4 [by substitution] y^2 = 2ab = 4c^2 d^2 = 4ef(e^2-f^2) = 4g^2 h^2 (g^4-h^4) > h^2 > 0 [by substitution and assuming y <> 0] Thus c^2 + h^4 = g^4 is a solution with 0 < |h| < |y|, which contradicts the assumption that y is minimal and nonzero. QED case I. II. x even, y odd, z odd Assume that x is minimal and nonzero. x = 2ab y^2 = a^2 - b^2 z^2 = a^2 + b^2 with a, b coprime, a odd, b even [known parametric solution] a = c^2 - d^2 b = 2cd c, d coprime and of opposite parity [follows from z^2 = a^2 + b^2] a = e^2 + f^2 b = 2ef e, f coprime and of opposite parity [follows from y^2 = a^2 - b^2] cd = ef c^2 - d^2 = e^2 + f^2 c, e odd, d, f even [combining the above] c = gh d = ij e = gj f = hi g, h, i, j pairwise coprime g, h, j odd, i even [follows from cd = ef, gcd(c,d) = gcd(e,f) = 1, g = gcd(c,e), etc.] g^2 h^2 - i^2 j^2 = g^2 j^2 + h^2 i^2 [by substitution] g^2 (h^2 - j^2) = i^2 (h^2 + j^2) [by rearrangement] 2g^2 = h^2 + j^2 2i^2 = h^2 - j^2 [follows from gcd(g,i) = gcd(h,j) = 1, g, h, j odd, i even] h^2 = g^2 + i^2 j^2 = g^2 - i^2 [by addition and subtraction] k = 2gi k^2 + j^4 = h^4 [by substitution] x = 2ab = 4cd(c^2-d^2) = 4ghij(g^2 h^2-i^2 j^2) = 2hjk(g^2 h^2-i^2 j^2) > k > 0 [by substitution and assuming x <> 0] Thus k^2 + j^4 = h^4 is a solution with 0 < |k| < |x|, which contradicts the assumption that x is minimal and nonzero. QED case II. Subject: Re: x^2 + y^4 = z^4 > Subject: Re: x^2 + y^4 = z^4 > >> the equation x^2 + y^4 = z^4 has no positive integer solutions. > >> Is the proof ... short enough for some kind soul to post it > > >An elementary proof is based on descent. That is, assume a > >solution exists in positive x, y, z, and show that another one can be > >found with smaller z. > Are you sure? > That's true of x^4 + y^4 = z^2 where infinite descent is used at > http://www.math.toronto.edu/mathnet/plain/questionCorner/ fermat4.html > However I see no way to modify the method used there to this problem. > ---- The impossibility of x^4 + y^4 = z^4 follows immediately from the impossibility of x^4 + y^4 = z^2 since z^4 = (z^2)^2, i.e. a solution to the first equation necessarily gives one to the second. The descent is used for x^4 + y^4 = z^2. Subject: Re: x^2 + y^4 = z^4 > >> the equation x^2 + y^4 = z^4 has no positive integer solutions. > >> Is the proof ... short enough for some kind soul to post it > >An elementary proof is based on descent. That is, assume a > >solution exists in positive x, y, z, and show that another one can be > >found with smaller z. > Are you sure? > That's true of x^4 + y^4 = z^2 where infinite descent is used at > http://www.math.toronto.edu/mathnet/plain/questionCorner/ fermat4.html > However I see no way to modify the method used there to this problem. descent methods for both x^4 + y^4 = z^2 and x^4 - y^4 = z^2 can be found in : http://www.mathpages.com/home/kmath288.htm -- philippe (chephip at free dot fr) Subject: Re: . The hardest of all hard facts . as in mister Peter's quote, there just haven't *been* any duplications of M-M-M, because of the massive say-so about their results; there are certainly anomalies in both results. > for this. D.C.Miller's refinement highlighted the anomolies > that MM had found > They found none. Miller's results could not be duplicated by anyone but > Miller with his apparatus. > It was dismissed because it could not be repeated. It was dismissed > because it was not verfiable. of course, a lot of this is moot, if we allow Cheeny to grind us into Tony's McCrusade (Usama's MacJihad). see my sig. --Give the World a Trickier Dick Cheeny -- out of office after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac Subject: Re: . The hardest of all hard facts . Dear Brian Quincy Hutchings: > as in mister Peter's quote, > there just haven't *been* any duplications of M-M-M, because > of the massive say-so about their results; > there are certainly anomalies in both results. There are *dozens* of documented MMX experiments with different apparati. Nothing *repeatable* has been found within their ability to resolve. Except for Peter's secular variation in the aether anisotropy, there had been nothing extraordinary. David A. Smith Subject: Re: . The hardest of all hard facts . > Hi Eleaticus, Re: How you think, Michelson-Morley and Kennedy-Thorndike do indeed fit > Galilean ( c + v ) physics , > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers. > That makes it: The hardest of all hard facts. > No experiment ever showed that the speed of light is > the same for all observers. Indeed any determination > of the one way speed of light can be used to demonstrate > the speed of light is NOT the same for all observers.... > A light----> B <-you > < ----------- L --------------> v m/s > Use syncronised clocks at A amd B to time how long it takes > light to travel a distance of L meters across the laboratory.. > Speed of light relative to the laboratory = L/ (tB - tA) = c > where 'tA' is the time at which the light left A > and 'tB' is the at which the light arrived at B > Now repeat the experiment while running towards B at v m/s > Note that 'in your frame of reference' the point B is moving , > so that the light must travel an extra distance = v * (tB - tA) > which is the distance B has moved as the light travels from > A to B. > Therefore: > Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) > = c + v > keith stein You had measured the relative velocity between a light ray front and a moving you with v velocity. Notice that the velocities of your two moving entities are c and v, measured in the Laboratory inertial frame. As c and v refer to the same inertial frame, you have the right to use ordinary vector algebra obtaining c+v, because Relative velocity can be >c and is NOT invariant (see my post with that title). If you measured the relative velocity in the inertial frame you is at rest you obtain c as the result, compatible with the second postulate of Special Relativity. ANY light continue having the same vacuum speed c for ANY observer AT REST in ANY inertial frame, The hardest of all hard facts . RVHG Subject: Re: . The hardest of all hard facts . > Hi Eleaticus, Re: How you think, Michelson-Morley and Kennedy-Thorndike do indeed fit > Galilean ( c + v ) physics , > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers. > That makes it: The hardest of all hard facts. > No experiment ever showed that the speed of light is > the same for all observers. Indeed any determination > of the one way speed of light can be used to demonstrate > the speed of light is NOT the same for all observers.... > A light----> B <-you > < ----------- L --------------> v m/s > Use syncronised clocks at A amd B to time how long it takes > light to travel a distance of L meters across the laboratory.. > Speed of light relative to the laboratory = L/ (tB - tA) = c > where 'tA' is the time at which the light left A > and 'tB' is the at which the light arrived at B > Now repeat the experiment while running towards B at v m/s > Note that 'in your frame of reference' the point B is moving , > so that the light must travel an extra distance = v * (tB - tA) > which is the distance B has moved as the light travels from > A to B. > Therefore: > Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) > = c + v > keith stein You had measured the relative velocity between a light ray front and a moving you with v velocity. Notice that the velocities of your two moving entities are c and v, measured in the Laboratory inertial frame. As c and v refer to the same inertial frame, you have the right to use ordinary vector algebra obtaining c+v, because Relative velocity can be >c and is NOT invariant (see my post with that title). If you measured the relative velocity in the inertial frame you is at rest you obtain c as the result, compatible with the second postulate of Special Relativity. ANY light continue having the same vacuum speed c for ANY observer AT REST in ANY inertial frame, The hardest of all hard facts . RVHG Subject: Re: . The hardest of all hard facts . > Hi Eleaticus, Re: How you think, > Michelson-Morley and Kennedy-Thorndike do indeed fit > Galilean ( c + v ) physics , > > All throughout the annals of history ... > No premise has been better tested than this premise: > The speed of light is the same for all observers. > > That makes it: The hardest of all hard facts. > No experiment ever showed that the speed of light is > the same for all observers. Indeed any determination > of the one way speed of light can be used to demonstrate > the speed of light is NOT the same for all observers.... > A light----> B <-you > < ----------- L --------------> v m/s > Use syncronised clocks at A amd B to time how long it takes > light to travel a distance of L meters across the laboratory.. > Speed of light relative to the laboratory = L/ (tB - tA) = c > where 'tA' is the time at which the light left A > and 'tB' is the at which the light arrived at B > Now repeat the experiment while running towards B at v m/s > Note that 'in your frame of reference' the point B is moving , > so that the light must travel an extra distance = v * (tB - tA) > which is the distance B has moved as the light travels from > A to B. > Therefore: > Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) > = c + v > keith stein > You had measured the relative velocity between a light ray front and a > moving you with v velocity. Notice that the velocities of your two > moving entities are c and v, measured in the Laboratory inertial > frame. As c and v refer to the same inertial frame, you have the right > to use ordinary vector algebra obtaining c+v, because Relative > velocity can be >c and is NOT invariant (see my post with that > title). If you measured the relative velocity in the inertial frame you is at rest you obtain c as the result, Perhaps you could show HOW you obtain c as the result, Mr. Hidalgo-Gato > compatible with the > second postulate of Special Relativity. ANY light continue having the > same vacuum speed c for ANY observer AT REST in ANY inertial frame, Note that the distance the light travelled relative in the inertial frame in which you is AT REST is L+ v * (tB - tA) as shown above, and your hardest of all hard facts is in fact a totally unjustified assumption. keith stein The hardest of all hard facts . > RVHG Subject: Re: . The hardest of all hard facts . > > Hi Eleaticus, Re: How you think, > Michelson-Morley and Kennedy-Thorndike do indeed fit > > Galilean ( c + v ) physics , > > > All throughout the annals of history ... > > No premise has been better tested than this premise: > > The speed of light is the same for all observers. > > > That makes it: The hardest of all hard facts. > > No experiment ever showed that the speed of light is > the same for all observers. Indeed any determination > of the one way speed of light can be used to demonstrate > the speed of light is NOT the same for all observers.... > > > > A light----> B <-you > < ----------- L --------------> v m/s > > Use syncronised clocks at A amd B to time how long it takes > light to travel a distance of L meters across the laboratory.. > > Speed of light relative to the laboratory = L/ (tB - tA) = c > where 'tA' is the time at which the light left A > and 'tB' is the at which the light arrived at B > > Now repeat the experiment while running towards B at v m/s > Note that 'in your frame of reference' the point B is moving , > so that the light must travel an extra distance = v * (tB - tA) > which is the distance B has moved as the light travels from > A to B. > > Therefore: > Speed of light relative to you = (L+ v * (tB - tA)) / (tB - tA) > = c + v > > > > keith stein > You had measured the relative velocity between a light ray front and a > moving you with v velocity. Notice that the velocities of your two > moving entities are c and v, measured in the Laboratory inertial > frame. As c and v refer to the same inertial frame, you have the right > to use ordinary vector algebra obtaining c+v, because Relative > velocity can be >c and is NOT invariant (see my post with that > title). If you measured the relative velocity in the inertial frame you is at rest you obtain c as the result, > Perhaps you could show HOW you obtain c as the result, Mr. Hidalgo-Gato > compatible with the > second postulate of Special Relativity. ANY light continue having the > same vacuum speed c for ANY observer AT REST in ANY inertial frame, > Note that the distance the light travelled relative in the inertial frame in > which you is AT REST is L+ v * (tB - tA) as shown above, and your hardest of all hard facts is in fact a totally unjustified assumption. > keith stein The hardest of all hard facts . > RVHG All the distances and times that you use are referred to the Laboratory inertial frame. Respect de inertial frame where you is at rest are valid different ones, every inertial frame has a different time and space. I refer you to Einstein's original June 30-1905 paper. In it he DEFINE what simultaneity is for two events that occur in different A and B space points. The following are part of Einstein's words: We have so far defined only an 'A time' and a 'B time'. We have not defined a common time for A and B, for the later cannot be defined at all unless we establish BY DEFINITION that the 'time' required by light to travel from A to B equals the 'time' it requires to travel from B to A. I recommend you to study in detail this fundamental paper, near to reach a century from written. You have all the right to reject Einstein's second postulate stating that vacuun light speed is the same c for all inertial frames, making your own theory to model Nature. But do not forget that there exist a huge quantity of experimental data that require explanation. The theory with the best performace win (until a better one is created), that is the rule in the scientific game. RVHG Subject: Re: . The hardest of all hard facts . >They found none. Miller's results could not be duplicated by anyone but >Miller with his apparatus. The following paper describes a number of experiments that seem to have detected motion relative to space or (something in space). http://arxiv.org/pdf/physics/0312082 The following abstract is from: The motion of the Solar System and the Michelson-Morley experiment M. Consoli and E. Costanzo Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/astro-ph/0311576 ---QUOTE--- Historically, the Michelson-Morley experiment has played a crucial role for abandoning the idea of a preferred reference frame, the ether, and for replacing Lorentzian Relativity with Einsteins Special Relativity. However, our re-analysis of the Michelson-Morley original data, consistently with the point of view already expressed by other authors, shows that the experimental observations have been misinterpreted. Namely, the fringe shifts point to a non-zero observable Earths velocity Vobs = 8 4 0 5 km s. Assuming the existence of a preferred reference frame, and using Lorentz transformations to extract the kinematical Earths velocity that corresponds to this Vobs , we obtain a real velocity, in the plane of the interferometer,v earth = 201 12 km s. This value is in excellent agreement with Millers calculated value Vearth = 203 8 km/s and suggests that the magnitude of the fringe shifts is determined by the typical velocity of the Solar System within our galaxy. This conclusion, which is also consistent with the results of all other classical experiments, leads to an alternative interpre- tation of the Michelson-Morley type of experiments. Contrary to the generally accepted ideas of last century, they provide experimental evidence for the existence of a preferred reference frame. This point of view is also consistent with the most recent data for the anisotropy of the two-way speed of light in the vacuum. ---END QUOTE--- The following paper proposes that motion relative to space (or something in space) should be detectable, because space is full of Vacuum condensates and ether-drift experiments M. Consoli, A. Pagano and L. Pappalardo Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/physics/0306094 The following is also interesting. Modern Michelson-Morley experiments and gravitationally-induced anisotropy of c M. Consoli Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/gr-qc/0306105 The abstract states: The recent, precise Michelson-Morley experiment performed by Muller et al. suggests a tiny anisotropy of the speed of light. I propose a quantitative explanation of the observed effect based on the interpretation of gravity as a density fluctuation of the Higgs condensate. Peter Subject: Re: . The hardest of all hard facts . Dear Peter: >They found none. Miller's results could not be duplicated by anyone but >Miller with his apparatus. > The following paper describes a number of experiments that seem to > have detected motion relative to space or (something in space). > http://arxiv.org/pdf/physics/0312082 > The following abstract is from: The motion of the Solar System and the Michelson-Morley experiment > M. Consoli and E. Costanzo > Istituto Nazionale di Fisica Nucleare, Sezione di Catania > http://arxiv.org/pdf/astro-ph/0311576 > ---QUOTE--- > Historically, the Michelson-Morley experiment has played a crucial > role for abandoning the idea of a preferred reference frame, the > ether, and for replacing Lorentzian Relativity with Einsteins Special > Relativity. However, our re-analysis of the Michelson-Morley original > data, consistently with the point of view already expressed by other > authors, shows that the experimental observations have been > misinterpreted. Namely, the fringe shifts point to a non-zero > observable Earths velocity Vobs = 8 4 10 5 km s. Assuming the > existence of a preferred reference frame, and using Lorentz > transformations to extract the kinematical Earths velocity > that corresponds to this Vobs , we obtain a real velocity, in the > plane of the interferometer,v earth = 201 112 km s. This value is in > excellent agreement with Millers calculated value Vearth = 203 18 > km/s and suggests that the magnitude of the fringe shifts is > determined by the typical velocity of the Solar System within our > galaxy. This conclusion, which is also consistent with the results of > all other classical experiments, leads to an alternative interpre- > tation of the Michelson-Morley type of experiments. Contrary to the > generally accepted ideas of last century, they provide experimental > evidence for the existence of a preferred reference frame. This point > of view is also consistent with the most recent data for the > anisotropy of the two-way speed of light in the vacuum. > ---END QUOTE--- > The following paper proposes that motion relative to space (or > something in space) should be detectable, because space is full of > Vacuum condensates and ether-drift experiments > M. Consoli, A. Pagano and L. Pappalardo > Istituto Nazionale di Fisica Nucleare, Sezione di Catania > http://arxiv.org/pdf/physics/0306094 > The following is also interesting. Modern Michelson-Morley experiments and gravitationally-induced > anisotropy of c > M. Consoli > Istituto Nazionale di Fisica Nucleare, Sezione di Catania > http://arxiv.org/pdf/gr-qc/0306105 > The abstract states: > The recent, precise Michelson-Morley experiment performed by Muller et > al. suggests a tiny anisotropy of the speed of light. I propose a > quantitative explanation of the observed effect based on the > interpretation of gravity as a density fluctuation of the Higgs > condensate. So your analysis of the literature has found that the purported anisotropy is always less-than-or-equal-to the smallest resolution of the recorded data, or that the anisotropy is a function of time. Surely the anisotropy decreased over the 100 years from your first citation to your last. Obviously you have discovered a new thorn for the aetherists. Good work! A secular change in the anisotropy of the aether! Amazing. How will they explain it. Surely they will not claim experimental error, since they bitch whenever the establishment does so. Us SRians are proud of your efforts. David A. Smith Subject: Re: . The hardest of all hard facts . >Hi Peter, You flip, > the Michelson-Morley experiment gave > results that were non-null, > due to the effect of air in the apparatus , >You're obviously wrong about that. Its true that Michelson reported a null result. However recently the Process Physics group at Flinders University in Australia got hold of the data from the Michelson-Morley experiments and when they re-analysed the data they concluded the result was non-null. They believe the reason Michelson reported a null result was because the fringe shifts he obtained were much smaller than he expected. This was due to the effect of the Lorentz contraction but because Michelson believed in Newtonian physics, he did not take this into account. However, the Australian physicists found that when they re-analysed the Michelson-Morley data to take into account the effects of both the Lorentz contraction and the reduced velocity of light through air, they obtained velocities relative to space (or something in space) that correlated with the rotation of the earth and its orbit round the sun. So they claim that M-M type experiments can detect motion relative to space/ether/spacetime foam/quantum foam, whatever you wish to call it, but there is an important proviso--The apparatus must contain a gas, because if the experiment is done with a vacuum interferometer, the effect of the motion is perfectly cancelled by the Lorentz contraction. The re-analysis is described in Michelson-Morley Experiments Revisited and the Cosmic Background Radiation Preferred Frame http://arxiv.org/pdf/physics/0205065 The following paper describes six other experiments that seem to have detected motion relative to space or (something in space). http://arxiv.org/pdf/physics/0312082 The above re-analysis by the Australia physicists is partially supported by an independent re-analysis by two Italian physicists. The following abstract is from: The motion of the Solar System and the Michelson-Morley experiment M. Consoli and E. Costanzo Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/astro-ph/0311576 ---QUOTE--- Historically, the Michelson-Morley experiment has played a crucial role for abandoning the idea of a preferred reference frame, the ether, and for replacing Lorentzian Relativity with Einsteins Special Relativity. However, our re-analysis of the Michelson-Morley original data, consistently with the point of view already expressed by other authors, shows that the experimental observations have been misinterpreted. Namely, the fringe shifts point to a non-zero observable Earths velocity Vobs = 8 4 0 5 km s. Assuming the existence of a preferred reference frame, and using Lorentz transformations to extract the kinematical Earths velocity that corresponds to this Vobs , we obtain a real velocity, in the plane of the interferometer,v earth = 201 12 km s. This value is in excellent agreement with Millers calculated value Vearth = 203 8 km/s and suggests that the magnitude of the fringe shifts is determined by the typical velocity of the Solar System within our galaxy. This conclusion, which is also consistent with the results of all other classical experiments, leads to an alternative interpre- tation of the Michelson-Morley type of experiments. Contrary to the generally accepted ideas of last century, they provide experimental evidence for the existence of a preferred reference frame. This point of view is also consistent with the most recent data for the anisotropy of the two-way speed of light in the vacuum. ---END QUOTE--- The following paper proposes that motion relative to space (or something in space) should be detectable, because space is full of Vacuum condensates and ether-drift experiments M. Consoli, A. Pagano and L. Pappalardo Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/physics/0306094 The following is also interesting. Modern Michelson-Morley experiments and gravitationally-induced anisotropy of c M. Consoli Istituto Nazionale di Fisica Nucleare, Sezione di Catania http://arxiv.org/pdf/gr-qc/0306105 Peter Subject: For what purpose ? Hi Peter, You mentioned, recently the Process Physics group at Flinders University in Australia got hold of the data from the Michelson-Morley experiments and when they re-analysed the data they concluded that the result was non-null . Why re-analyse the data from such an ancient and flawed experiment ? The Michelson-Morley experiment was good enough to suggest that the speed of light in a vacuum is the same for all observers. ( Which it is, to a very high precision ) The Kennedy-Thorndike experiment ruled out the Fitzgerald-Lorentz contraction hypothesis that objects experienced a compression in the direction of their motion relative to some third frame of reference. You suggest introducing a third frame of reference into relativity ... For what purpose ? 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 stein Subject: Letter to Prof Ullrich and others I am just curious: why are people like you interested in even acknowleding James Harris and his ilk ? This guy has been oopsing on sci.math for ages (at least several years) I see. Is it not worthwhile for everyone to just ignore him at this point ? There is real danger of course that someone would take him seriously, but I feel thats remote. It seems that he has taken to calling universities to complain. Lets not giving him any more attention. Without attention, he will shrivel up and disappear. Is there a history or an obvious reason that I am missing ? Subject: Re: Letter to Prof Ullrich and others > I am just curious: why are people like you > interested in even acknowleding James Harris and his ilk ? > This guy has been oopsing on sci.math for ages (at least several > years) I see. > Is it not worthwhile for everyone to just ignore him at this point ? > There is real danger of course that someone would take him seriously, > but I feel thats remote. People like you have come and gone, as your agenda is controlling other people's behavior. Why try? It's Usenet you know. Usenet is known for having people around who just post or reply, isn't that amazing! And in all that posting and replying there are people like yourself who try to control the process. > It seems that he has taken to calling universities to complain. > Lets not giving him any more attention. Without attention, he will > shrivel up and disappear. > Is there a history or an obvious reason that I am missing ? Yeah, the history of Usenet, and I guess you're a newbie, eh? Or you should have known enough not to have made the post you did. James Harris Subject: Re: Letter to Prof Ullrich and others > I am just curious: why are people like you > interested in even acknowleding James Harris and his ilk ? > This guy has been oopsing on sci.math for ages (at least several > years) I see. > Is it not worthwhile for everyone to just ignore him at this point ? > There is real danger of course that someone would take him seriously, > but I feel thats remote. > People like you have come and gone, as your agenda is controlling > other people's behavior. > Why try? > It's Usenet you know. Usenet is known for having people around who > just post or reply, isn't that amazing! Some do more than just post and reply. They build web sites documenting cranks! > And in all that posting and replying there are people like yourself > who try to control the process. > > It seems that he has taken to calling universities to complain. > Lets not giving him any more attention. Without attention, he will > shrivel up and disappear. > Is there a history or an obvious reason that I am missing ? > Yeah, the history of Usenet, and I guess you're a newbie, eh? > Or you should have known enough not to have made the post you did. > James Harris Subject: Re: Letter to Prof Ullrich and others Discussion, linux) > People like you have come and gone, as your agenda is controlling > other people's behavior. > Why try? > It's Usenet you know. Usenet is known for having people around who > just post or reply, isn't that amazing! Does this advice possess any relevance given your recent tantrums regarding Nora Baron's audacity to post in your threads? Nah, never mind. -- Even I, who know beyond doubt that my death will be caused by a silly girl, will not hesitate when that girl passes by. -- Merlin, as reported by John Steinbeck. Subject: Re: Letter to Prof Ullrich and others >> People like you have come and gone, as your agenda is controlling >> other people's behavior. >> Why try? >> It's Usenet you know. Usenet is known for having people around who >> just post or reply, isn't that amazing! >Does this advice possess any relevance given your recent tantrums >regarding Nora Baron's audacity to post in your threads? Hey, I know, I know! Call on me, come on, I know the answer! >Nah, never mind. ************************ Subject: Re: Letter to Prof Ullrich and others > Or you should have known enough not to have made the post you did. Speak for yourself. Your record of posting and defending ridiculous errors is legion. You should know better than to post any math at all. POSTER ACCURACY INDEX ---------------------------- JSH 0% MATHEMATICIANS 99% > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com Subject: Re: Letter to Prof Ullrich and others >I am just curious: why are people like you >interested in even acknowleding James Harris and his ilk ? >This guy has been oopsing on sci.math for ages (at least several >years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously, >but I feel thats remote. >It seems that he has taken to calling universities to complain. He's called universities with totally bogus complaints in the past, later admitting here that the reason he did so was to get someone (me, actually) to stop replying to his posts. Seems to me that it's important that he be absolutely certain that that sort of intimidation will not have the effect that he wants. (Because although the people here just laughed at the complaints someone elsewhere might not be so lucky.) >Lets not giving him any more attention. Without attention, he will >shrivel up and disappear. >Is there a history or an obvious reason that I am missing ? ************************ Subject: Re: Letter to Prof Ullrich and others >I am just curious: why are people like you >interested in even acknowleding James Harris and his ilk ? >This guy has been oopsing on sci.math for ages (at least several >years) I see. >Is it not worthwhile for everyone to just ignore him at this point ? >There is real danger of course that someone would take him seriously, >but I feel thats remote. >It seems that he has taken to calling universities to complain. > He's called universities with totally bogus complaints in the past, > later admitting here that the reason he did so was to get someone > (me, actually) to stop replying to his posts. Seems to me that it's > important that he be absolutely certain that that sort of intimidation > will not have the effect that he wants. (Because although the > people here just laughed at the complaints someone elsewhere > might not be so lucky.) David Ullrich is a tenured math professor at Oklahoma State University, and he made *public* statements that I thought reflected badly on him and on that university. He has been defensive about those statement for years, so it's easier for me to direct you to them. You can go to Google Groups, advanced search, at http://www.google.com/advanced_group_search?hl=en and put David Ullrich as the author, and racial slur in the search field. Or you can click on http://groups.google.com/groups?as_q=racial%20slur&safe=off&ie =ISO-8859-1&as _ugroup=sci.math&as_uauthors=David%20Ullrich&lr=&hl=en Oh yeah, in case you're wondering about what Ullrich finds so offensive that the subject of racial slur came to his mind, I said that he'd acted as my lapdog in an instance. That was YEARS ago. James Harris Subject: Re: Letter to Prof Ullrich and others > Oh yeah, in case you're wondering about what Ullrich finds so > offensive that the subject of racial slur came to his mind, I said > that he'd acted as my lapdog in an instance. > That was YEARS ago. Would you mind telling us what is the statute of limitations on your posting debacles? > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com Subject: Re: Letter to Prof Ullrich and others Bob > I am just curious: why are people like you > interested in even acknowleding James Harris and his ilk ? ... > Is it not worthwhile for everyone to just ignore him at this point ? Some responders have kicked the Harris habit, but others pick it up meanwhile. I'm been on and off the wagon for a couple of years, alas. And if, by a fluke of statistics, one of his threads gets no responses even for one day, he barrages the board with spew. > There is real danger of course that someone would take him seriously, > but I feel that's remote. I agree. > It seems that he has taken to calling universities to complain. > Lets not giving him any more attention. Without attention, he will > shrivel up and disappear. But Harris is good at getting attention, because he has given it years of practice, and cares about nothing else, as far as I can see. LH Subject: functional analysis--separable Hi all, I found the following proposition in a book without a proof. I have difficulty in proving it. Can anyone help? Suppose B is a separable Banach space. Let {x_n} be a countable dense subset of the unit circle C={x in B : ||x||=1}. How can I prove that: every element x in B can be written as: x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolute summable sequence. That's: summation[from 1 to infinity] {|a_n|} < infinity. Subject: Re: functional analysis--separable > I found the following proposition in a book without a proof. I > have difficulty in proving it. Can anyone help? > Suppose B is a separable Banach space. Let {x_n} be a countable > dense subset of the unit circle C={x in B : ||x||=1}. How can I prove > that: every element x in B can be written as: > x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolute > summable sequence. That's: > summation[from 1 to infinity] {|a_n|} < infinity. Hint: Let x be in B. Suppose you've approximated x nicely with a finite sum s of the sort under discussion. So x = s + (x-s) and |x-s| is small. Now approximate (x-s)/|x-s| by y in {x_n} as closely as you like. Then (x-s) - |x-s|*y will be even smaller. Continue ... Subject: Re: Silly question for someone with a big calculator. ... >4/3 = 1.333333... >24/17 = 1.411764... >816/577 = 1.414211... >941664/665857 = 1.414213... >If we let k represent that ratio for one of the exponents, the ratio for >the next in this list is (4k)/(k^2+2); it can be shown that the ratios >will converge to sqrt(2). ... > I thought your following line was neat: >1 = 3^2-2^2-2^2 = 17^2-12^2-12^2 = 577^2-408^2-408^2 = ... > [and] 1 = 665857^2 - 470832^2 - 470832^2 ... I missed the beginning of this thread so don't know what the question might be, but for approximation of sqrt(2), a term of a Taylor series will reduce the error quadratically. If a^2-2b^2=x and f(x)=(2b^2+x)^(1/2), f'(x)=1/(2(2b^2+x)^(1/2)), so a ~ s*(b+x/(4b)) [where s=sqrt(2)]. a b s-a/b s-a/(b+x/(4b)) 3 2 -.086 .0024 17 12 -.0025 .0000021 577 408 -2E-6 1.6E-12 665857 470832 1.6E-11 9E-25 -jiw Subject: Logic I'm stuck on these 2 problems. Represent the following compound propositions, each involving one or more of these three propositons, as symbolic expressions in the variables d,e,f. 1)e does not f unless d 2)d and e does not imply f Of coure I am told what these propositions are. I'm sorry but I can't even show you what I've done so far--I'm completely lost. Steven Subject: Re: Logic > I'm stuck on these 2 problems. > Represent the following compound propositions, each involving one or more of > these three propositons, as symbolic expressions in the variables d,e,f. > 1)e does not f unless d Does 1 this mean If d and e the f???? > 2)d and e does not imply f Does 2 mean not d or not e does imply f??? > Of coure I am told what these propositions are. > I'm sorry but I can't even show you what I've done so far--I'm completely > lost. > Steven Subject: Re: Logic > I'm stuck on these 2 problems. > Represent the following compound propositions, each involving one or more > of > these three propositons, as symbolic expressions in the variables d,e,f. > 1)e does not f unless d > Does 1 this mean If d and e the f???? Typo: Does 1 mean If d and e the f???? > 2)d and e does not imply f > Does 2 mean not d or not e does imply f??? > Of coure I am told what these propositions are. > I'm sorry but I can't even show you what I've done so far--I'm completely > lost. > Steven Subject: Re: Goldberg dual in other words, the fullerenes. your guess is ill-posed, since the fullerenes are all of 3-way vertices, their duals will be trigonated -- but the lengths for duals is not the same, although it's the same *number* of edges. it's interesting that the fullerenes can thus be modelled with trigona, and that Bucky never noticed that! > A Goldberg polyhedron has all hexagonal faces except for 12 pentagons > or 6 fourgons or 4 trigons, and are multi-symmetrical. A soccor ball > is a Goldberg polyhedron. Goldberg-like is the same but asymmetrical. > My guess is this. For 3(N-2) edge elements whose lengths are variable, > and where N is is an integer greater that 2, ther is a maximum > diameter polyhedron that can be contructed and a minimum diameter > polyhedron that can be constructed. The maximim diameter polyhedron > that can be built with 3(N-2) edges is a Goldberg Polyhedron like > structure. The minimum diameter sphere will be an omnitriangulated > icosahedral arrangement of the edge elements. of course, a lot of this is moot, if we allow Cheeny to grind us into Tony's McCrusade (Usama's MacJihad). see my sig. --Give the World a Trickier Dick Cheeny -- out of office after GIGA years. http://www.benfranklinbooks.com/ http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac Subject: Re: Goldberg dual > A Goldberg polyhedron has all hexagonal faces except for 12 pentagons > or 6 fourgons or 4 trigons, and are multi-symmetrical. A soccor ball > is a Goldberg polyhedron. Goldberg-like is the same but asymmetrical. Is there a name for an asymmetric Goldberg-like > polyhedron? What is the name of the dual of the Goldberg-like > polyhedron? > My guess is this. For 3(N-2) edge elements whose lengths are variable, > and where N is is an integer greater that 2, ther is a maximum > diameter polyhedron that can be contructed and a minimum diameter > polyhedron that can be constructed. The maximim diameter polyhedron > that can be built with 3(N-2) edges is a Goldberg Polyhedron like > structure. The minimum diameter sphere will be an omnitriangulated > icosahedral arrangement of the edge elements. > Additionally, even smaller spheres can be built by doubling up edges, > tripling up edges, etc. Subject: Suggestions requested for notation of functions such as sinc I mentioned the sine cardinal function (1 if x = 0 sinc(x) = ( ( sin(x)/x otherwise in a thread here not long ago. Since then it has appeared in other threads. Example 1. Rob Johnson, in the thread puzzle: GCDs of Infinite Set of Integer Pairs, showed a certain probability to be 1 - sinc(sqrt(6)). Example 2. In the Mathematica newsgroup thread how to explain this weird effect? Integrate, I noted that the integral of sin(m x) sin(n x) from 0 to 2 pi can be given nicely as pi(sinc(2 pi (m - n)) - sinc(2 pi (m + n))), which is valid for all m and n. OTOH, the expression given by Mathematica for the integral is not, literally, valid when m = n or m = -n. Besides sin(x)/x, there are other commonly occurring functions of the form f(x)/x having removable singularities at x = 0. I think it would be nice to have a consistent notation for the functions obtained once the singularities have been removed. Of course, some people will say that no special notation is needed since we can always just say something like Singularities are supposed to be understood as having been removed whenever possible. But that doesn't work very well with my computer algebra systems, for example. Another possibility which avoids giving a new name to such functions is to use limit notation. For example, g(x) = lim_{t -> x} sin(t)/t is the same as sinc(x). But I'm not fond of that alternative since it's not in closed form. The only special notation I've seen so far for such functions is based on analogy with sinc. The notation sinhc(x) is already in use, at least to some extent, for the function which is 1 if x = 0, sinh(x)/x otherwise. And at MathWorld, Eric Weisstein has introduced the notation tanc(x) for the function which is 1 if x = 0, tan(x)/x otherwise. But in those two instances, I feel that there is no real need for new notation since each of those functions can be expressed nicely in terms of sinc(x). Namely, instead of sinhc(x), we could always write sinc(i x), and instead of tanc(x), we could always write sinc(x)/cos(x). In other cases, however, when functions cannot be expressed handily in terms of sinc, I think that a special notation would be nice. Here's an example. For a spheroid having polar radius a and equatorial radius b, what's the surface area? Perhaps you'd say it's A = 2 pi b ( a Asin(d)/d + b ) where d = Sqrt(1 - (b/a)^2) and Asin denotes the principal-valued inverse sine function. But for the sphere itself (which I certainly consider to be a spheroid), that formula does not work, if taken literally, since 0/0 is encountered. However, if we have the function f(x) = 1 if x = 0, Asin(x)/x otherwise, then we may say, for any spheroid (prolate, sphere, or oblate), that the area is given by A = 2 pi b ( a f(d) + b ). What's a nice name for function f? Should we, by anaolgy with sinc, call it Asinc? (But then might not some misguided fellow misconstrue it as being the inverse of sinc instead? And BTW, how should we denote that inverse?) Am I correct in thinking that the name sine cardinal function and the abbreviation sinc are due to E. T. Whittaker? Why did he choose that name? Would names such as Asinc be based on false analogies with sinc? I would appreciate any thoughtful comments, suggestions for notation, etc. David Cantrell Subject: Re: Suggestions requested for notation of functions such as sinc > I mentioned the sine cardinal function > (1 if x = 0 > sinc(x) = ( > ( sin(x)/x otherwise > in a thread here not long ago. Since then it has appeared in other threads. > Example 1. Rob Johnson, in the thread puzzle: GCDs of Infinite Set of > Integer Pairs, showed a certain probability to be 1 - sinc(sqrt(6)). > Example 2. In the Mathematica newsgroup thread how to explain this weird > effect? Integrate, I noted that the integral of sin(m x) sin(n x) from 0 > to 2 pi can be given nicely as pi(sinc(2 pi (m - n)) - sinc(2 pi (m + n))), > which is valid for all m and n. OTOH, the expression given by > Mathematica for the integral is not, literally, valid when m = n or m = -n. > Besides sin(x)/x, there are other commonly occurring functions of the form > f(x)/x having removable singularities at x = 0. I think it would be nice to David, here's my take on this. Although a computer programming function to evaluate sin(x)/x, for example, would use the the definition you provided above since it would otherwise calculate 0/0, actually there is no singularity in sin(x)/x at x=0 because d[sin(x)]dx / (dx/dx) = cos(x)/1 which evaluated at x=0 is 1.0/1.0 = 1.0 and this is by definition how one resolves 0/0. KeithK > have a consistent notation for the functions obtained once the > singularities have been removed. Of course, some people will say that no > special notation is needed since we can always just say something like Singularities are supposed to be understood as having been removed > whenever possible. But that doesn't work very well with my computer > algebra systems, for example. Another possibility which avoids giving a new > name to such functions is to use limit notation. For example, > g(x) = lim_{t -> x} sin(t)/t > is the same as sinc(x). But I'm not fond of that alternative since it's not > in closed form. > The only special notation I've seen so far for such functions is based on > analogy with sinc. The notation sinhc(x) is already in use, at least to > some extent, for the function which is 1 if x = 0, sinh(x)/x otherwise. And > at MathWorld, Eric Weisstein has introduced the notation tanc(x) for the > function which is 1 if x = 0, tan(x)/x otherwise. But in those two > instances, I feel that there is no real need for new notation since each of > those functions can be expressed nicely in terms of sinc(x). Namely, > instead of sinhc(x), we could always write sinc(i x), and instead of > tanc(x), we could always write sinc(x)/cos(x). In other cases, however, > when functions cannot be expressed handily in terms of sinc, I think that a > special notation would be nice. Here's an example. For a spheroid having > polar radius a and equatorial radius b, what's the surface area? Perhaps > you'd say it's > A = 2 pi b ( a Asin(d)/d + b ) > where d = Sqrt(1 - (b/a)^2) and Asin denotes the principal-valued inverse > sine function. > But for the sphere itself (which I certainly consider to be a spheroid), > that formula does not work, if taken literally, since 0/0 is encountered. > However, if we have the function f(x) = 1 if x = 0, Asin(x)/x otherwise, > then we may say, for any spheroid (prolate, sphere, or oblate), that the > area is given by > A = 2 pi b ( a f(d) + b ). > What's a nice name for function f? Should we, by anaolgy with sinc, call it > Asinc? (But then might not some misguided fellow misconstrue it as being > the inverse of sinc instead? And BTW, how should we denote that inverse?) > Am I correct in thinking that the name sine cardinal function and the > abbreviation sinc are due to E. T. Whittaker? Why did he choose that name? > Would names such as Asinc be based on false analogies with sinc? > I would appreciate any thoughtful comments, suggestions for notation, etc. > David Cantrell Subject: Drawing subgroup lattices for research papers I'm working on a thesis (using MikTex and Winedt) that requires me to put in a lot of subgroup lattices. This results in several problems. For an early version of the paper, I made .bmps and cut and paste them in the appropriate spots, but now I want to put them in the actual code of the paper itself. My method was to import into Mathematica and export as .eps. First of all, the .eps files are huge. I tried to import a 3k grayscale gif and got a 400k eps file. Then, using graphicx I was able to get the picture to show up after Latexing. However, the graphic looks terrible. It is clearly something that is happening when I Latex the paper because I checked the eps file in Ghostview and though it was degraded some from the original bmp the subgroups were still recognizable. I've also tried downloading xypic and followed the instructions, but can't figure out how to get MikTex or WinEdt to recognize the new files. Does anyone either have a suggestion on getting what I have to work or a more tex-friendly way to draw lattices? Subject: Re: Drawing subgroup lattices for research papers > I'm working on a thesis (using MikTex and Winedt) that requires me to > put in a lot of subgroup lattices. This results in several problems. > For an early version of the paper, I made .bmps and cut and paste them > in the appropriate spots, but now I want to put them in the actual > code of the paper itself. My method was to import into Mathematica > and export as .eps. First of all, the .eps files are huge. I tried > to import a 3k grayscale gif and got a 400k eps file. Then, using > graphicx I was able to get the picture to show up after Latexing. > However, the graphic looks terrible. It is clearly something that is > happening when I Latex the paper because I checked the eps file in > Ghostview and though it was degraded some from the original bmp the > subgroups were still recognizable. > I've also tried downloading xypic and followed the instructions, but > can't figure out how to get MikTex or WinEdt to recognize the new > files. > Does anyone either have a suggestion on getting what I have to work or > a more tex-friendly way to draw lattices? xypic will do most things you can imagine, and some you can't. simpler but less flexible is paul taylor's diagrams package. diagrams is available from ctan archive. I don't understand what you mean by can't get winedt or miktex to recognize the new files, which new files, what are the error messages? one simple suggestion: have you updated the miktex installation after putting xypic in your path? Subject: Re: Drawing subgroup lattices for research papers The actual message is that it can't find xypic.sty. I tried pointing it there, but nothing happened. I read somewhere that there is a command in MikTex that allows you to update the file database that it will read, but I couldn't find anything. Do I need to just do a fresh install? > I'm working on a thesis (using MikTex and Winedt) that requires me to > put in a lot of subgroup lattices. This results in several problems. > For an early version of the paper, I made .bmps and cut and paste them > in the appropriate spots, but now I want to put them in the actual > code of the paper itself. My method was to import into Mathematica > and export as .eps. First of all, the .eps files are huge. I tried > to import a 3k grayscale gif and got a 400k eps file. Then, using > graphicx I was able to get the picture to show up after Latexing. > However, the graphic looks terrible. It is clearly something that is > happening when I Latex the paper because I checked the eps file in > Ghostview and though it was degraded some from the original bmp the > subgroups were still recognizable. > I've also tried downloading xypic and followed the instructions, but > can't figure out how to get MikTex or WinEdt to recognize the new > files. > Does anyone either have a suggestion on getting what I have to work or > a more tex-friendly way to draw lattices? > xypic will do most things you can imagine, and some you can't. simpler but > less flexible is paul taylor's diagrams package. diagrams is available > from ctan archive. > I don't understand what you mean by can't get winedt or miktex to > recognize the new files, which new files, what are the error messages? > one simple suggestion: have you updated the miktex installation after > putting xypic in your path? Subject: Re: Drawing subgroup lattices for research papers > The actual message is that it can't find xypic.sty. I tried > pointing it there, but nothing happened. I read somewhere that there > is a command in MikTex that allows you to update the file database > that it will read, but I couldn't find anything. Do I need to just do > a fresh install? I don't have MikTex anymore, but as I recall that there is some documentation on using kpathsea (the library that miktex uses to find things). I hope that helps. >>>I'm working on a thesis (using MikTex and Winedt) that requires me to >>>put in a lot of subgroup lattices. This results in several problems. >>>For an early version of the paper, I made .bmps and cut and paste them >>>in the appropriate spots, but now I want to put them in the actual >>>code of the paper itself. My method was to import into Mathematica >>>and export as .eps. First of all, the .eps files are huge. I tried >>>to import a 3k grayscale gif and got a 400k eps file. Then, using >>>graphicx I was able to get the picture to show up after Latexing. >>>However, the graphic looks terrible. It is clearly something that is >>>happening when I Latex the paper because I checked the eps file in >>>Ghostview and though it was degraded some from the original bmp the >>>subgroups were still recognizable. >>>I've also tried downloading xypic and followed the instructions, but >>>can't figure out how to get MikTex or WinEdt to recognize the new >>>files. >>>Does anyone either have a suggestion on getting what I have to work or >>>a more tex-friendly way to draw lattices? >>xypic will do most things you can imagine, and some you can't. simpler but >>less flexible is paul taylor's diagrams package. diagrams is available >>from ctan archive. >>I don't understand what you mean by can't get winedt or miktex to >>recognize the new files, which new files, what are the error messages? >>one simple suggestion: have you updated the miktex installation after >>putting xypic in your path? Subject: Re: Drawing subgroup lattices for research papers > I'm working on a thesis (using MikTex and Winedt) that requires me to > put in a lot of subgroup lattices. This results in several problems. > For an early version of the paper, I made .bmps and cut and paste them > in the appropriate spots, but now I want to put them in the actual > code of the paper itself. My method was to import into Mathematica > and export as .eps. First of all, the .eps files are huge. I tried > to import a 3k grayscale gif and got a 400k eps file. Then, using > graphicx I was able to get the picture to show up after Latexing. > However, the graphic looks terrible. It is clearly something that is > happening when I Latex the paper because I checked the eps file in > Ghostview and though it was degraded some from the original bmp the > subgroups were still recognizable. > I've also tried downloading xypic and followed the instructions, but > can't figure out how to get MikTex or WinEdt to recognize the new > files. I suggest that you post the same question at the comp.text.tex newsgroup. Best regards, Jose Carlos Santos Subject: help is needed!!! i want play whit MAPLE but i can not find sourse for it if any body know some sourses for it or any one can help and support me please say more thank you pad Subject: Re: help is needed!!! > i want play whit MAPLE but i can not find sourse for it > if any body know some sourses for it or any one can help and support > me please say more > thank you > pad As far as I know, the only one that's free (to students) is MuPad. And it may not be free any more. google. after a little search, i found dmoz open directory project, and then following its internal links to science:math:software discovered a whole pageload of relevant links. the big six are Derive, Maple, Mathcad, Mathematica, MATLAB, and MuPad. But there are links to apparently specialized software in lots of area, so if you're more interested in Algebraic Geometry or Algebra, there are links to software in those areas. Jon Miller Subject: Re: help is needed!!! > i want play whit MAPLE but i can not find sourse for it > if any body know some sourses for it or any one can help and support > me please say more You should contact the company; I guarantee that they'll be more than happy to help you out. Doug Subject: Re: help is needed!!! > i want play whit MAPLE but i can not find sourse for it > if any body know some sourses for it or any one can help and support > me please say more Subject: re:Axioms defining a finite field >Your example violates the last rule. Pay attention. > Posted Via Usenet.com Premium Usenet Newsgroup Services >---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** >---------------------------------------------------------- > http://www.usenet.com ...Huh?!? Whom/what does you comment apply to? Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc Subject: ATTENTION: Open dispute with my college about Procedures. Please read. My web site www.johncho.us has all the information. I am seeking people who are familiar with procedures. I am also considering getting a lawyer. Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. > My web site www.johncho.us has all the information. I am seeking people > who are familiar with procedures. I am also considering getting a lawyer. Why don't you put all this creative energy into studying? Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. >My web site www.johncho.us has all the information. I am seeking people >who are familiar with procedures. I am also considering getting a lawyer. The college is within its rights to require issuance of a 'W', and, in fact, may have a duty to do so under California law. If they reported your attendance for funding purposes, I'm pretty certain they are committed to reporting your attendance in other areas. What is the problem with the 'W'? It does not affect your GPA. My daughters were admitted to UCLA with several of them. There are lots of reasons for a 'W' - I think the majority of transfer students have them. I don't think it's worth your time or money to fight this particular battle. It's probably more appropriate to concentrate on your other classes. Hank Murphy speaking only for myself Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. >My web site www.johncho.us has all the information. I am seeking people >who are familiar with procedures. Did you bother to read any of the replies you got in the first thread you started on this topic? (If no: if you don't read replies why bother posting? If yes: If you read _those_ replies why would you think you're going to get anything out of this second post?) >I am also considering getting a lawyer. Good luck with that. Since you've given us absolutely no evidence that any school rules or procedures were violated I doubt you're going to find a lawyer interested in taking the case on a contingency basis. If you've got plenty of your own money I'm certain you'll be able to find lots of interested lawyers... ************************ Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. >My web site www.johncho.us has all the information. I am seeking people >who are familiar with procedures. > Did you bother to read any of the replies you got in the first thread > you started on this topic? (If no: if you don't read replies why > bother posting? If yes: If you read _those_ replies why would > you think you're going to get anything out of this second post?) Actually, I think it's becoming increasingly clear that he's just trolling. Nathan Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. i do not respond to many of the posts because my argument is with this school and not people of the message board. The ideas that the board gives www.versuslaw.com after that i will try to do another google search for statewide policies and procedures regarding academic progress Subject: Re: ATTENTION: Open dispute with my college about Procedures. Please read. > My web site www.johncho.us has all the information. I am seeking people > who are familiar with procedures. I am also considering getting a lawyer. As you insist on thinking legally, you ought to find out the official policy of your college and post all the relevant details here. Include all portions which refer to its expectations of you. If you just ask us for an uninformed opinion you will, IMHO, mainly get two kinds of answers, those of us who teach undergrad classes who think you should shut up and those who take undergrad classes and think you're hard done by. Subject: Re: Quadratics and transformation >> That is, if I have 3x^2 + 7x + 2, can I easily relate the coefficients >> and constant to the dilations, reflections and translations that this >> function represents compared to x^2 ? > Yes. 3(x^2 + 7/3*x) + 2, complete the square inside the parentheses. So > you end up with a(x-b)^2 + c. b is translation along the x-axis, c along > the y-axis, a is the dilation factor. If a is negative, that's a > reflection. > Jon Miller Jon, Thanks very much for this reply - it helped no end (as soon as I had refreshed myself on completing the square that is :) Having solved one of lifes little mysteries I shall now wander off to contemplate polynomials in general :) ... this, I suspect, will be rather more difficult but the fun is in the hunt :) Ivan. Subject: Re: Quadratics and transformation > It seems to me that I should be able to reasonably easily get a > general idea of the type(s) of transformations that have been applied > to x^2 just by looking at the finished quadratic. > That is, if I have 3x^2 + 7x + 2, can I easily relate the coefficients > and constant to the dilations, reflections and translations that this > function represents compared to x^2 ? > I have tried thinking about it as a combined function, that is, > f(x)=x^2, g[f(x)]=3x^2 + 7x + 2 and trying to find g(x). Well, I > haven't tried with this example but one I tried earlier turned into a > horrible mess and didn't give me what I wanted - I assumed that g(x) > is simply the inverse of g[f(x)] so maybe that's why it got so messy ? > It also seemed to be a lot of work! Transform both x and y First by scale then the translations > I'm aware that I could determine the turning point and work out the > translations and that the roots would indicate what horizontal > dilation has been applied. I expect that if I calculate two other > points I would also be able to determine reflection and vertical > dilation. Of course I could always graph it but I'm hoping to achieve > what I want from inspection of the description of the function (is > that the right terminology?). > Any and all help appreciated. > Ivan. If you only look at parabolas with v=u^2 as your ideal (1) Ax^2+Bx+C=y form By inspection A Scale A>0 positive orientation, i.e. parabola faces up A<0 negative orientation, i.e. parabola faces Down B/2A Horizontal translation C/A-B^2/4A^2= Vertical Translation {B^2-4AC} :: Discriminant Quadratic formula x = [-B+/-(B^2-4AC)^(1/2)]/2A Let u=U(x) and v=V(y) u^2=v Find U(x) And V(y) (1) Ax^2+Bx+C=y Divide by A A<>0 (2) x^2 +(B/A)x+(C/A)= y/A Subtract C/A from both sides (3)x^2 +(B/A)x = (y-C)/A Complete the Square (add (B/2A)^2 to both sides (4) x^2 +(B/A)x+B^2/4A^2= (y-C)/A+B^2/4A^2 (4a)(x+B/2A)^2= (y-C)/A+B^2/4A^2 LHS = {U(x)^2} RHS= V(y) (5) u=(x+B/2A) v=(y/A+(B^2-4AC)/4A^2) Standard form (x-h)^2=4p(y-k) Vertex (h,k) focus (h,k+p) Axis x = h directrix y= k-p General form Ax^2+Bx+C=y so h = -B/2A p = 1/4A k = (B^2-4AC)/4A Be careful when substituting A,B & C are all signed Carl -- If its Monday then I am a fool but not ignorant. Subject: Re: Quadratics and transformation < > Be careful when substituting A,B & C are all signed > Carl Thanks Carl, I appreciate the now snipped assistance. I think it's mostly just carelessness but whatever the reason I manage to mess up signage nearly every time! I suspect your advice was useful to me in the test I sat this morning :) Ivan. Subject: Simple idea, mathematics and common-sense I've had a time explaining some *very* simple mathematical ideas that lead to a few complexities, but it's been fruitful to explain, or try to explain, as I work to figure out why these simple ideas either excite derision, anger or confusion, and I think I have it figured out. I'm sure many of you are put off by mathematics, so I assure you up front that what I'll be talking about will mostly be *very* simple, and there will only be a few slightly complicated things at the end. First of all, I'm going to talk about a case where mathematicians gave up because they couldn't see something, and assumed that because they couldn't see something it didn't exist! You know how with simple quadratics like x^2 + 3x + 2, it's easy enough to see factors of 2 in the roots? I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. However, if it's something like x^2 + 7x + 2, you can use the quadratic formula and get the roots to find x = (-7 +/- sqrt(41))/2 and who can see factors of 2 in that thing? There's something else important here which is the ambiguity of the square root operator, which may sound complicated but it's easy to demonstrate with another root of a quadratic: (1+sqrt(9))/2 and you may think, silly, why show sqrt(9) when sqrt(9) = 3, but yeah, that's *one* of its solutions, as sqrt(9) = -3 as well, so you have *two* numbers (1+sqrt(9))/2 = 2 or -1 as either solution will work. I've had people argue with me that by definition (really by convention) you take the positive root. But imagine the world of Contrary. On Contrary the mathematicians for some odd reason *by definition* take the negative! Is mathematics really changing depending on such decisions. Imagine you've forgotten that you can resolve sqrt(9) to 3 or -3, so you write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 and mercifully discover that you can get rid of that square root sign by adding them together, or multiplying them together. Like adding them gives 1, and multiplying them together gives (1 - 9)/4 = -2 and your mathematicians scratched their heads and contemplated such numbers, and decided that there was *no way* to understand factors of 2 of numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 except as to consider them to be unique factors of 2, in some kind of mysterious way. But wait, that's not a problem here, of course, because you can just evaluate the square root, but look back now at x^2 + 7x + 2, where x = (-7 +/- sqrt(41))/2 and consider that you *cannot* resolve sqrt(41) in any way that will help you here with this question. Sure, you can write it out in decimal format with a lot of numbers after the decimal place. My computer tells me that sqrt(41) approximately equals 6.4031242374328486864882176746218, but of course, you can keep going out to infinity trying just to see sqrt(41). If you drop some of those numbers (to make it easier) and move those decimal places, to get 64031242374, squaring gives 4099999999957933155876, which looks VERY CLOSE to 41 * 10^20, and you can see what our approximations actually are. We approximate irrational numbers like square roots by finding some REALLY BIG natural number, and moving the decimal place to the left. However, you STILL don't have a simple idea of where factors of 2 go, like you had with the easy and you now might think comforting example of (x^2 + 3x + 2) = (x+2)(x+1). Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyze because we really, really like integers, but to get an integer you have to use (-7 - sqrt(41))/2, as then you can add them together to get -7 and multiply them to get 2, but mathematicians could not figure out a way to handle such numbers in less than pairs! That's important. Mathematicians could never figure out a way to handle such numbers in less than pairs. So some of them decided that what they couldn't see, wasn't meaningful. It'd be kind of like the weird mathematicians, who couldn't evaluate sqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 deciding that where factors of 2 resided was a mystery to them. But we *can* evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) which leaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, and some mathematicians have decided that that's it. For them, that's all you need to know. Here are these numbers where we can't evaluate the square root. Sorry, no go there, they decided, you're stuck, isn't it obvious? You see they decided that the limit on what they could *see* was a limit on what could mathematically exist for those examples--irational roots--where they could not see. As far as those mathematicians were concerned, end of story. But that's where my story begins. My mathematical research is about how the kind of patterns you see with integers, like with (x^2 + 3x + 2) = (x+2)(x+1) where one root has all the meaningful factors of 2 *continues* into realms where we can't see it directly because we can't get past those danged irrationals coming in at least pairs! I found an *indirect* way of looking, which involves putting expressions like these polynomials I've shown here, but more complicated into a special but VERY simple mathematical tool. For example, a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) can be peered into, by figuring out that its roots can be considered in the following mathematical structure: (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are the roots of that complicated looking cubic that I started with, and I've managed to place them in a structure opposite a polynomial that has 49 as a factor. In one sense that's easy. You can take just about any expression like that, focus on some factor of its last term, and build something like what I did. And it being so easy may explain some of the problems I'm having with people taking it seriously! What I figured out though is that what mathematicians couldn't see before can now be logically seen, and in fact, the way you divide off that 49 is to get something like (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22 but some vocal mathematicians don't like the idea as it contradicts what they've been taught, and believed because by their thinking and training, you can't get around the barrier. Notice that I picked two roots arbitrarily and in fact even if you solve for the roots you can't look and tell *which* roots should be divided by 7. All you know is that two of them should be, but can never know *which* two. Which makes my job of explaining harder, but it's just that ambiguity thing coming back into place. We like integers. The math doesn't care as it just handles numbers. We can't look *directly* at the roots, but we can use logic to consider them. What we can't see directly here *does* exist because the mathematics says it does. To some extent, my problems are with people and mathematicians who trust logic less than what they've been taught and their own personal sense of what makes since. In a nutshell that's the basis for the arguments. I say that just because we can't see the factors of irrational roots like we can with rational ones our limitation is not a mathematical constraint, while some people, including a lot of very obsessive posters, refuse to acknowledge the possibility, fighting for their old view. Is it important? Actually, it is important if mathematicians have an error in their thinking. It may be the case that things thought to have been proven in mathematics, really haven't. It's quite possible that any number of arguments claimed to be proofs assumed that what they couldn't see, could not exist, as this issue has been around for over a hundred years. Yes, I can talk it all out rigorously and in a heavily mathematical format, but I hope that giving you some idea what all the fighting is really about, will help in understanding what's going on, as well as why I don't just capitulate. Mathematicians may think that seeing is believing, but I think in mathematics, logic is king. James Harris Subject: Re: Simple idea, mathematics and common-sense Yes, I can talk it all out rigorously and in a heavily mathematical > format, James, I have been following your threads for some time now and I have seen on more occasions than I can count a request for you to do just this. I seem to recall that when you have deigned to acknowledge these requests it wasn't ...[talked] out rigourously and in a heavily mathematical format it was more of a diatribe against mathematicians and the evil society that they control. I have seen you claim that clearly stated definitions are wrong, that (as you did in this post) there is some inherent ambiguity and that professional mathematicians (as well as the very capable amatuers) are too stupid to understand but I have *NEVER* seen you post anything that even I would consider to be rigourous. Having now said that you can do this, when can I expect to see it? I'm sure that if I have any difficulty then either you or somebody else will be able to clarify. Ivan. Subject: Re: Simple idea, mathematics and common-sense I know I shouldn't try, but the light... it's so beautifu...zap! >There's something else important here which is the ambiguity of the >square root operator It is not ambiguous. Sqrt(x) (or x^(1/2) or the notation with the square root symbol) all define the principal square root (provided x is a positive real number). So Sqrt(9) = 3. It is not -3 even though -3 is a square root of 9. Just read about it on the following link (You could also go to the library and read a decent math-book there): http://mathworld.wolfram.com/SquareRoot.html Subject: Re: Simple idea, mathematics and common-sense > There's something else important here which is the ambiguity of the > square root operator, Which is easily resolved by stating what one means by the sqrt symbol. If one states that sqrt(x) requires that both x and sqrt(x) be non-negative reals, there is no ambiguity. If one allows that for non-zero complex numbers x, sqrt(x) can mean either of the complex numbers whose square equals x, THEN it is ambiguous. It is customary to presume the first meaning whenever possible, i.e., whenever x is a non-negative real. > which may sound complicated but it's easy to > demonstrate with another root of a quadratic: > (1+sqrt(9))/2 > and you may think, silly, why show sqrt(9) when sqrt(9) = 3, but yeah, > that's *one* of its solutions, as sqrt(9) = -3 as well, so you have > *two* numbers > (1+sqrt(9))/2 = 2 or -1 Only if you reject the customary interpretation and refuse to resolve the ambiguity in a civilized way. But no one has ever accused JSH of civility. > as either solution will work. Only if, contrary to custom, one assumes the second definition. Since JSH is so fervently anti-math, no one expects him to follow customary usage, but equally, those who are willing to follow customary usage should be very cautious about following JSH anywhere. I've had people argue with me that by > definition (really by convention) you take the positive root. As it is a conventional definition, it is really by both. > But > imagine the world of Contrary. That is certainly JSH's world. > On Contrary the mathematicians for > some odd reason *by definition* take the negative! Is mathematics > really changing depending on such decisions. Then the standard symbol would become -sqrt(x), no problem. > Imagine you've forgotten that you can resolve sqrt(9) to 3 or -3, Forgetting elementary properties of arithmetic is a favorite ploy of JSH. If done creatively, it can produce interesting ideas, but JSH is singularly uncreative in any mathematical sense, and his variations on this theme seems always to lead him into idiotic error. > so > you write these numbers like (1+sqrt(9))/2 and (1-sqrt(9))/2 and > mercifully discover that you can get rid of that square root sign by > adding them together, or multiplying them together. > Like adding them gives 1, and multiplying them together gives > (1 - 9)/4 = -2 > and your mathematicians scratched their heads and contemplated such > numbers, and decided that there was *no way* to understand factors of > 2 of numbers like > (1+sqrt(9))/2 and (1-sqrt(9))/2 > except as to consider them to be unique factors of 2, in some kind of > mysterious way. What seems like a mystery to JSH is simple arithmetic to everyone else. > But wait, that's not a problem here, of course, because you can just > evaluate the square root, but look back now at x^2 + 7x + 2, where > x = (-7 +/- sqrt(41))/2 > and consider that you *cannot* resolve sqrt(41) in any way that will > help you here with this question. It is certainly possible to add (-7 + sqrt(41))/2 to (-7 - sqrt(41))/2 geting -7 or to multiply them getting (49-41)/4 = 2, so no further resolution is required. > Sure, you can write it out in decimal format with a lot of numbers > after the decimal place. My computer tells me that sqrt(41) > approximately equals 6.4031242374328486864882176746218, but of course, > you can keep going out to infinity trying just to see sqrt(41). But sinice the expression 6.4031242374328486864882176746218 is just as artificial, and less accurate, than sqrt(41), why bother? Decimal notation is a convenience, not a necessity, for representation of numbers. The fact that some numbers are not so representable is merely an inconvenience, not a disaster. > But that's where my story begins. [fairy story snipped] > Mathematicians may think that seeing is believing, but I think in > mathematics, logic is king. What JSH thinks about how mathematicians think is irrelevant to anything in the real world. If logic is king, then JSH has shown himself to be the court fool. > James Harris Subject: Re: Simple idea, mathematics and common-sense Discussion, linux) > First of all, I'm going to talk about a case where mathematicians gave > up because they couldn't see something, and assumed that because they > couldn't see something it didn't exist! You know, I've *seen* mathematicians say something like this. Let me see if I can recall what it was. Oh yeah, it was this. ,----[ <3c65f87.0402140856.593bc13f@posting.google.com> ] | What I've shown in a rather simple way is that only such simple and | direct results will work, so if you don't *see* the non-unit factors | between two irrational algebraic integers, then you don't have any in | the ring of algebraic integers! `---- Wait. My mistake. That wasn't a mathematician. It was you. -- Well, I don't claim to be an expert, in fact I am a fry cook with a national burger chain, but I have solved many differential and partial differential equations numerically. --C. Bond Subject: Re: Simple idea, mathematics and common-sense > I've had a time explaining some *very* simple mathematical ideas that > lead to a few complexities, Your ignorance, incompetence, and psychosis are not of interest to the world at large. Quite the contrary. You are not even an interesting laughingstock. Hey stooopid loud troll James Always in error, never in doubt! Harris, put up or shut up. James Harris, King of the Primes! Where are your sceptor and crown, delusional James Harris, your regal clothes? Is a $10,000 prize no questions asked too small to justify your submission of two little prime numbers? Or are you a psychotic impotent gelding? http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net You know how with simple quadratics like x^2 + 3x + 2, it's easy > enough to see factors of 2 in the roots? > I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. > However, if it's something like x^2 + 7x + 2, you can use the > quadratic formula and get the roots to find > x = (-7 +/- sqrt(41))/2 > and who can see factors of 2 in that thing? Anyone can, if they look at it in the right way. If you denote the roots by x_1 and x_2, then since x_1 and x_2 satisfy x_1 * x_2 = 2, the roots are both factors of 2. I'll admit that in one sense one can't see the factors of 2, but if you use the equation that the x's satisfy then you see the factors of 2 immediately. The problem becomes trickier when you have a number like a = 1 + sqrt(-5) Now I know that that number divides 6, since (1 + sqrt(-5))(1 - sqrt(-5)) = 6 so a will have some factor, w, that divides 6. Thus, we're looking for some algebraic integer w that divides both a and 2. That would be easy if everything in sight were an integer: we could simply use w = gcd(a, 2), and in the integers it's easy to calculate the gcd. Well, it turns out that gcds exist in the algebraic integers as well, but it's by no means easy to calculate them. Furthermore, it often happens that we can't expect to get an easy solution to, say gcd(a, 2); instead, we often have to look for gcd(a^k, 2^k), for some integer k. This is where the class number comes in, but that'll be a matter for later discussion. Now we're lucky in this case. Trying k = 2 (which is the class number of Q(sqrt(-5)), we see that a^2 = (1 + sqrt(-5))^2 = -4 + 2sqrt(-5) and, of course 2^2 = 4 Now it's truly obvious in this case that 2 will divide both a^2 and 4, so sqrt(2) will divide both a and 2, so the factor of 2 we're looking for in 1 + sqrt(-5) happens to be sqrt(2). That's not the only one, of course: -sqrt(2), or sqrt(-2), or -sqrt(-2) will all work equally well, as will infinitely many others. The point here is that while there will always be a way to find the desired factors, it won't always be easy to do so. As an exercise, you might want to show that a common factor of 3 and 1 + sqrt(-5) is sqrt(2 - sqrt(-5)), among others. I found an *indirect* way of looking, which involves putting > expressions like these polynomials I've shown here, but more > complicated into a special but VERY simple mathematical tool. > For example, > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > can be peered into, by figuring out that its roots can be considered > in the following mathematical structure: > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are the roots of that complicated looking cubic that I > started with, and I've managed to place them in a structure opposite a > polynomial that has 49 as a factor. > In one sense that's easy. You can take just about any expression like > that, focus on some factor of its last term, and build something like > what I did. And it being so easy may explain some of the problems I'm > having with people taking it seriously! > What I figured out though is that what mathematicians couldn't see > before can now be logically seen, and in fact, the way you divide off > that 49 is to get something like > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 Unfortunately, no. The whole point of my example was to demonstrate that in general a factor on the right will be split evenly (in some sense) among the a's. In this case, the 49 on the right will almost always turn out to be w_1 * w_2 * w_3, where w_i are not units and each w_i will be a divisor of a_i and 49. Rick Subject: Re: Simple idea, mathematics and common-sense O > I found an *indirect* way of looking, which involves putting > expressions like these polynomials I've shown here, but more > complicated into a special but VERY simple mathematical tool. > For example, > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > can be peered into, by figuring out that its roots can be considered > in the following mathematical structure: > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where the a's are the roots of that complicated looking cubic that I > started with, and I've managed to place them in a structure opposite a > polynomial that has 49 as a factor. > In one sense that's easy. You can take just about any expression like > that, focus on some factor of its last term, and build something like > what I did. And it being so easy may explain some of the problems I'm > having with people taking it seriously! > What I figured out though is that what mathematicians couldn't see > before can now be logically seen, and in fact, the way you divide off > that 49 is to get something like > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 Prove that is true. I think the many counter examples to this might be a hindrance. The only thing you are claiming is that if (f(x)+a)(g(x)+b) is divisible by a, it must be that f(x) is divisible by a for all x where f and g are just two functions from the underlying ring to itself, with no particular special propertes that ensure divisibility, and nothing special about the ring either. Complete and utter crap. It may well be that there is some function that will do what you want, but the one you've got ain't it. There is still the x(x+1) counter example in Z. You claim that's because Z is a UFD, well, let j be the root of any square free integer where Z[j] is not a UFD, such exist. Then x(x+1)(x+j) is divisible by two again, but you can't tell presume it will divide the same factor always. Subject: Re: Simple idea, mathematics and common-sense > O > >> I found an *indirect* way of looking, which involves putting >> expressions like these polynomials I've shown here, but more >> complicated into a special but VERY simple mathematical tool. > >> For example, > >> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > >> can be peered into, by figuring out that its roots can be considered >> in the following mathematical structure: > >> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > >> 49(300125 x^3 - 18375 x^2 - 360 x + 22) > >> where the a's are the roots of that complicated looking cubic that I >> started with, and I've managed to place them in a structure opposite a >> polynomial that has 49 as a factor. > >> In one sense that's easy. You can take just about any expression like >> that, focus on some factor of its last term, and build something like >> what I did. And it being so easy may explain some of the problems I'm >> having with people taking it seriously! > >> What I figured out though is that what mathematicians couldn't see >> before can now be logically seen, and in fact, the way you divide off >> that 49 is to get something like > >> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = > >> 300125 x^3 - 18375 x^2 - 360 x + 22 > > Prove that is true. I think the many counter examples to this might be a > hindrance. > The only thing you are claiming is that if > (f(x)+a)(g(x)+b) is divisible by a, it must be that f(x) is divisible by a > for all x where f and g are just two functions from the underlying ring to > itself, with no particular special propertes that ensure divisibility, and > nothing special about the ring either. > Complete and utter crap. It may well be that there is some function that > will do what you want, but the one you've got ain't it. > There is still the x(x+1) counter example in Z. You claim that's because Z > is a UFD, well, let j be the root of any square free integer where Z[j] is > not a UFD, such exist. Then x(x+1)(x+j) is divisible by two again, but you > can't tell presume it will divide the same factor always. correction x(x+1)(x+j)(x+j+1) is divisible by two always. Subject: Re: Simple idea, mathematics and common-sense [snip James' math lessons for those less magnificent than himself] > But wait, that's not a problem here, of course, because you can just > evaluate the square root, but look back now at x^2 + 7x + 2, where > x = (-7 +/- sqrt(41))/2 > and consider that you *cannot* resolve sqrt(41) in any way that will > help you here with this question. > Sure, you can write it out in decimal format with a lot of numbers > after the decimal place. My computer tells me that sqrt(41) > approximately equals 6.4031242374328486864882176746218, but of course, > you can keep going out to infinity trying just to see sqrt(41). Whoa! What do you mean to see sqrt(41)? You are now talking about a *decimal expansion* of sqrt(41). Do you think you have to go out to infinity with 0.333333333... to see 1/3? How about going out to infinity with 1.0000000... to see unity? > If you drop some of those numbers (to make it easier) and move those > decimal places, to get 64031242374, squaring gives > 4099999999957933155876, which looks VERY CLOSE to 41 * 10^20, and you > can see what our approximations actually are. Yawn. [snip interminable, rambling which attempts to demonstrate the genius of JSH, but which actually confirms his idiocy] > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com Subject: Re: Simple idea, mathematics and common-sense Yes, I can talk it all out rigorously and in a heavily mathematical > format, but I hope that giving you some idea what all the fighting is > really about, will help in understanding what's going on, as well as > why I don't just capitulate. You've never demonstrated any ability to think or write mathematics rigorously. In fact just the opposite seems to be true of your ramblings Subject: Re: Simple idea, mathematics and common-sense >I've had a time explaining some *very* simple mathematical ideas that >lead to a few complexities, but it's been fruitful to explain, or try >to explain, as I work to figure out why these simple ideas either >excite derision, anger or confusion, and I think I have it figured >out. It's very simple. This is one of the many things that you never seem to grasp, no matter how many times it's explained, no matter how many people explain it: The reason people contradict you is because you're simply wrong. The reason people find you hilarious is because no matter how many times you finally admit you were wrong, the next time you have something to say you always insist that you understand the math better than _every_ professional mathematician on the planet. (Sometimes that's just implicit in the things you say, sometimes you actually say so explicitly.) The reason people get angry with you is that you're such a ing asshole. >I'm sure many of you are put off by mathematics, so I assure you up >front that what I'll be talking about will mostly be *very* simple, >and there will only be a few slightly complicated things at the end. And condescending to boot. >First of all, I'm going to talk about a case where mathematicians gave >up because they couldn't see something, and assumed that because they >couldn't see something it didn't exist! >You know how with simple quadratics like x^2 + 3x + 2, it's easy >enough to see factors of 2 in the roots? >I mean, it's just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. >However, if it's something like x^2 + 7x + 2, you can use the >quadratic formula and get the roots to find >x = (-7 +/- sqrt(41))/2 >and who can see factors of 2 in that thing? >[...] >Like adding them gives 1, and multiplying them together gives >(1 - 9)/4 = -2 >and your mathematicians scratched their heads and contemplated such >numbers, and decided that there was *no way* to understand factors of >2 of numbers like >(1+sqrt(9))/2 and (1-sqrt(9))/2 >except as to consider them to be unique factors of 2, in some kind of >mysterious way. >[...] >Now then numbers like (-7 + sqrt(41))/2 defy our ability to analyze >because we really, really like integers, but to get an integer you >have to use (-7 - sqrt(41))/2, as then you can add them together to >get -7 and multiply them to get 2, but mathematicians could not figure >out a way to handle such numbers in less than pairs! >That's important. Mathematicians could never figure out a way to >handle such numbers in less than pairs. >So some of them decided that what they couldn't see, wasn't >meaningful. >It'd be kind of like the weird mathematicians, who couldn't evaluate >sqrt(9), and were looking at (1+sqrt(9))/2 and (1-sqrt(9))/2 deciding >that where factors of 2 resided was a mystery to them. But we *can* >evaluate sqrt(9), but we *cannot* really evaluate sqrt(41) which >leaves a mystery with (-7 + sqrt(41))/2 and ((-7 - sqrt(41))/2, and >some mathematicians have decided that that's it. >For them, that's all you need to know. Here are these numbers where >we can't evaluate the square root. Sorry, no go there, they decided, >you're stuck, isn't it obvious? >You see they decided that the limit on what they could *see* was a >limit on what could mathematically exist for those examples--irational >roots--where they could not see. As far as those mathematicians were >concerned, end of story. >[...] >What I figured out though is that what mathematicians couldn't see >before can now be logically seen, and in fact, the way you divide off >that 49 is to get something like >(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 a_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 >but some vocal mathematicians don't like the idea as it contradicts >what they've been taught, and believed because by their thinking and >training, you can't get around the barrier. >[...] >We can't look *directly* at the roots, but we can use logic to >consider them. What we can't see directly here *does* exist because >the mathematics says it does. >To some extent, my problems are with people and mathematicians who >trust logic less than what they've been taught and their own personal >sense of what makes since. >In a nutshell that's the basis for the arguments. >I say that just because we can't see the factors of irrational roots >like we can with rational ones our limitation is not a mathematical >constraint, while some people, including a lot of very obsessive >posters, refuse to acknowledge the possibility, fighting for their old >view. >Is it important? >Actually, it is important if mathematicians have an error in their >thinking. >It may be the case that things thought to have been proven in >mathematics, really haven't. >It's quite possible that any number of arguments claimed to be proofs >assumed that what they couldn't see, could not exist, as this issue >has been around for over a hundred years. >Yes, I can talk it all out rigorously and in a heavily mathematical >format, but I hope that giving you some idea what all the fighting is >really about, will help in understanding what's going on, as well as >why I don't just capitulate. >Mathematicians may think that seeing is believing, but I think in >mathematics, logic is king. What a truly awesome load of crap. _You_ are the person who's been making ridiculous statements about how factors that you can't see cannot exist. Hint: Just because you don't understand something does not imply that it's not understood by mathematicians. This really is a new twist. For newcomers, what's traditional is something like this: James: Black is white. Someone: Uh, no, black is black and white is white - they're actually different colors entirely. [someone else inserts quibble about whether they're really _colors_...] James: It can be hard for me to explain why black is white, because you think if you didn't learn it in school it can't be right. Someone: Huh? Black is white. James: That's what people thought until the Object Ring was invented. Someone: This is ridiculous. Why are you ignoring the proofs that people have showed you that black is not white? James: OFF!!!!!!!! Your replies are NOT wanted - how many times do I have to say that? Someone: Here's an extremely simple proof that black is not white: James: I've contacted your university, informing them of your academic fraud. I'm considering hiring a lawyer. Someone: Uh, right. That really doesn't change the fact that black is not white. James: Liar. Doesn't anyone here care about mathematical Truth? And so on for months or years, until finally James: Well, I was wrong, black is not white. No big deal. That's what usually happens - then a day or so later the cycle starts again, this time a debate over whether red is blue. But this time we have a new twist: James: See, mathematicians have thought that black was white. It's kind of hard to see why black is not white - I'll try to explain it without a lot of math. The reason mathematicians have thought that black is white is just that that's the way they were trained... ************************ Subject: Re: Simple idea, mathematics and common-sense > James: See, mathematicians have thought that black was white. It's kind of > hard to see why black is not white - I'll try to explain it without a lot > of math. The reason mathematicians have thought that black is white is > just that that's the way they were trained... I have been following, on and off, the postings of Mr. Harris for years. Like you, I used to feel infuriated, for I couldn't believe that anyone could be so thick. Now that I have a bit more experience on the behavior of mentally disturbed individuals, I have come to the conclusion that Mr. Harris is a poor lunatic, for whom sci.math is the main means to live his madness. I guess we should feel compassionate about his regrettable predicament. He is very annoying all right but, what can be expected from a deranged individual? Subject: Re: Simple idea, mathematics and common-sense >> James: See, mathematicians have thought that black was white. It's kind of >> hard to see why black is not white - I'll try to explain it without a lot >> of math. The reason mathematicians have thought that black is white is >> just that that's the way they were trained... > I have been following, on and off, the postings of Mr. Harris for years. >Like you, I used to feel infuriated, for I couldn't believe that anyone >could be so thick. I haven't been infuriated by him in a long time - these days he's just entertainment. >Now that I have a bit more experience on the behavior >of mentally disturbed individuals, I have come to the conclusion that Mr. >Harris is a poor lunatic, for whom sci.math is the main means to live his >madness. Could be. > I guess we should feel compassionate about his regrettable predicament. I don't see why. Supposing for the sake of argument he's a poor lunatic, that doesn't give him the right to behave the way he does. >He is very annoying all right but, what can be expected from a deranged >individual? ************************ Subject: JSH: Prime concept in Object Ring As a note to myself, one of the reasons I found myself continually going back to the Decker quadratic was I realized I hadn't worked it all out for polynomials that weren't of a special class. That is, I'd focused on one particular type of polynomial as that's where I came into this area, but the Decker example caused me to focus on other areas. Apparently most polynomials are prime-like in that you can't separate off non-unit object factors of their constant terms within the ring, but have to go into the operator space. In retrospect, it's kind of obvious. Now I can move forward to the story of my work versus actively working on my own understanding, which hopefully will produce some progress here. Doing active research while trying to explain it as you go along is kind of difficult. James Harris Subject: Re: JSH: Prime concept in Object Ring > As a note to myself, Notes to yourself should only be published posthumously. Subject: Re: Prime concept in Object Ring > Doing active research while trying to explain it as you go along is > kind of difficult. Why not arrive at a useful result first, *prove* it, and *then* explain it? Skip Subject: Re: JSH: Prime concept in Object Ring > As a note to myself, one of the reasons I found myself continually > going back to the Decker quadratic was I realized I hadn't worked it > all out for polynomials that weren't of a special class. That is, I'd > focused on one particular type of polynomial as that's where I came > into this area, but the Decker example caused me to focus on other > areas. Hmm, so you were wrong for one type of polynomial, but dammit your right for the general one! > Apparently most polynomials are prime-like in that you can't separate > off non-unit object factors of their constant terms within the ring, > but have to go into the operator space. Eh? What the buggery does that mean? > In retrospect, it's kind of obvious. something is certainly > Now I can move forward to the story of my work versus actively working > on my own understanding, which hopefully will produce some progress > here. > Doing active research while trying to explain it as you go along is > kind of difficult. > James Harris Subject: Re: JSH: Prime concept in Object Ring > As a note to myself, Next time just use a PostIt on the refrigerator. > one of the reasons I found myself continually > going back to the Decker quadratic was I realized I hadn't worked it > all out for polynomials that weren't of a special class. The other reason is that you are too stupid to understand the errors in your own work. > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com Subject: Re: errors in an argument by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBHn07123; >> The way it seems to me (and I'm not talking as an expert in aerodynamics >> either) is that even for extremely small and light creatures, the ability >> to develop a functional limb that will allow it even the smallest >> maneuverability by mutation benefit is unlikely. Remember that it has to >> be aerodynamic to some extent, and to be evolutionary beneficial. It's >> very hard for me to imagine how such a thing would look like. >Imagine this: >http://www.animalnetwork.com/critters/profiles/ flyingsquirrel/default.a spKeep in mind that embryos of many species typically have webbing during >development; it would not take much of a mutation for the webbing to >persist. >A little research also turns up an explanation of feathers that has >nothing to do with their eventual use in flight. >-- This looks like a flawed argument. Assuming evolution, we are already at a stage that allows the appearance of wings or semi wings, so this is circular. However, the existence of similar limbs in fish like someone else noted here combined with the huge advantage even the slightest ability for such maneuverability can sound plausible. My problem is more general. Too many times we get a very complicated and fine tuned mechanism that would require a major jump in order to have any effectiveness. Fish that use electricity to stun and kill opponents is a remarkable one. The amount of electricity generated, and the coordination involving it is very big. It looks like it would take a big number of mutations to achieve it, with any partial mutation or a small effect totally useless. A functional yet small weapon of this kind is even punishing given the amount of energy needed for it, and lack of effective results. Subject: Re: errors in an argument > My problem is more general. Too many times we get a very complicated and > fine tuned mechanism that would require a major jump in order to have any > effectiveness. Fish that use electricity to stun and kill opponents is a > remarkable one. The amount of electricity generated, and the coordination > involving it is very big. The major jump is only in your mind. > It looks like it would take a big number of mutations to achieve it, with > any partial mutation or a small effect totally useless. A functional yet > small weapon of this kind is even punishing given the amount of energy > needed for it, and lack of effective results. Two words: active sensor http://www.flmnh.ufl.edu/fish/tropical/JSA/gymno.htm Next time, do your own research. http://www.google.com/search?q=evolution+%22electric+eel%22 -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W Subject: Re: errors in an argument >My problem is more general. Too many times we get a very complicated and fine tuned >mechanism that would require a major jump in order to have any effectiveness. Fish that >use electricity to stun and kill opponents is a remarkable one. The amount of electricity >generated, and the coordination involving it is very big. It looks like it would take a big >number of mutations to achieve it, with any partial mutation or a small effect totally >useless. A functional yet small weapon of this kind is even punishing given the amount of >energy needed for it, and lack of effective results. I think you should ask these questions on the newsgroup talk.origins (or some other newsgroup that talks about evolution/biology). I'm sure there are several people there who can explain this to you (to some extend). Subject: functional analysis--separable by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBEv07013; Hi all, I found the following proposition in a book without a proof. I have difficulty in proving it. Can anyone help? Suppose B is a separable Banach space. Let {x_n} be a countable dense subset of the unit circle C={x in B : ||x||=1}. How can I prove that: every element x in B can be written as: x= summation[from 1 to infinity] {a_n x_n} where a_n is a absolute summable sequence. That's: summation[from 1 to infinity] {|a_n|} < infinity. Subject: Cavemen by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KEBHw07133; Where can I find pics of cavemens language e.g. A pic of a wall inprint Subject: Re: Cavemen > Where can I find pics of cavemens language e.g. A pic of a wall inprint Is it a mathematical language? If not, not here. Subject: Re: Cavemen > Where can I find pics of cavemens language e.g. A pic of a wall inprint This one comes close: http://www.jmilne.org/bitted.html :D Subject: Re: Cavemen > Where can I find pics of cavemens language e.g. A pic of a wall inprint Not on sci.math, I can tell you that. Subject: Re: Cavemen > Where can I find pics of cavemens language e.g. A pic of a wall inprint > Not on sci.math, I can tell you that. No binaries, but maybe an ASCII rendering. To bring it on topic, van der Waerden in _Geometry and Algebra in Ancient Civilizations_ quotes a translation from something ancient (how's that for not consulting primary sources?) as if you do it this way [ruler and compass construction, with no measurements] you won't have to worry about the gods rejecting your sacrifice. Jon Miller Subject: Re: Logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KErjr10530; Hint for (1) - Unless means the same thing as if not. (2) you should be able to work out on your own. >> I'm stuck on these 2 problems. >> Represent the following compound propositions, each involving one or >more >> of >> these three propositons, as symbolic expressions in the variables d,e,f. >> 1)e does not f unless d >> Does 1 this mean If d and e the f???? >Typo: Does 1 mean If d and e the f???? >> 2)d and e does not imply f >> Does 2 mean not d or not e does imply f??? >> Of coure I am told what these propositions are. >> I'm sorry but I can't even show you what I've done so far--I'm >completely >> lost. >> Steven >> >> >> Subject: Re: Simple question regarding frequency detection by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KErjR10534; >I am writing an application to detect a particular frequency(viz. 4.5 >Khz) in an input audio signal. I have written the code to capture the >sound using a microphone; but I have to properly design a digital filter >which will return TRUE as soon as it detects sound of this frequency in >the input signal. Also my project demands that I have to detect the >signal by looking at a very small number of samples e.g. 44. So I will >apply the digital filter on a window of 44 samples of the input signal >repeatedly till I get a high response indicating that I have got the >reqd. frequency. >I know DFT/FFT is an option but are there any better algorithms I can >use for my project. Please suggest the best algorithm that will do my >work. I am writing the code in C++ so if anyone can point me to existing >code on the net that will be really helpful. >Thanks, >Atri. The problem is a bit more involved than you indicate. What is the desired probability of detection? What is the maximum allowable probability of false detection? What is the minimum signal-to-noise ratio at the input? etc. etc. phil Subject: Re: JSH: Non-uniqueness of factorization > To be fair I think Outlook does that. Anything behind a : at the start of > a >> subject line is treated as if it were Re: and is removed. I'd suggest > switching >> to [JSH] which is the convention in other groups and seems to work with > all >> newsreaders. > I'm using Outlook now, and it did not do that. It left the JSH: in there That's almost encouraging, but Microsoft is so abysmal when it comes to Usenet standards, I am still not persuaded that Outlook does something sensible. Maybe Outlook does this: If a subject starts with xxx: blah blah blah, then strip out xxx: and put Re:. Since you replied to Re: JSH: blah blah blah, it would have stripped out Re: and then put Re:, making no effective change. But if you had replied to JSH: blah blah, it would have chosen Re: blah blah as a subject line. I don't know if this is right. I'm just guessing, based on Microsoft's bad track record (like using different replacements for Re: depending on the language of the user, RFC standards be damned). So, can you check? Select the root of this thread and compose a followup (without posting) and let us know what the subject line says. Also, I wonder if it matters whether you use Outlook or Outlook Express. -- My proof has been checked very thoroughly, both by me and others. Those others apparently decided that they would not believe the proof was correct, but cannot support that position using mathematics. But hey, they're just human beings. --JSH, prover of Fermat's Last Thm Subject: Re: JSH: Non-uniqueness of factorization >> To be fair I think Outlook does that. Anything behind a : at the start of > a >> subject line is treated as if it were Re: and is removed. I'd suggest > switching >> to [JSH] which is the convention in other groups and seems to work with > all >> newsreaders. > > I'm using Outlook now, and it did not do that. It left the JSH: in there > That's almost encouraging, but Microsoft is so abysmal when it comes > to Usenet standards, I am still not persuaded that Outlook does > something sensible. > Maybe Outlook does this: If a subject starts with xxx: blah blah blah, > then strip out xxx: and put Re:. Since you replied to Re: JSH: blah > blah blah, it would have stripped out Re: and then put Re:, > making no effective change. > But if you had replied to JSH: blah blah, it would have chosen Re: > blah blah as a subject line. > I don't know if this is right. I'm just guessing, based on > Microsoft's bad track record (like using different replacements for Re: depending on the language of the user, RFC standards be > damned). So, can you check? Select the root of this thread and > compose a followup (without posting) and let us know what the subject > line says. You are right. I tried it. In this thread Re:JSH:blahblah does not change, but in another thread with subject JSH:blahblah, the JSH is removed and replaced with RE, thus changing JSH:blahblah to RE:blahblah KeithK > Also, I wonder if it matters whether you use Outlook or Outlook > Express. > -- My proof has been checked very thoroughly, both by me and others. > Those others apparently decided that they would not believe the proof > was correct, but cannot support that position using mathematics. But > hey, they're just human beings. --JSH, prover of Fermat's Last Thm Subject: Re: JSH: Non-uniqueness of factorization > That's almost encouraging, but Microsoft is so abysmal when it comes > to Usenet standards, I am still not persuaded that Outlook does > something sensible. > Maybe Outlook does this: If a subject starts with xxx: blah blah blah, > then strip out xxx: and put Re:. Since you replied to Re: JSH: blah > blah blah, it would have stripped out Re: and then put Re:, > making no effective change. > But if you had replied to JSH: blah blah, it would have chosen Re: > blah blah as a subject line. > I don't know if this is right. I'm just guessing, based on > Microsoft's bad track record (like using different replacements for Re: depending on the language of the user, RFC standards be > damned). So, can you check? Select the root of this thread and > compose a followup (without posting) and let us know what the subject > line says. That is correct, at least for OE6. The latest RFC (2822) in section 3.6.5 describing the subject field, mentions that Re: *may* (emphasis mine) be used but that is neither required nor to be expected. The same section cautions about building long strings of such add-on text - because the field is unstructured and meant for human eyes, not programs. I have seen the strings 'Reply:, Forward:, FWD:, FW: and many others. I think MS is just erring on the side of caution by deleting a string ending with :. One cannot fault them for not knowing who JSH is because, after all, he's a fan of Java, not C#. Subject: Re: JSH: Non-uniqueness of factorization <87ad3den4w.fsf@phiwumbda.org <25rZb.64909$KV5.49312@nwrdny01.gnilink.net Discussion, linux) > I think MS is just erring on the side of caution by deleting a > string ending with :. One cannot fault them for not knowing who > JSH is because, after all, he's a fan of Java, not C#. I guess we disagree on what constitutes caution. Mangling subject lines is not cautious. The fact is that, JSH aside, there are plenty of opportunities in which one wants to start a thread with a subject of the form, Blah: blah blah blah. That the MS coders couldn't think of any isn't surprising. They couldn't consider that (on rare occasions), lines might begin like this. Consequently, users of some of (which?) products of MS software cannot read this sentence. Admittedly, it is not often that a line begins with begin , but it is still ridiculous that MS coders chose to parse every such occurrence as a mime enclosure, regardless of whether the headers indicated there was an enclosure or not. I suppose they were being cautious and ensuring that they compensated for any mime enclosures that were incorrectly constructed. Or something. -- We are happy that you agree that customers need to know that Open Source is legal and stable, and we heartily agree with that sentence of your letter. The others don't seem to make as much sense, but we find the dialogue refreshing. -- Linus Torvalds to Darl McBride Subject: Re: JSH: Non-uniqueness of factorization > I think MS is just erring on the side of caution by deleting a > string ending with :. One cannot fault them for not knowing who > JSH is because, after all, he's a fan of Java, not C#. > I guess we disagree on what constitutes caution. Mangling subject > lines is not cautious. The fact is that, JSH aside, there are plenty > of opportunities in which one wants to start a thread with a subject > of the form, Blah: blah blah blah. > That the MS coders couldn't think of any isn't surprising. They > couldn't consider that (on rare occasions), lines might > begin... Ouch, the mime parsing error is a really horrible thing. I can't fathom that one. They must be working around some horrible kludge that requires it. I'll end this OT thread by saying that we do not disagree on the meaning of 'caution'. I was commenting on MS and I *did* say that they 'erred'! FWIW, I think user agents should leave the subject field alone. --Stan Gula Subject: Re: JSH: Non-uniqueness of factorization : J. Edgar Harris: >so you can factor *every* integer as 2j(j+1) or j(2j+1) with j an >integer in the ring of integers. > Huh???? Bizarre, isn't it? >The guy is SCUM. And he's fooled you repeatedly, and that's why I Again Harris is talking about himself. I wonder if he can keep it up much longer, before something really breaks. > Hint: When you call a person SCUM for making mathematical > comments you sound like a ing moron. He sounds like one for the same reason that ducks sound like ducks. Thanks D.U. for the tip about Free Agent. Someday I'll write a customizable packet driver for Windows, with anti-crank capabilities. Oughta be worth a fortune. LH Subject: Re: JSH: Non-uniqueness of factorization by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JKcdt19789; >That's why I hate posters like Dik Winter and Nora Baron because >they're so adept at hiding things. >Like notice how they used the example of >x(x+1)/2 >always being an integer in the ring of integers? >That's because the ring of integers is a unique factorization domain, >so you can factor *every* integer as 2j(j+1) or j(2j+1) with j an >integer in the ring of integers. just curious - how would you factor 1, 5, 6, 7, 8, or 9 in the form 2j(j + 1) or j(2j + 1) ??? and a few others. >Integers are a unique factorization domain!!! yup. >Now how many of you know that but listened to those posters? I dunno. It's hard to count us. >That's what's so frustrating!!! it is? >Dik Winter and Nora Baron are the two most capable, most able at >little tricks like that, where they hide their arguments in stuff that >you should catch, like that x(x+1)/2 is always true in integers >because integers have uniqueness of factorization. I thought uniqueness had something to do with primes. this sounds like sour grapes. >But algebraic integers do NOT! No ... but if A is an alg integer, is it true or not that A(A+1)/2 is also? I dunno. Prob not. who says one way or the other? >The ideas of Dik Winter and Nora Baron do NOT pass the smell-test. they smell OK to me. unliek some others. >Ultimately they're relying on confusion, and that you don't pay >careful attention so that they can sneak in little tricks like >x(x+1)/2 in integers. how is it a trick? isn't it true? >Without uniqueness of factorizaton Dik Winter can't give a reason why >for any particular w_1(x) and w_2(x), and notice he doesn't even >try!!! Dik gave explicit definitions. How is unique factorization involved? and anyway, what do you mean? a reason why what? >Why should he? why should he what? >As long as I'm supposedly just some nut, you are some nut. no doubt on that. and you're just willing to >attack my ideas without wondering about the underlying mathematics, as >long as a dumb device like x(x+1)/2 in the ring of integers get's past >you, why should he worry? where did he use this device? >Meanwhile he keeps up his own personal attacks with a webpage harping >on an old abandoned argument of mine from YEARS ago on x^4 + y^4 = >z^4. >The guy is SCUM. pond-scum? And he's fooled you repeatedly, and that's why I admittedly he is damn good at math and clearly you are not. what else is new??? Mr. Don't squeeze the charmin' Whipple >James Harris Subject: Re: Graph Theory Textbook Originator: dwildstr@euclid.ucsd.edu (Jake Wildstrom) The Prophet Jack Huizenga known to the wise as huizenga@uchicago.edu, opened the Book of Words, and read unto the people: > I'm an undergraduate, and this summer I will be participating in an >REU for discrete math and combinatorics. I am looking for a good >graph theory textbook to learn the basics from. I have seen the books >that Dover (which of course are very cheap) has to offer on the >subject, but unfortunately they seem too elementary. Is the Springer >GTM Graph Theory by Reinhard Diestel any good? Any suggestions are >highly welcomed. I learned from Diestel and found it quite readable and useful. The other books mentioned on this thread are by and large also excellent, but I thought I'd throw my support to Diestel's work. +-------------------------------------------------------------+ | D. Jacob Wildstrom -- Math monkey and freelance thinker | | Graduate Student, University of California at San Diego | | A mathematician is a device for turning coffee into | | theorems. -Alfred Renyi | +-------------------------------------------------------------+ The opinions expressed herein are not necessarily endorsed by the University of California or math department thereof. Subject: Re: Graph Theory Textbook > I'm an undergraduate, and this summer I will be participating in an > REU for discrete math and combinatorics. I am looking for a good > graph theory textbook to learn the basics from. I have seen the books > that Dover (which of course are very cheap) has to offer on the > subject, but unfortunately they seem too elementary. Is the Springer > GTM Graph Theory by Reinhard Diestel any good? Any suggestions are > highly welcomed. > Thanks, > Jack I like Bondy and Murty. Also Harary's book is a classic. You might also consider Graver and Watkins _Combinatorics with and Emphasis on the Theory of Graphs_. For applications Fred Roberts's _Discrete Mathematical Models_ is fun. Subject: Re: Graph Theory Textbook Is the Springer > GTM Graph Theory by Reinhard Diestel any good? Any suggestions are > highly welcomed. Yes definitely great; BTW you can find an online copy of the book (google will help you) so that you can make up your mind by yourself. -- Julien Santini Subject: Re: Graph Theory Textbook > I'm an undergraduate, and this summer I will be participating in an > REU for discrete math and combinatorics. I am looking for a good > graph theory textbook to learn the basics from. I have seen the books > that Dover (which of course are very cheap) has to offer on the > subject, but unfortunately they seem too elementary. Is the Springer > GTM Graph Theory by Reinhard Diestel any good? Any suggestions are > highly welcomed. > Thanks, > Jack [Obvious disclaimer: There are many good books, I've listed my personal favourites]. I like Graphs and Digraphs by Chartrand and Lesniak (3rd edition) very much. Other good introductory books, in addition to those Jim and Tim have already mentioned, are Graph Theory by Harary (an old classic but still amazingly relevant and accessible), Introductory Graph Theory by Chartrand and the more recent Graph Theory by Merris and Graph Theory by Douglas West. Also, I haven't yet read much of Diestel's book but what I have is excellent. HTH, Felix. Subject: Re: Graph Theory Textbook > I'm an undergraduate, and this summer I will be participating in an > REU for discrete math and combinatorics. I am looking for a good > graph theory textbook to learn the basics from. I have seen the books > that Dover (which of course are very cheap) has to offer on the > subject, but unfortunately they seem too elementary. Is the Springer > GTM Graph Theory by Reinhard Diestel any good? Any suggestions are > highly welcomed. There is a Dover book by Claude Berge on Graph Theory which is definitely not elementary, even for an experienced student in graph theory. Berge's Dover book: http://www.amazon.com/exec/obidos/tg/detail/-/0486419754/qid= 1077169399/sr=1 -1/ref=sr_1_1/104-7198216-1259942?v=glance&s=books An old book good for introductory - yet serious and real - graph theory is the textbook by Bondy and Murty, but this might be a little old and not too available, but I've heard they have created a new edition. Of course, there are many fields of graph theory... if you are approaching it from the computer science point, you might like Algorithmic Graph Theory... by Martin Golumbic. J Subject: What's the computation time of this Gradient optimization method, O(kN) or O(kN^2)? Dear All, I am trying to figure out the complexity level of the gradient method for minimizing an objective function. Assume J is the scalar objection function of N vectors, each vector is of k dimension. Now we are trying to find a k-dimension vector x such that J is minimized, e.g. J is the sum of squares of the distances from N points in R^k space to a line segment,which is connected by a known vector a and the unknown point x in R^k space. So given this definition of J, the way we did is to calculate dJ/dx = 0 and find the vector x. However, the expression dJ/dx is not usually in explicit form, i.e, x can not directly be obtained and have to be done by some iterative approach. Then for this kind of optimization problem, how to define the computational complexity? It is O(kN) or O(kN^2)? And how to verify this? Thanks a lot for your point. Fred Subject: Statistical independence test for continuous variables I need to know what are the available statistical independence tests for continuous variables. I know that the correlation is a test for linear dependence for continuous variables, but I was wondering if there were others. Thanks. Subject: bound on partitions of subsets Given a set S of n elements, i need to choose a subset S' of these elements and then choose a partitioning of S'. Can someone tell me a compact asymptotic upper bound on the number of ways this can be done? Anything that is better than (2^n)*(B_n). {B_n is the nth bell number}. Essentially a bound on the series sigma (nC_i B_i) If not, could someone direct me to a nice compact bound on the bell number B_n. Lovasz's result is too complicated for a humble computer scientist like me :-) Thanks. ~Deepak Subject: Re: Teaching philosophy Notwithstanding the draft I submitted in this thread, and others I might improvise, I am serious about the question of how to write this kind of statement. As I mentioned, I think I now have some idea of the genre. What I need to know now is how many yards of paper to use. Ignorantly, Allan Adler ara@zurich.ai.mit.edu ************************************************************** ************** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * ************************************************************** ************** Subject: Re: Teaching philosophy > Notwithstanding the draft I submitted in this thread, and others I > might improvise, I am serious about the question of how to write > this kind of statement. As I mentioned, I think I now have some idea > of the genre. What I need to know now is how many yards of paper to use. Having been asked to perform this silly exercise several times in past years, I've developed two documents: one, about two pages in length, that points out in broad strokes why I'm so wonderful in general, and one, about 6 pages in length that points out in particular why I'm so wonderful in introductory courses, in courses for a general audience, in courses for upper-level students, and in supervising research. This is a somewhat light-hearted description, but it's not too far from the truth. After a number of requests, one gets a certain facility at responding. The important thing to keep in mind is that within a very broad range anything you say will be acceptable. When we are trying to hire a new faculty member, our department asks for a statement on teaching, expecting (and usually receiving) the longer of the above-described versions. All we're really looking for is some facility in English and no evidence that the person will be an obvious mismatch, like If my students don't understand a concept after several attempts at explanation, I have no qualms about smacking them around. I am speaking English perfectly free. I expect students to concentrate on my course to the exclusion of everything else. Avoid obvous gaffes like these (two of which I've actually read), write with sincerity and (if you can do it) verve and you'll have nothing to worry about on this score. Rick Subject: Re: Teaching philosophy > Notwithstanding the draft I submitted in this thread, and others I > might improvise, I am serious about the question of how to write > this kind of statement. As I mentioned, I think I now have some idea > of the genre. What I need to know now is how many yards of paper to > use. Whatever else might be said, at least some people who will read the statement will be concerned with how _well-written_ it is. Such people come later in the loop, such as the dean and the people on the search who are outside the department. Nevertheless, such documents are mostly ways to _lose_ interviews, so if the question is how to write, then the answer is very well. Not just with exacting grammar, but also engagingly. In some sense, this is a sample of your teaching style in that it shows your mastery of English, your communications skills, and how well you read your audience. So in one sense, the topic is moot, as if they had just given you an essay question so you could show off your writing ability. Given that, how many yards of paper does it take to put your audience off? I'm guessing about 1.5 is the correct number of pages and that 3 pages would choke the entire committee. Bart Subject: Ping Jenny Wagner. You here? -- G.C. Subject: Infinite Galois Groups Let F be the field of lengths constuctible by compass and straighthedge. What is G(F/Q)? Let K be the field generated by the square roots of the prime numbers. What is G(K/Q)? What is G(F/K)? Subject: Re: Infinite Galois Groups Adjunct Assistant Professor at the University of Montana. >Let F be the field of lengths constuctible by compass and >straighthedge. What is G(F/Q)? Well, formally, it is the inverse limit of the Galois Groups of the finite Galois subextensions, with the connecting morphisms being the corresponding quotient maps. It should be fairly complicated, since it will include the subextensions given by the 2^n-th and 3^n-th roots of all integers. >Let K be the field generated by the square roots of the prime numbers. >What is G(K/Q)? What is G(F/K)? This is the increasing union of Q subset Q(sqrt(2)) subset Q(sqrt(2),sqrt(3)) subset Q(sqrt(2),sqrt(3),sqrt(5)) subset ... In each case, the relative Galois group is Z/2Z, and the Galois gropu of the extension Q(sqrt(2),...,sqrt(pn)) over Q where pn is the n-th prime is equal to (Z/2Z)^n (you would need to prove this; see for example Section 6.7 in Lang's Algebra, 3rd edition). The maps going down are ... -> Z_2 x Z_2 x Z_2 --> Z_2 x Z_2 --> Z_2 --> 1 where each map chops off the last coordinate. The inverse limit is an infinite product of copies of Z_2. -- ============================================================== ======== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ============================================================== ======== Arturo Magidin magidin@math.berkeley.edu Subject: Re: Infinite Galois Groups >Let F be the field of lengths constuctible by compass and >straighthedge. What is G(F/Q)? > Well, formally, it is the inverse limit of the Galois Groups of the > finite Galois subextensions, with the connecting morphisms being the > corresponding quotient maps. > It should be fairly complicated, since it will include the > subextensions given by the 2^n-th and 3^n-th roots of all integers. I think I am missing something here. Why the 3^n-th roots? It is not possible to construct a segment the length of the cube root of a given segment with compasses and straightedge. Darren >Let K be the field generated by the square roots of the prime numbers. >What is G(K/Q)? What is G(F/K)? > This is the increasing union of > Q subset Q(sqrt(2)) subset Q(sqrt(2),sqrt(3)) subset > Q(sqrt(2),sqrt(3),sqrt(5)) subset ... > In each case, the relative Galois group is Z/2Z, and the Galois gropu > of the extension > Q(sqrt(2),...,sqrt(pn)) over Q > where pn is the n-th prime is equal to (Z/2Z)^n (you would need to > prove this; see for example Section 6.7 in Lang's Algebra, 3rd > edition). The maps going down are > ... -> Z_2 x Z_2 x Z_2 --> Z_2 x Z_2 --> Z_2 --> 1 > where each map chops off the last coordinate. The inverse limit is an > infinite product of copies of Z_2. > -- > ============================================================== ======== It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > ============================================================== ======== > Arturo Magidin > magidin@math.berkeley.edu Subject: Re: e is transcendental by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KJtax05002; >> Since e^[iPi]=cosPi+isinPi >> or , e^[iPi]=-1+i[0] >> then there are two solutions here, to the given equatio: > >> A) e^[ipi]=-1 the real part solution and > >> B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio. >No, the conclusion from e^[iPi]=-1+i[0] is >Re(e^[iPi]) = Re(-1+i[0]) = -1 AND >Im(e^[iPi]) = Im(-1+i[0]) = 0 >-- >Daniel W. Johnson >panoptes@iquest.net >http:// members.iquest.net/~pano ptes/039 53 36 N / 086 11 55 W Daniel, Just saw Your message. I, thank You. This is it. Your help is great, and resolves my question completely. I, thank You all. Panagiotis Stefanides Subject: walking on a grate.. hello everybody, suppose I have a grate, N x M. how many paths are from node (0,0) to (N-1,M-1) eg. from one corner to the diagonal-opposite one? allowable paths are only with zero or positive change in node's coordinate. how can I compute this? I would appreciate any suggestion and help.. cheers, n. Subject: finite measure Let m be a finite measure on B(R), the borel sigma field generated by the real line. I have to prove that there are at most countably many x in R (atoms) such that m(x)>0. Obviously, I want to end the proof with something like SUM m(x) --> oo since a sum of uncountably many non-zero terms always diverges. But the problem is, I don't know exactly what sort of contradiction to set up. An uncountable collection of points in R always forms some sort of interval (a,b), right? Let I be the (uncountable) indexing set of these atoms. Then UNION (over I) x_i = (a,b) (say), so m( UNION (over I) x_i ) = m(a,b) < oo, so I'd like to show that the left-hand side = oo, but I can't do anything with an uncountable union since a measure only has countable additivity (for disjoint sets...points in this case). It seems like I need more tools than just the measure itself (since only countable operations are defined for it)...how can I proceed? Subject: Re: finite measure > An uncountable collection of points in R always forms some sort of > interval (a,b), right? Certainly not. Look at the Cantor set, or the irrationals. Subject: Re: finite measure > Let m be a finite measure on B(R), the borel sigma field generated by > the real line. I have to prove that there are at most countably many x > in R (atoms) such that m(x)>0. Isn't it the case that each atom x has an interior? If so there is a rational number in it. Since there are only a countable number of rationals there must be a countable number of atoms. Bob Kolker Subject: Re: finite measure > Isn't it the case that each atom x has an interior? No, consider the unit point mass at the origin. Subject: Re: finite measure > Let m be a finite measure on B(R), the borel sigma field generated by > the real line. I have to prove that there are at most countably many x > in R (atoms) such that m(x)>0. Use the classical argument: E = Union({x in X; m(x)>1/n}; n in N*), where each set of this union is measurable and contains finitely many elements (because m is totally finite). Subject: Re: there is no such thing as infinity Darryl L. Pierce,,, wibbled: >> And can be written in exponentially more obfuscated a manner than even C >> or assembler... :) > Assembler? Bah!! Raw machine language burnt into a ROM is the only way > to find the largest number. > Ah, I remember when I first started programming. All I had was a really hot > needle and had to write my code directly to the SIMMs. It was a major leap > forward when I upgraded and could just type copy con file.exe... When I started, SIMMs hadn't been invented. We had people who had seen so many punch cards and papertapes they could read the holes by eye. I set the bootstrap instructions using an array of toggle switches on one mainframe we had. Subject: Re: Partial-Sum -> Some Primes > Let a(1) = 1; > Let a(m) be the lowest yet unpicked positive integer such that: > sum{k=1 to m} k* a(k) is a prime. ... > a(k) : 1, 2, 4, 3, 6, 5, 12, 7,... > Question: > Is this a permutation of the positive integers? ... That is, choose a(m) so that m*a(m) + sum{k=1 to m-1} k*a(k) is prime and a(m) not equal a(k) if k= x/2 + y/2 (where >= is greater than or equal, in the event of ambiguity). (Is that all?) Euler's formula was no help with the original question, as it suggested an infinite number of possible tetrahedra, simply by adding k to the vertex count and k to the edge count, which comes by dividing an edge in two or more, which isn't a legal move. Thanks in advance, -- OWD [20 Feb '04] Subject: Re: Eulerian Polyheron Question > How many forms of n-hedra are there for arbitrary n? For example, > there is only one possible tetrahedron, i.e. various collineations of > the simplex. There are two possible pentahedra: the triangular prism > (and various distortions thereof) and the square pyramid. There are > at least (I'm not totally sure) four possible hexahedra: the pseudo-parallelepiped as it might be called, with quadrilateral > faces and triangular vertices; the pentagonal pyramid; the triangular > bi-pyramid; and a fourth solid formed by dividing a cube [0,1]^3 in > half by intersecting it with z >= x/2 + y/2 (where >= is greater than > or equal, in the event of ambiguity). (Is that all?) Euler's > formula was no help with the original question, as it suggested an > infinite number of possible tetrahedra, simply by adding k to the > vertex count and k to the edge count, which comes by dividing an edge > in two or more, which isn't a legal move. > Thanks in advance, > OWD [20 Feb '04] See Counting Polyhedra at http://home.att.net/~numericana/data/polyhedra.htm and http://home.att.net/~numericana/data/polycount.htm Hugo Pfoertner Subject: Re: walking on a grate.. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KLYl713563; >hello everybody, >suppose I have a grate, N x M. how many paths are from node (0,0) to >(N-1,M-1) eg. from one corner to the diagonal-opposite one? >allowable paths are only with zero or positive change in node's coordinate. >how can I compute this? I would appreciate any suggestion and help.. >cheers, >n. You should be able to calculate this. Take advantage of symmetry. Start 1 position away from the end and work backwards. Good luck phil Subject: Re: When is a finite field a cyclic field? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KLmoh14652; >>I don't intend to introduce new notion,but for the moment >>let's call a finite field F(+,*) a cyclic field if >>the abelian group (F,+) is cyclic.(We know that (F0,*) is always cyclic when F is finite).When is (F,+) also cyclic? >>May I ask Derek Holt who gave such a nice and clear solution >>to a recent problem for finite fields,to give answer to this? >>Thank you! >Yes you may ask! The answer is when |F| is prime. >the smallest positive integer p such that p.1 = 0 is prime, but then >it follows that p.a = 0 for all a in F, and so (F,+) cannot be cyclic >unless |F| = p. >Derek Holt. Thank you Derek,thank you Larry for your replies! They are very helpful! Best wishes, Cron Subject: Re: Pi gallons problem - any ideas? > [...] >>Perhaps a hint is needed. Here a well known mathematical formula, >>along with a lesser known one: > >>V1 = Ah/3 (r^3 - 1) / (r - 1), and > >>V2 = Ah/3 (r^(3/2) - 1) / (r - 1). > >>V1 is the volume of a frustum (not frustrum -- oops) with height = h, >>area of bottom circle = A, and ratio of radii of top circle to bottom >>circle = r. > >>V2 is the volume of that subset of the frustum below a plane tangent to >>the top and bottom circles at opposite points: i.e., the volume of water >>that would remain in a frustum if it were tipped over until the water >>just touched the bottom circle. > No takers? Okay, here's the answer: > [...] > I didn't really search, once you gave the tipped over hint... > I searched instead how to proove the > V2 = Ah/3 (r^(3/2) - 1) / (r - 1) > formula ... I didn't succeed.... > (V1 is easy, thanks) This is something that I derived myself once. I haven't seen it anywhere else, but if it's not well known, I'm sure it's at least known. I've cross-posted this to sci.math, hoping that it will be found interesting there. The issue here is proving the formula V = (pi a^2 h/3) (r^(3/2) - 1) / (r - 1), where V is the volume of the subset of a (right circular) frustum below a plane tangent to the top and bottom circles at opposite points, r = b/a is the ratio of the top to bottom radii, and h is the height. Here's a picture for those of you using monospace fonts: radius of upper circle = b o--------------------_o | o _/o | o _/.o | Side view o _/..o | <- height = h of frustum o A /...o | o _/....o | o _/.....o | o/______o | _/ o.....o <-- radius of lower circle = a _/ _ o...o _/ H_o.o / o An easy way to prove this formula is to compute the volume of the shaded region. The shaded region forms a cone (not a right circular cone, but an oblique, elliptical cone), so its volume is (1/3) base x height. Here, base is just the area of the ellipse, and height is the distance H from the apex of the cone to the cutting plane. The area of an ellipse can be written as (pi/4) A B, where A is the length of the major axis, and B, the length of the minor axis. Thus, the volume of the shaded region is V(shaded) = (pi/12) A B H But we can simplify this by noting that T = A H/2 is the area of the shaded *triangle* pictured above, so V(shaded) = (pi/6) T B. The area T is simply T = a (h + h'), where h' is the distance from the apex to the lower circle. By similarity, h':a = (h+h'):b, so h' = h a/(b-a), whence T = h ab/(b-a). The minor axis B has length 2 sqrt(ab). This can be seen by considering the midplane between the top and bottom circles. The midplane intersects the original frustum in a circle of radius (a+b)/2. The center of the ellipse lies in the midplane, at a distance (b-a)/2 from the center of this circle. Thus we have a right triangle with hypotenuse (a+b)/2, and two adjacent sides of length (b-a)/2 and B/2, which yields B = 2 sqrt(ab). Therefore V(shaded) = (pi/6) h ab/(b-a) 2 sqrt(ab) = (pi h/3) (ab)^(3/2) / (b-a). To find the desired area V, we now subract the volume of the little cone V(little cone) = (pi h'/3) a^2 = (pi h/3) a^3 / (b-a), to get V = (pi h/3) ((ab)^(3/2) - a^3) / (b-a), = (pi a^2 h/3) (r^(3/2) - 1) / (r - 1). QED. Note that this proof assumes b > a. The formula also holds for the a < b case (which can be seen by subtracting the above from the volume of the entire frustum). For a = b, there is a removable singularity which, when removed, gives the obvious answer pi a^2 h/2 (i.e., half the cylinder). -Jim Ferry Subject: Re: How many ways to put 5 balls into 500 ordered cups? > You are given 500 numbered cups and five identical balls. Any cup can > hold up to five balls. How many ways can you put the five balls into > the 500 cups? > As a warm-up, let's try smaller sets of ordered cups: > Cups Arrangements > ==================== > 1 1 > 2 6 > 3 21 > 4 56 > 5 126 > 6 252 > 7 462 > 8 792 > 9 1287 > 10 2002 For what its worth! I can't explain it. Arrangements = Binomial Cofficient(nCm), where n = number of cups +4; m = 5 Example; for C=5, n = C+4 = 9; 9C5 = 126 A = 504C5 = 265,661,562,600 for 500 cups. regards B. Subject: Re: How many ways to put 5 balls into 500 ordered cups? Originator: jeyadev@kaveri >You are given 500 numbered cups and five identical balls. Any cup can >hold up to five balls. How many ways can you put the five balls into >the 500 cups? >As a warm-up, let's try smaller sets of ordered cups: > .... >I have a brute force solution using an SQL query from hell (my >specialty), which is not quite the same thing as a formula and proof. > I cannot remember enough Combinatorics and Finite Differences to make >the next step. ARRRGH! >It is bothering me enough that I will offer a copy of my next book >(TREES & HIERARCHIES IN SQL) as a prize for the best solution. Have you considered the following: How many terms are there in (x1 + x2 + ...... + xn)^m ?? -- Surendar Jeyadev jeyadev@wrc.xerox.bounceback.com Remove 'bounceback' for email address Subject: Re: How many ways to put 5 balls into 500 ordered cups? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1JN2H932606; Yes, I took that into account in my step (2). Unfortunatley, my method is ineffiecient. I now remember deriving the efficient way to this problem a few years ago, but I had forgotten about that, and gave an answer in haste. Have a good day. >>> SUMMARY: > (1) Find the ways to partition the number 5: > 5 > 4 1 > 3 2 > 3 1 1 > 2 2 1 > 2 1 1 1 > 1 1 1 1 1 << >The gimmick is that the cups are ordered, so (2,3) is not the same as >(3,2) fo the two cup case and so forth upt to 500. Subject: Help needed from a group theorist by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1KMShs18156; An exponent E(G) of a group G is the lcm of all the orders o(g), g in G.We have that G is cyclic if and only if E(G)=|G|. Does anybody have an idea how can we find all n such that the exponent of Z_n*,the group of units of Z/Zn is 2? E(Z_n*)=2 n=? The solution is n=2,3,4,6,8,12,24 ,but I don't have a clue where did this come from. Subject: Re: 'erf' function in C > A 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. -- Dr.B.Voh ------------------------------------------------------ Applied Algorithms http://sdynamix.com Subject: Ideas for course on great ideas in (theoretical) CS? I have been coerced into teaching a Honors course the Fall (mostly for non-CS freshman/sophomore Honors students). My idea was to do some Great Ideas/Problems/Puzzles etc. from computer science -- emphasis on theory/algorithms and related areas like graph theory/combinatorics. Of course, the honors college wants a syllabus in one week! I looked at the book, Great Ideas in CS and though a nice book, seems a bit light on the theory side ... given that I want to focus on theory to keep me interested. I saw the course/web site at CMU Great Ideas in Theoretical Computer Science and may use that as a guide for some of the course. Examples of some things I might discuss (besides a couple weeks on basics/definitions/history) include Towers of Hanoi, Byzantine Generals, voting problems, maybe a gentle discussion of interactive proofs, prisoner's dilemma, game of life, primality testing, graph coloring ... anything that can be discussed in a day or so to folks with no CS background, yet which has some theory component to it ... stuff that is surprising or counter-intuitive is all the better ;) Anyway, if anyone has any suggestions for material/topics that I might cover, I would most appreciate it. Any pointers would be accessible to students would be great (I have a couple) would be great. Thanks in advance. If I get enough interest, I will post a syllabus (or a link to one) to comp.theory in a week or so. Chip Klostermeyer Subject: Math notation question: '(' I just ran across a math question in a review and I'm totally stumped by the answer. Then I realized that I may have been misreading the notation. The statement was: There exist x, y, and z that are nonzero. Such that 1 ( y>x and xy=z. Does 1 ( y>x mean that both y & x are less than 1? Looking at the answer that is the only thing that makes sense to me. I just don't understand the notation. Thanks for help in advance. Ben.. Subject: integral Hey, can anyone help me in integrating from 0 to infinity e^(-(x^2))dx? Thanks. -Greg R. Subject: Re: integral > ...integrating from 0 to infinity e^(-(x^2))dx? Let I be the number we want. I^2 = (integral from 0 to infinity e^(-(x^2))dx) times (integral from 0 to infinity e^(-(y^2))dy) = double integral (0 to infinity in x and y)[e^-(x^2+y^2)]dy*dx (change to polar coordinates) = double integral(0 <= theta <= pi/2)(0 <= r < infinity) of the function: e^(-r^2) times r*dr*d(theta) infinity = (-1/2)e^(-r^2)| times pi/2 0 = pi/4. Since I^2 = pi/4, I = (1/2)sqrt(pi). Subject: Re: integral > Hey, > can anyone help me in integrating from 0 to infinity e^(-(x^2))dx? > Thanks. > -Greg R. I'd suggest trying a power series. David Moran