mm-4159 === Subject: Re: decomposition of signals You can use what are known as joint-time freq analysis methods to >deal with problems of this time, I believe. The DSP group at Rice Univ >has an entire website devoted to this. The basic kernel of the idea >used is not so hard to understand and there are some decent texts, >internet tutorials, etc that give you the level of detail that you'd >need. Good lead! I found this: http://www.dsp.rice.edu/software/optkernel.shtml while poking around the Rice website. In order to find the `best' TFR (Time Frequency Representation) for any given signal, our research focuses on optimization-based kernel design. Optimal-kernel TFRs adapt to the signal at hand in order to extract the maximum possible time-frequency information. John === Subject: Re: decomposition of signals You can use what are known as joint-time freq analysis methods to >deal with problems of this time, I believe. The DSP group at Rice Univ >has an entire website devoted to this. The basic kernel of the idea >used is not so hard to understand and there are some decent texts, >internet tutorials, etc that give you the level of detail that you'd >need. Good lead! I found this:http://www.dsp.rice.edu/software/optkernel.shtml > while poking around the Rice website. > In order to find the `best' TFR (Time Frequency Representation) for > any given signal, our research focuses on optimization-based kernel > design. Optimal-kernel TFRs adapt to the signal at hand in order to > extract the maximum possible time-frequency information. > John Richard Baraniuk at Rice and D. Jones at UIUC are the two people who did a great deal of the seminal work in the area of joint-time frequency analysis (JTFA). Leon Cohen at Hunter Univ has written some very nice general reviews and has done important theoretical work in this area (in fact, the types of distributions you probably need are given the generic name the Cohen Class). There is also a very clear explanation of JTFA methods in the dissertation of Flemming Pedersen which can be found on the Web (which is where I started to learn about JTFA). GL, M === Subject: Re: decomposition of signals bit better such as the positive gabor transform but each one seems to have its own short commings. I'm not sure if there is some inherent reason why the uncertainty principle applies or not but I can't see why. It seems to me the problems are due to the decomposition/transformation rather than something in the signal itself. The reason I say this is because if I make a signal such as sin(w*t)*R(t) where R(t) is some rectanglular function(but say could have a smooth decay at the boundaries) then it has a definite fequency for most of the support of R(t) but in the fourier transform the frequency is not as definite. I can't see why the fourier transform(really, STFT) is so unresolving with w.r.t. frequency in this example as it seems it should be able to determine the frequency exactly when it is well defined. I'm going to try to play around some more and see what I can find out... just hope I'm not wasting my time. BTW, essentially as I said before, if I have a signal such as f(t) = sum(Ak*sin(wk*t + Ok)*R((t - bk)/ak)) then I would expect that I could find out the constants exactly(with numerical precision) by some decomposition. I would also expect that if I did have an algorithm that found the constance for f(t) represented above that it would work well for functions that were not exactly represented in finite terms. What I'm currently thinking is that I could use the fourier transform/spectrum to approximately determine the constants and then try to refine by minimizing some error. I think though that the algorithm would not be very efficient ;/ Anyways, Jon === Subject: name of a metric Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } where x=(x1,x2) y=(y1,y2) Dan === Subject: Re: name of a metric Hi! Have you tried the Dictionary of Distances? http://www.elsevier.com/wps/find/bookdescription.cws_home/708465/description #description > Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan > === Subject: Re: name of a metric [Humberto] > Have you tried the Dictionary of Distances? it. I don't have access to it here. If someone else has a copy and wouldn't mind taking a look, that would be much appreciated. http://www.worldcat.org/isbn/0444520872 http://www.amazon.com/dp/0444520872 Dan Greenhoe (OP) > Hi! Have you tried the Dictionary of Distances? http://www.elsevier.com/wps/find/bookdescription.cws_home/708465/desc... > Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan === Subject: Re: name of a metric Bytes: 1552 > Does anyone know of a standard name for this metric > in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan If (x,y) belongs to the intersection B((0,0),1) with the first quadrant, then: sqrt(x)+sqrt(y)<1 <=> x+y+2sqrt(xy)<1 <=> 2sqrt(xy)<1-x-y <=>4xy<1+x^2+y^2-2x-2y+2xy<=>0 [Fernando Revilla] > If (x,y) belongs to the intersection B((0,0),1) with the > first quadrant, then: > sqrt(x)+sqrt(y)<1 <=> x+y+2sqrt(xy)<1 <=> 2sqrt(xy)<1-x-y > <=>4xy<1+x^2+y^2-2x-2y+2xy<=>0 The boundary related to this open ball in the first > quadrant is a parabola, so I propose to name this metric, > parabolic metric. > --- > Of course 0 0 in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan If (x,y) belongs to the intersection B((0,0),1) with the > first quadrant, then: sqrt(x)+sqrt(y)<1 <=> x+y+2sqrt(xy)<1 <=> 2sqrt(xy)<1-x-y <=>4xy<1+x^2+y^2-2x-2y+2xy<=>0 quadrant is a parabola, so I propose to name this metric, > parabolic metric. P.S. Alternatively I propose the name Achimota's > metric in honour to the first person who asked for it. > It sounds well and exotic. > Fernando. === Subject: Re: name of a metric <19403448.1183832428210.JavaMail.jakarta@nitrogen.mathforum.org> [Fernando Revilla] > P.S. Alternatively I propose the name Achimota's > metric in honour to the first person who asked for it. > It sounds well and exotic. I want to say that this metric is _not_ my idea. I found it in Dr. Timothy S. Norfolk's unpublished paper When Does a Metric Generate Convex Balls? which can be downloaded from his website at http://www.math.uakron.edu/~norfolk/ It is an interesting paper --- take a look if you have time. Dan Greenhoe (original poster) P.S. Achimota is the name of a town in Ghana, West Africa, just north of the capital, Accra. It is a great place with great people and where I did have the very great privilege of living for some time some time ago teaching some math to some high school students. If you want to name a metric or anything else in honor of the people of Achimota, that would be great. On Jul 8, 2:19 am, fernando revilla in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan If (x,y) belongs to the intersection B((0,0),1) with the > first quadrant, then: sqrt(x)+sqrt(y)<1 <=> x+y+2sqrt(xy)<1 <=> 2sqrt(xy)<1-x-y <=>4xy<1+x^2+y^2-2x-2y+2xy<=>0 quadrant is a parabola, so I propose to name this metric, > parabolic metric. P.S. Alternatively I propose the name Achimota's > metric in honour to the first person who asked for it. > It sounds well and exotic. > Fernando. === Subject: Re: name of a metric <19403448.1183832428210.JavaMail.jakarta@nitrogen.mathforum.org> [Fernando Revilla] > P.S. Alternatively I propose the name Achimota's > metric in honour to the first person who asked for it. > It sounds well and exotic. I want to say that this metric is _not_ my idea. I found it in Dr. Timothy S. Norfolk's unpublished paper When Does a Metric Generate Convex Balls? which can be downloaded from his website at http://www.math.uakron.edu/~norfolk/ It is an interesting paper --- take a look if you have time. Dan Greenhoe (original poster) P.S. Achimota is the name of a town in Ghana, West Africa, just north of the capital, Accra. It is a great place with great people and where I did have the very great privilege of living for some time some time ago teaching some math to some high school students. If you want to name a metric or anything else in honor of the people of Achimota, that would be great. On Jul 8, 2:19 am, fernando revilla in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan If (x,y) belongs to the intersection B((0,0),1) with the > first quadrant, then: sqrt(x)+sqrt(y)<1 <=> x+y+2sqrt(xy)<1 <=> 2sqrt(xy)<1-x-y <=>4xy<1+x^2+y^2-2x-2y+2xy<=>0 quadrant is a parabola, so I propose to name this metric, > parabolic metric. P.S. Alternatively I propose the name Achimota's > metric in honour to the first person who asked for it. > It sounds well and exotic. > Fernando. === Subject: Re: name of a metric Of course 0 on Sat, 07 Jul 2007 05:20:49 -0700 > Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) Dan > That's a strange metric, as translation of the two points along either axis changes the value of d(x,y). Did you mean d(x,y) = sqrt(|x2-x1|) + sqrt(|y2-y1|) ? Or perhaps you were thinking of the Manhattan metric, which can be represented d_m(x,y) = |x2-x1|+|y2-y1| (Manhattan is noted for its gridlike, regular street map; the Manhattan metric therefore represents the shortest path along that map, and not only for integral x1,y1,x2,y2.) And of course the standard Euclidean metric is d_s(x,y) = sqrt((x2-x1)^2 + (y2-y1)^2) -- #191, ewill3@earthlink.net Windows. Multi-platform(1), multi-tasking(1), multi-user(1). (1) if one defines multi as exactly one. -- === Subject: Re: name of a metric <1qi4m4-18t.ln1@sirius.tg00suus7038.net> [The Ghost In The Machine] > Did you mean ... the Manhattan metric > d_m(x,y) = |x2-x1|+|y2-y1| > And of course the standard Euclidean metric is > d_s(x,y) = sqrt((x2-x1)^2 + (y2-y1)^2) Both the Manhattan metric and Euclidean metric are generated by norms and therefore their balls in R^n are convex. However the metric d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } where x=(x1,x2) y=(y1,y2) is a little bit unusual because it is _not_ generated by a norm and also its unit ball in R^2 is _not_ convex. In fact, when plotted on the traditional Euclidean plane, it looks like the diamond shaped Manhattan metric with its sides pushed in. Dan Greenhoe (original poster) On Jul 7, 9:58 pm, The Ghost In The Machine > In sci.math, Daniel J. Greenhoe > where x=(x1,x2) y=(y1,y2) Dan That's a strange metric, as translation of the two points along either > axis changes the value of d(x,y). Did you mean d(x,y) = sqrt(|x2-x1|) + sqrt(|y2-y1|) ? Or perhaps you were thinking of the Manhattan metric, which can > be represented d_m(x,y) = |x2-x1|+|y2-y1| (Manhattan is noted for its gridlike, regular street map; the Manhattan > metric therefore represents the shortest path along that map, and not > only for integral x1,y1,x2,y2.) And of course the standard Euclidean metric is d_s(x,y) = sqrt((x2-x1)^2 + (y2-y1)^2) -- > #191, ewi...@earthlink.net > Windows. Multi-platform(1), multi-tasking(1), multi-user(1). > (1) if one defines multi as exactly one. -- === Subject: Re: name of a metric On Sat, 07 Jul 2007 05:20:49 -0700, Daniel J. Greenhoe fed this fish to the penguins: >Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) > Euclidean metric? G. Rodrigues === Subject: Re: name of a metric >On Sat, 07 Jul 2007 05:20:49 -0700, Daniel J. Greenhoe > fed this fish to the penguins: >Does anyone know of a standard name for this metric in R^2? >> d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } >> where x=(x1,x2) y=(y1,y2) > >Euclidean metric? No, this is not the Euclidean metric at all. Even when we change it to the presumed one wanted, i..e., d(x,y) = sqrt(|x1-x2|) + sqrt(|y1-y2|). I don't know a name, but Rudin discusses such metrics at the end of the first chapter of his Functional Analysis text. Taken from there, here are some interesting results about R under this metric: - The only convex open sets are the empty set and R itself. - The only continuous linear map to R^n under the usual metric (more generally, to any locally convex topological vector space) is the 0 map, i.e., the map which is constantly 0. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: name of a metric Bytes: 3212 > On Sat, 07 Jul 2007 05:20:49 -0700, Daniel J. Greenhoe >> fed this fish to the penguins: > Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) >> Euclidean metric? > > No, this is not the Euclidean metric at all. Even when we change it > to the presumed one wanted, i..e., d(x,y) = sqrt(|x1-x2|) + sqrt(|y1-y2|). I don't know a name, but Rudin discusses such metrics at the end of > the first chapter of his Functional Analysis text. Taken from there, > here are some interesting results about R under this metric: - The only convex open sets are the empty set and R itself. > - The only continuous linear map to R^n under the usual metric (more > generally, to any locally convex topological vector space) is the 0 > map, i.e., the map which is constantly 0. in haste before rushing off to another commitment. First of all, I still assume Greenhoe intended to write my version. His version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. Secondly, while the (corrected) metric is not the Euclidean metric, it induces the same topology. For letting B indicate disks under the 1/2-metric and B2 disks under the Euclidean metric, B(x, 2^(-3/4) a) is a subset of B2(x, 2^(-3/2) a), which is a subset of B(x, a) for any positive a. At least this is so modulo my algebra. So all the stuff I took from Rudin, while still interesting, does not apply here. He discusses the space of all integrable functions f on [0,1] such that integral(x=0..1, sqrt(|f|)) is finite. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: name of a metric <468FAA0A.6050501@netscape.net> <4690147C.3060903@netscape.net> On Jul 7, 5:32 pm, Stephen J. Herschkorn version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. According to the original poster, d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) so x1 = x2 = 0 and y1 = y2 = 1. Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 d is a metric on R^2 because 1. d(x,y) is defined for all vectors x and y in R^2, 2. d(x,y) >= 0, and 3. if d(x,y) = 0 then x1=y1 and x2=y2 so x=y. Dave === Subject: Re: name of a metric > On Jul 7, 5:32 pm, Stephen J. Herschkorn version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. According to the original poster, > d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, > where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) > so x1 = x2 = 0 and y1 = y2 = 1. Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 d is a metric on R^2 because > 1. d(x,y) is defined for all vectors x and y in R^2, > 2. d(x,y) >= 0, and > 3. if d(x,y) = 0 then x1=y1 and x2=y2 so x=y. Dave > also d(x,y)= d(y,x) [which above is] and d(x,z) <= d(x,y) + d(y,z) === Subject: Re: name of a metric >On Jul 7, 5:32 pm, Stephen J. Herschkorn in haste before rushing off to another commitment. >>First of all, I still assume Greenhoe intended to write my version. His >>version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. > >According to the original poster, > d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, > where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) >so x1 = x2 = 0 and y1 = y2 = 1. Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 Sheesh! That's the one I meant; my corrected version was the incorrect one. I was thrown by another post: >In sci.math, Daniel J. Greenhoe > That's a strange metric, as translation of the two points along either >axis changes the value of d(x,y). > I don't see how this is so. I think both Ghost and I were confusing our x's and y's. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: name of a metric In sci.math, Stephen J. Herschkorn on Sat, 07 Jul 2007 19:56:39 -0400 <46902837.2060609@netscape.net>: >On Jul 7, 5:32 pm, Stephen J. Herschkorn version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. According to the original poster, >> d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, >> where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) >>so x1 = x2 = 0 and y1 = y2 = 1. >>Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 > > Sheesh! That's the one I meant; my corrected version was the > incorrect one. I was thrown by another post: >In sci.math, Daniel J. Greenhoe >>That's a strange metric, as translation of the two points along either >>axis changes the value of d(x,y). I don't see how this is so. I think both Ghost and I were confusing our > x's and y's. > If d(x,y) = sqrt(|x1-y1|) + sqrt(|x2-y2|) then d((0,0),(1,1)) = 0 but d((0,1),(1,2)) = 2 and d((0,4),(1,5)) = 4 Clearly this particular metric is not translation-invariant. -- #191, ewill3@earthlink.net 640K ought to be enough for anybody. - allegedly said by Bill Gates, 1981, but somebody had to make this up! -- === Subject: Re: name of a metric <468FAA0A.6050501@netscape.net> <4690147C.3060903@netscape.net> <46902837.2060609@netscape.net> <87n5m4-j52.ln1@sirius.tg00suus7038.net> On Jul 7, 7:20 pm, The Ghost In The Machine > If d(x,y) = sqrt(|x1-y1|) + sqrt(|x2-y2|) then d((0,0),(1,1)) = 0 > but d((0,1),(1,2)) = 2 > and d((0,4),(1,5)) = 4 Clearly this particular metric is not translation-invariant. But when you look at the OP's definitions of x1,x2,y1,y2, you see that d((0,0), (1,1)) = 2 because x1=0, x2=0, y1=1, y2=1 d((0,1), (1,2)) = 2 because x1=0, x2=1, y1=1, y2=2 d((0,4), (1,5)) = 2 because x1=0, x2=4, y1=1, y2=5 Dave === Subject: Re: name of a metric <468FAA0A.6050501@netscape.net> <4690147C.3060903@netscape.net> <46902837.2060609@netscape.net> <87n5m4-j52.ln1@sirius.tg00suus7038.net> [The Ghost In The Machine] > Clearly this particular metric is not translation-invariant. Tim Norfolk in his unpublished paper When Does a Metric Generate Convex Balls? (http://www.math.uakron.edu/~norfolk/) points out that if a metric d on a set X satisfies 1. d(x+z,y+z)=d(x,y) (translation invariant) 2. d(ax,ay) = ad(x,y) (homogeneous) then the balls in the space (X,d) are _convex_. It is interesting to note that balls in the metric space being discussed are _not_ convex. Dan Greenhoe (original poster) On Jul 8, 8:20 am, The Ghost In The Machine > In sci.math, Stephen J. Herschkorn > <46902837.2060...@netscape.net>: >>On Jul 7, 5:32 pm, Stephen J. Herschkorn in haste before rushing off to another commitment. First of all, I still assume Greenhoe intended to write my version. His >version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. >According to the original poster, >> d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, >> where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) >>so x1 = x2 = 0 and y1 = y2 = 1. >Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 Sheesh! That's the one I meant; my corrected version was the > incorrect one. I was thrown by another post: >>In sci.math, Daniel J. Greenhoe >That's a strange metric, as translation of the two points along either >>axis changes the value of d(x,y). I don't see how this is so. I think both Ghost and I were confusing our > x's and y's. If d(x,y) = sqrt(|x1-y1|) + sqrt(|x2-y2|) then d((0,0),(1,1)) = 0 > but d((0,1),(1,2)) = 2 > and d((0,4),(1,5)) = 4 Clearly this particular metric is not translation-invariant. -- > #191, ewi...@earthlink.net > 640K ought to be enough for anybody. > - allegedly said by Bill Gates, 1981, but somebody had to make this up! -- === Subject: Re: name of a metric >In sci.math, Stephen J. Herschkorn ><46902837.2060609@netscape.net>: >On Jul 7, 5:32 pm, Stephen J. Herschkorn in haste before rushing off to another commitment. >>First of all, I still assume Greenhoe intended to write my version. His >>version, taken literally, is not a metric, for d((0,0), (1,1)) = 0. >According to the original poster, > d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, > where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) >so x1 = x2 = 0 and y1 = y2 = 1. Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| = 2 Sheesh! That's the one I meant; my corrected version was the >>incorrect one. I was thrown by another post: >In sci.math, Daniel J. Greenhoe > That's a strange metric, as translation of the two points along either >axis changes the value of d(x,y). >I don't see how this is so. I think both Ghost and I were confusing our >>x's and y's. > >If d(x,y) = sqrt(|x1-y1|) + sqrt(|x2-y2|) then d((0,0),(1,1)) = 0 >but d((0,1),(1,2)) = 2 >and d((0,4),(1,5)) = 4 Clearly this particular metric is not translation-invariant. No, you (as many) have not been reading carefully. >According to the original poster, > d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| }, > where x=(x1,x2)=(0,0) y=(y1,y2)=(1,1) >so x1 = x2 = 0 and y1 = y2 = 1. Then d((0,0), (1,1)) = sqrt{ |0-1| } + sqrt{ |0-1| }= 2 > I also noted at one point that if (as I also thought) the proposed function were as you said, it would not be a metric, since d((0,0),(1,1)) = 0. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: name of a metric On Sat, 07 Jul 2007 10:58:18 -0400, Stephen J. Herschkorn fed this fish to the penguins: >On Sat, 07 Jul 2007 05:20:49 -0700, Daniel J. Greenhoe >> fed this fish to the penguins: >Does anyone know of a standard name for this metric in R^2? d(x,y) = sqrt{ |x1-y1| } + sqrt{ |x2-y2| } > where x=(x1,x2) y=(y1,y2) >Euclidean metric? > >No, this is not the Euclidean metric at all. Even when we change it to >the presumed one wanted, i..e., d(x,y) = sqrt(|x1-x2|) + sqrt(|y1-y2|). I don't know a name, but Rudin discusses such metrics at the end of the >first chapter of his Functional Analysis text. Taken from there, here >are some interesting results about R under this metric: - The only convex open sets are the empty set and R itself. >- The only continuous linear map to R^n under the usual metric (more >generally, to any locally convex topological vector space) is the 0 >map, i.e., the map which is constantly 0. Sorry, not paying attention. G. Rodrigues === Subject: Re: name of a metric <468FAA0A.6050501@netscape.net> Sorry, not paying attention. You are not the only one. Everyone seems to have missed the OP's unusual definitions of x1, x2, y1, and y2, and assumed the points were (x1,y1) and (x2,y2) instead of (x1,x2) and (y1,y2). Dave === Subject: Re: All solutions manuals are in pdf format or Microsoft Word format!! All solutions manuals are inpdfformat or Microsoft Word format. I have tried my best to put up with the collections of my solutions > manual. > Here is the list that only part of my collections: > I will put some time to make up the full list daily. > If you cant find the solution manual listed below, don't give up, feel >freeto email me To get the solutions manual, just email me with a price you want to > get the solutions manual you want. > To see samples, just email me. My email is edisonee(at)gmail.com Solutions Manuals .NET 2.0 Interoperability Recipes by Bruce Bukovics > A Course in Game Theory by Martin J. 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Shankar > Investment Analysis and Portfolio Management by Reilly Brown > Linear Algebra and its Applications 3rd Ed. by Lay > Linear Algebra- Hide quoted text - - Show quoted text -... read more ? Hydraulics in Civil and Environmental Engineering 4th Ed. by Andrew Chawick === Subject: Counting vertex-independent paths If I have an undirected graph G with two non-adjacent vertices x and y, is the following a valid algorithm for counting the maximum number of pairwise vertex-independent paths from x to y? Step 1. Pick at random one of the shortest paths from x to y. Step 2. Delete the all the vertices of the path chosen in step 1, except x and y, from the graph. Step 3. Repeat steps 1 and 2 until x and y are no longer connected. Step 4. Report the number of iterations of steps 1 and 2 as the max number of vertex-independent paths from x to y. It's clear that if step 1 were to pick a path that is not shortest, then step 2 could delete too many vertices, giving a number in step 4 that is too low. The question is whether a shortest path is a sufficient condition in step 1 to be sure of getting the highest possible number of iterations. My intent is to use Menger's Theorem to count the minimum number of vertices whose removal disconnects x and y. --Mike Amling === Subject: Re: Counting vertex-independent paths If I have an undirected graph G with two non-adjacent vertices x and > y, is the following a valid algorithm for counting the maximum number of > pairwise vertex-independent paths from x to y? > Step 1. Pick at random one of the shortest paths from x to y. > Step 2. Delete the all the vertices of the path chosen in step 1, > except x and y, from the graph. > Step 3. Repeat steps 1 and 2 until x and y are no longer connected. > Step 4. Report the number of iterations of steps 1 and 2 as the max > number of vertex-independent paths from x to y. It's clear that if step 1 were to pick a path that is not shortest, > then step 2 could delete too many vertices, giving a number in step 4 > that is too low. The question is whether a shortest path is a > sufficient condition in step 1 to be sure of getting the highest > possible number of iterations. > My intent is to use Menger's Theorem to count the minimum number of > vertices whose removal disconnects x and y. Shortly after posting, I found a counterexample to the proposed algorithm: | | | O-O-O---Y Two vertices must be removed to disconnect X and Y, and removal of the unique shortest path disconnects X and Y. D'oh! Now the question becomes, what's a good algorithm for finding the smallest number of vertices whose removal disconnects x and y. I just need the number, not a set of such vertices. --Mike Amling === Subject: Re: Counting vertex-independent paths Bytes: 2954 If I have an undirected graph G with two non-adjacent vertices x and > y, is the following a valid algorithm for counting the maximum number of > pairwise vertex-independent paths from x to y? > Step 1. Pick at random one of the shortest paths from x to y. > Step 2. Delete the all the vertices of the path chosen in step 1, > except x and y, from the graph. > Step 3. Repeat steps 1 and 2 until x and y are no longer connected. > Step 4. Report the number of iterations of steps 1 and 2 as the max > number of vertex-independent paths from x to y. It's clear that if step 1 were to pick a path that is not shortest, > then step 2 could delete too many vertices, giving a number in step 4 > that is too low. The question is whether a shortest path is a > sufficient condition in step 1 to be sure of getting the highest > possible number of iterations. My intent is to use Menger's Theorem to count the minimum number of > vertices whose removal disconnects x and y. Shortly after posting, I found a counterexample to the proposed > algorithm: X---O-O-O > | | | > O-O-O---Y Two vertices must be removed to disconnect X and Y, and removal of > the unique shortest path disconnects X and Y. D'oh! Now the question becomes, what's a good algorithm for finding the > smallest number of vertices whose removal disconnects x and y. I just > need the number, not a set of such vertices. You are looking for a minimum cut here. There are several algorithms for solving this problem; see http://www2.toki.or.id/book/AlgDesignManual/BOOK/BOOK4/NODE167.HTM . --- Christopher Heckman === Subject: !!!!!!!!ASPECTS OF BIO-TECHNOLOY!!!!!!!!!... !!!!!!!!ASPECTS OF BIO-TECHNOLOY!!!!!!!!!... -------- MORE DETAILS ABOUT RNA AND DNA.... DNA,RNA Related Bio Technology; Bio Technology Oriended Concepts; Biological functions; Genral Process RNA; CLICK HERE TO OBTAIN MORE DETAILS ABOUT RNA AND DNA http://kishamisher.com/ === Subject: Re: An exact 1-D summation challenge - 12 - (go and surpass all CASs) after using the integral representation of harmonic(n), interchanging summation and integration and a substition we get Pi^2 /2 - 8*Catalan. Andreas === Subject: Re: An exact 1-D summation challenge - 12 - (go and surpass all CASs) Sorry, the result I gave was for n = 0..infinity), which is Pi^2 /2 - 8 * Catalan. The one for n = 0..infinity) is Pi^2 /2 . The error happened while translating to Mathematica syntax... Andreas === Subject: Re: Mathematica's update! DA> Mathematica 6.0.1 is now available for free DA> download to all Mathematica 6.0.0 users. DA> This update features approximately 200 DA> minor improvements [... details ...] http://www.wolfram.com/mathematica DA> P.S. Get the inside story on Mathematica 6.0.1 DA> and learn more about our development cycle from DA> Software Technology Director Peter Overmann on DA> the Wolfram Blog http://blog.wolfram.com/2007/07/mathematica_601_arrives.html Bravo, Wolfram Research!! Keep her steady! ;) Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing > (copied from a MathGroup post) Mathematica 6.0.1 is now available for free download to all > Mathematica 6.0.0 users. This update features approximately 200 > minor improvements, including: * Enhanced automatic and manual proxy settings for internet > connectivity > * More extensive documentation on legacy Mathematica functions > * Significantly improved MathLink performance on Mac and Unix systems > * Improved performance of various Import and Export converters > * Faster ListPlot, ListPlot3D and Plot3D for large numbers of points > * Improved help system initialization > * Extensions to automatic file type recognition for Import > * Complete support for scaled FontSize values > * Enhancements to Table, CSV, TSV, and MathML import > * Evaluate Notebook menu item, and Magnification submenu > * Additional Mathematica function examples and tutorials > * Updates to curated data All Mathematica 6.0.0 users are encouraged to install this > update. For more information, please contact Customer Service at > i...@wolfram.com or visit:http://www.wolfram.com/mathematica P.S. Get the inside story on Mathematica 6.0.1 and learn more > about our development cycle from Software Technology Director > Peter Overmann on the Wolfram Blog > (http://blog.wolfram.com/2007/07/mathematica_601_arrives.html). === Subject: Re: Does anybody know how to evaluate this summation of alternating series? > Hi all, I am facing a series for which I would like to obtain the sum. It looks like > the following I=sum((-1)^k*a_k*b_k, k from 0 to 1000), where a_k's involve binomial coeffcients at order roughly linear with the ^^^^^^^^^^^^^^^^^^^^^^^ What does that mean? > growth of k. Hence a_k's can be extremely large, beyond the capability of > super computers; Nonsense. > however, b_k is around 10^(-10), and finally the summation I will fall into > [0, 1]. If b_k stays roughly constant, whether around 10 ^(-10) or 1 or 10^(+10), the series is divergent because the necessary condition for you have a finite sum, not an infinite series. Try to be more precise. Thus it looks like only the last 10s of digits of a_k that are really > important. And any error in computing the last a few digits of a_k*b_k will > lead to huge error and lead to wrong results. These are statements without any apparent basis in mathematics. But then, your problem has not been defined sufficiently to establish whether such statements are correct or incorrect. For example, what are the last a few digits? The last 5 out of 15? The last 200 out of 1000? > Could anybody give some suggestions on how to handle such series summations > involving large binomial coefficients... Not likely unless the questions are better stated. ??? > --- mecej4 === Subject: Rate of convergence of Chebyshev approximation For every absolutely continuous function on [-1, 1] the sequence of interpolating polynomials constructed on Chebyshev nodes converges to f(x) uniformly. What is the rate of convergence for this statement? Jamchid === Subject: Re: Rate of convergence of Chebyshev approximation For every absolutely continuous function on [-1, 1] the sequence of > interpolating polynomials constructed on Chebyshev nodes converges to > f(x) uniformly. What is the rate of convergence for this statement? > The rate of convergence strongly depends on the smoothness of f(x). === Subject: An exact 1-D summation challenge - 14 - (go and surpass all CASs) Hello computer algebra buffs, None of the modern computer algebra systems can calculate this. Is there a Summation Doc to invent the steps bringing the exact value of this sum sum((-1)^n*(2*n)!*(Psi(1/2+n)+Psi(1+n)-Psi(5/4+n)-Psi(7/4+n)) ? Best wishes, Vladimir Bondarenko Considered harmful by Richard Fateman, UC Berkley VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing === Subject: Re: An exact 1-D summation challenge - 14 - (go and surpass all CASs) Vladimir Bondarenko : > Hello computer algebra buffs, None of the modern computer algebra systems can calculate this. Is there a Summation Doc to invent the steps bringing the exact > value of this sum sum((-1)^n*(2*n)!*(Psi(1/2+n)+Psi(1+n)-Psi(5/4+n)-Psi(7/4+n)) ? Best wishes, Vladimir Bondarenko Considered harmful by Richard Fateman, UC Berkley > VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC > http://maple.bug-list.org/ Maple Bugs Encyclopaedia > http://www.CAS-testing.org/ CAS Testing You are apisteytos my friend! Dimitris === Subject: oxygen concentration in a room I realize this is an analytical chemistry question, but thought someone my help with the math. I need some help in trying to calculate the percentage of oxygen existing in a closed room. This is partly just for my interest, but there is a practical side to my question also. I live in a recently built house that is one of these super-insulated structures with little outside air exchange except what is forced via fans. During the winter months I live primarily in a couple of rooms closed off from the main house to save on heating bills. I have equipment that very accurately measures the carbon dioxide level in real time, that is located in my main living area. I live at 3000 feet elevation, and the house is all electric; no oxygen consuming heating, not even a wood buring stove. I don't have any way of directly measuring o2 in the house. The measured co2 in my living area ranges from approximately 500 ppm to over 2000 ppm. The higher concentrations generally occur during the winter months when I am spending more of my time indoors, and am preserving the heat by running the forced air intake fans less. (This will be changing in the future, as I will be forcing air from a solar heated porch into the house during some winter daylight hours) My question is, from knowing the co2 concentration (and humidity percent also) in a closed space, how can I calculate the oxygen concentration? I know that the o2 conentration of normal dry air is around 20.95 percent, but suspect that the relationship is not perfectly direct between the increase in co2 and the decrease of 02. Any help on this math problem would be appreciated. Jim === Subject: Re: oxygen concentration in a room- > I realize this is an analytical chemistry question, but thought someone my > help with the math. I need some help in trying to calculate the percentage of oxygen existing in > a closed room. This is partly just for my interest, but there is a > practical side to my question also. I live in a recently built house that is one of these super-insulated > structures with little outside air exchange except what is forced via fans. > During the winter months I live primarily in a couple of rooms closed off > from the main house to save on heating bills. I have equipment that very accurately measures the carbon dioxide level in > real time, that is located in my main living area. I live at 3000 feet > elevation, and the house is all electric; no oxygen consuming heating, not > even a wood buring stove. I don't have any way of directly measuring o2 in > the house. The measured co2 in my living area ranges from approximately 500 ppm to over > 2000 ppm. The higher concentrations generally occur during the winter > months when I am spending more of my time indoors, and am preserving the > heat by running the forced air intake fans less. (This will be changing in > the future, as I will be forcing air from a solar heated porch into the > house during some winter daylight hours) My question is, from knowing the co2 concentration (and humidity percent > also) in a closed space, how can I calculate the oxygen concentration? I > know that the o2 conentration of normal dry air is around 20.95 percent, but > suspect that the relationship is not perfectly direct between the increase > in co2 and the decrease of 02. Any help on this math problem would be appreciated. Jim > Complete oxidation of carbohydrates yields as many CO2 molecules as it takes O2 molecules. In formula: Cm H2n On + mO2 -> mCO2 + nH2O. In any gases, equal numbers of molecules correspond to equal volumes (assuming ideal gases or mixtures of ideal gases at identical temperatures and pressures). In the human organism any carbohydrate is ultimately transformed into glucose (C6 H12 O6), which yields CO2 and H2O when oxidised. Therefore, assuming that the human organism burns only carbohydrates during metabolism, one exhales the same volume of CO2 as one inhales O2. The ratio of CO2 exhalation to O2 inhalation is known as the respiratory quotient. See http://en.wikipedia.org/wiki/Respiratory_quotient . Depending upon your diet it varies from 0.7 for fat-only food to 1.0 for carbohydrates-only food. When one sits down at rest one breathes about 0.5 litres of air (so-called tidal volume) about 20 times per minute. Large individual differences exist of course, but one thing is almost constant: exhaled air that was used once contains about 14 volume percents of O2 and 6 volume percents of CO2 (assuming carbohydrates-only food) Now you can calculate what happens to CO2 and O2 concentrations in your room over the hours and what equilibrium state is approached given a specific degree of ventilation. To answer your specific question, with no ventilation one has O2 concentration = 20% minus CO2 concentration (always assuming a respiratory quotient of 1.0). For the general situation just find and solve some easy differential equations. === Subject: Re: oxygen concentration in a room Whatever the concentration, it was enough to retard George W. Bush. (And harelip everybody on Bear Creek) === Subject: size of class Bytes: 748 what is the size of the class of all sets? === Subject: Re: size of class Bytes: 1381 > what is the size of the class of all sets? http://planetmath.org/encyclopedia/Class.html R === Subject: Re: size of class > what is the size of the class of all sets? When you use the all in this context, you are on very shaky ground. === Subject: Re: size of class > what is the size of the class of all sets? How do you define the size of a class? Jose Carlos Santos === Subject: Re: PARADISE LOST: Debunking Cantor's theory Snipping again: I want to keep your Argument in this case close > together... > I don't think my distractions are irrelevant. I am offering simple logic: > 1. The vase can only become empty by balls being removed. Or by never having had any balls in. Yes. >> It can't become empty if it's already empty... > OK. We agree (I was only trying to be terminology-tolerant) > 2. The vase is never empty at any time before noon. > Correct > 3. No balls are removed at noon. > Correct (Though it might be slightly more careful to say There is no ball of > which the removal time is noon.) > 4. Therefore, at no time by noon does the vase become empty. > Follows perfectly *in real life*, which doesn't include infinite urns. > Doesn't follow in the mathematical argument which this story is > supposed to model. >> I am following it. It uses a Zeno time-machine to implement its >> absurdity. If there is time in your gedanken, then you should be able to >> tell me at which moment your vase becomes empty, and how. Yes, the vase becomes empty at noon. Because balls are moved at noon? > Well, if the time AT which something >> ... but nothing >> happened AT noon, so everything happened before noon. Right? > Well, that's the question. What does it mean to say Nothing happens > AT noon? Consider a few other examples. >> No balls are numbered such that they will be either inserted or removed >> at noon, since you have no infinitely-numbered balls. > (a) A (normal!) urn has a ball materialised inside it for all times t > where t= noon. In this example, at any time t before noon, there is a ball in the > urn, and nothing happens at this t. Does anything happen AT noon? Is there a ball in the urn at tea- > time? >> If it is out of the vase for t>=noon, and in before, then it has to have >> been removed AT noon. Something happened at that moment. OK: do you notice that this stuff about things happening AT a certain > time is a strictly weaker terminology than the mathematical statements > we can make. Consider the slight variant: (a') A (normal!) urn has a ball materialised inside it for all times t > where t <= noon, and no balls in for t > noon. In this case the ball is present AT noon, yet the time at which the > emptying occurs has to be noon. (Think about running a video > backwards: surely that could not change the time of an event?) Therefore, any argument that tries to solve the _mathematical_ > question by arguments from real life about things happening AT this or > that time is likely to be invalid. > Time is an inherent part of the gedanken, the trick that allows one to imagine finishing N. That's the source of the paradox. > (b) A (normal!) urn and a collection of balls, one for each natural > number (a pofnat with a two-ended decimal representation, that hasn't > become infinite, gone mouldy, or otherwise lost its normal integer > properties). The ball numbered n is inserted in the urn at noon - 2/n minutes and > removed at noon - 1/n minutes (physical specification), more precise > mathematical specification: it is present for t such that noon-2/n <= > t < noon-1/n Oops! Those weren't quite the intended values. Let's make it simple, > and use version (c) (c) A (normal!) urn and a collection of balls, one for each natural > number (a pofnat with a two-ended decimal representation, that hasn't > become infinite, gone mouldy, or otherwise lost its normal integer > properties). All of the balls are placed in the urn at 9:30 on the morning of the > experiment. In normal mathematical terminology this means there is an > infinite set of balls (because you can count them, but the process > never stops). The ball numbered n is removed at noon - 1/n minutes (physical > specification), more precise > mathematical specification: it is present for t such that t < noon-1/n. Now how many balls are left at tea-time? Tea time doesn't exist. Noon is a one point compactification at oo. What's outside the projectively extended real line? That's about as sensible a question. You could both tell us what your razor-sharp intuition tells you is > the Correct Answer, then you could comment on the following > application of your Argument: 1. The vase can only become empty by balls being removed. Well, the balls _are_ being removed. 2. The vase is never empty at any time before noon. Right. 3. No balls are removed at noon. > Correct > Though it might be slightly more careful to say There is no ball of > which the removal time is noon. 4. Therefore, at no time by noon does the vase become empty. Right, so some balls must be left in the vase at tea-time. It seems a rather bizarre conclusion, doesn't it? I think the entire Zeno-clock gedanken with N, while clever, is inherently paradox-prone, and invalid in surmising a distinct end to a process with no distinct end. >> I don't see the point of answering a set of questions for each >> situation. Suffice it to say that I think the use of the Zeno timer on N >> gives a false impression of having fully completed N without having >> touched infinite values. Hmm. But here the set we are talking about includes only pofnats. So > are you saying we can't somehow include all of the pofnats unless we > touch something else? Brian Chandler > http://imaginatorium.org > Can you finish an unending sequence? Really? === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2585 . >> I am following it. It uses a Zeno time-machine to implement its >> absurdity. If there is time in your gedanken, then you should be able to >> tell me at which moment your vase becomes empty, and how. Yes, the vase becomes empty at noon. Because balls are moved at noon? Because every ball inserted before noon is removed before noon. This is just another Zeno-like paradox that shows that infinite processes tend to deceive common sense approaches. === Subject: Re: PARADISE LOST: Debunking Cantor's theory >> At any finite time before noon, you are at a finite step in the process >> with a finite but growing number of balls in the vase. Your process >> completes AT noon, but nothing happens AT noon because all balls are >> inserted and removed at finite times BEFORE noon. The vase is not empty >> before noon, and nothing happens at noon, therefore it is not empty at noon. > By the same reasoning if we just add 1 ball at each time -1/2^n : > The vase does not contain an infinite number of balls before noon, and > nothing happens at noon, therefore it does not contain an infinite > number of balls at noon. > Do you agree with this? >> Well, yes, I think you've heard me say this. >> You have a countably infinite set of balls in there, but not actually >> infinite, and perhaps this is the best example of why that is so. There >> is no moment at which it can achieve actual infinity, even with Zeno's >> games. > I didn't mention countable vs actual (sic) infinity. > If we apply your earlier reasoning : > The vase does not contain a (countably) infinite number of balls > before noon, and nothing happens at noon, therefore it does not > contain a (countably) infinite number of balls at noon. > Do you agree with this? Or does this reasoning no longer seem valid? >> That would also seem to follow, in a sense. So there is a countable infinity of balls in the vase at noon, and > there is not a countable infinity of balls in the vase at noon? > It would appear that assuming noon exists in the experiment leads to a contradiction. >> To flesh it out, at every >> moment finitely before noon, the vase has a finite but growing number of >> balls. At noon, nothing is presumed to happen. One may say there is a >> finite number of balls, but no finite number will suffice as that count. >> There is some finite but unbounded set of balls. Even with the case >> where we remove ball 10n-9 at iteration n, and have balls obviously >> left, we can only say, in formulaic terms, that for n iterations, we >> have 9n balls left. Perhaps you could tell us the label on one of those balls left over at > noon? > If you inserted omega, then omega/10+1 through omega remain. > Do you agree that at noon, after all insertions and removals are > completed, there is no ball labelled '1' in the vase? > Do you agree that at noon, after all insertions and removals are > completed, there is no ball labelled '2' in the vase? > Do you agree that at noon, after all insertions and removals are > completed, there is no ball labelled '3' in the vase? >> Yes, however, when the last ball was removed and the vase became empty, >> since only one ball is removed at a time, what ball was that? n? Ball >> n+1 through 10n remain. > There was no last ball removed. There was no last iteration before > noon. There is no largest natural number. >> No kidding. Really? You learn something new every day.... > Do you agree that at noon, after all insertions and removals are > completed, there is no ball labelled with a natural number in the vase? >> I don't believe the formulation of the problem in that way is at all >> helpful. At every finite time before noon, more iterations are to occur, >> but at noon no iterations occur. So, there is no time all insertions >> and removals are completed. > At noon, not all insertions and removals are completed? Which ones > remain? > How about one minute past noon? >> At noon, everything that was going to happen has happened, ... > So you agree that at noon all actions have been completed. I will ask > my question again : > Do you agree that at noon, after all actions have been completed, > there is no ball labelled with a natural number in the vase? > Maybe this time you will be courteous enough not to dodge the question > with an irrelevent distraction. >> I don't think my distractions are irrelevant. I am offering simple logic: >> 1. The vase can only become empty by balls being removed. >> 2. The vase is never empty at any time before noon. >> 3. No balls are removed at noon. >> 4. Therefore, at no time by noon does the vase become empty. >> Please indicate which stage of that logic is incorrect. >> Tony 1, 2, 3 and 4 do not imply 5. The vase is not empty at noon. Also, please define the word become in a mathematical sense. > x becomes y at z means for ty and for t=z x=y. empty means 0 balls the vase becomes empty at noon means 11:59 balls>0 and t=noon -> balls=0. >> ... but nothing >> happened AT noon, so everything happened before noon. Right? Before >> noon, at every measurable point not yet up to noon, there remained balls >> in the vase, a growing number, faster and faster. Then AT noon, nothing >> happened. How many balls remain? >> What happens after noon is undefined until you specify how your real >> line wraps around... > No. The behavior of each and every ball involved in the problem is > precisely defined. It is inserted at some time before noon and removed > at some later time that is also before noon. Therefore, no ball is in > the vase at noon. Therefore, there are no balls in the vase at noon, > or at any time after noon. No response? I've heard that line of logic, but it doesn't make sense. The problem seems to derive from trying to pretend you have completed N and not gone beyond the finite, when there is no such end to the set. Here is how I solved this problem for myself, some two years ago. Yes, > it was originally counter-intuitive. Let 0 be noon and -1 be 1 minute before noon. Let A(n,t) be 1 if ball n is in the vase at time t, 0 if the ball n is > not in the vase at time t. Let S(t) be the number of balls in the vase at time t. S(t) = The sum over N of A(n,t). Let B(n) be the time the nth ball is added to the vase = -(1/ > (2^(floor((n-1)/10))). Let C(n) be the time the nth ball is removed from the vase = -(1/ > (2^(n-1))). Note that B(n) and C(n) are always < 0 for natural n. Now A(n,t) = 1 if B(n) <= t < C(n), 0 otherwise. The value of A(n,0) is 0 for any n as C(n) is < 0. S(0) = The sum of A(n,0) over N = sum of 0 over N = 0. Which step of this is incorrect or an innacurate modelling of the > problem as originally described? > The concept that you can actually reach noon in this experiment without inserting infinitely-numbered balls, really. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 4827 > The vase does not contain a (countably) infinite number of balls > before noon, and nothing happens at noon, therefore it does not > contain a (countably) infinite number of balls at noon. > Do you agree with this? Or does this reasoning no longer seem valid? >> That would also seem to follow, in a sense. So there is a countable infinity of balls in the vase at noon, and > there is not a countable infinity of balls in the vase at noon? > It would appear that assuming noon exists in the experiment leads to a > contradiction. Assuming that TO understands the problem is what leads to a contradiction. > To flesh it out, at every >> moment finitely before noon, the vase has a finite but growing number of >> balls. At noon, nothing is presumed to happen. One may say there is a >> finite number of balls, but no finite number will suffice as that count. >> There is some finite but unbounded set of balls. Even with the case >> where we remove ball 10n-9 at iteration n, and have balls obviously >> left, we can only say, in formulaic terms, that for n iterations, we >> have 9n balls left. Perhaps you could tell us the label on one of those balls left over at > noon? > If you inserted omega, then omega/10+1 through omega remain. But nobody inserted omega. x becomes y at z means for ty and for t=z x=y. empty means 0 balls the vase becomes empty at noon means 11:59 balls>0 and > t=noon -> balls=0. That's it! > ... but nothing >> happened AT noon, so everything happened before noon. Right? Before >> noon, at every measurable point not yet up to noon, there remained balls >> in the vase, a growing number, faster and faster. Then AT noon, nothing >> happened. How many balls remain? >> What happens after noon is undefined until you specify how your real >> line wraps around... > No. The behavior of each and every ball involved in the problem is > precisely defined. It is inserted at some time before noon and removed > at some later time that is also before noon. Therefore, no ball is in > the vase at noon. Therefore, there are no balls in the vase at noon, > or at any time after noon. No response? I've heard that line of logic, but it doesn't make sense. Any other analysis makes less sense. > The problem > seems to derive from trying to pretend you have completed N and not gone > beyond the finite, when there is no such end to the set. Just take it one ball at a time. === Subject: Re: PARADISE LOST: Debunking Cantor's theory >> Until you count to omega, you have not counted omega naturals. NeN > Which version of set theory is TO trying to sell in which any set is > allowed to a member of itself. >> sum(n=1->x: 1)=x. > How does than justify NeN? >> Okay, excuse me. I was kinda quoting Ross in that. I suppose it's more >> properly |N|eN. :) Since by definition, for all n in N, n+1 in N; it follows that |N|+1 > in N; and so on. So N contains strictly more elements than |N| by > your unusual measure of more? Yes, trying to pin a size on a set with no bound gets tricky, doesn't it? If N has a size, then its size is its largest element by obvious induction. It has no largest element. Therefore it has no size. Aleph_0 is a phantom, a number only in the sense that that is what it does to one's mind, numbs it. Toodles - Tony === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 3124 > Yes, trying to pin a size on a set with no bound gets tricky, doesn't > it? Depends on one's definition of size. Cantor's cardinalities do not have that problem, at least given AoC. if Cardinality s size then for any sets A and B, Card(A) <= Card(B) if and only if there is an injection from A to B. Equality then requires injections both ways. I.e., a bijection. And Card(A) < Card(B) requires no injection from B to A. > If N has a size, then its size is its largest element by obvious > induction. Not by any mathematically valid form of induction. > It has no largest element. Therefore it has no size. That a set be ordered is irrelevant to whether it has a size using the Cantor definition of cardinality for size, and unordered sets never have last elements, even when finite. > Aleph_0 is a phantom, a number only in the sense that that is what it > does to one's mind, numbs it. It only numbs minds inherently prone to numbness, like TO's. It stimulates other minds, like those who has a taste for honest mathematics. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > On Thu, 28 Jun 2007 21:14:01 -0400, Tony Orlow from what mathematicians understand by trees, he should call them > something else. Possibly TOrees. > > Sure. Given the set of edges, which is a relation between the set of >> nodes and itself, specifying an edge is the same as specifying a node. >> The data structure is the set of nodes, and the set of child relationships. > This can even be formalized as: > Let E and V be sets, where E contains a special element, e0. >> That would be the root edgenode, identifiable, presumably, by some n in >> N, corresponding to an e_n in E? >> What is V? V is a an arbitrary set; just as E is an arbitrary, non-empty set. The > elements of V and E can be anything you like; if you like you can call > the elements of V nodes and the elements of E edges. I am > specifying a certain relationship between the elements of the set V > and the elements of the set E. If two sets meet these and the > following constraints then I call the resulting structure a tree. > Let > produces: E -> V be a function, and producedBy: E{e0} -> V be a > function. >> Each produces 2, so produces is not one function. produceby is a >> function. See why I refer to this as a guessing game? I am forced to guess what > you mean formally from your verbal descriptions. Feel free to supply a formal description of what you mean. > Here, as below, I am using 'E' as the existential quantifier E e_0 E e_n -> E e_2n+1 E e_n -> E e_2n+2 With the normal definitions of arithmetic on naturals, this describes the tree. > Suppose we have the following situation: > (i) for all e in E{e0}, not(produces(e) = producedBy(e)) > (ii) produces is a bijection. > However, if we only specify (E, V, produces) as you did; and also > assert that produces() is a bijection to some set V of nodes, then > we don't have a model of a tree: instead, we simply have two > equinumerous sets. That was certainly the first reason why define > edges as that which produces a node gets you mocked. We need a lot / > more/ than that. >> All I asserted was that the addition of a single root edge would make >> it possible to combine the notion of node and edge as the geometrical >> unit of the tree. This is not all that different than the usual approach: there are > nodes (some set, /not/ neccessarily a set of /naturals/), and then > there is a data structure which represents relationships between the > elements we choose to call nodes. In the usual definition, this > relationship is specified by ordered pairs of nodes. That's fine. The fact that we tend to call that set of pairs of nodes the set of > edges is only for convenience; we can simply continue to call that > the the set of relationships between nodes, or even an accompanying > set of ordered pairs of from the set of nodes. Thus, we have only > one geometrical unit in the usual definition of the tree (and graphs > in general): the node. Okay, and edges are equinumerous with nodes. > Of course, the parent of the edgenode needs to be part >> of the edgenode, to fully describe the tree, informationally. What does part of mean? > Included within the tree unit definition. >> So, one >> can think of the node-edge-node as the basic descriptive unit, and in >> that context, the edge is superfluous and it becomes a node pair. That's >> indicated in the tree diagram as a geometrical connection, at a >> particular branching point, either right or left. So you then require /two/ sets: E (which I will call the set of edge- > nodes), and a set of relationships between edges-nodes. I would > formalize this second set as a set of ordered pairs (e_1, e_2) of > edges-nodes, where e_1 is a parent edge-node to e_2; but perhaps you > have a different idea in mind. No, that's fine. The whole point was that edges are equal to nodes, to within a difference of 1, anyway. What do you call that accompanying set? I provisionally suggest > calling that set: pseudo-edges. I would further suggest calling the > set of edge-nodes: pseudo-nodes. The reason for my suggestions may or > may not become clear eventually. > Or we can just drop that sidebar. >> I supposed it depends what you're doing, but I really don't see anything >> remotely wrong with considering the edge-node pair as a unit of the tree. Does it capture the same structure? We need to know how edge-node > relationships are to be described first. > They are described as a logical implication of existence, I suppose, the parent node implying the existence of the child nodes. As above: E e_0 E e_n -> E e_2n+1 E e_n -> E e_2n+2 > For example, here is a node: I will call it O = {n}, where n is some > natural number, say for argument, n = 1234567. > What is the parent node P = {m} of O? What natural number is m? How > can I deduce this from /only/ the data structure {{0}, {1}, {2}, ...}? >> Uh, the parent node is int(n/2), [for n>1]. >> No comment? Really no need. It's not like I don't understand how to guess the > relationship between nodes in some particular tree. Instead, my comments as usual are about /how/ you go about describing > your ideas, from a mathematical standpoint. For example, my exact > question was: How can I deduce this from /only/ the data structure {{0}, {1}, > {2}, ...}? Your answer was Uh, the parent node is int(n/2), for n>1. My comment would be: I don't think that is a reasonable answer to my > {2}, ...}. > I don't think I ever asserted that a set of nodes fully described a tree, so maybe that question was a little like asking when I stopped beating my wife. I told you how you can identify the parent of each numbered edge in the scheme we're discussing. >> See? > Not really. If you really want information reduction, then simply > identify the entire tree by the set {0}. How is that different from > what you are claiming? >> For one thing, I don't know what you're claiming. > I am claiming that we can identify the infinite binary tree by the set > {0}. This saves a tremendous amount of redundancy; since there is > really only one thing we mean by /the/ infinite binary tree, up to > re-labelling of nodes, This even keeps the set finite! >> I suppose you could, but then, what do you label the tree containing >> only the root node? Keeping in mind that this is a hyperbolic version of how your > description comes across, I might label it {1} if I felt so > inclined. But really, the whole /the/ infinite tree aspect of my argument > isn't key; it's just another method to try to get you t see what you > aren't being explicit about. The punchline is that /the/ infinite > binary tree is /the/ equivalence class of /all/ sets meeting a certain > definition, not some /single/ set which meets this definition. But this line of argument seems to be causing more confusion than > clarity, so I will drop it. > Sigh. Yeah, there is more information required besides simply the fact >> that there is one edge for every node and vice versa, in direct >> geometrical contact. One needs to know the parent of every node, except >> the root, if it exists (it need not). I said that every edge could be >> identified with a single node, and did >> not require that two nodes be specified, for each EDGE. Yes, that's /exactly/ my point. tweaking the definition of edge in this way /requires/ a /new/ > specification for the relationships between these new edge-node > entities. Since you hadn't previously described how that relationship > is to be represented, we needed more before your tweak resulted in > anything which we could agree on as being the same thing as the > infinite binary tree. It was an addition of one single edge, just to make edges and nodes the same. That's all. It's not important. That information is usually contained in the /definition/ of an edge > as an ordered pair of nodes. Change the definition of edge, and > you have to come up with a new way of describing those relationships. > I did not >> pretend, as you seem to claim, that this constituted a full description >> of the tree. I'll thank you to call the scarecrows off. You're making my >> hippopotamus edgy. Well, it's usual in this forum that if you say X can be done, then > people will ask you to prove it. That's what can be done means in > this context; and the onus is on the claimer that what you end up with > still has the same structure that you are trying to describe. Your off- > hand comment failed to do this. Okay, we cannot possibly combine a node and an edge into a single entity, because that's just not how it's done. Okay. I got it. Sorry about your hungry, hungry hippo. Here's some delicious sweet > hay. > Yummy. He's much calmer now, but I think it's giving him the farts. >> If I ask > you, but given a node, what is its parent node? you will say, uh, > the parent node is (some particular, previously unspecified function > of the node). The /set/ was already supposed to completely represent > the tree. >> What are you blathering on about? I gave examples below where we have nodes and ordered pairs of nodes: > e.g., {1,2,3} and {(1,2), (1,3)}. Suppose we have three edges-nodes {1,2,3}. What is the other set > that makes these edges into a tree, and not just a set of 3 numbers? > Until that is specified, it is just 3 numbers and an idea. > The fact there exist parent child relationships between n and 2n+1 and between n and 2n+2. Of course, given this definition of the tree, most subsets of N will give us disconnected subtrees rather than a single one. {1,2,3} looks like: 1 2 / 3 >> If each edge is simply a member of one node, > But it isn't. :-) >> Says you, because says them. >> it only requires one of that countably infinite set of numbers >> to specify the edge. > That's fascinating. And how would you determine the OTHER node, the > one the edge connects to your given node? >> Duh! It's the parent of the node you specify. Maybe you're the one that >> needs to check the medication. I mean, how obvious is that? > Yes, obviously the parent node of O is the parent node of O. Also, > obviously the kangaroo node of O is the kangaroo node of O; and the > avocado pit of node O is the avocado pit of node O. >> Oh shaddup. And shaddup shuttin up. > Hyuk-yuk. Glad you get the joke. But it's funny, 'cause it's true. >> You probably want to ask Lester about truth. :) Talking with Lester is boring. All he /really/ wants is to provoke a > fist fight. I usually drink at a different bar. I was joking anyway But that information is /not/ contained in your information > reduction plan via modelling the tree as a set of singleon naturals > {{0}, {1}, {2}, ..., etc.}. >> It's contained in the fact that for each node/edge pair there is one >> ordered pair of child edge/node pairs at the next level. > I don't see that information in the set {{0}, {1}, {2}, ...}. Could > you point it out to me? >> It's a {0,1}'s worth of information, not N's worth. Perhaps so. But the question I asked was: could you point out to me, > in the set of node-edge pairs {{0}, {1}, {2}, ...}, how I can deduce that for each node-edge-pair, there is one ordered > pair of child edge-node-pairs at the next level? > Using the formulas above. > It /is/ contained in the structure {{0,1}, {0,2}, {1,3}, {1,4}, {2,5}, > {2,6}, ..., etc.}; which is why the latter is preferred to the former. >> That's overkill. > . As you like. > If you must /define/ every property explicitly in this way, you > haven't reduced the information inherent in the representation; you've > simply moved it from one place (the data structure) to another (a > series of additional properties claimed as axioms). >> Sorry, but I don't see one finite rule as being an equivalent amount of >> information as an infinite set of natural numbers. > It's not a question of finite or infinite. > Here is a set of edges over the set of nodes {3, 13, 27}: > {(13,3), (13,27)} > You can see it satisfies the definition of being a tree. > Here is a set of TO-edges over the set of nodes {3, 13, 27}: > {{3}, {13}, {27}} > Does this satisfy the definition of being a tree? I can't tell. >> It requires the parent function only, a sparse one, to be sure. >> Yesno? Yes, I absolutely agree; just as in the usual definition, we /require/ > a set of relationships (it's not formally speaking a /function/: not > every element has a parent). Please, feel free to show that set in your system. > S = {{n,p}: neN ^ (p=2n+1 v p=2n+2)} That's your set of ordered pairs. > Similarly, I can /deduce/ many other properties from the usual data > structure as /theorems/: there are no loops in the graph, the graph is > totally connected, and so on. In your system, these properties / > cannot/ be deduced from {{0}, {1}, ...}; they are instead /additional/ > axioms that you must supply as information, as part of your supposed > information reduction plan. >> Incorrect. If the parent is always a smaller number than the child, then >> it follows that there are no loops. > Here is a set of edges over the set of nodes {3, 13, 27, 53}: > {(3,13), (13,53), (3,27), (27, 53)} > You can see that every parent is a smaller number than the child, and > there are no loops in this set of (directed) edges and nodes. > Here is a set of TO-edges over the set of nodes {3, 13, 27, 53}: > {{3}, {13}, {27}, {53}} > Is every parent a smaller number than the child? Does this set of TO- > edges and nodes contain a loop? I can't tell. > It /is/ true, though, in both cases that TO-edges are equinumerous to > nodes, so that is one advantage. > Does these examples make my critique any clearer? > You've made it quite clear. You think I said that everything was simply >> a set of edges/nodes in a tree and nothing more, when what I said was >> that a tiny tweak to the definition of an edge to make a root makes an >> edge one and the same unit as a node. And it was pointed out at that time that changing the definition of > edges based entirely on nodes and defined as an ordered pair of > nodes, to a definition of edges as the same thing as nodes, removes a > piece of information about nodes which form the graph we call a tree: > the relation a is a parent node to b, which is the /set/ we are > referring to when we talk about edges in the usual definition. So it's not a tiny tweak; it's more like a total wreck :). Except that parent-child relationship can be captured arithmetically to describe the edges. > I never said that described the >> tree in its entirety. I agree that an extra bit needs to be specified in >> addition to the node number, to denote which child edge is being >> chosen, but an additional natural number is not necessary. Is it? An additional natural number was /never/ neccessary in the original > scheme. Instead, we have nodes = naturals. And then we have some set > of ordered pairs of naturals. There is no particular /requirement/ > that we call that set a set of edges, but we usually do call them > edges for convenience. To be generous, you are attempting to replace this system with edge- > nodes = naturals. But then you need EN-relations = ordered pairs of > edge-nodes. For example, say edge-node 1 has two edge-node children, 2 > and 3; that would be the set of ordered pairs {(1, 2), (1, 3)}, which > you could say in words as: the parent of edge-node 2 is edge-node 1 > and the parent of edge-node 3 is edge-node 1. WIthout that > additional set, all you have is a set of edge-nodes. The resulting ordered pair of the sets of edge-nodes and the set of > edge-node-relationships ({1, 2, 3}, {(1, 2), (1, 3)}) looks mighty familiar. Note that it has three edge-nodes, and two > edge-node relationships. That's why your claim that you have > eliminated redundancy is ridiculous when one knows the /actual/ > usual definition of nodes and edges. > My claim had more to do with trying to formulate the relations between objects in the tree. The real point of all of this is that every pair of edges produces a branching in the tree, which is a path, so that there are only half as many paths as nodes+1. >> With the simple rule that node n has 2n+1 and 2n+2 as children, and the >> usual rules of arithmetic, a countably infinite number of naturals no >> longer require specification. > I don't really get your point here. Surely, the natural numbers don't > require specification; they are already specifed by the usual notion > of the naturals. > In any case, would you accept E = {(n,m) : n, m in N and m = ceiling(n/ > 2)} as a valid specification of a set of ordered pairs of nodes, said > nodes being labelled with the naturals, i.e., V = N? Would (E,V) still > be overkill as a way of representing an infinite binary tree? >> No, that would be okay, and works, because that allows for exactly two >> children for every node, each unique within all nodes. That sounds >> reasonable, and equivalent to what I suggested above. It /might/ be equivalent, if you specified that set of edge-node > relationships as a /part/ (along with a set of edge-nodes) of your / > definition/ of a tree. You still haven't done that; although I have > suggested various approaches above. Assuming for the moment that we mean the usual definition of > uncountable, it is hard to see how your argument can be > distinguished from: The binary representation of the paths specifies > any structure having an uncountable number of elements. >> I think that can be true... Until a structural relationship between elements is actually > specified, it's a matter of guessing, and not one of knowing. Sure, /if/ I know the structure, I can certainly map the elements in a > way that preserves the structure; but specifying those mapping which > preserve the structure is /exactly/ the part of the problem which / > needs/ to be specified. >> Well, Chas, that why I was saying, oh so long ago, that bijections >> between subsets of reals that have one or the other out of quantitative >> order are not to be trusted. I think that < is related to e in ways >> that haven't been properly explored yet, for whatever reasons. Trust is developed by learning /exactly/ what is meant mathematically > by the symbols x < y in various different /contexts/. The fact that > you often don't understand the exact meanings in these contexts > doesn't mean that there aren't simple, exact, mathematical meanings > available. There are many such equivalent sets of pairs. > What makes them equivalent? They satisfy certain definitions, > independently of any re-labellings: > (i) for all edges e = (e1, e2) in E, not (e1=e2) > (ii) for all edges e = (e1, e2) and f = (f1, f2) in E, if e2 = f2, > then e1 = f1 > (iii) exists a unique node v0 in V such that for all edges e = > (e1, e2) in E, not (e1 = v0) > (iv) for all edges e = (e1, e2) in E, the set of edges {f = (f1, > f2) : f1 = e2} has exactly two elements > (v) for all nodes v in V, exists an edge e = (e1, e2) in E such > that v = e1. > (i): (E,V) is a simple graph. (i), (ii): (E,V) is a forest; (i), (ii), > (iii): (E,V) is a tree; (i), (ii), (iii), (iv): (E,V) is a binary > tree; (i), (ii), (iii), (iv), (v): (E,V) is an infinite binary tree. > Individual elements of E and V may have certain other relationships > beyond those noted above. So what? >> E . >> E x -> E x0 >> E x -> E x1 Absolutely no idea what you're trying to say here as a response to the > above. Is E supposed to be the set E in my paragraph? Existential quantifier. I am producing string representations of paths here. > It boils down to the rule I stated, which eliminates the need to >> enumerate the pairs explicitly. The pairs are of the form {n,m) where >> neN and m=2n+1 or m=2n+2. > It's certainly one such numbering one could use. There are many other > numberings that seem to yield pretty much the same sort of thing. By > specifying the set of edges on an arbitrary set of nodes, there's no > need to keep track of such boring details: one need only /show/ that > some pair of sets S = (E,V) meets the definition of a tree. Then you > have said: S is a tree, in the following way: .... >> Sure. {n,b} where n e N and b e {0,1}. Sigh. After all of the above, why do you imagine that {n,b} where n e > N and b e {0,1} is sufficient to /show/ that something (what? the > set {0,2}?) is a tree, in the following way...? Never mind. Tony === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46808d82@news2.lightlink.com> <572dna3hDIwLAh3bnZ2dnUVZ_u-unZ2d@comcast.com> <46812bda@news2.lightlink.com> <4681496d@news2.lightlink.com> <4683fa9c@news2.lightlink.com> <46845cdd@news2.lightlink.com> <46853961@news2.lightlink.com> <46879459@news2.lightlink.com> <468c0341@news2.lightlink.com> <468fd687@news2.lightlink.com On Thu, 28 Jun 2007 21:14:01 -0400, Tony Orlow from what mathematicians understand by trees, he should call them > something else. Possibly TOrees. > > Sure. Given the set of edges, which is a relation between the set of >> nodes and itself, specifying an edge is the same as specifying a node. >> The data structure is the set of nodes, and the set of child relationships. > This can even be formalized as: > Let E and V be sets, where E contains a special element, e0. >> That would be the root edgenode, identifiable, presumably, by some n in >> N, corresponding to an e_n in E? >> What is V? V is a an arbitrary set; just as E is an arbitrary, non-empty set. The > elements of V and E can be anything you like; if you like you can call > the elements of V nodes and the elements of E edges. I am > specifying a certain relationship between the elements of the set V > and the elements of the set E. If two sets meet these and the > following constraints then I call the resulting structure a tree. > Let produces: E -> V be a function, and producedBy: E{e0} -> V be a > function. >> Each produces 2, so produces is not one function. produceby is a >> function. See why I refer to this as a guessing game? I am forced to guess what > you mean formally from your verbal descriptions. Feel free to supply a formal description of what you mean. Here, as below, I am using 'E' as the existential quantifier E e_0 > E e_n -> E e_2n+1 > E e_n -> E e_2n+2 With the normal definitions of arithmetic on naturals, this describes > the tree. > Not to me. As far as I can tell, it is is an overwrought way of saying the equivalent for each natural m, there exists an entity named 'e_m'. I.e., it is logically equivalent to: E e_0 E e_n -> E e_(n+1) You appear to /want/ to define something like what is called an L- system (or Lindenmayer-system); but this again requires a lot more than what you have stated above. I think you would find L-systems /very/ interesting; check out the beautiful book The Algorithmic Beauty of Plants. It is available on- line at http://algorithmicbotany.org/papers/ Scroll to the bottom of the page for the book in PDF format. It will change the way you look at even the ugliest weeds growing by the side of the road. But anyway, even an L-system based on rules you seem to allude to above does not correspond to the infinite binary tree. It's the usual problem: you cannot acheive an infinite set simply by adding 1 element again and again without further axioms or rules. > The fact that we tend to call that set of pairs of nodes the set of > edges is only for convenience; we can simply continue to call that > the the set of relationships between nodes, or even an accompanying > set of ordered pairs of from the set of nodes. Thus, we have only > one geometrical unit in the usual definition of the tree (and graphs > in general): the node. Okay, and edges are equinumerous with nodes. > No, in the usual system, nodes and edges (node-relationships) are / not/ equinumerous. Consider the tree ({a,b,c}, {(a,b), (a,c)}). >> Of course, the parent of the edgenode needs to be part >> of the edgenode, to fully describe the tree, informationally. What does part of mean? Included within the tree unit definition. ... the parent of the edgenode needs to be included within the tree unit definiton of the edgenode...? > So, one >> can think of the node-edge-node as the basic descriptive unit, and in >> that context, the edge is superfluous and it becomes a node pair. That's >> indicated in the tree diagram as a geometrical connection, at a >> particular branching point, either right or left. So you then require /two/ sets: E (which I will call the set of edge- > nodes), and a set of relationships between edges-nodes. I would > formalize this second set as a set of ordered pairs (e_1, e_2) of > edges-nodes, where e_1 is a parent edge-node to e_2; but perhaps you > have a different idea in mind. No, that's fine. The whole point was that edges are equal to nodes, to > within a difference of 1, anyway. > And the whole point of the response is that that makes absolutley no sense in the usual definitions of node, edge, tree. In any case, nodes cannot be defined to be equal to ordered pairs of nodes in a definition which is not circular. > What do you call that accompanying set? I provisionally suggest > calling that set: pseudo-edges. I would further suggest calling the > set of edge-nodes: pseudo-nodes. The reason for my suggestions may or > may not become clear eventually. Or we can just drop that sidebar. > I supposed it depends what you're doing, but I really don't see anything >> remotely wrong with considering the edge-node pair as a unit of the tree. Does it capture the same structure? We need to know how edge-node > relationships are to be described first. They are described as a logical implication of existence, I suppose, the > parent node implying the existence of the child nodes. As above: E e_0 > E e_n -> E e_2n+1 > E e_n -> E e_2n+2 > Once something exists, it simply exists. The method of the proof of its existence does not, by itself, add additonal structure to that object. The above is equivalent to saying: Exists an element e_0. If an element e_n exists for some natural n, then the element e_(n+1) exists. {2}, ...}? Your answer was Uh, the parent node is int(n/2), for n>1. My comment would be: I don't think that is a reasonable answer to my > {2}, ...}. I don't think I ever asserted that a set of nodes fully described a > tree, so maybe that question was a little like asking when I stopped > beating my wife. I told you how you can identify the parent of each > numbered edge in the scheme we're discussing. > You claimed something about edges in a tree, if we redefined edges in a certain way. When it was pointed out to you that the result no longer defined a tree, you complained that it /could/. How? Sigh. Yeah, there is more information required besides simply the fact >> that there is one edge for every node and vice versa, in direct >> geometrical contact. One needs to know the parent of every node, except >> the root, if it exists (it need not). I said that every edge could be >> identified with a single node, and did >> not require that two nodes be specified, for each EDGE. Yes, that's /exactly/ my point. tweaking the definition of edge in this way /requires/ a /new/ > specification for the relationships between these new edge-node > entities. Since you hadn't previously described how that relationship > is to be represented, we needed more before your tweak resulted in > anything which we could agree on as being the same thing as the > infinite binary tree. It was an addition of one single edge, just to make edges and nodes the > same. That's all. It's not important. > Of course, none of this is important in itself, since statements you make such as the above communicate no mathematical sense. Do you want to learn why? > That information is usually contained in the /definition/ of an edge > as an ordered pair of nodes. Change the definition of edge, and > you have to come up with a new way of describing those relationships. > I did not >> pretend, as you seem to claim, that this constituted a full description >> of the tree. I'll thank you to call the scarecrows off. You're making my >> hippopotamus edgy. Well, it's usual in this forum that if you say X can be done, then > people will ask you to prove it. That's what can be done means in > this context; and the onus is on the claimer that what you end up with > still has the same structure that you are trying to describe. Your off- > hand comment failed to do this. Okay, we cannot possibly combine a node and an edge into a single > entity, because that's just not how it's done. Okay. I got it. > No, you can easily do so. But what you end up with is exactly the same definition: a set of elements, and a set of ordered pairs of elements, meeting certain rules. Whether you call the elements in question nodes or edges or edge-nodes or compact subspaces of a Hausdorff space makes no difference whatsoever; just as whether you call the set of ordered pairs of nodes edges or edge-node cuplings or the subset relation makes no difference whatsoever. Get it? The confusion arises when you call the set of elements edges, and the set of ordered pairs of elements edges. Then you have mathematical gibberish. > Sorry about your hungry, hungry hippo. Here's some delicious sweet > hay. Yummy. He's much calmer now, but I think it's giving him the farts. > Stanky! >> If I ask you, but given a node, what is its parent node? you will say, uh, > the parent node is (some particular, previously unspecified function > of the node). The /set/ was already supposed to completely represent > the tree. >> What are you blathering on about? I gave examples below where we have nodes and ordered pairs of nodes: > e.g., {1,2,3} and {(1,2), (1,3)}. Suppose we have three edges-nodes {1,2,3}. What is the other set > that makes these edges into a tree, and not just a set of 3 numbers? > Until that is specified, it is just 3 numbers and an idea. The fact there exist parent child relationships between n and 2n+1 and > between n and 2n+2. So you claim there is a set {(n,m) : (m = 2+1) or (m = 2n+2)}. That set is usually called the set of edges. Of course, given this definition of the tree, most > subsets of N will give us disconnected subtrees rather than a single > one. {1,2,3} looks like: 1 2 > / > 3 I'm sure it looks like that /to you/. Mathematics is a way of formalizing what looks like means, so we can agree that we actually see the same thing. > Perhaps so. But the question I asked was: could you point out to me, > in the set of node-edge pairs {{0}, {1}, {2}, ...}, how I can deduce that for each node-edge-pair, there is one ordered > pair of child edge-node-pairs at the next level? Using the formulas above. > So your system is identical to the usual definition; You simply choose to call the set normally called nodes as edge-nodes, and you have no particular name for the set usually called the set of edges. That's why I called your sets pseduo-nodes and pseudo-edges. /Names/ are unimportant. /Definitions/ are important. If you call what people usually call a car a tire, and then justify it by saying that way, a car only has one tire; and it still rolls on four thingies, expect some flack. > Please, feel free to show that set in your system. S = {{n,p}: neN ^ (p=2n+1 v p=2n+2)} That's your set of ordered pairs. > Excellent! Minor terminology nitpick: {n,p} = {p,n}; sets have no order. If p = n, then {n,p} = {p,n} = {n} = {p}. But (n,p) is not equal to (p,n) unless n = p; and (p,p) is not equal to (p). In an ordered pair, order is important. The convention is to use {} for sets, and () for ordered pairs (or triples, and so on). And that meets the definition of the set usually called edges in what is usually called a tree. Congratulations! The ordered pairs of sets (N, S) meets the usual definition of a tree. Your definition is identical to the usual, except you call different things different names. But in this definition, for finite trees at least, it is not the case that the set of elements is the same as the set of element-element relationship pairs. In a finite tree at least, there are still more nodes/edges/edges-nodes than there are edges/ (unamed entity)/edge-node relationships; so your original motivation was without merit. You've made it quite clear. You think I said that everything was simply >> a set of edges/nodes in a tree and nothing more, when what I said was >> that a tiny tweak to the definition of an edge to make a root makes an >> edge one and the same unit as a node. And it was pointed out at that time that changing the definition of > edges based entirely on nodes and defined as an ordered pair of > nodes, to a definition of edges as the same thing as nodes, removes a > piece of information about nodes which form the graph we call a tree: > the relation a is a parent node to b, which is the /set/ we are > referring to when we talk about edges in the usual definition. So it's not a tiny tweak; it's more like a total wreck :). Except that parent-child relationship can be captured arithmetically to > describe the edges. > It cetainly /can/ be; as you have done above. But it's never the case that we can simply equate the former set with the set of relationships (e.g., equate nodes with edges, so they have the same number). That's where your statements became incoherent. edge-node-relationships ({1, 2, 3}, {(1, 2), (1, 3)}) looks mighty familiar. Note that it has three edge-nodes, and two > edge-node relationships. That's why your claim that you have > eliminated redundancy is ridiculous when one knows the /actual/ > usual definition of nodes and edges. My claim had more to do with trying to formulate the relations between > objects in the tree. But that's /already been done/, ya maroon! We assume you know the definitions in play for edges, nodes, and so on. You simply showed that you didn't know what those definitions were; and compounded this lack of knowledge by proposing some poorly thought out substitutes. > The real point of all of this is that every pair of > edges produces a branching in the tree, which is a path, so that there > are only half as many paths as nodes+1. > Since you didn't know the usual formal definitions of node and edge, why should I now imagine that you know the formal definition of path? beyond those noted above. So what? >> E . >> E x -> E x0 >> E x -> E x1 Absolutely no idea what you're trying to say here as a response to the > above. Is E supposed to be the set E in my paragraph? Existential quantifier. I am producing string representations of paths here. > To reduce confusion, I would suggest that you don't use the mathematical shorthand until you learn to use it properly. If you want to communicate the above, simply say What about these as rules? There exists a string . If there exists a string x, then there exists the string x + 0 If there exists a string x, then there exists the string x + 1 It doesn't become more mathematical by replacing there exists a string with E; and it's just confusing unless the context is clear. Firstly, as noted above, existence is not sufficient to determine structure. It simply determines existence. You have the nodes, you do not have the edges until you define them /explicitly/. For example, the above rules are equivalent to: The empty string exists. The string 0 exists The string 1 exists If strings x and y exist, then the string x + y exists. A string exists in your system if, and only if, there exists a corresponding string in the latter system prepended with the character .. The resulting set of strings arise in many situations and is sometimes called the set of strings freely generated by {0, 1}. They are not trees; they are just sets of strings. Any other structure on or relationship between these strings must be /explicitly stated/. Secondly, even if you were using, say, an L-system to produce this tree, there's not enough in the above (by itself) to prove the existence of a string with an infinite number of 1's in it. Every string whose existence can be proven from these rules will have a finite number of 1's in it. For these two reasons, it is /not/ isomorphic to the infinite binary tree. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46808d82@news2.lightlink.com> <572dna3hDIwLAh3bnZ2dnUVZ_u-unZ2d@comcast.com> <46812bda@news2.lightlink.com> <4681496d@news2.lightlink.com> <4683fa9c@news2.lightlink.com> <46845cdd@news2.lightlink.com> <46853961@news2.lightlink.com> <46879459@news2.lightlink.com> <468c0341@news2.lightlink.com> <468fd687@news2.lightlink.com> Bytes: 2750 For these two reasons, it is /not/ isomorphic to the infinite binary > tree. > Why do you try to veil clear thoughts by complicated pictures? If all real numbers exist, then they can be represented by binary sequences such that equal initial segments are written only once. That is a binary tree. The only difference to a Cantor list is, that in the binary tree all real numbers are included by definition. This means a diagonal number cannot be outside of the tree. As every separation requires exactly one point of separation, the separations and points of separation are equinumerous. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 3874 For these two reasons, it is /not/ isomorphic to the infinite binary > tree. Why do you try to veil clear thoughts by complicated pictures? If all > real numbers exist, then they can be represented by binary sequences > such that equal initial segments are written only once. If 'equal initial segments' can be written only once, then WM's set of reals consists of precisely two numbers, and no more. > That is a > binary tree. A binary tree with precisely two paths. > The only difference to a Cantor list is, that in the > binary tree all real numbers are included by definition. This means a > diagonal number cannot be outside of the tree. Another difference from a Cantor list is that, if one does no limit the tree to two paths as WM seems to ant to do, there are for any given list more paths unlisted than listed. As every separation requires exactly one point of separation And every such 'separation', separates infinitely many paths from infinitely many other paths. WM's delusion that at any one node in an infinite tree only one lath branches off in either direction is foolish and willfully wrong. In fact from any node in an infinite tree as many paths branch in either direction as there are paths in the whole tree. I.e., rom any node in the icomplete infinite binary tree, as many paths go through that node spring from the root node. > separations and points of separation are equinumerous. But each 'separation' separates an uncountably infinite sets of paths from another uncountably infinite set of paths, not just one path from one other path. > WM again displays for all the world to see his total lack of comprehension. === Subject: Re: PARADISE LOST: Debunking Cantor's theory >> Why do you try to veil clear thoughts by complicated pictures? [WM] >> Well, everything should be made as simple as possible, but not simpler. (A. Einstein) F. -- E-mail: infosimple-linede === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468CA756.10208@et.uni-magdeburg.de> <468E6775.4000601@et.uni-magdeburg.de> But still fictitious. And what do you mean by numerically accessable? > WM will claim that not even the naturals are all numerically > accessable. And isn't it so? Have you ever had access to all numbers between 10^10^1000 and 130^120^38890, which is a vanishingly small interval compared to all naturals. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2444 > But still fictitious. And what do you mean by numerically accessable? > WM will claim that not even the naturals are all numerically > accessable. And isn't it so? Have you ever had access to all numbers between > 10^10^1000 and 130^120^38890, which is a vanishingly small interval > compared to all naturals. > WM supposes that access to a number comes only from being able to write it out exactly in decimal notation on a piece of paper, or some physical equivalent. What he is doing is called accounting, not mathematics, at least by mathematicians. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468CA756.10208@et.uni-magdeburg.de> <468E6775.4000601@et.uni-magdeburg.de> But still fictitious. And what do you mean by numerically accessable? > WM will claim that not even the naturals are all numerically > accessable. And isn't it so? Have you ever had access to all numbers between > 10^10^1000 and 130^120^38890, which is a vanishingly small interval > compared to all naturals. > Betcha can't name an inaccessible natural. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468CA756.10208@et.uni-magdeburg.de> <468E6775.4000601@et.uni-magdeburg.de> WM will claim that not even the naturals are all numerically > accessable. And isn't it so? Have you ever had access to all numbers between > 10^10^1000 and 130^120^38890, which is a vanishingly small interval > compared to all naturals. > Betcha can't name an inaccessible natural. That's why it is inaccessible! === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2655 > But still fictitious. And what do you mean by numerically accessable? > WM will claim that not even the naturals are all numerically > accessable. And isn't it so? Have you ever had access to all numbers between > 10^10^1000 and 130^120^38890, which is a vanishingly small interval > compared to all naturals. > Betcha can't name an inaccessible natural. That's why it is inaccessible! > That one cannot get there from here does not mean that there is no there there. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468c4cb6@news2.lightlink.com> <468CAB55.5050407@et.uni-magdeburg.de> <468E6B8F.7010303@et.uni-magdeburg.de> an axiom system, which requires only faith in the reliability of formal > logic. But 0 < n for every positive n in N does not belong to your logic? Or how can you conclude from Card(N) = omega on Sum(N) = 0? === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2757 You are the one acting on faith. I only go by what logic can deduce > from an axiom system, which requires only faith in the reliability > of formal logic. But 0 < n for every positive n in N does not belong to your logic? As 0 is not a member of what WM accepts as N, the inequality is meaningless without a larger context. In what system is WM comparing 0 with members of N? > Or how can you conclude from Card(N) = omega on Sum(N) = 0? I do not argue that Card(N) = omega implies Sum(N) = 0. Dik argues that since SUM_{n in N} n is not defined, it can be defined much as one chooses without harm. His definition, while harmless is also pretty much useless, so would normally simply be withdrawn or allowed to die for lack of use. Definitions that persist do so normally only because they are found to be useful. But WM has concentrated so much of his energies on Dik's definition that he has given it a life it would ordinarily never have had. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468c4cb6@news2.lightlink.com> Irrational numbers that cannot be defined by > less bits than their binary representation requires, cannot get > written and will never get written. Those numbers are most of what you > consider to be your real numbers. Alas they cannot be distinguished, > so they are not different. If one knows that x is between 0 and 1 and that y is between 3 and 4, > then x and y can be distinguished from each other even if nothing else > is, or can be, known about them. So they are different! That remains to be figured out. You first have to relate x to such an unaddressable number and y to another one. But the real problem is this: If you know that x and y both are between 0 and 1, but you don't know which one is larger. How can you tell then which is larger or how to distinguish x and y otherwise? === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 3421 Irrational numbers that cannot be defined by > less bits than their binary representation requires, cannot get > written and will never get written. Those numbers are most of what you > consider to be your real numbers. Alas they cannot be distinguished, > so they are not different. If one knows that x is between 0 and 1 and that y is between 3 and 4, > then x and y can be distinguished from each other even if nothing else > is, or can be, known about them. So they are different! That remains to be figured out. You first have to relate x to such an > unaddressable number and y to another one. If one already knows of reals x and y that x < 1 and 3 < y, one knows enough to know that they are different from each other. But the real problem is this: If you know that x and y both are > between 0 and 1, but you don't know which one is larger. How can you > tell then which is larger or how to distinguish x and y otherwise? You cannot do any better when x and y are naturals between, say, 1 and 4. You still cannot distinguish x from y. But it is just a part of WM's general foolishness to say, as he does above, that one has to have an exact numerical representation for each of two real numbers in order to distinguish between them. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468CA3CC.5010707@et.uni-magdeburg.de> <468E6678.7030708@et.uni-magdeburg.de> systems whose axioms allow the proving of both a statement and its > negation are not of much use, so such systems are not much in favor.. ZFC is of no use at all. Why are so many believers supporting it? === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2344 Anything can be declared an axiom, but it has been fund that axiom > systems whose axioms allow the proving of both a statement and its > negation are not of much use, so such systems are not much in favor.. ZFC is of no use at all. Why are so many believers supporting it? That ZFC is of no use to WM only shows WM's weaknesses, and does not show that ZFC has any. Until WM can make explicit his own set of axioms (things he accepts without mathematical proof) he is in no position to object to the systems adopted by those who can. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468c4cb6@news2.lightlink.com> Bytes: 2471 > Just stick with good old-fashioned integers. > The finiteness of the universe is enough to show that the > integers than can be written in decimal notation cannot be put together > in an ordered list. Surely. Correct. There is a limit. If all agree, we can use the ink available to write just one single integer with 10^100 bits, or we can write 10^50 integers with 10^50 bits each. Then we have gone as far as one can go. And there is no drop of ink remaining for any diagonal. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 2958 Just stick with good old-fashioned integers. > The finiteness of the universe is enough to show that the > integers than can be written in decimal notation cannot be put together > in an ordered list. Surely. Correct. There is a limit. If all agree, we can use the ink available > to write just one single integer with 10^100 bits, or we can write > 10^50 integers with 10^50 bits each. Then we have gone as far as one > can go. And there is no drop of ink remaining for any diagonal. > Poor WM, who is obliged to write out everything in order to be able to think about it. Those of us with better minds can think without writing, and can imagine the whole from rules which will generate the whole, even if it would take infinitely many steps to do so. Thus: There is a set N which (1) contains {} as a member (2) for each of its members x also has (x / {x}) as a member (3) is a subset of every set satisfying (1) and (2). === Subject: Re: PARADISE LOST: Debunking Cantor's theory <468B8769.3010601@et.uni-magdeburg.de> <468c4cb6@news2.lightlink.com> The finiteness of the universe is enough to show that the > integers than can be written in decimal notation cannot be put together > in an ordered list. Surely. Correct. There is a limit. If all agree, we can use the ink available > to write just one single integer with 10^100 bits, or we can write > 10^50 integers with 10^50 bits each. Then we have gone as far as one > can go. And there is no drop of ink remaining for any diagonal. > Poor WM, who is obliged to write out everything in order to be able to > think about it. Those of us with better minds can think without writing, and certainly also write without thinking > and can imagine > the whole from rules which will generate the whole, even if it would > take infinitely many steps to do so. Thus: > There is a set N which > (1) contains {} as a member > (2) for each of its members x also has (x / {x}) as a member > (3) is a subset of every set satisfying (1) and (2). Remember my model M7 allowing for strings of seven symbols used in the tiny toy universe. Some of these strings include the symbol ... and the tiny intelligences populating this universe think they have introduced infinity by that. === Subject: Re: PARADISE LOST: Debunking Cantor's theory Bytes: 3957 > Just stick with good old-fashioned integers. > The finiteness of the universe is enough to show that the > integers than can be written in decimal notation cannot be put together > in an ordered list. Surely. Correct. There is a limit. If all agree, we can use the ink available > to write just one single integer with 10^100 bits, or we can write > 10^50 integers with 10^50 bits each. Then we have gone as far as one > can go. And there is no drop of ink remaining for any diagonal. > Poor WM, who is obliged to write out everything in order to be able to > think about it. Those of us with better minds can think without writing, and certainly also write without thinking We usually leave that sort of thing to WM and his ilk. and can imagine > the whole from rules which will generate the whole, even if it would > take infinitely many steps to do so. Thus: > There is a set N which > (1) contains {} as a member > (2) for each of its members x also has (x / {x}) as a member > (3) is a subset of every set satisfying (1) and (2). Remember my model M7 allowing for strings of seven symbols used in the > tiny toy universe. Some of these strings include the symbol ... and > the tiny intelligences populating this universe think they have > introduced infinity by that. As WM's models exist only in MathUnRealism, no mathematician need be concerned with them. But a model very like the one for N above exists in almost every viable set theory. So those who deny it, bar themselves from much important mathematics. === Subject: General solution to equation I need to solve the following equation for n, and I'm having a lot of trouble. (s*(1+t)^n-s)*(1-z) = (s*(1+x)^n-(s+(s*d*(1-(1+x)^n)/-x)))*(1-c)+ (s*d*(1-(1+x)^n)/-x) Any help finding a general solution for n would be much appreciated. It would even be helpful if someone could tell me that it's not possible to find a general solution so I stop banging my head against === Subject: Re: General solution to equation grove.matthew@gmail.com a .8ecrit : > I need to solve the following equation for n, and I'm having a lot of > trouble. (s*(1+t)^n-s)*(1-z) = (s*(1+x)^n-(s+(s*d*(1-(1+x)^n)/-x)))*(1-c)+ > (s*d*(1-(1+x)^n)/-x) Any help finding a general solution for n would be much appreciated. > It would even be helpful if someone could tell me that it's not > possible to find a general solution so I stop banging my head against > The 's' factor appears everywhere and may be removed. Let's note L= 1-z and R= (c-1) (1-d/x) + d/x then your equation is simply (if L not 0 that is z not 1) : ((1+t)^n - 1) = R/L ((1+x)^n - 1) I'll define a function of n (with r= R/L) : f(n)= ((1+t)^n - 1) - r ((1+x)^n - 1) I'll suppose that t and x are different and not -1 and that you are searching a real positive solution for n A simple solution of f(n)= 0 is of course n=0 Let's study the behavior of f(n) for positive values of n. The derivative of f(n) is f'(n)= ln(1+t) (1+t)^n - r ln(1+x) (1+x)^n f'(n)= 0 iff ln(1+t)/ln(1+x) = r ((1+x)/(1+t))^n that is n0= ln(ln(1+t)/ln(1+x)/r)/ln((1+x)/(1+t)) verifies f'(n0)=0 n0 may be complex, real and negative or real and positive. Since f is smooth, f(0)= 0 and since limit_{n->+oo} f(n)= +/- oo you may obtain a positive solution for f(n)= 0 only in the last case (n0 positive real). In this case you may solve f(n)= 0 by iterations for example using Newton's method : Start with a 'large enough' value of n (say 2 n0) and compute g(n)= n - f(n)/f'(n)= n - (((1+t)^n - 1) - r ((1+x)^n - 1))/(ln(1+t) (1+t)^n - r ln(1+x) (1+x)^n) replace n by g(n) and continue until required precision Should this method be unstable for your t, x, r values then try one of the methods proposed here : http://en.wikipedia.org/wiki/Root-finding_algorithm Hoping it helped, Raymond === Subject: Re: General solution to equation <469025f3$0$5089$ba4acef3@news.orange.fr grove.matt...@gmail.com a ?crit : I need to solve the following equation for n, and I'm having a lot of > trouble. (s*(1+t)^n-s)*(1-z) = (s*(1+x)^n-(s+(s*d*(1-(1+x)^n)/-x)))*(1-c)+ > (s*d*(1-(1+x)^n)/-x) Any help finding a general solution for n would be much appreciated. > It would even be helpful if someone could tell me that it's not > possible to find a general solution so I stop banging my head against The 's' factor appears everywhere and may be removed. Let's note L= 1-z and R= (c-1) (1-d/x) + d/x then your equation is > simply (if L not 0 that is z not 1) : > ((1+t)^n - 1) = R/L ((1+x)^n - 1) I'll define a function of n (with r= R/L) : > f(n)= ((1+t)^n - 1) - r ((1+x)^n - 1) I'll suppose that t and x are different and not -1 and that you are > searching a real positive solution for n A simple solution of f(n)= 0 is of course n=0 Let's study the behavior of f(n) for positive values of n. The derivative of f(n) is > f'(n)= ln(1+t) (1+t)^n - r ln(1+x) (1+x)^n f'(n)= 0 iff ln(1+t)/ln(1+x) = r ((1+x)/(1+t))^n > that is n0= ln(ln(1+t)/ln(1+x)/r)/ln((1+x)/(1+t)) verifies f'(n0)=0 > n0 may be complex, real and negative or real and positive. Since f is smooth, f(0)= 0 and since limit {n->+oo} f(n)= +/- oo you > may obtain a positive solution for f(n)= 0 only in the last case (n0 > positive real). In this case you may solve f(n)= 0 by iterations for example using > Newton's method : Start with a 'large enough' value of n (say 2 n0) and compute > g(n)= n - f(n)/f'(n)= n - (((1+t)^n - 1) - r ((1+x)^n - 1))/(ln(1+t) > (1+t)^n - r ln(1+x) (1+x)^n) > replace n by g(n) and continue until required precision Should this method be unstable for your t, x, r values then try one > of the methods proposed here :http://en.wikipedia.org/wiki/Root-finding algorithm Hoping it helped, > Raymond === Subject: Re: General solution to equation Bytes: 1516 >I need to solve the following equation for n, and I'm having a lot of >trouble. (s*(1+t)^n-s)*(1-z) = (s*(1+x)^n-(s+(s*d*(1-(1+x)^n)/-x)))*(1-c)+ >(s*d*(1-(1+x)^n)/-x) Any help finding a general solution for n would be much appreciated. Surely you must be joking. Don't waste your time even trying. quasi === Subject: =?UTF-8?Q?Derivative_=E2=80=93_what_is_wrong_in_my_calculation_=3F?= I started learning calculus. I am in the derivatives stage. I am following the book Calculus Demystified. However I don't understand the result of the derivative of [ e^x^2]. ( ^ means raised to the power ) In my calculation I did this: If (f .90 g)(x) = e^x^2 and f(s) = s^2 Considering that ((f .90 g)(x)) =[ df/ds(g(x))].d/dx(f(x)) then ((f .90 g)(x)) = 2(e^x) . (e^x) = 2e^2x But the book says that the resulting derivative is 2x.e^x^2. If the book is correct, could someone explain how this derivative was calculated ? === Subject: =?UTF-8?Q?Re:_Derivative_=E2=80=93_what_is_wrong_in_my_calculation_=3F?= === Subject: =?UTF-8?Q?Re:_Derivative_=E2=80=93_what_is_wrong_in_my_calculation_=3F?= === Subject: =?UTF-8?Q?Re:_Derivative_=E2=80=93_what_is_wrong_in_my_calculation_=3F?= tedhill >I am following the book Calculus Demystified. However >I don't understand the result of the derivative of [ >e^x^2]. y=e^(x^2); x^2=f(x), y=e^f(x) y'=e^f(x)*f'(x) y'=2x*e^(x^2) Vincenzo Librandi === Subject: =?utf-8?q?Re:_Derivative_=E2=80=93_what_is_wrong_in_my_calculation_=3F?= > I started learning calculus. I am in the derivatives stage. I am following the book Calculus Demystified. However I don't understand the result of the derivative of [ e^x^2]. You should write this as e^(x^2), to distinguish it from (e^x)^2 = e^(2x). ( ^ means raised to the power ) In my calculation I did this: If (f .90 g)(x) = e^x^2 and f(s) = s^2 Considering that ((f .90 g)(x)) =[ df/ds(g(x))].d/dx(f(x)) then ((f .90 g)(x)) = 2(e^x) . (e^x) = 2e^2x This would have been correct if, by e^x^2, you meant (e^x)^2 = e^(2x). The book meant e^(x^2). See, now, why brackets are important? But the book says that the resulting derivative is 2x.e^x^2. If the book is correct, could someone explain how this derivative was calculated ? (d/dx) f[g(x)] = f'[g(x)] * g'(x), where f'[g(x)] means f'(y) = df(y)/ dy is computed first, then y = g(x) is substituted in later. Thus, (d/ dx)exp(x^2) = (d/dx)(x^2)*exp(x^2), because (d/dy)exp(y) = exp(y). Thus, answer = 2*x*exp(x^2). R.G. Vickson === Subject: Re: Ko0kso0t Alert and TommY whining: Re: Have you been slandered on the Internet? Expires: 6/30/2007 Keywords: X On Sat, 7 Jul 2007 20:55:13 +0800, in alt.astronomy, Tom Potter > On Fri, 6 Jul 2007 19:54:09 +0800, in alt.astronomy, Tom Potter > but I'm just trolling to try to get some > enterprising person to set up a web site > to ferret out the names of the anonymous bigots on > the Internet who attack folks, and sell the data to the folks libeled, > so they can sue or kill > as they think best. >> Let's see, Tom. In the past you've advocated cyberstalking >> in the form of posting personal information on the web >> and harassing people at their place of work. Now you're >> implicitly advocating aiding dangerous lunatics who >> would kill someone because their feelings were hurt >> on a newsgroup. >> If your dream comes true and there's a website >> with the name of those who attack folks, >> I can think of one person's name who will >> certainly be on it. >> Am I the only one who's reminded of the classic >> Twilight Zone with the guy who wants to >> shrink all the evil people (with the predictable >> ending)? >> -jc jc, considering my age and my wishes for posterity, >I would have no problems being shrunk if >a large number of sociopaths >were shrunk with me. I was disappointed to see that jc thinks that >making a effort to expose the names and employers >of anonymous bigots and sociopaths is stalking and a negative thing. How do you feel about the laws >that make child molesters and sex offenders register, jc? I must point out that the people that >jc calls dangerous lunatics are those folks >who are the first to get fed up with >sociopaths abusing them or their fellow man. >I used to play pickup basketball games >a couple of times per week, and periodically >some sociopath would drift into the group, >and ruin the fun and comradeliness by cursing, >yelling, throwing and kicking the ball, pushing >etc. and it was always good to see one of these >dangerous lunatics step forth and take the lead >in getting rid of the sociopath. >> How's that for projection? >> Tom seems to be a well trained Group person. Potter sees those out >> side his current group of choice, as socipaths. Most class groups are >> trained from an early age to distrust outsiders and those that are >> different. Tom's group is equal is a nothing more than a Caste >> System. Said he doesn't see the attempt to exclude outsiders from the >> group and the refuse comradery for what it is, a form of bigotry. >When sociopathic groups escalate their abuses >of a society and do not integrate peacefully with the society. >conditions for the non-sociopaths deteriorate until >dangerous lunatics come forth and put their >names, resources and asses on the line. What jc calls dangerous lunatics at the social level >are called anti-bodies at the organic level. >> I would point out that sociopathy is not a generally group activity. >> Group activity is contradictory to the diagnosis of sociopathy. >> Sociopaths are generally loners. >> Is there any doubt... Crackpotter is a projectionist. Potter seems to >> projects his own social sort comings onto others. I am pleased to see that Bob Officer comprehends that >my method is to project the bigotry and sociopath of people >back into bigots and sociopaths, >with the hope they will comprehend >that they are seeing a reflection of themselves. No Potter, those other people are neither bigots nor sociopath. When I say you seem to be projecting, that means you are taking your own faults and seeing them in other people, even if those faults do not exist in those people. >Some people get all bent out of shape >when they see how ugly they are. See, You see other people as ugly, that could also be seen as another instant of projection. Your method was found 1st year psychology texts 40 years ago. Projection is a method which people ignore they own faults and see them in others. I do suggest you visit a good mental health professional. Visit that health professional soon, if you are seeing everyone other than you are bigots and sociopaths And ugly people. -- Ak'toh'di === Subject: You're the one that needs to see the Shrink for being fixated on Deco's pecker. Bytes: 1928 > I do suggest you visit a good mental health professional. You're the one that needs to see the Shrink for being fixated on Deco's pecker. Your Pal, HJ === Subject: Re: Have you been slandered on the Internet? As can be seen from the post listed below, > and reading the newsgroups, > about 10 or 15 bigots and sociopaths > try to control the opinions, ideas, and information > expressed on the Internet, Nobody controls information on the internet. As you've shown, any > idiot can post whatever nonsense he or she wants. No one > would *dream* of stopping you, because you're so damn entertaining > to ridicule. > Nobody _officially_, as in a government or formal administrative body. But the Internet community has it's own dynamic that controls it, which includes not least this sick form of entertainment you derive that make more civilized people like myself cringe. Deriving pleasure from ridicule, which ultimately causes suffering of the ridiculed, to me is a high moral wrong. When I see, say, an invalid argument, I point out it's invalidity, and stick to that logic for the debate. If I cannot convice the person to drop it, I simply decide to quit the debate. I don't bother ridiculing as I do not feel it is morally right to enjoy someone suffering -- a world where *that* happens will not be one that will have peace, and we *need* peace. The Internet, by the way, provides a good monitor for the condition of the world. Since there is a lack of government controls, the people then express themselves more freely and thus one can see what they are about. And it shows that many of them are not civilized, which in turn therefore helps explain why our world is so screwed up as well as showing the depth of the problem. Without a governing body, the people then would have to govern themselves -- and it shows that their standards are quite low. That in turn reflects the condition of the society, since the society affects the people and the people affect the society. Hence it explains a lot of world problems and provides a monitor on the condition of the world. And that is another reason I choose to use the moral codes I use, since by doing that I am contributing, even if only a small piece, to making things just a bit better. Every bit of good contributes a small piece to this process. === Subject: Re: Have you been slandered on the Internet? > On Tue, 3 Jul 2007 13:55:11 -0500, in alt.astronomy, Herb Martin > TommY Crackpotter is a USDA Choice, Grade A #1 level kook. He's > been > infesting sci.phusics with his pseudoscience spews for years now. >>Well, it would be easy enough to determine that in sci.physics since >>that >>is a (hard) science subject not easily subject to opinion as are >>political >>and religious subjects. >> Bob Officer's problem is that he is all >> bent out of shape because I point out that >> General Relativity is a Tower of Babel > Spamming Potter 1:69 again, huh? > BTW, how is this relevant to a news group about the San Antonio Spurs? 100,000 sperm to choose from, and The_Man was the fastest. Talk about bad luck. -- Tom Potter *** Time Magazine Person of the Year 2006 *** *** May 2007 Anti-Bigot Award *** http://home.earthlink.net/~tdp http://no-turtles.com http://www.frappr.com/tompotter http://photos.yahoo.com/tdp1001 http://spaces.msn.com/tdp1001 http://www.flickr.com/photos/tom-potter http://tom-potter.blogspot.com -- === Subject: Re: Have you been slandered on the Internet? > Unfortunately, as I cannot generate the same Quality posts >> as you and hanson, I have to rely on Quantity. Truer words were never spoken, Tom. >>I am pleased to see that PD agrees with me >>that John C and hanson are very creative >>thinkers and writers, >>and make high Quality posts. >>I urge everyone to read their posts, >>and observe how they handle sociopaths. Quality, eh? Art Deco has just retired from boy hunting. I'll go ahead and predict stone silence from all of the Gay AUKers > over Deco out of control Gayness. You'll need a body mitt where you're going, Demon Deco! Say did the boys in High School call you Mr. Chicken? Wrong kinda Indian from Bob-Squaw but I'm sure that Deco will just > Luv him because he's much younger than ancient Bob Duckie-Puss and > we both know how Deco loves the younger set. Heck, I ain't never seen no Gay Injun before! Deco lucky him got um > Bitch-Boi sqwah! The same kind of son-of-a-bitches that tell you to suck Art Deco's > dick. Are you paid to suck Deco's dick? Fat s do sweat a lot! I try not to be in the same house with Chubettes. Good slurp, you faggot. Not as much as Injun Bob sucks Deco! [...] Well, I've carefully read John C's posts, tried to grasp the inner > meaning, and after careful reflection, upon his content and style, > here's what I have to say: > Go suck my dick, Potter, you flaming fat . I don't know what Archie's problem is, but I'll bet it's hard to pronounce. -- Tom Potter *** Time Magazine Person of the Year 2006 *** *** May 2007 Anti-Bigot Award *** http://home.earthlink.net/~tdp http://no-turtles.com http://www.frappr.com/tompotter http://photos.yahoo.com/tdp1001 http://spaces.msn.com/tdp1001 http://www.flickr.com/photos/tom-potter http://tom-potter.blogspot.com -- === Subject: Re: Have you been slandered on the Internet? Bytes: 4062 >> Unfortunately, as I cannot generate the same Quality posts >> as you and hanson, I have to rely on Quantity. Truer words were never spoken, Tom. >>I am pleased to see that PD agrees with me >>that John C and hanson are very creative >>thinkers and writers, >>and make high Quality posts. >>I urge everyone to read their posts, >>and observe how they handle sociopaths. Quality, eh? Art Deco has just retired from boy hunting. I'll go ahead and predict stone silence from all of the Gay AUKers > over Deco out of control Gayness. You'll need a body mitt where you're going, Demon Deco! Say did the boys in High School call you Mr. Chicken? Wrong kinda Indian from Bob-Squaw but I'm sure that Deco will just > Luv him because he's much younger than ancient Bob Duckie-Puss and > we both know how Deco loves the younger set. Heck, I ain't never seen no Gay Injun before! Deco lucky him got um > Bitch-Boi sqwah! The same kind of son-of-a-bitches that tell you to suck Art Deco's > dick. Are you paid to suck Deco's dick? Fat s do sweat a lot! I try not to be in the same house with Chubettes. Good slurp, you faggot. Not as much as Injun Bob sucks Deco! [...] Well, I've carefully read John C's posts, tried to grasp the inner > meaning, and after careful reflection, upon his content and style, > here's what I have to say: > Go suck my dick, Potter, you flaming fat . I don't know what Archie's problem is, > but I'll bet it's hard to pronounce. LOL === Subject: Re: Ko0kso0t Alert and TommY whining: Re: Have you been slandered on the Internet? <030720070808593417%erfc@caballista.org> <468bb57c$0$10851$88260bb3@free.teranews.com> <468e2697$0$16315$88260bb3@free.teranews.com >> but I'm just trolling to try to get some >> enterprising person to set up a web site >> to ferret out the names of the anonymous bigots on >> the Internet who attack folks, > and sell the data to the folks libeled, >> so they can sue or kill >> as they think best. Let's see, Tom. In the past you've advocated cyberstalking > in the form of posting personal information on the web > and harassing people at their place of work. Now you're > implicitly advocating aiding dangerous lunatics who > would kill someone because their feelings were hurt > on a newsgroup. If your dream comes true and there's a website > with the name of those who attack folks, > I can think of one person's name who will > certainly be on it. Am I the only one who's reminded of the classic > Twilight Zone with the guy who wants to > shrink all the evil people (with the predictable > ending)? -jc jc, considering my age and my wishes for posterity, > I would have no problems being shrunk if > a large number of sociopaths > were shrunk with me. > No, I'm pretty sure it would just be you. > I was disappointed to see that jc thinks that > making a effort to expose the names and employers > of anonymous bigots and sociopaths is stalking and a negative thing. How do you feel about the laws > that make child molesters and sex offenders register, jc? > Just so we're all calibrated properly, are you saying that being rude to you on the internet is kind of like being a child molester? > I must point out that the people that > jc calls dangerous lunatics are those folks > who are the first to get fed up with > sociopaths abusing them or their fellow man. > I think most sane people would be comfortable classifying as a dangerous lunatic a person who would kill another because their feelings were hurt on the internet. > I used to play pickup basketball games > a couple of times per week, and periodically > some sociopath would drift into the group, > and ruin the fun and comradeliness by cursing, > yelling, throwing and kicking the ball, pushing > etc. and it was always good to see one of these > dangerous lunatics step forth and take the lead > in getting rid of the sociopath. > How many did they kill? -jc > When sociopathic groups escalate their abuses > of a society and do not integrate peacefully with the society. > conditions for the non-sociopaths deteriorate until > dangerous lunatics come forth and put their > names, resources and asses on the line. What jc calls dangerous lunatics at the social level > are called anti-bodies at the organic level. Your pal, -- > Tom Potter *** Time Magazine Person of the Year 2006 *** -- === Subject: Re: Ko0kso0t Alert and TommY whining: Re: Have you been slandered on the Internet? <030720070808593417%erfc@caballista.org> <468bb57c$0$10851$88260bb3@free.teranews.com> <468e2697$0$16315$88260bb3@free.teranews.com > but I'm just trolling to try to get some >> enterprising person to set up a web site >> to ferret out the names of the anonymous bigots on >> the Internet who attack folks, > and sell the data to the folks libeled, >> so they can sue or kill >> as they think best. Let's see, Tom. In the past you've advocated cyberstalking > in the form of posting personal information on the web > and harassing people at their place of work. Now you're > implicitly advocating aiding dangerous lunatics who > would kill someone because their feelings were hurt > on a newsgroup. If your dream comes true and there's a website > with the name of those who attack folks, > I can think of one person's name who will > certainly be on it. Am I the only one who's reminded of the classic > Twilight Zone with the guy who wants to > shrink all the evil people (with the predictable > ending)? -jc jc, considering my age and my wishes for posterity, > I would have no problems being shrunk if > a large number of sociopaths > were shrunk with me. No, I'm pretty sure it would just be you. I was disappointed to see that jc thinks that > making a effort to expose the names and employers > of anonymous bigots and sociopaths is stalking and a negative thing. How do you feel about the laws > that make child molesters and sex offenders register, jc? Just so we're all calibrated properly, are you saying that > being rude to you on the internet is kind of like > being a child molester? I must point out that the people that > jc calls dangerous lunatics are those folks > who are the first to get fed up with > sociopaths abusing them or their fellow man. I think most sane people would be comfortable > classifying as a dangerous lunatic a person > who would kill another because their feelings > were hurt on the internet. I used to play pickup basketball games > a couple of times per week, and periodically > some sociopath would drift into the group, > and ruin the fun and comradeliness by cursing, > yelling, throwing and kicking the ball, pushing > etc. and it was always good to see one of these > dangerous lunatics step forth and take the lead > in getting rid of the sociopath. In you're complaining up pickup basketball games being antisocial, you obviously never played pickup volleyball. It's the most sociopathic game in the universe, next to Rollerball. How many did they kill? -jc When sociopathic groups escalate their abuses > of a society and do not integrate peacefully with the society. > conditions for the non-sociopaths deteriorate until > dangerous lunatics come forth and put their > names, resources and asses on the line. What jc calls dangerous lunatics at the social level > are called anti-bodies at the organic level. Your pal, -- > Tom Potter *** Time Magazine Person of the Year 2006 *** -- - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Ko0kso0t Alert and TommY whining: Re: Have you been slandered on the Internet? >> How do you feel about the laws >> that make child molesters and sex offenders register, jc? Just so we're all calibrated properly, are you saying that >being rude to you on the internet is kind of like >being a child molester? Zing. Much more concise than my refutation. === Subject: Re: Have you been slandered on the Internet? Bytes: 2991 > Art Deco has just retired from boy hunting. I'll go ahead and predict stone silence from all of the Gay AUKers > over Deco out of control Gayness. You'll need a body mitt where you're going, Demon Deco! Say did the boys in High School call you Mr. Chicken? Wrong kinda Indian from Bob-Squaw but I'm sure that Deco will just > Luv him because he's much younger than ancient Bob Duckie-Puss and > we both know how Deco loves the younger set. Heck, I ain't never seen no Gay Injun before! Deco lucky him got um > Bitch-Boi sqwah! The same kind of son-of-a-bitches that tell you to suck Art Deco's > dick. Are you paid to suck Deco's dick? Fat s do sweat a lot! I try not to be in the same house with Chubettes. Good slurp, you faggot. Not as much as Injun Bob sucks Deco! [...] Well, I've carefully read John C's posts, tried to grasp the inner > meaning, and after careful reflection, upon his content and style, > here's what I have to say: > Go suck my dick, Potter, you flaming fat . Very good synthesis! However you threw that sharpened dart at the wrong person. Ciao, HJ === Subject: Re: y^2=4x+3 On Jul 6, 9:29 pm, Vincenzo Librandi Why y^2=4x+3 (for x,y, integer) does not have solutions ? Haven't you been messing around with these kinds of problems long enough not to dwell on such an utterly trivial result? Because it is equivalent to (y - 1)(y + 1) = 2.(2x + 1), whereas the left side is a product of two integers of the same parity. Happy now? John R Ramsden === Subject: Re: y^2=4x+3 Why y^2=4x+3 (for x,y, integer) does not have solutions ? >>Haven't you been messing around with these kinds of >>problems long enough not to dwell on such an utterly >>trivial result? >>Because it is equivalent to (y - 1)(y + 1) = 2.(2x + >>1), whereas the left side is a product of two integers >>of the same parity. >>Happy now? I don't like it at all. not to make irony, and you read the continuation. I attend your contribution for others. See 2k+3=p (p=prime number) Vincenzo Librandi === Subject: Re: y^2=4x+3 Ok for all. My provocation was not turned, in reality, in order to know if y^2=4x+3 it had or it did not have solutions, but simply in order to introduce the study of the triangular matrices. Every shape of the type ax+b can be transposed in a matrix triangular where first diagonal values x supply us i for which the squared expression ax+b = square. In the shape 4x+3 the first diagonal does not exist and, therefore, no value of x supplies a square. Example: 4k+3 with the triangular matrix: - 3,- -,8,- 6,-,15,- -,13,-,24,- 9,-,22,-,35,- and so on To not exist the first diagonal therefore 4k+3 isn't never square, and 4k+3=p (p=odd prime number) if and only if K isn't belong to matrix k=0, p=3 k=1, p=7 k=2, p=11 k=3 belong to matrix, k=4, p=19 k=5, p=23 and so on for all prime p and p+4. ----- 2k+3 with the triangular matrix 3 6,11 9,16,23 12,21,30,39 15,26,37,48,59 18,31,44,57,70,83 and so on with k=(3,11,23,39,59,83,...,) first diagonal 2k+3= odd square(9,25,49,81,121,169,...,) The more important thing for this triangular matrix is that for every k that does not belong to the matrix has the succession of the prime odd numbers (3,5,7,11,13,...,) k=0, p=3 k=1, p=5 k=2, p=7 k=3 belong to matrix k=4, p=11 k=5, p=13 and so on Ask, a large part of them, the comment. Tanks, Vincenzo Librandi vincenzo.librandweoz@alice.it === Subject: Re: y^2=4x+3 <22372175.1183791629434.JavaMail.jakarta@nitrogen.mathforum.org> On 7 Jul., 08:59, Vincenzo Librandi in order to know if y^2=4x+3 it had or it > did not have solutions, but simply in order > to introduce the study of the triangular > matrices. Every shape of the type ax+b can > be transposed in a matrix triangular where > first diagonal values x supply us i for which > the squared expression ax+b = square. > In the shape 4x+3 the first diagonal does not > exist and, therefore, no value of x supplies > a square. Example: > 4k+3 with the triangular matrix: > - > 3,- > -,8,- > 6,-,15,- > -,13,-,24,- > 9,-,22,-,35,- > and so on Sorry, it was totally incomprehensible to me how this matrix is related to the expression 4x+3. For example, how do you obtain a[5,2]=13? To not exist the first diagonal > therefore 4k+3 isn't never square, and 4k+3=p (p=odd prime number) > if and only if > K isn't belong to matrix > k=0, p=3 > k=1, p=7 > k=2, p=11 > k=3 belong to matrix, > k=4, p=19 > k=5, p=23 > and so on > for all prime p and p+4. ----- 2k+3 with the triangular matrix > 3 > 6,11 > 9,16,23 > 12,21,30,39 > 15,26,37,48,59 > 18,31,44,57,70,83 > and so on with k=(3,11,23,39,59,83,...,) first diagonal > 2k+3= odd square(9,25,49,81,121,169,...,) The more important thing for this triangular > matrix is that for every k that does not belong > to the matrix has the succession of the prime > odd numbers (3,5,7,11,13,...,) Do you mean: If 2k+3 is prime then 2k+3 is not a square? k=0, p=3 > k=1, p=5 > k=2, p=7 > k=3 belong to matrix > k=4, p=11 > k=5, p=13 > and so on LOL! Good time to stop. k=6, p=15 Ask, a large part of them, the comment. Tanks, > Vincenzo Librandi > vincenzo.librandw...@alice.it === Subject: Re: y^2=4x+3 > Ok for all. My provocation was not turned, in reality, > in order to know if y^2=4x+3 it had or it > did not have solutions, but simply in order > to introduce the study of the triangular > matrices. Every shape of the type ax+b can > be transposed in a matrix triangular where > first diagonal values x supply us i for which > the squared expression ax+b = square. > In the shape 4x+3 the first diagonal does not > exist and, therefore, no value of x supplies > a square. Example: > 4k+3 with the triangular matrix: > - > 3,- > -,8,- > 6,-,15,- > -,13,-,24,- > 9,-,22,-,35,- > and so on >>Sorry, it was totally incomprehensible to me how this >>matrix is related to the expression 4x+3. >>For example, how do you obtain a[5,2]=13? triangular matrix=A(m,n) then 4k+3=(2m+1)*(2n+1) k=m*n+[(m+n-1)/2] m=5, n=2, A(5,2)=5*2+3=13. You understand ? To not exist the first diagonal > therefore 4k+3 isn't never square, and 4k+3=p (p=odd prime number) > if and only if > K isn't belong to matrix > k=0, p=3 > k=1, p=7 > k=2, p=11 > k=3 belong to matrix, > k=4, p=19 > k=5, p=23 > and so on > for all prime p and p+4. ----- 2k+3 with the triangular matrix > 3 > 6,11 > 9,16,23 > 12,21,30,39 > 15,26,37,48,59 > 18,31,44,57,70,83 > and so on with k=(3,11,23,39,59,83,...,) first diagonal > 2k+3= odd square(9,25,49,81,121,169,...,) The more important thing for this triangular > matrix is that for every k that does not belong > to the matrix has the succession of the prime > odd numbers (3,5,7,11,13,...,) >>Do you mean: If 2k+3 is prime then 2k+3 is not a >>square? What ? Not, not !! if k isn't belong to matrix, then 2k+3 is prime. k=0, p=3 > k=1, p=5 > k=2, p=7 > k=3 belong to matrix > k=4, p=11 > k=5, p=13 > and so on >>LOL! Good time to stop. >>k=6, p=15 NOT LOL k=6 is belong to matrix then 2k+3=2*6+3=15 isn't prime instead k=7, 2k+3=17 k=8, 2k+3=19 k=9 is belong to matrix then 2k+3 isn't prime k=10, 2k+3=23 k=11 is belong to matrix, not prime k=12 is belong to matrix, not prime k=13, 2k+3=29 and so on You undertand ? Vincenzo Librandi === Subject: Re: y^2=4x+3 <24714265.1183796019422.JavaMail.jakarta@nitrogen.mathforum.org> On 7 Jul., 10:13, Vincenzo Librandi Ok for all. My provocation was not turned, in reality, > in order to know if y^2=4x+3 it had or it > did not have solutions, but simply in order > to introduce the study of the triangular > matrices. Every shape of the type ax+b can > be transposed in a matrix triangular where > first diagonal values x supply us i for which > the squared expression ax+b = square. > In the shape 4x+3 the first diagonal does not > exist and, therefore, no value of x supplies > a square. Example: > 4k+3 with the triangular matrix: > - > 3,- > -,8,- > 6,-,15,- > -,13,-,24,- > 9,-,22,-,35,- > and so on >>Sorry, it was totally incomprehensible to me how this >>matrix is related to the expression 4x+3. >>For example, how do you obtain a[5,2]=13? triangular matrix=A(m,n) then 4k+3=(2m+1)*(2n+1) > k=m*n+[(m+n-1)/2] > m=5, n=2, A(5,2)=5*2+3=13. > You understand ? So you mean A(m,n) = k if (2m+1)(2n1+1)=4k+3, A(m,n) = undefined (or -) otherwise? Especially, A(n,n) = k if (2n+1)^2 = 4k+3. Is your theory supposed to give an alternative proof that no perfect square is of the form 4k+3? > To not exist the first diagonal > therefore 4k+3 isn't never square, and 4k+3=p (p=odd prime number) > if and only if > K isn't belong to matrix > k=0, p=3 > k=1, p=7 > k=2, p=11 > k=3 belong to matrix, > k=4, p=19 > k=5, p=23 > and so on > for all prime p and p+4. ----- 2k+3 with the triangular matrix > 3 > 6,11 > 9,16,23 > 12,21,30,39 > 15,26,37,48,59 > 18,31,44,57,70,83 > and so on with k=(3,11,23,39,59,83,...,) first diagonal > 2k+3= odd square(9,25,49,81,121,169,...,) The more important thing for this triangular > matrix is that for every k that does not belong > to the matrix has the succession of the prime > odd numbers (3,5,7,11,13,...,) >>Do you mean: If 2k+3 is prime then 2k+3 is not a >>square? What ? Not, not !! if k isn't belong to matrix, then > 2k+3 is prime. k=0, p=3 > k=1, p=5 > k=2, p=7 > k=3 belong to matrix > k=4, p=11 > k=5, p=13 > and so on >>LOL! Good time to stop. >>k=6, p=15 > NOT LOL k=6 is belong to matrix then Hm, it looks like I fell for > k=(3,11,23,39,59,83,...,) cause I misread your post. But apparently with A(m,n) = k if 4k+3=(2m+1)(2n+1) your claim about primes is more than trivial. === Subject: Re: y^2=4x+3 So you mean A(m,n) = k if (2m+1)(2n1+1)=4k+3, A(m,n) = >undefined (or -) otherwise? >Especially, A(n,n) = k if (2n+1)^2 = 4k+3. n=[-2-/+sqrt(16k+12)]/4 isn't A(n,n) Is your theory supposed to give an alternative proof that no perfect square is of the form 4k+3? Yes, no exist the first diagonal to triangular matrix: - 3,- -,8,- 6,-,15,- -,13,-,24,- 9,-,22,-,35,- and so on. More interesting remains the generated triangular matrix from 2k+3, that to want to study better than how much from made me. The problem is in characterizing, after some passage, if k it belongs or it does not belong to the triangular matrix. Vincenzo Librandi === Subject: Re: y^2=4x+3 <70703.1183839820088.JavaMail.jakarta@nitrogen.mathforum.org> On 7 Jul., 22:23, Vincenzo Librandi So you mean A(m,n) = k if (2m+1)(2n1+1)=4k+3, A(m,n) = >undefined (or -) otherwise? >Especially, A(n,n) = k if (2n+1)^2 = 4k+3. n=[-2-/+sqrt(16k+12)]/4 isn't A(n,n) Of course n isn't A(n,n). You spelled out that A(n.n) is ((2n+1)^2-3)/ 4 if that's an integer and undefined otherwise. > Is your theory supposed to give an alternative proof > that no perfect square is of the form 4k+3? Yes, > no exist the first diagonal > to triangular matrix: > - > 3,- > -,8,- > 6,-,15,- > -,13,-,24,- > 9,-,22,-,35,- > and so on. OK, that's your strategy, but where is the proof? ou have to show that ((2n+1)^2-3)/4 is never an integer, and for the proof to be new, this must not rely on odd^2==1 mod 8 More interesting remains the generated > triangular matrix from 2k+3, that to want > to study better than how much from made me. > The problem is in characterizing, after some > passage, if k it belongs or it does not belong > to the triangular matrix. Vincenzo Librandi === Subject: Re: y^2=4x+3 Why y^2=4x+3 (for x,y, integer) does not have solutions ? Look at the two cases: y even (=2z), y odd (=2z+1), in each case see what 4x is equal to. === Subject: Re: Object Convolution I finally found it. This operation is also called Minkowski product when the product is taken as the complex product. See for instance page 6 of http://mae.ucdavis.edu/~farouki/covering.pdf He does not detail the line segment convolution but does mention some applications which sound impressive. -tpg1183842159s === Subject: Re: Object Convolution > In toying with a geometrical calculus I've come across an interesting procedure that is easily computed. I'm calling it object convolution but I am wondering if this has already been developed. The description: Two geometrical objects can be joined via an arithmetic product operation. > Each object can be regarded as a set of points: > O1 = p11, p12, p13, ..., p1n > and likewise for O2 a series of points p21 ... p2m. The resultant object > O3 = p11 p21, p11 p22, p11 p23, ... p11 p2m, > p12 p21, p12 p22, p12 p23, ... p12 p2m, > p13 p21, p13 p22, p13 p23, ... p13 p2m, > ... , > p1n p21, p1n p22, p1n p23, ... p1n p2m In the complex plane line segments will yield areas with interesting boundaries. This geometric product could have some relation to spatial curvature interpretations whose constituents are flat. Under the polysign construction such geometric products can be carried out in any dimension. It would seem that if this math were already built it would be built on the complex plane. What area of mathematics is this? I don't see these geometrical objects as functions. Though they are nearby they are simpler. Their continuous form still lends themselves to this computation. For instance a point set which is connected (meaning that paths or solids may be approximated) can maintain a coherent image in its approximated form. This is a simplistic construction that relies upon a geometric arithmetic product. Not much to it. Is this overlooked in existing mathematics? Point products in real and complex spaces are common enough. This is merely a generalization of that concept. -tpg20070624 I've published some graphics on my website: http://bandtechnology.com/ObjectConvolution/index.html These graphics are not a complete survey but I am fairly sure that they expose enough for most to see what is going on. The four vertex form has endpoints which are the endpoint products of the segments. It's quite a diverse operator since it can hide within its own coverage. I'd like to get this put in terms of relativity theory but I'm afraid that it is just pure geometry until then. -tpg1183808556s === Subject: the great whitening <468bb56f$0$10851$88260bb3@free.teranews.com> <468cd4ce$0$16340$88260bb3@free.teranews.com> <468e269b$0$16315$88260bb3@free.teranews.com> <468f85eb$0$16290$88260bb3@free.teranews.com> and all disturbances disappeared so clean harmonious with the common background of choice so white... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: the great whitening Bytes: 2610 > and all disturbances disappeared so clean > harmonious with the common background of choice so white... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar > You'll Wonder Where de Sturbants Went When You Brush Yo' Brain wif Pepsodent! My Momma done tol' me, when I wuZ in sho't pants, Use plenty o' toothpaste, git after dem 'Bants, Reach over de palate and on up de nose, And scrub out de places where Sturbances goes. Reach inter de brain an' be scrub' it up right Or de Sturbants will gives you dem Blues In The Night!a -- -------(m+ ~/:o)_| I do not negotiate for half my baby back, Solomon. http://scrawlmark.org === Subject: Re: the great whitening <468bb56f$0$10851$88260bb3@free.teranews.com> <468cd4ce$0$16340$88260bb3@free.teranews.com> <468e269b$0$16315$88260bb3@free.teranews.com> <468f85eb$0$16290$88260bb3@free.teranews.com and all disturbances disappeared so clean > harmonious with the common background of choice so white... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Only Ariel brings you that bluey-whiteness. Unless you live in France, in which case it might bring you a yellowy- whiteness. When visiting family in the UK, I used to wonder why all UK newsreaders had purple noses. Was it the stress of the job? Conversation with a Sony technician in their TV factory in Viladecavalls provided the necessary clue: they retune the colours of their TVs according to country exported to. A UK white has to have blue in it to be perceived as a 'good' white, he explained, whereas in France, a good white must have a tinge of yellow. Each country has its own ideal colour configuration. Needless to say, everyone in a Sony factory wears dove-grey uniforms, works between dove-grey walls, eats dove-grey sandwiches, and dreams dove-grey dreams. (0F0F0F, IIRC) === Subject: (geek) the great whitening <468bb56f$0$10851$88260bb3@free.teranews.com> <468cd4ce$0$16340$88260bb3@free.teranews.com> <468e269b$0$16315$88260bb3@free.teranews.com> <468f85eb$0$16290$88260bb3@free.teranews.com > and all disturbances disappeared so clean > harmonious with the common background of choice so white... Only Ariel brings you that bluey-whiteness. Unless you live in France, in which case it might bring you a yellowy- > whiteness. When visiting family in the UK, I used to wonder why all UK > newsreaders had purple noses. Was it the stress of the job? > Conversation with a Sony technician in their TV factory in > Viladecavalls provided the necessary clue: they retune the colours of > their TVs according to country exported to. A UK white has to have > blue in it to be perceived as a 'good' white, he explained, whereas > in France, a good white must have a tinge of yellow. Each country has > its own ideal colour configuration. Needless to say, everyone in a Sony factory wears dove-grey uniforms, > works between dove-grey walls, eats dove-grey sandwiches, and dreams > dove-grey dreams. (0F0F0F, IIRC) self-flagellation brings me furthur from my pain struggle to decontaminate fictions of a lifetime without chains bite another chicken head smile at the crowd and understand that truth is dead when the clapping gets too loud ((..)) the ringmaster my master shines the spotlight away the tent folds in the breeze my eyes turn from the day ##..## when the circus geek died last week was it just another act? the puppets cried the games all sighed the children stood and laughed (*..*) music of the insane terrorises trailer walls a sky melts into night the geek creeps down bent halls mirrors cast an eerie light rats hide in their lairs one more day of carnivals one more life despairs ~~..~~ when the circus geek died last week was it just another act? the puppets cried the games, they sighed the children stood and laughed &&..^ the camp has left the trash and the emptiness memories for the children a dead man sad and prideless the midgets are riding first class now there's plenty of room for them someone somewhere stops a moment and remembers a horrible man ..--.. and when the circus geek died last week was it just another act? the puppets cried the games all sighed the in' children laughed -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: (geek) the great whitening <468bb56f$0$10851$88260bb3@free.teranews.com> <468cd4ce$0$16340$88260bb3@free.teranews.com> <468e269b$0$16315$88260bb3@free.teranews.com> <468f85eb$0$16290$88260bb3@free.teranews.com and all disturbances disappeared so clean > harmonious with the common background of choice so white... Only Ariel brings you that bluey-whiteness. Unless you live in France, in which case it might bring you a yellowy- > whiteness. When visiting family in the UK, I used to wonder why all UK > newsreaders had purple noses. Was it the stress of the job? > Conversation with a Sony technician in their TV factory in > Viladecavalls provided the necessary clue: they retune the colours of > their TVs according to country exported to. A UK white has to have > blue in it to be perceived as a 'good' white, he explained, whereas > in France, a good white must have a tinge of yellow. Each country has > its own ideal colour configuration. Needless to say, everyone in a Sony factory wears dove-grey uniforms, > works between dove-grey walls, eats dove-grey sandwiches, and dreams > dove-grey dreams. (0F0F0F, IIRC) self-flagellation brings me > furthur from my pain struggle to decontaminate > fictions of a lifetime without chains bite another chicken head smile at the crowd and understand that truth is dead > when the clapping gets too loud ((..)) the ringmaster > my master > shines the spotlight away the tent folds in the breeze my eyes turn from the day ##..## when the circus geek > died last week > was it just another act? the puppets cried > the games all sighed > the children stood and laughed (*..*) music of the insane > terrorises trailer walls a sky melts into night > the geek creeps down bent halls mirrors cast an eerie light rats hide in their lairs one more day of carnivals > one more life despairs ~~..~~ when the circus geek > died last week > was it just another act? the puppets cried > the games, they sighed > the children stood and laughed &&..^ the camp has left > the trash and the emptiness memories for the children a dead man > sad and prideless the midgets are riding first class now > there's plenty of room for them someone somewhere stops a moment > and remembers a horrible man ..--.. and when the circus geek > died last week > was it just another act? the puppets cried the games all sighed the in' children laughed -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar- Looking beyond the formatting, this has quite a nice rhythm. I can imagine it set to music; either an Eno-style electronic mix, or an Aznavour clone and a lone piano. === Subject: Re: (geek) the great whitening <468bb56f$0$10851$88260bb3@free.teranews.com> <468cd4ce$0$16340$88260bb3@free.teranews.com> <468e269b$0$16315$88260bb3@free.teranews.com> <468f85eb$0$16290$88260bb3@free.teranews.com> Bytes: 4814 > and all disturbances disappeared so clean > harmonious with the common background of choice so white... Only Ariel brings you that bluey-whiteness. Unless you live in France, in which case it might bring you a yellowy- > whiteness. When visiting family in the UK, I used to wonder why all UK > newsreaders had purple noses. Was it the stress of the job? > Conversation with a Sony technician in their TV factory in > Viladecavalls provided the necessary clue: they retune the colours of > their TVs according to country exported to. A UK white has to have > blue in it to be perceived as a 'good' white, he explained, whereas > in France, a good white must have a tinge of yellow. Each country has > its own ideal colour configuration. Needless to say, everyone in a Sony factory wears dove-grey uniforms, > works between dove-grey walls, eats dove-grey sandwiches, and dreams > dove-grey dreams. (0F0F0F, IIRC) self-flagellation brings me > furthur from my pain struggle to decontaminate > fictions of a lifetime without chains bite another chicken head smile at the crowd and understand that truth is dead > when the clapping gets too loud ((..)) the ringmaster > my master > shines the spotlight away the tent folds in the breeze my eyes turn from the day ##..## when the circus geek > died last week > was it just another act? the puppets cried > the games all sighed > the children stood and laughed (*..*) music of the insane > terrorises trailer walls a sky melts into night > the geek creeps down bent halls mirrors cast an eerie light rats hide in their lairs one more day of carnivals > one more life despairs ~~..~~ when the circus geek > died last week > was it just another act? the puppets cried > the games, they sighed > the children stood and laughed &&..^ the camp has left > the trash and the emptiness memories for the children a dead man > sad and prideless the midgets are riding first class now > there's plenty of room for them someone somewhere stops a moment > and remembers a horrible man ..--.. and when the circus geek > died last week > was it just another act? the puppets cried the games all sighed the in' children laughed -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Plath without tears. === Subject: Snark Theorem Equivalent to the 4CT. The snark theorem (paraphrased) says the statement that all snarks are non-planar is equivalent to the 4CT. The exsitence of a planar snark would prove the 4CT to be false; ie 1, A snark in not 3 edge colorable 2. Therefore the snark is cannot be 4 face colorable, and the 4CT is false. But is the absence of a planar snark sufficient to prove the 4CT true? Bill J === Subject: Re: Snark Theorem Equivalent to the 4CT. > The snark theorem (paraphrased) says the statement that > all snarks are non-planar is equivalent to the 4CT. The exsitence of a planar snark would prove the 4CT to be false; ie 1, A snark in not 3 edge colorable > 2. Therefore the snark is cannot be 4 face colorable, and the 4CT is false. But is the absence of a planar snark sufficient to prove the 4CT true? If some snark S happens to be planar, then that graph cannot be 3-edge colored. This means that the regions of the dual of S cannot be colored with 4 colors. That would mean the 4CT is false. If the 4CT is false, there is a counterexample G. Let H be the triangulated version of G (in case G isn't); then H isn't 4-colorable, either. The dual of H is a 3-regular graph which cannot be 3-edge colored (because a 3-edge coloring would produce a 4-coloring of H). The dual of H is 2-connected, since a bridge of the dual of H corresponds to a loop in H. Thus the dual of H would be a snark, which is also planar. Thus, proving there is no planar snark is equivalent to the 4CT, so it is both necessary and sufficient for the 4CT to be true. --- Christopher Heckman === Subject: Re: Snark Theorem Equivalent to the 4CT. Bytes: 2823 The snark theorem (paraphrased) says the statement that > all snarks are non-planar is equivalent to the 4CT. The exsitence of a planar snark would prove the 4CT to be false; ie 1, A snark in not 3 edge colorable > 2. Therefore the snark is cannot be 4 face colorable, and the 4CT is false. But is the absence of a planar snark sufficient to prove the 4CT true? If some snark S happens to be planar, then that graph cannot be 3-edge > colored. This means that the regions of the dual of S cannot be > colored with 4 colors. That would mean the 4CT is false. If the 4CT is false, there is a counterexample G. Let H be the > triangulated version of G (in case G isn't); then H isn't 4-colorable, > either. The dual of H is a 3-regular graph which cannot be 3-edge > colored (because a 3-edge coloring would produce a 4-coloring of H). > The dual of H is 2-connected, since a bridge of the dual of H > corresponds to a loop in H. Thus the dual of H would be a snark, which > is also planar. Thus, proving there is no planar snark is equivalent to the 4CT, so > it is both necessary and sufficient for the 4CT to be true. --- Christopher Heckman I am confused! Let C be the 2-connected cubic dual of H. If C is not 3 edge colorable, then C is a snark. If C is a snark and C is planar, then the 4CT is false If C is a snark and C is non-planar, then the 4CT is false. Bill J