mm-296 == Subject: Re: Totally stumped, linear algebra 5x5 matrix determinantI get2 -1 0 4 1> 0 2.5 -1 -4 -1.5> 0 0 -0.6 4.6 1.6> 0 0 0 29.6667 16.6667> 0 0 0 0 -1.55056Multiplication of diagonals yields 138.I tried (for about 1 hour tonight) to get my matrix reduced to the waythat you have it. And no luck thus far. However, one question(actually verification).When using Gaussian elimination and you factor out a number from a rowby dividing that row by the number, is it imperitave to save thatfactored value for the end and then multiply the diagonal product bythat row(s)?For example, if during the row operations, I factor out a 2 and a 3 from2 different rows. At the very end when I multiply the diagonal values,do I also need to make sure I multiply that answer by 6? My textbookis quite weak in explaining that but they did have one example (with noexplanation) where they factored out a 16 from a row and then they endedup multiplying the final answer by 16 again to account for the factthat 16 was factored out earlier. The factoring out of 16 was theirconvenience to end up with a diagonal of (1)(-1)(-1)(1)(10) which theygot 160 for the final answer. I know that with Gaussian elimination to solve a system, if I factoredout a number from a row, I did not need to save that factored valuewhen computing my results vector. > Maybe you could take a look at these online lectures:http://ocw.mit.edu/OcwWeb/Mathematics/18- 06Linear-AlgebraFall2002/VideoLectures/index.htmThank you so much for this link. I had no idea this exis. I will bespending alot of time here!!!thanks again!!!Confused. ==== Subject: Re: Totally stumped, linear algebra 5x5 matrix determinant> I get2 -1 0 4 1> 0 2.5 -1 -4 -1.5> 0 0 -0.6 4.6 1.6> 0 0 0 29.6667 16.6667> 0 0 0 0 -1.55056Multiplication of diagonals yields 138.> I tried (for about 1 hour tonight) to get my matrix reduced to the way> that you have it. And no luck thus far. However, one question> (actually verification).When using Gaussian elimination and you factor out a number from a row> by dividing that row by the number, is it imperitave to save that> factored value for the end and then multiply the diagonal product by> that row(s)?Yes.If A -> B by swapping rows of A then det(B) = -det(A);If A -> B by multiplying a row of A by r then det(B) = r*det(A)[So, if you divide a row of A by r, i.e. multiply by 1/r, det(B) = (1/r)*det(A)]If A -> B by adding a multiple of one row of A to another row thendet(A) = det(B).-- ==== Subject: Re: Explaining factorizations, relevanceDirk Van de moortel the exploits and important work of *past* mathematicians who have> done much to advance knowledge. However, it is problematic if... an idiot like you decides to try to understand some of that work?> Indeed, very true. No argument on that one. ==== Subject: Re: Explaining factorizations, relevance> Mathematicians today are inheritors of a rather deserved glory from> the exploits and important work of *past* mathematicians who have done> much to advance knowledge. However, it is problematic if> mathematicians today can rely on social support garnered from the> efforts of their predecessors to hide acts of wrongdoing today.> Here I'll outline the basic mathematics underlying a find of mine of a> strange error in an esoteric branch of mathematics called algebraic> number theory.> To understand posts on this subject, it will help if you understand> the basics of factorizations.> 1. Factorizations are just ways of looking at factors of some> product.> For instance, 2(3) is a factorization of 6, which tells you that 6 is> a product of 2 and 3.> That factorization is in an integer domain, which simply means that> you have only integers, so a factorization like 5(6/5) = 6, is not> valid in integers, though it is valid in some other domain like> rational numbers that has 1/5 as a member.> 2. Algebra allows a generalization of factorizations beyond specific> number examples, for instance> (x+2)(x+3) = x^2 + 5x + 6> which is true without regard to what domain x is in, so you can let x> be an integer, or it could be a fraction.> 3. If x is an integer, then you're guaranteed to be in the integer> domain, which logically follows from the fact that x is added to and> multiplied times other integers, as addition and multiplication are> what are called ring operations by mathematicians.> 4. Algebraic integers are roots of monic polynomials with integer> coefficients e.g. x^2 + 5x + 6, as monic just means that leading> coefficient is a factor of 1 in the integer domain.> Not surprisingly, if x is an algebraic integer, then x^2 + 5x + 6 is> an algebraic integer, and the factorization> (x+2)(x+3) = x^2 + 5x + 6 is in the algebraic integer domain.> What I've done is consider factorizations of polynomials, where the> factors are not polynomials.> 5. In an attempt at questioning an important conclusion a Rick> Decker, a professor at Hamilton College, made a post with his own> example of such a factorization, and I've used it partly because it's> a quadratic and easier to handle.> That factorization is> (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).> Notice that it is a factorization because you have factors> (5a_1(x) + 7) and (5a_2(x) + 7)> which multiply to give the product 7(25x^2 + 30x + 2).> Algebra allows me to make the substitution a_2(x) = b_2(x) - 1 to get> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> where the purpose is to balance out constant factors, which here are> numbers, on both sides of the equation.> Importantly, with that substitution, at x=0, a_1(0) = b_2(0) = 0, as I> emphasize that the 7 and 2 visible on the left multiply to give 7(2)> on the right.> That is, when you multiply through by 7 on the right side you have> 7(25)x^2 + 7(30)x + 7(2)> where the 7(2) is the product of the factors visible on the left in> (5a_1(x) + 7)(5b_2(x) + 2).> I have to elaborate to that level of detail because for mathematicians> to fool you about my discovery, you have to be confused on that simple> point.> Integers are rather easy to work with so now I can use them to> determine what is forced upon the variables.> Notice that multiplying out gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 => 175 x^2 + 210 x + 14.> That's important because I now attempt to eliminate the factor of 7> from both sides, and to do so while remaining in the domain of> algebraic integers.> 6. The logical conclusion is that the only way that might possibly> allow that is> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> as consider another possibility, like> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) => 25x^2 + 30x + 2> which gives the factorization> sqrt(7)(2/sqrt(7)) = 2> as multiplying out gives> 25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5b_2(x) + 2 => 25 x^2 + 30 x + 2> and the two factors are> (5a_1(x)/sqrt(7) + sqrt(7))> and> (5b_2(x)/sqrt(7) + 2/sqrt(7))> which internally have sqrt(7) and 2/sqrt(7) as the factors of 2.> But that second factor cannot be in the domain of algebraic integers> as 2/sqrt(7) is not an algebraic integer.> Generalizing, for any factorization of 7, with algebraic integers f_1> and f_2, where f_1 f_2 = 7, and f_1 is assumed to be a factor of> (5a_1(x) + 7) you will have> (5b_2(x)/f_2 + 2/f_2)> as a factor, and if 2/f_2 is not an algebraic integer, then that> factor is not in algebraic integers.> 7. Logically that follows from the requirement that for a> factorization to be in a particular domain, its factors must be in> that domain as well.> Like before, with 2(3) = 6, in the domain of integers that is valid,> while 5(6/5) = 6 is not valid in the domain of integers.> So if you're trying to remain in the domain of algebraic integers then> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> is the only possibility.> Mathematicians, however, for many years believed that if the roots of> a quadratic like> a^2 - (x - 1)a + 7(x^2 + x)> are not integers, then there's no way to figure out where factors of 7> might go between those roots, but they believed that they did split up> in some way such that each root would have some factor of 7 in the> ring of algebraic integers.> My work proves that wrong with simple algebraic principles by using> non-polynomial factors, which is the step mathematicians did not take.> Now algebraic integers have been defined to be the roots of monic> polynomials with integer coefficients, which is actually something of> an arbitrary definition.> Mathematicians found it useful for over a hundred years without fully> understanding its limitations, like how you can be forced out of that> domain as I have shown.> It could be a small mistake, but by trying to deny and hide it,> today's mathematicians are making a HUGE deal out of it, and bringing> their own credibility into question.> They are damaging their discipline, by not telling the truth fully and> honestly.> I've shown what is mathematically correct. It's been that way from> the beginning as the correct mathematics does not change.> Today's mathematicians gain nothing by hiding from the truth. After> all, in mathematics, when you're wrong, you're just wrong.> James, Admit it. You don't care about math; you just care about fame andfortune. Get a life, crank. ==== Subject: rounding integers===if you round -3.5 to 1dp, what is the answer?does 'round up' mean further right on the number line... giving -3or does it mean further from zero... giving -4what's the protocol.i await your answers ==== Subject: Re: rounding integers>mbaiou>if you round -3.5 to 1dp, what is the answer?-3.5 is already rounded to one decimal place.If you mean to one significant digit, then whether the answer is -3 or -4 depends on your definition of rounding. Some authors define it one way, some define it the other.If this is for a course, the only answer is to ask your instructor which method s/he uses.-- 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: complex powersThanks for your responses, and the book reference. I don't have a clear picture of what exponentiation by a complex number implies yet, but I'll keep working at it.mathedman, I think that there is always some way of visualizing a particular mathematical problem -- otherwise math would be reduced to just a formal system, and we would be no better at forming theorems than a well-programmed computer. It is our intuition that lets humans do the mathematical feats we do, and I think it reasonable to say a large part of intuition arises from visualization.BTW, the question came up because my prof gave us a review of a book entitled something like An Imaginary Tale: the History of i, and in it, it was sta that i^i is real. I can show that formulaicly, but I wan to be able to visualize it.Alex-- ==== Subject: Ability In MathematicsTO THE ALT MATHEMATICS UNDERGRAD NEWSGROUPI have an enqiry which may be of interest to this newsgroup:Firstly, I have to say that,personally,I have always found Mathematicsdifficult.It has hindered my understandings of Astronomy and I failedthe British 11+ examination BECAUSE of it and so,consequently, wentto a British Secondary Modern School as opposed to the moreacademic British Grammar School.Even though I had understandingand competence in every other subject studied both at school and,later, college.So, in at least two crucial ways, Mathematics has beena very significant subject in my life.So,given all this, have you any ideas as to why, for some people,Mathematics ability comes so naturally whilst other people findit so difficult?Clearly some people see patterns in Mathematicswhilst others,NO MATTER HOW MUCH THEY TRY, cannot.Why is this so? Brian Devonald ==== Subject: Re: Ability In Mathematics> So,given all this, have you any ideas as to why, for some people,> Mathematics ability comes so naturally whilst other people find> it so difficult?Clearly some people see patterns in Mathematics> whilst others,NO MATTER HOW MUCH THEY TRY, cannot.> Why is this so?What a depressing question, when you ask it. What do you mean, specifically,by mathematical ability?-- ==== Subject: Re: Ability In Mathematics> TO THE ALT MATHEMATICS UNDERGRAD NEWSGROUPI have an enqiry which may be of interest to this newsgroup:Firstly, I have to say that,personally,I have always found Mathematics> difficult.It has hindered my understandings of Astronomy and I failed> the British 11+ examination BECAUSE of it and so,consequently, went> to a British Secondary Modern School as opposed to the more> academic British Grammar School.Even though I had understanding> and competence in every other subject studied both at school and,> later, college.So, in at least two crucial ways, Mathematics has been> a very significant subject in my life.So,given all this, have you any ideas as to why, for some people,> Mathematics ability comes so naturally whilst other people find> it so difficult?Clearly some people see patterns in Mathematics> whilst others,NO MATTER HOW MUCH THEY TRY, cannot.Why is this so?Why does Michael Jordan have a better jump shot than me? I've always wondered about that. When I was young I loved to play basketball. But I wasn't very good. Not fair. ==== Subject: Re: Ability In MathematicsBRIAN@devonaldspace.freeserve.co.uk asks:>So,given all this, have you any ideas as to why, for some people,>Mathematics ability comes so naturally whilst other people find>it so difficult?Clearly some people see patterns in Mathematics>whilst others,NO MATTER HOW MUCH THEY TRY, cannot.>Why is this so?> Brian DevonaldLack of repea study.Some people are simply less mentally format for Mathematics than otherpeople, and will need much more study of the same material in order to learn itadequately. This is the answer for at least SOME people who find Math to bedifficult, but possibly not the answer for everybody who finds it difficult. G C ==== Subject: Re: Ability In Mathematics>So,given all this, have you any ideas as to why, for some people,>Mathematics ability comes so naturally whilst other people find>it so difficult?Clearly some people see patterns in Mathematics>whilst others,NO MATTER HOW MUCH THEY TRY, cannot.>Why is this so?Why are any talents not distribu equally to all members of the human race? That's just how it is. Some people are better at math than others, just as some are better at baseball than others.That said, I honestly believe that for the great majority of applied math courses, native math ability is far less important for success that good work habits.Gran, I see only the students at a two-year college. But for every one student who really seems to be trying hard yet can't get it, I see a dozen who never turn in the homework, never ask a question in or out of class, never visit the (free!) tutoring center, etc.Abraham Lincoln was talking about law school not mathematics when he said Always bear in mind that your own resolution to succeed, is more important than any other one thing; but I believe the same applies to math -- and I tell my students that.-- 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: integral problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UIL0l30916; === Hi Jane,Perhaps the question is asking you to prove that the function, f, isperiod in 1? (It could be that the function is period in less than 1).Take the derivative of both sides and use the Fundemental Theorem ofCalculus, and you're done!i.e.d/dt (int({x}..{x+1}) f(t) dt) = 0f(x + 1) - f(x) = 0f(x+1) = f(x)and this is the definition of periodicity.MM. ==== Subject: integral problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UIL0p30904; === Hi Jane,what about f(x) = 0, if x in R Zf(x) = x, if x in Z Maybe f is supposed to be continuous on R ?Yeah,you are right-i hadn't write that f is continious on R-it's myerror-escuse me.So,now i think that it must be right.Thanks!!! ==== Subject: Strange result X-no-archive: yesX-archive: no I'm taking a course on numerical methods and was trying to solve the equation 0 = e^{-x} - x. When using the fixpoint iteration formula x = - ln x I got strange results. Naturally that formula would give strange results since for a start value x_0 > 0 => x_1 < 0 and therefore x_2 is undefined(or complex). The iteration did something really strange though it didn't just diverge or converge to a complex number. It ended up alternating in a complex conjuga pair. Anyone know why this happens? I couldn't find anything at mathworld.wolfram.com on complex logarithms. Only found complex exponentiation and could not figure it out from there. Help apprecia!-- Sigblock empty. By choice. ==== Subject: Somebody please help me. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UNWRX26724; === Can anybody figure out this riddle? What has two eyes and lives inone? I've been trying for a week now and it's driving me crazy.Doesanybody have any idea. Please help.