mm-188 === Subject: 9 digit number combinationI have a number combination that is minimum of 4 digits and maximum is8 digits and can be 0-9. The numbers can be used mutiple times. Myquestion is what are the combinations you can use. -- tml === Subject: Re: 9 digit number combination<>[0-8 digits combinations] - [0-3 digits combinations]-- tml === Subject: Re: 9 digit number combinationSo this means you have can go from 0000 - 99999999> I have a number combination that is minimum of 4 digits and maximum is> 8 digits and can be 0-9. The numbers can be used mutiple times. My> question is what are the combinations you can use.-- tml === Subject: Re: math>At 3:20 pm a jeweler set three antique clocks to the correct time. The>next afternoon at 3:20,she found that one clock was correct, one clock>was two minutes fast and the other was 2 minutes slow. At those rates,>how long will it take for all three clocks to show 3:20 againLet's see. Since one is going forward and one backwards at the samepace I think I can ignore one and still get the right answer.Basically one clock has to race completely around the dial to meet theoriginal 3:20. There are 720 minutes in a day. @ 2 minutes per day itwould take 360 days. But we know the jeweler has already made oneobservation, so I'd say 359 days.I'm I way off?JD-- tml === Subject: New jokes on mathNew jokes on math added on www.bymath.com-- tml === Subject: De'ning percent in terms of ratioHi --I'm proofreading translations of a math curriculum from English toanother language. The abovementioned phrase appears in a list ofskills the student should possess at a certain point in the program,and I want to make sure I have the meaning right.Does this mean that given a percentage, the student can restate it inthe form of a ratio -- e.g., turning 20% into 1:5? Or am I totallyoff the === Subject: Re: De'ning percent in terms of ratio> Hi -->> I'm proofreading translations of a math curriculum from English to> another language. The abovementioned phrase appears in a list of> skills the student should possess at a certain point in the program,> and I want to make sure I have the meaning right.>> Does this mean that given a percentage, the student can restate it in> the form of a ratio -- e.g., turning 20% into 1:5?Yes. Speci'cally a% is the ratio a:100, although the ratio may reduce, eg20:100 = 1:5.IIRC the English percent is derived from the Latin per centum, meaningper 100, or one part in 100, or some such.-- Darrell-- tml === Subject: Re: De'ning percent in terms of ratioI'm proofreading translations of a math curriculum from English toanother language. The abovementioned phrase appears in a list ofskills the student should possess at a certain point in the program,and I want to make sure I have the meaning right.> >Does this mean that given a percentage, the student can restate it inthe form of a ratio -- e.g., turning 20% into 1:5? Yes. Speci'cally a% is the ratio a:100, although the ratio may reduce, eg> 20:100 = 1:5. IIRC the English percent is derived from the Latin per centum, meaning> per 100, or one part in 100, or some such.> DarrellGood points. I've always taught percents as you mention, explainingwhat per and cent means, and that the equation a/100 = b/c isinvolved, somehow someway in a percentage problem. I have them writethis equation down in the process of any percentage problem.Especially, so as to see what of the given information is to replacec, as to what corresponds to the 100 in the denominator. Then they cansee through using algebra how and why of usually means multiply:Solving for instance for b yields, in words, b is a percent of c.It seems to me that the above provides a nice algebraic framework andapproach to percentage problems, an approach that could be used in apre-algebra framework quite easily.Paul-- tml === Subject: Re: De'ning percent in terms of ratioIn English, your interpretation is correct. You should have no trouble with itno matter what language you are translating. >>I'm proofreading translations of a math curriculum from English to>another language. The abovementioned phrase appears in a list of>skills the student should possess at a certain point in the program,>and I want to make sure I have the meaning right.>>Does this mean that given a percentage, the student can restate it in>the form of a ratio -- e.g., turning 20% for your help.>>-->Steven>>-- -- tml === Subject: Re: Limit that I can't 'nd the answerI am curious what might be a relevant homework problem if this one isa challenge?Was there's a discussion about sin(x)/x as x->0?Alexander Bogomolnyhttp://www.cut-the-knot.org-- tml === Subject: Re: Limit that I can't 'nd the answergo tohttp://lzkiss.net'rms.com/cgi-bin/igperl/igp.pl#cut and paste this ïprogram' onto the empty textarea:ZOOM(-pi,0,pi,1,1);graph(coord;func=sin(x)/x;);and press submit.-- tml === Subject: mechanicsA boy is tobogganing down a slope inclined at 25* to the horizontal.The ristances to his motion amounts to 15N. By modelling the toboggan-- tml === Math and just recently graduated fromCentral Washington University. I am currently residing in the SeattleArea, Kirkland Washington and would love to tutor again. I lovetutoring kids, and I seem to do very well at it!! I have tutoringexperience and the children I have tutored always seem to understandthe way I teach math. If you know of anyone who needs a math tutorplease let me know, the websit that I was referred to has apparentlymoved, I do not know if you are still placing tutors but just in caseI thought this was worth === Subject: TI-84I wanted to get some idea about the evolution of graphing calcs in highschool classrooms.When I was in high school the TI-82 was the latest and greatest....I haveone of the 'rst ones with the Yellow printing on it. Lately is has startedto turn it self off whenever it wants to (it's going on 10 years old now).Not sure what the problem is, but it got me into looking at new calcs. Iwill begin teaching math in the Spring 2005 semester, and I know all thehigh school students are required these days to have a graphing calc.How long does it usually take for a new model to make its way intoclassrooms? I missed the whole TI-83 life cycle since I was out of the mathloop for a while.Is it safe to get the TI-84 when it comes out in the summer, or will most of my students still be using === the TI-83? -- tmlSubject: Re: JSH: GOT IT!!! Algebraic integers check proof> Well I kept 'ddling at it, and now have on my blog the complete proof> that there's a problem with the ring of algebraic integers by checking> what happens with the roots of x^2 - x + 42 and y^2 + by - 7, with an> algebraic integer b. It's all at http://mathforpro't.blogspot.com/ which is easier to keep up with than posting on Usenet having to do> updates and corrections in a thread. Now I can write another paper (other one is STILL at the math journal> where it's been since Augus, but that's not necessarily strange as it> can take months). > James HarrisYour central claim is that (1 + sqrt(-167))/2 is coprime to 7.Given that [(1 + sqrt(-167))/2] * [(1 - sqrt(-167))/2] = 42are you claiming (a) (1 - sqrt(-167))/2 is divisible by 7or (b) there exist algebraic integers r and s such that 7 divides r*s, 7 is coprime to r, 7 does not divide s? -William HughesP.S. both (a) and (b) are false === Subject: Help With Pre Calc Problem support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 08:19:05 -0500 Can someone tell me how to do this problem? Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmeticsequence. I looked in the back and saw that the answer was (-3/2), but I am notsure how they got that. === Subject: Re: Help With Pre Calc Problem Tonight!!!> Can someone tell me how to do this problem? Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmetic> sequence. I looked in the back and saw that the answer was (-3/2), but I am not> sure how they got that. > If they must be *consecutive* terms, as listed, for some arithmetic sequence, then the differences between consecutive terms must all be equal and (2x+1) - (x+3) = (5x+3) - (2x+1).If they need not be consecutive, for any arithmetic sequence, the problem has no unique solution. === Subject: Re: Help With Pre Calc Problem someone tell me how to do this problem? >>Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmetic>sequence. What is an arithmetic sequence? It is one in which there is the same difference between each term and the next.So the difference between the 'rst and second terms must equal the difference (in the same direction) between the second and third terms.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: Help With Pre Calc Problem someone tell me how to do this problem?>> Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmetic> sequence.>> I looked in the back and saw that the answer was (-3/2), but I am not> sure how they got that.In the arithmetic sequence a_0 = k, a_1 = k + b, a_2 = k + 2b, what area_1 - a_0 and a_2 - a_1? Indeed, if a_n = k + b*n, what isa_n - a_(n-1)?Now, in your problem, 'nd an appropriate linear equation satis'ed by vast majority of Iraqis want to live in a peaceful, free world.And we will 'nd these people and we will bring them to justice. - George W. Bush (Washington DC, Oct 27 === Subject: Re: JSH: Attention in mathematics> It really was a fruitful weekend for me as the argument I've worked> out hits the question of shared factors between algebraic integers> head-on by working to try and 'nd some shared factors between the> roots of x^2 - x + 42 = 0 and y^2 - by - 7 = 0. Of course, x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b2 + 28))/2. So I simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b2 + 28))z/2, so z = (1 + sqrt(-167))/(b + sqrt(b2 + 28)) and I can then show that the problem is that sqrt(-167), as you can't> eliminate it to get a monic polynomial with integer coef'cients that> has z as a root. One way of looking at it is that the splitting 'elds of the> quadratics are too different, and no algebraic integer b can bridge> that difference. The proof is up at my blog: http://mathforpro't.blogspot.com/ and I like it because it shows posters like Dik Winter and Arturo> Magidin for what they are.> True enough. Your claim amounts to saying that one root,(1 + sqrt(-167))/2, of x^2 - x + 42 is coprime to 7. That means the other root, (1 - sqrt(-167))/2,must be divisible by 7 in the algebraic integers. You can easily show this is false in several different ways.Hence a contradiction. Conclusion: the proof on your blog is wrong. Thereare several wrong steps, but in particular, your assumption that b^2 + 28 = -167*n^2for some integer n, is not justi'ed.Nora B.[nonsense about ïhuman intention' as an element of factorization deleted] === Subject: Re: JSH: Attention in mathematicsX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>It really was a fruitful weekend for me as the argument I've worked>out hits the question of shared factors between algebraic integers>head-on by working to try and 'nd some shared factors between the>roots ofBizarre de'nition of fruitful. Looks to the rest of us like what you accomplished was to continue to ignore the severalarguments that you're wrong (some of which are so simpleeven you could understand them!)>[...]>>However, it took *intent* to hide, and in fact Dik Winter and posters>like him are arguing for a mathematics that is deliberately setting>out to hide factors, which is a fascinating position.Uh, no, those guys have never said anything this stupid.Mathematics does not deliberately do things - people dothings deliberately, and mathematics is not a person.Now _your_ assertion, made a few times, that factorsyou can't see cannot exist - _that_ is a fascinatingposition.>[...]>>To believe Dik Winter, and Decker himself for that matter, you have to>believe that there are some functions multiplied times those factors>on the left that are *hidden* from you quite deliberately, let alone>understand why they're the ones picked.>>[...]>>Notice that people like Winter never give a *reason* that makes sense,>as they put out theories that they can't prove.>>[...]>>If you believe that they're functions, where the actual function has>been hidden by mathematics intent on deceiving you, then you're in the>Dik Winter camp.>>[...]>>It occurs to me that a poster like Dik Winter or Arturo Magidin can>believe in a deceitful mathematics bent on fooling human beings,>showing conscience intent and caring about what each and every one of>you thinks.Except for the misquotations, outright lies and slander you makesome good points here...>But I believe that mathematics is consistent.>James Harris>Decker Quadratic Source Information>--------------------->Recently Rick Decker, a professor at Hamilton College, apparently>trying to refute my research came up with a quadratic example, which I>like because it's a quadratic, and easier to manipulate than the>cubics I've used before.>>If you wish to see his original post here are some headers which also>show that he posts from === Hamilton College:>>Subject: Re: Mathematical consistency, courage>>Decker put forward the quadratic>>(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >>where his a's are roots of >>a^2 - (x - 1)a + 7(x^2 + x).He didn't put forward this quadratic, he actually _said_something about it. Why don't you ever say what he said?************************David C. Ullrich === Subject: Re: JSH: Attention in mathematics It occurs to me that a poster like Dik Winter or Arturo Magidin can> believe in a deceitful mathematics bent on fooling human beings,> showing conscience intent and caring about what each and every one of> you thinks.And here you are again, once more attacking Prof. Magidin because youknow he won't reply to you to defend himself. You really are just asniveling little coward with delusions of manhood, aren't you?-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Attention in mathematicsWayne Brown Winter or Arturo Magidin canbelieve in a deceitful mathematics bent on fooling human beings,showing conscience intent and caring about what each and every one ofyou thinks.>> And here you are again, once more attacking Prof. Magidin because you> know he won't reply to you to defend himself. You really are just a> sniveling little coward with delusions of manhood, aren't you?>> -- > Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise> fwbrown@bellsouth.net | if you're good enough. Otherwise you give> | your pelt to the trapper.> e^(i*pi) = -1 -- Euler | -- John Myers Myers, SilverlockI hope Magidin does reply to this, though he won't. I don't think he shouldhave given into James' requests because James doesn't own these threads andcan't stop someone from posting. James is just a man (I use that termloosely) who acts like an immature brat.David Moran === Subject: Re: JSH: Attention in mathematics Adjunct Assistant Professor at the University of message>> >>It occurs to me that a poster like Dik Winter or Arturo Magidin can>believe in a deceitful mathematics bent on fooling human beings,>showing conscience intent and caring about what each and every one of>you thinks.>> And here you are again, once more attacking Prof. Magidin because you>> know he won't reply to you to defend himself. You really are just a>> sniveling little coward with delusions of manhood, aren't you?>>I hope Magidin does reply to this, though he won't. I don't think he should>have given into James' requests because James doesn't own these threads and>can't stop someone from posting.I did not give in, I offered; when I made the offer originally,James had not yet begun telling people to intercourse off. And Ireally do think that describing James's statements of over a yearlater as requests is... highly inaccurate.-- = It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: JSH: Attention in mathematics> It really was a fruitful weekend for me as the argument I've worked> out hits the question of shared factors between algebraic integers> head-on by working to try and 'nd some shared factors between the> roots of x^2 - x + 42 = 0 and y^2 - by - 7 = 0. Of course, x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b2 + 28))/2. So I simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b2 + 28))z/2, so z = (1 + sqrt(-167))/(b + sqrt(b2 + 28)) and I can then show that the problem is that sqrt(-167), as you can't> eliminate it to get a monic polynomial with integer coef'cients that> has z as a root.When b = -6, one 'nds that z = (1=sqrt(-167))/2, which, as a root ofyour own equation x^2 - x + 42 = 0, is one of those allegedly impossible algebraic integers. One way of looking at it is that the splitting 'elds of the> quadratics are too different, and no algebraic integer b can bridge> that difference.I think it is the split between JSH's mental processes and the requirements of standard mathematics that cannot be bridgesd. The proof is up at my blog: http://mathforpro't.blogspot.com/ and I like it because it shows posters like Dik Winter and Arturo> Magidin for what they are.And, if that falsehood above is any example, it shows JSH for what he is, too.Dik and Arturo shine even more brightly by comparison . === Subject: Re: Attention in mathematicsJames fruitful weekend for me as the argument I've worked....> There's a problem though, if I decide to multiply by the w's set as> the roots of that quadratic, I have to do it *deliberately*, which is> to say, it takes human intention, a deliberate action to split that 7> up into a particular way.>Wow. There's my problem. I keep doing it *accidentally*. Or on theother hand, could it just be spontaneous 'ssion, without anyhuman intervention? Hmm...Skip === Subject: Re: Jessica Vivian Valois - June 17th 1984Please do not feed the trolls.Simply add Daryl's address to your kill'le, ignore or whatever, and his annoying counterpart Thomas, who doesn't seem to realise his pedo-boy accusations create just as much A L O I S>>22 1 12 15 9 19 = 78 > [...] *> Post some math (if you know any) and leave the stupid crap numerology > at home. earle> * === Subject: Consider Dik Winter's claimsI don't know how many of you know about Dik Winter so I thought I'dmake a quick post to highlight this particular poster, who has made ithis business to put up negative web pages about my *old* work, as i't were pertinent to continually push my old mistakes. He is a ratherobsessive person who also has often replied to my posts on Usenet, soI thought I'd point out how he has lead some of you astray by beinglazy.Consider the Decker quadratic example (reference at bottom).Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Dik Winter has rather feverishly and repeatedly asserted that thereexists varying algebraic integer functions w_1(x) and w_2(x), suchthatw_1(x) w_2(x) = 7and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) andw_2(x) as factors, respectively.However, the trouble is that it doesn't take much effort to directlytest that out by assuming their existence, and I want you to payattention here and consider that Dik Winter is either lazy, stupid, orcontemptuous of your intelligence. Maybe he 'gured that none of youwould ever actually bother to check him.Now then, introducing f_1(x) and f_2(x) as the other factors of thea's, you havew_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, soa_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5.Now it matters to use the fact that the a's are roots ofa^2 - (x - 1)a + 7(x^2 + x)as the sum of the a's is (x-1), and they multiply to give 7(x^2 + x).But adding a_1(x) to a_2(x) now gives(w_1(x) f_1(x) + w_2(x) f_2(x) -14)/5 = (x-1), sow_1(x) f_1(x) + w_2(x) f_2(x) -14 = 5x - 5, which isw_1(x) f_1(x) + w_2(x) f_2(x) = -5x + 9.But if those function exist, they aren't stuck in the ring ofalgebraic integers, so now consider Z[1/5], and let x=2/5, which givesw_1(2/5) f_1(2/5) + w_2(2/5) f_2(2/5) = 7but f_1(x) f_2(x) = 25x^2 + 30x + 2, sof_1(2/5) f_2(2/5) = 18, which is, of course, coprime to 7, even inZ[1/5].Notice that now it's implied that *both* w_1(2/5) and w_2(2/5) have afactor of 7, that is sqrt(7). But looking back ata^2 - (x - 1)a + 7(x^2 + x), I havea^2 - (2/5 - 1)a + 7(4/25 + 2/5) = 0, which isa^2 + 3a/5 + 7(14)/25 = 0so there is a contradiction.(Can you 'gure out the resolution?)That Dik Winter. He's something else. However, I'm sure that someonewho makes such simple mistakes could hardly have fooled all of you.Right?James HarrisDecker Quadratic Source Information---------------------Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he posts from Hamilton === College:Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: Consider Dik Winter's claims > I don't know how many of you know about Dik Winter so I thought I'd > make a quick post to highlight this particular poster, who has made it > his business to put up negative web pages about my *old* work, as if > it were pertinent to continually push my old mistakes. He is a rather > obsessive person who also has often replied to my posts on Usenet, so > I thought I'd point out how he has lead some of you astray by being > lazy.I was? > Consider the Decker quadratic example (reference at bottom). > > Decker put forward the quadratic > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > where his a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x). > > Dik Winter has rather feverishly and repeatedly asserted that there > exists varying algebraic integer functions w_1(x) and w_2(x), such > that > > w_1(x) w_2(x) = 7 > > and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and > w_2(x) as factors, respectively.Yup, note that w_1(x) and w_2(x) are de'ned in terms of the gcd function. > However, the trouble is that it doesn't take much effort to directly > test that out by assuming their existence, and I want you to pay > attention here and consider that Dik Winter is either lazy, stupid, or > contemptuous of your intelligence. Maybe he 'gured that none of you > would ever actually bother to check him. > > Now then, introducing f_1(x) and f_2(x) as the other factors of the > a's, you have > > w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, > > so > > a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. > > Now it matters to use the fact that the a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x) > > as the sum of the a's is (x-1), and they multiply to give 7(x^2 + x). > > But adding a_1(x) to a_2(x) now gives > > (w_1(x) f_1(x) + w_2(x) f_2(x) -14)/5 = (x-1), so > > w_1(x) f_1(x) + w_2(x) f_2(x) -14 = 5x - 5, which is > > w_1(x) f_1(x) + w_2(x) f_2(x) = -5x + 9.+5x+9 I would say. > But if those function exist, they aren't stuck in the ring of > algebraic integers, so now consider Z[1/5], and let x=2/5, which givesHow wrong you are. They *are* stuck to the algebraic integers becausethey are de'ned in terms of the gcd as it operates on the algebraicintegers.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Consider Dik Winter's claims> I don't know how many of you know about Dik Winter so I thought I'd> make a quick post to highlight this particular poster, who has made it> his business to put up negative web pages about my *old* work, as if> it were pertinent to continually push my old mistakes. He is a rather> obsessive person who also has often replied to my posts on Usenet, so> I thought I'd point out how he has lead some of you astray by being> lazy. Consider the Decker quadratic example (reference at bottom). Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Dik Winter has rather feverishly and repeatedly asserted that there> exists varying algebraic integer functions w_1(x) and w_2(x), such> that w_1(x) w_2(x) = 7 and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and> w_2(x) as factors, respectively. However, the trouble is that it doesn't take much effort to directly> test that out by assuming their existence, and I want you to pay> attention here and consider that Dik Winter is either lazy, stupid, or> contemptuous of your intelligence. Maybe he 'gured that none of you> would ever actually bother to check him. Now then, introducing f_1(x) and f_2(x) as the other factors of the> a's, you have w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, so a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. Now it matters to use the fact that the a's are roots of a^2 - (x - 1)a + 7(x^2 + x) as the sum of the a's is (x-1), and they multiply to give 7(x^2 + x). But adding a_1(x) to a_2(x) now gives (w_1(x) f_1(x) + w_2(x) f_2(x) -14)/5 = (x-1), so w_1(x) f_1(x) + w_2(x) f_2(x) -14 = 5x - 5, which is w_1(x) f_1(x) + w_2(x) f_2(x) = -5x + 9.> Dumb mistake. Oh well. Isn't the 'rst time.James Harris === Subject: Re: Consider Dik Winter's claimsOriginator: a@shell3.shore.net (a)>> I don't know how many of you know about Dik Winter so I thought I'd>> make a quick post to highlight this particular poster, who has made it>> his business to put up negative web pages about my *old* work, as if>> it were pertinent to continually push my old mistakes. He is a rather>> obsessive person who also has often replied to my posts on Usenet, so>> I thought I'd point out how he has lead some of you astray by beingLed, not lead.>> lazy.>> Consider the Decker quadratic example (reference at bottom).>> Decker put forward the quadratic>> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >> where his a's are roots of >> a^2 - (x - 1)a + 7(x^2 + x).>> Dik Winter has rather feverishly and repeatedly asserted that there>> exists varying algebraic integer functions w_1(x) and w_2(x), such>> that>> w_1(x) w_2(x) = 7>> and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and>> w_2(x) as factors, respectively.>> However, the trouble is that it doesn't take much effort to directly>> test that out by assuming their existence, and I want you to pay>> attention here and consider that Dik Winter is either lazy, stupid, or>> contemptuous of your intelligence. Maybe he 'gured that none of you>> would ever actually bother to check him.>> Now then, introducing f_1(x) and f_2(x) as the other factors of the>> a's, you have>> w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, >> so>> a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5.>> Now it matters to use the fact that the a's are roots of>> a^2 - (x - 1)a + 7(x^2 + x)>> as the sum of the a's is (x-1), and they multiply to give 7(x^2 + x).>> But adding a_1(x) to a_2(x) now gives>> (w_1(x) f_1(x) + w_2(x) f_2(x) -14)/5 = (x-1), so>> w_1(x) f_1(x) + w_2(x) f_2(x) -14 = 5x - 5, which is>> w_1(x) f_1(x) + w_2(x) f_2(x) = -5x + 9.>>Dumb mistake. Oh well. Isn't the 'rst time.Does this mean Dik Winter is still lazy, stupid, or contemptuousof my intelligence, or was that a dumb mistake on your part too? === Subject: Re: Consider Dik Winter's claims> Dumb mistake. Oh well. Isn't the 'rst time.It isn't even the 'rst time today!Bump the Oops! counter again. (You may need order a new Oops! counter lever since this one's wearing out.)> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Consider Dik Winter's claims> I don't know how many of you know about Dik Winter so I thought I'd> make a quick post to highlight this particular poster, who has made it> his business to put up negative web pages about my *old* work, as if> it were pertinent to continually push my old mistakes. You should also do some posts about me. After all, I am the one who ABSOLUTELY SMASHED your feeble efforts at a prime counting algorithm by producing one that was ONE THOUSAND TIMES FASTER. It always is a pleasure to show you up as the stupid, arrogant imbecile that you are. === (Can you 'gure out the resolution?)Sure. It's very simple. You screwed up again, James Harris.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: Consider Dik Winter's claims> I don't know how many of you know about Dik Winter so I thought I'd> make a quick post to highlight this particular poster, who has made it> his business to put up negative web pages about my *old* work, as if> it were pertinent to continually push my old mistakes. He is a rather> obsessive person who also has often replied to my posts on Usenet, so> I thought I'd point out how he has lead some of you astray by being> lazy.JSH calling anyone else obsessive is excessive hutzpah. === Subject: differential equation support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 12:13:56 -0500 I'm having issues with this extremely simple differential equation. Idon't have a clue what I'm doing wrong.dP/dt = k*P^(1/2)with initial conditions dP/dt = 20 with t = 0and P = 100 at t = 0I keep ending up with the wrong answer (derivs don't match).Help is appreciated.Patrick H === Subject: Re: differential equation question I'm having issues with this extremely simple differential equation. I> don't have a clue what I'm doing wrong. dP/dt = k*P^(1/2)> with initial conditions dP/dt = 20 with t = 0> and P = 100 at t = 0 I keep ending up with the wrong answer (derivs don't match).> Help is appreciated.Separate variables: dP ------- = k dt sqrt(P)Integrate each side: 2*sqrt(P) = k*t + cWhence: P = ((k*t + c)/2)^2 . . . . . . (*)In which put P = 100 and t = 0 to get: c = 20 dP k*t + 20 -- = k * -------- dt 2In which put dP/dt = 20 and t = 0 to get: k = 2Whence from (*): P = (t + you're e-mailing me. === Subject: Re: differential equation question> I'm having issues with this extremely simple differential equation. I> don't have a clue what I'm doing wrong. dP/dt = k*P^(1/2)> with initial conditions dP/dt = 20 with t = 0> and P = 100 at t = 0 I keep ending up with the wrong answer (derivs don't match).> Help is appreciated. Patrick H> t = integral 1/(k*P^(1/2) dP === Subject: Re: differential equation question> dP/dt = k*P^(1/2)> with initial conditions dP/dt = 20 with t = 0> and P = 100 at t = 0How can dp/dt = 20 when there's no dependence on t?The solution if you separate the variables gives:P =((kt + c) / 2) ^ 2Hence c = 20But, somethings not quite right. Have you quoted the question correctly.Perhaps we wanted dp2/dt2?Kev === Subject: Re: differential equation questionPatrick H with this extremely simple differential equation. I> don't have a clue what I'm doing wrong.>> dP/dt = k*P^(1/2)> with initial conditions dP/dt = 20 with t = 0> and P = 100 at t = 0>> I keep ending up with the wrong answer (derivs don't match).> Help is appreciated.Without a clue as to what you're doing, I haven't the foggiest notion whatyou're doing wrong. Show your work.Here's a guess: you're calculating your derivative wrong. You need tomultiply by 1/2*k (chain rule).2nd guess: you're forgetting to subtract 1 from the exponent when youcalculate the derivative, thus ending up with 100 in the numerator ratherthan the denominator.Any further guesses get even wilder. Like you integrated wrong, putting 2in the denominator instead of 1/2 or something like that. Or you forgot howto simplify complex fractions. But how can we know without even a hint asto what you did?Jon Miller === Subject: JSH: Why am I so important?I just made a couple of posts, one of which had an obvious sign error.Now then, probably quite a few posters will gleefully leap upon thoseposts, as if it's such a big deal.Why?Why do I see webpages scattered across the Internet dedicated topushing negatives about me?Why am I so important to these people?Given the fact that I've been posting for years, and making mistakesfor years, it stands to reason that posters who so happily reply to meas if it matters are strangely deluded.What changes?It's a bizarre drama, where I'm this guy, who posts a lot and has ablog.For some odd reason, there are all these other people, like DikWinter, Erik Max Francis and Rick Decker who have these webpages,which always 'nd some way to insult me.Now sure, maybe they get their jollies doing that, but why keep at itfor YEARS.Look at Decker's webpage, and notice the date.Look at Dik Winter's webpages and notice the dates to which he'sreferring.Clearly I'm VERY IMPORTANT to these people!!!Oh well, I'll just keep posting, and I guess these people will keepreplying.James Harris === Subject: Re: Why am I so important?James Harris posts, one of which had an obvious sign error.Apparently not-so-obvious.Doug === Subject: Re: JSH: Why am I so important?yeah; it comprizes a region of my brain,only slightly smaller than the one labeled SEXwith all sentient beings. I've never triedto get professional help, before this, butI may begin to duly consider it! but that shouldn't interferewith The Interdimensional Geophysical Year,(Decade Two). > It's a bizarre drama, where I'm this guy, who posts a lot and has a > Clearly I'm VERY IMPORTANT to these people!!! anyway, the Protocol of the Elders of Kyoto is on,as of Dec.12 (LA(TRIBco)Times in November --on the Chicago BoT, I think), so thatthe oilcos' free market has a new pricing tool;have you heard any thing about this?... the dystributionis mainly tied-up with Dutch and British interests,where the Common Market is more radical than Californiaon these enviro matters -- except for the bull**** republican policyof not getting oil & gas off of our own coasts! thus quoth:A COORDINATED WALL ST. ASSAULTWho Killed U.S. Nuclear Power? by Marsha Freeman --zero-dimensional holograms.utterly compressed information.DemConK2 cloning malfunction:(Citizen John-mick) Kane/Bore/Gush/Nadir;LaRouche et al versus Fowler et al, ï96 --Supreme Decision, March 27, 2000. but, hey;the Voting Rights Act is upfor republication in ï07!--Give the World a Trickier Dick Cheeny -- out of of'ce after GIGA years.http://www.benfranklinbooks.com/http://larouchepub.com/ www.rand.org/publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanachttp://www.wlym.com/PDF-68 soros.html === Subject: Re: JSH: Why am I so important? Why am I so important to these people?You aren't; that's the whole point. It's your own highly exaggeratedsense of self-importance that makes it so much fun for so many peopleto point out your unimportance.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Why am I so important to these people?Easy. You are among the least competent mathematicians and most obnoxioushuman beings to ever populate the planet. You make a 'ne example to showour kids what *not* to be.> James Crank Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: JSH: Why am I so important?As an anodyne to ennui only! === Subject: Re: I am having dif'culty understanding following.>> If f(z)=cos(z/z^2), then> z=0 is a pole of order 2.> &> If g(z)=e^z/(z-1)^3, then> z=1 is a removable singular pt of g(z).>> I read above from book,> but, I don't know why...-->> If z=1 is a removable singular pt of g(z),> then lim_z->1 g(z) must exist.> and if z=0 is a pole of order 2, then> lim_z->0 z^2*cos(z/z^2) must converge to nonzero, nonin'nity value.> but, It seems to me, both limit doesn't exist.>> Could somebody explain me why z=0 is a pole of order 2> and why z=1 is a removable singular pt of g(z) ?I assume you copied something wrong. cos(z/z^2) = cos (1/z) has anessential singularity at z=0. cos(z)/z^2 has a pole of order 2 at z=0.Usually, the de'nition of pole is that f(z) has a pole of order n at a ifz^n*f(z) is analytic at z=a (removing removable singularities, of course).However, it's easier to see if you expand your function in a Laurent series.In this case, (assuming cos(z)/z^2 is your function), f(z) = 1/z^2*(1 -z^2/2! + z^4/4! - . . .) = 1/z^2 - 1/2! + z^2/4! - . . . . Since thelargest negative exponent is 2, the function has a pole of order 2.I don't know what you meant in the other one. g(z)= e^z/(z-1)^3 goes toe/oo^3 (dontcha just love that notation?) as z->1, so it has a pole of order3.Jon Miller === Subject: Dik Winter's claims revisited, dependency issueI let it get personal, and made an earlier post with a silly signmistake. Then I just got really pissed, and made some posts. Buthere's what I feel is the proper approach, with a focus on an implieddependency that Dik Winter never bothered to handle.Consider the Decker quadratic example (reference at bottom).Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Dik Winter has repeatedly asserted that there exists varying algebraicinteger functions w_1(x) and w_2(x), such thatw_1(x) w_2(x) = 7and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) andw_2(x) as factors, respectively.Now then, introducing f_1(x) and f_2(x) as the other factors of thea's, you havew_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, soa_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5.Now it matters to use the fact that the a's are roots ofa^2 - (x - 1)a + 7(x^2 + x),as solving that quadratic, and picking a_1(x) for the positive signroot givesa_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so(w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2sow_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, sof_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x),which implies a relationship between f_1(x) and w_1(x), butf_1(x) f_2(x) = 25x^2 + 30x + 2and it is arbitrary that 7 was the multiple before as it could havebeen 11 or 13 or any of an in'nity of other numbers, and w_1(x) andw_2(x) are themselves not determined, so it's a spurious appearance ofdependency.That's the little detail that Dik Winter never bothered to address.Now then, if f_1(x) is in fact independent of w_1(x), then how do youaccount for the appearance of a dependency inf_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)?If you try to push the issue that it isn't so required consider whathappens if you try to divide through by w_1(x), as then you havef_1(x) = (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x))and the problem is that all of the w's need to go away. So assumingthat a_1(x) has w_1(x) as a factor means, introducing g(x), that youhavea_1(x) = w_1(x) g(x), sof_1(x) = (5g(x)/2+ 2 w_2(x))which indicates that f_1(x) is dependent on w_2(x).However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them asan in'nity of functions will work, while f_1(x) is independent ofw_1(x) and w_2(x) sincef_1(x) f_2(x) = 25x^2 + 30x + 2.The point is you can multiply some polynomial like 25x^2 + 30x + 2, byanything you choose. There's no way that it's locking into it thatfunctions that are factors of 7 are required.The simplest answer is that w_1(x) either equals 7, or it equals 1,and checking with the 'rst givesf_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/14 + 2)/w_1(x)implying that 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x))) has a factor thatis 14, which is false in the ring of algebraic integers, so Dik Winterand his cohorts falsely seize on that to hide the issue with their ownclaims.They've gone so far that Keith Ramsay even claimed to have posted asolution for the w's, but how did he pick w's from in'nity?Given that his polynomial is of degree 22, it's quite possible that hejust chose a really big polynomial!Remember the issue here is f_1(x) and it's *lack* of a dependency withw_1(x).The problem for Dik Winter and his buddies is that you have twoindependent equations:w_1(x) w_2(x) = 7andf_1(x) f_2(x) = 25x^2 + 30x + 2To try and dispute my results, posters like Dik Winter or Nora Baronsimply skip past mathematical consequences of their claims, like theindependence between those equations.I'm still pissed. But there's not a lot I can do when people like DikWinter can so easily get away with basic problems in their claims, ona newsgroup that doesn't seem to give a damn about mathematical truth.James Harris === Subject: Re: Dik Winter's claims revisited, dependency issue> I let it get personal, and made an earlier post with a silly sign> mistake. Then I just got really pissed, and made some posts. But> here's what I feel is the proper approach, with a focus on an implied> dependency that Dik Winter never bothered to handle. Consider the Decker quadratic example (reference at bottom). Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Dik Winter has repeatedly asserted that there exists varying algebraic> integer functions w_1(x) and w_2(x), such that w_1(x) w_2(x) = 7 and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and> w_2(x) as factors, respectively.> Your description is accurate. I agree with Dik on this. We haveboth given explicit de'nitions of w_1(x) and w_2(x). See below.> Now then, introducing f_1(x) and f_2(x) as the other factors of the> a's, you have w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, so a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. Now it matters to use the fact that the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), as solving that quadratic, and picking a_1(x) for the positive sign> root gives a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 so w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x), which implies a relationship between f_1(x) and w_1(x), but f_1(x) f_2(x) = 25x^2 + 30x + 2> Correct. First, a_1(x) is a root of[*] a^2 - (x - 1)*a + 7*(x^2 + x).Thus a_1(x) is clearly dependent on x. Note however that itis also dependent on 7: if you replace 7 by, say, 17, thenthen a_1(x) is not a root of [*] but instead is a root of a^2 - (x - 3)*a + 17. Now, back to a_1(x) when 7 is used: w_1(x) is also a functionof both x and 7, because when 7 is used, you get w_1(x) = GCD(a_1(x), 7).and when 17 is used, you would get w_1(x) = GCD(a_1(x), 17). Note that when you replace 7 by 17, a_1(x) also changes;therefore w_1(x) does also. What you call f_1(x) is: f_1(x) = (5 a_1(x) + 7)/w_1(x),which is ALSO clearly dependent on x, but also clearly dependenton 7: again, if you replace 7 by 17, you would get a differentvalue for a_1(x), w_1(x), AND f_1(x). If you want all thesefunctions to be distinguished for situations other than 7, you really should write them as a_1(x, 7), a_1(x, 17), f_1(x, 7), f_1(x, 17), etc..> and it is arbitrary that 7 was the multiple before as it could have> been 11 or 13 or any of an in'nity of other numbers, and w_1(x) and> w_2(x) are themselves not determined, so it's a spurious appearance of> dependency.> No - it's not spurious at all. If you replace 7 by something else, everything changes. For example, with 17, the originalassumption would be[**] (5 a_1(x) + 17)*(5 a_2(x) + 17) = 17*(25*x^2 + 30*x + 2).There is no reason to think that the same a_1(x) will work for17 as for 7; the factorizations are quite different. [For onething, the CONSTANT terms are different.] Similarlyfor w_1(x) and f_1(x): they will be different for 17 than theyare for 7, etc.. > That's the little detail that Dik Winter never bothered to address.> Why should he? He was considering only the 7 factorization.You appear to think that a_1(x), etc., will not change when you replace 7 by other numbers. Clearly from the left side of[**] above, that is not going to be true.> Now then, if f_1(x) is in fact independent of w_1(x), then how do you> account for the appearance of a dependency in> f_1(x) is NOT independent of w_1(x). You *de'ned* it as f_1(x) = (5 a_1(x) + 7)/w_1(x).> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)? If you try to push the issue that it isn't so required consider what> happens if you try to divide through by w_1(x), as then you have f_1(x) = (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x)) and the problem is that all of the w's need to go away. So assuming> that a_1(x) has w_1(x) as a factor means, introducing g(x), that you> have a_1(x) = w_1(x) g(x), so f_1(x) = (5g(x)/2+ 2 w_2(x)) which indicates that f_1(x) is dependent on w_2(x).> It is, of course. w_1(x) and w_2(x) are highly dependentalso since their product is 7.> However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them as> an in'nity of functions will work, Not for 'xed values like 7. There are a lot of constraints. For example, for 7, a_1(x) is a root of a^2 - (x - 1) + 7*(x^2 + x),so a_1(x) is restricted to only being one of two possible numbers.Then since the right de'nition of w_1(x) is w_1(x) = GCD(a_1(x), 7), w_1(x) is determined once you have chosen a_1(x). Why? Becausethe GCD function is well-de'ned (to within units) in the ringof algebraic integers.> while f_1(x) is independent of> w_1(x) and w_2(x) since f_1(x) f_2(x) = 25x^2 + 30x + 2. The point is you can multiply some polynomial like 25x^2 + 30x + 2, by> anything you choose. There's no way that it's locking into it that> functions that are factors of 7 are required.> Not right - see above. If you replace 7 by something else, like 17,all of a_1(x), w_1(x), and f_1(x) change. They are all linked together by the equation f_1(x) = (5 a_1(x) + p)/w_1(x),where p could be 7 or 17, etc.> The simplest answer is that w_1(x) either equals 7, or it equals 1,> and checking with the 'rst gives f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/14 + 2)/w_1(x)> That simplest answer however does not give algebraic integercoef'cients as you well know, and that is what you wanted in the 'rst place. It's simple, but it is the wrong factorization.Things are not always simple! > implying that 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x))) has a factor that> is 14, which is false in the ring of algebraic integers, so Dik Winter> and his cohorts falsely seize on that to hide the issue with their own> claims. They've gone so far that Keith Ramsay even claimed to have posted a> solution for the w's, but how did he pick w's from in'nity?> He didn't. There is a well-de'ned algorithm for computingthe w's, starting with the number (1 + sqrt(-167))/2. If youhad replaced 7 with 17, the starting number would have been(-1 + sqrt(-407))/2 and the result would be different.> Given that his polynomial is of degree 22, it's quite possible that he> just chose a really big polynomial!> Actually I think it had degree 11. I think he used a programto do the computation, and I think it 'nds the lowest-degreepolynomial that works. I believe it computes powers of anideal, ^M, until it 'nds an M where that idealis principal. It is not a matter of human choice, just a programmable algorithm.> Remember the issue here is f_1(x) and it's *lack* of a dependency with> w_1(x).> See above. They are quite dependent, and both changeif you replace 7 by other numbers.> The problem for Dik Winter and his buddies is that you have two> independent equations: w_1(x) w_2(x) = 7 and f_1(x) f_2(x) = 25x^2 + 30x + 2 To try and dispute my results, posters like Dik Winter or Nora Baron> simply skip past mathematical consequences of their claims, like the> independence between those equations.> We focussed on the 7 case because that was Decker's originalexample and your own main interest. Other cases are similar*but not the same*. We never claimed they were.> I'm still pissed. Too bad. In fact, seeing the number of threads you have started todayon this, and their content, it looks to me like you are frantic,perhaps even dangerously upset. You need to calm down about this. We are not evil conspirators trying to rob you of yourrightful credit. We are just ordinary people trying to show youwhere you are making mistakes. Nora B.> But there's not a lot I can do when people like Dik> Winter can so easily get away with basic problems in their claims, on> a newsgroup that doesn't seem to give a damn about mathematical truth. > James Harris === Subject: Re: Dik Winter's claims revisited, message>> I'm still pissed. But there's not a lot I can do when people like Dik> Winter can so easily get away with basic problems in their claims, on> a newsgroup that doesn't seem to give a damn about mathematical truth.>Actually Jimmy, *nobody* gives a about mathematical truth (exceptfor you.) You are quite obviously the only person on Earth or possiblyin the entire universe who does. Guess you'll just have to learn to livewith that terrible loneliness. === Subject: Re: Dik Winter's claims revisited, dependency issueX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>I let it get personal, and made an earlier post with a silly sign>mistake. Then I just got really pissed, and made some posts. Yeah, those were some posts, all right.But what the heck, it's not the 'rst time you've insisted onsounding like a lunatic in front of the entire planet, probablywon't be the last.>But>here's what I feel is the proper approach, with a focus on an implied>dependency that Dik Winter never bothered to handle.Idiot.>[...]>>To try and dispute my results, posters like Dik Winter or Nora Baron>simply skip past mathematical consequences of their claims, like the>independence between those equations.>>I'm still pissed. But there's not a lot I can do when people like Dik>Winter can so easily get away with basic problems in their claims, on>a newsgroup that doesn't seem to give a damn about mathematical truth.ing lying slanderous imbecile.>James Harris************************David C. Ullrich === Subject: Re: Dik Winter's claims revisited, dependency issue> Now then, if f_1(x) is in fact independent of w_1(x), then how do you> account for the appearance of a dependency in f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)?JSH is arguing that because f_1(x)*f_2(x) = 7, it must be false that f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x).That is the same as arguing that because 9*16 = 144 it must be false that 9 = 63/7.In fact, all that f_1(x)*f_2(x) = 7 means is that f_2(x) = 7/f_1(x) for whatever value f_1(x) might have.It appears as though JSH's attempts at reasoning get more irrational as the evidence against him gets more overwhelming. === Subject: Re: Dik Winter's claims revisited, dependency issue> (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 so w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2Oh? === Subject: Re: Dik Winter's claims revisited, dependency issue> I let it get personal, and made an earlier post with a silly sign> mistake.Nevertheless, you continue to behave as if you are infallible. Bump theOops! counter.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: help solve this support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 14:39:03 -0500 my 'rst is in douhnuts but not in pizzamy second is in bubbles but not in toothpastemy third is in kiss but not in branston pickle ??any ideas === Subject: 2 Stage support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 14:51:47 -0500 I have to prove that q(k+1) = q(k) + q(k) * (((1/2) Omega dt) + ((1/8)Omega^2 dt^2)) is the discrete numerical integration solution for the differentialequation of the quaternion parameters dq/dt = 0.5 Omega qusing the 2-stage Runge-Kutta method.dt is the time interval, or data rate in this case, and Omega is a 4by 4 matrix of rotation rates around the x, y and z axes (but I don'tthink that really matters for this exercise).I have used the Runge-Kutta method as followsX1 = x(n)Xi = x(n) + h Sum(j=1,i-1)(a_ij f(tn + c_jh, Xj)) i=2,...,sx(n+1) = x(n) + h Sum(j=1,i-1)(b_i f(tn + c_1h, Xi))Because this is the 2 stage version s = 2, a_21 = 1/2, b_1 = 0, b_2 =1, c_1 = 0 and c_2 = 1/2. I have assumed that the interval h = 1 hencethe form above becomesX(1) = x(n)X(2) = x(n) + 1/2 f(t(n),X(1))x(n+1) = x(n) + f(t(n) + 1/2,X(2))Now the problem I am having is when I start plugging in the function Idon't seem to come up with the correct solution. I have got so farX(1) = q(n)X(2) = q(n) + 1/2 f(dt,q(n)) = q(n) + 1/2 * 1/2 Omega dt q(n) = q(n) + q(n) (1/4 Omega dt)x(n+1) = q(n) + f(dt + 1/2, q(n) + q(n) (1/4 Omega dt)) = q(n) + ...My question is am I going about this correctly? If so, how do I solvethe last step as I do not seem === Invention or Discovery I have Goooooooogled and there is some history on this question but it's've years ago now ( I knew Usenet had gone downhill) so: are mathematicaltheorems inventions or discoveries?cheersdd === Subject: Re: Invention or Discovery>> are mathematical theorems inventions or discoveries?They are simply stepping stones to true enlightenment. When one day weUNconver (neither discover nor invent) the ultimate single equation whichexplains everything!:)Kev === have Goooooooogled and there is some history on this question but it's> 've years ago now ( I knew Usenet had gone downhill) so: are mathematical> theorems inventions or discoveries?.... This question is a hoary perennial, but anyway here's a repeat of my post to sci.math under the heading Re: discovered or invented? on the 27th July 1999.> Was math discovered or invented?.... > You'll never get general agreement on such philosophical questions,> but FWIW here are some ideas I've gradually developed. Our early childhood thinking includes intuitions of quantity, size> and shape. More mature thought about those things could become much> better organized after the invention of writing about 3000 B.C. When> people worked with abstracted concepts such as the number 4 (rather than> just four 'ngers or four sheep) you might describe their thinking as> mathematical. A similar attitude may have persisted with the Greeks, who> seem to have viewed geometrical objects as idealized physical objects> (e.g. a line thinner than any physical tool can draw). Since the late 19th century we've had modern axiomatic theories (in> which e.g. a line is unde'ned, but lines have to satisfy certain> axioms). It's often said that in principle anybody could study the> consequences of any consistent set of axioms whatever, but we know that in> practice most of those would be a waste of time. Why? And what about the> well-known question of why our abstract theories are so often applicable> to the real world? This seems to me not at all mysterious. We're not gods. Our> imaginations are both developed and restricted by our experiences in the> real world. The abstract structures which interest us are usually> in§uenced indirectly by reality, even if it's hidden at the back of our> minds. Pure mathematical theories built by human thought are very likely> to be relevant to the real world in some way, just because we and our> limited minds inhabit that world. So what about the original question? Technically, we're free to> _invent_ any axiomatic system we like, then _discover_ its consequences. > For example, the invention of the group axioms posed the problem of> discovering their consequences, including such discoveries as the Monster> group. But that philosophically simple view doesn't do justice to the> history or psychology of it all. A lot was known about groups before the> modern axioms were thought of, so those axioms weren't just somebody's> whim. That's where my earlier paragraphs come in. It was hard work to> discover what was the best thing to invent. :-) Ken Pledger.