mm-23 In a fifty-year-old math paper about integrals of hermite polynomialsthe author uses this strange symbol (-m)_{k}, i.e. lower indexnotation, for what appears to be something like binomial coefficients(m, k not necessarily integers).Is that correct? Or what else does it mean? It's probably NOT binomial coefficients if my own calculations arecorrect. But it's definition for integers appears to be some quotientof factorials.Heiko Gimperlein > In a fifty-year-old math paper about integrals of hermite polynomials> the author uses this strange symbol (-m)_{k}, i.e. lower index> notation, for what appears to be something like binomial coefficients> (m, k not necessarily integers).Is that correct? Or what else does it mean?> It's probably NOT binomial coefficients if my own calculations are> correct. But it's definition for integers appears to be some quotient> of factorials.I would guess that (x)_k denotesx(x-1)(x-2)...(x-k+1).Does that make sense in this paper?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen > In a fifty-year-old math paper about integrals of hermite polynomials> the author uses this strange symbol (-m)_{k}, i.e. lower index> notation, for what appears to be something like binomial coefficients> (m, k not necessarily integers).It's probably NOT binomial coefficients if my own calculations are> correct. But it's definition for integers appears to be some quotient> of factorials.Might be a rising or falling factorial power. For integer exponent, mto the rising power k is the product of the k factors m, m + 1, m + 2,..., m + k - 1, and m to the falling power k is the product of the kfactors m, m - 1, ..., m - k + 1. For k zero, the empty product is 1as usual; for k negative, define so as to keep certain simple laws ofexponents. Factorial of the natural number m is either 1 to m risingor m to m falling.I learnt this from Graham, Knuth, Patashnik: Concrete Mathematics.-- Jussi > 1. what is the magnitude of the set of all groups?There is no set large enough to contain all groups.> 2. what is the magnitude of the set of all sets?There is no set large enough to contain all sets.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. If I did not make a mistake calculating the first few terms by hand,here is a recursively-defined sequence which is not in the EIS yet.a[0] = 1;and for m >= 1, a[m] = sum{j=0 to a[m-1](mod m)} a[j]Ascii-art: a[m-1](mod m) --- a[m] = / a[k] --- k=0And, 0 <= a[m-1](mod m) <= m-1.The sequence begins (maybe):1, 1, 2, 4, 1, 2, 4, 9, 2, 4, 9, 30, 15,...What can be said about this sequence? Does it have a closed-form (ienonrecursive) representation?Also, other sequences can be based on the same idea: a partial sumsomehow involving earlier terms, BOTH in the general term and in thelimit of the indexes used in the sum.Leroy Quet presentation ofmultiplication of complex numbers as a free convention : (a,b)*(c,d) =(ac-bd, ad+bc) is insufficient because also (a,b)*(c,d) = (ac-2bd,ad+bc) will conform to the laws of algebra. As you showed itsnecessary to introduce the concept of metrics.L.Rodriguez ESB Consultancy is pleased to announce the latest release of ESBCalc Pro - a full featured, scientific calculator for Win32 Platforms.http://www.esbconsult.com/esbcalc/ esbcalcpro.htmlESBCalc Pro is an Enhanced Scientific Calculator for Win32 Platforms with Infix Notation, Full Exponential Notation support, Brackets, Scientific Functions (Trigonometric, Hyperbolic, Logarithmic - including Base 10, Base 2 & Natural - plus more), Memory, Paper Trail, Result Help. Now also includes Floating Decimal Point, optional start in last position and many other enhancements to the User Interface and Calculator Engine.Grab a trial version today :)-- ESB Consultancy, http://www.esbconsult.comHome of ESBPCS, ESBStats, ESBPDF Analysis & ESBCalcKalgoorlie-Boulder, Western Australia(TeamND, TeamOE, Addict Support, eLists.org Management) Every morning here in Singapore I catch a bus, which takes me to atrain, which takes me to work. The buses are not scheduled, andarrive at essentially random times, every 15 minutes or so.Often I will just miss the bus, and end up having to wait fifteenminutes for the next one. This is rather annoying when it happens.My urge is to hurry out to the bus stop. I figure, this strategywill never get me to work any later, and might possibly get me there15 minutes sooner, if I just catch a bus I would have otherwisemissed. (Possibly even more, if I end up catching the subsequent train, which also comes at random times.)However, looking at the problem mathematically, I can only concludethat my hurrying is wasted. Getting to the bus stop an average ofone minute sooner will get me to work, on the average, one minutesooner -- any way you look at it.While not a stunning tour de force of mathematics, this result isinteresting to me in that it runs counter to my instinct, namely, to run out to the stop.Are there any other angles to this problem to consider?Come to think of it, I just remembered something that's possiblyrelated. Once I took a defensive driving course, to get the pointsfor a speeding ticket taken off my license. The instructor, tryingto make a point about speeding, claimed that going five miles anhour faster was pretty pointless, in that the slower cars willcatch up to you at the next stop light. Mathematically speaking,is there any truth to this? Does the introduction of random stops along a route have an equalizing tendency between fast and slowdrivers, or does it (as is my suspicion) really penalize the slowdrivers, spreading out the pack so to speak?--Mark >Every morning here in Singapore I catch a bus, which takes me to a>train, which takes me to work. The buses are not scheduled, and>arrive at essentially random times, every 15 minutes or so.>Often I will just miss the bus, and end up having to wait fifteen>minutes for the next one. This is rather annoying when it happens.>My urge is to hurry out to the bus stop. I figure, this strategy>will never get me to work any later, and might possibly get me there>15 minutes sooner, if I just catch a bus I would have otherwise>missed. (Possibly even more, if I end up catching the subsequent >train, which also comes at random times.)>However, looking at the problem mathematically, I can only conclude>that my hurrying is wasted. Getting to the bus stop an average of>one minute sooner will get me to work, on the average, one minute>sooner -- any way you look at it.>While not a stunning tour de force of mathematics, this result is>interesting to me in that it runs counter to my instinct, namely, >to run out to the stop.15 minutes difference with probability 1/15 _might_ be of much moresignificance to you than 1 minute difference (e.g. if the 15 minutescould get you fired for being late). It's called having a nonlinear utility function.>Come to think of it, I just remembered something that's possibly>related. Once I took a defensive driving course, to get the points>for a speeding ticket taken off my license. The instructor, trying>to make a point about speeding, claimed that going five miles an>hour faster was pretty pointless, in that the slower cars will>catch up to you at the next stop light. Mathematically speaking,>is there any truth to this? Does the introduction of random stops >along a route have an equalizing tendency between fast and slow>drivers, or does it (as is my suspicion) really penalize the slow>drivers, spreading out the pack so to speak?Neither. It increases the variance of your trip time.On the other hand, it dilutes the relative benefit of speeding, i.e.if you increase your speed by a factor 1+f, although you multiply thetime spent moving by 1/(1+f), you don't change the average time spentwaiting for lights to change(*), so the percentage change in overall trip time is not as much.(*) Well, that's not _quite_ true, because at higher speed you're a bit more likely to run a yellow light.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > Every morning here in Singapore I catch a bus, which takes me to a> train, which takes me to work. The buses are not scheduled, and> arrive at essentially random times, every 15 minutes or so. Often I will just miss the bus, and end up having to wait fifteen> minutes for the next one. This is rather annoying when it happens. My urge is to hurry out to the bus stop. I figure, this strategy> will never get me to work any later, and might possibly get me there> 15 minutes sooner, if I just catch a bus I would have otherwise> missed. (Possibly even more, if I end up catching the subsequent> train, which also comes at random times.) However, looking at the problem mathematically, I can only conclude> that my hurrying is wasted. Getting to the bus stop an average of> one minute sooner will get me to work, on the average, one minute> sooner -- any way you look at it. While not a stunning tour de force of mathematics, this result is> interesting to me in that it runs counter to my instinct, namely,> to run out to the stop. Are there any other angles to this problem to consider? Come to think of it, I just remembered something that's possibly> related. Once I took a defensive driving course, to get the points> for a speeding ticket taken off my license. The instructor, trying> to make a point about speeding, claimed that going five miles an> hour faster was pretty pointless, in that the slower cars will> catch up to you at the next stop light. Mathematically speaking,> is there any truth to this? Does the introduction of random stops> along a route have an equalizing tendency between fast and slow> drivers, or does it (as is my suspicion) really penalize the slow> drivers, spreading out the pack so to speak?One way of decreasing your journey time is to make your brakes much morepowerful.You hurtle towards red stop lights at top speed, hoping the light willchange. If it does, you carry on at top speed without slowing down. If itstays red you slam on your brakes a couple of meters from the stop light andcome to rest without going past the red light.-- Clive Toothhttp://www.clivetooth.dk > Every morning here in Singapore I catch a bus, which takes me to a> train, which takes me to work. The buses are not scheduled, and> arrive at essentially random times, every 15 minutes or so.What's the train's schedule then? Do you mean a subway that comesevery so often, or do you mean a commuter that goes in fifteen minuteor longer intervals? Namely, that is my case. First, I take a bus(actually, a subway, but it is easier to talk about buses here), whichgoes every 6-10 minutes, and then I take a commuter train that goesevery 15-20 minutes. If I miss a particular bus, I may or may not missmy train. It is the train schedule that decides, not the bus.> Often I will just miss the bus, and end up having to wait fifteen> minutes for the next one. This is rather annoying when it happens.What is really annoying is to miss the bus by a second or two, andthen miss the train only by a second or two. However, the total lossis not the sum of the both waiting times, it is only the longer of thetwo. If I had catched the bus, I would have catched the train, and itonly means the fifteen or twenty minutes that is between the trains;the ten minutes wait for the bus doesn't count. I mean, should not becounted.> My urge is to hurry out to the bus stop. I figure, this strategy> will never get me to work any later, and might possibly get me there> 15 minutes sooner, if I just catch a bus I would have otherwise> missed. (Possibly even more, if I end up catching the subsequent > train, which also comes at random times.)Apparently your train is much more frequent than the bus, then. >One way of decreasing your journey time is to make your brakes much more>powerful.>You hurtle towards red stop lights at top speed, hoping the light will>change. If it does, you carry on at top speed without slowing down. If it>stays red you slam on your brakes a couple of meters from the stop light and>come to rest without going past the red light.Especially effective if your destination is a hospital...Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > lot of talk no action> haven't seen any replies with the answer next to the phone number yet!> its not that difficult, this whole town of 100,000 people are all in on it. No one responded to your first post because you are a COMPLETE ING> LOON, and it made no sense. > the others have clarified it,> its going up your optic nerve but the signals not registering.> 1000s of 1000s of witnesses to I'm the truman, all have known> for ___over_1_year___, i'm getting impatient it hasn't gone public yet> so what can I do? make sense yet? you got anything to lose> by replying 'yes i checked your witnesses they didn't know any truman',> you got anything lose 'yes these 5 all said its real, you R the truman'.HercHerc,What you say if the following townsville numbers didn't have a cluewhat on earth you were talking about:4750 0700 4031 1107 4753 44444779 0233 4760 1380 4781 4182 4781 4301 > What you say if the following townsville numbers didn't have a clue what on earth you weretalking about: Let's take a look at these numbers (just out of curiosity):> 4750 0700> Great Barrier Reef Marine Park Authority - ermits Officer - Research 4031 1107> NuShape Foods Outlet 4753 4444> Australian Institute of Marine Science 4779 0233> Townsville Vet Clinic 4760 1380> Alamo Rent-A-Car - Townsville Airport 4781 4182> James Cook University - School of Psychology 4781 4301> James Cook University - Distance Education Interesting what you can come up with using Yahoo! for just a couple of> minutes, eh? :) Have a nice day.interesting what reading a post will reveal, 0700, 1107, 4444?so don't phone businesses with non business enquiries.I guarantee several of these will say Yes The Truman lives in Townsvilleand Yes we can hear him. I can post up more random numbers fromthe phone book or scan a few pages and upload that if its not enoughwitnesses. Try is anything weird happening in Townsville the last year?if you want to guarantee they're not being sarchastic. 47482160> 47788360> 47792822> 47290642> 47254486> 47855847> 47491445> 47230018> 47785779> 47861843> 47772731> 47720161> 47714484> 47211929> 47753611> 47210420> 47235886> 47790345> 47831124The comments you will get will be exactly like the posts to aus.tv,the only people admitting they were from townsville ALL shared thesame sarchasm approach, direct confirmation.Herc >I pinpoint my affection for someone I state looks like Hollywood>Trumans next romance costar! Any chance you could use your super-powers to write a sentence that> actually parses in English?>>I noticed that one, how about :Prefix with 61 7, reply here next to the number if anyone confirms or>>denies it. 47482160>> 47788360>> 47792822Come on Alan, you have NO evidence I'm not the Truman, back up your>>claim by proving my witnesses are bogus. 50 cents.Someone in Australia do it for 20 cents. 100,000 people torment me for>>over 1 year every day I go out, it will take another 5 years to go>>public unless someone gives me 1 minute of benefit of the doubt. >>Laurie Holden is Hollywoods model for Eve, the perfect 10, she was 21>>when I last saw her 5 years ago, 5 years of the Truman company>>tormenting me, the last 2 its been in public. They will honestly>>squeeze every bit of value out of me they can, one of intel just said>>then yeah no joking.If this was a video converence link you'd hear the Truman company>>constantly in the background showing off their laser to sound satelite>>spy systems.Its actually on topic for all the groups, a variant of my myriad of>>proposals to each for benefit of the doubt that a man *can* be unique.Ahem, with all due respect sir, just what the are you talking> about?He believes he lives the Truman Show.Saaaaaaaaay WHAT?!?Yes.>>No, really.Really, really?Very really. Fruity as a nutcake.>>This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........Indeed - That movie was terrible ;)I know. Crazy people these days. No taste...-- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion____________________________________________________ ____________You are not a fiscal conservative when the deficit runs to $400 billion on your watch. I don't believe that saying that ATHEISTS are gutless, is accurate. They> are like white washed tombs, clean on the outside and with dead man's> bones on the inside. :-D.Sparky, given your history of posting and demonstrated lack of braincells, being just about anything is still better than being you...-- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion____________________________________________________ ____________You are not a fiscal conservative when the deficit runs to $400 billion on your watch. >>I pinpoint my affection for someone I state looks like Hollywood>Trumans next romance costar! Any chance you could use your super-powers to write a sentence that> actually parses in English?>>I noticed that one, how about :Prefix with 61 7, reply here next to the number if anyone confirms or>>denies it. 47482160>> 47788360>> 47792822Come on Alan, you have NO evidence I'm not the Truman, back up your>>claim by proving my witnesses are bogus. 50 cents.Someone in Australia do it for 20 cents. 100,000 people torment me for>>over 1 year every day I go out, it will take another 5 years to go>>public unless someone gives me 1 minute of benefit of the doubt.>>Laurie Holden is Hollywoods model for Eve, the perfect 10, she was 21>>when I last saw her 5 years ago, 5 years of the Truman company>>tormenting me, the last 2 its been in public. They will honestly>>squeeze every bit of value out of me they can, one of intel just said>>then yeah no joking.If this was a video converence link you'd hear the Truman company>>constantly in the background showing off their laser to sound satelite>>spy systems.Its actually on topic for all the groups, a variant of my myriad of>>proposals to each for benefit of the doubt that a man *can* be unique. Ahem, with all due respect sir, just what the are you talking> about?He believes he lives the Truman Show. Saaaaaaaaay WHAT?!? Yes.>No, really. Really, really? Very really. Fruity as a nutcake.>This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........ Indeed - That movie was terrible ;) I know. Crazy people these days. No taste...That movie just went into my emotional side and how they control me,half of all released movies base their themes, plots, or scripts on the collectivespying on my entire life. Mel Gibson plays one part of me, Robert Denirotells me YOU HAVE A GIFT, you'd be amazed when you see how1000s of plots all broadcast to you seperately Show is..............can you guess?The Pretender >Re: ATHEISTS are gutless No, you are NOT on topic for alt.atheism. You call we atheists> gutless for not doing your job for you and, on top of it, you aren't> even an atheist! Please see a psychiatrist. You need medical help.when 10,000 posts per year from this group say noone has presentedempirical evidence for the existence of god then I can assume thisis on topic since there are currently over 100,000 witnesses, 20 randomphone numbers of those are presented in this post.Herc aka the star You are the one who believes in the 'witnesses'. It's up to you to> prove that they are real. It's not up to us to spend money to prove> that you are right . That's your responsibility.>this is public communications forum, its not presenting empirical datayou wanted all along, its some kind of holy water pouring out of your monitor.proving real means the public forum moves off a purely information mediumto examine the claim. there is no sequence of words i can write on a forumto make an atheist open his mind.my responsibility to what exactly! Im god, 100,000 people know it, now what?if you want to keep spouting there is no known evidence for god then rejoiceall your life, atleast until the tvs tell you what fools you are.if you want to keep claiming noone has presented evidence then you are liars.Herc > you have no right to speakAh. Obviously you are a True Christian.socode > this is public communications forum, its not presenting empirical > data you wanted all along, its some kind of holy water pouring > out of your monitor.Man cannot live on LSD alone.socode They are like white washed tombs, clean on>>the outside and with dead man's bones on the inside. :-D. These and other fine sayings can be found in my latest book 1001>> dumbass glib preacherisms for your church's marquee. Look for it in>> your local Christian bookstore.you have no right to speak, your TJ Hooker links are down,>hang your head in shame....Well crap. tj-hooker.com is no more. Poo.---john@thecodezone.com http://www.shatnerology.com year from this group say noone has presented> empirical evidence for the existence of god then I can assume this> is on topic since there are currently over 100,000 witnesses, 20 random> phone numbers of those are presented in this post. Herc aka the starThe second greatest error in reasoning is mistaking evidence for proof. Thegreatest error is mistaking testimony for evidence.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com says...>I pinpoint my affection for someone I state looks like Hollywood>Trumans next romance costar! Any chance you could use your super-powers to write a sentence that> actually parses in English?>>I noticed that one, how about :Prefix with 61 7, reply here next to the number if anyone confirms or>>denies it. 47482160>> 47788360>> 47792822Come on Alan, you have NO evidence I'm not the Truman, back up your>>claim by proving my witnesses are bogus. 50 cents.Someone in Australia do it for 20 cents. 100,000 people torment me>>for over 1 year every day I go out, it will take another 5 years to>>go public unless someone gives me 1 minute of benefit of the doubt.>>Laurie Holden is Hollywoods model for Eve, the perfect 10, she was 21>>when I last saw her 5 years ago, 5 years of the Truman company>>tormenting me, the last 2 its been in public. They will honestly>>squeeze every bit of value out of me they can, one of intel just said>>then yeah no joking.If this was a video converence link you'd hear the Truman company>>constantly in the background showing off their laser to sound>>satelite spy systems.Its actually on topic for all the groups, a variant of my myriad of>>proposals to each for benefit of the doubt that a man *can* be>>unique.Ahem, with all due respect sir, just what the #### are you talking> about?He believes he lives the Truman Show.Saaaaaaaaay WHAT?!?Yes.No, really.Really, really?Very really. Fruity as a nutcake.This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........Indeed - That movie was terrible ;)I know. Crazy people these days. No taste...Maybe the fruitcake works for the phone company? ;-)Dunno. But we sure need a better class of trolls and kooks. We're scrapingthe bottom of the barrel lately...-- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion____________________________________________________ ____________You are not a fiscal conservative when the deficit runs to $400 billion on your watch. Let's list your various disorders... ok, superiority complex,schizophrenia, acute paranoia and random dribbling. Atleast, that's what the cameras show... whoops, I've said toomuch!In case you can't tell, I was being sarcastic!Sarcastic!shape! It spelled this...> TOP POST: Interesting, the groups this nut chose to post in...> I guess he thinks that's where all the smart people are, and the postapart from alt.business I have 100s of ontopic posts in each group , some> over years, its more the popularity and diversity of the groups to demonstrate> the subject heading.seems to want to prove something to these smart people; something smart> people think is dumb.prove is what I have in mind, 100,000 witness proof at your fingertips| prefix with 61 7 reply here next to the number if anyone confirms or> denies it> |> | > 47482160> | > 47788360> | > 47792822> | > 47290642> | > 47254486> | > 47855847> | > 47491445> | > 47230018> | > 47785779> | > 47861843> | > 47772731> | > 47720161> | > 47714484> | > 47211929> | > 47753611> | > 47210420> | > 47235886> | > 47790345> | > 47831124all you need is a 50c international phone call and 6 words> DOES THE TRUMAN LIVE IN TOWNSVILLE?any results positive and negative post up here, cmon people dont> let the truman suffer years longer than he has to, I've served for your> media long enough, and Adam and Eve are both getting old.Herc aka the star >20 random phone numbers from the 100,000 people who all know>Adam of the bible, The Truman lives in Townsville Australia for over 1 year.prefix with 61 7 reply here next to the number if anyone confirms or denies itAm I alone in not having a clue what you are talking about?1) Who/what the is The Truman?2) Adam (of the bible) is a fictitious character in a book.3) Townville, Australia is significant because...?In what way will a phone call to Australia prove/disprove whatever itis you're trying to prove/disprove?Hey, I'm Bilbo Baggins. Call (almost) anyone in the UK and they willtell you that they've heard of Bilbo Baggins. That proves I'm BilboBaggins, formerly of Bag End.... right?Red Celtaa#883--Boy, I've never seen an issue so divisive. It's like a civil war,isn't it? Even amongst my friends, who are all very intelligentpeople, they are totally divided on abortion. Some of my friends, forinstance, think these pro-life people are annoying idiots. Others ofmy friends think these pro-life people are evil s.How are we going to come to a consensus?You want to hear the arguments around my house. They're annoying!They're idiots!They're evil! They're s!Brothers, sisters come together! Can't we once just join hands andthink of them as evil annoying idiot s? >>I pinpoint my affection for someone I state looks like Hollywood>Trumans next romance costar! Any chance you could use your super-powers to write a sentence that> actually parses in English?>>I noticed that one, how about :Prefix with 61 7, reply here next to the number if anyone confirms or>>denies it. 47482160>> 47788360>> 47792822Come on Alan, you have NO evidence I'm not the Truman, back up your>>claim by proving my witnesses are bogus. 50 cents.Someone in Australia do it for 20 cents. 100,000 people torment me for>>over 1 year every day I go out, it will take another 5 years to go>>public unless someone gives me 1 minute of benefit of the doubt.>>Laurie Holden is Hollywoods model for Eve, the perfect 10, she was 21>>when I last saw her 5 years ago, 5 years of the Truman company>>tormenting me, the last 2 its been in public. They will honestly>>squeeze every bit of value out of me they can, one of intel just said>>then yeah no joking.If this was a video converence link you'd hear the Truman company>>constantly in the background showing off their laser to sound satelite>>spy systems.Its actually on topic for all the groups, a variant of my myriad of>>proposals to each for benefit of the doubt that a man *can* be unique. Ahem, with all due respect sir, just what the are you talking> about?He believes he lives the Truman Show. Saaaaaaaaay WHAT?!? Yes.>No, really. Really, really? Very really. Fruity as a nutcake.>This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........ Indeed - That movie was terrible ;) I know. Crazy people these days. No taste...That movie just went into my emotional side and how they control me,> half of all released movies base their themes, plots, or scripts on the collective> spying on my entire life. Mel Gibson plays one part of me, Robert Deniro> tells me YOU HAVE A GIFT, you'd be amazed when you see how> 1000s of plots all broadcast to you the Truman Show is..............can you guess?> Jackass?-Alanaa#1608 BAAWA > Let c be the well-ordering at hand and let x be the c-least element of> the set of undefinable[1] real numbers. This is your definition of> x, right?But this definition has a parameter in it, namely c. This parameter> is undefinable in the same sense that x was, I think. So, it> doesn't seem like you've really defined x, since you can't write down> what you mean by c.> Right, this also occured to me after some hard thinking. So we haven'tfound a well-ordering of the real numbers so far simply because it isimpossible to construct one!This can be generalized to the following result: let X be anuncountable set, P(X) its power set. The axiom of choice postulatesthe existence of a mapping f:P(X)->X such that for every M subset Xit is f(M)in M. But by the same argument we can prove that we willnever find any such f, no matter what X is. This does not contradictits existence of course, we assume that all these undefinable realnumbers <3F26B9A2.1A2063D6@btinternet.com> <87u192fkfw.fsf@phiwumbda.localnet> >> Let c be the well-ordering at hand and let x be the c-least element of>> the set of undefinable[1] real numbers. This is your definition of>> x, right?>>> But this definition has a parameter in it, namely c. This parameter>> is undefinable in the same sense that x was, I think. So, it>> doesn't seem like you've really defined x, since you can't write down>> what you mean by c.>Right, this also occured to me after some hard thinking. So we haven't> found a well-ordering of the real numbers so far simply because it is> impossible to construct one!> This can be generalized to the following result: let X be an> uncountable set, P(X) its power set. The axiom of choice postulates> the existence of a mapping f:P(X)->X such that for every M subset X> it is f(M)in M. For M non-empty, of course.> But by the same argument we can prove that we will never find any> such f, no matter what X is. I suppose it depends on what one means by find, but this doesn'tseem right.Let X be an uncountable ordinal (so X is well-ordered by assumption).Define f : P(X){0} -> X to be the function taking a non-empty subsetM of X to its least element. It appears to me that we have found anf satisfying your condition.The problem is not in describing the function f. It lies indescribing all of the elements of P(X).-- Jesse HughesTime and again, history has shown that people who think their beliefstrump reality lose, and lose badly. Luckily, I don't have to listento you. -- James Harris on reality avoidance > Herc is mad as a hatter. Now, he's trying to end atheism. That> isn't going to gain him any kind of sympathy in alt.atheism.>when your newsreader starts calling it alt.apatheism then you'll know I'm onlineHerc Can anyone provide me with the citation of Cantor's paper where he > Can anyone provide me with the citation of Cantor's paper where heGeorg Cantor, Ueber eine Eigenschaft des Inbegriffs aller reellenalgebraischen Zahlen, J. f. Math. 77 (1874), 258-262. > So we look for a solution near t = 32/11. Solving numerically,>> the maximum is at approximately t = 2.9096004696658838724, where>> f = 2.5980706142146652690. The time in hr:min:sec is>> approximately 2:54:34.5616907971819 approximately.>but you already knew that! from rec.puzzles 2 months ago! Glad to see somebody's paying attention. Although this is similar to the rec.puzzles question> optimized is different, and the answer is therefore slightly different:> the answer in that thread was approximately 2:54:34.520548.>approximately 2:54:34.520548approximately 2:54:34.5616907971819Maximum seperation angle can be considered a heuristic for maximum area then.Herc >approximately 2:54:34.520548>approximately 2:54:34.5616907971819.04 seconds doesn't sound like much, but it's been the margin of victory for Olympic medals. (I don't know why that seemed relevant)>Maximum seperation angle can be considered a heuristic for maximum area then.Yes.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 approximately 2:54:34.520548>approximately 2:54:34.5616907971819> .04 seconds doesn't sound like much, but it's been the margin of> victory for Olympic medals. (I don't know why that seemed relevant)Maximum seperation angle can be considered a heuristic for maximum area then. Yes.That's the number one reason my cousin said he didn't take up cyclingseriously after testing the highest VO2 level in Australia Institute of Sports,and at most events he gets urged by most everyone to do so....I'm not training full time to get a silver medal!Herc I present the following integer sequence as something original, as faras I know, although if it's already old hat, I should appreciate beingtold so. Like the unfortunately not-quite-perfect Perrin sequence, q.v.,this one is hoped to constitute a primality assay by means of thedivisibility or not of any sequence term by its index or argument.The twin evils potentially afflicting any would-be prime-testingalgorithm are, naturally, false negatives and false positives. Now, thesequence at hand interestingly appears to work for both positive andnegative indices to a degree. To first appearances (but this is only aguess) the problem here is going to be false positives, not falsenegatives, exactly as with Perrin. However, provided that two factscould be proven, but only in that case, then one could add or subtractthe terms associated with equal positive and negative indices withassurance of thus obtaining a new sequence that would precisely mirrorthe distribution of the primes. Those two conditions obviously are: (1) no false negatives in either the positive or negative domain of thesequence, and (2) the false positives of the two halves of the domainnever occur at equal distances from zero. But I admit at once thatcalling for proof of propositions of this sort in the realm of empiricalnumber studies, even if they should chance to be true, smacks of thefabled prescription for catching birds just salt their tails.Without further ado, the recursion rule is T(n) + T(n+3) = T(n+5), thefirst few terms, starting with the zeroth, being 5, 0, 2, 0, 2. Whereasthe heuristic inspiration for Perrin is 2+3=5, the present sequence nodstoward the first typical twin primes, i.e., those flanking a multiple of6, namely 5 and 7.Should this sequence eventually come to seem worth the effort, thenperhaps some gifted analyst might be inspired to undertake to forge thecorresponding continuous function, heroically indifferent to the tormentand the sweat. The next task up would be a proof, or else an expandedquest for counterexamples.It will be interesting to see whether the odd number indices withfalse-positive terms occur in a discernible pattern. If so, they couldbe ignored right along with all even indices.Any comments would be appreciated. Situation:----------Let P and Q be simple polygons. Assume, they intersect in anon-degenerate way, i.e. no vertex of one polygon is on theborder of the other.Idea:-----I'm interested in a lower bound on the number of holes in Pcup Q. I want to get it by counting holes coming from partsof the borders of P and Q.Formally:---------Split the border of P into disjoint polylines p_i in a waythat all p_i are either convex or concave and the totalchange of direction when walking along each p_i is less thanpi. Split the border of Q into q_j the same way.(I don't know whether all these conditions on p_i and q_jare necessary. But this is what I have.)Let n_{ij} be the number of enclosed regions between p_i andq_j that are on the outer side (concerning P or Q,respectively) of the bounding parts of p_i and q_jQuestion:---------Prove or disproveP cup Q has at least sum_{ij} n_{ij} holes.I know:-------Every hole counting for n_{ij} contains a hole of P cup Q.(That's easy because a point close to an intersection of p_iand q_j cannot be covered by P cap Q.)However the same hole of P cup Q might be enclosed byseveral (p_i,q_j).I hope that it's possible to understand my question thoughseveral exact defintions are missing. I will ask moreprecisely and/or modify the question if necessary.Any help will be appreciated. >(That's easy because a point close to an intersection of p_i>and q_j cannot be covered by P cap Q.) This should be P cup Q. Sorry. In sci.math, ChrisCoaster<77a787ce.0307300907.3fc86650@ height of a chord through a given> circle, but beyond that I don't know how to go about it.Edia = 7,926(miles) or 41,849,280(feet)> Ecirc = 24,900(miles) or 131,472,000(feet)Eradius = 1/2 dia = 3963miles or 20,924,640feet.What is formula or procedure to find the answer? I'm learning> disabled so please explain it as to a 10 year old.> ChrisIf you understand trigonometry, you'll understand this explanationreasonably enough, because that's what's needed to solvethis particular problem.Take the Earth's radius to be 6.378 * 10^6 m, or 6,378 km.(Scientists and I prefer metric, for various reasons.This is slightly higher than your value. Feel free torework the computations if you want to; they work wellenough in miles as it turns out but remember to convertthe result at the end to feet or inches.)We're assuming a perfectly circular Earth for the purposeof this argument; obviously that's not the case but theerrors won't affect the amount much.This means that the circumference of the Earth will be2 * pi * 6.378 * 10^6 = 4.007 * 10^7 m, or 40,070 km.One mile on the Earth's surface therefore subtends1.609344 / 6378 radian = 0.0002523 radian. (1 mile =5280 feet x 12 inches/ft x 0.0254 m/in x 1/1000 km/m.Bear in mind the circumference is 2 pi radians. Mathematicianslove radians as 1 radian along a circle's circumference is thesame distance as the radius. :-) It turns out a radianis 180 / pi degrees, or about 57.3.)Cut that triangle in half lengthwise and you get 2 very thinright triangles. The rise of the surface is the distancebetween the base of each right triangle and the arc.The base is the cosine of half the angle times the radius,or 6.378 * 10^6 * cos(0.0002523) = 6377999.796959 m.The difference is then about 0.203 m or 20.3 cm -- lessthan a foot. This means that you'll be able to see yourfriend standing a mile away although you may not be able tosee his shoetops, assuming sufficiently powerful binocularsand a nice flat surface, such as the ocean. :-) )-- #191, ewill3@earthlink.netIt's still legal to go .sigless. I'm looking for a good modern book on differential geometry, ascomprehensive as possible, but also good for self-study (I know I'masking too much!). I've been suggested Lee's Introduction to SmoothManifolds.Does anybody have it? can you comment on it?Any suggestions, opinions, comments, etc. appreciated.nojb. > Q: What's the highest number in group theoryA: a belianQ: What's purple and commutes?A: an abelian grape-- http://hertzlinger.blogspot.com if f(n) = n!/( (n/2)! 2^n) )> then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he > reconsider.Ross, consider this, which you may easily verify yourself: f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. Thus f(2n+2) > f(2n) > 1 for every integer n > 2, > and f(2n) gets further away from 1 as n increases,> and, in fact, diverges towards +oo.f(n) = n! / ( (n/2)! 2^n )f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) )> f(2n) = (2n)! / ( n! 2^(2n) )f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2)> = (2n+1)(2n+2) / ( (n+1) 4 )> = (2n+1)2/4> = n + 1/2Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than> f(2n). That seems to be a counterexample.My point is that f(2n+2) is (n+1/2) times greater than f(2n). There is no integer n for which (n+1/2) is infinite.(2n+2) / 2n = 1 + 1/n> lim (1 + 1/n) = 1But this is irelevant. What one concludes from my result is that f(2n) > f(2)*n! for all n > 1.And since n! -> +oo, amd f(1) <> 0. f(2n) -> +oo.I'm more interested in contradictions to the steps of the proof, yet I> still have to figure out how to respond to the proferred> counterexample, unless you do.RossMy counterexample proves your argument contains at least one error. An acknowledgement of that fact would be in order. The problem of where those errors occur is all yours. Ross has claimed that if f(n) = n!/( (n/2)! 2^n) )> then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he > reconsider.Ross, consider this, which you may easily verify yourself: f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. Thus f(2n+2) > f(2n) > 1 for every integer n > 2, > and f(2n) gets further away from 1 as n increases,> and, in fact, diverges towards +oo.f(n) = n! / ( (n/2)! 2^n )f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) )> f(2n) = (2n)! / ( n! 2^(2n) )f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2)> = (2n+1)(2n+2) / ( (n+1) 4 )> = (2n+1)2/4> = n + 1/2Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than> f(2n). That seems to be a counterexample.My point is that f(2n+2) is (n+1/2) times greater than f(2n). There > is no integer n for which (n+1/2) is infinite.> (2n+2) / 2n = 1 + 1/n> lim (1 + 1/n) = 1But this is irelevant. What one concludes from my result is that > f(2n) > f(2)*n! for all n > 1.And since n! -> +oo, amd f(1) <> 0. f(2n) -> +oo.I'm more interested in contradictions to the steps of the proof, yet I> still have to figure out how to respond to the proferred> counterexample, unless you do.RossMy counterexample proves your argument contains at least one error. > An acknowledgement of that fact would be in order. The problem of where those errors occur is all yours.I understood your point, of f(2n+2)/f(2n).What we're looking for in this as a test of possible convergence isthat that quotient should be equal to or have a limit evaluation ofone.What I want to make clear is the application of that as a test or as acounterexample that the same reasoning would imply the divergence ofStirling's identity, which is proven by different means.So I think I have a proof that doesn't depend on successive termsdemonstrating equality, and I look to a well-known proof of a similaridentity and see that its successive terms don't approach equality. Iwant to resolve that as a line of argument against the result, andstart by saying it would invalidate another unrelated result, which isknown to be true and accurate. I'm saying the successive termquotient argument would be invalid for the same reaons it is invalidagainst the well-known result, Stirling's identity.Now, I haven't verified Stirling's identity, to be frank it is beyondmy current knowledge and would require weeks.Here's something I like to consider, where I assumelim n-> oo n! / ( (n/2)! 2^n) = 1It's that (n!)^2 / ( ((n/2)!)^2 2^2n ) = 1(n!)^4 / ( ((n/2)!)^4 2^4n ) = 1(n!)^8 / ( ((n/2)!)^8 2^8n ) = 1etcetera, fromn! / ( (n/2)! 2^n) = ( (n/2)! 2^n) / n!That also leads me to thinksqrt(n!) / ( sqrt((n/2)!) 2^(n/2) ) = 1that isx E Z | lim n->oo ( n! ^(2^x) ) / ( ((n/2)!)^(2^x) (2^n)^(2^x) ) = 1It's a good argument, the quotient argument, I don't yet know how tocounter it directly, so I counter it indirectly by showing thecounterexample against a well-known good result. If the argumentaffects the F/E identity proof, then it would affect the Stirlingidentity proof.So my response is I understand that and am considering it, and don'tconsider it to affect the rationale of the proof, tell it toStirling.I'm happy because it, the argument, doesn't represent a contradictionto the progression of the proof.About the factorial/exponential identity proof, I think there are gapsin the proof, that is to say, I haven't fully demonstrated all theaspects of the proof. It's only been put forth on this thread onsci.math within the past several days, I imagine it will get furtherreview.So, please attack the proof with logical, rational arguments so I cansee if it holds.Ross > Got it!!! Relatively minor correction finishes the argument.See below...No, it didn't. However, the exercise wasn't useless.James Harris > And as I pointed out before there have been people who have been> attacking algebra itself in posts. Now it still bothers me that I was> the one person standing up to defend algebra itself from what I> noticed.The image of JSH as a stand up guy defending any part of mathemetics from the attacks of others is hysterical. > Got it!!! Relatively minor correction finishes the argument. See below... No, it didn't. However, the exercise wasn't useless. James HarrisYes, it was. Your proof has been completely refuted. It is beyond repair.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com Suppose y^3+3y = 2 and ry + 2s = 1 for algebraic integers r,s.> Then> 1 = y r + s 2> = y r + s y(y^2+3)> = y(r + s (y^2+3))Clearer: y|2 => y|(y,2)=1 => y unitIn general if y^n + ... + c = 0 then y|c so (y,c) = (y), which is (1) iff y is a unit. i.e. y is coprime to its norm iff y is a unit. -Bill Dubuque > My paper does rest fully on the following rather basic lemma:Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c> is a factor of the contant term P(0), while r=g-c, so it varies if g> varies. That simple lemma not only anchors my paper, and my proof of Fermat's> Last Theorem, it can be used in lots of places. Now I'll use it with> the roots of y^3 + 3y - 2 I find its roots fascinating. > and the three roots are y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}As simple as they are you can't just take the roots and *look* at themto tell whether or not any one of them is coprime to 2. It is justimpossible.Yet I work that out and still I keep coming back, as the simplicityand elegance of those roots keeps calling me.Sure I know that the cuberoot is ambiguous, so depending on how youtake the cuberoot, you shift between the solutions, which is why youcan't prove whether or not one of them is coprime to 2, but still itseems like there should be some way to see by looking.But there isn't.Given the impossibility of proving that none of them are coprime to 2,you'd think that people objecting would sit back, but nope, they claimGalois Theory eliminates a possibility.But that can't be. Galois Theory is based on that shifting. Actually, they can neither prove nor disprove the assertion with thetools they're using.> where the polynomial I'm interested in is P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}.Letting g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, I have c=1, so r=g-1. (Some may worry about my choice for c, I'll come back to that later.)Now using the substitution g=r+1, with x=sqrt(2), with my roots, I> have y_1 = r+1 - 1/(r+1) y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1)so I have y_1 = (r^2 + 2r)/(r+1) y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1)So once again I found myself drawn to that simplicity, looking forsome way, some how to make the math definite to the eye--to make iteliminate a possibility on sight.But it doesn't. You can't look at those roots and tell.There's just no way to limit the math in that way, and intellectuallyI know why:The math has the flexibility to handle several possibilities. Forsome polynomials, none of roots would be coprime to integer factors ofthe constant term, while for others, like y^3 + 3y - 2, several arecoprime.But you have to use advanced techniques to prove it.Still something keeps drawing me back to stare at those roots, lookingfor what cannot be found--a limitation that I can see.It's like in physics where you have quarks that can't be pulled outone by one. You can determine they're there experimentally in pairsor threesomes. You can bounce things off them, but you can't *see*them.And there's such a need to see.James Harris >> Got it!!! Relatively minor correction finishes the argument.>> >> See below...No, it didn't. Well duh, of _course_ it didn't - your fixes _never_ actually fix theproblem. Everyone else realized that long ago.> However, the exercise wasn't useless.>James Harris************************David C. Ullrich > My paper does rest fully on the following rather basic lemma: Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c> is a factor of the contant term P(0), while r=g-c, so it varies if g> varies. That simple lemma not only anchors my paper, and my proof of Fermat's> Last Theorem, it can be used in lots of places. Now I'll use it with> the roots of y^3 + 3y - 2 I find its roots fascinating.You are again answering your own posts with narcissistic monologues instead of answering your critics. Ifyou had any integrity, you would deal with the challenges not carry on conversations with yourself. Talk toa mirror if you want, but why use a newgroup to talk to yourself.> and the three roots are y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} As simple as they are you can't just take the roots and *look* at them> to tell whether or not any one of them is coprime to 2. It is just> impossible.You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root isa divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2.> Yet I work that out and still I keep coming back, as the simplicity> and elegance of those roots keeps calling me.Didn't someone 'call you home' once? If so, how come you came back?> Sure I know that the cuberoot is ambiguous, so depending on how you> take the cuberoot, you shift between the solutions,What the hell does that mean?> which is why you> can't prove whether or not one of them is coprime to 2,Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in thering of algebraic integers. This is because the quotient equals the product of the other two roots and theproduct of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2.> but still it> seems like there should be some way to see by looking. But there isn't.So what? If mathematician's could prove everything by 'looking' they wouldn't need theorems and lemmas.> Given the impossibility of proving that none of them are coprime to 2,Hold it. This is proven by noting that the product of the roots, y_1 y_2 y_3, equals 2 and that each of theroots divides 2 in the ring of algebraic integers -- meaning the quotient is an algebraic integer. Hencenone of the roots are coprime to 2. QED> you'd think that people objecting would sit back, but nope, they claim> Galois Theory eliminates a possibility.I don't.> But that can't be. Galois Theory is based on that shifting.> Actually, they can neither prove nor disprove the assertion with the> tools they're using.Galois Theory isn't necessary in this case. 'Basic algebra' does the trick. You wouldn't want to attack'basic algebra', would you?> So once again I found myself drawn to that simplicity, looking for> some way, some how to make the math definite to the eye--to make it> eliminate a possibility on sight. But it doesn't. You can't look at those roots and tell.As has been shown above, it isn't necessary to 'look at those roots and tell'. None of them is coprime to2. You have been refuted, defeated, vanquished, humiliated, crushed, conquered, exposed and undone. Takedown your false proof and acknowledge your errors.> There's just no way to limit the math in that way, and intellectually> I know why: The math has the flexibility to handle several possibilities. For> some polynomials, none of roots would be coprime to integer factors of> the constant term, while for others, like y^3 + 3y - 2, several are> coprime.None of them are coprime. The product of all the roots, y_1 y_2 y_3, is equal to 2. Each root divides 2producing an algebraic integer quotient. Hence, in the ring of algebraic integers, none of the roots, count'em again!, none is coprime to 2. Your claim is false.> But you have to use advanced techniques to prove it.You can't prove your claim at all, because it is false.> Still something keeps drawing me back to stare at those roots, looking> for what cannot be found--a limitation that I can see.The limitation is between your ears, not in your eyes.> It's like in physics where you have quarks that can't be pulled out> one by one. You can determine they're there experimentally in pairs> or threesomes. You can bounce things off them, but you can't *see*> them.No, it isn't like quarks. It is like an LSD trip.> And there's such a need to see. James HarrisOpen your eyes. You have been shown the way. To paraphrase Samuel Johnson: I can give you a solution, but Ican't give you an understanding.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com flexibility to handle several possibilities. For>some polynomials, none of roots would be coprime to integer factors of>the constant term, while for others, like y^3 + 3y - 2, several are>coprime.(I assume your talking about _algebraic integers_.) The constant term of y^3 + 3y - 2 is -2. If Y is _any_ root of y^3 + 3y - 2, then (-2)/Y is root of x^3 + 3x^2 + 4. That's a monic polynomial, so (-2)/Y is an algebraic integer, and therefore Y divides -2 in the ring of algebraic integers. So Y in _not_ coprime to -2.>But you have to use advanced techniques to prove it.No, you don't. Solving cubics is not an advanced technique.>Still something keeps drawing me back to stare at those roots, looking>for what cannot be found--a limitation that I can see.It's like in physics where you have quarks that can't be pulled out>one by one. You can determine they're there experimentally in pairs>or threesomes. You can bounce things off them, but you can't *see*>them.If that's not seeing (in physics), then what is?>And there's such a need to see.Yes. And (almost) everybody _does_.-- Thomas Wasell | I'd love to go out with you, but I'm having all mywasell@bahnhof.se | plants neutered. |> Sure I know that the cuberoot is ambiguous, so depending on how you|> take the cuberoot, you shift between the solutions,||What the hell does that mean?it means that he's much closer to actually understanding somethingabout mathematics in this post than he has been in just about any ofhis other posts. he's still horribly mixed up, but there's animportant truth that he almost seems close to understanding here.|> But that can't be. Galois Theory is based on that shifting.|> Actually, they can neither prove nor disprove the assertion with the|> tools they're using.||Galois Theory isn't necessary in this case.galois theory is highly relevant and perhaps crucial here.|> It's like in physics where you have quarks that can't be pulled out|> one by one. You can determine they're there experimentally in pairs|> or threesomes. You can bounce things off them, but you can't *see*|> them.||No, it isn't like quarks. It is like an LSD trip.no, it's in fact very much like quarks, and in a surprisingly deepway. but yes, it's also like an lsd trip.-- > I've written a paper and it is currently at a journal.> And I'm waiting to hear further from them.Why wait? Everyone already knows what's going to happen. They'regoing to reject it, and you're going to make excuses. Why not startrationalizing your failure *now* and get a head start?-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock |> Sure I know that the cuberoot is ambiguous, so depending on how you> |> take the cuberoot, you shift between the solutions,> |> |What the hell does that mean? it means that he's much closer to actually understanding something> about mathematics in this post than he has been in just about any of> his other posts. he's still horribly mixed up, but there's an> important truth that he almost seems close to understanding here.Why not enlighten the rest of us then? Evidently James isn't quite up toit, and you could perform a service by identifying the 'important truth healmost seems close to understanding'.> |> But that can't be. Galois Theory is based on that shifting.> |> Actually, they can neither prove nor disprove the assertion with the> |> tools they're using.> |> |Galois Theory isn't necessary in this case. galois theory is highly relevant and perhaps crucial here.It may be relevant, but it certainly isn't necessary or crucial in provingthat the roots of a monic polynomial with integer coefficients are divisorsof the constant term within the ring of algebraic integers.> |> It's like in physics where you have quarks that can't be pulled out> |> one by one. You can determine they're there experimentally in pairs> |> or threesomes. You can bounce things off them, but you can't *see*> |> them.> |> |No, it isn't like quarks. It is like an LSD trip. no, it's in fact very much like quarks, and in a surprisingly deep> way. but yes, it's also like an lsd trip.Please provide an in-depth explanation. Why leave the less informed hangingwhen you are able to clarify why it is that proving/disproving coprimalitywithin the ring of algebraic integers is like quarks?> -->Looking forward to your follow-up, as I assume your post was not a 'hit andrun' contribution.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com >> My paper does rest fully on the following rather basic lemma:>>> Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c>> is a factor of the contant term P(0), while r=g-c, so it varies if g>> varies. That simple lemma not only anchors my paper, and my proof of Fermat's>> Last Theorem, it can be used in lots of places. Now I'll use it with>> the roots of>>> y^3 + 3y - 2 I find its roots fascinating.>> and the three roots are>>> y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3}>>> y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 ->>> (-1-sqrt(-3))/2(1+sqrt(2))^{1/3}>>> y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 ->>> (-1+sqrt(-3))/2(1+sqrt(2))^{1/3}As simple as they are you can't just take the roots and *look* at them>to tell whether or not any one of them is coprime to 2. It is just>impossible.No, it is not impossible. It is amazingly trivial, as Keith Ramsaypointed out right after you posted them:NONE OF THEM ARE COPRIME TO 2. THEY ARE ALL DIVISORS OF 2:y_1*(y_1^2+3) = 2y_2*(y_2^2+3) = 2y_3*(y_3^2+3) = 2.Surely you agree with all three of those expressions, since they areall roots of y^3 + 3y - 2.So they are all divisors of 2, and since none of them are units, theyare none of them coprime to 2. Simple, nay, trivial. [.snip.]>Given the impossibility of proving that none of them are coprime to 2,it follows that red is blue, 0 is 1, and I am the Pope (who,therefore, is not catholic).>you'd think that people objecting would sit back, but nope, they claim>Galois Theory eliminates a possibility.Not even Galois theory is needed here. ->THIS<- one is trivial. [.snip.] a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan [.snip.]Nitpicking...The first time you made the point, you correctly stated:>You don't have to just look. The observation that y_1 y_2 y_3 = 2 is sufficient to prove that each root is>a divisor of 2 in the ring of algebraic integers and, since none of these is a unit, none are coprime to 2. [.snip.]Later, you said, in several different versions, things like:>Oh yes, you can. The product of the roots, y_1 y_2 y_3 equals 2. Hence, each of the roots divides 2 in the>ring of algebraic integers. This is because the quotient equals the product of the other two roots and the>product of two algebraic integers is an algebraic integer. Hence none of the roots are coprime to 2.The last statement only follows once you add since none of them is aunit [in the ring of algebraic integers]. such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan consider the functional equation a^f(x)=f(x+1) with f(0)=0, a>1for x between zero and one I have noticed that one solution for f(u)is very nearly (1-a^u)/(1-a) for u between zero and one. I call thisinterested in this conjecture. A particular case for a =exp(1) wouldgive f(1/2)=1/2. The true value is f(1/2)=.4985628 for the canonicalsolution. says...If we consider the functional equation a^f(x)=f(x+1) with f(0)=0, a>1>for x between zero and one I have noticed that one solution for f(u)>is very nearly (1-a^u)/(1-a) for u between zero and one.That equation doesn't pin down f at all between 0 and 1.Let's define the function T(a,n,x) (where a and x are real,but n is a nonnegative integer) T(a,0,x) = x T(a,n+1,x) = a^T(a,n,x)Then your equation for f implies that for absolutelyany function g(x), we can find a solution f such that f(x) = T(a,floor(x),g(frac(x)))where floor(x) = largest integer that is less than x, andfrac(x) = x - floor(x).If x is between 0 and 1, then f(x) = g(x)so f(x) is completely arbitrary in this region.--Daryl McCulloughIthaca, NY Functions and set theory doubton information theory and now have to document it. Exposing my ideas usingmathematical language, I have come up with the need to identify abijection wich is something like this: A -----> (B, C) where C is an integer in the [0, X[B]) range Do you know if there is any special name for this or if I can callit Uncle Eddie's bijection? I ask because computing being my area ofknowledge, I don't know if somebody has already baptized this thing.How math about mathYou just did, every 3 to 8 hours new posts in most bulletin board about math in a couple hours and your post will show up.Hercps this single webpage is the most powerful webpage of all,its like a computerised memory, any post you can vaguely rememberjust type what you know about it here and the post will spring backfrom the > |-|erc says...> I don't know what you are talking about, but if HH is a computer>> program, then we can prove that it isn't equal to Halt.>>//----------->//A 01L B>//A 11R B>//B 01R C>//B 10L E>//C 00R D>//C 10L A>//D 01L A>//D 10R D>//E 01L HALT>//E 10L C>//------------->This is HH, I'm pretty sure its the bona fide Halt algorithm :)This is actually TM number 9320303603822 where the transition sequence>above is mapped to the number, where state HALT is 0, L is 0, R is 1, and the>later transitions are more significant, all the smaller size TMs with under 5>states>were added to that number. So 01LB is (1*12+0*6+2) * 24^0 = 14.>i.e there are (2*2*6)^10 5 state TMs.>You can give it any parameter in unary, and the claim is it will output the>value for Halt for that number by terminating to the right with either 01 or>00. Well, there is an easy proof that the claim is false. Remember, by definition> of Halt, if n is the Godel number of a program, then Halt(n) = 0 if that program> halts on input n, and Halt(n) = 1, if it never halts on input n. So, given your program HH, we define a new function g(n) by g(n) = if HH(n) = 0, then loop forever> else output 0 We compute the Godel number k of g, and consider the computation> g(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, and> otherwise, Halt(k) = 1. So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above,> we know that g(k) only halts if HH(k) is nonzero. So in this> case, it is clear that Halt(k) != HH(k), so HH does not compute> Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g,> we know that the only way for g(k) to fail to halt is if HH(k)=0.> So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly> compute Halt.>Which is it 1 or 2? You did a specific proof for pi/4, why can't you disprove HH?not convincing in the least when you have an actual program and use aconceptual proof for all programs. If your Halting proof is valid then youcan demonstrate it for the program given. The flaw in your proof isexactly the same I've been pointing out the last 3 posts, try to think aheadwhere your argument is leadingI said the observable options for g(k) are different to the options listed so:Observered Your 'conclusion'g(k) halted : g(k) haltsg(k) hasn't halted : g(k) haltsg(k) hasn't halted : g(k) never haltsYou then substantiate this by concluding we know the values of f. ????This is a different example, think again.Like I said 10 posts ago, Halt not being computable is a different meaning tono TM possibly calculates Halt, as long as the TM is unidentifiable, such as HHwhich does not exhibit any characteristics to differentiate it from Halt.Its been trialled for 10^12 iterations with continually changing pattern, after youprove it does or doesn't halt then you can proceed with your examinationof the values of g(k). Good luck!Herc |-|erc says...> So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above,>> we know that g(k) only halts if HH(k) is nonzero. So in this>> case, it is clear that Halt(k) != HH(k), so HH does not compute>> Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g,>> we know that the only way for g(k) to fail to halt is if HH(k)=0.>> So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly>> compute Halt.>>Which is it 1 or 2?I don't know. But if you know that either A or B holds,and you know that A implies C, and you also know that Bimplies C, then you can conclude that C holds.For example: Either it will rain tomorrow, or it won't.If it doesn't rain tomorrow, then I will water my garden.If it does rain tomorrow, then it will rain on my garden.I don't know whether it will rain or not, but I don't need to knowwhich is the case to know that my garden will get water.>You did a specific proof for pi/4, why can't you disprove HH?I did. My proof works for any program whatsoever.>not convincing in the least when you have an actual program and use a>conceptual proof for all programs.If I prove something about all programs, and then you give mea particular program, then it is necessarily true about thatparticular program. That's the whole point of proving generalstatements.>If your Halting proof is valid then you>can demonstrate it for the program given.I did.>I said the observable options for g(k) are different to the options listed so:Observered Your 'conclusion'g(k) halted : g(k) halts>g(k) hasn't halted : g(k) halts>g(k) hasn't halted : g(k) never haltsYou then substantiate this by concluding we know the values of f. ????I don't need to *know* whether g(k) will ever halt or not. I knowthat *either* it will halt, or it won't. In either case, I can provethat f does not compute the halting program.>This is a different example, think again.Like I said 10 posts ago, Halt not being computable is a different meaning to>no TM possibly calculates HaltYou were wrong when you said it 10 posts ago, and you are still wrong.Halt not being computable means exactly that no TM can possiblycalculate Halt.--Daryl McCulloughIthaca, NY > |-|erc says...>> So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above,>> we know that g(k) only halts if HH(k) is nonzero. So in this>> case, it is clear that Halt(k) != HH(k), so HH does not compute>> Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g,>> we know that the only way for g(k) to fail to halt is if HH(k)=0.>> So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly>> compute Halt.>>Which is it 1 or 2? I don't know. But if you know that either A or B holds,> and you know that A implies C, and you also know that B> implies C, then you can conclude that C holds. For example: Either it will rain tomorrow, or it won't.> If it doesn't rain tomorrow, then I will water my garden.> If it does rain tomorrow, then it will rain on my garden.> I don't know whether it will rain or not, but I don't need to know> which is the case to know that my garden will get water.>Then why did you provide a specific proof for the known computable pi/4?g(k) hasn't haltedNo, there is a third observable: the definition of g. If we lookat the definition of g, we can see that if f(k)=0, then g(k) willnever halt, and that if f(k)!=0, then g(k) will halt....We can look at the definition of g and we can look at the value of f(k).That confirmationn of your Halting proof is completely irrelevant then!A only implies C because we know why A holds, your proof makes assumptions.>You did a specific proof for pi/4, why can't you disprove HH? I did. My proof works for any program whatsoever.it works for programs that are known to be computable.the conclusion is that we cannot know a function to compute theHalting problem, nothing more.If your Halting proof is valid then you>can demonstrate it for the program given. I did.I said the observable options for g(k) are different to the options listed so:Observered Your 'conclusion'g(k) halted : g(k) halts>g(k) hasn't halted : g(k) halts>g(k) hasn't halted : g(k) never haltsYou then substantiate this by concluding we know the values of f. ???? I don't need to *know* whether g(k) will ever halt or not. I know> that *either* it will halt, or it won't. In either case, I can prove> that f does not compute the halting program.Here are the specifics for your proof applied to HH:Well, there is an easy proof that the claim is false. Remember, by definitionof Halt, if n is the Godel number of a program, then Halt(n) = 0 if that programhalts on input n, and Halt(n) = 1, if it never halts on input n.So, given your program HH, we define a new function g(n) by g(n) = if HH(n) = 0, then loop forever else output 0** forall(n) HH(n) has not yet terminated, we DONT KNOW if HH(n) = 0 or not, so regardless_of_the_value of HH(n), g(n) will loop waiting for HH to give a result.We compute the Godel number k of g, and consider the computationg(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, andotherwise, Halt(k) = 1.So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above, we know that g(k) only halts if HH(k) is nonzero. So in this case, it is clear that Halt(k) != HH(k), so HH does not compute Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, we know that the only way for g(k) to fail to halt is if HH(k)=0. So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly compute Halt.Its option 2, g(k) hasn't finished running yet, its waiting for the value of HH(k)in its subprogram. Halt(k) is not known. The only way for g(k) to notto halt is if HH(k) = 0, ***or**** HH(k) hasn't finished computing yet.No contradiction with Halt(k) = HH(k).Herc > I'm going by the one on the net at> http://home.ddc.net/ygg/etext/godel/godel3.htm> if that helps. It seems more or less legit to my eyes.That's the one, but it doesn't render well outside of the Microsoftuniverse.>> Then there are primitive-recursive functions (just recursive in>> the paper). These are number-theoretic functions, taking numbers as>> arguments and returning numbers. These functions are not objects of the>> formal system P, so Godel does not assign Godel numbers to functions.>> There are no FUNCTIONS, but there are formulas (lower case) that represent>> functions in the formal system in a precise sense explained in the paper.>OK this is good because I was never sure exactly how to approach> these. Sometimes they seemed to be like a computer function as you> described, taking numbers as> arguments and returning numbers but other times it seemed less> clear. Does this include all those n Gl x type bits numbered 1 - 45> that Godel terms functions(relations).Yes, I think so. Those are mathematical objects.> If that is so when we see function 1: x/y as part of a formula or> another function do we replace it with a 1 or 0 to indicate the truth> or falsity of the statement.No.> I seem to recall this being done> elsewhere in the paper. Or do we replace it with (Ez)(z <= x & x => y.z) and then work out the Godel number from these symbols.> That seems> different from the number in, number out concept but would that be> one of the formulas (lower case) that represent> functions in the formal system in a precise sense explained in the> paper that you mentioned.Yes, exactly. The formula z <= x & x = y*z REPRESENTS the relationz= x/y in system P. Actually, the <= symbol is probably not in P'salphabet but is a shorthand a defined relation, so you would have toreplace it by its definition, for example ((Ew)(w+z = x)) & x = y*z.It's pretty easy to find a representing formula for a relation, becausesystem P follows Principia Mathematica and it can talk about sets,ordered pairs, relations, functions, finite sequences etc. All you haveto do is transcribe your mathematical definition in the notation of P.Thus every mathematical object in the chain (1--45) has a representingformula.> Jesus this is getting complicated.No . It's not over, either. To proceed further you have to provethat your formula formally represents the original relation. And *that*means that, whenever you substitute the numerals of numbers that obey therelation, system P has a proof of the substituted formula; but if yousubstitute the numerals of number that don't obey the relation, sytemP has a proof of the negation of the substituted formula. For example,3= 6/2 and 4 != 6/2; sure enough, P proves ((Ew)(w+sss0 = ssssss0)) & sssss0 = ss0 * sss0and P also proves ~(((Ew)(w+ssss0 = ssssss0)) & ssssss0 = ss0 * ssss0)You have to convince yourself that this works in the general case.Note that this notion of representability does not assume that P isconsistent. If P is inconsistent, any formula represents everyrelation. Exercise.This notion of representability is crucial in the proof that the number17 Gen r is an UNDECIDABLE number. Neither a THEOREM nor the NEGATIONof a THEOREM.> Here is one of my main problems. Godel uses formulas such as> Q(x,y): ~{B[Sb(y 19|z(y))]}> Godel later talks about a relation sign q for this formula (by which I> presume he means what has become known as a Godel number)I think Q() is a mathematical relation, not a formula. The lower-case q,the relation sign, is either the formula, or the Godel number thereof.If it is a Godel number, Godel will probably call it a RELATION SIGN,not a relation sign.> meaning that> he has worked out the Godel number for Q. But in doing so, when we get> to the part of Q(x,y) that is Sb(y 19|z(y)) and we try to follow the> above steps we find that y is a free variable and hence not a FORMULA,That's ok, y was already a metamathematical variable in Q(x,y). BothQ() and Sb() are mathematical objects. What you're looking at is adefinition of Q() in terms of Sb() and z() and the x PROVES yrelation (was that Bw() ? should be number 44).> and therefore Sb(y 19|z(y)) is undefined as you point out and this> makes it impossible to find the Godel number q for Q(x,y) which is> used later in the proof.No, it's defined, in mathematics. Sb(y, 19|z(y)) is a number-theoreticfunction of one variable.The function Sb(), or the relation t= Sb(x,y,z), has a representingformula with 4 free variables. The function z(), or if you prefer therelation t= z(y), also has a representing formula, this one with twofree variables.By composing the two formulas appropriately, you get a representingformula for the function Sb(y, 19|z(y)). You then need to combine thatwith the representing formula of x PROVES y to obtain a formula q withtwo variables that represents the mathematical relation Q(x,y).> Would this be where we use the formula> representation of the fuction to represent Sb(y 19|z(y)) in the formal> version of Q(x,y). I could see how this might get around the problem> if it is possible but it seems that Sb() is sometimes a FORMULA and> other times not quite the same and there is no explicit instruction as> to which to use, unless I've missed saomething.Again, read the paper slowly. I think Sb() is always anumber-theoretic function, but you have to make sure.> [ ... ] That is pretty much how I see the process going also if Sb() is> treated as a number in number out formula.Function, not formula. This nitpicking really matters.> [ how to code P(y) Ax ~(x PROVES Sb(y 19|z(y)) ) ]> OK so we can write out Exists t = Sb(y 19|z(y)) to get around the> problem? And presumably we expand the recursive formula definiton of> Sb(y 19|z(y)) in terms of the other functions it is defined as and we> get a loooooong string of formal symbols (Big Mess) that we can> Godelize into a number.which is p.> and then when we substitute in the numeral of> p to this formula we are replacing the occurences of y in the big> mess.Exactly. The resulting formula has Godel number r.> [ ... ] Which would mean that it is> not like a computer function where you can make recursive calls to the> functionYou're correct, this is not like recursive calls. The whole point is to*not* have any recursion. Godel constructs a sentence that says somethingof its own Godel number. If you perform the following computation,blah blah, the result has property blah. So you perform the indicatedcomputation and, sure enough, the result is the Godel number of If youperform the following computation, blah, blah, the result has propertyblah. A self-referential sentence. Indirectly self-referential,without any infinite regress, but the representability theorem isstrong enough to make that stick.> a lot of things in this paper a lot clearer though it seems i'm still> a little behind.But that's normal. how can I show that the polynomial>T^8 - 40 T^6 + 352 T^4 - 960 T^2 + 576 in Q[T]>is irreducible over Q ist?>(According to MuPAD (an algebra system) that is the case, but I need a >calculation (that is a proof.)).>PS: I got the polynomial by the following MuPAD-calculation>(In case it helps...):fac := [ T+a*sqrt(2)+b*sqrt(3)+c*sqrt(5)> $ a in [-1,1] $ b in [-1,1] $ c in [-1,1] ];>p := _mult ( i $ i in fac );>f:= expand (p);>g := poly (f, [T], Dom::Rational);> A much different approach denotes the eight real roots by r1, r2, ..., r8. All of these are bounded in absolute valueby sqrt(2) + sqrt(3) + sqrt(5) < 1.5 + 1.8 + 2.3 < 6.The numerical value of the polynomial is prime when T = 11.If the polynomial f(T) has a factorization f1(T) * f2(T)where f1, f2 are monic of degrees d1 and d2 and integer coefficients, then |f1(11)| >= (11 - 6)^d1 |f2(11)| >= (11 - 6)^d2 .Unless d1 = 0 or d2 = 0, both |f1(11)| and |f2(11)| exceed 1.But these are integers whose product is a prime. (f(5) = -37799 is the negative of a prime -- you may be able to use T = 5 rather than T = 11 in a variation of this proof.)-- Spammers: I don't want a small digital camera to post photos of a large, lowweight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI Integrate { sin^2(x) + cos^2(x)*ln(cos(x)) } / { 0.25*sin^2(2x) }indefinitely wrt x.(where sin^2(x) refers to sin(x) squared, etc.)It would be of immense help if anyone could integrate the abovefunction. The reason is because I am currently working on someintegration techniques, and I have tried them on many integrals andthey all seem to work. However, every integral which I have tried myown developed techniques on could be solved by some other techniqueanyway (ie substitution or by parts or some other standard method).Therefore, I tried to think of an integral which could not becalculated anayltically using any technique I know of, and came upwith the above nasty looking integral. My own developed techniques cansolve it, but no other technique I am aware of can solve it, unless Iam missing something obvious.If someone can solve this integral(giving their method if possible)then that would greatly help my current research. I am aware thatpeople frequently post integral requests onto this group so Iapologise in advance for any annoyance and thank anyone who tries tohelp me. > Integrate { sin^2(x) + cos^2(x)*ln(cos(x)) } / { 0.25*sin^2(2x) }> indefinitely wrt x. (where sin^2(x) refers to sin(x) squared, etc.) It would be of immense help if anyone could integrate the above> function. The reason is because I am currently working on some> integration techniques, and I have tried them on many integrals and> they all seem to work. However, every integral which I have tried my> own developed techniques on could be solved by some other technique> anyway (ie substitution or by parts or some other standard method). Therefore, I tried to think of an integral which could not be> calculated anayltically using any technique I know of, and came up> with the above nasty looking integral. My own developed techniques can> solve it, but no other technique I am aware of can solve it, unless I> am missing something obvious. If someone can solve this integral(giving their method if possible)> then that would greatly help my current research. I am aware that> people frequently post integral requests onto this group so I> apologise in advance for any annoyance and thank anyone who tries to> help me.This looks quite straightforward.int [{sin^2(x) + cos^2(x)*ln(cos(x)) } / { 0.25*sin^2(2x)} ] dxstep 1. Use the double angle formula sin(2x) = 2 sin(x) cos(x)= int [{sin^2(x) + cos^2(x)*ln(cos(x)) } / { sin^2(x)cos^2(x) } ] dxstep 2. Break the fraction and simplify= int [ 1 / cos^2(x) + { ln(cos(x)) } / sin^2(x) ] dxstep 3. The first part is easy; integrate the second by parts:= int [1 / cos^2(x)] - ln(cos(x))*cot(x) - int [ tan(x)*cot(x) ] dxAre you sure this is not a homework problem? > I want to ask if the Fourier Series Expansion is unique to each > function?>>> For continuous bounded functions f on [-Pi, Pi), yes.>>> Even for L^1 functions if one identifies>> function equal Lebesgue almost everywhere.measures, for which you can assert actual uniqueness, and beyond that are >distributions, for which the same can be said. Hell, why stop there?I was getting tired.> Let P be the space of trigonometric polynomials (just a vector space,> no topology). Let P' be the space of all linear functionals on P. Then> every element of P' has a unique Fourier series expansion... (and now> _every_ trig series has become a Fourier series, so this is as far> as we can go. I'm gonna be famous for this, the world's most general> notion of Fourier series (on the circle).)Fine, but what bearing does this have on the uniqueness claims being made? > I want to ask if the Fourier Series Expansion is unique to each >> function?> For continuous bounded functions f on [-Pi, Pi), yes.> Even for L^1 functions if one identifies> function equal Lebesgue almost everywhere.measures, for which you can assert actual uniqueness, and beyond that are >>distributions, for which the same can be said. >>> Hell, why stop there?I was getting tired.That's not good enough, sorry.>> Let P be the space of trigonometric polynomials (just a vector space,>> no topology). Let P' be the space of all linear functionals on P. Then>> every element of P' has a unique Fourier series expansion... (and now>> _every_ trig series has become a Fourier series, so this is as far>> as we can go. I'm gonna be famous for this, the world's most general>> notion of Fourier series (on the circle).)Fine, but what bearing does this have on the uniqueness claims being made???? Of course I was being a little silly, but I don't follow thequestion - the relevance is the same as for L^1 functions,measures and distributions: An element of P' is determinedby its Fourier series.************************David C. Ullrich integral... please help me judge:a) Does it exist/converge at all?b) Is it possible to write the explicit form out? c) If it is integrable then how to evaluate it?The integral is:Integrate[exp(-0.25*w^2)*exp( _i_ *w*t)/(-w^2+ _i_ *w+1), w from -infto inf, t is parameter]My bag of tricks such as residue theory, change variables, etc. failedfor this integral... please help me out of the swamp!-Losmnd all,I am facing with the following crazy integral... please help me judge:a) Does it exist/converge at all?> b) Is it possible to write the explicit form out? > c) If it is integrable then how to evaluate it?The integral is:Integrate[exp(-0.25*w^2)*exp( _i_ *w*t)/(-w^2+ _i_ *w+1), w from -inf> to inf, t is parameter]Yes, it converges, at least when t is real. Indeed, for each real w and t,we have|exp(-0.25*w^2)*exp( _i_ *w*t)/(-w^2 + _i_ *w + 1)| = = exp(-0.25*w^2)/sqrt(w^4-w^2+1) <= 1/sqrt(w^4 - w^2 + 1),and the integral of 1/sqrt(w^4 - w^2 + 1) when w goes from -infinityto +infinity obviously converges.Jose Carlos Santos integral... please help me judge:>a) Does it exist/converge at all?>b) Is it possible to write the explicit form out? >c) If it is integrable then how to evaluate it?>The integral is:>Integrate[exp(-0.25*w^2)*exp( _i_ *w*t)/(-w^2+ _i_ *w+1), w from -inf>to inf, t is parameter]>My bag of tricks such as residue theory, change variables, etc. failed>for this integral... please help me out of the swamp!For (a), the answer is trivially yes, assuming that _i_is the imaginary unit. The terms multiplying exp(-0.25*w^2)are uniformly bounded.As for evaluating it, one can write w = 2z + 2_i_t, obtaining 2*exp(-t^2)*int exp(-z^2)/Q(z) dz,where Q is a quadratic. The limits are rather odd, but the integral is unchanged if the limits are moved to the real line. Write Q in terms of partial fractions,and use that the integral of the exp(-z^2)/(z+c) is thecomplex error function if c is not real. If c is purely imaginary, it is the proportional to the ratioof the tail error function to the density.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University > Integrate[exp(-0.25*w^2)*exp( _i_ *w*t)/(-w^2+ _i_ *w+1), w from -inf> to inf, t is parameter] Do you mean i*(w+1) or i*w+1? (not that it makes that much of a difference) > My bag of tricks such as residue theory, change variables, etc. failed> for this integral... please help me out of the swamp! Small bag of tricks try Fourier analysis. Also Abs[Integral[F(x)dx]]<=Integral[Abs[F(x)]dx] so your integral exists |I found that all metal means can also have a proportional reduction of|its rectangles like Phi's' golden rectangle.|Then with the input of another poster (John Burglund) who found that x|can be any value greater than zero that is plugged into this equation|will do the same thing and have proportional reducing rectangles as|well!This is a (known) general property of rectangles where the ratio betweenthe lengths of the sides is given by a continued fraction.|Where x can = transcendental or irrational or rational or|integer values.I found the following difficult to read:|Another fact also is if a rational x of n decimal length like as an|example x = .141. Then its decimal length (n) = 3 is plugged into the|equation the outcome will be m = an irrationalOk so far.|and 1/m = m's decimal|expansion after (n) decimal length in m and 1/m.Apparently what you mean here is that the decimal expansions of m andof 1/m are the same after the first n decimal places. The way you haveit written made it seem like you were saying either that 1/m = m, whichisn't true, or that 1/m was the portion of the decimal expansion of m.|This holds true for Phi and the metal means where the match of m = 1/m|decimal expansion is from the first decimal place because x = an|integer with zero decimal places.This is true for a very simple reason; m = x + 1/m, so if x has a decimalexpansion which stops at a given point, the decimal expansions of m and1/m will agree beyond that point.Keith Ramsay >I have a system of PDEs: >f_x = g_t>f_t = (c^2)g_x>f= f(x,t)>g= g(x,t)>c=c(t)> I don't have any particular c(t) in mind; just so long as it's > NOT constant. > How can I find general solutions to the PDEs for f, g and c?If you don't really care what c(t) is, you can replace the second equation > by (f_t/g_x)_x = 0, i.e. f_{xt} g_x = f_t g_{xx}, and define > c(t) = sqrt(f_t/g_x) (of course you'll want f_t/g_x > 0, but hopefully> that will be true at least in some region).Maple 9 then finds three families of solutions. Two are not of interest > because f_t or g_x is 0, but the third may be of some interest although> it's not very general:f(x,t) = c1 x + c3 x^2/2 + F1(t), g(x,t) = c2 + c1 t + (c3 t + c4) x> where c^2(t) = F1'(t)/(c3 t + c4). Here c1, c2, c3, c4 are arbitrary> constants and F1 is an arbitrary function; of course we want > F1'(t)/(c3 t + c4) > 0, which would cause a singularity at t=-c4/c3 if > c3 <> 0 unless F1'(-c4/c3) = 0. Thus with c3 = 1, c4 = 0, > F1(t) = t^2/2 + t^4/2 + t^6/6, we get solutionsf(x,t) = c1 x + x^2/2 + t^2/2 + t^4/2 + t^6/6> g(x,t) = c2 + c1 t + x t> c(t) = 1+t^2I also found another interesting family of polynomial solutions:> c(t) = c[0] + c[1] t> 2 3 2 4> f(x, t) = 2/3 b[2] c[1] x t + 1/2 a[2] c[1] t 2> + 2 c[0] c[1] b[2] x t 2 3 2> + (4/3 c[0] c[1] a[2] + 1/3 b[1] c[1] ) t + a[2] x 2 2 2> + 2 c[0] b[2] x t + (c[0] a[2] + c[0] c[1] b[1]) t 2> + a[1] x + c[0] b[1] t + a[0]> 2 4 3 2> g(x, t) = 1/6 b[2] c[1] t + 2/3 c[0] c[1] b[2] t + b[2] x 2 2> + 2 a[2] x t + c[0] b[2] t + b[1] x + a[1] t + b[0]> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2Wow, Robert! That's amazing. I was able to guess f(x,t) = c1 x + c3 x^2/2 + F1(t) g(x,t) = c2 + c1 t + (c3 t + c4) xbut your 4th order polynomial solution is truly awesome.I have never used anything like Maple or similar programs but I'mguessing that you assumed a general 4th order polynomial solution forf and g and then allowed Maple to solve for (f_t/g_x)_x = 0. Is thatright?Do you have any insights about other solutions to these equations?Eugene Shuberthttp://www.everythingimportant.org >Wow, Robert! That's amazing. I was able to guess >f(x,t) = c1 x + c3 x^2/2 + F1(t) >g(x,t) = c2 + c1 t + (c3 t + c4) x>but your 4th order polynomial solution is truly awesome.>I have never used anything like Maple or similar programs but I'm>guessing that you assumed a general 4th order polynomial solution for>f and g and then allowed Maple to solve for (f_t/g_x)_x = 0. Is that>right?IIRC I assumed polynomials fourth order in x and in t for f and g and afourth order polynomial in t for c(t), substituted in to the equations andsolved the mess of equations you get for the coefficient of each x^i t^j.Actually, I could have done better:assume f(x,t) = f0(x) + x f1(t) + f2(t) [where we can take f0(0)=0] g(x,t) = g0(x) + x g1(t) + g2(t) [ where say g0(0) = 0]The first equation says f0'(x) + f1(t) = x g1'(t) + g2'(t)Then we must have f0'(x) = a1 x + a0, g1'(t) = a1, a0 + f1(t) = g2'(t).The second equation says x f1'(t) + f2'(t) = c(t)^2 (g0'(x) + g1(t))so g0'(x) = b1 x + b0, f1'(t) = c(t)^2 b1, f2'(t) = c(t)^2 (b0 + g1(t)).Now g0(x) = b1 x^2/2 + b0 x f0(x) = a1 x^2/2 + a0 x g1(t) = a1 t + a3 f2(t) = int c(t)^2 (b0 + a1 t + a3) dt + a4 f1(t) = b1 int c(t)^2 dt + a5 g2(t) = a0 t + int f1(t) dt + a6This solution is actually a superposition of the following:(corresponding to a0) f(x,t) = x, g(x,t) = t(corresponding to a1) f(x,t) = x^2/2 + int c(t)^2 t dt, g(x,t) = x t(corresponding to a3 and b0) [ seems I didn't have an a2 ] f(x,t) = int c(t)^2 dt, g(x,t) = x(corresponding to a4) f(x,t) = 1, g(x,t) = 0(corresponding to a5) f(x,t) = x, g(x,t) = t(corresponding to a6) f(x,t) = 0, g(x,t) = 1(corresponding to b1) f(x,t) = x int c(t)^2 dt, g(x,t) = x^2/2 + int(int c(t)^2 dt) dtRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 Robert,I can't tell you how delighted I am with your answer.I'm guessing that you tried a solution of the form:f(x,t) = f0(x) + x f1(t) + f2(t)g(x,t) = g0(x) + x g1(t) + g2(t)and then realized you had uncovered a superposition of severalsolutions because of the linearity off_x = g_t f_t = (c^2)g_x If any other ideas occur to you related to this problem (anything atall), or if you think of more solutions that may be added to thegeneral form, please let me know.Eugene Shuberthttp://www.everythingimportant.org >I have a system of PDEs: >f_x = g_t>f_t = (c^2)g_x>f= f(x,t)>g= g(x,t)>c=c(t)Another useful family of solutions isf(x,t) = - i/k exp(i k x) g1'(t) g(x,t) = exp(i k x) g1(t)where g1(t) + k c(t)^2 g1(t) = 0Of course you can take real and imaginary parts to get real solutionsinvolving sin(kx) and cos(kx).It may be hard to solve the DE for g1 explicitly if c(t) is given, butif you don't really care what c(t) is you can choose the g1 firstand then get c(t); of course you want g1(t)/g1(t) < 0 so c willbe real. One nice example is c(t) = exp(rt), g1(t) = a1 J_0(exp(rt)/r) + a2 Y_0(exp(rt)/r)where J_0 and Y_0 are Bessel functions of the first and second kind. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > (corresponding to a0)> f(x,t) = x, g(x,t) = t(corresponding to a5)> f(x,t) = x, g(x,t) = tIn the joy of reading your post carefully (making sure every line was as delightful as it appeared) I didn't notice this duplication. I realize it's not a problem. So a5=0.Eugene Shuberthttp://www.everythingimportant.org Faggot ass trolls fight it out, get ur pay per view herewww.fisting.com > As for the mention of Jews, that, too, is easily explained.> rears its pin-shaped head, Jews top the list as an obvious target> group. As noted jackass Fred Reed admits, his random clicking on> websites turned up no hard evidence that Jews had taken over> American research activities, but that's because Jews typically> disguise themselves using false names to better infiltrate and> destroy American society. Heck, points out Fred, even a guy named> Miller could be a secret Jew! Fred probably had a few dozen> paragraphs on the Protocols of the Elders of Zion written up for the> whenever you burn a flag, a kitten dies public service> advertisements.It sure sounded like he was complimenting Jews and Asians.It sure sounded like he was complimenting Jews and Asians.I believe I said the racism was thinly disguised not so blatantlyobvious even Joseph Hertzlinger can spot it.Of course he was complimenting Asians. As any good racist knows,Asians are generally smarter and more industrious than Caucasians.compliments Jews, unless Sally Chen is supposed to be Jewish. IfFred was complimenting the Jewish people, it was only to point outtheir cleverness in disguising their last names from real Americansand in hiding behind the liberal taboo against forcing students tostate their religion to get into Harvard.If you don't spot anything off-key about phrases like read like a NewDelhi phone book, colonies of Indians, and Jews are doing lotsmore than their share of research or the endless cited lists of nameschosen for their foreign sound, you have my congratulations. Yourcandy-coated dream world must be a beautiful place to live and work!Hell, don't you see anything odd about the fact that all of Fred'sevidence involves lists of mostly Indian and Asian-sounding names(with a few Arabic and Greek ones tossed in), yet for some strangereason Jews make a unwarranted and jarring appearance in paragraphthree?If I was babbling on about poverty in India and started interjectingthings like not only is India highly populated, but there sure are alot of Jews around these days and speaking of crowded trains, haveyou noticed how many people in the entertainment industry seem to beJews, would it zip right over your head?Anyway, I call Ban on Politics, at least for me. I'm done with thisthread.-- Kevin >> It sure sounded like he was complimenting Jews and Asians....>compliments Jews, unless Sally Chen is supposed to be Jewish. If>Fred was complimenting the Jewish people, it was only to point out>their cleverness in disguising their last names from real Americans>and in hiding behind the liberal taboo against forcing students to>state their religion to get into Harvard.The fact that they got into Harvard without any special assistance isa compliment.>If you don't spot anything off-key about phrases like read like a New>Delhi phone book, colonies of Indians, and Jews are doing lots ^^^^^^^^^^^^^^^^^^^>more than their share of research or the endless cited lists of names ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^Sounds like a compliment.lojbab-- lojbab lojbab@lojban.orgBob LeChevalier, Founder, The Logical Language Group(Opinions are my own; I do not speak for the organization.)Artificial language Loglan/Lojban: http://www.lojban.org >Why is there something called Abstract Algebra when algebra already >is abstract without calling some particular kind of algebra, >Abstract?Abstract or modern algebra is called abstract or modern because it's moreabstract and modern than algebra. >Are any of those other algebras in reach for someone who currently >maintains some Intermediate Algebra knowledge, and who has only faint >memories of Calculus (three semesters) with which he struggled many >many years ago?Here have a go at it. Set theory will be of greater comfort thancalculus. 3.1.3. Definition. A group (G,*) is a nonempty set G together with a binary operation * on G such that the following conditions hold: (i) Closure: For all a,b in G, a*b, written ab, is in G. (ii) Associativity: For all a,b,c in G, a(bc) = (ab)c. (iii) Identity: There's an identity element e in G such that ea = a = ae, for all a in G. (iv) Inverses: For each a in G there's inverse element a^-1 in G with aa^-1 = e = a^-1 a. a(b + c) = ab + ac. (v) Additive identity: R contains an additive identity 0, such that for all a in R, a + 0 = a and 0 + a = a. (vi) Additive inverses: For each a in R, the equation a + x = 0 has a solution x in R, the additive inverse of a, denoted by -a. The commutative ring R is called a commutative ring with identity when R contains an identity element 1 /= 0 with for all a in R, 1a = a.Also, from your knowledge of algebra give some examples of a ring.Acknowledgment: www.math.niu.edu/~beachy/abstract_algebraAbstract Algebra, 2nd Edition 1996, by John A. Beachy---- In sci.math, Steven:> You have n towels of different sizes placed randomly into a single pile(> after doing your laundry). The one move you are allowed to make is to take a> stack of towel off the top of the pile and place this stack onto the bottom> of the pile. The goal is to get the towels into a pile from the largest> sized towel to the smallest sized towel. What is the least number of moves> required?> This is an odd way of doing laundry; are you sure you're notmisstating the problem? For example, one could hypothesizetwo piles A and B, with an initial state of all towels in arandom order on A, and operations allowing movement ofsome towels from the top of A to the bottom of A or B, and fromthe top of B to the bottom of A or B. One can therefore solvethe problem without much difficulty by simply doing a variantof a pick sort; rotate the towel in A just over the smallesttowel into A's bottom then pick up the single smallest towel andput it on B; then rotate the towel over the next-smallesttowel in A to A's bottom and pick that towel and shove it underthe one towel in B, and so on.A variant of the above would rotate towels from the bottomto the top or simply pulling out towels from the middleof A, which is probably the simplest solution anyway(although the least interesting).Still another variant would pull chunks out of the middle,using a single pile. This is indeed a pick sort and issolvable in at most N moves, and can take at least Nmoves if the pile starts up upside down. (Of coursein that case one might simply flip the entire pile instead.)Other users have commented at some length as to what onedoes with the pile after removal (if not flipped, theuser is simply rotating the accumulator pile; if flipped,the problem is solvable in at most N moves); this is probablywhat you had in mind.Still another slightly more interesting variant wouldallow a user to move single towels into 3 piles, withthe requirement that no towel ever be placed over anothersmaller towel (except during the initial removal from thedryer, perhaps). This is in fact a variant of the classicHanoi Towers puzzle and may require at most 2^N-1 moves.A variant of that would allow pickup of an ordered pilefrom the original stack, as opposed to single towels.Of course it would be a very inconvenient method ofdoing laundry... :-)Just remember the quarters for the washing machine.... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. You got a point there. I for my part hate the reverse numbering of thesigngroups (J. P. Olivier uses it).On the other hand, sometimes you got to think new.I suggest my alternativ notation by two good reasons:1) By the same reason that letters are used in algebra.2) I introduce the idea of counting the functions of the stems,referring to letters are thereby far more conveinent.For instance in Evans numbering the frequency of sign 2 is 19. In thealphabetic notation you got A frequency 19.Moreover my kind of notation forms minatures (thumbnails) of thesigngroups. Evans: A01) 18 1 13 12 2. Hagen: A01) ZUTBA(g).Hagen>> SNIP>> So if any other editor can offer me to get my discovery launched in a>> periodical, cosidering it must be a quick and honest co-operation, he>> shall be very much welcome to contact me about a publication of my>> alphabetic notation of the signs.May I give you an advice, Ole ?.. >If you are re-publishing somewhere the content of your book, could you>use the NUMBERING of the SIGNS adopted by almost every specialist>since Evans proposed it ?..>Your own alphabetic notation of the signs is a real problem for>those who want to understand your work !.. (I am talking by http://www.gvdnet.dk/~hagen/faistos.htm are 5 things you can find at www.geocities.com/erniespage... I am having trouble understanding this problem. How can a sequence ofintegers such as {1,2,3,1,2,3,.......} have 1, 2, and 3 as limits? Ithought that the limit of a sequence was unique. How does | x_n - L | < e,for n > N, work if the sequence just keeps cycling? Shouldn't the distancebe ever decreasing as n --> oo ? If say 3 is the limit, then| x_n - 3 | < e will hold only for 3, not 2 or 1.Maybe this is a bad example because it is monotonically increasing, but myconfusion is in any sequence which has repeating elements.TIALurch I am having trouble understanding this problem. How can a sequence of>integers such as {1,2,3,1,2,3,.......} have 1, 2, and 3 as limits? It doesn't. It has them as limit points, which is something different.> I>thought that the limit of a sequence was unique. When it exists, a limit exists. However, some people say that x_0 is alimit point of the sequence if there is a SUBsequence that convergesto x_0. So maybe that's what is confusing you? what I accept as reality. --- Calvin (Calvin and Hobbes) am having trouble understanding this problem. How can a sequence of> integers such as {1,2,3,1,2,3,.......} have 1, 2, and 3 as limits?It doesn't. It has 1, 2, and 3 as subsequential limits. >>Draw a path consisting of connected straight>>line-segments within the nonagon, so that each segment starts/ends on>>the vertexes of the 9-gon, and where each vertex is visited exactly>>once, and where the 1st and last segment are connected.>>Now the path must be such that there are a total of exactly 13 crossings >>inside the nonagon.>> This rule generates the right sequence:>> Starting at an arbitrary vertex, visit alternately the second and third vertex>> from the last, when you find that you would be revisiting a vertex, skip to the>> next empty one.--les ducs d'Enron!>http://members.tripod.com/~american_almanacas there are only three (well-known) star possibilites,>other than the simple enneagon:>skip every-other vertex (nine total crossings);>skip over two (three trigona, each line crosses four others);>skip over three (each line crosses six others, but>combinatorics isn't my specialty; I just drew them .-)And which of these gives exactly 13 crossings?-- Patrick Hamlyn posting from Perth, Western AustraliaWindsurfing capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com) > I don't grok the difficulty,> as there are only three (well-known) star possibilites,> other than the simple enneagon:> skip every-other vertex (nine total crossings);> skip over two (three trigona, each line crosses four others);> skip over three (each line crosses six others, but> combinatorics isn't my specialty; I just drew them .-)>Draw a path consisting of connected straight>line-segments within the nonagon, so that each segment starts/ends on>the vertexes of the 9-gon, and where each vertex is visited exactly>once, and where the 1st and last segment are connected.>[...] such that there are a total of exactly 13 crossings>inside the nonagon.This rule generates the right sequence:> Starting at an arbitrary vertex, visit alternately the second and > third vertex from the last, when you find that you would be > revisiting a vertex, skip to the next empty one.it's correct, there are 467 other permutations (with 1 fixed inplace) that generate 13 crossings, eg 135268497, 139752846, etc.most of which probably ignore your rule. Here are the numbersof permutations that give various crossing counts, per program: 9. 2 10. 54 11. 108 12. 252 13. 468 14. 450 15. 198 16. 576 17. 396 18. 168 19. 144 20. 162 21. 90 23. 18 24. 18 27. 2The program considers the 3106 permutations that start with1 and have no adjacencies (like 1 next to 9 or 2) but it doesnot exclude paths followed in reverse, so for example it distinguishes 2 9-crossing and 2 27-crossing paths even though they look the same when drawn.1. Is there an easy way (aside from enumeration) to showthat no 22- and 25-crossing paths exist?2. In a 23-crossing path I drew, I don't see offhand how to relate the permutation (148372596) to line intersections since some steps that skip 3 nodes cross 5 lines, and others cross 6.3. There is a straightforward way to tell from a permutation whether the path has an axis of symmetry (for example, 136824795 does, 135268497 doesn't).Is there an easy way to tell if all paths of a givencrossing count are symmetric?-jiw why fix top-posting? often, it just means that one hasto use the scroll-down, instead of immediately seeingwhat the reply is, thus engaging the conceptof context (or memory) in the OP and his reply. > skip every-other vertex (nine total crossings);>skip over two (three trigona, each line crosses four others);>skip over three (each line crosses six others, but>combinatorics isn't my specialty; I just drew them .-)And which of these gives exactly 13 crossings?--les ducs d'Enron!http://members.tripod.com/~american_almanac In sci.math, Colin someone can settle this argument for me. I've been> wondering if your chances of winning the lottery increase over time if> you keep the same numbers for each draw rather than picking random> numbers.My thinking is this -If you were to flip a coin 10 times, you should end up with roughly 5> heads and 5 tails in the results. Now suppose that your strategy is> to always call heads. Using this logic you should win 50% of the> time.If on the other hand your strategy was to randomly call heads or> tails, you introduce the possibility of always calling the wrong> result. So over 10 flips its possible to get a score of 0 correct> calls. Calling the same side should discount this posibility and> should therefore increase your chance of winning over time.If the coin is perfect it makes no difference. It is also possibleto get all tails if one always chooses heads, for example.The issue is independent versus dependent events. Flipping acoin is an independent event. Drawing a ball without replacementfrom a bowl is a dependent event, as subsequent draws aredependent on that draw.Going back to the lottery, my friends argument> is that your chances of winning will always be> 13 million to 1 no matter what you do.Depends on the lottery setup. Of course I'm assuming 13 millionis a C(n,r) number of some sort; the California SuperLotto Plusfor example uses 5 balls picked without replacement numbered1 to 47 (with inexact ordering) and 1 mega ball numbered 1 to 27,yielding C(47,5) * 27 = 41416353 combinations. The lotteryonly becomes interesting (E(play) > 1) if the jackpotgoes over this number when measured in dollars (at $1 a play).(As of this writing it's $52M. Play responsibly, people. :-) )It turns out 6 balls numbered 1 to 34 picked without replacementand without regard to order gives one 1344904 possibilities.In a fair number of lotteries (California among them) one isrequired to share jackpots; this skews the expected resultsslightly, as everyone has their system, usually basedon dates of birth (translation: numbers from 1-30). Thereforethe numbers 1-30 are disproportionately represented in thetickets prior to the draw. However, it does not affect theprobability of one winning a single draw, merely the expected jackpot.If a ball comes up 3 times in 3 successive draws it is replaced;assuming perfect balls these replacements are random and alsodo not affect the probability. Of course balls aren't perfect. :-)> I accept that, but what happens over time? My view is that if you did> the lottery 13 million times with the same numbers, then in a perfect> world you should win once.If the lotto is done once a day 13 million days = about35 1/2 millennia. Good luck. (You'll need it.) Nor areyou guaranteed to win.The probability of you winning a single draw is P = 1/13M.The probability of you not winning that draw is 1 - P.The probability of you not winning N draws is (1 - P)^N.The probaiblity of you winning at least once in N draws is1 - (1 - P)^N.If we pick P as 1/13M and N as 13M, this value turns outto be 0.6321 (63.21%). It's certainly not certainty,mostly because lotto draws can come up more than once inthat sequence of 13 million. If you're real lucky yournumbers come up more than once.It doesn't matter whether you change your numbers or not.> But if you choose different numbers for> each draw then you are effectively resetting your chances of winning> over time in every draw...Is there any truth in this?No, not if the system is honest and truly random. Draws areindependent events from each other (the individual ballsin each draw are dependent events, requiring some care, butthat's where C(n,r) comes in, as I did above).TIA,Colin-- #191, ewill3@earthlink.netIt's still legal to go .sigless. >Like that to a lesser degree. Mine mapped to another unique integer,> like>>yours not a prime, but rather lower. My instructor (of course that was> in>>uni) still gave me good marks, so no scars today. Marks? Marks can leave scars, depending on the implement used to make> the mark. Personally I give grades and never mark my students. Good marks leave no marring? Or at least no scarring. Sometimes mediocre marks lead to scaring, and bad> marks can lead to scarring. In some cases, I'm sure I left more marks onmy> instructors than they left on me. And I didn't even go to uni in Germany.> OTOH, marrying can lead to marring, and even scarring. Mine has only left> good marks so far. No right marks of any kind.>Mediocre marks lead to marring, Sucky scores lead to scarring.It's both obvious *and* alliterative. > I don't know what you are talking about, but if HH is a computer> program, then we can prove that it isn't equal to Halt.>//-----------//A 01L B//A 11R B//B 01R C//B 10L E//C 00R D//C 10L A//D 01L A//D 10R D//E 01L HALT//E 10L C//-------------This is HH, I'm pretty sure its the bona fide Halt algorithm :)This is actually TM number 9320303603822 where the transition sequenceabove is mapped to the number, where state HALT is 0, L is 0, R is 1, and thelater transitions are more significant, all the smaller size TMs with under 5 stateswere added to that number. So 01LB is (1*12+0*6+2) * 24^0 = 14.i.e there are (2*2*6)^10 5 state TMs.You can give it any parameter in unary, and the claim is it will output thevalue for Halt for that number by terminating to the right with either 01 or 00.Herc |-|erc says...> I don't know what you are talking about, but if HH is a computer>> program, then we can prove that it isn't equal to Halt.>>//----------->//A 01L B>//A 11R B>//B 01R C>//B 10L E>//C 00R D>//C 10L A>//D 01L A>//D 10R D>//E 01L HALT>//E 10L C>//------------->This is HH, I'm pretty sure its the bona fide Halt algorithm :)This is actually TM number 9320303603822 where the transition sequence>above is mapped to the number, where state HALT is 0, L is 0, R is 1, and the>later transitions are more significant, all the smaller size TMs with under 5>states>were added to that number. So 01LB is (1*12+0*6+2) * 24^0 = 14.>i.e there are (2*2*6)^10 5 state TMs.>You can give it any parameter in unary, and the claim is it will output the>value for Halt for that number by terminating to the right with either 01 or>00.Well, there is an easy proof that the claim is false. Remember, by definitionof Halt, if n is the Godel number of a program, then Halt(n) = 0 if that programhalts on input n, and Halt(n) = 1, if it never halts on input n.So, given your program HH, we define a new function g(n) by g(n) = if HH(n) = 0, then loop forever else output 0We compute the Godel number k of g, and consider the computationg(k). By definition of Halt, if g(k) halts, then Halt(k) = 0, andotherwise, Halt(k) = 1.So, consider the two possibilities: 1. g(k) halts, and Halt(k) = 0. But by definition of g above, we know that g(k) only halts if HH(k) is nonzero. So in this case, it is clear that Halt(k) != HH(k), so HH does not compute Halt. 2. g(k) never halts, and Halt(k) = 1. Buy by definition of g, we know that the only way for g(k) to fail to halt is if HH(k)=0. So Halt(k) = 1 and HH(k) = 0, so again HH does not correctly compute Halt.--Daryl McCulloughIthaca, NY Is there an algorithm or method to find the minimal number of simplexes in a triangulation of the n-cube (by n dimensional simplexes) ? I should say,it's not a generalization of Morley's theorem, buta *completion*, using all possible combinationsof interior, exterior, supplementary & cyclic (?) anglesof the trigon. has anyone shown a similar propertyof the tetrahedron? > http://planetmath.org/encyclopedia/ CorollaryOfMorleysTheorem.html--les ducs d'Enron!http://members.tripod.com/~american_almanac Brian Quincy Hutchings> this is covered with a fabulous generalization> in _Complex Geometry_ by a Chinese guy,> an Numbers and Geometry_Liang-shin Hahnhttps://enterprise.maa.org/ecomtpro/timssnet/books/ Spectrum.cfmMorley's figure appears on the front cover, and the author also bumps offthis problem:http://planetmath.org/encyclopedia/ NapoleonsTheorem.htmlwhich I worked out only yesterday. There's not much new under the sun :)LH > It occurred to me a while back that the claims against factors I'd> used for certain factorizations would apply to *any* non unit factor,> but I've hesitated in putting that forward, but now I might as well. Consider y^3 + 3y - 2 and its three roots. I've been arguing that> only one of those roots is not coprime to 2, but now imagine that as> others have argued, they each have some algebraic integer in common> with 2, and let's call those factors f_1, f_2, and f_3.Never mind the factors. Just consider the roots: y_1, y_2 and y_3.[snip irrelevant jabber about unnecessary quantities: f_1, etc.]Each and every root of the polynomial y^3+3y-2 divides 2 in the ring ofalgebraic integers. None of the roots are units, hence none are coprime to2. In fact,y_1(y_1^2+3) = 2, so 2/y_1 is an algebraic integer.y_2(y_2^2+3) = 2, so 2/y_2 is an algebraic integer.y_3(y_3^2+3) = 2, so 2/y_3 is an algebraic integer.Therefore each root divides 2 in the ring of algebraic integers so *NONE*,count 'em *NONE* of the roots of y^3+3y-2 are coprime to 2.QED(Better luck next time!)--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com > It occurred to me a while back that the claims against factors I'd> used for certain factorizations would apply to *any* non unit factor,> but I've hesitated in putting that forward, but now I might as well.Consider y^3 + 3y - 2 and its three roots. I've been arguing that> only one of those roots is not coprime to 2, but now imagine that as> others have argued, they each have some algebraic integer in common> with 2, and let's call those factors f_1, f_2, and f_3.Now given a root r_1, you have that r_1/f_1 should be a unit.Here's that word *should* again. Prove it or don't use it.Generalizing, use r = fw, for a root, and substitute to get f^3 w^3 + 3fw - 2 = 0,Which is what type of what?so you can divide off f, and get f^2 w^3 + 3w - 2/f.Which gives you a different type of what?That is non-monic and not in Q as f is not rational.A non-monic WHAT?However, no amount of algebraic manipulations can turn f^2 w^3 + 3w - 2/finto a polynomial with integer coefficients.It's a polynomial? Why didn't you say so? Where'd w come from? For that matter, if you start with a non-monic, what's the big deal about ending up with a non-monic?For one thing, for any f, like f_1, you need f_2 or f_3 as well to get> back to an integer.Basically there is NO way to find a polynomial with integer> coefficients let alone a monic one with integer coefficients to prove> that one of the roots of f^2 w^3 + 3w - 2/fis a unit.What does this have to do with the original polynomial?And in fact, as what should be the unit root is NOT expressible as the> root of a monic polynomial with integer coefficients, it is not an> algebraic integer.You have NOT proven this statement. You will have to back up a lot before this is accepted. You've not even shown an attempt to construct it's monic polynomial... or f for that matter.Several posters have been getting away with casting doubt on values> for f that I've used that follow from advanced factorization> techniques, but they've managed to hide the fact that if you divide> off *any* non unit factor of 2 from the roots you run into the same> thing.That is, their own argument traps them.Do you even understand the arguments?Do you even respond to the arguments?Here's one for en&lr=&ie=UTF-8&selm=3f1af19c%241_2%40newsfeed&rnum=52The only conclusion that can be drawn is that what should be algebraic> integer units are not, and now the challenge that has been directed my> way, can go elsewhere as those who wish to dispute that conclusion can> do a simple thing:Give the monic polynomial with integer coefficients. You have changed the topic of discussion with the above work and don't appear to realize it.By their own claims at least you know the last coefficient should be 1> or -1.Given the polynomial y^3 + 3y - 2, one of these people should be able> to give a monic polynomial with integer coefficients that define what> should be units when you divide off factors in common with 2 from the> roots, if they're telling the truth.What makes you think the results will be units? 2/sqrt(2) is not a unit, yet sqrt(2) is a factor of 2.-- Will Twentyman > It occurred to me a while back that the claims against factors I'd> used for certain factorizations would apply to *any* non unit factor,> but I've hesitated in putting that forward, but now I might as well.Consider y^3 + 3y - 2 and its three roots. I've been arguing that> only one of those roots is not coprime to 2, but now imagine that as> others have argued, they each have some algebraic integer in common> with 2, and let's call those factors f_1, f_2, and f_3.Now given a root r_1, you have that r_1/f_1 should be a unit.> No. The roots of the polynomial y^3 + 3y - 2 are all *divisors* of 2in the ring of algebraic integers. That is a trivial observation fromthe fact that the product y1 y2 y3 is equal to 2.Here, let me show you: Let y1, y2, y3 be the roots of the polynomial. Then, the polynomial factors as follows: y^3 + 3y - 2 = (y - y1)(y - y2)(y - y3) = y^3 - (y1 + y2 + y3) y^2 + (y1 y2 + y2 y3 + y3 y1) y - y1 y2 y3.Equating equal powers of y (since this is a factorization as apolynomial in y), you get this: 1 = 1 0 = y1 + y2 + y3 3 = y1 y2 + y2 y3 + y3 y1 2 = y1 y2 y3Thus, the product of the roots is 2. In other words, 2 is a multipleof each of the three roots.Your claim (that r1 an algebraic integer that is a common divisor ofa root (y1, say) and 2, should lead to y1/r1 = a unit) does not follow,for the simple reason that y1^(1/2), picking either square root of y1,divides both y1 and 2 (since y1 itself divides 2), and is an algebraicinteger, since the ring of algebraic integers is closed under solutionof monic polynomials, including, for instance, P(x) = x^n - A, where Ais an arbitrary algebraic integer.> Generalizing, use r = fw, for a root, and substitute to get f^3 w^3 + 3fw - 2 = 0,> If I read this correctly, you're taking an arbitrary number f,and considering the polynomial (in x): f^3 x^3 + 3 f x - 2Is that correct? So, the roots of this polynomial are the quotients y1/f, y2/f, y3/f.So what? If f is an arbitrary algebraic number, the roots are algebraicnumbers. If f is an algebraic integer, again the roots are going to bealgebraic numbers. They won't be algebraic integers, unless f is adivisor of the y's. However, in that case, if f divides all the y's,we see that f^3 divides 2, and f^2 divides 3 (see the expansion of theproduct of monomials (y - y1)(y - y2)(y - y3) to see that 3 is equal tothe elementary symmetric polynomial of degree 2 in the yi's).> so you can divide off f, and get f^2 w^3 + 3w - 2/f.That is non-monic and not in Q as f is not rational.So you say, now. That wasn't in your original descriptionHowever, no amount of algebraic manipulations can turn f^2 w^3 + 3w - 2/finto a polynomial with integer coefficients.> However, if f divides all three of the yi's, then the whole polynomialcan be divided by f^3 *in the ring of algebraic integers*, yieldinga monic polynomial with coefficients in A, the ring of algebraicintegers. Standard techniques (which I would need to consult closelyto be able to apply correctly, not being an algebraist) will producea monic polynomial of somewhat higher degree, but finally with integercoefficients, for which w would be a root.> For one thing, for any f, like f_1, you need f_2 or f_3 as well to get> back to an integer.Basically there is NO way to find a polynomial with integer> coefficients let alone a monic one with integer coefficients to prove> that one of the roots of f^2 w^3 + 3w - 2/fis a unit.Several questions present themselves: What sort of number is f? If f is not a unit in the ring of algebraic integers, there is no general guarantee that the roots are even algebraic integers, so you must *not* be talking about units in the ring of algebraic integers. So, what ring are you referring to?And in fact, as what should be the unit root is NOT expressible as the> root of a monic polynomial with integer coefficients, it is not an> algebraic integer.> WHAT UNIT ROOT?I'll relent a little, and suppose that you really meant this number fto be the quotient yi/ri, where ri is a factor in common between yi and2. I'll then note that one *could* take fi = yi, to get w = 1.I'll bet that *even you* can produce a monic polynomial with integralcoefficients that has x=1 as a root.> Several posters have been getting away with casting doubt on values> for f that I've used that follow from advanced factorization> techniques, but they've managed to hide the fact that if you divide> off *any* non unit factor of 2 from the roots you run into the same> thing.> What same thing?> That is, their own argument traps them.The only conclusion that can be drawn is that what should be algebraic> integer units are not, and now the challenge that has been directed my> way, can go elsewhere as those who wish to dispute that conclusion can> do a simple thing:Give the monic polynomial with integer coefficients. > See below.> By their own claims at least you know the last coefficient should be 1> or -1.> See below.> Given the polynomial y^3 + 3y - 2, one of these people should be able> to give a monic polynomial with integer coefficients that define what> should be units when you divide off factors in common with 2 from the> roots, if they're telling the truth.> Here we go: let f = y1. Then you get q(w)= y1^3 w^3 + 3 y1 w - 2.But this polynomial is not irreducible: Define: u = sqrt(2) + 1; v = sqrt(2) - 1. q3 = 3(v^(1/3) - u^(1/3)) + 2; q2 = 0 q1 = 3(v^(1/3) - u^(1/3)) q0 = -2.Then q(x) = q3 x^3 + q2 x^2 + q1 x + q0.Further, since q(1) = 0, (x-1) divides q(x): Define r2 = 3(v^(1/3) - u^(1/3)) + 2 r1 = 3(v^(1/3) - u^(1/3)) + 2 r0 = 2.Then: r(x) = r2 x^2 + r1 x + r0and q(x) = r(x)*(x-1).So, yes, I can find a polynomial with integer coefficients that has thatso-mysterious unit root: x - 1.James HarrisDale. [...] Basically there is NO way to find a polynomial with integer> coefficients let alone a monic one with integer coefficients to prove> that one of the roots of f^2 w^3 + 3w - 2/f is a unit [...]> no amount of algebraic manipulations can turn [it] into a polynomial> with integer coefficients [...]w is by your construction a unit algebraic integer. As you know,this is true iff there EXISTS a polynomial of a certain form havingw as a root. This does not imply that every polynomial having w asa root must necessarily be somehow transformable into the required(minimum) polynomial by some amount of algebraic manipulations(whatever that might mean). Indeed, as you noticed, it is trivialto derive all sorts of other random polynomials having w as a root,most all of which prove to be completely worthless in this regard.-Bill Dubuque > ... a buncha stuff ... > Here we go: let f = y1. Then you get> I neglected to state *which* root of x^3 + 3x - 2 I took here, so thislittle note is intended to provide that value. Let: u = sqrt(2) + 1 v = sqrt(2) - 1then y1 = u^(1/3) - v^(1/3).The following formulas were derived using that value (together withMacsyma for the symbolic procesing + a little intelligence in doingsubstitutions I couldn't figure just how to convince Macsyma to do)For the record, the remaining two roots: y2 = z u^(1/3) - zbar v^(1/3) y3 = zbar u^(1/3) - z v^(1/3)give rise to different factorizations of their respective polynomials(where x is replaced by yi x, and the factor (x - 1) exposed as I didbelow). All of the coefficients of the factorization I've derived areelements of the splitting field, and so when the Galois group acts onthe splitting field, the factorization determined by one choice of yigets mapped to the factorization determined by another root, based onthe root that yi gets mapped to. Briefly, the formulas given for thecoefficients can be written in terms of the root y1. Replacing y1 foranother value, y2 or y3, yields the factorization for the other case.(Please pardon my awkward wording; I'm trying to make an even margin)> q(w)= y1^3 w^3 + 3 y1 w - 2.But this polynomial is not irreducible: Define: u = sqrt(2) + 1; v = sqrt(2) - 1. q3 = 3(v^(1/3) - u^(1/3)) + 2; q2 = 0 q1 = 3(v^(1/3) - u^(1/3)) q0 = -2.Then q(x) = q3 x^3 + q2 x^2 + q1 x + q0.Further, since q(1) = 0, (x-1) divides q(x): Define r2 = 3(v^(1/3) - u^(1/3)) + 2 r1 = 3(v^(1/3) - u^(1/3)) + 2 r0 = 2.Then: r(x) = r2 x^2 + r1 x + r0and q(x) = r(x)*(x-1).So, yes, I can find a polynomial with integer coefficients that has that> so-mysterious unit root: x - 1.> James Harris> Dale.> According to KASH, the following factorization works for any of the yi: Let: q(x) = yi^3 x^3 + 3 yi x - 2 Define: r(x) = yi^3 x^2 + yi^3 x - 2 Then q(x) = r(x) (x-1).Macsyma agrees: r(x)*(x-1) = yi^3 x^3 + (2 - yi^3) x - 2recall q(x) = yi^3 x^3 + 3 yi x - 2so we see q(x) - r(x)*(x-1) = (yi^3 + 3 yi - 2) xbut yi is a root of x^3 + 3 x - 2, so that coefficient is zero.Thus q(x) = r(x)*(x-1), for any of the yi's.Dale > It occurred to me a while back that the claims against factors I'd> used for certain factorizations would apply to *any* non unit factor,> but I've hesitated in putting that forward, but now I might as well.Consider y^3 + 3y - 2 and its three roots. I've been arguing that> only one of those roots is not coprime to 2, but now imagine that as> others have argued, they each have some algebraic integer in common> with 2, and let's call those factors f_1, f_2, and f_3.Now given a root r_1, you have that r_1/f_1 should be a unit.Generalizing, use r = fw, for a root, and substitute to get f^3 w^3 + 3fw - 2 = 0,so you can divide off f, and get f^2 w^3 + 3w - 2/f.That is non-monic and not in Q as f is not rational.However, no amount of algebraic manipulations can turn f^2 w^3 + 3w - 2/finto a polynomial with integer coefficients.> Someone else pointed this out also. Why not letf1 = r1, f2 = r2, f3 = r3? In that case, w1 = r1/f1 = 1, w2 = r2/f2 = 1, and w3 = r3/f3 = 1.The polynomial you are after is then S(w) = (w - 1)^3.It has integer coefficients.The leading coefficient is 1.The constant term is -1.All of its roots are units.All of its roots are of the form ri/fi,where fi is a divisor of 2.> For one thing, for any f, like f_1, you need f_2 or f_3 as well to get> back to an integer.Basically there is NO way to find a polynomial with integer> coefficients let alone a monic one with integer coefficients to prove> that one of the roots of f^2 w^3 + 3w - 2/fis a unit.> With the choice described above, this equationis r1^2 * w^3 + 3*w - 2/r1,which is a polynomial in w, but it does not have integer coefficients. But what you say is wrong.Here w = 1. It *IS* possible to find a monic polynomialwith integer coefficients such that one of the rootsof Q(w) = f^2 w^3 + 3w - 2/f is a unit. Why? Because w = 1 is one of the rootsof Q(w), and S(w) = (w - 1)^3 fits exactly the description of the polynomial that you saycannot be found.> And in fact, as what should be the unit root is NOT expressible as the> root of a monic polynomial with integer coefficients, it is not an> algebraic integer.> You are confused. See above.> Several posters have been getting away with casting doubt on values> for f that I've used that follow from advanced factorization> techniques, but they've managed to hide the fact that if you divide> off *any* non unit factor of 2 from the roots you run into the same> thing.That is, their own argument traps them.> Not at all.> The only conclusion that can be drawn is that what should be algebraic> integer units are not, and now the challenge that has been directed my> way, can go elsewhere as those who wish to dispute that conclusion can> do a simple thing:Give the monic polynomial with integer coefficients. > S(w) = (w - 1)^3.> By their own claims at least you know the last coefficient should be 1> or -1.> Check.> Given the polynomial y^3 + 3y - 2, one of these people should be able> to give a monic polynomial with integer coefficients that define what> should be units when you divide off factors in common with 2 from the> roots, if they're telling the truth.> Seems straightforward: see above. Nora B.James Harris >It occurred to me a while back that the claims against factors I'd>used for certain factorizations would apply to *any* non unit factor,>but I've hesitated in putting that forward, but now I might as well.Consider y^3 + 3y - 2 and its three roots. I've been arguing that>only one of those roots is not coprime to 2, but now imagine that as>others have argued, they each have some algebraic integer in common>with 2, and let's call those factors f_1, f_2, and f_3.In fact, f1, f2, and f3 are associates of the roots, because EACH ROOTIS A NON-UNIT DIVISOR OF 2. As is obvious from the fact that theproduct of all 3 equals 2, and the polynomial is irreducible withconstant term different from 1 or -1.>Now given a root r_1, you have that r_1/f_1 should be a unit.That's correct. For THIS polynomial>Generalizing, use r = fw, for a root, and substitute to get f^3 w^3 + 3fw - 2 = 0,so you can divide off f, and get f^2 w^3 + 3w - 2/f.That is non-monic and not in Q as f is not rational.This does not follow; for example, f=1 or f=-1 would be valid.>However, no amount of algebraic manipulations can turn f^2 w^3 + 3w - 2/finto a polynomial with integer coefficients.This again is false. Let x^n + ... + a1*x + a_0 be the minimalirreducible polynomial of f. Since f is a unit, it follows that a_0 =1 or a_0 = -1. Let f=f_1, ..., f_n be all the roots of this polynomialNote that f_1*...*f_n = -a_0, and f_1^2*...*f_n^2 = 1.So now simply take the product of all the polynomialsf^2 w^3 + 3w - 2/ff_2^2 w^3 + 3w - 2/f_2...f_n^2 w^3 + 3w - 2/f_nmultiply them together, and lo and behold, get a monic polynomial withinteger coefficients that has the element you want as a root.>For one thing, for any f, like f_1, you need f_2 or f_3 as well to get>back to an integer.Basically there is NO way to find a polynomial with integer>coefficients let alone a monic one with integer coefficients to prove>that one of the roots of f^2 w^3 + 3w - 2/fis a unit.That's because the root there is not supposed to be a unit, it'ssupposed to be w, which is the NON-unit common divisor of r and 2,remember? As it happens, it is just an associate of r. The one that issupposed to be a unit is f, not r. [.snip.]>Given the polynomial y^3 + 3y - 2, one of these people should be able>to give a monic polynomial with integer coefficients that define what>should be units when you divide off factors in common with 2 from the>roots, if they're telling the truth.Sigh. Let r_1,r_2,r_3 be the three roots. The unit you want iscalled 1, the monic polynomial with integer coefficients thatdefines it is reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan wish to critique the dumbasses that respond to my trolls.Dumbass Norba,Dumbass Arturblow,Dumbass CBond,Dumbass Ullruch from the University of whatever,etcThey all suck horse dick >Let g be in G and choose e such that ge = g. Now Let h be any element>of G and let h = cg. Then he = (cg)e = c(ge) = cg = h so e is a right>identity. Now, for any g in G choose g' so that gg' = e and choose h so that>(g'g)h = e note g'g = g'(gg')g and so e = (g'g)h = g'(gg')gh =>(g'g)(g'g)h = (g'g)e = g'g. That is, g'g = gg' = e.Finally, with gg' = e, eg = (gg')g = g(g'g) = ge = g so e is two sided.Is it more straightforward to construct a left and right identityand then write e = fe = f? To complete, find gG = e, Hg = e,G = HgG = H, so any inverse is double-sided.-- A. Cairns (cola@spiny.at.org) It took me a while, but I've figured a few things out, and I think itcovers why a few people are so dedicated in replying to my posts:They are dumbasses and they don't know a troll when they see it.I speak of Norba, Arturo, Ulicsh, Cbond, Fishfry, and a few otherdumbassesOne wonders how much I have to ing troll till they stop respondingto me? >A recent report shows high-pitched rapidly repeated subliminal>messages in a large number of television broadcasts. For the most>part the messages are of a political nature, insisting such things as>Bush is good and The war in Iraq is going well. (...)>following message was deciphered repeatedly by audio analysts in>broadcasts from around the nation:>Cantor knows all. Cantor cannot be disproved. Bush's economic>platform is based on Cantor's theorems and is therefor invincible. >Cantor is omniscient. Bush must be reelected to save the nation from>Terrorism>As most people (I think) who read the above post will have seen, thereis another post in this NG that contains interesting details about(supposed ?) arguments of George W. B+++ where he 'uses' Cantor'sdiscoveries about infinite sets, quite specific. For those who didn't,this is the subject line of it:Bush unveils new economic plans based on the plans based onthe Theorems of CantorI wander about the arguments it contains: somebody able to conceivethem must at least have understood a substantial (elementary) part ofCantor's theory of infinite sets, even if the arguments are specious,supposing it's serious. So my question is: do people quoting thesearguments or just having read about here think that the US presidenthas that minimal understanding ? If yes, do you (=these people) thinkhe has actually told these arg's and if yes, if he is serious withthem or what is he trying to do mentioning them ? Or is it possiblethat the arg's actually come from people with deeper math. knowledgeand understanding than just 95 % of western people ? In the last case,they must be ironic, perhaps showing by putting to the extremewhat Bush is trying to do or make people believe. How about this last? >>A recent report shows high-pitched rapidly repeated subliminal>>messages in a large number of television broadcasts. For the most>>part the messages are of a political nature, insisting such things as>>Bush is good and The war in Iraq is going well. But what is>>puzzling some political scientists is the occasional subliminal>>message regarding the credibility of a certain mathematician. The>>following message was deciphered repeatedly by audio analysts in>>broadcasts from around the nation:> And an even more recent report shows that text-based usenet groupscontain almost un-noticable (2-point or less) messages hidden inup to 10% of the messages containing similar propaganda.As an example, this message contains the subliminal messageThis is absurd - disregard it.See - it works.Martin Cohen Placing n points in the unit circle withsmallest pairwise distance d will alwayssatisfy d <= Delta(n)where Delta(n) is chosen minimal with this property(*).In a discussion in de.rec.denksport Alfred Pfeifferremarked that Delta(6) = Delta(7)and we thought about Delta(n) = Delta(n+k). (1)On a large scale, Delta(n) will be monotonicallydecreasing. But locally it may stay constant fora while. Not only for k=1 steps but maybe for many more, depending on n. So I proposed theAlfred-Pfeiffer function APf(n) defined by k = APf(n)is the largest nonnegative integer such that (1)holds. We have APf(6) = 1.When APf(n) is large, then we may interpret that asa necessarily blown-up situation, whereas APf(n+1)= APf(n)-1 demonstrates, that there is less room now.Question #1: Is Apf(n) bounded?Question #2: how about online references for Delta(n)?Hmmm... while rereading my posting, I see something,which answers Q1 with No. Whenever the optimal placementallows for regular 6-gons, then we may place anotherpoint in the middle - just as in the case n=6. So I am quitesure, that APf(n) is increasing beyond any linit.Nonetheless I will press the Send-Button NOW.(*) I like math and I like English. But sometimesit seems to me they don't like each other, especiallywhen I tell them to be friends. So I'd be happy forany corrections of what I tried to explain as care-fully as I could (within 3 minutes or so).Rainer Rosenthalr.rosenthal@web.de >Placing n points in the unit circle with>smallest pairwise distance d will always>satisfy d <= Delta(n)where Delta(n) is chosen minimal with this >property(*).In a discussion in de.rec.denksport Alfred Pfeiffer>remarked that Delta(6) = Delta(7)>I do not quite follow. Does not Delta(n) = the length of the side of a regular n-gon inscribed in the circle? If so, Delta(n) is strictly decreasing. Or am I missing something obvious?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu > Placing n points in the unit circle with> smallest pairwise distance d will always> satisfy d <= Delta(n)where Delta(n) is chosen minimal with this > property(*).I believe there's more literature on this when it's framed ascircle packing -- e.g.,http://mathworld.wolfram.com/CirclePacking.html-- | Jim Ferry | Center for Simulation |+------------------------------------+ of Advanced Rockets || http://www.uiuc.edu/ph/www/jferry/ +------------------------+| jferry@[delete_this]uiuc.edu | University of Illinois | Placing n points in the unit circle with>smallest pairwise distance d will always>satisfy d <= Delta(n)where Delta(n) is chosen minimal with this >property(*). Delta(6) = Delta(7) > Does not Delta(n) = the length of the side of a > regular n-gon inscribed in the circle? For n=6 we have Delta(6) = 1, i.e. the length ofthe regular 6-gon inscribed in the unit circle.Yes.But for n=7 we can just place the seventh point inthe center ... Oh wait maybe my fault was ... ermI think I should have said:>Placing n points in the unit *disc* with ...Sorry. Just as I tried to say: Could someone please beso kind and formulte my quetion correctly? But possiblydo not have any certain view regarding APf(n), becauseit's far too late and those 6-gons, I was talking about,are most probably nonsense. Large sets of points willform a lot of equilateral triangles but not hexagons.Well, but I still think this to be an interestingquestion. (And yes - I still work on n-prime sets.)Rainer Rosenthalr.rosenthal@web.de >I posted a few weeks back and am still puzzled. I want to confirm my>suspicion>that the following system has a unique nonzero solution (or find a>counterexample). Let A be n by n matrix with positive entries and>largest magnitude eigenvalue > 1 .>Show that there is exactly one nonzero x=(x1,x2,...,xn) with all>entries>between 0 and 1 such that: >1 - e^{-(Ax)_1} = x1 1 - e^{-(Ax)_2} = x2 >.>.>.>1 - e^{-(Ax)_n} = xn where (Ax)_k is kth entry of vector Ax, >(Ax)_k = A_{k1} x1 + A_{k2} x2 + ... + A_{kn} xn Write the system of equations as F(x) = x.Let v be the eigenvector for eigenvalue lambda > 1. Ifx >= epsilon v, then A x >= epsilon lambda v, and all(F(x))_k >= 1 - exp(-lambda epsilon v_k) = lambda epsilon v_k + O(epsilon^2)Thus for sufficiently small epsilon, if x >= epsilon v then F(x) >= epsilon v. The compact set S = {x: epsilon v_j <= x_j <= 1 for all j}is mapped into itself by F. Since S is homeomorphic to a ball,by Brouwer's Fixed Point Theorem there is at least one fixed point in S. I don't know about uniqueness, though.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > I am trying to create a wind study and one of the things involved is> computing the difference between two given angles in degrees. The> problem is that if for example the degrees are 350 and 10, the> difference should be 20, not 340. I need to be able to do this using> just standard functions and nothing fancy, because i need to be able> to put this into a MS Excel worksheet. Trig functions and i think the> floor/ceiling function are fine. John > I am trying to create a wind study and one of the things involved is> computing the difference between two given angles in degrees. The> problem is that if for example the degrees are 350 and 10, the> difference should be 20, not 340. I need to be able to do this using> just standard functions and nothing fancy, because i need to be able> to put this into a MS Excel worksheet. Trig functions and i think the> floor/ceiling function are fine.>This should do the trick for angles (expressed as degrees)sitting in cells A1 and A2: = ACOS( COS( PI()*(A1-B1)/180 ) )*180/PI()Dirk Vdm : I am trying to create a wind study and one of the things involved is: computing the difference between two given angles in degrees. The: problem is that if for example the degrees are 350 and 10, the: difference should be 20, not 340. I need to be able to do this using: just standard functions and nothing fancy, because i need to be able: to put this into a MS Excel worksheet. Trig functions and i think the: floor/ceiling function are fine.You could try=(A1-B1)-ROUND((A1-B1)/360;0)*360This even preserves the sign of the difference.- Niilo > : I am trying to create a wind study and one of the things involved is> : computing the difference between two given angles in degrees. The> : problem is that if for example the degrees are 350 and 10, the> : difference should be 20, not 340. I need to be able to do this using> : just standard functions and nothing fancy, because i need to be able> : to put this into a MS Excel worksheet. Trig functions and i think the> : floor/ceiling function are fine. You could try =(A1-B1)-ROUND((A1-B1)/360;0)*360 This even preserves the sign of the difference.*Much* cheeper than mine. Nice :-)My version of Excel uses a colon in the ROUND function.To always have a positive, take ABS( ) of the expression.Dirk Vdm Let a anb b be two angles between 0 and 360(assume a >= b). Then if a - b isno more than 180 your answer is a - b. If a - b > 180 then your answer is360 - (a - b). In your example the answer is 360 - (350 - 10) = 360 - 340 =20. If the restriction a anb b are two angles between 0 and 360 is liftedthen the answer would be calculated as follows: 1st calculate (a - b) -360*n where n is the integer such that 0<= (a - b) - 360*n <= 360. If thisnumber is less than or = to180 then this number is your answer. If thisnumber is more than 180 then simply subtract this number from 360 and youhave your answer.> I am trying to create a wind study and one of the things involved is> computing the difference between two given angles in degrees. The> problem is that if for example the degrees are 350 and 10, the> difference should be 20, not 340. I need to be able to do this using> just standard functions and nothing fancy, because i need to be able> to put this into a MS Excel worksheet. Trig functions and i think the> floor/ceiling function are fine. John |Regarding the recent thread about the Peano postulates and their|potential to establish a rich axiomatic foundation for number|theory:||Are they really insufficient? I was surprised at that suggestion.It depends upon what you're asking your foundations to do. If you just needrich, they are rich. One can make a research career in mathematics doingnothing but deriving consequences from them.|If Peano axioms don't do the job then, what currently does do|the job? What is the typical way in which mathematicians|formally prove theorems about number theory? Or what is the|preferred method of formalization?We discussed two functions of an axiom system, which could be calledformal and informal systems. A formal system is designed to make itpossible at least in principle to write a computer program which wouldcheck proofs written in the formal system for correctness. On the otherhand a lot of the axiom systems in mathematics are simply lists ofpropositions which are to be assumed in some context. You are assumed tobe able to understand what the propositions mean. (If you can't understandwhat an axiom means, then it doesn't do you much good to try to assume it'strue, does it?) One can be doing both of these activities at the same time,producing a formal system where you also have an interpretation for whatthe terms being manipulated are supposed to mean, but they are twodistinct services being supplied by the axioms.Number theorists typically are not especially familiar with formalizationsof any kind. We learn some basic principles in an informal way and go fromthere. Why not consider that adequate? On the informal side, we havePeano's axioms as originally stated:|The Peano postulates could mean several things. They could refer to|Peano's original postulates, which by modern standards are slightly vague.Are they? I thought Peano had pretty definitely used a set of axioms whichwere informal in that he didn't give deductive rules, and because in orderto use the axiom of induction, you have to know what a property is. Onehas a function, the successor function S(x) which is an undefined term,which is assumed to satisfy the axioms: 1. There is a unique element 0 which is not a successor. 2. Two elements having the same successor are equal. 3. If P is a property which holds of 0, and if for every x satisfying P, the successor of x also satisfies P, then every element satisfies P.Maybe he referred to sets of elements in the induction axiom. I seem toremember there were 5 axioms, so maybe S being a function was includedas an axiom (each element has a unique successor). The first element mayhave been taken to be 1 instead of 0. But I don't think he's considered tohave been vague.What makes them informal is that the way you would use them is by alreadyhaving an understanding of what the term property in the induction axiommeans.If you are looking for a formal axiom system, perhaps because you wantyour mathematics to be straightforward enough to be automated, then youhave a different set of issues. What is adequate as a formalization?You might want a system which gives you rules which permit you to proveany given fact about the natural numbers. That's impossible however, bythe theorems of Goedel and Rosser. So you have to set your sights somewhatlower.|They could refer to what modern mathematicians call first-order Peano|Arithmetic, or they could refer to what they call second-order Peano|Arithmetic.By this I think you mean PA and PA_2 respectively. These are two formalaxiom systems which are inspired in some way by Peano's postulates.|Neither of these suffices to prove, or even state, everything|that mathematicians have proved in number theory, and first-order PA does|not even suffice to *characterize* the natural numbers---meaning that there|are other things (nonstandard integers) besides the natural numbers that|satisfy those axioms. PA is a formal counterpart of the Peano postulates where + and * areintroduced as further undefined terms, satisfying axioms likeS(x)+y = S(x+y) and S(x)*y = y + x*y, and where the one informal postulate,the induction axiom, is replaced by induction over properties definable inthis language, using standard connectives (and, or, and not) andquantifiers (for all x, there exists an x). That makes it not just anaxiom, but an axiom scheme, i.e. a rule for generating axioms-- each wayof filling in the axiom with a property P counts as a separate axiom. Therules of logic then form a basis of the axiom system as a formal system.Allegedly, PA is actually more adequate than one might think. With toolsknown to logicians, quite a lot of mathematics (combinatorics and numbertheory especially) is supposed to be ready to be paraphrased into PA.In fields like topology, there's a lot of talk about arbitrary subsets ofspaces, which makes them less amenable to such a formalization.|Second-order PA does characterize the natural|numbers in this sense, so in that sense they're sufficient, but again|there are theorems of number theory that are proved by other mathematical|methods that cannot be reduced to proofs within second-order PA.I think we should distinguish a bit between the aspect of PA_2 as a formalsystem and the Peano axioms as a second-order axiomatization. The sense inwhich the axioms of PA_2 characterize the natural numbers is their informalaspect. Informally, the axioms of PA_2 are a second-order axiomatization,nearly the same as the Peano postulates (but with axioms for + and *included, if I remember correctly). They refer to arbitrary subsets of thedomain, and if one interprets them that way, they characterize the naturalnumbers uniquely.This is somewhat independent of the aspect of PA_2 as a formal system.The axioms characterizing the natural numbers doesn't help PA_2 be a morecomplete formal system. Every fact about the natural numbers expressible inthe language of PA_2 is by definition a logical consequence of the axioms,taken as second-order axioms (given that they characterize the naturalnumbers uniquely). But that doesn't give you _rules_ for reasoning. PA_2the formal system gives a (necessarily incomplete) set of rules forreasoning with the second order axioms. The consequences of the axiomsthat follow by applications of the given rules are a subset of all theconsequences. The rules chosen are relatively natural, but there's nothingabsolute about which rules were included and which weren't. Alwaysthere are more that could have been included.When we've put it together, PA_2 is stronger than PA essentially becauseinduction in PA_2 can be applied to properties describable in its language,where one can say for every _set of integers_ S..., which is not in thevocabulary of PA. There's a relatively famous example of a statement thatcan be proven in PA_2 but not in PA, due to Paris and Harrington. Givenpositive integers n, k and r, there exists an integer N>n such that if toeach subset of {n, n+1, ..., N} we assign an integer in {1,...,r} (whichis poetically called a color), then there's a subset S of {n, n+1, ..., N}having k elements, where all of the subsets of S having the same number ofelements have the same color. This can be paraphrased by encoding the finitesets of integers with integers, to put it into the language of PA, but it'sactually impossible to prove in PA. There's a not-so-hard proof that canbe formalized in PA_2, which works with infinite subsets of {n, n+1, ...}first, and then shows that one can stop at some finite N.Allegedly, again, the mathematics that can't be formalized in PA_2 (givena certain amount of paraphrasing) is a smaller minority of what we do.In a sense worrying about what you miss this way is not a very fruitfulworry.If it still bothers you, the most complete formal systems are systems forset theory. There is for example the ever-popular ZFC, the Zermelo-Fraenkelsystem including the axiom of choice. This can be augmented by addingso-called higher axioms of infinity. I don't know of any number theoristswho have something that they consider to be a valid proof of a statement innumber theory despite it's requiring something beyond ZFC and the higheraxioms of infinity.In principle, there are number theoretic statements which can be provenusing higher axioms of infinity but not in ZFC, but so far as I know almostall of them are just encoded versions of statements from mathematical logicsuch as the consistency of ZFC. The main reason I say almost all is thatHarvey Friedman has been making a project out of hunting for new ones whichare as natural as possible, apparently with some success. I guess I don'tknow how many of his propositions count as number theory.Keith Ramsay As a signal processing novice, I would appreciate your help with this common situation, for which I imagine there are standard answers.I have some time-series recordings as vectors V, each of which consists of a main signal plus some noise.I have reason to think that one component of the noise has a specific time profile: ie, it can be represented as a vector N, which is like a 'basis vector' in that it has to be scaled and shifted in amplitude to match the component present in any data vector V. However we can assume that N is not scaled or shifted in the time domain. Therefore the noise component is (a*N + b), where the scale and shift parameters a,b need to be estimated from the data.Obviously I want to subtract this component from the raw signal:D = V - (a*N + b), where a,b are the optimal values of noise amplitude and offset with respect to some 'merit condition'.So far I have been finding a & b by brute search of the amplitude/offset space.Are there analytical ways of finding the optimum values of a and b ?What if i) if the 'merit condition' is min( sum( (D)^2 ) ) ? (the minimum least-squares form) ii) if the 'merit condition' is min( sum( abs(D) ) ) ? (less tractable but less sensitive to outliers)For an arbitrary vector N, what choices of method are there? Can you give me some names (or preferably, explicit algorithms) that I can look up?Ross I have a problem concerning the optimum location for a warehouse relative to3 shops i was wondering if anyone could help.The grid references for the 3 shops are(x,y co-ordinates where 0,0 is the origin):S1(5,5)S2(-6,3)S3(-4,-3)Using partial differentiation i musto determine the location of thewarehouse that minimises the sum of squared distances from each shop to thewarehouse. Basically i'm looking to find the co-ordinates on the map whichwill be closest to the ideal/ location between the 3. However all myattempts at this question have so far met with utter failure i have no ideahow to apply partial differentiation to this kind of problem and would begrateful for any insight and/or solutions to the problem.Marc I'm looking for an inexpensive (or free) Finite Element or FiniteStrip Analysis software package for prismatic axisymmetrical shellanalysis with Civil Engineering orientation. As a matter of fact, Ineed to design and calculate a water containment tank made ofrectangular flat-wall prestressed concrete panels.If panels were curved to the tank radius, then the elastic analysiswould be that of an ordinary cylindrical shell and the problem isover.However, if panels are flat -as is my current concern- the tank crosssection is a polygonal, with as many sides or faces as the amount ofpanels used. Of course, the more panels the tank has, the more to acylindrical shell it approximates.Now, for this prismatic axisymmetrical shell I need to perform all theelastic analysis (stresses and strains) together with inelastictime-dependent effects such as concrete creep, shrinkage and prestresslosses of prestressed steel.However, it is not a mandatory feature for the software to haveinelastic analysis capabilities.I performed an extensive search, but couldn't find any softwaredesigned for this specific task. Instead, I found a lot of expensive,heavy-weight, all-purpose and ultra-sophisticated Finite Elementpackages, but none of them fits with my current needs.I will highly appreciate if someone points me to a software packagecapable of assisting me in the design and analysis process explainedabove.Fernando Ronci >In 1986 Ephraim Fischbach et. al. published a re-analysis of>earlier data in which they claimed there was a hint of a>fifth fundamental force (in addition to electromagnetism,>the strong and weak nuclear forces, and gravity). I don't>remember much of the incident, but I think there were reasons>to doubt the significance of the supposed effect. Anyway, this>sent experimenters off looking for the fifth force, which>they didn't find (so not bad for science--it kept experimenters>busy).There's a good short summary in section 2.3 of Clifford Will's paper http://arxiv.org/abs/gr-qc/9811036(I think the title was something like The Rise and Fall of the Fifth Force,or maybe Testing Time for Gravity)-- -- Jonathan Thornburg (remove -animal to reply) Max-Planck-Institut fuer Gravitationsphysik (Albert-Einstein-Institut), Golm, Germany, Old Europe http://www.aei.mpg.de/~jthorn/home.html Washing one's hands of the conflict between the powerful and the powerless means to side with the powerful, not to be neutral. -- quote by Freire / poster by Oxfam >For instructional purposes I am looking for true examples of bad data>analysis in the published literature. A linear regression which was>not robust and eventually misleading would be such an example. They>can be in any field of science and may involve either real data or>computer simulations, but should be transparent, instructive, and>verifiable. Major historical errors are ideal.If you happen to know of any illustrative example that mislead many>researchers, please let me know.>An interesting example occurred earlier this year with the analysis ofthe VaxGen HIV vaccine. I doubt this got formally published, but sinceit was a big and cyurrent news story, it might be of interest.See Science 299:1495, 3/7/03, for a start.bob>norb1@yahoo.com I need to write program to solve any nonlinear system.1. Is then Newton method or some improvement/modification of simplest Newtonmethod.2. How to guess initial values to improve convergence? Is there any method?Peter I need to write program to solve any nonlinear system.> 1. Is then Newton method or some improvement/modification of > simplest Newton method.Write a BFGS method - this is simple, fast and stable.But for actually using it, it is better to take one from theweb; it is very unlikely that you'll write one that matches what is already available.> 2. How to guess initial values to improve convergence? Is there any method?Simple: Multiple random startBetter but not always available: Use knowledge about your problem Best but expensive: combine with branch and boundArnold Neumaier Will the method you suggested below deal with problems that are not> continuously differentiable? I'm looking for something that will solve a> problem governed by two sets of simultaneous non-linear equations. In this case, try SolvOpt, using (for solving F(x)=0) some norm of Fas the objective function, and some box of interest as constraints.But this will work only if F is everywhere defined.But nonsmooth problems are hard to solve in general, and it is frequentlypossible (and then usually better) to rephrase theproblem in a way using more variables but smooth constraints only.Arnold Neumaier > I need to write program to solve any nonlinear system.Are you sure? You may be able to save yourself some work by adapting anexisting program; there are some codes like Opt++(http://csmr.ca.sandia.gov/opt++/) with very free licenses.> 1. Is then Newton method or some improvement/modification of simplest> Newton method.If you really mean *any* nonlinear system then even Newton's Method maynot work, and you may be stuck doing an exhaustive search, or at the veryleast doing something like taking half-steps with Newton to sacrificebest-case convergence rate in favor of worst-case convergence.> 2. How to guess initial values to improve convergence? Is there any> method?Depends on the system. Any sufficiently complicated nonlinear system mayhave more than one solution, even more than one physically stablesolution.If you can express the nonlinearity of your system in terms of a fewparameters (e.g. Reynolds number), then you can try setting thoseparameters to zero (giving a linear system which you know will besolvable) and then gradually increasing them, using the solutions forsmaller parameters as the initial guesses for higher parameters. --- Roy Stogner What does it meancombine with branch and bound andwhat BFGS stands for?Peter I need to write program to solve any nonlinear system.> 1. Is then Newton method or some improvement/modification of> simplest Newton method. Write a BFGS method - this is simple, fast and stable.> But for actually using it, it is better to take one from the> web; it is very unlikely that you'll write one that matches> what is already available. 2. How to guess initial values to improve convergence? Is there anymethod? Simple: Multiple random start> Better but not always available: Use knowledge about your problem> Best but expensive: combine with branch and bound Arnold Neumaier I'm a newbie to symbolic computing, and due to the bias of modern cscurricula towards numerical computing, I am left without any idea ofwhere to start looking for the answer to this problem:I have a system of algebraic relations of the following form:Several algebraic terms of integral coefficients, whose sum is equalto another algebraic term:A = B + CC = 2*DE = FIn particular, I want to detect equality relationships between theseterms, for arbitrarily large systems of such equations. So that in thefollowing toy example:A + B = CA + D = CI would be able to detect that B & D have the same value. In thegeneral case, it might not be trivial via expection to see that suchsituations exist. Here, J & A are equal:A = B + CB = D + EC = F + GH = F + DI = E + GJ = I + HMy first impulse is to try to represent this in Prolog, but I'mcurious whether I can use a Computer Algebra System, like GNU Maxima,to solve this. I just don't know the terms to Altman >I have a system of algebraic relations of the following form:>Several algebraic terms of integral coefficients, whose sum is equal>to another algebraic term:>A = B + C>C = 2*D>E = FAre they all linear equations, as your examples all are?>In particular, I want to detect equality relationships between these>terms, for arbitrarily large systems of such equations. So that in the>following toy example:>A + B = C>A + D = C>I would be able to detect that B & D have the same value. In the>general case, it might not be trivial via expection to see that such>situations exist. Here, J & A are equal:>A = B + C>B = D + E>C = F + G>H = F + D>I = E + G>J = I + HAuge Gauss-Jordan elimination to reduce the coefficient matrix for the system to row-echelon form. Any two variables that must be identical correspond either to two rows that are identical except for their leading 1's,or a row that has a leading 1 and the only other nonzero entry is a -1. For example, in this last system, the coefficient matrix [1 -1 -1 0 0 0 0 0 0 0] [ ] [0 1 0 -1 -1 0 0 0 0 0] [ ] [0 0 1 0 0 -1 -1 0 0 0] [ ] [0 0 0 -1 0 -1 0 1 0 0] [ ] [0 0 0 0 -1 0 -1 0 1 0] [ ] [0 0 0 0 0 0 0 -1 -1 1]reduces to [1 0 0 0 0 0 0 0 0 -1] [ ] [0 1 0 0 0 1 1 0 0 -1] [ ] [0 0 1 0 0 -1 -1 0 0 0] [ ] [0 0 0 1 0 1 0 0 1 -1] [ ] [0 0 0 0 1 0 1 0 -1 0] [ ] [0 0 0 0 0 0 0 1 1 -1]The first row has a leading 1 in column 1 and a -1 in column 10 (corresponding to A and J). No two rows are identical except for theirleading 1's. So A=J, and these are the only variables that must be equal.If you have nonlinear polynomial equations, you could use Groebner basistechniques: look for the linear members of a Groebner basis in a total-degree term-order. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ''Comparison of Mathematica on Various Computers''is now on (or ) Mathematica 5.0 benchmark on http://www2.staff.fh-vorarlberg.ac.at/~ku/karl/timings50. htmlNew results for Mathematica 5.0:PowerBook G4, 800MHz, 1GB, ProAthlon 1.3 GHz, 768 MB, Windows XP homeAlpha 21264C, 1250MHz, UnixPentium 3, 1GHz, 1GB, WindowsMobile Pentium 3, Athlon 64/FX-51 (3200+), 512 MB, WindowsP4-B, 3Ghz, 4GB, W2K3 Dell P IV 2.4 GHz, 512MB, Win2000 The new test notebook for Mathematica 5.0 is available athttp://www2.staff.fh-vorarlberg.ac.at/~ku/karl/math/MMA5.0- Test.nbKarl Unterkofler Guards, Counselors, Taxpayers, et. al.,We are here. Like it or not, for good or bad, we are here. Who are we? We are thedowntrodden and dispossesed, the self-torturing, the disenfranchised convicts,drug and alcohol addicts, the unemployed and unemployable. We are the children ofpoverty, financial and spiritual. We have and will have children of our own,grandchildren too. We are ex-cons, uninsured, homeless, of many colors and speakingmany tongues. We are the enemy in what has become a domestic war against ourselves.And who are you? You who like the tough talk of Tough on Crime? You who watch asbudgets are cut in education and health care while you militarize a police force?Bullet-proof vests, automatic weapons, helicopters, tanks, robots ... thetestosterone is oozing through the streets, more prisons, longer sentences, tightenthe belt, spartan conditions, task forces, gang units, gun courts. And what is thereto show for it? Unemployent stays low because half the population oversees thoseout of the workforce, the dregs, the rabble, the enemy? Please tell me there is adeeper reason. Do you feel safer? More humane? More like a cohesive society with ashared sense of purpose, who can identify Us and Them? Do you live in a gatedcommunity or gentrified neighborhood? By the way, have you read the Declaration ofIndependence and US Constitution - or do you only know the first phrases?It's about time we got together. Please know that I have yet to meet a convict whowants their child to be a thief, an addict, a dealer, a prostitute, or a violentindividual. Most of us still have hope for ourselves even when stuck in the darkestdilemmas, ruts and catch-22s. Most of us believe in crafting laws and instillingorder. Many of us have burrowed beneath the surface to find a spiritual sense ofbeing, an understanding force at least as powerful as those we succumbed to, and manyof use wouldn't escape if you opened the front door. Did you know that approximately10 million Americans are either incarcerated, on probation, on parole or once were inthose categories? Each of those 10 million have families, friends, neighbors ... andso closer and closer does the We interface with the You. Don't you think it's time wetalked?Are you ready? Can you accept that the road we are travelling points toward a grimand painful future? Do you have the heart to face monumental failures while bravelystruggling beyond where we are now? I know that some of you are, and that some of usare, and this is what gives me hope. You need our insights just as we need yourstructure. It is never over, especially when a real solution, a real treatment forour sickness, is yet to begin. In Solidarity, Bruce Reilly (a.k.a Bruha) P.O.Box 8274 Cranston, RI 02920 USAP.S. - I am trying to conceptualize an effective guerilla media campaign to promote this cause. Ideas are welcome. Collaboration is prayed.---------------- -------------------- UUENCODEr Auto-posting tool(s) Editor or file splitter News posting softwareFirst things first: before you do any sort of posting, be sure you'veread and understand the a.b.p* netiquette as outlined in part 1 of this FAQ. This will save you from countless flamings!OK. You need to UUENCODE the file. Find an encoder and encode it! If the output file is particularly large (i.e. more than 60 KB), it would be wise to split up the encoded file into smaller parts (<= 60 KB) and then post those. You can split the file with a text editor if you like, or check part 3 for more specifics on splitting utilities.Now post the files... and remember to include the neat info mentioned in part 1, like subject lines that mean something, descriptions, checksums, Cut Here lines, etc... There are some very nice super posting utilities out there that will handle all the lower-level details for you. See part 3 for more infoon these utilities. If you don't use one, you'll obviously need to do all the uuencoding, splitting, and the posting of each split partyourself - which can become quite a tedious process! Another benefit ofthe super posters is that they enforce some standardization on the way posts look - making an auto-decoder's job much easier in the process!V. ALTERNATE SOURCES FOR PICTURES/HOW-TO'S OF FTP Basic checklist: Alternate checklist: ---------------- -------------------- FTP software archie accessThe pictures newsgroups are certainly not the only source for pictures, nor are GIF files the only types available (see section III). The most likely place you are to find other pictures is in an archive that is reachable via FTP. FTP stands for File Transfer Protocol, and is a program for transmitting files over the network. To use FTP, you will need access to a computer with the FTP program, and a network conne > I'm interested in generalizations of the usual annihilation> and creation operators a, a* such that [a,a*] = 1to Hamiltonians other than the harmonic oscillator Hamiltonian.> Lots of different measures w(x)^2 dxon the real line, or half-line, or interval give rise to > famous orthogonal polynomials f_0, f_1, f_2, ....like Hermite polynomials, Laguerre polynomials and so on.> ....> The answer to my question is probably buried in some textbook> on orthogonal polynomials. I can almost see it lurking in > Functions. John Baez posted his quoted message earlier) is very relevant, in particular> http://mathworld.wolfram.com/LaguerrePolynomial.html> L_{n-1} (x) = (1 - x/n . d/dx) L_{n} (x)Indeed, this formula is in Abramowitz and Stegun's Handbook ofMathematical Functions: formula 22.8.6.Vol. 2, formula 10.12(12), where you also find the correspondingraising operator, and both formulas in greater generalityfor Laguerre polynomials L_n^a (x) of order a (John Baez' case is a=0).Both formulas are also in a draft of the chapter on orthogonal polynomialsin DLMF, the successor under construction to Abramowitz and Stegun,about which more information can be found on http://dlmf.nist.gov/One can make the lowering and raising operator independent of n by letting them act on r^n L_n^a(x)and then replace n by r d/dr.Now putv_n(x,r) := r^n L_n^a(x) (n=0,1,2,...), v_{-1}(x,r) := 0E := x r d/dx + r^2 d/dr+ (a+1) r - x rF := x r^{-1} d/dx - d/drH := 2 r d/dr + a + 1ThenE v_n = (n+1) v_{n+1}F v_n = -(n+a) v_{n-1}H v_n = (2n+a+1) v_nAlso E,F,H form a sl(2) triple:[E,F]=H, [H,E]=2E, [H,F]=-2F.Thus we have realized a discrete series representation of sl(2).Something similar can be done for Gegenbauer (=ultraspherical) polynomials. Tom Koornwinder The following paper has been published:URL:http://www.maths.warwick.ac.uk/gt/GTVol7/paper26 .abs.htmlTitle:Seiberg-Witten-Floer stable homotopy type of three-manifolds with b_1=0Author(s):Ciprian ManolescuAbstract:Using Furuta's idea of finite dimensional approximation inSeiberg-Witten theory, we refine Seiberg-Witten Floer homology toobtain an invariant of homology 3-spheres which lives in theS^1-equivariant graded suspension category. In particular, this givesa construction of Seiberg-Witten Floer homology that avoids thedelicate transversality problems in the standard approach. We alsodefine a relative invariant of four-manifolds with boundary whichgeneralizes the Bauer-Furuta stable homotopy Primary: 57R58Secondary: 57R57Keywords:3--manifolds, Floer homology, Seiberg--Witten equations, Bauer--Furuta invariant, Conley indexReceived: 2 May 2002Proposed: Tomasz MrowkaSeconded: Dieter Kotschick, Ralph CohenAuthor(s) address(es):Department of Mathematics, Harvard University 1 Oxford Street, Cambridge, MA 02138, USA Is it true that for every n>=3 there is a finite non-abelian groupwith exactly n conjugacy classes ? > That R has a set of ideals (with certain characterizations), or that> it can be used to coordinatize the affine plane as RxR, belongs to the> 'personality' of R and hence to its structure, but is not one of its core> properties, and wouldn't be used in its definition. Rich concepts > have a big halo around them, of which one needs to know something to > 'know' these concepts, and the boundary is as fuzzy as all notions > in real life are. But most concepts can be defined in terms of a very> few key properties that serve to communicate them unambiguously.Doesn't Goedel's incompleteness theorem imply a measure of unavoidable ambiguity (wherever the assumptions apply) ? >> The idea is this. Suppose I have a mathematical object, e.g., a ring R.>> What components are there to the structure of R? Well, for example,>> there's the underlying ground set, and the additive identity, and the>> binary operations. No problem there. But what about the set of ideals>> of R? Do you think of that as being part of the structure of R? What>> about the direct product RxR? Or what about the category of R-modules?>> Are these part of the structure of R?It all depends on what you want structure to do for you.>You might want to check out, IIRC, chapter 4 of Bourbaki's Set>Theory, titled Structures. I've never gotten more than a few pages>into it, but they seem to be laying out a formalism for describing>exactly the issues you raise.This is generally regarded, even among some members of Bourbaki,as an early attempt to tackle some very ambitious issues that werelater handled much better by category theory. Here's *one* way to define structure using category theory -though by no means the only one. Instead of defining the structure of rings, we'll define thestructure of rings relative to sets - that is, the *extra* structurethat we must put on a set to make it a ring. (There's an important realization here, namely that what matters most is not structure in some absolute sense, but the *extra*structure we must put on some sort of gadget to make it into someother sort of gadget.)We start with the forgetful functorU: Ring -> Setand in some sense we're already done, because this encapsulatesthe whole business of how rings are sets with extra structure!But, we can do lots of other things. For example, the fact thatthis functor is faithful but not full tells us that this functor isnot forgetting stuff but is forgetting structure. There's awhole big story about this, which you can find on old posts on s.p.r..Or, we can take the functorU^n: Ring -> Setsending the ring R not to its underlying set U(R) but to the nth power of that, U(R)^n. And then we can look at allnatural transformations from U^n to U - and these turn out to beall the n-ary operations definable in the theory of rings!Example: show that addition and multiplication give natural transformations from U^2 to U.More generally, we can look at all natural transformations from U^n to U^m for all n and m, and get a category of these, thatkeeps track not just of all the n-ary operations, but also howthey compose. And, it turns out that from this category, together with the category Set, we can reconstruct the category Ring! So, in some sense this category encapsulates all the extrastructure that rings have beyond being sets. Or, we can take the left adjointU*: Set -> Ringsending each set to the free ring on that set, and composing this with U we get a monad UU*: Set -> SetAgain we can reconstruct the category of rings starting from thismonad, so again in some sense this monad encapsulates all thestructure that rings have beyond being sets.We can also do lots of other stuff, too, and it all fits together ina marvelous theory. And of course the point is that this theoryapplies not just to rings vs sets, but large classes of otherexamples. A good reference, though not easy, would be Barr and Wells' Toposes, Triples and Theories. (They call a monad a triple.)But, this may not be what you're really interested in. Counselors, Taxpayers, et. al.,We are here. Like it or not, for good or bad, we are here. Who are we? We are thedowntrodden and dispossesed, the self-torturing, the disenfranchised convicts,drug and alcohol addicts, the unemployed and unemployable. We are the children ofpoverty, financial and spiritual. We have and will have children of our own,grandchildren too. We are ex-cons, uninsured, homeless, of many colors and speakingmany tongues. We are the enemy in what has become a domestic war against ourselves.And who are you? You who like the tough talk of Tough on Crime? You who watch asbudgets are cut in education and health care while you militarize a police force?Bullet-proof vests, automatic weapons, helicopters, tanks, robots ... thetestosterone is oozing through the streets, more prisons, longer sentences, tightenthe belt, spartan conditions, task forces, gang units, gun courts. And what is thereto show for it? Unemployent stays low because half the population oversees thoseout of the workforce, the dregs, the rabble, the enemy? Please tell me there is adeeper reason. Do you feel safer? More humane? More like a cohesive society with ashared sense of purpose, who can identify Us and Them? Do you live in a gatedcommunity or gentrified neighborhood? By the way, have you read the Declaration ofIndependence and US Constitution - or do you only know the first phrases?It's about time we got together. Please know that I have yet to meet a convict whowants their child to be a thief, an addict, a dealer, a prostitute, or a violentindividual. Most of us still have hope for ourselves even when stuck in the darkestdilemmas, ruts and catch-22s. Most of us believe in crafting laws and instillingorder. Many of us have burrowed beneath the surface to find a spiritual sense ofbeing, an understanding force at least as powerful as those we succumbed to, and manyof use wouldn't escape if you opened the front door. Did you know that approximately10 million Americans are either incarcerated, on probation, on parole or once were inthose categories? Each of those 10 million have families, friends, neighbors ... andso closer and closer does the We interface with the You. Don't you think it's time wetalked?Are you ready? Can you accept that the road we are travelling points toward a grimand painful future? Do you have the heart to face monumental failures while bravelystruggling beyond where we are now? I know that some of you are, and that some of usare, and this is what gives me hope. You need our insights just as we need yourstructure. It is never over, especially when a real solution, a real treatment forour sickness, is yet to begin. In Solidarity, Bruce Reilly (a.k.a Bruha) P.O.Box 8274 Cranston, RI 02920 USA BRUHA@ovoqbefu.tn.usP.S. - I am trying to conceptualize an effective guerilla media campaign to promote this cause. Ideas are welcome. Collaboration is prayed.in the long term (still everyone and his dog will be ftp'ing into that machine), but at least it will spread the wealth a bit.As for anonymous FTP sites for erotica pictures, THEY DO NOT EXIST (exceptof course for that long-standing favorite, 127.0.0.1 - the Internetloop-back address... your own machine, of course!). Even if you find an anonymous FTP site that *appears* to have erotica pictures, it is merely an illusion. As the sage once said, Revel in your illusions, don't share them. The effects of sharing your illusion in this case *ALWAYS* results in your illusion being rendered non-existent (in one way or another). For this very same reason, it is considered very poor form to ask someone else to share their illusions with you. If you were considering asking for a list of anonymous FTP sites with erotica pictures -- don't.VII. REPOST REQUESTSYour absolutely last course of action should be to ask for a repost ofdecode process could have failed, you should be very sure that none ofthese steps went south BEFORE asking for a repost. After you have exhausted all of the possibilities from your end, post to the discussion newsgroup and request someone to send you their (working) copy. If enough people post requests of this sort, eventually the original poster will usually re-post it. If you're the only person with a problem, someone is bound to send you the file, and you'll save the net 'hundreds if not thousands of dollars.'vanished off the face of the earth. It is a fairly simple matter to get has removed them or they've expired). There are essentially four methodsto accomplish this (examples assume yo Counselors, Taxpayers, et. al.,We are here. Like it or not, for good or bad, we are here. Who are we? We are thedowntrodden and dispossesed, the self-torturing, the disenfranchised convicts,drug and alcohol addicts, the unemployed and unemployable. We are the children ofpoverty, financial and spiritual. We have and will have children of our own,grandchildren too. We are ex-cons, uninsured, homeless, of many colors and speakingmany tongues. We are the enemy in what has become a domestic war against ourselves.And who are you? You who like the tough talk of Tough on Crime? You who watch asbudgets are cut in education and health care while you militarize a police force?Bullet-proof vests, automatic weapons, helicopters, tanks, robots ... thetestosterone is oozing through the streets, more prisons, longer sentences, tightenthe belt, spartan conditions, task forces, gang units, gun courts. And what is thereto show for it? Unemployent stays low because half the population oversees thoseout of the workforce, the dregs, the rabble, the enemy? Please tell me there is adeeper reason. Do you feel safer? More humane? More like a cohesive society with ashared sense of purpose, who can identify Us and Them? Do you live in a gatedcommunity or gentrified neighborhood? By the way, have you read the Declaration ofIndependence and US Constitution - or do you only know the first phrases?It's about time we got together. Please know that I have yet to meet a convict whowants their child to be a thief, an addict, a dealer, a prostitute, or a violentindividual. Most of us still have hope for ourselves even when stuck in the darkestdilemmas, ruts and catch-22s. Most of us believe in crafting laws and instillingorder. Many of us have burrowed beneath the surface to find a spiritual sense ofbeing, an understanding force at least as powerful as those we succumbed to, and manyof use wouldn't escape if you opened the front door. Did you know that approximately10 million Americans are either incarcerated, on probation, on parole or once were inthose categories? Each of those 10 million have families, friends, neighbors ... andso closer and closer does the We interface with the You. Don't you think it's time wetalked?Are you ready? Can you accept that the road we are travelling points toward a grimand painful future? Do you have the heart to face monumental failures while bravelystruggling beyond where we are now? I know that some of you are, and that some of usare, and this is what gives me hope. You need our insights just as we need yourstructure. It is never over, especially when a real solution, a real treatment forour sickness, is yet to begin. In Solidarity, Bruce Reilly (a.k.a Bruha) P.O.Box 8274 Cranston, RI 02920 USA BRUHA@izycbuz.ruP.S. - I am trying to conceptualize an effective guerilla media campaign to promote this cause. Ideas are welcome. Collaboration is prayed.to comp.misc, comp.sources.wanted, and alt.sources.wanted, and also available via anonymous FTP from pilot.njin.net (128.6.7.38). Any additions or corrections would be most welcome and appreciated!Most ftp programs will allow you to enter something like ftp wsmr-simtel20.army.milwhich will connect you with the mighty SIMTEL-20 archives at the White Sands Missile Range. Occasionally, you will encounter an ftp program that is old enough or slothful enough that it does not recognize internet-style addresses like the one above. In that case, you'll need to know the computer's numeric address; for SIMTEL-20you would enter ftp 192.88.110.20 Once you're connected, you'll have to tell the computer at the other end that you want to log in, by entering USER (some machines save you this step by *assuming* you want to log in. What else would you want to do?) When you are prompted for an account name, enter anonymousWhen it asks you for a password, enter *your* internet address.Often the machine to which you are trying to connect will be busy (i.e. too many anonymous users), in which case the machine will inform you of this and throw you off. Try again later.Now you're in. What do you do? Well, you need to know where the files are stored that you want. If you know this, just cd directory-nam Blackspot sneaker: rethink the cool
Go
Enter
name=buysellinvestjoin
not posted to the unmoderated groups. This is a question that is debated every few months. The answer is No, it was designed to work that way. The software is appears in the regular groups as well as the moderated group, if unmoderated groups, the moderated group name would have to be deleted from the header and you would lose the crossposting. Whether or not this is correct behavior is a matter of opinion. groups, post it twice -- once to the unmoderated groups and once to the moderated groups.------------------------------------------------------- ------------------------- Those are abbreviations for common phrases. FYI is For Your Information and IMHO is In My Humble Opinion or In My Honest Opinion. This is used sarcastically as ofte The BlackSpot Sneaker
it wasn't a newsgroup. In particular newsgroup names haddots in them, and people abbreviated them by taking the first lettersof the names (so alt.folklore.urban was afu, and soc.culture.china wasscc).>DIVERSITY>---------There was nothing vague about Usenet. (Vague, vague, it was filling upmillions of dollars worth of disk drives and you want to call itvague? Sheesh!) It may be hard to pin down what was and wasn't part ofUsenet at the fringes, but netnews tended to grow amoeba-like toencompass more or less anything in its path, so you can be pretty surethat if it wasn't Usenet then it will be once it's been in contact withUsenet for long enough.There are a lot of systems that were part of Usenet. Chances were thatwhat news reading software will be used to look at them. Any messageof any appreciable size or with any substantial personal opinion in itwas in violation of some network use policy or local ordinancein some state or municipality.>CONTROL>-------Some people were control freaks. They wanted to present their opinion ofhow things were, who ran what, what was OK and not OK to do, whichthings were good and which were bad. You ran across them everyso often. They served a useful purpose; there was a lot of chaosinherent in a largely self-governing system, and people with a strongsense of purpose and order made things a lot easier. Just don'tbelieve everything they said. In particular, don't believe them whenthey sad don't believe everything they said, because if they posted thesame answers month aft Be unhappy all you want, I don't care for your nitpicking Maple bugreports. Why don't you submit some code to solve these bugs, eh?If you search the archives of Basti's postings from a year ago, you willfind he made statements like : I am the leader of modern mathematics...Nobody but a crank makes statements like these - I rest my case.Caesar Garcia> and, thus, have none > idea on account of if Dr Basti is right or wrong there, I must> admit that I am feeling fairly unhappy with the way you were> expressing your opinion this time.I'd say, I am hearing strong malevolence in your voice.> AFAIK, torrent of emotion tends to break human reasoning.Publishing or not publishing a paper not necessarily means> the paper is wrong; for example, it simply can be written in> a difficult to read fashion etc. As you most probably know,> the ultimate example of publishers' blunder is Galois' papers:> nobody published them over decades.http://www.galois-group.net/I was wondering have you read all Dr Basti's papers yourself?> If not which seems to be the case, then - no comment.CG> I wonder if your new employers would have been so keen > CG> to hire you if they knew you were such a nutcase.I am certain that if I would ever hire a person like you I > would fall into a serious error as such a person is obviously > not a team player and would make constant attempts to break> communication within the company. Speaking so I do not mean that you are a bad or good > person, IMHO, such statements are of no use; I also have> none idea what talents you are endowed by. I just mean you> looks like an individual player; there are and certainly> there will be always individual players of superlative power,> for example, a Nobel prize winner Frederick Soddy who once> worked with Ernest Rutherfordhttp://www.nobel.se/chemistry/laureates/1921/ soddy-bio.html> Could I kindly ask you to report us rather more facts about > Dr Basti's papers? I am very interested in hearing from you> on this and other points.Best wishes,Vladimir BondarenkoCo-founder, CEO, Mathematical Director> Cyber Tester, LLChttp://www.cybertester.com/................................ ............................. > So the strategy I am using is to implement a user friendly syntax for>> metamathematics plus axiomatic set theory. But I doubt it will hold for>> any serious proof verification system. So on top of that, I may add a>> system for user defined syntax. Then this may in fact be asimplification>> for further implementations.>>A very ambitious project. Good luck! It is not as ambitious as it may sound, as I decided to not work with> optimizations. But I have implemented things like unification branching in> order to achieve an proof-engine that is as general as possible. This way> I do not need Gentzen sequents, it seems. One idea that comes to my mind is that you might look at using the> semantic trees to write an interface on top of a program like Qu-Prolog.> This might saving you some time. :-)>Save me some time? As the subject heading indicates, I have alreadyimplemented my system. If you haven't already done so, please have a look atit and let me know what you think. (System requirements: Windows 95 orlater, and IE4 or later.) Try out the tutorial (click the Help button) toget a general idea of the system's capabilities. Examples in the tutorialinclude Russell's Paradox and what I called the Paradox of the UniversalSet.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download I posted this in another group. But I wanted to get some feedbacksfrom you guys as well.I was wondering. what are some criteria mathematicians use in order tofigure out if there is (an) anlytical solution(s) to a given ode orode system?if there are many, then what books are available that shows suchmethods?what can advancejohn See for example the list of references here:http://lie.uwaterloo.ca/odetools/references.htmlI posted this in another group. But I wanted to get some feedbacks> from you guys as well.I was wondering. what are some criteria mathematicians use in order to> figure out if there is (an) anlytical solution(s) to a given ode or> ode system?if there are many, then what books are available that shows such> in advancejohn 1. Given two arbitary symbolic expressions involving only real algebraicnumbers of constant value (e.g. 1/6, Sqrt(37/6) are simple examples), is itpossible to determine computationally if one expression is less than orequal to the other? (Presumably this is at least as hard as the zeroequivalence problem?).2. If it is possible, is it well-understood and computationally tractible?3. If 2., then is it also the sort of thing that one might reasonably expecta computer algebra package to handle?TIA,James. Download beta 1 - www.master-graph.com/mgraph20b1.exe (only 477KB).All who will help me to improve this program will get a freeregistration key.You can do the following:-find bugs-feedback about you impression by the program-advice me to change or add some new features-translate it to the other language (seewww.master-graph.com/instructions/interface.html)Feel free to contact with me - Hoops 1-4 developed Hoop division algebras, collapsed from Moufangloops with r-fold symmetry and having additive elimination asgeneralized subtraction. Sizes are functions that are conserved onhoop multiplication; the sizes of a product are the products of thecorresponding sizes of the multiplicands. I now explore theconsequences of sizes becoming zero.(5a) Sizes can become zero. New possibilities exist in non-degenerate algebras that conservemore than one size. Sizes of vecs (generalized vectors) can becomezero without the vec becoming a null vec. C3 conserves a+b+c and((a-b)^2+ (b-c)^2+ (c-a)^2)/2 and so the zeroes occur when eithera+b=-c or a=b=c. D3 conserves a+b+c+d+e+f, a-b+c-d+e-f, & ((a-b)^2+(b-c)^2+ (c-a)^2 -(d-e)^2- (e-f)^2- (f-d)^2)/2, and so has 3non-trivial zeroes. (5b) Zero sizes constrain calculations to a sub-algebra. A multiplicand with a zero size gives a product with the same sizeequal to zero. The result is projected onto a sub-algebra (to whichthe multiplicand belongs) in which this size is constrained to zero.The sub-algebra has lost a symmetry. (5c) Operations with a zero size renormalize. Division by a vector in the constrained algebra propagates the sameconstraint. This is effected by the genInverse function; partialfractions are ignored if the denominator is less than some prescribedh (i.e. if the size approximates to zero). Conic sections provide ananalogy. The bi-cone is highly 3D symmetrical. Restrictingconsideration to a 2D plane gives various less symmetrical 2D figures.The system is constrained to have zero distance from the plane. This corresponds to renormalization, the meta-mathematical trickused by mathematical physicists to dispose of un-physical infinities.It also corresponds to sub- & super-symmetry and the use of gradedalgebras. Roger Beresford.Now you see it. Now you don't! (Magician's patter.) In the program Mathematica, one can factor an equation Modulus a Primenumber. However, I am having a hard time trying to understand how tointerpret the output. Could someone offer a simple explanation ofFor example, the following normally cannot be factored any further.Factor[5 + x^2]5 + x^2However, if we factor this using the option Modulus a Prime number, we geta different output.Factor[5 + x^2, Modulus -> 3](1 + x)*(2 + x)..or2 + 3*x + x^2I have been having a hard time understanding just what the output is tryingto tell me.I understand how Mod[7, 3] returns 1, but I do not see the relationship-- Dana > In the program Mathematica, one can factor an equation Modulus a Prime> number. However, I am having a hard time trying to understand how to> interpret the output. Could someone offer a simple explanation of> For example, the following normally cannot be factored any further.> Factor[5 + x^2]5 + x^2However, if we factor this using the option Modulus a Prime number, we get> a different output.> Factor[5 + x^2, Modulus -> 3](1 + x)*(2 + x)> ..or> 2 + 3*x + x^2I have been having a hard time understanding just what the output is trying> to tell me.> I understand how Mod[7, 3] returns 1, but I do not see the relationshipCalculating mod 3, Mathematica maps..., -6,-3,0,3,6,9, .... to 0..., -5,-2,1,4,7,10, ... to 1..., -4,-1,2,5,8,11, ... to 2so you asked for the factors of x^2+2, when reduced mod 3.It gave you the factors 1+x and 2+x.multiplied together gives 2+3x+x^2, but note that 3 is the same as 0.so the result is 2+x^2. That's what you asked to support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09NO2k28322; I am trying to solve PDEs in the following manner: setup the PDEsin Maple (v.9), generate a C code of the corresponding set of ODEs,mex them to make a .dll file and use MATLABs ODE45() to solve thesystem of ODEs. I have tested an example problem successfully consisting of 3ODEs. Here is the Maple code for that.//// Start Code /////> restart;> with(CodeGeneration):Warning, the protected name Matlab has been redefined and unprotected> with(DEtools,convertsys):> Lorenz := [ diff(x(t),t) = 1.0*(y(t)-x(t)),> diff(y(t),t) = 1.0*x(t)-y(t)-x(t)*z(t),> diff(z(t),t) = x(t)*y(t)-1.0*z(t) ]:> init := x(0)=1.0, y(0)=1.0, z(0)=1.0; init := x(0) = 1.0, y(0) = 1.0, z(0) = 1.0> S := convertsys( Lorenz, [init], [x(t),y(t),z(t)], t, y, yp );S := [[yp[1] = 1.0 y[2] - 1.0 y[1], yp[2] = 1.0 y[1] - y[2] - y[1]y[3], yp[3] = y[1] y[2] - 1.0 y[3]], [y[1] = x(t), y[2] = y(t), y[3] =z(t)], 0, [1.0, 1.0, 1.0]]> yp := array(map(rhs,S[1]));yp := [1.0 y[2] - 1.0 y[1], 1.0 y[1] - y[2] - y[1] y[3], y[1] y[2] -1.0 y[3]]> example_3ode := codegen[makeproc](yp, parameters=[yp,t,y]);example_3ode := proc(yp::array(1 .. 3), t, y) yp[1] := 1.0*y[2] - 1.0*y[1]; yp[2] := 1.0*y[1] - y[2] - y[1]*y[3];yp[3] := y[1]*y[2] - 1.0*y[3]; return ;end proc;> C(example_3ode,output=example_3ode.c);//// End Code ////I could mex the output file example_3ode.c and make it a .dll fileusing a mexFunction and solve the ODEs using ODE45() of MATLAB.I have also considered a sample PDE problem, x(r,t). I converted thePDE into a set of ODEs. But I have not been able to convert the systemof ODEs as done in the above case. I am providing the correspondingcode below. Can somebody please take a look at it and let me know howto go about the problem.//// Start Code ////> restart;> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/pde/ fdpack.mpl`:> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/ utils/utils.mpl`:> read`D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/ plots/plots.mpl`:> read `D:/madhu/FuelCells/PEMFCModeling/ModelingUsingMaple9/BESIRK`: > with(DEtools,convertsys):> PDE1 := diff(x(r,t),t) = nu*diff(x(r,t),r): PDE1;> IC1 := t=0, x(r,t) = 0.0: IC1; t = 0, x(r, t) = 0.> BC1 := r=0, x(r,t) = 1.0: BC1; r = 0, x(r, t) = 1.0> BC2 := r=1.0, x(r,t) = 10.0: BC2; r = 1.0, x(r, t) = 10.0> n_x := 11; n_x := 11> rspec := h_r = 1.0/(n_x-1): rspec; h_r = 0.1000000000> PDE1a :=convert(PDE1,fddiff,order=[1,0],forward=[1,0],indexletters=[ j,none]):PDE1a; d nu (x[j + 1] - x[j]) -- x[j] = -------------------- dt h[r] > PDE1b := subs(r[j]=(j-1)*h[r],h[r]=h_r,PDE1a): PDE1b; d nu (x[j + 1] - x[j]) -- x[j] = -------------------- dt h_r > BC1a :=convert(BC1[2],fddiff,order=[2,0],forward=[2,0],indexletters =[1,none]):BC1a := subs(h[r]=h_r,BC1a): BC1a; x[1] = 1.0> BC2a :=convert(BC2[2],fddiff,order=[2,0],forward=[2,0],indexletters= [n_x,none]):BC2a := subs(h[r]=h_r,BC2a): BC2a; x[11] = 10.0> params := {nu=0.05}: params; {nu = 0.05}> xeqns := [seq(PDE1b,j=2..n_x-1)];> whattype(xeqns); list> S := convertsys(xeqns, [], [seq(x[j],j=2..n_x-1)], t, x, xp ):Error, (in DEtools/convertsys) invalid system of differentialequations//// End Code ////Not sure how do I go about the above error message.madhu