mm-1529 === Subject: Sketches of trig functions Could you check if my workings are right, please? f(x) = 5 sin 3x.bc, 0 <= x <= 180 a) Sketch the graph of f(x)n indicating the value of x at each point where it intersects the x-axis. I drew a graph with the sin curve stretched between 5 and -5 on the y-scale and squashed by 1/3 along the x-axis. It intersects the x-axis at 0, 60, 120 and 180 .bc b) Write down the coordinates of all the maximum and minimum poits of f)x) Not too sure about this as axis different scales. MAX (30,5); (150,5) MIN (90,-5) Is that it? c) Calculate all the values of x for which f(x) = 2.5 5sin3x = 2.5 let y = 3x ... 5 siny = 2.5 .... siny = 0.5 ... y = 30.bc y= 3x = 30.bc ... x = 10.bc Soloutions: 10.bc and 170.bc Is that it? I doubt it... Jo === Subject: Re: Sketches of trig functions --------------------------------------------------------------------- > Could you check if my workings are right, please? > f(x) = 5 sin 3x.bc, 0 <= x <= 180 Ok, a) and b) > c) Calculate all the values of x for which f(x) = 2.5 > 5sin3x = 2.5 > let y = 3x ... 5 siny = 2.5 .... siny = 0.5 ... y = 30.bc > y= 3x = 30.bc ... x = 10.bc Since 0 <= x <= 180, 0 <= y <= 520 So find all the values of y. y = 30, 150, 390, 510 > Soloutions: 10.bc and 170.bc > Is that it? I doubt it... No, you missed two. Did you not look at graph and draw horizontal line thru (0, 2.5) to see what's happening? === Subject: formulas by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBSE0FC14833; last day of school before break I had a math test on formulas.I did fine today because I had my student refrence book but my teacher said next time that we wont have that option . Help me please I've always done well in school and I realy dont want to stop now. === Subject: Re: Basic Math, Help Me Please? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBSE0Fp14827; what do you need help with in math I can help I'm not an expeprt but I do know the basic's ( pre algebra , subtraction ,additon , divison, multiplacation. === Subject: Failed exams in the horizon! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBSE0Ef14758; By completing the square, find in terms of k the roots of the equation: x^2 + 2kx - 7 = 0 The best I can do is (x + k)^2 - k^2 -7 = 0 And if that is ok, then what?... Help Jo === Subject: Re: Failed exams in the horizon! >> The best I can do is (x + k)^2 - k^2 -7 = 0 >> And if that is ok, then what?... Assume a*x^2 + k*x - c = 0 then the perfect square would look like this: a( x + k/2 )^2 - (k^2) / 4 + c In your case: x^2 + 2kx - 7 = 0 (x + (2k)/2 )^2 - (2k)^2 /4 -7 (Note the term k^2 /4 instead of just k^2, i.e. you forgot to divide by 4 -- Ronny Mandal === Subject: Re: Failed exams in the horizon! > By completing the square, find in terms of k the roots of the > equation: > x^2 + 2kx - 7 = 0 > The best I can do is (x + k)^2 - k^2 -7 = 0 > And if that is ok, then what?... What you doing? Solving for x? Then add k^2 + 7 to both sides, ie start to isolate x. (x + k)^2 = k^2 + 7 Can you take it from there? === Subject: Re: Surrogate factoring, revisited > For some time now I've been exploring ideas having to do with what I > call surrogate factoring, where you try to factor one number by instead > factoring another, which I call the surrogate. Which, in this case, turns out to be (A^4)(T^2), where you state in the paper that you have no idea how to determine A and T. > Does it work? > I don't know. But that didn't stop you from stating, in the conclusion: So, incredibly, the factoring problem is a simple one, where you can shift the factorization of one integer into a question of factoring another, which I call the surrogate. The use of a non-monic quadratic is clearly key here. So the factoring problem is now solved. > The theory is intriguing enough to talk about, and just > to be sure I did send the paper to some experts, as well as some people > in the US government. I hope those US government folks don't get too upset when they realize you lied to them about solving the factoring problem. Lying to the government about matters of National Security isn't any sort of crime, is it? === Subject: Re: Looking for love or just some fun. Please contact me........................................ 2G85 message <131399941352984064@alice.com>: > Hi > Toh tarolih wicosa das fitigoba mayamumesep . I wouldn't be interested but there's a hot 22-year-old on this group that speaks the same language. Perhaps you should get in touch. -- Paul Townsend Pair them off into threes Interchange the alphabetic letter groups to reply === Subject: Re: Pierre de Fermat:1605-2005 ( 400 years Anniversary) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j07JtOR13815; Hi stefan, Yes,Officially you are Right.My memory registrated 1605 because the date of his birthday is controversial.As you can see in another posting of mine with the title First part I said that he was Anyway,I still encourage anybody which likes theory of numbers and likes to defend Fermat's reputation as a great mathematicien to try to find the whole proof as indicated in this posting. george ghiata P.S.It looks that I have to be wrong or believed to be wrong to recieve any reply to my postings.Exept MO reply to my >> This year will be 400 years from the time Pierre de Fermat was born. >Fermat was born in 1601. >-- >Stefan Holm >They're mutant! They're deformed! They're funky! === Subject: Re: To mr.Michael Orion About Fermat's proof by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j07JtQb13859; Hi Will, It looks that it is posted now. george ghiata >> Hi Michael, >> I 'd like to post Fermat's Proof of Last theorem as soon as I get >> the e-mail address of this discussion group.Do you know if that is >> possible and what is the policy about this? >> I sent a posting like this to Sci.math.num-analiys discussion group >> and is not posted. >sci.math.num-analysis is, I believe, a moderated group. The moderator >may not have felt that your post was legitimate and axed it, or simply >not gotten to it yet. >-- >Will Twentyman >email: wtwentyman at copper dot net === Subject: Re: Logarithmic equation with base 1/2 >>log_a (x) is an increasing function when a > 1. When 0 < a < 1, >>log_a (x) is a decreasing function. And I should have remembered: log_(1/2) (x) = log_10(x) / log_10(1/2) The denominator is negative constant and the numerator is increasing everywhere; therefore the function is decreasing everywhere. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 === Subject: Re: integral on curve > how do i solve this with integral on curve: > area of the plane y-5z=1 that bounded between x=1-z^2,x=z^2-1 is.. It'dbeeasiertoexplainthisifyouusednospacesatall,insteadofnospacesformath. === Subject: 0 times infinity Hi When working with the extended real number system, we define 0.infinity = 0 Why? This appears to be in contradiction to the definitions 1/0 = infinity and a/infinity = 0 Surely, we should simply state that 0.infinity is undefined. Phil === Subject: Re: 0 times infinity reply-type=response I'm may be getting out of my depth here, but do we have to make a distinction between a countable and an uncountable infinity. By countable, I mean for example the sum of an infinite series. By uncountable, I mean for example the union of all the single points in the interval [0,1). In this case, it seems reasonable that the length of a single point is 0, but the uncountable infinite sum of the lengths of all the points in the interval [0,1) is 1. This implies (to me) that 0*oo can be equal to 1. But as I say, I'm getting out of my depth. Phil === Subject: Re: 0 times infinity >I'm may be getting out of my depth here, but do we have to make a >distinction between a countable and an uncountable infinity. >By countable, I mean for example the sum of an infinite series. >By uncountable, I mean for example the union of all the single points in the >interval [0,1). In this case, it seems reasonable that the length of a >single point is 0, but the uncountable infinite sum of the lengths of all >the points in the interval [0,1) is 1. This implies (to me) that 0*oo can be >equal to 1. But as I say, I'm getting out of my depth. Uncountable sums do not exist at all. For countable sums there is a definition using limit, but this definition relies on the ability to enumerate the sum members. So we can only talk about countable sums, an as I've explained in previous post, such sum of zeros is equal to zero. Also, length of the point is also an undefined term. If you mean the standard Lebesque measure M, then, really M({x})=0, M( [0,1) ) = 1. But this statement is false: M( [0,1) ) = sum of M({x}) for each 0<=x<1. === Subject: Re: 0 times infinity them! > I'm may be getting out of my depth here, but do we have to make a > distinction between a countable and an uncountable infinity. > By countable, I mean for example the sum of an infinite series. > By uncountable, I mean for example the union of all the single points in the > interval [0,1). In this case, it seems reasonable that the length of a > single point is 0, but the uncountable infinite sum of the lengths of all > the points in the interval [0,1) is 1. This implies (to me) that 0*oo can be > equal to 1. But as I say, I'm getting out of my depth. > Phil Why choose a length of 1? Isn't this just the old k/n * n = k, let n -> oo, therefore 0*oo = k argument which demonstrates that 0*oo is indeterminate? (i.e. divide an interval [0,k] into n parts and let the parts get infinitely many and infinitely small) === Subject: Re: 0 times infinity > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > Surely, we should simply state that 0.infinity is undefined. I am afraid so... The simple equation 0*oo=X <==> X/oo = 0 <==> X*0 = 0 shows that X can be any real value. Is it that hard to believe? I like to think that the above equation has the dimension 0 and 1 variable. Logically if the dimension is smaller than the number of variables at your disposal, there is no unique solution! You can compare this with X + Y =1 (1 is randomly chosen), we don't have enough information to solve the equation, we have dimension 1( a line) and 2 variables. so our solution will always be a couple (X,Y) eg (0.5,0.5), but there are infinitely many solutions. Yet nobody will oppose me if I say the solution is undefined, because they will say it is defined by the above equation. Well, you can also look at our first equation like that... (infinite solutions of dimension 1) Alban De Schutter > Phil === Subject: Re: 0 times infinity > 0*oo=X <==> X/oo = 0 That only holds if infinity is supposed to be invertible in the extended reals. There really isn't any reason for that assumption. The definition 0*oo = 0 gives a fully consistent system (it's not a field, but that's really not a problem) which proves very useful in measure theory. -- Stefan Holm I called my insurance company and they don't cover losses to rifts in the space-time continuum. === Subject: Re: 0 times infinity Hello. Math programs like Mathematica call it Indeterminate. In[1]:= 0*Infinity Out[1]= Indeterminate -- Dana > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > Surely, we should simply state that 0.infinity is undefined. > Phil === Subject: Re: 0 times infinity >When working with the extended real number system, we define >0.infinity = 0 >Why? >This appears to be in contradiction to the definitions >1/0 = infinity >and >a/infinity = 0 >Surely, we should simply state that 0.infinity is undefined. In _some_ situations we define 0*infinity = 0, while in other situations we _do_ say that it is undefined. What's the context where you say we do define it to equal 0? (One situation where we say it = 0 is in integration theory, ie measure theory. That's because if f(x) = 0 we want to have the integral of f(x) from -infinity to infinity to equal 0, and we also want to say that if f = c times g, where g = 1 on a set E and 0 off E, then the integral of f is c times the measure of E.) >Phil ************************ David C. Ullrich === Subject: Re: 0 times infinity Measure theory is actually the case in point. I am trying to understand the construction of a non measureable subset of IR, i.e a non Borel set. I have seen several such constructions, mostly quite similar. For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf (Problem 6) or http://www.ma.umist.ac.uk/pas/341/not3b.pdf (I searched Google for construct non measurable borel). I must admit that I have difficulty understanding these constructions, but they all seem to implicitly assume 0 * oo = 0 which I am not really happy with. Phil === Subject: Re: 0 times infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LESN26329; I wish I could write it here, but it seems, I can't. The message posted over an hour ago isn't displayed. === Subject: Re: 0 times infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LERA26276; >Measure theory is actually the case in point. >I am trying to understand the construction of a non measureable subset of >IR, i.e a non Borel set. I have seen several such constructions, mostly >quite similar. >For example in http://www.maths.b ris.ac.uk/~maxsv/EDU/p3hw1.pdf (Problem 6) May be it's better to explain the method shown in Problem 6 itself. I've read it and I trink I've found what place there confused you. Since there are infinitely many Eq's we must have P(Eq)=0 !!!and so P(UEq)=0!!!... Here we are using the sigma-addiction of measure: P(UEq)=Sum{P(EQ)) where Sum is the sum of infinite sequence (sorry for the terms, but I don't know the English for math terms). By definition, such infinite Sum is a limit of finite sums of first n elements where n goes infinity: S = lim{n->oo}(Sn), Sn = P(E1)+P(E2)+...+P(En), Surely, any finite sum of zeros is zero ==> The limit of zeros is also zero ==> P(UEq)=0. By the way, no part of the analysis theory uses formulas like 0*oo=something or 1/0=something - it's short for well mathematically defined limits. for example lim(1/x) where x>0 and x->0 is infinity, so we shortly write 1/(+0) = oo but it means only the former statement. === Subject: Re: 0 times infinity >>Measure theory is actually the case in point. >>I am trying to understand the construction of a non measureable >subset of >>IR, i.e a non Borel set. I have seen several such constructions, >mostly >>quite similar. >>For example in href=http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf>http://www.maths. bris.ac.uk/~maxsv/EDU/p3hw1.pdf >(Problem 6) >May be it's better to explain the method shown in Problem 6 itself. >I've read it and I trink I've found what place there confused you. >Since there are infinitely many Eq's we must have P(Eq)=0 !!!and so >P(UEq)=0!!!... >Here we are using the sigma-addiction of measure: >P(UEq)=Sum{P(EQ)) where Sum is the sum of infinite sequence (sorry >for the terms, but I don't know the English for math terms). >By definition, such infinite Sum is a limit of finite sums of first n >elements where n goes infinity: >S = lim{n->oo}(Sn), Sn = P(E1)+P(E2)+...+P(En), >Surely, any finite sum of zeros is zero ==> The limit of zeros is also >zero ==> P(UEq)=0. Having just pointed out that something you say below is not true, I should also say that you have an excellent point above. In fact 0*infinity does not come up in _this_ argument at all - all that's used is 0 + 0 + ... = 0. >By the way, no part of the analysis theory uses formulas like >0*oo=something or 1/0=something - it's short for well >mathematically defined limits. >for example lim(1/x) where x>0 and x->0 is infinity, so we shortly >write 1/(+0) = oo but it means only the former statement. ************************ David C. Ullrich === Subject: Re: 0 times infinity Hello David, Hello Zruty, I guess you're right, but non measurable sets still bother me. What I would like is an extended number system of some kind, that includes values for the measure of these constructed sets. This would have to be a different kind of 0, lets say 0c, such that the countable sum 0c + 0c + ... would be a finite number. If you accept the idea of different kinds of infinity (countable, uncountable, ...), then different kinds of zero don't seem too outrageous. After all, non measureable sets are pretty outrageous themselves. But that is just my musing and its probably complete nonsense, and as I have already said, I'm out of my depth. Phil === Subject: Re: 0 times infinity >Hello David, Hello Zruty, >I guess you're right, but non measurable sets still bother me. >What I would like is an extended number system of some kind, that includes >values for the measure of these constructed sets. This would have to be a >different kind of 0, lets say 0c, such that the countable sum 0c + 0c + ... >would be a finite number. I'm afraid it will lead us to contradiction. Assume a function f defined as f(n)=0c, where n is natural number. Then Sum{f(k)} = 0c+0c+0c+...+0c+... = x = const. Let's define g(n) = {0c, if n=2k, {0, if n=2k+1. Then Sum{g(k)} = 0+0c+0+0c+... = what? 1) Sum = x. Then we need Sum{f(n)-g(n)} = x-x = 0, but f(n)-g(n) = g(n-1) and Sum{f(n)-g(n)} = x. 2) Sum = y=0 so we can rearrange the sum as 0 + 0 + 0 + 0c + 0 + 0 + 0 + 0c + ..., which, by similar argumentation, equals to z100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- === Subject: Re: 0 times infinity >Hello David, Hello Zruty, >I guess you're right, but non measurable sets still bother me. Well that's life - deal with it... >What I would like is an extended number system of some kind, that includes >values for the measure of these constructed sets. This would have to be a >different kind of 0, lets say 0c, such that the countable sum 0c + 0c + ... >would be a finite number. >If you accept the idea of different kinds of infinity (countable, >uncountable, ...), then different kinds of zero don't seem too outrageous. >After all, non measureable sets are pretty outrageous themselves. I don't see why. You're used to the idea that any subset of R has a measure, just because when you think subset of R your mental picture is of a fairly simple set, a few intervals plus or minus a sequence of points or something. Why should _every_ set have a measure? Two comments. First, if you really want to be bothered by something check out a thing called the Banach-Tarski paradox: A ball in R^3 can be cut into finitely many pieces in such a way that the pieces can be rearranged (by rigid motions) so their union is another ball of whatever size you want! A pea gets sliced into finitely many pieces which when rearranged give a ball the size of the Sun. Obviously impossible, right? Well, if we allowed uncountably many pieces instead of finitely many it would be obviously possible - just take the individual points and move each one farther from the center. Of course it _is_ obviously impossible with actual physical pieces of a physical pea - the moral is that subset of R is a more abstract concept than you thought. Second: All these strange things require the Axiom of Choice. It's possible to prove that if you don't assume the Axiom of Choice then it's impossible to prove that a non-measurable set exists (and hence it's certainly impossible to prove the B-T paradox). This means that any time you explicitly construct a set it's going to measurable - to get a non-measurable set some sort of non-explicit application of AC must have been used. (Why not take the B-T paradox as proof that AC is simply wrong? Well, we can't live without AC. And there's no actual contradiction involved with the B-T paradox - seems very strange at first but you get used to it. It's also possible to prove that if AC leads to an actual contradiction then it's also possible to get a contradiction from set theory without AC: while AC has some consequences that seem very curious it's provably not responsible for any actual contradition.) >But that is just my musing and its probably complete nonsense, and as I have >already said, I'm out of my depth. >Phil ************************ David C. Ullrich === Subject: Re: 0 times infinity > Hello David, Hello Zruty, > I guess you're right, but non measurable sets still bother me. > What I would like is an extended number system of some kind, that > includes values for the measure of these constructed sets. This would > have to be a different kind of 0, lets say 0c, such that the countable > sum 0c + 0c + ... would be a finite number. > If you accept the idea of different kinds of infinity (countable, > uncountable, ...), then different kinds of zero don't seem too > outrageous. After all, non measureable sets are pretty outrageous > themselves. > But that is just my musing and its probably complete nonsense, and as I > have already said, I'm out of my depth. I don't think you really want different kinds of zero. I suggest that you want infinitesimals (other than just zero itself). You might consider J. H. Conway's system of surreal numbers. It's a Field. Division by zero is not defined there. 0*x is always 0. There are plenty of different infinite elements, along with their multiplicative inverses, which are infinitesimal. A Google web search for surreal numbers should give you many excellent links. David Cantrell === Subject: Re: 0 times infinity >>Measure theory is actually the case in point. >>I am trying to understand the construction of a non measureable >subset of >>IR, i.e a non Borel set. I have seen several such constructions, >mostly >>quite similar. >>For example in href=http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf>http://www.maths. bris.ac.uk/~maxsv/EDU/p3hw1.pdf >(Problem 6) >May be it's better to explain the method shown in Problem 6 itself. >I've read it and I trink I've found what place there confused you. >Since there are infinitely many Eq's we must have P(Eq)=0 !!!and so >P(UEq)=0!!!... >Here we are using the sigma-addiction of measure: >P(UEq)=Sum{P(EQ)) where Sum is the sum of infinite sequence (sorry >for the terms, but I don't know the English for math terms). >By definition, such infinite Sum is a limit of finite sums of first n >elements where n goes infinity: >S = lim{n->oo}(Sn), Sn = P(E1)+P(E2)+...+P(En), >Surely, any finite sum of zeros is zero ==> The limit of zeros is also >zero ==> P(UEq)=0. >By the way, no part of the analysis theory uses formulas like >0*oo=something or 1/0=something - it's short for well >mathematically defined limits. Not true. That's true of elementary calculus, perhaps, but honest actual 0*infinity=)'s are definitely part of measure theory. >for example lim(1/x) where x>0 and x->0 is infinity, so we shortly >write 1/(+0) = oo but it means only the former statement. ************************ David C. Ullrich === Subject: Re: 0 times infinity >>By the way, no part of the analysis theory uses formulas like >>0*oo=something or 1/0=something - it's short for well >>mathematically defined limits. >Not true. That's true of elementary calculus, perhaps, but >honest actual 0*infinity=)'s are definitely part of measure >theory. >************************ >David C. Ullrich Can you please give me an example of statement in measure theory, where 0*infinity = smth us used in it's proof? I'm sure there isn't any. Everywhere where mathematicians use infinity, they assume that there is some limit. (In this particular case we had lim0 = 0) === Subject: Re: 0 times infinity >By the way, no part of the analysis theory uses formulas like >0*oo=something or 1/0=something - it's short for well >mathematically defined limits. >>Not true. That's true of elementary calculus, perhaps, but >>honest actual 0*infinity=)'s are definitely part of measure >>theory. >>************************ >>David C. Ullrich >Can you please give me an example of statement in measure theory, >where 0*infinity = smth us used in it's proof? I've given such an example several times in this thread. If f = c*1_E (so f = c on E and f = 0 off E) then int f d mu = c mu(E). That's true regardless of the values of c >= 0 and mu(E). If c = 0 and mu(E) = infinity then this requires that we set 0*infinity = 0. >I'm sure there isn't any. Everywhere where mathematicians use >infinity, they assume that there is some limit. Well you can be sure of whatever you want. It's hard for me to see how you can _know_ that something does not appear in _any_ standard text, since you haven't read every book in the world... Ok, I happen to have a copy of Folland Real Analysis here at home. On page 11 in my edition, possibly on another page in another edition (check the index for extended real number system) he says explicitly that he's going to honor the convention that 0*infinity = 0. >(In this particular case we had lim0 = 0) Actually, as has been pointed out, 0*infinity doesn't actually arise in this case at all, all that actually comes up is the sum 0 + 0 + ... . ************************ David C. Ullrich === Subject: Re: 0 times infinity >Measure theory is actually the case in point. >I am trying to understand the construction of a non measureable subset of >IR, i.e a non Borel set. I have seen several such constructions, mostly >quite similar. >For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf (Problem 6) >or http://www.ma.umist.ac.uk/pas/341/not3b.pdf >(I searched Google for construct non measurable borel). >I must admit that I have difficulty understanding these constructions, but >they all seem to implicitly assume >0 * oo = 0 >which I am not really happy with. It's just a definition. Why you're not happy with it is not so clear. You read the rest of my post, right? Say 1_E is the characteristic function of E (the function which equals 1 on E and 0 elsewhere.) Are you happy with both of the following? (i) int c 1_E dm = c m(E) if c is a constant (and m is a measure). (ii) int 0 dm = 0. Both (i) and (ii) seem like perfectly natural things that you should want to be true. But if we're going to have both (i) and (ii) hold we need to define 0*infinity = 0. This really has very little to do with the construction of a nonmeasurable set per se, 0*infinity = 0 comes up all the time in measure theory, hence all the time in analysis. >Phil ************************ David C. Ullrich === Subject: Re: 0 times infinity > Measure theory is actually the case in point. > I am trying to understand the construction of a non measureable subset of > IR, i.e a non Borel set. I have seen several such constructions, mostly > quite similar. > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf (Problem 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > (I searched Google for construct non measurable borel). > I must admit that I have difficulty understanding these constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. well if you think of infinity as the largest possible number you can think of, then thik of one larger... then that number whatever it is, when multiplied by zero will give zero, even if it is infinitely large, so yes crudely out '0 time infinity' is zero, as is anything > Phil === Subject: Re: 0 times infinity > Measure theory is actually the case in point. > I am trying to understand the construction of a non measureable subset of > IR, i.e a non Borel set. I have seen several such constructions, mostly > quite similar. > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > (I searched Google for construct non measurable borel). > I must admit that I have difficulty understanding these constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > well if you think of infinity as the largest possible number you can think > of, then thik of one larger... then that number whatever it is, when > multiplied by zero will give zero, even if it is infinitely large, so yes > crudely out '0 time infinity' is zero, as is anything > Phil Statements like this don't make any sense (not to me, anyway). The largest possible number you can think of is necessarily finite (and so is the number one larger), so essentially you are saying that infinity is finite. If infinity were finite then 0*oo would undoubtedly equal zero, but unfortunately it isn't. === Subject: Re: 0 times infinity > Measure theory is actually the case in point. > > I am trying to understand the construction of a non measureable > subset of > IR, i.e a non Borel set. I have seen several such constructions, > mostly > quite similar. > > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > well if you think of infinity as the largest possible number you can > think > of, then thik of one larger... then that number whatever it is, when > multiplied by zero will give zero, even if it is infinitely large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > Statements like this don't make any sense (not to me, anyway). The > largest possible number you can think of is necessarily finite (and so > is the number one larger), so essentially you are saying that infinity > is finite. If infinity were finite then 0*oo would undoubtedly equal > zero, but unfortunately it isn't. No I meant 'go one larger' figuratively rather than literally. Indeed people tend to think of infinity as being 'a number', when its a concept. To express ideas involving infinity really should be handled as limits and not as something like '0 * infinity = 0' , should really be re-expressed as '1/x tends to infinity as x tends to zero'. But having said that, anything multiplied by zero is always zero... even infinitely large numbers. === Subject: Re: 0 times infinity > Indeed > people tend to think of infinity as being 'a number', when its a concept. If someone uses infinity in a vague fashion, then it's reasonably to say that their infinity is just a concept. But this thread began by asking a question specifically in the context of the extended real number system. In such a system, infinity is a specific mathematical object, and may reasonably be considered to be a number > To express ideas involving infinity really should be handled as limits > and not as something like '0 * infinity = 0' , should really be > re-expressed as '1/x tends to infinity as x tends to zero'. Good luck expressing 0 * infinity = 0 in terms of limits! The function f(x, y) = x * y has an essential singularity at (x, y) = (0, infinity). David Cantrell === Subject: Re: 0 times infinity > Indeed > people tend to think of infinity as being 'a number', when its a concept. > If someone uses infinity in a vague fashion, then it's reasonably to say > that their infinity is just a concept. But this thread began by asking > a question specifically in the context of the extended real number > system. In such a system, infinity is a specific mathematical object, and > may reasonably be considered to be a number > To express ideas involving infinity really should be handled as limits > and not as something like '0 * infinity = 0' , should really be > re-expressed as '1/x tends to infinity as x tends to zero'. sorry that statement was a mistake. what i meant was that the vague term '0 * infinity = zero' should be replaced using limits, whereby 'x * zero = zero' as x tends to infinity. > Good luck expressing 0 * infinity = 0 in terms of limits! The function > f(x, y) = x * y has an essential singularity at (x, y) = (0, infinity). > David Cantrell === Subject: Re: 0 times infinity <0o8Ed.43$0n1.15@newsfe3-gui.ntli.net> > Measure theory is actually the case in point. > > I am trying to understand the construction of a non measureable > subset of > IR, i.e a non Borel set. I have seen several such constructions, > mostly > quite similar. > > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > > well if you think of infinity as the largest possible number you can > think > of, then thik of one larger... then that number whatever it is, when > multiplied by zero will give zero, even if it is infinitely large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > Statements like this don't make any sense (not to me, anyway). The > largest possible number you can think of is necessarily finite (and so > is the number one larger), so essentially you are saying that infinity > is finite. If infinity were finite then 0*oo would undoubtedly equal > zero, but unfortunately it isn't. > No I meant 'go one larger' figuratively rather than literally. Indeed > people tend to think of infinity as being 'a number', when its a concept. > To express ideas involving infinity really should be handled as limits and > not as something like '0 * infinity = 0' , should really be re-expressed as > '1/x tends to infinity as x tends to zero'. Well then, what is the limit of x*oo as x tends to zero? === Subject: Re: 0 times infinity > Measure theory is actually the case in point. > > I am trying to understand the construction of a non measureable > subset of > IR, i.e a non Borel set. I have seen several such > constructions, > mostly > quite similar. > > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > > well if you think of infinity as the largest possible number you > can > think > of, then thik of one larger... then that number whatever it is, > when > multiplied by zero will give zero, even if it is infinitely > large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > > Statements like this don't make any sense (not to me, anyway). The > largest possible number you can think of is necessarily finite (and > so > is the number one larger), so essentially you are saying that > infinity > is finite. If infinity were finite then 0*oo would undoubtedly > equal > zero, but unfortunately it isn't. > No I meant 'go one larger' figuratively rather than literally. > Indeed > people tend to think of infinity as being 'a number', when its a > concept. > To express ideas involving infinity really should be handled as > limits and > not as something like '0 * infinity = 0' , should really be > re-expressed as > '1/x tends to infinity as x tends to zero'. > Well then, what is the limit of x*oo as x tends to zero? you've sort of mixed nomenclatures there. as i've said the statement 'x * infinity' is meaningless anyway, so theres no point in arguing about it in the first place. If you mean what is the limit x* y as x tends to zero and y tends to infinity, then there isnt one. But , the limit of 'x * zero' as x tends to infinity is zero. which is the way that 'x * infinity' should meaningfully be expressed. === Subject: Re: 0 times infinity <0o8Ed.43$0n1.15@newsfe3-gui.ntli.net> > Measure theory is actually the case in point. > > I am trying to understand the construction of a non measureable > subset of > IR, i.e a non Borel set. I have seen several such > constructions, > mostly > quite similar. > > For example in http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > > well if you think of infinity as the largest possible number you > can > think > of, then thik of one larger... then that number whatever it is, > when > multiplied by zero will give zero, even if it is infinitely > large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > > Statements like this don't make any sense (not to me, anyway). The > largest possible number you can think of is necessarily finite (and > so > is the number one larger), so essentially you are saying that > infinity > is finite. If infinity were finite then 0*oo would undoubtedly > equal > zero, but unfortunately it isn't. > > No I meant 'go one larger' figuratively rather than literally. > Indeed > people tend to think of infinity as being 'a number', when its a > concept. > To express ideas involving infinity really should be handled as > limits and > not as something like '0 * infinity = 0' , should really be > re-expressed as > '1/x tends to infinity as x tends to zero'. > Well then, what is the limit of x*oo as x tends to zero? > you've sort of mixed nomenclatures there. as i've said the statement > 'x * infinity' is meaningless anyway, so theres no point in arguing about > it in the first place. > If you mean what is the limit x* y as x tends to zero and y tends to > infinity, then there isnt one. > But , the limit of 'x * zero' as x tends to infinity is zero. which is the > way that > 'x * infinity' should meaningfully be expressed. I assume your last line should read which is the way that 'infinity * 0' should meaningfully be expressed? If you allow 0*oo = lim {n->oo: 0*n} = 0 then you have to also allow, for x =/= 0 x*oo = lim {n->oo: x*n} = oo Now let x tend to zero and x*oo = oo, oo, oo and always oo, therefore 0*oo = oo. The point I'm making is that you can do what you like with these limit arguments... you can make 0*oo into whatever you want... === Subject: Re: 0 times infinity > > Measure theory is actually the case in point. > > I am trying to understand the construction of a non > measureable > subset of > IR, i.e a non Borel set. I have seen several such > constructions, > mostly > quite similar. > > For example in > http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > > well if you think of infinity as the largest possible number > you > can > think > of, then thik of one larger... then that number whatever it > is, > when > multiplied by zero will give zero, even if it is infinitely > large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > > Statements like this don't make any sense (not to me, anyway). > The > largest possible number you can think of is necessarily finite > (and > so > is the number one larger), so essentially you are saying that > infinity > is finite. If infinity were finite then 0*oo would undoubtedly > equal > zero, but unfortunately it isn't. > > No I meant 'go one larger' figuratively rather than literally. > Indeed > people tend to think of infinity as being 'a number', when its a > concept. > To express ideas involving infinity really should be handled as > limits and > not as something like '0 * infinity = 0' , should really be > re-expressed as > '1/x tends to infinity as x tends to zero'. > Well then, what is the limit of x*oo as x tends to zero? > you've sort of mixed nomenclatures there. as i've said the statement > 'x * infinity' is meaningless anyway, so theres no point in arguing > about > it in the first place. > If you mean what is the limit x* y as x tends to zero and y tends > to > infinity, then there isnt one. > But , the limit of 'x * zero' as x tends to infinity is zero. which > is the > way that > 'x * infinity' should meaningfully be expressed. > > I assume your last line should read which is the way that 'infinity * > 0' should meaningfully be expressed? > If you allow > 0*oo = lim {n->oo: 0*n} = 0 > then you have to also allow, for x =/= 0 > x*oo = lim {n->oo: x*n} = oo > Now let x tend to zero and x*oo = oo, oo, oo and always oo, therefore > 0*oo = oo. no, you're mixing nomenclatures again, interchanging the symbolism of limits with that of ordinary arithmetic. again you're treating infinity as if it is an arithmetic entity. it is not. > The point I'm making is that you can do what you like with these limit > arguments... you can make 0*oo into whatever you want... we'll have to agree to disagree on this one then :) But i still think that you are confusing yourself by thinking of infinity as being a numer per se. Its an idea or a concept, not a number, so that to write say 'x = infinity' is meaningless, but 'x tends to infinity' is not - as the former treats infinity as a number, the latter treats it as a concept. so when dealing with infinity, always think in terms of limits. === Subject: Re: 0 times infinity <0o8Ed.43$0n1.15@newsfe3-gui.ntli.net> > Measure theory is actually the case in point. > > I am trying to understand the construction of a non > measureable > subset of > IR, i.e a non Borel set. I have seen several such > constructions, > mostly > quite similar. > > For example in > http://www.maths.bris.ac.uk/~maxsv/EDU/p3hw1.pdf > (Problem > 6) > or http://www.ma.umist.ac.uk/pas/341/not3b.pdf > > (I searched Google for construct non measurable borel). > > I must admit that I have difficulty understanding these > constructions, but > they all seem to implicitly assume > 0 * oo = 0 > which I am not really happy with. > > well if you think of infinity as the largest possible number > you > can > think > of, then thik of one larger... then that number whatever it > is, > when > multiplied by zero will give zero, even if it is infinitely > large, so > yes > crudely out '0 time infinity' is zero, as is anything > > Phil > > > > Statements like this don't make any sense (not to me, anyway). > The > largest possible number you can think of is necessarily finite > (and > so > is the number one larger), so essentially you are saying that > infinity > is finite. If infinity were finite then 0*oo would undoubtedly > equal > zero, but unfortunately it isn't. > > No I meant 'go one larger' figuratively rather than literally. > Indeed > people tend to think of infinity as being 'a number', when its a > concept. > To express ideas involving infinity really should be handled as > limits and > not as something like '0 * infinity = 0' , should really be > re-expressed as > '1/x tends to infinity as x tends to zero'. > Well then, what is the limit of x*oo as x tends to zero? > > you've sort of mixed nomenclatures there. as i've said the statement > 'x * infinity' is meaningless anyway, so theres no point in arguing > about > it in the first place. > > If you mean what is the limit x* y as x tends to zero and y tends > to > infinity, then there isnt one. > > But , the limit of 'x * zero' as x tends to infinity is zero. which > is the > way that > 'x * infinity' should meaningfully be expressed. > > > > I assume your last line should read which is the way that 'infinity * > 0' should meaningfully be expressed? > If you allow > 0*oo = lim {n->oo: 0*n} = 0 > then you have to also allow, for x =/= 0 > x*oo = lim {n->oo: x*n} = oo > Now let x tend to zero and x*oo = oo, oo, oo and always oo, therefore > 0*oo = oo. > no, you're mixing nomenclatures again, interchanging the symbolism of limits > with that of ordinary arithmetic. again you're treating infinity as if it > is an arithmetic entity. it is not. > The point I'm making is that you can do what you like with these limit > arguments... you can make 0*oo into whatever you want... > we'll have to agree to disagree on this one then :) I guess so ... I'm still not convinced that 0*oo has a well-defined interesting discussion! matt === Subject: Re: 0 times infinity > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? Who's we? At http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html (which usually seems to be a fairly reliable site, give or take the odd typo) it says there is no consensus about what 0*oo should mean in extended real numbers. Some say it's undefined, others define it as zero as a convention which is useful in certain contexts. I thought about this once, and... Doing it in the most direct way possible: what is the limit of 0*n as n -> oo? As n increases indefinitely the value of 0*n is 0,0,0,0 ... and always 0, so you could argue that the limit is zero. On the other hand, using various other limit arguments you can make 0*oo whatever you like. Therefore I have always considered 0*oo to be undefined. I'm not sure if there's a right or wrong answer ... I think it's more a question of definition. If you want to define it some way that makes sense in the context of what you're doing then you're free to go ahead and do that... Maybe! === Subject: Re: 0 times infinity > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? because anything multiplied by zero is zero > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 'infinity' is not a number, and if you think of it and treat as a number you run into problems. it is a concept not a number. something like '1/0 = infinity' is mathematical nonsense really. > Surely, we should simply state that 0.infinity is undefined. > Phil === Subject: Re: 0 times infinity > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? > because anything multiplied by zero is zero > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > 'infinity' is not a number, and if you think of it and treat as a number you > run into problems. it is a concept not a number. something like '1/0 = > infinity' is mathematical nonsense really. Isn't infinity a number in the extended real number system? One does (limited) arithmetic with it. === Subject: Re: 0 times infinity > Hi > > When working with the extended real number system, we define > 0.infinity = 0 > Why? > because anything multiplied by zero is zero > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > 'infinity' is not a number, and if you think of it and treat as a number you > run into problems. it is a concept not a number. something like '1/0 = > infinity' is mathematical nonsense really. > Isn't infinity a number in the extended real number system? One does > (limited) arithmetic with it. No you cannot do any arithmetic with infinity - it is not a number. You start to use it as an ordinary algebraic symbol, and immediately you start on a flawed basis. It should only ever be used formally within the context of limits. === Subject: Re: 0 times infinity > > Hi > > When working with the extended real number system, we define > 0.infinity = 0 > Why? > > because anything multiplied by zero is zero > > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > > 'infinity' is not a number, and if you think of it and treat as a number > you > run into problems. it is a concept not a number. something like '1/0 > infinity' is mathematical nonsense really. > Isn't infinity a number in the extended real number system? One does > (limited) arithmetic with it. > No you cannot do any arithmetic with infinity - it is not a number. You > start to use it as an ordinary algebraic symbol, and immediately you start > on a flawed basis. > It should only ever be used formally within the context of limits. In measure theory one has things like infinity + c = infinity for c =/= -infinity. That's the sort of thing that I was referring to when I said that one does (limited) arithmetic with it. Your claim that infinity should only ever be used formally within the context of limits is just plain _wrong_ there are many other formally correct uses of infinity. === Subject: Re: 0 times infinity message > > Hi > > When working with the extended real number system, we define > 0.infinity = 0 > Why? > > because anything multiplied by zero is zero > > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > > 'infinity' is not a number, and if you think of it and treat as a number > you > run into problems. it is a concept not a number. something like '1/0 > = > infinity' is mathematical nonsense really. > > Isn't infinity a number in the extended real number system? One does > (limited) arithmetic with it. > No you cannot do any arithmetic with infinity - it is not a number. You > start to use it as an ordinary algebraic symbol, and immediately you start > on a flawed basis. > It should only ever be used formally within the context of limits. > In measure theory one has things like > infinity + c = infinity > for c =/= -infinity. > That's the sort of thing that I was referring to when I said that one > does (limited) arithmetic with it. > Your claim that infinity should only ever be used formally within the > context of limits is just plain _wrong_ there are many other formally > correct uses of infinity. i cant say i've ever studied measure theory, so cant really comment, but something like 'infinity + c = infinity' is true in a common sense way, and use of limits to express this statement would be unnecessary. But i still stand by my statement that the symbol of infinity should not be used in arithmetic statements. === Subject: Re: 0 times infinity message > > message > > Hi > > When working with the extended real number system, we define > 0.infinity = 0 > Why? > > because anything multiplied by zero is zero > > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > > 'infinity' is not a number, and if you think of it and treat as a > number > you > run into problems. it is a concept not a number. something like > '1/0 > = > infinity' is mathematical nonsense really. > > Isn't infinity a number in the extended real number system? One does > (limited) arithmetic with it. > > No you cannot do any arithmetic with infinity - it is not a number. You > start to use it as an ordinary algebraic symbol, and immediately you > start > on a flawed basis. > > It should only ever be used formally within the context of limits. > In measure theory one has things like > infinity + c = infinity > for c =/= -infinity. > That's the sort of thing that I was referring to when I said that one > does (limited) arithmetic with it. > Your claim that infinity should only ever be used formally within the > context of limits is just plain _wrong_ there are many other formally > correct uses of infinity. > i cant say i've ever studied measure theory, so cant really comment, but > something like 'infinity + c = infinity' is true in a common sense way, and > use of limits to express this statement would be unnecessary. > But i still stand by my statement that the symbol of infinity should not be > used in arithmetic statements. It depends on the system of arithmetic. The extended real numbers (which is what we're talking about here?) are a system (different from the ordinary finite reals) in which the symbols oo (or +oo and -oo, depending on which system) are added, and arithmetic operations are then DEFINED on them. The question is then how to define what the arithmetic operations should mean, so as to create a sensible, consistent and useful system... === Subject: Re: 0 times infinity > message > > message > > Hi > > When working with the extended real number system, we > define > 0.infinity = 0 > Why? > > because anything multiplied by zero is zero > > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > > 'infinity' is not a number, and if you think of it and treat > as a > number > you > run into problems. it is a concept not a number. something > like > '1/0 > = > infinity' is mathematical nonsense really. > > Isn't infinity a number in the extended real number system? > One does > (limited) arithmetic with it. > > No you cannot do any arithmetic with infinity - it is not a > number. You > start to use it as an ordinary algebraic symbol, and immediately > you > start > on a flawed basis. > > It should only ever be used formally within the context of > limits. > > In measure theory one has things like > > infinity + c = infinity > > for c =/= -infinity. > > That's the sort of thing that I was referring to when I said that > one > does (limited) arithmetic with it. > > Your claim that infinity should only ever be used formally within > the > context of limits is just plain _wrong_ there are many other > formally > correct uses of infinity. > i cant say i've ever studied measure theory, so cant really comment, > but > something like 'infinity + c = infinity' is true in a common sense > way, and > use of limits to express this statement would be unnecessary. > But i still stand by my statement that the symbol of infinity should > not be > used in arithmetic statements. > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different from > the ordinary finite reals) in which the symbols oo (or +oo and -oo, > depending on which system) are added, and arithmetic operations are > then DEFINED on them. > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... what the original poster was pointing out was that there is a contradiction in the terms '0 x inifinity = 0' and '1/0 = infinity' . All i pointed out was that this contradiction arises exactly because you are using infinity as a number when it is not. Start treating infinity as a number and you will produce inconsistencies. There is a proof of '1 = 2', I cant remember it offhand but the fallacy arises in diving both sides of an expression by zero. This is the line that produces the inconsistency. You cannot divide both sides of an expression by zero, in exactly the same sense that you cannot multiply anything by infinity, you will always produce contradictions. === Subject: Re: 0 times infinity > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different from > the ordinary finite reals) in which the symbols oo (or +oo and -oo, > depending on which system) are added, and arithmetic operations are > then DEFINED on them. > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... Well put, Matt. > what the original poster was pointing out was that there is a > contradiction in the terms '0 x inifinity = 0' and '1/0 = infinity' . But no contradiction arises if you know what you're doing! > All i pointed out was that this contradiction arises exactly because you > are using infinity as a number when it is not. I guess that depends on what you mean by using [something] as a number. If you think that all numbers behave the same way, you're wrong. Does a*c = b*c imply that a = b? Well, as long as c is a nonzero real, it does. The fact that the implication fails for c = 0 doesn't tell me that zero shouldn't be used as a number. Rather, it just tells me that the behavior of zero is exceptional here. When we use infinity as a number in some extended system, we must be aware that it, like zero, is a bit eccentric, so to speak. > Start treating infinity as a number and you will produce inconsistencies. Well, maybe _you_ would. But people who know what they're doing don't. > There is a proof of '1 = 2', I cant remember it offhand but the fallacy > arises in diving both sides of an expression by zero. This is the line > that produces the inconsistency. I'll grant you that that's the way it's is normally described. But that description is crucially inaccurate. Explanation: In a nutshell, it's often presented inaccurately as something like 0*1 = 0*2 Dividing by 0 on both sides, we get 1 = 2. Of course, we started with a true statement and produced a contradiction, and the way it's stated makes it _seem_ as though the culprit has to be dividing by 0. But that's not right. Let's go through the steps again _carefully_. And let's first assume we're dealing with a system in which nonzero/0 is defined somehow, but 0/0 is undefined, as is often the case. 0*1 = 0*2 If we divide both sides by 0, we get 0*1 0*2 --- = --- 0 0 Following order of operations then, we evaluate the numerators, obtaining 0 0 - = - 0 0 and so all we have is undefined = undefined. There's no contradiction. Of course, we would have gotten a contradiction if we had mistakenly assumed, after dividing both sides by 0, that the factors of 0 in Thus, it's not literally dividing by zero that causes a contradiction. There's nothing wrong with having a system, such as the one-point extension of the reals, in which x/0 = oo (unsigned) for nonzero x. You just need to be aware that 0 doesn't have a multiplicative inverse, and therefore > You cannot divide both sides of an > expression by zero, in exactly the same sense that you cannot multiply > anything by infinity, you will always produce contradictions. You will iff you don't know what you're doing. Hmm. Considering what this thread was originally about, perhaps I should also point out what happens in that classic seemingly-paradox-producing scenario, if we don't leave 0/0 undefined. If instead we define it, along with 0*oo, to be 0, then 0*1 = 0*2 Dividing both sides by 0, we get 0*1 0*2 --- = --- 0 0 Following order of operations, 0 0 - = - and so we have simply 0 = 0. No contradiction. 0 0 BTW, there's an altenative to following order of operations. We could instead have chosen to use the law (ab)/c = a(b/c) on either side, in either of two ways. For example, for (0*1)/0, we could have rewritten it either as 0*(1/0) or as 1*(0/0). But note that in any event, we obtain 0 on both sides of the equation. David Cantrell === Subject: Re: 0 times infinity > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different from > the ordinary finite reals) in which the symbols oo (or +oo and -oo, > depending on which system) are added, and arithmetic operations are > then DEFINED on them. > > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... > Well put, Matt. > what the original poster was pointing out was that there is a > contradiction in the terms '0 x inifinity = 0' and '1/0 = infinity' . > But no contradiction arises if you know what you're doing! > All i pointed out was that this contradiction arises exactly because you > are using infinity as a number when it is not. > I guess that depends on what you mean by using [something] as a number. > If you think that all numbers behave the same way, you're wrong. Does > a*c = b*c imply that a = b? Well, as long as c is a nonzero real, it does. > The fact that the implication fails for c = 0 doesn't tell me that zero > shouldn't be used as a number. Rather, it just tells me that the behavior > of zero is exceptional here. When we use infinity as a number in some > extended system, we must be aware that it, like zero, is a bit eccentric, > so to speak. > Start treating infinity as a number and you will produce inconsistencies. > Well, maybe _you_ would. But people who know what they're doing don't. > There is a proof of '1 = 2', I cant remember it offhand but the fallacy > arises in diving both sides of an expression by zero. This is the line > that produces the inconsistency. > I'll grant you that that's the way it's is normally described. But that > description is crucially inaccurate. Explanation: > In a nutshell, it's often presented inaccurately as something like > 0*1 = 0*2 > Dividing by 0 on both sides, we get 1 = 2. > Of course, we started with a true statement and produced a contradiction, > and the way it's stated makes it _seem_ as though the culprit has to be > dividing by 0. But that's not right. > Let's go through the steps again _carefully_. And let's first assume we're > dealing with a system in which nonzero/0 is defined somehow, but 0/0 is > undefined, as is often the case. nonzero/0 defined somehow? how exactly? all you can say about this is the result is infinitely large. And as for 0/0 surely thats 1 isnt it? > 0*1 = 0*2 > If we divide both sides by 0, we get > 0*1 0*2 > --- = --- > 0 0 > Following order of operations then, we evaluate the numerators, obtaining > 0 0 > - = - > 0 0 > and so all we have is undefined = undefined. There's no contradiction. > Of course, we would have gotten a contradiction if we had mistakenly > assumed, after dividing both sides by 0, that the factors of 0 in > Thus, it's not literally dividing by zero that causes a contradiction. > There's nothing wrong with having a system, such as the one-point extension > of the reals, in which x/0 = oo (unsigned) for nonzero x. You just need to > be aware that 0 doesn't have a multiplicative inverse, and therefore > You cannot divide both sides of an > expression by zero, in exactly the same sense that you cannot multiply > anything by infinity, you will always produce contradictions. > You will iff you don't know what you're doing. > Hmm. Considering what this thread was originally about, perhaps I should > also point out what happens in that classic seemingly-paradox-producing > scenario, if we don't leave 0/0 undefined. If instead we define it, along > with 0*oo, to be 0, then > 0*1 = 0*2 > Dividing both sides by 0, we get > 0*1 0*2 > --- = --- > 0 0 > Following order of operations, > 0 0 > - = - and so we have simply 0 = 0. No contradiction. > 0 0 > BTW, there's an altenative to following order of operations. We could > instead have chosen to use the law (ab)/c = a(b/c) on either side, in > either of two ways. For example, for (0*1)/0, we could have rewritten > it either as 0*(1/0) or as 1*(0/0). But note that in any event, we obtain > 0 on both sides of the equation. > David Cantrell === Subject: Re: 0 times infinity > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different > from the ordinary finite reals) in which the symbols oo (or +oo and > -oo, depending on which system) are added, and arithmetic > operations are then DEFINED on them. > > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... > Well put, Matt. > what the original poster was pointing out was that there is a > contradiction in the terms '0 x inifinity = 0' and '1/0 = infinity' . > But no contradiction arises if you know what you're doing! > All i pointed out was that this contradiction arises exactly because > you are using infinity as a number when it is not. > I guess that depends on what you mean by using [something] as a > number. If you think that all numbers behave the same way, you're > wrong. Does a*c = b*c imply that a = b? Well, as long as c is a nonzero > real, it does. The fact that the implication fails for c = 0 doesn't > tell me that zero shouldn't be used as a number. Rather, it just tells > me that the behavior of zero is exceptional here. When we use infinity > as a number in some extended system, we must be aware that it, like > zero, is a bit eccentric, so to speak. > Start treating infinity as a number and you will produce > inconsistencies. > Well, maybe _you_ would. But people who know what they're doing don't. > There is a proof of '1 = 2', I cant remember it offhand but the > fallacy arises in diving both sides of an expression by zero. This is > the line that produces the inconsistency. > I'll grant you that that's the way it's is normally described. But that > description is crucially inaccurate. Explanation: > In a nutshell, it's often presented inaccurately as something like > 0*1 = 0*2 > Dividing by 0 on both sides, we get 1 = 2. > Of course, we started with a true statement and produced a > contradiction, and the way it's stated makes it _seem_ as though the > culprit has to be dividing by 0. But that's not right. > Let's go through the steps again _carefully_. And let's first assume > we're dealing with a system in which nonzero/0 is defined somehow, but > 0/0 is undefined, as is often the case. > nonzero/0 defined somehow? The reason I said simply defined somehow is that _how_ it might be defined is completely irrelevant to the discussion which followed. > how exactly? Well, that depends on exactly which extended system you want to deal with. Since you didn't specify, I'll choose the simplest one in which this would all make sense: the system of extended nonnegative reals, [0, +oo]. There, we have x/0 = +oo for all nonzero x. Then, as to exactly how is +oo defined: Do you like to develop the nonnegative reals by using equivalence classes of appropriate rational sequences? If so, then in extending that system, we adjoin one more equivalence class. Its members are those rational sequences which increase without bound. That equivalence class is +oo. Or do you prefer to develop the nonnegative reals by using Dedekind cuts? If so, just let me know, and I'll be happy to show how to define +oo as a cut. [If you've never developed the reals from the rationals, then I suggest that this discussion is not right for you yet.] > all you can say about this is the result is infinitely large. No. In a specific extended system, such as [0, +oo], the infinite element is a _specific mathematical object_. > And as for 0/0 surely thats 1 isnt it? No, it surely is not! David Cantrell > 0*1 = 0*2 > If we divide both sides by 0, we get > 0*1 0*2 > --- = --- > 0 0 > Following order of operations then, we evaluate the numerators, > obtaining > 0 0 > - = - > 0 0 > and so all we have is undefined = undefined. There's no contradiction. > Of course, we would have gotten a contradiction if we had mistakenly > assumed, after dividing both sides by 0, that the factors of 0 in > Thus, it's not literally dividing by zero that causes a contradiction. > Rather, it's assuming incorrectly that factors of zero can be > There's nothing wrong with having a system, such as the one-point > extension of the reals, in which x/0 = oo (unsigned) for nonzero x. > You just need to be aware that 0 doesn't have a multiplicative inverse, > You cannot divide both sides of an > expression by zero, in exactly the same sense that you cannot > multiply anything by infinity, you will always produce > contradictions. > You will iff you don't know what you're doing. > Hmm. Considering what this thread was originally about, perhaps I > should also point out what happens in that classic > seemingly-paradox-producing scenario, if we don't leave 0/0 undefined. > If instead we define it, along with 0*oo, to be 0, then > 0*1 = 0*2 > Dividing both sides by 0, we get > 0*1 0*2 > --- = --- > 0 0 > Following order of operations, > 0 0 > - = - and so we have simply 0 = 0. No contradiction. > 0 0 > BTW, there's an altenative to following order of operations. We could > instead have chosen to use the law (ab)/c = a(b/c) on either side, in > either of two ways. For example, for (0*1)/0, we could have rewritten > it either as 0*(1/0) or as 1*(0/0). But note that in any event, we > obtain 0 on both sides of the equation. > David Cantrell === Subject: Re: 0 times infinity > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different > from the ordinary finite reals) in which the symbols oo (or +oo and > -oo, depending on which system) are added, and arithmetic > operations are then DEFINED on them. > > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... > > Well put, Matt. > > what the original poster was pointing out was that there is a > contradiction in the terms '0 x inifinity = 0' and '1/0 = infinity' . > > But no contradiction arises if you know what you're doing! > > All i pointed out was that this contradiction arises exactly because > you are using infinity as a number when it is not. > > I guess that depends on what you mean by using [something] as a > number. If you think that all numbers behave the same way, you're > wrong. Does a*c = b*c imply that a = b? Well, as long as c is a nonzero > real, it does. The fact that the implication fails for c = 0 doesn't > tell me that zero shouldn't be used as a number. Rather, it just tells > me that the behavior of zero is exceptional here. When we use infinity > as a number in some extended system, we must be aware that it, like > zero, is a bit eccentric, so to speak. > > Start treating infinity as a number and you will produce > inconsistencies. > > Well, maybe _you_ would. But people who know what they're doing don't. > > There is a proof of '1 = 2', I cant remember it offhand but the > fallacy arises in diving both sides of an expression by zero. This is > the line that produces the inconsistency. > > I'll grant you that that's the way it's is normally described. But that > description is crucially inaccurate. Explanation: > > In a nutshell, it's often presented inaccurately as something like > > 0*1 = 0*2 > Dividing by 0 on both sides, we get 1 = 2. > > Of course, we started with a true statement and produced a > contradiction, and the way it's stated makes it _seem_ as though the > culprit has to be dividing by 0. But that's not right. > > Let's go through the steps again _carefully_. And let's first assume > we're dealing with a system in which nonzero/0 is defined somehow, but > 0/0 is undefined, as is often the case. > nonzero/0 defined somehow? > The reason I said simply defined somehow is that _how_ it might be > defined is completely irrelevant to the discussion which followed. > how exactly? > Well, that depends on exactly which extended system you want to deal with. > Since you didn't specify, I'll choose the simplest one in which this would > all make sense: the system of extended nonnegative reals, [0, +oo]. There, > we have x/0 = +oo for all nonzero x. > Then, as to exactly how is +oo defined: > Do you like to develop the nonnegative reals by using equivalence classes > of appropriate rational sequences? If so, then in extending that system, > we adjoin one more equivalence class. Its members are those rational > sequences which increase without bound. That equivalence class is +oo. And how does this answer the original posters question exactly...? > Or do you prefer to develop the nonnegative reals by using Dedekind cuts? > If so, just let me know, and I'll be happy to show how to define +oo as a > cut. > [If you've never developed the reals from the rationals, then I suggest > that this discussion is not right for you yet.] > all you can say about this is the result is infinitely large. > No. In a specific extended system, such as [0, +oo], the infinite element > is a _specific mathematical object_. > And as for 0/0 surely thats 1 isnt it? why not? its consistent, as if 0/0 = 1, then 0 = 0 x 1 . dont see a problem there > No, it surely is not! > David Cantrell > 0*1 = 0*2 > > If we divide both sides by 0, we get > > 0*1 0*2 > --- = --- > 0 0 > > Following order of operations then, we evaluate the numerators, > obtaining > > 0 0 > - = - > 0 0 > > and so all we have is undefined = undefined. There's no contradiction. > > Of course, we would have gotten a contradiction if we had mistakenly > assumed, after dividing both sides by 0, that the factors of 0 in not. > > Thus, it's not literally dividing by zero that causes a contradiction. > Rather, it's assuming incorrectly that factors of zero can be > > There's nothing wrong with having a system, such as the one-point > extension of the reals, in which x/0 = oo (unsigned) for nonzero x. > You just need to be aware that 0 doesn't have a multiplicative inverse, > > You cannot divide both sides of an > expression by zero, in exactly the same sense that you cannot > multiply anything by infinity, you will always produce > contradictions. > > You will iff you don't know what you're doing. > > Hmm. Considering what this thread was originally about, perhaps I > should also point out what happens in that classic > seemingly-paradox-producing scenario, if we don't leave 0/0 undefined. > If instead we define it, along with 0*oo, to be 0, then > > 0*1 = 0*2 > > Dividing both sides by 0, we get > > 0*1 0*2 > --- = --- > 0 0 > > Following order of operations, > > 0 0 > - = - and so we have simply 0 = 0. No contradiction. > 0 0 > > BTW, there's an altenative to following order of operations. We could > instead have chosen to use the law (ab)/c = a(b/c) on either side, in > either of two ways. For example, for (0*1)/0, we could have rewritten > it either as 0*(1/0) or as 1*(0/0). But note that in any event, we > obtain 0 on both sides of the equation. > > David Cantrell === Subject: Re: 0 times infinity > nonzero/0 defined somehow? > The reason I said simply defined somehow is that _how_ it might be > defined is completely irrelevant to the discussion which followed. > how exactly? > Well, that depends on exactly which extended system you want to deal > with. Since you didn't specify, I'll choose the simplest one in which > this would all make sense: the system of extended nonnegative reals, > [0, +oo]. There, we have x/0 = +oo for all nonzero x. > Then, as to exactly how is +oo defined: > Do you like to develop the nonnegative reals by using equivalence > classes of appropriate rational sequences? If so, then in extending > that system, we adjoin one more equivalence class. Its members are > those rational sequences which increase without bound. That equivalence > class is +oo. > And how does this answer the original posters question exactly...? It doesn't; it wasn't intended to. I was answering _your_ question: how exactly? referring to how nonzero/0 would be defined in such a system. Of course, I could have simply said x/0 = +oo for nonzero x but you then might well have said something like But that's merely giving a _name_, +oo, to x/0. Unless you _define_ +oo, you haven't defined x/0. And so I was answering your question how exactly? by showing precisely how +oo may be defined in such a system. > Or do you prefer to develop the nonnegative reals by using Dedekind > cuts? If so, just let me know, and I'll be happy to show how to define > +oo as a cut. > [If you've never developed the reals from the rationals, then I suggest > that this discussion is not right for you yet.] > all you can say about this is the result is infinitely large. > No. In a specific extended system, such as [0, +oo], the infinite > element is a _specific mathematical object_. > And as for 0/0 surely thats 1 isnt it? > why not? its consistent, as if 0/0 = 1, then 0 = 0 x 1 . dont see a > problem there To see a problem with taking 0/0 = 1, let's consider the law of algebra a(b/c) = (ab)/c One would certainly like that law to apply, even when c = 0, if division by 0 is defined. But notice what happens if we take 0/0 to be 1 and try to apply that law: 2 = 2*1 = 2(0/0) = (2*0)/0 = 0/0 = 1 Contradiction! In general, anyone is going to get into trouble thinking that factors of Remember: Zero does not have a multiplicative inverse. > No, it surely is not! > David Cantrell === Subject: Re: 0 times infinity > nonzero/0 defined somehow? > > The reason I said simply defined somehow is that _how_ it might be > defined is completely irrelevant to the discussion which followed. > > how exactly? > > Well, that depends on exactly which extended system you want to deal > with. Since you didn't specify, I'll choose the simplest one in which > this would all make sense: the system of extended nonnegative reals, > [0, +oo]. There, we have x/0 = +oo for all nonzero x. > > Then, as to exactly how is +oo defined: > > Do you like to develop the nonnegative reals by using equivalence > classes of appropriate rational sequences? If so, then in extending > that system, we adjoin one more equivalence class. Its members are > those rational sequences which increase without bound. That equivalence > class is +oo. > And how does this answer the original posters question exactly...? > It doesn't; it wasn't intended to. I was answering _your_ question: how > exactly? referring to how nonzero/0 would be defined in such a system. Of > course, I could have simply said > x/0 = +oo for nonzero x > but you then might well have said something like But that's merely giving > a _name_, +oo, to x/0. Unless you _define_ +oo, you haven't defined x/0. > And so I was answering your question how exactly? by showing precisely > how +oo may be defined in such a system. > Or do you prefer to develop the nonnegative reals by using Dedekind > cuts? If so, just let me know, and I'll be happy to show how to define > +oo as a cut. > > [If you've never developed the reals from the rationals, then I suggest > that this discussion is not right for you yet.] > > all you can say about this is the result is infinitely large. > > No. In a specific extended system, such as [0, +oo], the infinite > element is a _specific mathematical object_. > > And as for 0/0 surely thats 1 isnt it? > why not? its consistent, as if 0/0 = 1, then 0 = 0 x 1 . dont see a > problem there > To see a problem with taking 0/0 = 1, let's consider the law of algebra > a(b/c) = (ab)/c > One would certainly like that law to apply, even when c = 0, if division > by 0 is defined. But notice what happens if we take 0/0 to be 1 and try to > apply that law: > 2 = 2*1 = 2(0/0) = (2*0)/0 = 0/0 = 1 Contradiction! > In general, anyone is going to get into trouble thinking that factors of > Remember: Zero does not have a multiplicative inverse. indeed it doesnt, and it is this reason why it cannot be treated as any other number. Just re-capping , all i've ever said on ths is in answering the OP, :- When working with the extended real number system, we define 0.infinity = 0 Why? This appears to be in contradiction to the definitions 1/0 = infinity and a/infinity = 0 there is no contradiction as all three statements are true. 0 . infinity = 0 , because anything multiplied by zero is zero (even an infinitely large number) 1/0 = infinity , because '1' divided by an infinitely small number (in other words zero) produces an infinitely large numer (infinity) a/infinity = 0, because any non zero number divided by an infinitely large number (infinity) produces an infinitely small number (zero) i see no contradiction and no question to answer. > No, it surely is not! > > David Cantrell === Subject: Re: 0 times infinity > message > > message > > Hi > > When working with the extended real number system, we > define > 0.infinity = 0 > Why? > > because anything multiplied by zero is zero > > > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > > 'infinity' is not a number, and if you think of it and treat > as a > number > you > run into problems. it is a concept not a number. something > like > '1/0 > = > infinity' is mathematical nonsense really. > > Isn't infinity a number in the extended real number system? > One does > (limited) arithmetic with it. > > No you cannot do any arithmetic with infinity - it is not a > number. You > start to use it as an ordinary algebraic symbol, and immediately > you > start > on a flawed basis. > > It should only ever be used formally within the context of > limits. > > In measure theory one has things like > > infinity + c = infinity > > for c =/= -infinity. > > That's the sort of thing that I was referring to when I said that > one > does (limited) arithmetic with it. > > Your claim that infinity should only ever be used formally within > the > context of limits is just plain _wrong_ there are many other > formally > correct uses of infinity. > i cant say i've ever studied measure theory, so cant really comment, > but > something like 'infinity + c = infinity' is true in a common sense > way, and > use of limits to express this statement would be unnecessary. > But i still stand by my statement that the symbol of infinity should > not be > used in arithmetic statements. > It depends on the system of arithmetic. The extended real numbers > (which is what we're talking about here?) are a system (different from > the ordinary finite reals) in which the symbols oo (or +oo and -oo, > depending on which system) are added, and arithmetic operations are > then DEFINED on them. ok even accepting infinity as an arithmetic entity in its own right then ANYTHING multiplied by zero can only be zero. So looking at this using standard arithmetic symbolism or thinking in terms of limits, then the question what does '0 x infinity' equal, is zero. > The question is then how to define what the arithmetic operations > should mean, so as to create a sensible, consistent and useful > system... === Subject: Re: 0 times infinity > > When working with the extended real number system, we define > 0.infinity = 0 > Why? > because anything multiplied by zero is zero > This appears to be in contradiction to the definitions > 1/0 = infinity Hmm. I must now wonder _which_ extended real system you're dealing with. Is it the one-point or the two-point extension? When one says just the extended real number system, it's most often the two-point extension, [-oo, +oo], in which 1/0 is normally left undefined. [It's left undefined in order to avoid having something like +oo = 1/0 = 1/(-0) = -(1/0) = -oo.] But in the one-point extension, R U {oo}, where the infinite element is unsigned, we do have 1/0 = oo. > and > a/infinity = 0 I can see why you say that there appears to be a contradiction. But there is not. Zero and the infinite element(s) do not have multiplicative inverses. Saying that 1/0 = oo in the one-point extension is _not_ saying that 1 = 0 * oo, for example. > 'infinity' is not a number, That depends on one's definition of number; whether by infinity one means something vague or, instead, a specific element of a certain mathematical system; etc. > and if you think of it and treat as a number you run into problems. Certainly true, if you don't know what you're doing. But OTOH, if you know what you're doing in dealing, say, with the one-point extension of the reals, you will not run into problems. > something like '1/0 = infinity' is mathematical nonsense really. No, it's hardly nonsense in, say, the one-point extension of the reals. Given definitions of reciprocation, 0 and oo, the statement 1/0 = oo is perfectly rigorous. > Isn't infinity a number in the extended real number system? One could reasonably choose to call it a number. But in any event, it's an element of the system. That's what's important. David Cantrell === Subject: Re: 0 times infinity > Hi > When working with the extended real number system, we define > 0.infinity = 0 > Why? > because anything multiplied by zero is zero The rule anything multiplied by zero is zero applies to finite numbers. As you yourself say, infinity is not a number, and it does not automatically follow that the rule extends to 0*oo. If you are going to use expressions like 0*oo then you must define what they mean, separately from the definition for finite numbers. Often this is done using limit arguments. > This appears to be in contradiction to the definitions > 1/0 = infinity > and > a/infinity = 0 > 'infinity' is not a number, and if you think of it and treat as a number you > run into problems. it is a concept not a number. something like '1/0 = > infinity' is mathematical nonsense really. > Surely, we should simply state that 0.infinity is undefined. > Phil === Subject: How to generate distribution functions from general functions? How to generate distribution functions from general functions? Hello If I have a function g: [0, T] -> R+ with g strictly decreasing g(T) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 [I think g right continous and continously differentiable is needed also] How would I go about generating f from f s.t. f: [0, S] -> R+ for S > 0 f strictly decreasing f(S) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 integral from 0 to S of F(x) dx = 1 f preserving the distribution of the density of g (this is sort of like a probability density function) OR How would I go about generating F from f s.t. F: [0, S] -> R+ for S > 0 F strictly increasing supremum(F'(x)) on x in [0, S] = 1 F(S) = M, M>=0 s.t. |M - 1| <= epsilon, for small epsilon >=0 F preserving the distribution of the density of g (this is sort of like a cumulative density function) f and F are related as f = 1st derivative of F F = indefinate integral on 0 to x of f I have U (in R+{0}) units of something and I want to give it out over time S according to a certain shape (e.g. sin(x), x in [0, pi/2]; -x + 1, x in [0, 1]; e^{-x}, x in [0, 6]). I am trying to take shapes in defined domains, stretch them over the required time period and scale them along the Y-axis so that their integral is 1 and that they preserve the distribution of the density on [0, T] over [0, S]. For f : I can integrate them over N (in Naturals) time steps ( = S/N) to determine what portion of the U units to give out at time n*S/N (1 <= n <= N, n in Naturals). For F : I can evaluate them over N (in Naturals) time steps ( = S/N). So that at time n+1 I would give out a portion F(x_{n+1)) - F(x_{n}). I worked out a specific case but my method will only work with 1st order polynomials. This method creates a 2 simultaneous equations with 2 unknows. With n-th degree polynomials it will create 2 simultaneous equations with n+1 unknowns (i.e. no closed solution). I included my working below. I need a generalized method for all functions that will give a closed solution. If you have a link that explains and/or walks through the building of the solution I would be really grateful. Is there also a solution for strictly increasing g, g(0) = m; strickly increasing f, f(0) = m; strickly decreasing F, F(0) = M? Please help Robby *************** Here is how I worked my specific case ... g(x) = -x+1, x in [0, 1] I wanted to stretch this over S = [0, 10]. I used the ratio of the maximums of the domains and applied this to x therefore replacing all x with (1/10)x giving ... h_{1}(x) = -(1/10)x + 1 Next I knew that I would need to add some terms in order to get the integral over 0 to 10 to be 1. h_{2}(x) = -(1/10)x + Ax + 1 + B Then I integrated this over 0 to 10 and set the result to 1 as required. (1) 50A + 10B + 5 = 1 => 50A + 10B = -4 Then I evaluated h_{2} at x = 10 and set the result to 0 as required. (2) 10A + B = 0 Solving (1) and (2) simutaneously gives A = 2/25 , B = -4/5 Giving f(x) = -(1/50)x + 1/5, x in [0, 10] and F(x) = -(1/100)x^2 + (1/5)x, x in [0, 10] I can use these functions as described above to get my portions of U units to give out at the required times. As you can see this is not a robust method and will only work with 1st degree polynomials. I think this might be able to be done with a change of variables but I am unsure of where to start. Please help. === Subject: Re: How to generate distribution functions from general functions? > How to generate distribution functions from general functions? > Hello > If I have a function g: [0, T] -> R+ with > g strictly decreasing > g(T) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 Huh? > [I think g right continous and continously differentiable is needed also] > How would I go about generating f from f s.t. Do you mean f from g? > f: [0, S] -> R+ for S > 0 > f strictly decreasing > f(S) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 If you're allowing m > 0 (don't understand whether you are or aren't), then you can't stipulate f(S) = m AND integral from 0 to S of f(x) dx = 1 (which I assume is what you want; see next comment) without changing the shape of the distribution. > integral from 0 to S of F(x) dx = 1 Do you mean integral from 0 to S of f(x) dx = 1? If not then I am lost (and your later example solution is wrong). > f preserving the distribution of the density of g > (this is sort of like a probability density function) > OR > How would I go about generating F from f s.t. Do you mean F from g? > F: [0, S] -> R+ for S > 0 > F strictly increasing > supremum(F'(x)) on x in [0, S] = 1 > F(S) = M, M>=0 s.t. |M - 1| <= epsilon, for small epsilon >=0 Huh? > F preserving the distribution of the density of g > (this is sort of like a cumulative density function) > f and F are related as > f = 1st derivative of F > F = indefinate integral on 0 to x of f Do you mean definite integral? > I have U (in R+{0}) units of something and I want to give it out over time > S according to a certain shape (e.g. sin(x), x in [0, pi/2] You said earlier that g is decreasing? Sin(x) in the range [0, pi/2] isn't. (See also comment below.) > ; -x + 1, x in > [0, 1]; e^{-x}, x in [0, 6]). > I am trying to take shapes in defined domains, stretch them over the > required time period and scale them along the Y-axis so that their integral > is 1 and that they preserve the distribution of the density on [0, T] over > [0, S]. > For f : I can integrate them over N (in Naturals) time steps ( = S/N) to > determine what portion of the U units to give out at time n*S/N (1 <= n <= > N, n in Naturals). > For F : I can evaluate them over N (in Naturals) time steps ( = S/N). So > that at time n+1 I would give out a portion F(x_{n+1)) - F(x_{n}). > I worked out a specific case but my method will only work with 1st order > polynomials. This method creates a 2 simultaneous equations with 2 unknows. > With n-th degree polynomials it will create 2 simultaneous equations with > n+1 unknowns (i.e. no closed solution). I included my working below. I need > a generalized method for all functions that will give a closed solution. If > you have a link that explains and/or walks through the building of the > solution I would be really grateful. Is there also a solution for strictly > increasing g, g(0) = m; strickly increasing f, f(0) = m; I don't understand why it matters if f or g are increasing or decreasing. To get a valid pdf out of it all that matters is g(x) >= 0 in the prescribed range? > strickly decreasing F, F(0) = M? If F is supposed to be a cumulative distribution function then it can't be decreasing. > Please help > Robby > *************** > Here is how I worked my specific case ... > g(x) = -x+1, x in [0, 1] > I wanted to stretch this over S = [0, 10]. I used the ratio of the maximums > of the domains and applied this to x therefore replacing all x with (1/10)x > giving ... > h_{1}(x) = -(1/10)x + 1 > Next I knew that I would need to add some terms in order to get the integral > over 0 to 10 to be 1. You can't just add some terms without changing the shape of the distribution. You just struck lucky here because g is linear and the terms you're adding preserve linearity. > h_{2}(x) = -(1/10)x + Ax + 1 + B > Then I integrated this over 0 to 10 and set the result to 1 as required. > (1) 50A + 10B + 5 = 1 > => 50A + 10B = -4 > Then I evaluated h_{2} at x = 10 and set the result to 0 as required. > (2) 10A + B = 0 > Solving (1) and (2) simutaneously gives > A = 2/25 , B = -4/5 > Giving > f(x) = -(1/50)x + 1/5, x in [0, 10] > and > F(x) = -(1/100)x^2 + (1/5)x, x in [0, 10] > I can use these functions as described above to get my portions of U units > to give out at the required times. As you can see this is not a robust > method and will only work with 1st degree polynomials. I think this might > be able to be done with a change of variables but I am unsure of where to > start. Please help. The best interpretation I can make of all this is that you want to horizontally scale g by a factor of S/T, and then vertically scale so as to make the integral over 0 to S equal to one (and hence construct a valid pdf). If that's the case then try f(x) = g(x*T/S)/Integral{x = 0->S: g(x*T/S)dx} For example, in your test case, g(x) = -x + 1 T = 1 S = 10 so f(x) = (-x*1/10 + 1)/Integral{x = 0->10: -x*1/10 + 1 dx} = (-x/10 + 1)/5 = -x/50 + 1/5 === Subject: Re: How to generate distribution functions from general functions? > supremum(F'(x)) on x in [0, S] = 1 should be > supremum(F(x)) on x in [0, S] = 1 > How to generate distribution functions from general functions? > Hello > If I have a function g: [0, T] -> R+ with > g strictly decreasing > g(T) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 > [I think g right continous and continously differentiable is needed also] > How would I go about generating f from f s.t. > f: [0, S] -> R+ for S > 0 > f strictly decreasing > f(S) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 > integral from 0 to S of F(x) dx = 1 > f preserving the distribution of the density of g > (this is sort of like a probability density function) > OR > How would I go about generating F from f s.t. > F: [0, S] -> R+ for S > 0 > F strictly increasing > supremum(F'(x)) on x in [0, S] = 1 > F(S) = M, M>=0 s.t. |M - 1| <= epsilon, for small epsilon >=0 > F preserving the distribution of the density of g > (this is sort of like a cumulative density function) > f and F are related as > f = 1st derivative of F > F = indefinate integral on 0 to x of f > I have U (in R+{0}) units of something and I want to give it out over > time S according to a certain shape (e.g. sin(x), x in [0, pi/2]; -x + 1, > x in [0, 1]; e^{-x}, x in [0, 6]). > I am trying to take shapes in defined domains, stretch them over the > required time period and scale them along the Y-axis so that their > integral is 1 and that they preserve the distribution of the density on > [0, T] over [0, S]. > For f : I can integrate them over N (in Naturals) time steps ( = S/N) to > determine what portion of the U units to give out at time n*S/N (1 <= n <= > N, n in Naturals). > For F : I can evaluate them over N (in Naturals) time steps ( = S/N). So > that at time n+1 I would give out a portion F(x_{n+1)) - F(x_{n}). > I worked out a specific case but my method will only work with 1st order > polynomials. This method creates a 2 simultaneous equations with 2 > unknows. With n-th degree polynomials it will create 2 simultaneous > equations with n+1 unknowns (i.e. no closed solution). I included my > working below. I need a generalized method for all functions that will > give a closed solution. If you have a link that explains and/or walks > through the building of the solution I would be really grateful. Is there > also a solution for strictly increasing g, g(0) = m; strickly increasing > f, f(0) = m; strickly decreasing F, F(0) = M? > Please help > Robby > *************** > Here is how I worked my specific case ... > g(x) = -x+1, x in [0, 1] > I wanted to stretch this over S = [0, 10]. I used the ratio of the > maximums of the domains and applied this to x therefore replacing all x > with (1/10)x giving ... > h_{1}(x) = -(1/10)x + 1 > Next I knew that I would need to add some terms in order to get the > integral over 0 to 10 to be 1. > h_{2}(x) = -(1/10)x + Ax + 1 + B > Then I integrated this over 0 to 10 and set the result to 1 as required. > (1) 50A + 10B + 5 = 1 > => 50A + 10B = -4 > Then I evaluated h_{2} at x = 10 and set the result to 0 as required. > (2) 10A + B = 0 > Solving (1) and (2) simutaneously gives > A = 2/25 , B = -4/5 > Giving > f(x) = -(1/50)x + 1/5, x in [0, 10] > and > F(x) = -(1/100)x^2 + (1/5)x, x in [0, 10] > I can use these functions as described above to get my portions of U units > to give out at the required times. As you can see this is not a robust > method and will only work with 1st degree polynomials. I think this might > be able to be done with a change of variables but I am unsure of where to > start. Please help. === Subject: Re: How to generate distribution functions from general functions? > supremum(F'(x)) on x in [0, S] = 1 should be > How to generate distribution functions from general functions? > Hello > If I have a function g: [0, T] -> R+ with > g strictly decreasing > g(T) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 > [I think g right continous and continously differentiable is needed also] > How would I go about generating f from f s.t. > f: [0, S] -> R+ for S > 0 > f strictly decreasing > f(S) = m, m>=0 s.t. |m - 0| <= epsilon, for small epsilon >=0 > integral from 0 to S of F(x) dx = 1 > f preserving the distribution of the density of g > (this is sort of like a probability density function) > OR > How would I go about generating F from f s.t. > F: [0, S] -> R+ for S > 0 > F strictly increasing > supremum(F'(x)) on x in [0, S] = 1 > F(S) = M, M>=0 s.t. |M - 1| <= epsilon, for small epsilon >=0 > F preserving the distribution of the density of g > (this is sort of like a cumulative density function) > f and F are related as > f = 1st derivative of F > F = indefinate integral on 0 to x of f > I have U (in R+{0}) units of something and I want to give it out over > time S according to a certain shape (e.g. sin(x), x in [0, pi/2]; -x + 1, > x in [0, 1]; e^{-x}, x in [0, 6]). > I am trying to take shapes in defined domains, stretch them over the > required time period and scale them along the Y-axis so that their > integral is 1 and that they preserve the distribution of the density on > [0, T] over [0, S]. > For f : I can integrate them over N (in Naturals) time steps ( = S/N) to > determine what portion of the U units to give out at time n*S/N (1 <= n <= > N, n in Naturals). > For F : I can evaluate them over N (in Naturals) time steps ( = S/N). So > that at time n+1 I would give out a portion F(x_{n+1)) - F(x_{n}). > I worked out a specific case but my method will only work with 1st order > polynomials. This method creates a 2 simultaneous equations with 2 > unknows. With n-th degree polynomials it will create 2 simultaneous > equations with n+1 unknowns (i.e. no closed solution). I included my > working below. I need a generalized method for all functions that will > give a closed solution. If you have a link that explains and/or walks > through the building of the solution I would be really grateful. Is there > also a solution for strictly increasing g, g(0) = m; strickly increasing > f, f(0) = m; strickly decreasing F, F(0) = M? > Please help > Robby > *************** > Here is how I worked my specific case ... > g(x) = -x+1, x in [0, 1] > I wanted to stretch this over S = [0, 10]. I used the ratio of the > maximums of the domains and applied this to x therefore replacing all x > with (1/10)x giving ... > h_{1}(x) = -(1/10)x + 1 > Next I knew that I would need to add some terms in order to get the > integral over 0 to 10 to be 1. > h_{2}(x) = -(1/10)x + Ax + 1 + B > Then I integrated this over 0 to 10 and set the result to 1 as required. > (1) 50A + 10B + 5 = 1 > => 50A + 10B = -4 > Then I evaluated h_{2} at x = 10 and set the result to 0 as required. > (2) 10A + B = 0 > Solving (1) and (2) simutaneously gives > A = 2/25 , B = -4/5 > Giving > f(x) = -(1/50)x + 1/5, x in [0, 10] > and > F(x) = -(1/100)x^2 + (1/5)x, x in [0, 10] > I can use these functions as described above to get my portions of U units > to give out at the required times. As you can see this is not a robust > method and will only work with 1st degree polynomials. I think this might > be able to be done with a change of variables but I am unsure of where to > start. Please help. === Subject: Obtaining a particular acceleration formula by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08Eh5d04919; Could somebody please outline the steps needed to obtain the following formula? acceleration=d/dx(0.5(v^2)) === Subject: Re: Obtaining a particular acceleration formula > Could somebody please outline the steps needed to obtain the following > formula? > acceleration=d/dx(0.5(v^2)) d/dx(0.5(v^2)) = dv/dx * d/dv(0.5(v^2)) by chain rule = dv/dx * v do the differentiation = dv/dx * dx/dt from definition of v = dv/dt by reverse chain rule = a from definition of a === Subject: Re: Obtaining a particular acceleration formula >Could somebody please outline the steps needed to obtain the following >formula? >acceleration=d/dx(0.5(v^2)) Acceleration is the time derivative of velocity. You seem to be taking the _space_ derivative of 0.5v^2. ('m not sure if you're actually doing that, but I'm guessing at the parentheses you omitted.) problem. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 === Subject: Re: Negative poewrs of j=jsqrt(-1) question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08Eh4P04904; >Active8 dixit: >>I saw -j in a denominator negated and moved to the numerator today. >>I may vaguely remember something about this. If I extrapolate what I >>know, >>j^4 = 1 >>j^3 = -j >>j^2 = -1 >>j^1 = j >>j^0 = 1 >>does >>J^(-1) = -j ?? >>That would justify 1/(-j) = j >So you had 1/(0-j)? The usual way to make the denominator a real >number is to multiply by the conjugate: >1/(0-j) * (0+j)/(0+j) = (0+j)/1 = j this method is not possible... j being an imaginary no., cannot be multiplied and divided just the way we cannot multiply & divide by zero. === Subject: Re: Negative poewrs of j=jsqrt(-1) question >So you had 1/(0-j)? The usual way to make the denominator a real >number is to multiply by the conjugate: >1/(0-j) * (0+j)/(0+j) = (0+j)/1 = j > this method is not possible... Of course it is. > j being an imaginary no., cannot be multiplied and divided What on earth are you talking about? Of course it can be. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Negative poewrs of j=jsqrt(-1) question >> does J^(-1) = -j ?? > Yes. Proof: > j^(-1) = 1/j = j^3/j^3 = j^3/1 = j^3 = -j There is mistake. > To put it bluntly but fairly, anyone today who doubts that the > variety of life on this planet was produced by a process of > evolution is simply ignorant -- inexcusably ignorant, in a world > where three out of four people have learned to read and write. > --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 Such abusive freedom of speech reveals the author's inexcusable ignorance. === Subject: Re: Negative poewrs of j=jsqrt(-1) question >> j^(-1) = 1/j = j^3/j^3 = j^3/1 = j^3 = -j >There is mistake. apologize for the typo; for some reason the spell checker didn't find it. :-) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 === Subject: Re: Negative poewrs of j=jsqrt(-1) question > j^(-1) = 1/j = j^3/j^3 = j^3/1 = j^3 = -j >>There is mistake. > apologize for the typo; for some reason the spell checker didn't find > it. :-) had it and now I've got two other ways to look at it. -- Mike === Subject: Re: Negative poewrs of j=jsqrt(-1) question >If by j, you mean i, the unit imaginary number, then >yes, 1/i = -i. Engineering students are often taught to call it j because i is electric current. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 === Subject: Re: Negative poewrs of j=jsqrt(-1) question >If by j, you mean i, the unit imaginary number, then >yes, 1/i = -i. > Engineering students are often taught to call it j because i is > electric current. I know :) I'm a physics and EE double-major. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Vectors and more vectors by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08Eh0L04712; Here's an exam question from last year which I can't work out: In this question the vectors i and j are horizontal unit vectors in the directions due east and due north respectively Two boats A and B are moving with constant velocities. Boat A moves with velocity 9j km/h. Boat B moves with velocity (3i + 5j) km/h a) Find the bearing on which B is moving --- So far I wonder about the meaning for the question of unit vectors (should I have to multiply by something?... and the fact that the units are in Kms. I think I can leave then as it is..I hope) The answer to this question hopefully is Angle= 59.03.bc so bearing = 30.96.bc But then the real problems startr: --- At noon, A is at the point O, and B is 10 Km due west of O. At time t hours after noon, the postion vectors of A and B relative to = are a Km and b Km respectively. b) Find expressions for a and b in terms of t, giving your answers in the form pi + qj --- The best I can guess (forgetting if I may the business about unit vectors) is that A is at 0i + 0j and that B is at 10i + 0j (all in Kms), and that at time t with v as above A will be 9t j + (r) and B(3i + 5j)t + 10i. If this makes sense, it still I think does not answer the question as for finding expressions... The question goes on: --- Find the time when B is due south of A --- Can't understand. I would think that B is always south of A. After all A only has one component (j) which is positive and will remain so increasingle with time. B starting at 5j is already south of A. Which shows that I don't understand the question. It goes on further but I'll wait fo.82r help on this before tackling the rest. Jo === Subject: Re: Vectors and more vectors > In this question the vectors i and j are horizontal unit vectors in > the directions due east and due north respectively > Two boats A and B are moving with constant velocities. Boat A moves > with velocity 9j km/h. Boat B moves with velocity (3i + 5j) km/h > a) Find the bearing on which B is moving > --- > So far I wonder about the meaning for the question of unit vectors > (should I have to multiply by something?... and the fact that the > units are in Kms. I think I can leave then as it is..I hope) Unit vectors are pure directions. They are of magnitude 1 (note the dimensionlessness of that). The units are part of what the unit vectors are multiplied by. So you don't talk about a unit vector of 1km/hr east. Rather, you talk about a unit vector of 1 in the east direction that is multiplied by the scalar 1km/hr. > The answer to this question hopefully is Angle= 59.03deg so bearing = > 30.96deg If by bearing the book actually means heading -- the angle between north and the direction the boat is moving, then yes, it is 30.96deg. By the way, you can get this in a single step. Consider the right triangle OTU formed by the points in velocity space O (0,0), T (3, 5), and U (5, 0). By the definition of heading, angle TOU is the heading. The side opposite angle TOU is of length 3km/hr and the side adjacent to angle TOU is of length 5km/hr. By the definition of tangent tan(angle TOU) = opposite/adjacent = (3km/hr)/(5km/hr) = 3/5 tan(angle TOU) = 3/5 angle TOU = arctan(3/5) angle TOU = 30.96deg In other words, you aren't obliged to find the standard angle (relative to the east axis -- like on usual unit circle charts in trig classes) and then convert it to an angle relative to north (though there's nothing wrong with doing that!). You can instead set things up to find the angle from north directly. Sometimes that may be easier, sometimes it may not. > At noon, A is at the point O, and B is 10 Km due west of O. At time t > hours after noon, the postion vectors of A and B relative to = are a > Km and b Km respectively. I am assuming relative to = is really supposed to read relative to O. > b) Find expressions for a and b in terms of t, giving your answers in > the form pi + qj > --- > The best I can guess (forgetting if I may the business about unit > vectors) is that A is at 0i + 0j and that B is at 10i + 0j (all in > Kms), and that at time t with v as above A will be 9t j + (r) and B(3i > + 5j)t + 10i. First, A is not at 0i+0j. You have no idea where A is. What you do know is that A is at 0i+9j *relative to O*. Now, since B's position is also given relative to O, you can treat O as the origin of the co-ordinate system you're working in and treat the positions of A and B as absolute positions. But the problem could have given the positions of A and B relative to different points, and then you'd have to reconcile that before proceeding. Second, it says B is 10km *west* of O. So relative to O, B's initial position is (-10km)i, not (10km)i. Aside from that, you basically have it though you didn't write the answers in the desired form. A: (9km/hr)t j + (0km i + 0km j) = 0km i + (9km/hr)t j In other words, p(t) = 0km and q(t) = (9km/hr)t B: [(3km/hr)t i + (5km/hr)t j] + 0km i - 10km j = (3km/hr)t i + [(5km/hr)t - 10km]j In other words, p(t) = (3km/hr)t and q(t) = (5km/hr)t - 10km > Find the time when B is due south of A > Can't understand. I would think that B is always south of A. B is always south of A. But B is not always *due south* of A. > all A only has one component (j) which is positive and will remain so > increasingle with time. B starting at 5j is already south of A. south is not the same as due south. B being due south of A means that if you stand on A and look south (i.e. 180deg from north), B is directly in your line of sight. Not just more to the south than A is, but directly in your line of sight. Hopefully that helps. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Vectors and more vectors by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmTU31387; I left this question aside because I had a Pure Maths exam yesterday, and now I came back to it because I have a mechanics exam tomorrow, and still can't understand a lot of this business of vectors!! Hope you can help me a little bit more: >Second, it says B is 10km *west* of O. So relative to O, B's initial >position is (-10km)i, not (10km)i. but I'm afraid I can't see it. If B is west of A it is on a graph, along the positive side of the x-axis, while A is at the origin. So why should it be (-10km)i ?? As for the second part of the question I am still at a loss. Could you please help? Jo >> In this question the vectors i and j are horizontal unit vectors in >> the directions due east and due north respectively >> Two boats A and B are moving with constant velocities. Boat A moves >> with velocity 9j km/h. Boat B moves with velocity (3i + 5j) km/h >> a) Find the bearing on which B is moving >> --- >> So far I wonder about the meaning for the question of unit vectors >> (should I have to multiply by something?... and the fact that the >> units are in Kms. I think I can leave then as it is..I hope) >Unit vectors are pure directions. They are of magnitude 1 (note the >dimensionlessness of that). The units are part of what the unit >vectors are multiplied by. So you don't talk about a unit vector of >1km/hr east. Rather, you talk about a unit vector of 1 in the east >direction that is multiplied by the scalar 1km/hr. >> The answer to this question hopefully is Angle= 59.03deg so bearing = >> 30.96deg >If by bearing the book actually means heading -- the angle between >north and the direction the boat is moving, then yes, it is 30.96deg. >By the way, you can get this in a single step. Consider the right >triangle OTU formed by the points in velocity space O (0,0), T (3, 5), >and U (5, 0). By the definition of heading, angle TOU is the heading. >The side opposite angle TOU is of length 3km/hr and the side adjacent >to angle TOU is of length 5km/hr. By the definition of tangent > tan(angle TOU) = opposite/adjacent = (3km/hr)/(5km/hr) = 3/5 > tan(angle TOU) = 3/5 > angle TOU = arctan(3/5) > angle TOU = 30.96deg >In other words, you aren't obliged to find the standard angle >(relative to the east axis -- like on usual unit circle charts in trig >classes) and then convert it to an angle relative to north (though >there's nothing wrong with doing that!). You can instead set things >up to find the angle from north directly. Sometimes that may be >easier, sometimes it may not. >> At noon, A is at the point O, and B is 10 Km due west of O. At time t >> hours after noon, the postion vectors of A and B relative to = are a >> Km and b Km respectively. >I am assuming relative to = is really supposed to read relative to O. >> b) Find expressions for a and b in terms of t, giving your answers in >> the form pi + qj >> --- >> The best I can guess (forgetting if I may the business about unit >> vectors) is that A is at 0i + 0j and that B is at 10i + 0j (all in >> Kms), and that at time t with v as above A will be 9t j + (r) and B(3i >> + 5j)t + 10i. >First, A is not at 0i+0j. You have no idea where A is. What you do >know is that A is at 0i+9j *relative to O*. Now, since B's position >is also given relative to O, you can treat O as the origin of the >co-ordinate system you're working in and treat the positions of A and >B as absolute positions. But the problem could have given the >positions of A and B relative to different points, and then you'd have >to reconcile that before proceeding. >Second, it says B is 10km *west* of O. So relative to O, B's initial >position is (-10km)i, not (10km)i. Aside from that, you basically >have it though you didn't write the answers in the desired form. >A: (9km/hr)t j + (0km i + 0km j) = 0km i + (9km/hr)t j >In other words, p(t) = 0km and q(t) = (9km/hr)t >B: [(3km/hr)t i + (5km/hr)t j] + 0km i - 10km j > = (3km/hr)t i + [(5km/hr)t - 10km]j >In other words, p(t) = (3km/hr)t and q(t) = (5km/hr)t - 10km >> Find the time when B is due south of A >> Can't understand. I would think that B is always south of A. >B is always south of A. But B is not always *due south* of A. >> all A only has one component (j) which is positive and will remain so >> increasingle with time. B starting at 5j is already south of A. >south is not the same as due south. >B being due south of A means that if you stand on A and look >south (i.e. 180deg from north), B is directly in your line of sight. >Not just more to the south than A is, but directly in your line >of sight. >Hopefully that helps. >-- >Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Vectors and more vectors >Second, it says B is 10km *west* of O. So relative to O, B's initial >position is (-10km)i, not (10km)i. > but I'm afraid I can't see it. If B is west of A it is on a graph, > along the positive side of the x-axis, while A is at the origin. So > why should it be (-10km)i ?? N | | | | | W----B-------A------------E | | | | | S That's why B is (-10km)i relative to A. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: indefinite integrals+elasticity(phy) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08Eh3p04872; please help me find the soln to these problems.. 1)find the indefinite integral of (1+cosa.cosx)/(cosa+cosx)dx i think it would become simpler if '1' is converted into (cos a)^2 + (sin a)^2. 2)A material has Poisson's ratio 1/x. if long. strain is x/10000, what is the %age increase in volume? (is it zero?) === Subject: Re: indefinite integrals+elasticity(phy) Harsh, Rational functions of trigonometric expressions can, with a little work, be transformed into standard rational functions. Make the substitution x=2*arctan(t). Then, dx=(2/(1+t^2))dt Also, cos(x) =cos(2arctan(t)) =(cos(arctan(t)))^2-(sin(arctan(t)))^2 =(1/(1+t^2))-(t/(1+t^2)) =(1-t^2)/(1+t^2) This substitution and a little more work will give you a rational expression in t, which you can integrate conventionally. Travis > please help me find the soln to these problems.. > 1)find the indefinite integral of (1+cosa.cosx)/(cosa+cosx)dx > i think it would become simpler if '1' is converted into (cos a)^2 + > (sin a)^2. > 2)A material has Poisson's ratio 1/x. if long. strain is x/10000, what > is the %age increase in volume? > (is it zero?) === Subject: Re: indefinite integrals+elasticity(phy) > Rational functions of trigonometric expressions can, with a little work, be > transformed into standard rational functions. > Make the substitution x=2*arctan(t). > Then, dx=(2/(1+t^2))dt > Also, > cos(x)=cos(2arctan(t)) > =(cos(arctan(t)))^2-(sin(arctan(t)))^2 > =(1/(1+t^2))-(t/(1+t^2)) > =(1-t^2)/(1+t^2) For a person who knows math, I expect you'd also know about how to present readable equations.That,alasyoudonot.Wouldyouuseamplespaces, so your equations would be easy to read, insteadofsocrammedtogether,they'rehardtoread? === Subject: Re: proving trig identities by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08Eh3n04847; use pythagoras theorem to any right triangle. ull get your answer. thank you harsh. === Subject: Computing limits by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08K2TZ29918; Perhaps a dumb question, but here it goes. When I'm computing a limit, for example: lim (2x^2+3x)/x Here I can divide by x and then I have a nice polynomial that allows me to simply plug in x=0 and get the result. How come I can divide by x? When calculating equations such as x^2+2x+5=0, I cannot divide by === Subject: Re: Computing limits > lim (2x^2+3x)/x > x->0 > Here I can divide by x and then I have a nice polynomial that allows > me to simply plug in x=0 and get the result. How come I can divide by > x? Good question. What you are essentially doing (conciousnessly or not) is saying since (2x^2+3x)/x and 2x+3 agree at all but one point (they are the same everywhere except where x=0) then both limits as x-->c are the same for some c. This is a property of limits. Realize we are not saying the functions are the same, but their _limits_ are. > When calculating equations such as x^2+2x+5=0, I cannot divide by > x. When is dividing by x allowed and when is not? As already noted, in solving an equation, you can multiply or divide by anything nonzero. Since x is a variable you do not yet necessarily know it is not 0, so one must be careful. In this case it is pretty obvious x!=0 since 5!=0. Although legal, this is not a good example since, as noted, it doesn't help you solve the equation. A better example: Let x+2+(5/x)=0 be the equation you want to solve, such that multiplying through by x will actually be helpful in solving it. The term 5/x implicitly states that x!=0. -- Darrell === Subject: Re: Computing limits When evaluating limits apporaching infinity with qoutient functions you can divide both the numerator and denominator by the highest power of x in the denominator. For your problem you can do a little factorization of the function that will make the solution clearer. lim_{x->0} (2x^2 + 3x)/x = lim_{x->0} x(2x + 3) / x Most of these types of limits have a factorization trick you can do. Another trick is to multiply by the completed square half of the numerator. (2x^2 + 3x)/x = [(2x^2 + 3x)/x] * (2x^2 - 3x)/(2x^2 - 3x) but this is normally use for quotient functions that envolve some sort of root. Have fun Robby > Perhaps a dumb question, but here it goes. When I'm computing a limit, > for example: > lim (2x^2+3x)/x > x->0 > Here I can divide by x and then I have a nice polynomial that allows > me to simply plug in x=0 and get the result. How come I can divide by > x? When calculating equations such as x^2+2x+5=0, I cannot divide by === Subject: Re: Computing limits > When I'm computing a limit, for example: > lim (2x^2+3x)/x > x->0 Taking the limit is finding the value the expression is approaching as x gets close to zero from both sides of zero but not at x = 0. And when x is not zero, the expression (2x^2+3x)/x is the same as 2x+3. > Here I can divide by x and then I have a nice polynomial > that allows me to simply plug in x=0 and get the result. It's a shortcut, an algorithm that allows you to compute the limit without thinking about or understanding what taking a limit is. > How come I can divide by x? When calculating equations > such as x^2+2x+5=0, I cannot divide by x. We are solving an equation, here, finding all of its roots. You can divide by x, in this case, also. But if one of the roots is zero, that information may be lost. > When is dividing by x allowed and when is not? The answer involves knowing what you are doing and not just following an algorithm (set of rules). === Subject: Re: Computing limits > Perhaps a dumb question, but here it goes. When I'm computing a limit, > for example: > lim (2x^2+3x)/x > x->0 > Here I can divide by x and then I have a nice polynomial that allows > me to simply plug in x=0 and get the result. How come I can divide by > x? Because you can always divide by x (though after doing so you have to exclude zero from the domain of the quotient). > When calculating equations such as x^2+2x+5=0, I cannot divide by > x. Yes you can. But it might not do you any good. If you divide your equation by x, you get the perfectly valid: x + 2 + 5/x = 0 But it doesn't do you any good. It's allowed when x is not equal to zero and not allowed when x is equal to zero. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Second Part:To MO,JO,Matted Man by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j08K2QA29789; This Second Part is a continuation of one of the messages posted on January 6 2005 (the one with a POST SCRIPTUM)having the titleFirst Part:ToMO,JA,Matted Man Fermat's Proof OfLast Theorem and Is given here AS a decisive HINT and I will use the same mathematical symbols(letters). Here IT IS: 1)Instead to choose X,Y,Z natural numbers(thatleads to the first End of proof,but the path of proof is the same for both Ends) we take X,Y,Z as being INTEGERS.The First Part does not change. 2)Here I will show What the Great Mathematicien (as EULer) Missed to find (First Year College Algebgra Manipulations- -sorry,no modesty Mr.Mathed Man) 3) Nr 3 Will show that, after this message has been posted. george ghiata === Subject: Re: Second Part:To MO,JO,Matted Man > This Second Part is a continuation of one of the messages posted > on January 6 2005 (the one with a POST SCRIPTUM)having the > titleFirst Part:ToMO,JA,Matted Man Fermat's Proof OfLast Theorem > and Is given here AS a decisive HINT and I will use the same > mathematical symbols(letters). > Here IT IS: > 1)Instead to choose X,Y,Z natural numbers(thatleads to the first > End of proof,but the path of proof is the same for both Ends) > we take X,Y,Z as being INTEGERS.The First Part does not change. > 2)Here I will show What the Great Mathematicien (as EULer) > Missed to find (First Year College Algebgra Manipulations- > -sorry,no modesty Mr.Mathed Man) > 3) > Nr 3 Will show that, after this message has been posted. > george ghiata George, Might I make a helpful suggestion here? If you have a serious proposal for a simple FLT proof and you want to air it via this newsgroup, then it would be hugely easier if you were to put the entire proof into one coherent posting. Without presuming to speak for anyone else, it's likely that many people will be so certain that your proof is incorrect that they will not be prepared to invest much time trying to understand it. Therefore it is important that it's very clearly presented. In my view the probability that your proof is correct is approximately zero, but if you do as I suggest then at some kind person might be able to point out the mistake that is (almost) certainly present. === Subject: Maths Problem - Need all the help I can get! Sorry for X-Posting but needing help quickly, hopefully someone can assist. The problem: Two voltages v1 and v2 are to be summed using an analogue circuit. v1 = 5 cos (5 pi t + (pi / 6)) v2 = 10 sin (5 pi t - (pi / 6)) Calculate the time at which the sum is first equal to 4 volts. I have added the two voltages and left the result in R = sin (wt + a) and was calculated in radians w = omega a = alpha My answer to the sum is 11.18 sin (5 pi t + 1.63) Now, how do I find the time when the voltage is first equal to 4 volts. Appreciate any help. TIA === Subject: Re: Maths Problem - Need all the help I can get! > Two voltages v1 and v2 are to be summed using an analogue circuit. > v1 = 5 cos (5 pi t + (pi / 6)) > v2 = 10 sin (5 pi t - (pi / 6)) > Calculate the time at which the sum is first equal to 4 volts. > I have added the two voltages and left the result in R = sin (wt + a) and > was calculated in radians > w = omega a = alpha > My answer to the sum is 11.18 sin (5 pi t + 1.63) > Now, how do I find the time when the voltage is first equal to 4 volts. Huh, you cannot finish? Solve 11.18 sin(5pi.t + 1.63) = 4 for t. It will have a number of values. I'd consider the smallest positive value for t, to be when the voltage is first 4 volts. === Subject: Re: Maths Problem - Need all the help I can get! > I have added the two voltages and left the result in R = sin (wt + a) and was calculated in radians > w = omega a = alpha > My answer to the sum is 11.18 sin (5 pi t + 1.63) > Now, how do I find the time when the voltage is first equal to 4 volts. > Appreciate any help. For the equation sin(wt+a) = A (*) your solution for t satisfies t = (2k pi + asin(A) - a) / w OR t = (2k pi + pi - asin(A) - a) / w, integer k, where asin(A) denotes the inverse of the restriction of sin to some properly chosen domain (e.g. the value given by the sin^{-1} button on your calculator). The above equations define two arithmetic sequences. Finding the smallest positive value in each of them and choosing the smaller of the two gives you the first positive value of t for which (*) is satisfied. -- Pouya D. Tafti http://grads.ece.mcmaster.ca/~pouya p dot d dot tafti at ieee dot org === Subject: Re: Maths Problem - Need all the help I can get! > I have added the two voltages and left the result in R = sin (wt + a) and > was calculated in radians > w = omega a = alpha > My answer to the sum is 11.18 sin (5 pi t + 1.63) > Now, how do I find the time when the voltage is first equal to 4 volts. > Appreciate any help. For the equation sin(wt+a) = A (*) your solution for t satisfies t = (2k pi + asin(A) - a) / w OR t = (2k pi + pi - asin(A) - a) / w, integer k, where asin(A) denotes the inverse of the restriction of sin to some properly chosen domain (e.g. the value given by the sin^{-1} button on your calculator). The above equations define two arithmetic sequences. Finding the smallest positive value in each of them and choosing the smaller of the two gives you the first positive value of t for which (*) is satisfied. -- Pouya D. Tafti http://grads.ece.mcmaster.ca/~pouya p dot d dot tafti at ieee dot org === Subject: Re: Maths Problem - Need all the help I can get! > Sorry for X-Posting but needing help quickly, hopefully someone can > assist. > The problem: > Two voltages v1 and v2 are to be summed using an analogue circuit. > v1 = 5 cos (5 pi t + (pi / 6)) > v2 = 10 sin (5 pi t - (pi / 6)) > Calculate the time at which the sum is first equal to 4 volts. > I have added the two voltages and left the result in R = sin (wt + a) > and was calculated in radians > w = omega a = alpha > My answer to the sum is 11.18 sin (5 pi t + 1.63) > Now, how do I find the time when the voltage is first equal to 4 > volts. > Appreciate any help. Solve the equation 11.18 sin (5 pi t + 1.63) = 4. There are an infinite number of solutions, but you can get the smallest positive one using the inverse sine function. -- Christopher Heckman === Subject: The Internet Paper Challeran Contest The Internet Paper Challenge ranking Contest for children. http://www5.ocn.ne.jp/~pachalle/pachalleEnglishpage.html You can challenge it, but you must solve yourself without computer programing. Time limit of 33th contest registration is the end of January 2005(Japan Standard Time). Please tell your family, friends, and students of your school this interesting contest puzzle as many as possible and enjoy to solve it each other. === Subject: a man by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LEQg26237; a man has a book worth 40,000 dollars, why does he deliberately destroy it? does any one know this if so please tell me... its been driving me === Subject: Re: a man He has two copies. Destroying one increases the value of the other. or He has not destroyed the book. An elegant rouse was constructed to deter any attempt of theft. If the book does not exist then there is no incentive to steal it. Robby >a man has a book worth 40,000 dollars, why does he deliberately > destroy it? > does any one know this if so please tell me... its been driving me === Subject: Re: a man > He has two copies. Destroying one increases the value of the other. > or > He has not destroyed the book. An elegant rouse was constructed to deter > any attempt of theft. If the book does not exist then there is no incentive > to steal it. > Robby >>a man has a book worth 40,000 dollars, why does he deliberately >>destroy it? >>does any one know this if so please tell me... its been driving me Or it is insured, and he finds out that it is a fake, so destroys it in a bid to collect the insurance rather then acknowledge that it is worthless. Personally I like the first answer (has two copies). Never been a fan f these extremely ambiguous questions though... === Subject: Re: a man > He has two copies. Destroying one increases the value of the other. > or > He has not destroyed the book. An elegant rouse was constructed to deter > any attempt of theft. If the book does not exist then there is no incentive > to steal it. > Robby >>a man has a book worth 40,000 dollars, why does he deliberately >>destroy it? >> >>does any one know this if so please tell me... its been driving me >> > Or it is insured, and he finds out that it is a fake, so destroys it in > a bid to collect the insurance rather then acknowledge that it is worthless. > Personally I like the first answer (has two copies). Never been a fan f > these extremely ambiguous questions though... Personally I think it was a book of riddles with no mathematical content, which the owner had bought intending to post to math newsgroups. He then realised what a dumb idea that was and destroyed the book as an act of penance. === Subject: A problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LER226295; If g1(x)=f(x) g2(x)=f(f(x)) . . . gn(x)=f(F(F(F...F(x)))...)) that f is reapeated n times AND :F(x)=cos(x) then lim gn(x)=? n->infinite === Subject: Re: A problem The function FixedPoint repeats the calculation until the results no longer change. It looks like it tends towards .739085 for all starting values between -1 & 1. In[1]:= fx[n_] := Cos[n] In[2]:= Table[N[FixedPoint[fx, n, 100]], {n, -1, 1, 1/10}] Out[2]= {0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085, 0.739085} HTH -- Dana Mathematica 5.1 > If g1(x)=f(x) > g2(x)=f(f(x)) > . > . > . > gn(x)=f(F(F(F...F(x)))...)) that f is reapeated n times > AND :F(x)=cos(x) > then lim gn(x)=? > n->infinite === Subject: Re: A problem >If g1(x)=f(x) > g2(x)=f(f(x)) > . > . > . > gn(x)=f(F(F(F...F(x)))...)) that f is reapeated n times >AND :F(x)=cos(x) >then lim gn(x)=? > n->infinite -1 <= cos(x) <= 1 0 <= cos(cos(x)) <= about .54 about .86 < cos(cos(cos(x))) <= 1 about .54 <= cos(cos(cos(cos(x)))) <= about .65 and so forth. What does that suggest? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 === Subject: Re: A problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A25Db16797; >If g1(x)=f(x) > g2(x)=f(f(x)) > . > . > . > gn(x)=f(F(F(F...F(x)))...)) that f is reapeated n times >AND :F(x)=cos(x) >then lim gn(x)=? > n->infinite Should be the function which sends every x to the constant c where c is the unique number satisfying cos(c) = c. This is based on the fact that cos(cos(x)) lies in [0, 1] for all x, and the fact that for x, y in [0, 1], |cos(x) - cos(y)| <= K|x - y| where K is the maximum value of |sin(c)| for c in [0, 1] (this follows from the mean value theorem). Since K < 1, we see that f: [0, 1] --> [0, 1] is a strict contraction, i.e., for x in [0, 1], |gn(x) - c| = |gn(x) - gn(c)| <= K^n --> 0 as n --> oo, and the conclusion follows. Todd Trimble === Subject: Re: First Part:To Mo,JA,Matted Man Fermat's Proof of Last theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LES126344; Hi Matt,Michael Orion ,Joseph A,and Matted Man and Prof Alf Poorten AN Out line Proof Of Fermat's proof of Last Theorem. > X^n+Y^n=Z^n ;natural numbers > X+Y-Z=B; X=B+Q; Y=B+P; Z=B+Q+P >EQ.A: (B+Q)^n+(B+P )^n=(B+Q+P)^n > If Z*X*Y is not divisible by n then 2*B+Q+P=U^n and Z=U*V > Q= q^n and P=p^n >Therefore 2*B=U^n-q^n -p^n >If Z Is divisible by n then 2*B+Q+P=[n^(n-1)]*U^n and Z=n*U*V > Q=q^n and P=p^n >Therefore 2*B=[n^(n-1)]*U^n -q^n-p^n >I am going to use the symbol W for both U^n and {n^(n-1)]*U^n >If we divide EQ.A by 2*B+Q+P we get: >EQ.G:{ B^(n-1) +(a1)*B^(n-2) +(a2)*B^(n-3)+...........+c*B+ > +[Q^n+P^n]/(Q+P)}=s*V^n > where (a1),(a2),(a3)........c are coeficients, and s is equal > 1 or n and > c=[n*Q^(n-1)+n*P^(n-1)-2*(Q^n+P^n)/(Q+P)]/(Q+P) >Now X^n-X +Y^n-Y=Z^n-Z +Z-X-Y > X+Y-Z=B=n*k >Now EQ.F: X^n-Q^n+Y^n -P^n -Z^n+W^n=(W^n -W ) +(-Q^n+Q)+(-P^n+P)+ > +(W-Q-P) > EQ.F can be changed to: > K*B*n=M*n^2 +(W-Q-P) >Therefore W-Q-P=2*B is divisible by n^2 >Therefore B is divisible by n^2 >In EQ.G If we take s=1 (case Z not dovisible by n) and substract on >both side 1 we get that (V-1) is divisible by n and > {[(Q^n+P^n)/(Q+P)]-1} is divisible by n^2 >Now if we extend the parantheses in the Eq.A we get: B^n + (1/2)*[n*(n-1)]*(B^n-2)*[Q^2+P^2-(Q+P)^2].+........ >........ +n*B*[Q^(n-1)+P^(n-1)-(Q+P)^(n-1)] +Q^n+P^n-(Q+P)^n=0 >Post-scriptum: The End Of FermaTs Proof Of Last THeorem shows that E is divisible by n (ONLY-no more) as much as n*B is divisible by n where E=[(X^n+Y^n)/(X+Y)-s*V^n and s=1 or s=n Therefore E can not be equal to O. Therefore Fermat Last theorem is true. > Here it is how to get to that End: >EQ.1):X^(n-1)-[X^(n-2)]*Y +[X^(n-3)]*Y^2........+Y^(n-1)-s*V^n=0 > We write the EQ.1)In two ways after we multiply it with Z: > Let's take X+Y=B+Z=W > Then we have: > 1. Z*{W^(n-1)-n*[W^(n-2)]*Y +[n*(n-1)/2]*[W^(n-3)]*Y^2.... ........-[n*(n-1)/2]*W*Y^(n-2)+n*Y^(n-1)}-Z*s*V^n=0 > 2. The same as 1.exept we substitute Y with X > We add-up now (1.+ 2.) Taking in account the information given before Post-scriptum > We get to the that End which proves Last theorem. > The margine is too narrow(across of page)for me to continuu. > That is the Outline of Fermat's Proof Last Theorem. > Created by Gheorghe Ghiata. Warning:You have to be very good at algebraic manipulations, as good as Mr.Prof.Andrew Wiles. You need First year College algebga. Today January 9 -2005 === Subject: Re: First Part:To Mo,JA,Matted Man Fermat's Proof of Last theorem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j07JtPZ13844; Yes,I have a simple proof of Fermat Last theorem What made you confused is that I was talking about a second simple proof . Another thing is that My simple proof can result as ending with two diferent arguments each one being correct.Depends on how far you want to go with the proof. When I said Matted Man I did not reffer to you. the issue. george ghiata >Well you've certainly confused me. >Do you or do you not claim to have a simple proof of Fermat's Last >Theorem? >Hint: the answer is yes or no. === Subject: Re: First Part:To Mo,JA,Matted Man Fermat's Proof of Last theorem > Yes,I have a simple proof of Fermat Last theorem Er ... cool! > What made you confused is that I was talking about a second simple > proof . > Another thing is that My simple proof can result as ending > with two diferent arguments each one being correct.Depends on how far > you want to go with the proof. > When I said Matted Man I did not reffer to you. No, I realise that. When you post to a newsgroup it's public and anyone can chip in with their two cents' worth ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LESJ26350; Hello Michael, Hello Matt, Today january 9 2005 I posted under the old title FirstPart-To MO,JA,Matted Man Fermat'Proof of Last theorem a kind of final Outline of Fermat's Last theorem. That is my last word about It. I can not go in detail with the End of It.It is simple but it takes a lot of space and it turns off the people. You can look at it or show the posting to anybody who thinks is good at algebraic manipulations and wants to proof me wrong or right.My Opinion? IT is correct everything What I said in the posting.I hope is going to be posted soon. Should be posted before you get to read this message. george ghiata -january 9-2005 -1:10PM EST > Hello Michael, > Hello Matt, Would you please post your follow-ups to the same thread, with the same subject line. Those who are interested can follow it more easily that way, and those who are _not_ interested don't have to keep killing an ever- proliferating set of threads. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ To put it bluntly but fairly, anyone today who doubts that the variety of life on this planet was produced by a process of evolution is simply ignorant -- inexcusably ignorant, in a world where three out of four people have learned to read and write. --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A5Fu931216; Hello Stan, I make it clear .Today January 6- 10:45 PM-(EAST) 2005 The RIGHT OUTLINE of Fermat's Proof of Last Theorem is presented under the posting titleFirs Part:To MO,JA,Matted MAN Fermat'Proof Of Last Theorem and it is the one posted on January 9 -05-12 :37 (sorry Michael-wrong E-mail the last one which I sent to you) george ghiata >> Hello Michael, >> Hello Matt, >Would you please post your follow-ups to the same thread, with the >same subject line. >Those who are interested can follow it more easily that way, and >those who are _not_ interested don't have to keep killing an ever- >proliferating set of threads. >-- >Stan Brown, Oak Road Systems, Tompkins County, New York, USA > http://OakRoadSystems.com/ >To put it bluntly but fairly, anyone today who doubts that the >variety of life on this planet was produced by a process of >evolution is simply ignorant -- inexcusably ignorant, in a world >where three out of four people have learned to read and write. > --Daniel Dennett, /Darwin's Dangerous Idea/ (1995), page 46 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A25D216801; HI,MO and MATT Read again this message > Today january 9 2005 I posted under the old title > FirstPart-To MO,JA,Matted Man Fermat'Proof of Last theorem TWO MESSAGES.THE SECOND ONE IS NOT POSTED YET. THE RIGHT ONE IS THE SECOND ONE((the latestof today). > IT IS the final Outline of Fermat's Last theorem. > That is my Last word about It. > I can not go in detail with the End of It.It is simple but > it takes a lot of space and it turns off the people. > You can look at it or show the posting to anybody who thinks is > good at algebraic manipulations and wants to proof me wrong or >right.My Opinion? IT is correct everything What I said in the > posting.I hope is going to be posted soon. >Should be posted before you get to read this message. > george ghiata -january 9-2005 -6:40 PM EST === Subject: Why POL.post.are posted fast and Math.ones very late? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j09LEP926211; WHY WHY WHY WHY? I posted my Math message 2 hours ago and the political ones are display imediately? Please let me know WHY? This disscussion group it looks that it goes doWn the tube. Why you do not Call it a alt.Political and Dating Service. discussion group? DOES anybody have a ANSWER? WHAT ARE the priorities OF this DIscussions Group? george ghiata === Subject: Re: Why POL.post.are posted fast and Math.ones very late? > WHY WHY WHY WHY? > I posted my Math message 2 hours ago and the political ones are > display imediately? > Please let me know WHY? > This disscussion group it looks that it goes doWn the tube. > Why you do not Call it a alt.Political and Dating Service. > discussion group? > DOES anybody have a ANSWER? > WHAT ARE the priorities OF this DIscussions Group? > george ghiata I'm guessing that you are posting through Google. If so, you may be working with two different systems, and therefor getting two different results. Google accesses usenet (with things like alt.math.undergrad) and is not always fast. There are also groups hosted by Google that are completely seperate (though still accessed in the same area on Google's page) and could be faster. Rule of thumb: when working with a free service, you don't always get instant results. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Why POL.post.are posted fast and Math.ones very late? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0C2ujB13457; Hi Will, The only thing which I know is that I use Explorer(microsoft)sometimes Is it that what you talk about? george ghiata >> WHY WHY WHY WHY? >> I posted my Math message 2 hours ago and the political ones are >> display imediately? >> Please let me know WHY? >> This disscussion group it looks that it goes doWn the tube. >> Why you do not Call it a alt.Political and Dating Service. >> discussion group? >> DOES anybody have a ANSWER? >> WHAT ARE the priorities OF this DIscussions Group? >> george ghiata >I'm guessing that you are posting through Google. If so, you may be >working with two different systems, and therefor getting two different >results. Google accesses usenet (with things like alt.math.undergrad) >and is not always fast. There are also groups hosted by Google that are >completely seperate (though still accessed in the same area on Google's >page) and could be faster. >Rule of thumb: when working with a free service, you don't always get >instant results. >-- >Will Twentyman >email: wtwentyman at copper dot net === Subject: Re: Why POL.post.are posted fast and Math.ones very late? > Hi Will, > The only thing which I know is that I use Explorer(microsoft)sometimes > Is it that what you talk about? > george ghiata No. Google collects two seperate categories of posts under one heading. One is their copy of usenet, which includes alt.math.undergrad. These posts are distributed across a number of different usenet providers, and seem to function somewhat slowly. Google also hosts groups that are not part of usenet, though they are presented the same way. These, I suspect, show up much quicker. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Why POL.post.are posted fast and Math.ones very late? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A25D716810; > WHY WHY WHY WHY? >I posted my Math message 2 hours ago and the political ones are >display imediately? Calm down. You don't know when the political posts are actually sent, do you? Why would you think they are treated differently? As far as I know, all messages are queued; the speed with which they appear depends on many factors beyond your control (protocols between servers, etc.). Sometimes there is considerable time lag. >Please let me know WHY? >This disscussion group it looks that it goes doWn the tube. >Why you do not Call it a alt.Political and Dating Service. Well, you're still a newbie -- the frequency of this sort of thing waxes and wanes. Ignore it if it bothers you. >discussion group? > DOES anybody have a ANSWER? >WHAT ARE the priorities OF this DIscussions Group? >george ghiata My advice: chill out. Listen to some Boney M instead of being glued to the computer. There are no particular priorities; people post whatever they want. I agree there are too many off-topic posts, but what are you going to do about it? Todd Trimble === Subject: Simplifying infinite convergent sum by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A25E816843; curves I'm working on, but am having trouble trying to reduce the infinite sum notation. The program works fine but it needs to do tens of thousands of calculations inside a loop and I was hoping the basic problem will reduce to a nice clean non-summation expression. Can anyone tell me if this can be simplified? d = distance s = limit (mostly interested in case where s is infinite) length = sum(k = 1, s, [( ( d/s + cos( 2 * pi * k / s) - cos( 2 * pi * (k - 1) / s))^2 + (sin(2 * pi * k / s) - sin(2 * pi * (k - 1) / s)^2)^.5] Two special cases I've found: if d = 0, as s->infinity, length-> 2 * pi if d = 2 * pi, as s-> infinity, length-> 8 Ken === Subject: Re: Simplifying infinite convergent sum > curves I'm working on, but am having trouble trying to reduce the > infinite sum notation. The program works fine but it needs to do tens > of thousands of calculations inside a loop and I was hoping the basic > problem will reduce to a nice clean non-summation expression. Can > anyone tell me if this can be simplified? > d = distance > s = limit (mostly interested in case where s is infinite) > length = sum(k = 1, s, [( ( d/s + cos( 2 * pi * k / s) - cos( 2 * pi * > (k - 1) / s))^2 + (sin(2 * pi * k / s) - sin(2 * pi * (k - 1) / > s)^2)^.5] > Two special cases I've found: > if d = 0, as s->infinity, length-> 2 * pi > if d = 2 * pi, as s-> infinity, length-> 8 > Ken Your brackets don't match up, but fixing them in the way that seems to be intended (and which agrees with your special case answers), I think your sum as s -> infinity is equivalent to the integral: length = Integral{x = 0->2*pi: Sqr((d / (2 * pi)) ^ 2 + 1 - d / pi * Sin(x)) dx} Unfortunately I can't find a closed-form solution for arbitrary d, though it is solvable for the special cases d = 0 and d = 2*pi, giving the values you indicate. Maybe someone else can solve the integral. Otherwise you're stuck with numerical methods. If you do the above integration numerically then your code may run quite a bit faster (you can precalculate d / (2 * pi)) ^ 2 + 1 and d / pi), but essentially you'd be doing the same as what you're doing at the moment. === Subject: Re: Simplifying infinite convergent sum > Can anyone tell me if this can be simplified? > d = distance > s = limit (mostly interested in case where s is infinite) > length = sum(k = 1, s, [( ( d/s + cos( 2 * pi * k / s) - cos( 2 * pi * > (k - 1) / s))^2 + (sin(2 * pi * k / s) - sin(2 * pi * (k - 1) / s)^2 > )^.5] Setting a and b to the appropriate desirables (d/s + cos a/s - cos b/s)^2 + (sin a/s - sin b/s)^2 = (d/s)^2 + cos^2 a/s + cos^2 b/s + 2d/s * cos a/s - 2d/s * cos b/s - 2.cos a/s * cos b/s + sin^2 a/s - 2.sin a/s * sin b/s + cos^2 b/s = (d/s)^2 + 2 + 2d/s * cos a/s - 2d/s * cos b/s - 2.cos a/s * cos b/s - 2.sin a/s * sin b/s = (d/s)^2 + 2 + 2d/s * 2.cos (a+b)/2s * cos (a-b)/2s - 2.sin (a+b)/s a + b = 2pi.(2k - 1); a - b = 2pi > Two special cases I've found: > if d = 0, as s->infinity, length-> 2 * pi > if d = 2 * pi, as s-> infinity, length-> 8 > Ken === Subject: Re: ellipse equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0A7WSo09726; How to find the general expression for the major axis and minor axis if the ellipse is inclined at an angle gamma to x-axis? === Subject: Re: ellipse equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmXA31572; (Andrew >> Can someone please tell me the general equation for an ellipse, with >> the centre at any point, and the major/minor axes in any direction? >> programming assignment). >> Andrew >Ellipses are one of the conic sections, which also include circles, >parabolas and hyperbolas. >The general Cartesian equation for a conic section is > ax^2 + bxy + cy^2 + dx + ey + f = 0, >where a,b,c,d,e,f are real numbers with some restrictions. >The type of conic section is determined by the relative values of a,b and c. >As I recollect, the conic is an ellipse when b^2 < 4ac. but you had better >check this out. >-- >V. Hancher >vmhjr@frii.com The above condition is necessary but NOT sufficient - it allows the possibility of both no real solution and the degenerate case of a single point. Using the same notation as above, you also need the condition that the determinant of the matrix [2a, b, d] [b, 2c, e] [d, e, 2f] is NOT equal to zero. Pretty simple to include this in your code. === Subject: Re: ellipse equation >How to find the general expression for the major axis and minor axis >if the ellipse is inclined at an angle gamma to x-axis? Apply the rotation formulas at > http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node9.html#SECTION01210 0 00000000000000 to the equation. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ Please don't mistake my anal retentiveness for actual affection. === Subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition <3fctt0h6cov9c142v3kq9hdos9h6doj3uv@4ax.com> >> Not to mention the USS Abraham Lincoln carrier group off-shore. > Sitting >> there desalinating 400,000 gallons of water a day, with the crew >> skipping showers to conserve potable water for the tsunami > survivors. The >> medical crew is ashore treating survivors and even the bakers are > making >> bread. Last report was that 10 of it's choppers were delivering > supplies. >> Of course this stuff doesn't cost anything. > What a commendable performance. Actually, approx 100 choppers are in action... I doubt that all other countries together (other than the locals) have 100 choppers available. John === Subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition >Not to mention the USS Abraham Lincoln carrier group off-shore. >>Sitting >there desalinating 400,000 gallons of water a day, with the crew >skipping showers to conserve potable water for the tsunami >>survivors. The >medical crew is ashore treating survivors and even the bakers are >>making >bread. Last report was that 10 of it's choppers were delivering >>supplies. >Of course this stuff doesn't cost anything. >>What a commendable performance. > Actually, approx 100 choppers are in action... I doubt that all other > countries together (other than the locals) have 100 choppers available. > John 100 choppers multiplied by what, $1000/hr in maintainence, delivery costs, personnel, and fuel... multiplied by how many days? 24h/day for two weeks could be 33.6 million USD. === Subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition <41D53B8F.31AFD0FD@msgid.michael.mendelsohn.de> <41D5FF6E.1BFB3F09@msgid.michael.mendelsohn.de> >> So far I have not noticed much by way of american planes mixed up with >> the European planes which are actually flying in the aid. >> > As a former Air Force member I have to disagree here Franz, respectfully. >> >> As a former AF member, how much of the US airlift capacity do you think >> is presently tied up supplying troops in Iraq? >> I don't know.. you tell me... I don't think the Air Force is critically >> overextended in deployable airlift capability, why don't we wait a couple months >> and see what is going on instead of slamming them a few days into the crisis. > I don't think they're critically overextended either, I'm just offering > the Iraq involvement as a visible reason why the air force might not do > as much relief work as usual - quite a legitimate reason I might add > (notwithstanding my opinion about the legitimacy of that war, but let's > not get into that, the outcome is predictable). Currently the US has approx 100 choppers and 13000 people working on relief (probably providing infrastructure for our own contributions along with French/EU contributions also.) The airfields/etc are already saturated, and it will be the US who really provides the infrastructure (even if it is also indirectly funded through the UN.) John === Subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition >> So far I have not noticed much by way of american planes mixed up with >> the European planes which are actually flying in the aid. >> > As a former Air Force member I have to disagree here Franz, respectfully. >> >> As a former AF member, how much of the US airlift capacity do you think >> is presently tied up supplying troops in Iraq? >> >> I don't know.. you tell me... I don't think the Air Force is critically >> overextended in deployable airlift capability, why don't we wait a couple months >> and see what is going on instead of slamming them a few days into the crisis. > I don't think they're critically overextended either, I'm just offering > the Iraq involvement as a visible reason why the air force might not do > as much relief work as usual - quite a legitimate reason I might add > (notwithstanding my opinion about the legitimacy of that war, but let's > not get into that, the outcome is predictable). > Currently the US has approx 100 choppers and 13000 people working on > relief (probably providing infrastructure for our own contributions along > with French/EU contributions also.) The airfields/etc are already > saturated, and it will be the US who really provides the infrastructure > (even if it is also indirectly funded through the UN.) That is a remarkable effort. Franz === Subject: derivative proof by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0AD4L204567; proof: A(x^2) + 2hxy + b(y^2) = 1 then prove that d^2(y)/dx^2 = (h^2 - ab) / (hx + by)^3 hp === Subject: Re: derivative proof >proof: >A(x^2) + 2hxy + b(y^2) = 1 then prove that d^2(y)/dx^2 = (h^2 - ab) / >(hx + by)^3 Try implicit differentiation. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ Please don't mistake my anal retentiveness for actual affection. === Subject: Re: derivative proof > proof: > A(x^2) + 2hxy + b(y^2) = 1 then prove that d^2(y)/dx^2 = (h^2 - ab) / > (hx + by)^3 Start by differentiating the equation by x to get 2ax + 2hy + 2hx.dy/dx + 2by.dy/dx = 0 dy/dx = -(ax + hy)/(hx + by) Thus d^2y/dx^2 = a lot of stuff involving a,h, x,y, dy/dx and d^2y/dx^2 So solve for d^2y/dx^2 using that and the above equation. === Subject: Re: derivative proof >> proof: >> A(x^2) + 2hxy + b(y^2) = 1 then prove that d^2(y)/dx^2 = (h^2 - ab) / >> (hx + by)^3 > Start by differentiating the equation by x to get > 2ax + 2hy + 2hx.dy/dx + 2by.dy/dx = 0 and again 2a + 2h.dy/dx + 2h.dy/dx + 2hx.d^2y/dx^2 + 2b(dy/dx)^2 + 2by.d^2y/dx^2 = 0 Use both equations to solve for d^2y/dx in terms of a,b,h,x,y eleminating dy/dx. > dy/dx = -(ax + hy)/(hx + by) > Thus d^2y/dx^2 = a lot of stuff involving a,h, x,y, dy/dx and d^2y/dx^2 > So solve for d^2y/dx^2 using that and the above equation. === Subject: Matrix Calculus-[Simple questions] [1] Let A be a 2 x 2 complex matrix such that A^{2005}=O. Then A^2 = ?? [2] Let m,n distinct positive integers >= 2 . Suppose that A is a m x n matrix, and B is a n x m matrix [with complex elements]. Then det(AB)det(BA)= ?? === Subject: Re: Matrix Calculus-[Simple questions] > [1] Let A be a 2 x 2 complex matrix such that A^{2005}=O. Then A^2 = ?? [...] You might want to look at -- Paul Sperry Columbia, SC (USA) === Subject: Re: Matrix Calculus-[Simple questions] by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0AKVHd14649; >[1] Let A be a 2 x 2 complex matrix such that A^{2005}=O. Then A^2 = ?? >[2] Let m,n distinct positive integers >= 2 . Suppose that > A is a m x n matrix, and > B is a n x m matrix [with complex elements]. >Then det(AB)det(BA)= ?? Re [2], assume that m < n. Then the rank of A and B can be at most m. Now BA is n x n, but its rank is at most m, since the rank of B is at most m and likewise for A. Since rank(AB) = m < n, AB must be singular, and det(BA) = 0. Therefore, det(AB)det(BA) = 0. - MO === Subject: Re: Matrix Calculus-[Simple questions] >[1] Let A be a 2 x 2 complex matrix such that A^{2005}=O. Then A^2 = ?? >[2] Let m,n distinct positive integers >= 2 . Suppose that > A is a m x n matrix, and > B is a n x m matrix [with complex elements]. >Then det(AB)det(BA)= ?? [1] There is theorem about matrices A: A^m=O. Besides other, it says that if A is n*n matrix then A^n=O. Here we have n=2 ==> A^2=O Its proof is rather hard, I can't reproduce it right now as a whole, and it's different to adjust it for our situation without proving the whole theorem. [2] Let's call Rank(A) = number of nonzero lines in the matrix, after it was transformed into the trapezium form: a11 a12 a13 ... a1k ... a1n 0 a22 a23 ... a2k ... a2n 0 0 a33 ... a3k ... a3n ... 0 0 0 ... amk ... amn 0 0 0 ... 0 ... 0 ... 0 0 0 ... 0 ... 0 Theorem: Rank(AB) <= Rank(A), Rank(AB) <= Rank(B) It's obvious, that Rank(A)<=m, Rank(a)<=n (if A is m*n matrix) So, if we assume in in our problem m>n, we have: Rank(AB) <= n It's also obvious, that if A is n*n matrix and Rank(A)> >> The only terrorist attacks on the UK have been from the >> IRA - >> financed for >> many years by the USA. > You mean Irish-Americans. Not a penny of my taxes ever > went to support the IRA. >> Then you do not realise that the organisations collecting >> for the IRA >> in your country are recognised charities with the usual tax >> concessions for charities, >> Yes, I know that. >> so in fact you are contributing indirectly? >> Does not follow. A tax not charged does not equal money >> given. It equals a disbursement from their total budget not >> made, not an increase in their total budget. Yes, I know it >> tracks back to a tax deduction for the donors, but I'm not >> giving them any money either. Ultimately, it amounts to a >> smaller total tax collected, > Surely compensated for by tax rates which are ever so slightly higher > than they might otherwise have been? That would require a proactive taxation scheme where tax rates are scaled to meet projected expenditures. We don't have that kind of system; tax rates are tied to income (which is why it's called Income Tax) and expenditures have to be taken from the total after it's collected. Hence the Conservative philosophy; the more income individuals and Corporations have, the greater the GNP, and the more taxes can be collected while actually lowering tax rates overall. The Extreme Liberal philosophy of tax the wealthy until they're as poor as street people makes no sense to me at all; that just minimizes your tax base to the point you _have_ to Nationalize the entire economy. Oh, now I get it. ;>) Of course overruns are common because politicians have to promise to spend tax money on stupid things in order to get elected, which is where National Debt comes in. Nobody but politicians and other con artists can spend money they don't have without believable credit, but then politicians get to write and certify their own credit ratings... I am constantly amused when anyone mentions balanced budget and congress in the same breath. >> not monies transferred from me >> to anyone. > I'm not so sure of that. You're essentially having politicians say Since we can't take from x to fill out our budget, we'll make up the difference by taking it from y. That still transfers no money from y to x. Why am I reminded of the Is zero even or odd thread? Mark L. Fergerson === Subject: software in university Last few years I've been out of contact with some things. Do university Mathematics courses now require students to use some software such as Maple or Mathematica for tests, or homework? Are these used for calculational efficiency or for something else? Do these software retard the students' efforts to learn the concepts and apply the mathematical steps and skills? Algebryonic [no cross-posting through compuserve. please excuse separate postings] === Subject: Re: software in university > Last few years I've been out of contact with some things. Do university > Mathematics courses now require students to use some software such as Maple or > Mathematica for tests, or homework? Are these used for calculational > efficiency or for something else? Do these software retard the students' > efforts to learn the concepts and apply the mathematical steps and skills? > Algebryonic > [no cross-posting through compuserve. please excuse separate postings] To answer your second question, it would be very difficult to use scilab without know ing what I was doing. I find that it reinforces my knowledge, I can double check things, (ie have I found the inverse of the matrix? For my assignments we are always required to show working, and always given bonus marks for confirming our solutions using 'technology' === Subject: Re: software in university > Last few years I've been out of contact with some things. Do university > Mathematics courses now require students to use some software such as Maple or > Mathematica for tests, or homework? Are these used for calculational > efficiency or for something else? Do these software retard the students' > efforts to learn the concepts and apply the mathematical steps and skills? > Algebryonic > [no cross-posting through compuserve. please excuse separate postings] We are asked to either buy MatLab, or to use the free clone called 'scilab'. This is at university level, and our assignments expect us to demonstrate that we can use them. Having been noticing that highschools students are being required to purchase graphics calculators...I find that interesting. === Subject: Re: software in university > Last few years I've been out of contact with some things. Do university > Mathematics courses now require students to use some software such as Maple or > Mathematica for tests, or homework? Are these used for calculational > efficiency or for something else? Do these software retard the students' > efforts to learn the concepts and apply the mathematical steps and skills? It depends a lot on the school, but they'll probably have labs available with the software. In my oppinion, the software is not a good idea, and graphing calculators are questionable. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Please help me find the answer.. > 2616, 3, 11, 10850, ________, 25, 2038 > What's the missing number in the sequence? > Not enough information. > -- > -- Geo. Michael Henry > No! Bad dog! I said sit! anonymous This is the Lenny Conundrum from Neopets. I haven't heard of there being a not enough info solution before, although I'm not pretending to be and expert on the LC. I'll post if I figure it out. Good luck. === Subject: Re: Diophantine approximation or something? > for two centuries that solutions are given by (x, y) = (1, 1) and > (13, 239). It was proved by Ljunggren that these are the only positive > integer solutions. The proof is exceedingly complicated. > The Ljunggren reference is given as W. Ljunggren, Zur Theorie der > Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo, No. 51 (1942). > Nice to know they were still publishing math in occupied Norway. There are more recent solutions simpler than Ljunggren's. These are described in Math Reviews in my prior post [1]. Google web/groups searching on Ljunggren 239 13 also turns up much. --Bill Dubuque === Subject: Sequences again by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0E1rOr09453; I hate this stuff... The sequence u_1, u_2, u_3..., where u_1 ia a given real number, is defined by u_n+1 = 2u_n -1 a) Describe the behaviour of the sequence for each of the cases u_1 = 1 and u_2 = 2 Ok! This was done. No problem b) The sequence x_1, x_2, x_3, ... is defined by x_n = u_n - 1 Show that x_n+1 = 2x_n, and hence express x_n in terms of x_1 and n. c) deduce that u_n = 2^(n-1)(u_1 -1) ---- I hope you can understand the notation. This subscripts totally defeat me. I have looked at the answer for b) and it seems you can use the n+1 subscript to add onto the u making a u_n+1 = 2u...!!! I would be very grateful< if you could try and explain the basics of these operations with subscripts, which are driving me crazy Jo