mm-1019 === Subject: Arc via bezier? Hi all, my head hurts. Im trying to draw an arc with a beziercurve given a centerpoint (x,y), radius r and angles alphaStart and alphaEnd I can draw a quarter circle with Const kappa = 0.5522847498307936 k = kappa * radius a.x = center_x a.y = center_y - radius a.x2 = center_x + radius a.y2 = center_y a.order = 2 a.controlx(0) = center_x + k a.controly(0) = center_y - radius a.controlx(1) = center_x + radius a.controly(1) = center_y - k and I think Ive figured out that for angles alphaStart and alphaEnd I need a.x = center_x + (radius * cos(alphaStart)) a.y = center_y - (radius * sin(alphaStart)) a.x2 = center_x + (radius * cos(alphaEnd)) a.y2 = center_y - (radius * sin(alphaEnd)) a.order = 2 a.controlx(0) = center_x + ??? a.controly(0) = center_y - ??? a.controlx(1) = center_x + ??? a.controly(1) = center_y - ??? but thats where my Ôschool maths leave me hanging. How can I calculate the x and y coordinates of the control points for this? Please reply directly to me. Markus -- Dr. Markus Winter Tel: 0064 (0)9 373 7599 (wait for message then type extension) 83960 |// (o o) -. .-. .-oOOo~(_)~oOOo-. .-. .-. .-. .-. .-. .-. .-. .-. .-. ||X||| /|||X||| /|||X||| /|||X||| /|||X||| /|||X||| /|||X|||/|||X |/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ |||X|||/ Ô `- `- `- `- `- `- `- `- `- `- `- `- `- `- === Subject: Re: Arc via bezier? > Hi all, > my head hurts. Im trying to draw an arc with a beziercurve given a > centerpoint (x,y), radius r and angles alphaStart and alphaEnd > I can draw a quarter circle with > Const kappa = 0.5522847498307936 > k = kappa * radius > a.x = center_x > a.y = center_y - radius > a.x2 = center_x + radius > a.y2 = center_y > a.order = 2 > a.controlx(0) = center_x + k > a.controly(0) = center_y - radius > a.controlx(1) = center_x + radius > a.controly(1) = center_y - k > and I think Ive figured out that for angles alphaStart and alphaEnd I need > a.x = center_x + (radius * cos(alphaStart)) > a.y = center_y - (radius * sin(alphaStart)) > a.x2 = center_x + (radius * cos(alphaEnd)) > a.y2 = center_y - (radius * sin(alphaEnd)) > a.order = 2 > a.controlx(0) = center_x + ??? > a.controly(0) = center_y - ??? > a.controlx(1) = center_x + ??? > a.controly(1) = center_y - ??? > but thats where my Ôschool maths leave me hanging. > How can I calculate the x and y coordinates of the control points for this? Beziers can only approximate circles (and conic sections generally). Id have thought that a reasonable approximation would be to ensure that the Ôhalf-way point ends up on the circle half way between alphaStart and alphaEnd. To that end, look up de Casteljaus algorithm. You presently only have one degree of freedom (assuming youve fixed the control vectors angles to be the correct tangent at the endpoints, so all you have to do is fix that middle point. Id stick it in a symbolic algebra package, the expressions could get a bit noisy if done by hand. > Please reply directly to me. Im in a newsreader, I cant send mail from it. Anyway, ask here, get answer here is one of the Usenet mottos. Phil -- Unpatched IE vulnerability: DNSError folder disclosure Description: Gaining access to local security zones Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html === Subject: Quotient Space help Hi all, Im trying to learn about Quotient Spaces and am doing exercises in Quotient Space chapter and have come across one which I dont know how to do...maybe someone can help? I think this is a difficult topic, but it takes getting used to. Anyway, let X = R^2 {(0,0)}. Define the equivalence relation on X by saying x ~ y if and only if there exists a non-zero element t in R(the real numbers) such that x = ty. Let X* be the collection of equivalence classes in the quotient topology. What is X* homeomorphic to? I was thinking it would be easier to view R^2 {(0,0)} as the complex plane without the origin, in which case our equivalence relation beomes w ~ z if and only if there exists a non-zero element t in R suh that w = tz, where w,z are in the complex field. Would this help? I would really appreciate it if someone could give me a step by step proof of what X* is homeomorphic to because it would help me understand these spaces a lot more I think. Any help appreciated, John === Subject: Re: Quotient Space help > Hi all, > Im trying to learn about Quotient Spaces and am doing exercises in Quotient > Space chapter and have come across one which I dont know how to do...maybe > someone can help? I think this is a difficult topic, but it takes getting > used to. > Anyway, let X = R^2 {(0,0)}. Define the equivalence relation on X by > saying x ~ y if and only if there exists a non-zero element t in R(the real > numbers) such that x = ty. Let X* be the collection of equivalence classes > in the quotient topology. What is X* homeomorphic to? I was thinking it > would be easier to view R^2 {(0,0)} as the complex plane without the > origin, in which case our equivalence relation beomes w ~ z if and only if > there exists a non-zero element t in R such that w = tz, where w,z are in the > complex field. Would this help? Well, would it? Youre the one trying to grasp the nature of the quotient space; if that model makes things more comprehensible to you, then it helps. If it leaves you in the same state of puzzlement as before, then its really a bit of a no-op. > I would really appreciate it if someone could give me a step by step proof > of what X* is homeomorphic to because it would help me understand these > spaces a lot more I think. Its useful to do several things. First, understand what the equivalence classes look like. In this situation, those are lines through the origin of R^2, with that origin deleted. Note that each is the union of two copies of R+, the set of positive reals. Second, its helpful to ones understanding to be able to identify a canonical representative of each equivalence class, in such a way that neighboring equivalence classes correspond to neighboring representatives. Its more easily understood if those representatives can be identified with elements of their equivalence classes, but that sometimes cant be done. Continuing with this example, the fact that the equivalence classes are pairs of homeomorphs of the real line can be used to note that those two halves are equivalence classes of a smaller relation: (x,y) ~ (u,v) if there is a positive t such that (x,y) = t(u,v). Considering this smaller relation, note that one readily (well, once you see what the answer is, its pretty quick) identifies a canonical representative of each of the smaller equivalence classes, by taking the intersection of each ray with the unit circle: identify (x,y) with (u,v) where (u,v) = (x,y)/sqrt(x^2 + y^2). That choice of representative is easily seen to be a continuous selection: its simply the radial projection of the punctured plane to the unit circle. Once you can visualize the projection of the punctured plane to the unit circle, the rest of the problem is not too difficult. Note that the two halves of the original decomposition are mapped to antipodal points on the circle. The map of the unit circle to itself via the map z |--> z^2 (thinking of z as a complex number of unit magnitude) similarly has the property that antipodal points (z & -z) are mapped to the same value. The full quotient map is then identifiable as the composition of two maps: the first is the map R^2 {(0,0)} --> S^1 by radial projection, and the second is the mapping S^1 --> S^1 by z|--> z^2. H : R^2 {(0,0)} --> S^1 --> S^1. Note finally that the pre-image of each point of the final copy of S^1 consists of the desired set { t(x,y) | t != 0 }. Thus, this composition factors through your desired quotient space X: P R^2 {(0,0)} ------> S^1 _ /| / Q / H / / _| / X The fact that the pre-image, under P, of each point of S^1 is a single equivalence class of your original relation means that the map H is one- to-one. The fact that P is a surjection implies H is a surjection. An inverse map can be constructed as follows: given z in S^1, take the pair w_1, w_2 of solutions of w^2 = z; then construct the rays R1, R2 where Ri = { tw_i | t > 0 in R }. The value of the inverse map is then the equivalence class formed by the union of R1 and R2. The proof that this map is well-defined, continuous, and an inverse to H will be left as an exercise. > Any help appreciated, > John Dale === Subject: Limits I have a continuous function defined for all x where the properties are increasing f, concave down, f(5) = 2 and f Ô(5) = 1/2. For lim x > -infin f(x) to exist, does the gradient have to stay below 1? Im thinking it does otherwise when f(x) = - infin, x > -infin and therefore isnt defined for all x. Feel free to straighten me out on this if Im wrong. Phil Holman === Subject: Re: Polygon names Ive been using trigona as the plural, as well, for some time. 12-gon is dodecagon, 20-gon is icosagon. vertices are apices are the corners; this gives an alternative defintion for polyhedra, based on their number of (identical) vertices: polyasteron. >A 12 sided polygon is called a ????? (dodecagon?) >A 20 sided polygon is called a ????? (vigintigon? or vigesagon?) > Wot r vertices plz help me --les ducs dEnron! === Subject: Re: JSH: My lapdogs James@MSN used to be an impostor; are you all? > I learned a lot from that exchange years ago. --les ducs dEnron! === Subject: Re: JSH: My lapdogs well said, I think, monsieur ZZB; the Copenhagen school has also been formalized as Fuzzy Logic; FL is just the intersection of probability and 2-value logic; rather, combinatorics. Alan Sokal rides again; is he Black? > Play may play with dice, but he certainly doesnt > play with the Fourier stooges of probability theory. --les ducs dEnron! === Subject: Product of n consecutive integers Where can I find proof of the theorem the product of n consecutive integers can never be a perfect power of an integer? === Subject: Re: Product of n consecutive integers > Where can I find proof of the theorem the product of n consecutive > integers can never be a perfect power of an integer? Thats not entirely accurate. (Im going to assume you are excluding first powers, as you surely are.) (-2)*(-1)*0*1*2*3=0^4, for instance. The text above was written by Richard Carr in sci.math date: 2000/04/15 Richard Carr (carr@cpw.math.columbia.edu) Tapio === Subject: Re: Product of n consecutive integers >Where can I find proof of the theorem the product of n consecutive >integers can never be a perfect power of an integer? Heres the reference to the original 1975 paper by Erdos and Selfridge: Erdos, P.; Selfridge, J. L. The product of consecutive integers is never a power. Illinois J. Math. 19 (1975), 292--301. -- Erick === Subject: Re: Product of n consecutive integers >Where can I find proof of the theorem the product of n consecutive >integers can never be a perfect power of an integer? > Heres the reference to the original 1975 paper by Erdos and Selfridge: > Erdos, P.; Selfridge, J. L. > The product of consecutive integers is never a power. > Illinois J. Math. 19 (1975), 292--301. > -- Erick Don === Subject: Re: Product of n consecutive integers > Where can I find proof of the theorem the product of n consecutive > integers can never be a perfect power of an integer? I gave you the reference a few days ago: Erd.9as, P.; Selfridge, J. L. The product of consecutive integers is never a power. Illinois J. Math. 19 (1975), 292--301. Wassamatta? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Product of n consecutive integers > Where can I find proof of the theorem the product of n consecutive > integers can never be a perfect power of an integer? > I gave you the reference a few days ago: > Erd.9as, P.; Selfridge, J. L. > The product of consecutive integers is never a power. > Illinois J. Math. 19 (1975), 292--301. > Wassamatta? Sorry, I must have missed it. I dont get on the internet everyday. === Subject: Re: Science is a human activity (was: Python syntax in Lisp and Scheme) 3QLpj-NoP*NzsIC,boYU]bQ]Hy<#4ga3$21: >Its certainly true that mathematicians do not _write_ >proofs in formal languages. But all the proofs that Im >aware of _could_ be formalized quite easily. Are you >aware of any counterexamples to this? Things that >mathematicians accept as correct proofs which are >not clearly formalizable in, say, ZFC? > I am not claiming that it is a counterexample, but Ive always met > with some difficulties imagining how the usual proof of Eulers > theorem about the number of corners, sides and faces of a polihedron > (correct terminology, BTW?) could be formalized. Also, however that > could be done, I feel an unsatisfactory feeling about how complex it > would be if compared to the conceptual simplicity of the proof itself. Which one do you think is the usual proof? http://www.ics.uci.edu/~eppstein/junkyard/euler/ Anyway, this exact example was the basis for a whole book about what is involved in going from informal proof idea to formal proof: http://www.ics.uci.edu/~eppstein/junkyard/euler/refs.html#Lak -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Google Calculator It even handles factorials. E.g. 5 ! = 120 But even more amazing... it handles factorials of the reals: E.g. 1.2 ! = 1.10180249 3.5 ! = 11.6317284 Here are some other neat resutls I discovered with some experimentation: pi/2 pi / 2 = 1.57079633 2*phi 2 * the golden ratio = 3.23606798 gamma + 1 Eulers constant + 1 = 1.57721566 (1+2i)/2 (1 + 2i) / 2 = 0.5 + i log(-1)+2i log(-1) + (2 * i) = 3.36437635 i === Subject: Re: Google Calculator > Perhaps it should be noted that, > if you type in -2^2 , it then displays -2^2 = 4 , > Nothing particularly unusual about that: I didnt say that there was anything particularly unusual about it. But I think it should be noted because it violates an order of operations which is almost universally accepted. I seriously doubt that you would be able to find any mathematics text or any computer algebra system which would say that -2^2 should be evaluated as +4. > bash-2.05a$ bc > bc 1.06 > Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. > This is free software with ABSOLUTELY NO WARRANTY. > For details type `warranty. > -2^2 > And C/C++ also both adopt the convention of binding unary minus > expressions very tightly. > whereas, > if you type in 0-2^2 , it then displays 0-(2^2) = -4 . > Unary operator Ô- and binary operator Ô- are not the > same operator, they neednt even use the same symbol > (but always seem to in any grammer Ive encontered). IIRC, different symbols are used for them in APL and J. And that, IMO, is laudable. David > Even in the grammer of a language that evaluates Ô-2^2 > to be -4, youll see that unary Ô- and binary Ô- are > different operators. Its just a question of the > operator precedence of the unary one that varies. > Phil === Subject: Re: Google Calculator > Perhaps it should be noted that, > > if you type in -2^2 , it then displays -2^2 = 4 , > Nothing particularly unusual about that: > I didnt say that there was anything particularly unusual about it. > But I think it should be noted because it violates an order of operations > which is almost universally accepted. I seriously doubt that you would be > able to find any mathematics text or any computer algebra system which > would say that -2^2 should be evaluated as +4. > bash-2.05a$ bc > bc 1.06 > Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. > This is free software with ABSOLUTELY NO WARRANTY. > For details type `warranty. > -2^2 > 4 bc runs on my computer. It seems to have as good a model of Z as any other computer algebra system I have (Mathematica, Maxima, and Pari/GP). So guess what - Ill propose bc as a computer algebra system, albeit a rather limited one, which says that -2^2 should be evaluated as +4. You never saw that one coming, I bet. Im sure its a given that _ x^2 is interpreted as _ (x)^2 So why is negation so different from conjugation? Why can conjugation bind more tightly than exponentiation, but your negation is forced to bind quite weakly? > And C/C++ also both adopt the convention of binding unary minus > expressions very tightly. > whereas, > > if you type in 0-2^2 , it then displays 0-(2^2) = -4 . > Unary operator Ô- and binary operator Ô- are not the > same operator, they neednt even use the same symbol > (but always seem to in any grammer Ive encontered). > IIRC, different symbols are used for them in APL and J. > And that, IMO, is laudable. I believe that Ive seen different renderings in some typesetting conventions too. (Then again Ive seen books that use both groups of 3 digits and groups of 5 digits for rendering numbers of different lengths, and others that have used different notation for multiplication (concatenated versus an explicit symbol), so I dont think that all books can be held up as demonstrating standards to aspire to.) Phil -- Unpatched IE vulnerability: mhtml wecerr CAB ßip Description: Delivery and installation of an executable Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/48.html === Subject: Re: Google Calculator > Im sure its a given that > x^2 > is interpreted as > _ > (x)^2 > So why is negation so different from conjugation? > Why can conjugation bind more tightly than exponentiation, > but your negation is forced to bind quite weakly? Perhaps so that 0 - x^2 = -x^2 ? === Subject: Re: Google Calculator > Im sure its a given that > _ > x^2 > > is interpreted as > _ > (x)^2 > > So why is negation so different from conjugation? > Why can conjugation bind more tightly than exponentiation, > but your negation is forced to bind quite weakly? > Perhaps so that 0 - x^2 = -x^2 ? Which is fair, but it relies on the historical coincidence of both of those operations, and they are completely distinct operations, having the same symbol. In this medium. As David points out, and Im unable to verify this, never having used APL or C, its not true in all media. Does negation really correspond to subtraction for you? When you see -2 Do you really interpret that in your head as 0-2 ? There are others who view -2 as the-negative-unit * 2, i.e. unary minus represents a multiplication by -1, rather than a subtraction from zero. For those, +/- represents multiplication by either unit, which I think is an eminently fair view. I tend to prefer to view -2 as atomic, a number in Z, and not as the subtraction of two numbers in N. To this end, I bind negation higher than I bind multiplication. This is also the convention in bc and C. And its not just some algoloids that have this property, Lisp does too: -2 and (* -2 -2) are valid expressions, but 0-2 and (* 0-2 0-2) arent. Id wager that Forth would similarly barf on such infix expressions. Phil -- Unpatched IE vulnerability: NavigateAndFind file proxy Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ PU-Content.HTM Exploit: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ PU-MyPage.htm === Subject: Re: Google Calculator > I tend to prefer to view -2 as atomic, a number in Z, and not as > the subtraction of two numbers in N. I prefer to view -2 as the equivalence class of natural number pairs, {(m,n): m in N, n in N, m + 2 = n}. God made the natural numbers, all else is the work of man. === Subject: Re: Google Calculator > Does negation really correspond to subtraction for you? No, its the other way around. Subtraction corresponds to negation. > When you see > -2 > Do you really interpret that in your head as > 0-2 No. When I see a - b, I interpret it as a + (-b). > There are others who view -2 as the-negative-unit * 2, i.e. > unary minus represents a multiplication by -1, rather than a > subtraction from zero. For those, +/- represents multiplication > by either unit, which I think is an eminently fair view. I agree. > I tend to prefer to view -2 as atomic, a number in Z, and not as > the subtraction of two numbers in N. So do I. > To this end, I bind negation > higher than I bind multiplication. This is also the convention in > bc and C. What are the roots of -x^2 + 4 = 0? > And its not just some algoloids that have this > property, Lisp does too: -2 and (* -2 -2) are valid expressions, > but 0-2 and (* 0-2 0-2) arent. Id wager that Forth would > similarly barf on such infix expressions. Lisp parses the - in -2 as part of the number, not as an operator. You cant write -x, for example; you have to write (- x) instead, where the - is now clearly an operator. The only thing distinguishing unary minus from ordinary subtraction is the number of arguments supplied. In fact, there is no need to stop with just one or two: an expression like (- a b c d e f) means a - b - c - d - e - f in infix notation. Lisp and forth happen to share the property that the ordinary concept of operator precedence does not apply. Both are strange choices of a language to argue for a particular operator precedence. All of your arguments support the view that unary minus is more fundamental than subtraction. I entirely agree. But it doesnt change the fact that -x^2 + 4 = 0 has the roots x = +/-2, not x = +/-2i. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Google Calculator > Lisp and forth happen to share the property that the ordinary concept of > operator precedence does not apply. Both are strange choices of a > language to argue for a particular operator precedence. Deliberately chosen to demonstrate the inclusion of the Ô- at the tokeniser stage (and as you say, specifically when prefixing a number). I tried to avoid the -x^2 issue, as you also had noticed, as I was offered the -2^2 bait as well, for which its far easier to pull in lexing arguments. (This thread was about lexing though.) [ -x^2 ] For -x^2, C deserts me, as there is no exponentiation, but bc still honours the disfavoured precedence: x=2 -x^2 4 So, reasonable or not, there is grounds to claim that unless theres some context (i.e. we know were working in Mathematica, GP, or whatever) the notation could be considered ambiguous, and therefore people should be prepared to disambiguate if the question of notation arises. And that, your honour, is the case for the defence. Phil -- Unpatched IE vulnerability: Alexa Related Privacy Disclosure Description: Unintended disclosure of private information when using the Related feature Reference: http://www.secunia.com/advisories/8955/ Reference: http://www.imilly.com/alexa.htm === Subject: Re: Google Calculator > Perhaps it should be noted that, > > if you type in -2^2 , it then displays -2^2 = 4 , > > Nothing particularly unusual about that: > I didnt say that there was anything particularly unusual about it. > But I think it should be noted because it violates an order of > operations which is almost universally accepted. I seriously doubt > that you would be able to find any mathematics text or any computer > algebra system which would say that -2^2 should be evaluated as +4. > bash-2.05a$ bc > bc 1.06 > Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. > This is free software with ABSOLUTELY NO WARRANTY. > For details type `warranty. > -2^2 > 4 > bc runs on my computer. It seems to have as good a model of Z > as any other computer algebra system I have (Mathematica, Maxima, > and Pari/GP). So guess what - Ill propose bc as a computer > algebra system, albeit a rather limited one, which says that > -2^2 should be evaluated as +4. > You never saw that one coming, I bet. True, I never saw that one coming. But describes bc as a numerical processing language. If thats an accurate description, then bc is clearly not a computer algebra system, not even a rather limited one. > Im sure its a given that > x^2 > is interpreted as > _ > (x)^2 Of course. If instead we had wanted the conjugate of the square, we would have written ___ x^2 > So why is negation so different from conjugation? Convention, together with the fact that the vinculum is itself a grouping symbol. > Why can conjugation bind more tightly than exponentiation, > but your negation is forced to bind quite weakly? Convention... BTW, I never said that I _like_ that convention. If I had my way, negation and subtraction would have distinct symbols, and unary operations would always have precedence over binary operations. David > And C/C++ also both adopt the convention of binding unary minus > expressions very tightly. > > whereas, > > if you type in 0-2^2 , it then displays 0-(2^2) = -4 . > > Unary operator Ô- and binary operator Ô- are not the > same operator, they neednt even use the same symbol > (but always seem to in any grammer Ive encontered). > IIRC, different symbols are used for them in APL and J. > And that, IMO, is laudable. > I believe that Ive seen different renderings in some > typesetting conventions too. > (Then again Ive seen books that use both groups of 3 > digits and groups of 5 digits for rendering numbers > of different lengths, and others that have used different > notation for multiplication (concatenated versus an > explicit symbol), so I dont think that all books can > be held up as demonstrating standards to aspire to.) > Phil === Subject: Re: MKC Set Theories > To sum up so far: > --There is one axiom that theories of collections proposed since the > 1900s all share: Standard Extensionality. > AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality There are constructive set theories which are not extensional. There is some material on this in Michael Beesons Foundations of Constructive Mathematics. Roger Jones === Subject: Re: MKC Set Theories >>To sum up so far: >>--There is one axiom that theories of collections proposed since the >>1900s all share: Standard Extensionality. >> AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality > There are constructive set theories which are not extensional. > There is some material on this in Michael Beesons Foundations of > Constructive Mathematics. There are also so called intensional set theories (by Goodman and from North-Holland 1985. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: MKC Set Theories > > >>To sum up so far: >>--There is one axiom that theories of collections proposed since the >>1900s all share: Standard Extensionality. >> AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality > > > There are constructive set theories which are not extensional. > There is some material on this in Michael Beesons Foundations of > Constructive Mathematics. > There are also so called intensional set theories (by Goodman and > from North-Holland 1985. Im not aware of these. Perhaps Michael Potter wasnt either. In any event, maybe someone familiar with these will want to shed some light. --John === Subject: Re: MKC Set Theories >To sum up so far: >--There is one axiom that theories of collections proposed since the >1900s all share: Standard Extensionality. > AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality >> There are constructive set theories which are not extensional. >> There is some material on this in Michael Beesons Foundations of >> Constructive Mathematics. >There are also so called intensional set theories (by Goodman and >from North-Holland 1985. You guys are just trying to confuse the issue with facts. ************************ David C. Ullrich === Subject: Re: MKC Set Theories >--There is one axiom that theories of collections proposed since the >1900s all share: Standard Extensionality. >> AxAy[Az(z in x <-> z in y) -> x=y] Standard Extensionality >> There are constructive set theories which are not extensional. >> There is some material on this in Michael Beesons Foundations of >> Constructive Mathematics. Aatu Koskensilta >There are also so called intensional set theories (by Goodman and >from North-Holland 1985. > You guys are just trying to confuse the issue with facts. I completely agree. The issue is about theories where things that are defined by whats in them, alone, and things that (in addition to what they have in them) are also defined from other inputs, can peacefully co-exist. That is not what is going on in constructivism. === Subject: elliptic integral of first kind hello, i have a solution to a physics problem that involves the elliptic integral function of the first kind K(z) Namley the problem is determining the potential of a finite disc in the z=0 plane carrying constant charge. now the potential i get is phi = 2/(pi*r) * [ r^2*E(z^2) + (a^2-r^2)*K(z^2) ] [1] where z = a/r, a=radius of disc, r=radial coord, E is the elliptic integral of the second kind. differentiating w.r.t r i get the electric field as e(r) = 2/pi * [ E(z^2) - K(z^2) ] [2] now what i wish to do is look at what happens with both [1] and [2] when i take the limit as r approaches the circumference of the disc i.e. as r->a or z->1. it would *appear* that [1] is bounded as the term infront of K(z^2) goes to zero as r->a. however in [2] im not sure what happens as K(z) has a branch point at z=1. so what does [2] do in this limit? do i need analytic continuation? if so how does one apply it to K(z) at z=1? perhaps i have made a mistake? i dont think so as i have derived the potential various ways, and in all representations / derivations the electric field ends up not being defined at the circumference of the disc. Is this a well known fact (it must be in potential theory). how does one get around it? Where can i find a derivation of the 3D potential of a disc in the plane with a constant distribution of charge on its surface that avoids these issues? cheers moth I have emailed one of the authors, and he was kind enough to provide me with the c++ source code. So it looks like I have found a solution to the problem... (If I get the MATLAB program going, I can provide the source code if anyone is interested) Jaco > decoding algorithm for correcting both erasures and errors of > Reed-Solomon codes (appeared in IEEE Transactions on Communications, > In section IV, the authors present an algorithm, (Algorithm 2), to > decode both erasures and errors, using an algorithm based on the > Euclidean algorithm together with Berlekamp-Massey algorithm. > I have tried to implement the pseudocode in MATLAB, and used Example 1 > to verify if my implementation works. The output of my implementation > does not correspond at all with the output given in the example. My > first problem is that the polynomial OMEGA^(b) is not computed > according to the pseudocode. > In the pseudocode, OMEGA^(b)(x) is multiplied by x, in line 41, but > when I look at the output given in Example 1, the code is not > multiplied by x. > Did anyone implement this algorithm? Any suggestions would be greatly > appreciated > Jaco Versfeld === Subject: Re: Constant weight codes in GF(2^m)? My Prof has a copy of the book mentioned and I will definitly try to Your time and effort is greatly appreciated Jaco > I hope that this is not too far off topic? I am interested in > constructing constant weight codes using elements from a Galois group > GF(2^m)? How can I construct such codes? In the literature that I > have found up to now, the authors only consider binary constant weight > codes. > Any comments, suggestions and/or pointers to literature will be > greatly appreciated > Jaco Versfeld === Subject: Re: Is this a valid proof? The proof strikes me as correct. I have a Ph.D. in math and Ive seen many similar proofs. However, Im quite rusty (I earned the degree in 1973 and I havent worked as a mathematician since), so dont take my word as gospel! I just like to lurk in math newsgroups every now then. :-) .... Bob === Subject: Re: Magidin is too many > /~~~ > / | /~~| `=` |/~~ > /_/ | `` | | `` |/~~ > / | | | | `` | > { <| | | | | > | | > | / . . ` / > | . . / > . / > . / > / > / > | | > | | > --les ducs dEnron! Larouche is a jailbird and his groupies are nuts. Get lost. === Subject: Re: Magidin is too many > > > I have no idea who is competent to judge JSHs work. > > Which prompts the question once again: If you have no idea who is > qualified to even judge JSHs work, how do you conclude he is a > genius? What evidence has persuaded you that James is brilliant? Is > it merely that mathematicians reject his claims? >> The perp in the hotseat is YOU, O Epitome of the Flaming-As-Inholian! > > Sorry, youve become incoherent. > become? :-) A picture is worth a thousand words: Finger This! ___ / ~ |___/| | | | = | | | | . | _ | ` ` | / | /~~| `=` |/~~ /_/ | `` | | `` |/~~ / | | | | `` | { <| | | | | | | | / . . ` / | . . / . / . / / / | | | | | } | | | . | === Subject: Re: Magidin is too many as a graph can be illustrative, dont sell yourself out on being able to handle the new math of JSH, once anyone can pin him down on his terminology. like, the example that he gave (in my own notation) of P(m) = 2(mm+2m+1) = 2(am+1)(bm+1) = (Am+2)(bm+1), then substituting zero for m, and ... what was the point of that? if you can tell where the factor of two went, then maybe you can decide. almost *any* thing is better than Nick Atnite Bourbaki! > I have no idea who is competent to judge JSHs work. > A picture is worth a thousand words: --les ducs dEnron! === Subject: Re: Formulae for Latin squares of size 2^n >> For size 2^n, an obvious type of Latin squares has its >> element in row i and column j (counting from 0) given by >> s(i,j) = i + j mod 2^n >> (We consider those that result from row and/or column >> permutations of the Latin square to belong to the same type.) >> One can easily verify that there is another type whose >> elements may be described as follows: >> t(i,j) = 2*i*j + i + j mod 2^n >> Are there additional types having elements that could be >> in advance. >The multiplication table of any group (lots of these >of order 2^n as evidenced in another recent thread). The number of distinct groups of order p^n for any prime p is known to be about p^e(n), where e(n) is of order 2n^3/27. More precisely, e(n) is known to satisfy 2n^3/27 - an^2 <= e(n) <= 2n^3/27 + bn^(8/3) for constants a and b. The lower bound comes from considering only special p-groups, which are p-groups with Z(G) = [G,G] and both Z(G) and G/Z(G) elementary abelian. It is a plausible conjecture that almost all finite groups (in the usual sense) are special 2-groups, but the known results have their error terms in the exponents, so they are not nearly precise enoguh to prove anything like that. Derek Holt. === Subject: Re: Formulae for Latin squares of size 2^n > For size 2^n, an obvious type of Latin squares has its > element in row i and column j (counting from 0) given by > s(i,j) = i + j mod 2^n > (We consider those that result from row and/or column > permutations of the Latin square to belong to the same type.) > One can easily verify that there is another type whose > elements may be described as follows: > t(i,j) = 2*i*j + i + j mod 2^n > Are there additional types having elements that could be > in advance. Would using the product of Latin squares help? If A is a Latin square with m rows and B is a Latin square with n rows, then C=AxB is a Latin square with mn rows and defined by (using index sets 0..n-1 and 0..m-1 respectively) the formula c_{n*j+i, n*l+k}=n*b_{jl}+a_{ik} (may be I reversed the roles of A and B, look this up). It seems to me that as you have ways of building some small Latin squares, you can generate many larger ones by the product construction (order matters, i.e. AxB is not the same as BxA), as the above formula surely is easy to code. Another nice property of the product construction is that it preserves orthgonality: if A is orthogonal to A and B orthogonal to B, then AxB is orthogonal to AxB. Jyrki Lahtonen, Turku, Finland === Subject: Re: Formulae for Latin squares of size 2^n > Would using the product of Latin squares help? > If A is a Latin square with m rows and B is a Latin square with > n rows, then C=AxB is a Latin square with mn rows and defined > by (using index sets 0..n-1 and 0..m-1 respectively) the formula > c_{n*j+i, n*l+k}=n*b_{jl}+a_{ik} > (may be I reversed the roles of A and B, look this up). > It seems to me that as you have ways of building some small > Latin squares, you can generate many larger ones by the product > construction (order matters, i.e. AxB is not the same as BxA), > as the above formula surely is easy to code. > Another nice property of the product construction is that > it preserves orthgonality: if A is orthogonal to A and > B orthogonal to B, then AxB is orthogonal to AxB. Yes. One can indeed build a large Latin square from that technique. (I recall now a post of Terry Ritter in sci.crypt quite a time ago in which he discussed a example of the nature you indicated.) M. K. Shen === Subject: Re: OLD HP CALUCLATORS > You mean the ones with algebraic notation and no parentheses? (Yes, they > really exist.) Or with the (seemingly always) insufficiently deep nesting. > Ive exceeded 9 levels of parentheses with algebraic (left-to-right) entry, > but I vaguely remember *once* exceeding a 4-level stack on an RPN > calculator. I couldnt even tell you the circumstances. I might be > misremembering -- or remembering an error I made. > I get around the algebraic problems by (drum roll, please) working from the > inside out and either using stored memory locations or writing down the > intermediate results. I also claim that anyone who can actually use an > algebraic calculator for difficult calculations (like mortgage payment > calculations) does the equivalent (well, maybe not, since it can be done > with about 5 levels of parentheses). But after a while, you forget which > level of parentheses youre in, so you have to be making notes of some sort > on a sheet of paper. > Oh, I forgot algebraic notation but no algebraic hierarchy. So 1+2*3=9. > Aaaaaauuuuuuuggggggghhhhhhhhh! Usually with no parentheses. > Just curious, I dont have a calculator with parentheses, but if you > have one, could you test this out and tell how many seconds > and how many keystrokes it takes to calculate this hypothetical > thing: > (9+7)(8-3)(5+4) > ------------------ + (6-4)(3-1) > (9+2)(9-4)(7+3) > Of course youre not allowed to use your knowledge of simple > arithmetic ;-) > With RPN (and without practicing first) it this takes 39 keystrokes > and approx. 13 seconds. And without using memory other than the > implicit stack. > How does this go with an algebraic with parentheses? Surely I dont have to conclude that it cant be done? I wouldnt be surprised though ;-) Can anyone with an algebraic parentheses calculator please try this? Im really interested in the number of keystrokes (exact) and seconds (approx.) it takes... Dirk Vdm === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > In sci.math, Mark Nudelman >> Would you point out where he makes this assumption? I repeat the >> entire proof, expanded a bit, with line numbers added for your >> convenience: >> [1] 9=9 >> [2] 9=9.9999999...-.9999999... >> [3] 9(1)=10(.9999999...)-.9999999 >> [4] Let x = .9999999... and substitute in [3] >> [5] 9(1) = 10x - x >> [6] 9(1) = 9(x) >> [7] Therefore x=1 >> In which line is the assumption .99999... = 1 used? > [3]. The possibility of an infinite borrow generates headaches. Going from [2] to [3] merely assumes that 10(.99999...) = 9.9999.... This is indeed problematic and needs to be proven, as does the assumption that 9 = 9.99999...- 0.999999 in going from [1] to [2], but I dont see that either of these steps uses the assumption that .99999... = 1. --Mark === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, >> 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >>One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company > > .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end shit comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end shit comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas