mm-3849 === Subject: Re: preimage of 0 and C^1 class functions > Herman Rubin nous a r.8ecemment amicalement signifi.8e : >> The World Wide Wade nous a rcemment amicalement signifi : > Closed sets A. > f(x) = dist(x,A)^2 is C^1, right? >> Almost. It's C^1 at the hard points, like points of A. >> It's not differentiable at midpoints of the component >> intervals of the complement of A, but that's easily >> fixed. > Although no one's ever actually done it. >> Rather easy : >> Any open subset of R dfferent from R is a countable union of >> distincts intervals ]a,b[ or ]-inf,a[, or ]a, +inf[. >> on ]a, b[ : f(x) = dist(x,A)^2 [dist(x,A) - 3(b-a)/4] >> on ]-inf, a[ or ]a, +inf[ : f(x) = dist(x,A)^2 > Even easier, in arbitrary metric spaces, to get a C^oo function > if this is meaningful and the metric is reasonable: > f(x) = exp(-1/d(x,A)) if x not in A, f(x) = 0, x in A. > I'm afraid you have the same problem than A.S. Niel, as noticed by David > You are C^oo on A (and boundaries) but you are differentiable at > midpoints of the component intervals of the complement of A. > if A = ]-oo, -1] U [1, +oo[, your proposal is : > f(x) = exp(-1/(1-|x|)) if x not in A, f(x) = 0, x in A. > f(x) is not differentiable in 0. This suggests the question: How nonsmooth can d(x,A) be in higher dimensions? === Subject: Re: preimage of 0 and C^1 class functions > This suggests the question: How nonsmooth can d(x,A) > be in higher dimensions? This question leads to quite a bit of serious mathematics. For example, much of Elias M. Stein's book Singular Integrals and Differentiability Properties of Functions revolves around smoothness properties of the distance function to a closed set. Related to this, or maybe not but it's something I find neat, is the following result: Theorem: The set of points a fixed distance from a given closed set in R^n has Lebesgue measure zero. Proof: It is easy to show that the set in question has no points of Lebesgue density. In fact, it satisfies at each of its points what in potential theory is called an external sphere condition (in particular, the set is porous in a very strong way). Hence, the set has measure zero by the Lebesgue density theorem. This result and its short proof is due to Erd.9as (yes, *that* Erd.9as). Paul Erd.9as, Some remarks on the measurability of certain sets, Bulletin of the American Mathematical Society 51 (1945), 728-731. http://www.emis.de/cgi-bin/MATH-item?0063.01269 Dave L. Renfro === Subject: Re: preimage of 0 and C^1 class functions > Related to this, or maybe not but it's something I find neat, > is the following result: > Theorem: The set of points a fixed distance from a given > closed set in R^n has Lebesgue measure zero. I meant to point out that this holds even for very irregular shaped closed sets whose boundaries have positive measure. The result isn't all that interesting for run-of-the-mill closed sets in R^n. Dave L. Renfro === Subject: Re: preimage of 0 and C^1 class functions >> Related to this, or maybe not but it's something I find neat, >> is the following result: >> Theorem: The set of points a fixed distance from a given >> closed set in R^n has Lebesgue measure zero. >I meant to point out that this holds even for >very irregular shaped closed sets whose boundaries >have positive measure. You might also want to point out that the fixed distance had better ought to be greater than 0... Lee Rudolph === Subject: Re: preimage of 0 and C^1 class functions <44867523$0$873$ba4acef3@news.orange.fr> <4486f376$0$890$ba4acef3@news.orange.fr> Theorem: The set of points a fixed distance from a given >> closed set in R^n has Lebesgue measure zero. > You might also want to point out that the fixed distance > had better ought to be greater than 0... Hehe...Incidentally, when the fixed distance is positive, the set not only has Lebesgue n-measure zero, it's Hausdorff dimension is n-1. In fact, I believe the set has sigma-finite Hausdorff (n-1)-measure. Dave L. Renfro === Subject: Re: preimage of 0 and C^1 class functions > The World Wide Wade nous a r.8ecemment amicalement signifi.8e : Closed sets A. f(x) = dist(x,A)^2 is C^1, right? >> Almost. It's C^1 at the hard points, like points of A. >> It's not differentiable at midpoints of the component >> intervals of the complement of A, but that's easily >> fixed. > Although no one's ever actually done it. > Rather easy : This was a bit of a joke. The midpoints business keeps coming up; it's trivial conceptually, but I find it kind of humorous. > Any open subset of R dfferent from R is a countable union of distincts > intervals ]a,b[ or ]-inf,a[, or ]a, +inf[. > on ]a, b[ : f(x) = dist(x,A)^2 [dist(x,A) - 3(b-a)/4] > on ]-inf, a[ or ]a, +inf[ : f(x) = dist(x,A)^2 Well that's not just fixing up the midpoints, which would require dabbling in some smooth non-analytic stuff. Also your function is not asymptotic to d(x,A)^2 at the boundary points. One that is would be to define f(x) = [x(1-x)]^2 on [0,1] and then set g(x) = (b-a)^2*f((x-a)/(b-a)) for all open intervals (a,b) in the complement of A. === Subject: Re: I was wrong all along >> This happens to you over and over, >> and when you feel you are not sick anymore you start posting again. >> Just stay on your meds, and read an algebra book in large letters. > You failed to notice that this was not the real James Harris, right?! May seem so, but........ perhaps James posts this way when on his meds. [ Has JSH really won? the Troll has a constant crowd.] === Subject: Re: stat question mean x1 and mean x2 with standard deviations of s1 and s2. What is the >standard deviation of the ratio x1/x2? >This is a real problem. If x1 < x2 and x1 and x2 are normally >distributed, it seems clear that the ratio would NOT be normally >distributed. What would we call the distribution? Plotting it out it >looks log normal but I can't prove this. > If your assumptions are correct, then x1 and x2 are not > independent, so it is difficult to give an answer. In any > case, it would be difficult, and I do not even believe > possible, for the ratio to have a mean, let alone a > variance. It cannot be lognormal. OK I can accept this last fact. Let me restate my original question about the ratios with an example. If I can get my head around this example the real problem can be tackled. Let x1 be the number of disintegrations in 5 min for some radioactive isotope. s1 will be the sq rt of known dpm times 5 min. Let x2 be the number of disintegrations in 20 min so s2 wil be sq rt of known dpm times 20 min. I repeat this measure N times sometimes for shorter or longer periods of time but that time for x1 is always 1/4 the time for x2. For each pair the ratio of x1/x2 is computed. It seems obvious that the average of ratios will approach exactly 1/4. Why can't there be a std dev associated with that ratio that is related to s1 and s2? The actual problem that I am working on involves a question in molecular evolution. As to the specific question: Is the protein molecular clock a stochastic clock? The data I am looking at are ratios, hence the original question. have. Mike Syvanen === Subject: Re: performance enhancing drugs that increase mathematical potency i don't know because i did not train as hard as barry and did not take performance enhancers. Instead of math, I went out to the real world an made hundreds of millions after my phd. > Some of us, like barry bonds, are already in the hall of fame category > without any help. However, a little boost is all it takes for us to > climb up to the Issac newton category. Who wants to bath in the > medicrity of being a mere nobel laureate when we can surpass issac > newton? > You are just as good a mathematician as Barry Bonds? You done your > parents proud. > -- > If .999... = 1 then (.999...)/1 should equal 1 > let's see > (.999...)/1 = .999... > [Therefore] .999... still=/= 1 -- An astonishing proof by S. Enterprize === Subject: Re: Renormalization of Equality > http://hdebruijn.soo.dto.tudelft.nl/jaar2006/margriet.pdf > Gist of the argument. Define the renormalization of an equality (a = b) > as the outcome of the function exp(-((a-b)/sigma)^2/2) . The closer to > 1 the outcome, the better the equality. (sigma) is the uncertainity in > the data. A renormalized equality is reflexive, symmetric & (transitive > in some sense). It converges to an exact equality for sigma -> 0 . > But, our theory about Idealization, Materialization and Renormalization > is no longer just another theory. > It has become a technology now! A technology that can be used e.g. for > detecting the differences in two pictures that are nearly the same (as > in some puzzles for children, called zoekplaatjes in Dutch): > http://hdebruijn.soo.dto.tudelft.nl/fototjes/gezocht.htm > Here is the program that detects the differences in two pictures which > are nearly the same, together with its Delphi (Pascal) sources: > http://hdebruijn.soo.dto.tudelft.nl/jaar2006/MARGRIET.ZIP I haven't tried your programs, but I'm wondering what you mean by nearly the same. Does your program detect the parts of one picture that may be a translation, rotation, magnification, slight distortion, or color modification of parts of another picture? That would be really cool if it did. === Subject: Re: Renormalization of Equality > I haven't tried your programs, but I'm wondering what you mean by > nearly the same. Does your program detect the parts of one picture > that may be a translation, rotation, magnification, slight distortion, > or color modification of parts of another picture? That would be > really cool if it did. The program maps one picture upon the other by translation and (slight) rotation. Since the zoekplaatjes (- what is the English term? -) are supposed to be the same, no distortion or magnification is involved: http://www.puzzels.bambamscorner.nl/zoekplaat/index.php But there _is_ a difference in colors at the same spot, even if the two pictures are supposed to be the same there. That's the place where the fuzzy equality comes in: exp (-A^2/2) where A = (color1 - color2)/sigma I have ideas about how to implement the cool things you mentioned, but I'm still in search for a suitable application to demonstrate some ... Han de Bruijn === Subject: how to train intuition hi Is there a way to train intuition? (seriously) ie, I need a way to beat (train) my brain to see it. Can someone help spot the deficiency in my training. for example, I can clearly see, 1. if a==b mod p then f(a)==f(b) mod p (for all f in Z[x]) 2. x^(p-1)-1==0 (mod p) (for all x in Z) but I can't in my life see (or train myself to see) 3. gcd(x^(p-1)-1, f(x)) is the product of all factors, such that f(x)==0 mod p I think is probably related to my deficiency in seeing phism or equivalence or some such, because whenever I see the two for all statements, I know something is up, but I just can't deduce further to see the third statement. I ran into this kind of situations a lot of times when I was a student, but never find a way around it. Since now I have sometime on my hand, I like to see if I can cure myself by doing lots of exercises in this particular kind of deduction. Ie, when you see double for all statements, what must follow? Is there exercise problems that I can train myself to always honestly see this kind of deduction. I mean I've done enough to even recgonize the situation itself but can't be freed from it. I'll find more examples when I've found my old textbooks. === Subject: Re: how to train intuition > hi > Is there a way to train intuition? (seriously) > ie, I need a way to beat (train) my brain to see it. > Can someone help spot the deficiency in my training. > for example, I can clearly see, > 1. if a==b mod p then f(a)==f(b) mod p (for all f in Z[x]) > 2. x^(p-1)-1==0 (mod p) (for all x in Z) > but I can't in my life see (or train myself to see) > 3. gcd(x^(p-1)-1, f(x)) is the product of all factors, such that > f(x)==0 mod p > I think is probably related to my deficiency in seeing phism or > equivalence or some such, because whenever I see the two for all > statements, I know something is up, but I just can't deduce further > to see the third statement. I ran into this kind of situations a lot of > times when I was a student, but never find a way around it. Since now I > have sometime on my hand, I like to see if I can cure myself by doing > lots of exercises in this particular kind of deduction. Ie, when you > see double for all statements, what must follow? > Is there exercise problems that I can train myself to always honestly > see this kind of deduction. I mean I've done enough to even recgonize > the situation itself but can't be freed from it. I'll find more > examples when I've found my old textbooks. Well I can only speak for myself. I see these things. Seriously a human can only handle about 3 groups of 3 elements at once. Our ability to do logic is therefor pretty weak. But our visual cortex is unique - the most powerful in the animal kingdom. We can visually distinguish thousands of entities. Therefore I would recommend you train yourself to find some sort of visual representation of such expressions. Then describe what you see and translate it to math. -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/ === > Bob answered my question better than i expected and > So, your method sounds like a version of > the oldest wheels.All of the criticisms apply, > as well as its uselessness in the configurations > which can be visually inspected as the > star constellations. > It is really nice of You to answer my question. So i'm much obliged to > You and want to thank You. ideas new to you -- which is the hardest thing for most people to do. > So visual inspection is the best method of finding clusters and all the > algorithms give no efficient help, using a computer or not. I really > would recommend not to hide this in statistical journals and to set up > a website about this, so we can all be more educated about this. That's hardly necessary. I dare say EVERYONE in the ASA's statistical computing section and statistical graphics section know this idea already. That goes with Berkson's Interrocular Traumatic Test applied to the use of PP and QQ plots for normality instead of all the chi-square, Kolmogorov-Smirnov, Ron-Barcadi, and dozens of other analytic tests for distributional goodness of fit. The visual graphics will hit you between the eye -- which is the axiom known to all statistical graphics people. > Sometimes a No-result is as important as a result (and especially seen, > how many effort is still put into this field (and brain-fat burned)). > And this might be nice for people, not getting jobless and replaced by > computers. > Hero My Computer Generated Aid for Cluster Analysis in Communications of the ACM (1973) but presented in greater details in Brian Everitt's book, Graphical Techniques for Multivariate Data, NY: North Holland, 1978. is particularly useful for showing the before and after displays of a cluster analysis, so that one can SEE whether anything was actually accomplished or not. The plotting technique was available in the Cluster Analysis and Factor Analysis sections of several well known stat. packages. I assume they are still available. > PS I'm nearly convinced. But i like to say ,,but and to argue, so > if You and/or someoneelse like to....: > Every cluster-method has its counter-examples, this method can not > detect. This is true for visual inspection too. > As an illustration: a constellation of stars can look like a cluster, > lets say ,,cancer or the ,,pleiades. Measurement of distance > can give extra information, so that one star of ,,cancer actually > belongs more to a cluster together with the ,,pleiades, than to the > other stars of ,,cancer. Now this is easily corrected by depicting > the stars in 3D. The star constellations are what they looked like to the star gazers of thosands of years ago. They are the apparent distances on the surface of the visual sphere rather than the real distances in 3-space. So, in the Star constallation clustering in 2-D surface is KNOWN to be different from any kind of clustering in the real 3-D space. The constellations are ancient folk's visual imagination on what the BRIGHTEST stars sweem to form certain shapes, when visually connected: The Big Dipper and Little Dipper are two well-known examples. The Pleiades, or Seven Sister cluster is a very tight VISUAL cluster of 7 stars visible among hundreds of other stars. > Now the counter-example: Let's assume we have collected data about a > thousand planets ( in 2030). And just for this example we avoid space > dimensions, but we have albedo, mass, speed of rotation, surface > temperature, and six other physical measures from them. Now in > displaying these data in 2 or 3-D-graphs might be as misleading about > clusters, as the impression we get about one constellation like > ,,cancer. But by calculating over let's say nine physical > properties, we will a get a cluster in some range of these data. This > does not mean, there is an inner bound of these properties in the range > of the cluster, but it's like a proposal to check by physics. And this > cluster could not be found by visually examination of a graphic, > displaying temperature as a length, albedo as a length perpendicular, > mass as a third length also at right angle, and then the same done with > three other properties and so on , wouldn't it? The astronomical clustering of stars into galaxies and super-galaxies are 3-D clusterings of stars in the expanding universe (or 4-D clustering in the astro-physicists' view of the universe). The kind of clustering cluster analysts talk about are the kinds 4-year olds know how to do in 2-D, and they wish they knew how to do it in high dimensions. -- Bob. === > The star constellations are what they looked like to the star gazers > of thosands of years ago. They are the apparent distances on > the surface of the visual sphere rather than the real distances in > 3-space. > So, in the Star constallation clustering in 2-D surface is KNOWN > to be different from any kind of clustering in the real 3-D space. And displaying physical properties as length might lead to an even deceptive picture. Look at the Hertzsprung-Russell-diagramm, add a third dimension, let's say magnetic strength to it and You might get a different view. > The kind of clustering cluster analysts talk about are the kinds 4-year > olds know how to do in 2-D, and they wish they knew how to do it in > high dimensions. I nearly cannot believe You in this, but i'll sure keep it in mind. Still remains the question, how nature does cluster recognition with eyes. May be it's just simple looking for the greatest density of points, deciding if this is one center or more, also when shrinking the spheres around the points the first seperation into different sets occur, deciding if this is two parts or more. And then with one of the alternatives going to a pattern library in the brain and compare. Hero. === Subject: Re: Question > ... > > I noted that the Mersenne prime project says that it > > takes about a month for a single test of primality, but of course they > > are dealing with *very* big numbers. > Indeed. The time is still exponential in the size of the number. Erm? That depends on whether by the number, you mean the exponent of two, or the number whose primality is being tested. Former - true; latter - false. Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: Re: Curtis' Miracle Octad Generator Victor, Have anything happen yesterday? haha === Subject: Mathematica code for an exponential sum graph I recently came across something I did several years ago that might be of interest to others. I used the code below as part of a practice with Mathematica worksheet for a calculus class once, and a lot of them thought it was neat. Rick Mabry still (then, and now) a novice with Mathematica. Push Enter after each of the following lines. Clear[whorl] wh[n_]=n^(4/3)/5.+3n^(2/3); whorl[0]=0; whorl[n_]:= whorl[n]=whorl[n-1]+Exp[2*Pi*I*wh[n]]; xy[z_]:={Re[z],Im[z]} nn=20000; w=Table[xy[whorl[n]],{n,1,nn}]; ListPlot[w,AspectRatio->Automatic,PlotRange->All]; Dave L. Renfro === Subject: Re: Mathematica code for an exponential sum graph > I recently came across something I did several > years ago that might be of interest to others. > I used the code below as part of a practice with > Mathematica worksheet for a calculus class once, > and a lot of them thought it was neat. Rick Mabry > still (then, and now) a novice with Mathematica. > Push Enter after each of the following lines. > Clear[whorl] > wh[n_]=n^(4/3)/5.+3n^(2/3); > whorl[0]=0; > whorl[n_]:= > whorl[n]=whorl[n-1]+Exp[2*Pi*I*wh[n]]; > xy[z_]:={Re[z],Im[z]} > nn=20000; > w=Table[xy[whorl[n]],{n,1,nn}]; > ListPlot[w,AspectRatio->Automatic,PlotRange->All]; > Dave L. Renfro A very nice plot indeed! There is an EXCELLENT book called Modern Differential Geometry of Curves and Surfaces with Mathematica, Second Edition by Alfred Gray. He has all of these beautiful curves and includes Mathematica code for them (maybe the code is posted somewhere). There are many excellent functions and canned examples and this book is a reminder of the tru beauty of Mathematics (I think it is an amazing piece of work). Flip === Subject: Re: Mathematica code for an exponential sum graph > A very nice plot indeed! > There is an EXCELLENT book called > Modern Differential Geometry of Curves and > Surfaces with Mathematica, Second Edition > by Alfred Gray. > He has all of these beautiful curves and includes > Mathematica code for them (maybe the code is > posted somewhere). I doubt the code is there because we (well, mostly Rick) came up with this ourselves. I wanted to reproduce the graph that Noam D. Elkies has at http://www.math.harvard.edu/~elkies/M259.02/whorls.html However, it turned out that a slightly different choice of parameters resulted in a better looking Mathematica graph, so the formula Elkies uses is not the same as the one in the Mathematica code I posted. Dave L. Renfro === Subject: Largest numbers are Fun! Friendly and Fun: Largest numbers are friendly and fun. What is a largest number? I will tell show you with an example. The largest number in this application is 6 - one, two, three, four, five, six. Wasn't that easy? One Fun thing and One Interesting thing: First, we will remember that Largest numbers occur only in their particular applications - like the one above! The next fun thing to note is that we cannot count a Largest number in, or out, of its application. So, we cannot say, 'there is one largest number in this application'. Why is this? This is another fun thing about largest numbers: If we count a largest number in an application then we are using another application. However, and this bit is interesting, if we step outside of the application where the largest number lives, and start a new application so that we can count this largest number, then all we end up counting is a numeral. Why is this? - I will tell you. It is because a number is only a number in its application. That bit was a little hard, but was certainly interesting and fun. An Interesting thing and an Exciting thing: If you have read the fun and interesting things above, then we can sum them up (with number's permission, of course!) into an exciting new thing about Largest numbers: Here it is: Largest numbers cannot be counted! There. I think that is very interesting and exciting. Don't you? But look! There is another interesting thing that Largest numbers do: if we cannot count Largest numbers, then Largest numbers don't live in collections, sets, groups, or families. Well, would you believe it? I certainly did not when I started writing this. A Sad thing, a Happy ending and an Interesting thing: But now I am sad. I don't know where Largest numbers live. I am sad because I think they do not have a home and have no friends. Why it seems they just appear and disappear, according to their applications... but wait.. a clue! Of course! Largest numbers live 'in their apllications' Yes, of course! This means that we have yet another interesting thing, but this time it's not about Largest numbers but about applications. What can it be.. It seems that 'applications' are completely independent of collections, sets, groups, and families. Well.. I think we shall come back to that one at another time. Don't you? More Things? Are there any other games that we can play with Largest numbers? I am sure there are many, and I can't wait to try! But first, I must post these fun and interesting things about Largest numbers, because I am so excited! So! Here they are! === Subject: Re: Largest numbers are Fun! > There is another interesting thing that Largest numbers > do: if we cannot count Largest numbers, then Largest numbers don't live > in collections, sets, groups, or families. Well, would you believe it? > I certainly did not when I started writing this. Why is this? There are plenty of uncountable things that exist in, for example, sets. I guess you just can't count on uncountable things to account for much, counter to common thought. :-) === Subject: Re: Largest numbers are Fun! John Jones a .8ecrit : > It is because a number is only a number in its application. Seriously...Does anyone understand what an application is in this context?? Cody === Subject: Re: Largest numbers are Fun! Phew, the happy ending was a relief. > Friendly and Fun: > Largest numbers are friendly and fun. What is a largest number? I will > tell show you with an example. The largest number in this application > is 6 - one, two, three, four, five, six. Wasn't that easy? > One Fun thing and One Interesting thing: > First, we will remember that Largest numbers occur only in their > particular applications - like the one above! The next fun thing to > note is that we cannot count a Largest number in, or out, of its > application. So, we cannot say, 'there is one largest number in this > application'. Why is this? This is another fun thing about largest > numbers: If we count a largest number in an application then we are > using another application. However, and this bit is interesting, if we > step outside of the application where the largest number lives, and > start a new application so that we can count this largest number, then > all we end up counting is a numeral. Why is this? - I will tell you. > It is because a number is only a number in its application. That bit > was a little hard, but was certainly interesting and fun. > An Interesting thing and an Exciting thing: > If you have read the fun and interesting things above, then we can sum > them up (with number's permission, of course!) into an exciting new > thing about Largest numbers: Here it is: Largest numbers cannot be > counted! There. I think that is very interesting and exciting. Don't > you? But look! There is another interesting thing that Largest numbers > do: if we cannot count Largest numbers, then Largest numbers don't live > in collections, sets, groups, or families. Well, would you believe it? > I certainly did not when I started writing this. > A Sad thing, a Happy ending and an Interesting thing: > But now I am sad. I don't know where Largest numbers live. I am sad > because I think they do not have a home and have no friends. Why it > seems they just appear and disappear, according to their > applications... but wait.. a clue! Of course! Largest numbers live 'in > their apllications' Yes, of course! This means that we have yet > another interesting thing, but this time it's not about Largest numbers > but about applications. What can it be.. It seems that 'applications' > are completely independent of collections, sets, groups, and families. > Well.. I think we shall come back to that one at another time. Don't > you? > More Things? > Are there any other games that we can play with Largest numbers? I am > sure there are many, and I can't wait to try! But first, I must post > these fun and interesting things about Largest numbers, because I am so > excited! So! Here they are! === Subject: Re: Largest numbers are Fun! : Friendly and Fun: : Largest numbers are friendly and fun. What is a largest number? I will : tell show you with an example. The largest number in this application : is 6 - one, two, three, four, five, six. Wasn't that easy? Let me know what you're smoking and where you got it; I've got to get myself some. Justin === Subject: Largest numbers are Fun! Friendly and Fun: Largest numbers are friendly and fun. What is a largest number? I will show you with an example. The largest number in this application is 6 - one, two, three, four, five, six. Wasn't that easy? One Fun thing and One Interesting thing: First, we will remember that Largest numbers occur only in their particular applications - like the one above! The next fun thing to note is that we cannot count a Largest number in, or out, of its application. So, we cannot say, 'there is one largest number in this application'. Why is this? This is another fun thing about largest numbers: If we count a largest number in an application then we are using another application. However, and this bit is interesting, if we step outside of the application where the largest number lives, and start a new application so that we can count this largest number, then all we end up counting is a numeral. Why is this? - I will tell you. It is because a number is only a number in its application. That bit was a little hard, but was certainly interesting and fun. An Interesting thing and an Exciting thing: If you have read the fun and interesting things above, then we can sum them up (with number's permission, of course!) into an exciting new thing about Largest numbers: Here it is: Largest numbers cannot be counted! There. I think that is very interesting and exciting. Don't you? But look! There is another interesting thing that Largest numbers do: if we cannot count Largest numbers, then Largest numbers don't live in collections, sets, groups, or families. Well, would you believe it? I certainly did not when I started writing this. A Sad thing, a Happy ending and an Interesting thing: But now I am sad. I don't know where Largest numbers live. I am sad because I think they do not have a home and have no friends. Why, it seems they just appear and disappear, according to their applications... but wait.. a clue! Of course! Largest numbers live 'in their apllications' Yes, of course! This means that we have yet another interesting thing, but this time it's not about Largest numbers but about applications. What can it be.. It seems that 'applications' are completely independent of collections, sets, groups, and families. Well.. I think we shall come back to that one at another time. Don't you? More Things? Are there any other games that we can play with Largest numbers? I am sure there are many, and I can't wait to try! But first, I must post these fun and interesting things about Largest numbers, because I am so excited! So! Here they are! === Subject: Andre Weil and his three conjectures For me, the Weil conjectures have to do with counting points on an algebraic variety V as the finite field F_q (q=p^n, where p is some prime) varies. Algebra and algebraic geometry were not my specialty in grad. school. I know that the first two conjectures were proven by Grothendieck, but that he didn't prove the third Weil conjecture (zeta function-type relation). The 3rd conjecture was proven by Pierre Deligne. What amazes me is that Weil could come up with these three conjectures in the first place, knowing how hard they were to prove. Are there any accounts of how Weil came to formulate these conjectures? David Bernier === Subject: Re: Andre Weil and his three conjectures > For me, the Weil conjectures have to do with counting points on an > algebraic variety V as the finite field F_q > (q=p^n, where p is some prime) varies. > Algebra and algebraic geometry were not my specialty in grad. school. > I know that the first two conjectures were proven by Grothendieck, but > that he didn't prove the third Weil conjecture (zeta function-type > relation). > The 3rd conjecture was proven by Pierre Deligne. > What amazes me is that Weil could come up with these three conjectures > in the first place, knowing how hard they were to prove. > Are there any accounts of how Weil came to formulate these conjectures? > David Bernier The short answer seems to be, by analogy with the Riemann zeta function and the Riemann hypothesis. By the late 1940's Weil had already established a different sort of analogy to the Riemann hypothesis, one for algebraic curves (algebraic function fields). http://en.wikipedia.org/wiki/Weil_conjectures In fact the case of curves over finite fields had been proved by Weil himself, finishing the project started by Hasse's theorem on elliptic curves over finite fields. The conjectures were natural enough in one direction, simply proposing that known good properties would extend. Their interest was obvious enough from within number theory: they implied the existence of machinery that would provide upper bounds for exponential sums, a basic concern in analytic number theory. === Subject: Re: Indra's Net of Pearls > I am not sure about the following, but it appears that Indra's Net, as a > fractal at least, is much more complex than Wada's triangle, since there is > an infinite 3D grid of pearls which gets reflected on each ball. Not just > three. I have placed one more close-up of the central Indra Pearl here, with the viewer positioned at (0.5,0.5,-0.5) http://misc.virtualcomposer2000.com/indra.jpg (125 KB) This was taken with the net containing 4913 pearls. Its complexity has increased phenomenally. -- Ioannis === Subject: Re: Indra's Net of Pearls oooOOOOOooooo! This is cool. Hey, Ioannis, here's some code for ya.... Brian ------------------------------ #include colors.inc // Standard colors library #include shapes.inc // Commonly used object shapes #include textures.inc // LOTS of neat textures. Lots of NEW textures. global_settings { max_trace_level 50 max_intersections 128 } //camera camera { location < 2, -2, 2> up y look_at <0, 0, 0> angle 25 } // Light source light_source { <0, 350, 0> color White } //floor plane { y, -3 pigment { checker color Green color Yellow } finish {Shiny} } sky_sphere { pigment { gradient y color_map { [ 0.5 color CornflowerBlue ] [ 1.0 color MidnightBlue ] } scale 2 translate -1 } } //Wada object #declare R = sqrt(2)/2; sphere { <-0.5,-0.5, 0.5>, R pigment {color Red} finish {Mirror} } sphere { < 0.5, 0.5, 0.5>, R pigment {color Green} finish {Mirror} } sphere { <-0.5, 0.5,-0.5>, R pigment {color Blue} finish {Mirror} } sphere { < 0.5,-0.5,-0.5>, R pigment {color Yellow} finish {Mirror} } -- http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html Quake predictions: http://www.skywise711.com/quakes/EQDB/index.html Sed quis custodiet ipsos Custodes? === Subject: Re: Indra's Net of Pearls > oooOOOOOooooo! This is cool. > Hey, Ioannis, here's some code for ya.... > Brian [snip code for brevity] Yeap. That's it. Looks beautiful. I can see the spheres. :-) -- Ioannis === Subject: Re: need help with differential equations > Or, numerically integrate matrix de, > d/dt Phi(t) = K*Phi(t) > on [0, t] with Phi(0) = I. The solution is exp(K*t). > What you say is true, but if the OP wanted numerical > results, then they might be better off numerically > integrating the first order system rather than the > matrix ODE. It would seem so, however the matrix exponential lets you generate the solution recursively, x(k+1) = Phi*x(k) for k = 0,1,... far more efficiently than direct numerical integration of the ode. Had it not been for these reasons, the Holy Grail of matrix exponentials - see, Nineteen dubious ways to calculate... twenty five years later - would have long since turned into a historical footnote. --- sdx http://www.sdynamix.com === Subject: Re: ulam's conjecture > does anyone know what it the status of this well-known problem? > my friend has just told me that he thinks he proved it. > http://mathworld.wolfram.com/UlamsConjecture.html > is there any connection between the class of NP-complete probems and > ulam's conjecture? I think it's also known as the Graph Reconstruction Conjecture. To the best of my knowledge, it is an open problem. No efficient algorithm is known for deciding whether two given graphs are isomorphic. The truth of Ulam's Conjecture would give you a way to test graph isomorphism, but it's not clear to me that it would give you an efficient way. I gladly yield to anyone who actually knows something about graph theory. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Distinct perfect squares with equal abundancy indices <20061203.1149653947558.JavaMail.jakarta@nitrogen.mathforum.org>, > Suppose gcd(A^2, B^2) = k, k >= 3. Then k divides A^2, and k also divides > B^2, which gives rise to the equations: > (a) A^2 = kx, for some positive integer x. > (b) B^2 = ky, for some positive integer y. > Does it follow that k is a perfect square? gcd(A^2, B^2) = (gcd(A, B))^2. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Calculus XOR Probability <446c8091$0$24200$88260bb3@free.teranews.com> <44732f25$0$24324$88260bb3@free.teranews.com> <447352c9$0$24287$88260bb3@free.teranews.com> <44748ee9$0$24320$88260bb3@free.teranews.com> I suggest you stick to cardinality when speaking of the standard theory, and >> not claim it has anything to do with the size of the set or number of elements. >> While you're at it, you have a term, transfinite. Why don't you stick to that >> too, instead of claiming omega has anything to do with actual infinity? David R Tribble said: >> Well, transfinite means beyond any finite, which is the same >> phrase you use for infinite, right? And omega is larger than any >> finite number, so it's infinite, even by your definition, right? > Except that I don't consider omega to be larger than any finite, since it's a > member of a set of finite values. Omega is not a member of N. Since N contains all the finite naturals, and omega is not in N, omega is not a finite natural. QED. Perhaps you were thinking of some other member of N, maybe your Alpha? > Remember, I reject the von Neumann ordinal > construction as having any bearing on the set of naturals, and instead consider > the naturals to be characaterized by the identity function between element > count and value. As Ross says, N e N. Isn't that right Ross? :) Then your set theory has no relation to standard arithmetic. === Subject: Re: Calculus XOR Probability <446c8091$0$24200$88260bb3@free.teranews.com> <44732f25$0$24324$88260bb3@free.teranews.com> <447352c9$0$24287$88260bb3@free.teranews.com> <44748ee9$0$24320$88260bb3@free.teranews.com> construction as having any bearing on the set of naturals, and instead consider > the naturals to be characaterized by the identity function between element > count and value. As Ross says, N e N. Isn't that right Ross? :) Hi Tony, I believe it is so, N E N, that the set N contains as an element itself. I know that is somewhat counterintuitive. Consider the Russell paradox, the set contains an element that is not proscribed by the predicate, P(x) => x is a finite non-negative integer, and compare to P(x) <=> x is a finite non-negative integer, the singular and dual implication. have Goedelian incompleteness. It is said that Paris and Kirby have a result that some true statements about the natural integers are provable only where there are infinite elements of the natural integers. C'est ca. The von Neumann ordinals are not a bad way to look at the natural integers, as an embodiment or form of the natural integers. Some mechanistic operations upon the natural integers are simple, having few necessary mechanistic operations on the elements, as von Neumann ordinals, and the results are adequate, ie exact. While that is so, there are a variety of other constructions of ordinals, with as well varying usages in the mechanistic and symbolic senses. For example, a notion of the null axiom theory is that powerset is order type is successor, that the ordinals are the powersets, representing combinatoric explosion in natural induction. Tony, a variety of others besides you and I obviously consider some unit infinity to be a suitable member of a number system, for where its application _is_ both intuitive and analytically perfect. For example, there was regular usage of such a notion for several thousands of years. Ross === Subject: Re: Calculus XOR Probability <446b791f$0$24231$88260bb3@free.teranews.com> David R Tribble said: >> My original question was how do you propose generating all the >> infinite subsets of N (of which there are an uncountable number of >> them). You still haven't come even close to answering that. > Actually I have many times, but you're completely resistant to the answer. Well, you'v certainly given many answers, but all of them were wrong, which might explain my resistance to them. > It takes an infinite number of iterations to bridge those gaps. You insist that > any sequence can only be countably long, but there is no justification for that > position. Obviously, if you start at 0, it takes ...0101 steps to get to > ...0101. You want to get to ...1010? Add another ...0101 steps to that. If your > set starts at 0, the odds are twice as many subsets from 0 as the evens. Right, you apply a countable number of increments to get to any particular subset. You just said so, it takes n steps to get there. That's what we mean by countable. You're counting each step, or increment, as you go. > You can only get to an infinite number from a finite through incrementation if you > apply an infinite number of increments, in which case you must reach an > infinite number. Isn't that close to answering that question? Does it sound > familiar? Yes, you've said this a jillion times, but no, it's not close to answering the question. You're saying that by applying a countable number of increments, that we somehow get an uncountable number of increments. === Subject: Re: Calculus XOR Probability Again, this is a typical Cantorian no largest finite argument for infinity. > The reason that argument is repeatedly used is no secret, It's because > it's *correct*, easy to follow and utterly convincing to anyone capable > of following a mathematical argument. Theorem. The number of finite naturals, |N|, is infinite. Proof. Assume that |N| is finite. Then there exists a finite natural m such that every member b in N is less than m, so b < m for all b in N. Since m is a finite natural, it is a member of N. Since m is a member of N, it is either equal to or less than the largest member g of N, so m <= g. So there is a member g in N such that g >= m, but b < m for all b in N, which is a contradiction. So our assumption is false, and |N| is not finite, so it is infinite. QED. === Subject: p-Laplacian Just wondering if people ever look at the p-Laplacian when p=1 or when p -> 1^+ ( p goes to 1 from right) ? Any references ? craig === Subject: Re: Discretization on PDEs > I'm currently studying partial differential equations (numerical methods) > and I understand that generally, problems may be too hard to solve > directly so you need to approximate them. In order to do so, you > discretize the problem and create an iterative method... But, is there a > way to approximate the problem without discretizing it? I am no expert, just a student in this as well, but I don't think so. The discretization reduces an infinite problem into a finite problem. That is, you only look for solution on these grid points. Even with Galerkin method, where you seek a solution as a linear combination of basis functions, you have to pick a finite subspace of the infinite subspace to develop an algorithm that will terminate in finite time. === Subject: Re: does a mean vector exist? 06/06/2006 >If there are infinitely many vectors, extending to an integral to get >a mean value ought to be analogous to the extension of a finite sum >to an integral. No, because you might have a countable set of vectors, in which case there is no reasonable way to define a mean value. >But there is a crucial difference: We usually consider and define >mean value of a single function over an interval! >Without some notion of an interval, I am not clear as to how to >extend... Well, if instead of taking a set of vectors you take a vector-valued function on a real interval, then you can integrate. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: infinite integral help Integral( exp( -abs(x0-x)) * exp( -abs(x) ) dx from -inf to +inf) It seems like splitting the integral up to remove the absolute values would help, but I can only split at 0 or x0, and not both. So it seems I will still have one factor with an absolute value. In case it helps, the answer I'm expecting this to be is (1 + |x|)exp(-abs(x)) === Subject: Re: infinite integral help > Integral( exp( -abs(x0-x)) * exp( -abs(x) ) dx from -inf to +inf) > It seems like splitting the integral up to remove the absolute values would > help, but I can only split at 0 or x0, and not both. What's the problem? Suppose x0 > 0. Integrate over (-oo, 0), (0, x0), (x0, oo) separately. > So it seems I will > still have one factor with an absolute value. Nope. > In case it helps, the answer I'm expecting this to be is (1 + > |x|)exp(-abs(x)) Well no, the answer has no x's in it. Maybe some x0's ... === Subject: Re: infinite integral help >> Integral( exp( -abs(x0-x)) * exp( -abs(x) ) dx from -inf to +inf) >> It seems like splitting the integral up to remove the absolute values >> would >> help, but I can only split at 0 or x0, and not both. > What's the problem? Suppose x0 > 0. Integrate over (-oo, 0), (0, > x0), (x0, oo) separately. I didn't include the (0, x0) originally, but when I added this, them I see === Subject: Re: line eq'n > In the slope-point formula of a line y - y1 = m(x - x1), what exactly > is the difference between y and y1 (and similarly x and x1). > Secondly, if you have a question where you're given two specific points > and have to determine the line, the procedure is to sub the points into > the slope formula to get the slope, and then re-sub the slope and one > of the points into the slope formula, and then rearrange to get the > equation. Let's say we have a slope of 8 and the point (3, 5). Then > the equation is 8 = (y - 5)/(x - 3), which rearranges to y = 8x - 19. > But shouldn't there be a restriction on x for the equation, namely x > =/= 3? But wouldn't the line be continuous, and so how can there be > any restriction? Besides, I've never seen any restriction associated > with linear equations? Generally, restrictions apply only to those values of x that will cause y to be undefined. As you can see, when the equation is rearranged; there are no restrictions. === Subject: Re: line eq'n > In the slope-point formula of a line y - y1 = m(x - x1), what exactly > is the difference between y and y1 (and similarly x and x1). Textbooks > I've seen tend to describe (x1, y1) and any particular point, but it > seems to me that (x, y) is also any particular point. Think of (x1, y1) as some _specific_ point (and hence x1 and y1 are constants) and think of (x, y) as a _general_ point (and hence x and y are variables). > Secondly, if you have a question where you're given two specific points > and have to determine the line, the procedure is to sub the points into > the slope formula to get the slope, and then re-sub the slope and one > of the points into the slope formula, and then rearrange to get the > equation. Let's say we have a slope of 8 and the point (3, 5). Then > the equation is 8 = (y - 5)/(x - 3), which rearranges to y = 8x - 19. > But shouldn't there be a restriction on x for the equation, namely x > =/= 3? A good question! Yes, those two equations aren't quite equivalent. Nonetheless, the second equation, in slope-intercept form, is correct, valid for all x. There's a nice way to avoid the problem you perceive: Given two specific distinct points, use the equation (y - y1)(x2 - x1) = (x - x1)(y2 - y1) It is always valid, even when the line is vertical. > But wouldn't the line be continuous, and so how can there be > any restriction? Besides, I've never seen any restriction associated > with linear equations? Right. David === Subject: Re: line eq'n <20060607192033.799$G7@newsreader.com A good question! Yes, those two equations aren't quite equivalent. > Nonetheless, the second equation, in slope-intercept form, is correct, > valid for all x. > There's a nice way to avoid the problem you perceive: Given two specific > distinct points, use the equation > (y - y1)(x2 - x1) = (x - x1)(y2 - y1) > It is always valid, even when the line is vertical. derived this equation from the slope definition, no? In that case, it had to start with (x2 - x1) and (x - x1) in the denominator position, i.e., (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) In that case, the equation would have the restrictions x2 =/= x1, and x1 =/= x. Thus, the problem that x =/= x1 remains (and thus be discontinuous at x1), and it wouldn't work for a vertical line at all because in a vertical line all x are equal. Please correct me if I'm wrong. === Subject: Re: line eq'n > A good question! Yes, those two equations aren't quite equivalent. > Nonetheless, the second equation, in slope-intercept form, is correct, > valid for all x. > There's a nice way to avoid the problem you perceive: Given two specific > distinct points, use the equation > (y - y1)(x2 - x1) = (x - x1)(y2 - y1) > It is always valid, even when the line is vertical. > derived this equation from the slope definition, no? Not at all. That equation can be derived directly from noting that the vectors [x2-x1, y2-y1] and [x-x1, y-y1] , in 3-space, must be parallel if (x,y) is to be on the line, so that their vector product analogue, (y - y1)(x2 - x1) - (x - x1)(y2 - y1) must be zero. This is quite independent of whether or not any slope is defined for the line, and works quite nicely for lines parallel to the y axis or in any direction whatsoever. Note that for a line parallel to the y-axis one has x1 = x2, but y1 <> y2, so the equation reduces to merely x = x1, with y arbitrary. > In that case, it > had to start with (x2 - x1) and (x - x1) in the denominator position, > i.e., > (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) > In that case, the equation would have the restrictions x2 =/= x1, and > x1 =/= x. Thus, the problem that x =/= x1 remains (and thus be > discontinuous at x1), and it wouldn't work for a vertical line at all > because in a vertical line all x are equal. Please correct me if I'm > wrong. You are and I did. === Subject: Re: line eq'n > A good question! Yes, those two equations aren't quite equivalent. > Nonetheless, the second equation, in slope-intercept form, is correct, > valid for all x. > There's a nice way to avoid the problem you perceive: Given two > specific distinct points, use the equation > (y - y1)(x2 - x1) = (x - x1)(y2 - y1) (*) > It is always valid, even when the line is vertical. > derived this equation from the slope definition, no? That could certainly be _part_ of the derivation. But ultimately, it's not important where (*) came from; that's because (*) is _not quite equivalent_ to, say, your (#) below. As you correctly said, (#) has restrictions; but my (*) does not. Perhaps we should consider a simpler example of the same sort of thing. Suppose that, for an answer to some problem, you were to get x/x. And you say to yourself: Hmm... x/x is not defined when x = 0, and yet I _know_, because of , that the answer must be 1 when x = 0. Well then, there's nothing wrong with saying, assuming that is actually a good reason, that we know that the answer must be 1 if x = 0; x/x otherwise. But then that answer _simplifies_ to just 1, _regardless of x_. > In that case, it > had to start with (x2 - x1) and (x - x1) in the denominator position, > i.e., > (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1) (#) > In that case, the equation would have the restrictions x2 =/= x1, and > x1 =/= x. Thus, the problem that x =/= x1 remains (and thus be > discontinuous at x1), and it wouldn't work for a vertical line at all > because in a vertical line all x are equal. Please correct me if I'm > wrong. You're wrong if you're saying that my (*) doesn't work when the line is vertical. Consider a simple example. Suppose we want an equation of the line through the points (5, 2) and (5, 6). Substituting in (*), we get (y - 2)(5 - 5) = (x - 5)(6 - 2) (y - 2) 0 = (x - 5) 4 0 = (x - 5) 4 0 = x - 5 5 = x David === Subject: Re: Combinatorial Probability Question >My apologies for reposting this. >Combinatorial Probability Question >Suppose there are M objects to be placed in N bins. What is the >probability distribution of the number of objects in the bin with the >largest number of objects? If K is the number, we say we want to find >P(K, M, N) for fixed M, N. >Clearly the number of possible ways of distributing the >objects is (M + N - 1)!/(M!(N - 1)!). Indeed. This is the number of nonnegative integer solutions to x_1 + x_2 + ... + x_N = M, indicating that you consider the bins to be distinguishable. >We have >P(M, M, N) = N/((M + N - 1)!/(M!(N - 1)!)) Not so fast, as Mr. Eastham points out. The correct probability is that of placing every object into the same bin, which is N^(1-M). Let's calculate the cumulative distribution function first. In that case, what we're looking is the probability that in distributing M objecting into N bins, every bin contains at most K objects. The number of ways to do that is the coefficient of x^M in the generating function F(x) = [1+x+...+x^K]^N = [1-x^{K+1}]^N/[1-x]^N. Expanding this into a series (sums over nonnegative indeces) F(x) = Sum_k[ C(N,k)(-1)^k(x^{K+1})^k ]Sum_m[ C(N+m-1,N-1)x^m ], so that the requisite coefficient should be a_M(K) = Sum_k[ (-1)^k C(N,k) C(N+M-k(K+1)-1,N-1) ]. (Note that the terms in the sum are zero for k > min{N,M/(K+1)}.) And so the cumulative probability is Q(K, M, N) = a_M(K)/[ C(M+N-1,N-1) ], while the distribution is P(K, M, N) = Q(K, M, N) - Q(K-1, M, N), K > 0. This might simplify, and there may be an easier method altogether, but I haven't spent much time on this. --- Stan Liou === Subject: Re: Combinatorial Probability Question Stan Liou nous a r.8ecemment amicalement signifi.8e : > Let's calculate the cumulative distribution function first. In that > case, what we're looking is the probability that in distributing M > objecting into N bins, every bin contains at most K objects. The > number of ways to do that is the coefficient of x^M in the > generating function F(x) = [1+x+...+x^K]^N = [1-x^{K+1}]^N/[1-x]^N. Oh yes, very nice ! I did not know this method. > Expanding this into a series (sums over nonnegative indeces) > F(x) = Sum_k[ C(N,k)(-1)^k(x^{K+1})^k ]Sum_m[ C(N+m-1,N-1)x^m ], > so that the requisite coefficient should be > a_M(K) = Sum_k[ (-1)^k C(N,k) C(N+M-k(K+1)-1,N-1) ]. > (Note that the terms in the sum are zero for k > min{N,M/(K+1)}.) I do agree > And so the cumulative probability is > Q(K, M, N) = a_M(K)/[ C(M+N-1,N-1) ], > while the distribution is > P(K, M, N) = Q(K, M, N) - Q(K-1, M, N), K > 0. > This might simplify, and there may be an easier method altogether, > but I haven't spent much time on this. Very nice. -- Patrick === Subject: Re: SF: Easy math, theory behind the solution | | > But all that money means that soon enough you will be facing more | > people, and then you will fall. | | But the time that you could be spending spinning your wheels futilely you | could be working, earning money, getting an education, etc. A Cajun man wants a job, but the foreman won't hire him until he passes a little math test. Here is your first question, the foreman said. Without using numbers, represent the number 9 Without numbers? The Cajun says, Dat is easy. And proceeds to draw three trees. What's this? the boss asks? Ave you got no brain? Tree and tree and tree make nine, says the Cajun Fair enough, says the boss. Here's your second question. Use the same rules, but this time the number is 99 The Cajun stares into space for a while, then picks up the picture that he has just drawn and makes a smudge on each tree. Ere you go. The boss scratches his head and says, How on earth do you get that to represent 99 Each of da trees is dirty now. So, it's dirty tree, and dirty tree, and dirty tree. Dat is 99 The boss is getting worried that he's going to actually have to hire this Cajun, so he says, All right, last question. Same rules again, but represent the number 100 The Cajun stares into space some more, then he picks up the picture again and makes a little mark at the base of each tree and says, Ere you go. One hundred The boss looks at the attempt. You must be nuts if you think that represents a hundred The Cajun leans forward and points to the marks at the base of each tree and says, A little dog come along and crap by each tree. So now you got dirty tree and a turd, dirty tree and a turd, and dirty tree and a turd, which makes one hundred. So, when I start? -- Ed. ----------------------------------------------------- hex->bin->b64 79DC1E8ABD40B42A2671AB2D0E8B4D7A === Subject: Re: SF: Easy math, theory behind the solution But all that money means that soon enough you will be facing more > | > people, and then you will fall. > | But the time that you could be spending spinning your wheels futilely you > | could be working, earning money, getting an education, etc. > A Cajun man wants a job, but the foreman won't hire him until he passes a > little math test. Here is your first question, the foreman said. Without > using numbers, represent the number 9 Without numbers? The Cajun says, > Dat is easy. And proceeds to draw three trees. Weird, in the version I know (which is exactly the same bar this change), it's an Irishman. Do Cajun folk talk like Irish folk, then? -- Larry Lard Replies to group please === Subject: Re: SF: Easy math, theory behind the solution >> | >> | > But all that money means that soon enough you will be facing more >> | > people, and then you will fall. >> | >> | But the time that you could be spending spinning your wheels futilely you >> | could be working, earning money, getting an education, etc. >> A Cajun man wants a job, but the foreman won't hire him until he passes a >> little math test. Here is your first question, the foreman said. Without >> using numbers, represent the number 9 Without numbers? The Cajun says, >> Dat is easy. And proceeds to draw three trees. > Weird, in the version I know (which is exactly the same bar this > change), it's an Irishman. Do Cajun folk talk like Irish folk, then? H's are hardly audible in any accent, really. Writing that joke, I would have considered Scottish. -- Andrew Poelstra < http://www.wpsoftware.net/blog > To email me, use apoelstra at the above address. I know that area of town like the back of my head. === Subject: Re: SF: Easy math, theory behind the solution > Easier than giving the surrogate factoring equations that must work is > explaing why they must. > This goes straight into the top 10, surely, because it's not just that > it's really funny, but also that it demonstrates just how ill-conceived > James' attitude to how maths works is. > ARRG... people gotta stop replying to him... he's not a scientist, he's > not that smart, well he is enough to get you guys all revved up... Heal thyself. You yourself just replied to James a few days ago. === Subject: Re: SF: Easy math, theory behind the solution Instead of explaining this over and over, don't you think it would save you a little time if you stepped through your method factoring some simple two-digit composite like 15 or 21? === Subject: Re: SF: Easy math, theory behind the solution > Instead of explaining this over and over, don't you think it would save > you a little time if you stepped through your method factoring some > simple two-digit composite like 15 or 21? The method works flawlessly. The problem in a nutshell is that in order to factor a number T you have to have g_1 and g_2, where (wait for it) g_1 * g_2 = T. That's right--it factors correctly *if* you already have the factors. Rick === Subject: Re: SF: Easy math, theory behind the solution you a little time if you stepped through your method factoring some > simple two-digit composite like 15 or 21? > The method works flawlessly. The problem in a nutshell is that in > order to factor a number T you have to have g_1 and g_2, where > (wait for it) g_1 * g_2 = T. > That's right--it factors correctly *if* you already have the factors. > Rick He's lying. Look at the stock market. The world is treading water waiting on the result of this debate. The answer is, I found an easy solution to the factoring problem. Math people lie all the time. They're lying about my solution. RSA is broken. James Harris === Subject: Re: SF: Easy math, theory behind the solution >> Instead of explaining this over and over, don't you think it would save >> you a little time if you stepped through your method factoring some >> simple two-digit composite like 15 or 21? >> The method works flawlessly. The problem in a nutshell is that in >> order to factor a number T you have to have g_1 and g_2, where >> (wait for it) g_1 * g_2 = T. >> That's right--it factors correctly *if* you already have the factors. >> Rick > He's lying. > Look at the stock market. > The world is treading water waiting on the result of this debate. > The answer is, I found an easy solution to the factoring problem. > Math people lie all the time. They're lying about my solution. > RSA is broken. > James Harris Note: Whan James says people lie, it more than likely means he doesn't understand what they're talking about. Dave === Subject: Re: SF: Easy math, theory behind the solution Discussion, linux) Instead of explaining this over and over, don't you think it would save >> you a little time if you stepped through your method factoring some >> simple two-digit composite like 15 or 21? >> The method works flawlessly. The problem in a nutshell is that in >> order to factor a number T you have to have g_1 and g_2, where >> (wait for it) g_1 * g_2 = T. >> That's right--it factors correctly *if* you already have the factors. >> Rick > He's lying. > Look at the stock market. > The world is treading water waiting on the result of this debate. Are you suggesting that the stock market is reacting to your sci.math posts? That is, that at least *some* investors are aware of your surrogate factoring research and this awareness partly explains their investment behavior? -- Jesse F. Hughes There's a thrill that's gone that I'll probably not have in quite the same way again. After all, FLT was a unique animal, and we had a great dance. -J.S. Harris on proving Fermat's last theorem === Subject: Re: SF: Easy math, theory behind the solution explaing why they must. >Consider that given >w + x - 2z = f_1 >w + 3x + 2y + 2z = f_2 >w + x + y + z = g_1 >3w + x - y - 3z = g_2 > using the 4 linear equations you get >z = (g_2 - g_1 + f_2 - 3f_1)/4 > Lets examine that, shall we? > z = (3w + x - y - 3z - w - x - y - z +w +3x +2y +2z - 3w - 3x +6z)/4 > z = (0w + 0x + 0y + 4z) / 4 > z = 4z / 4 > z = z So what? > You zero out the other terms to find out that z is equal to itself. How could > the world have overlooked your brilliance. >So why do I make a post like this? > Rather than repeatedly showing us your equations, how about you actually use > them to factor something? === Subject: Re: SF: Easy math, theory behind the solution >>Easier than giving the surrogate factoring equations that must work is >>explaing why they must. >>Consider that given >>w + x - 2z = f_1 >>w + 3x + 2y + 2z = f_2 >>w + x + y + z = g_1 >>3w + x - y - 3z = g_2 >> With these particular equations it turns out that if you solve for z >>using the 4 linear equations you get >>z = (g_2 - g_1 + f_2 - 3f_1)/4 >> Lets examine that, shall we? >> z = (3w + x - y - 3z - w - x - y - z +w +3x +2y +2z - 3w - 3x +6z)/4 >> z = (0w + 0x + 0y + 4z) / 4 >> z = 4z / 4 >> z = z >So what? z = (g_2 - g_1 + f_2 - 3f_1)/4 and if g_1 g_2 = T, your target, then it is obvious enough that part of the solution for z is a difference of factors of the target How exactly does the equation z = z help us find z? === Subject: Re: SF: Easy math, theory behind the solution >>Easier than giving the surrogate factoring equations that must work is >>explaing why they must. >>Consider that given >>w + x - 2z = f_1 >>w + 3x + 2y + 2z = f_2 >>w + x + y + z = g_1 >>3w + x - y - 3z = g_2 >> With these particular equations it turns out that if you solve for z >>using the 4 linear equations you get >>z = (g_2 - g_1 + f_2 - 3f_1)/4 >> Lets examine that, shall we? >> z = (3w + x - y - 3z - w - x - y - z +w +3x +2y +2z - 3w - 3x +6z)/4 >> z = (0w + 0x + 0y + 4z) / 4 >> z = 4z / 4 >> z = z >So what? > z = (g_2 - g_1 + f_2 - 3f_1)/4 > and if g_1 g_2 = T, your target, then it is obvious enough that part of > the solution for z is a difference of factors of the target > How exactly does the equation z = z help us find z? It doesn't. It shows that the equation was (probably) solved correctly for z. --- Christopher Heckman === Subject: Presidential Aproval Raitings & The Stock Market Have you noticed a link between the presidents aproval ratings and the stock market? I have noticed that as the aproval ratings go down the interest rates have been going up. I don't think the president is trying to hurt the economy I just feel it is inevetable that lower aproval ratings aren't good for the economy. If we all gave bush two thumbs up I think it could really help even the average American. === Subject: Probability Incidence of TB in the US is 5 cases per 10,000. A test for TB is 99% accurate, that is it gives 1% of false positive (positive results for a healthy person) and 1% of false negative (negative result for a sick person). Find the probability that a person who tested positively actually has TB. Why is the probability less than 5%? === Subject: Re: Probability > Incidence of TB in the US is 5 cases per 10,000. A test for TB is 99% > accurate, that is it gives 1% of false positive (positive results for a > healthy person) and 1% of false negative (negative result for a sick > person). Find the probability that a person who tested positively actually > has TB. > Why is the probability less than 5%? consider 10,000 cases, of which 5 cases have TB, 9,995 without. Of the 5 TB cases 4.95 (positive) .05 (negative) Of the 9,995 no-TB cases 99.95 (positive) 9,900.05 (negative) The probability that a person tested positive actually had TB is 4.95/(99.95+4.95) = 0.04719. The above is the consequence of Bayes' Theorem on conditional probability. The tabular form in COUNTS make it much easier to see the reason without using any conditional probability of Bayes's formula. Standard material in a First Course in Probability and Statistics. -- Bob. === > For the sake of humanity please stop replying to this troll!!!!!!!!!! This > is really pathetic. I killfiled JSH a long time ago. The problem is that (as someone else pointed out) you still get the replies, which are almost as annoying. The better strategy would be for people replying to Harris to add the three letters JSH into the subject field. People completely uninterested in anything to do with JSH (such as me) can create message rule to not download anything with these three letters in the subject (as I have). If everybody could remember to do this, everybody would be happy (except possibly JSH himself). If you want to reply to this post, please remove the JSH from the subject or I won't see it - a discussion about how to completely ignore JSH is about the only conversation I want to read on this subject ... Peter Webb === >> For the sake of humanity please stop replying to this troll!!!!!!!!!! >> This is really pathetic. > I killfiled JSH a long time ago. The problem is that (as someone else > pointed out) you still get the replies, which are almost as annoying. > The better strategy would be for people replying to Harris to add the > three letters JSH into the subject field. People completely uninterested > in anything to do with JSH (such as me) can create message rule to not > download anything with these three letters in the subject (as I have). If > everybody could remember to do this, everybody would be happy (except > possibly JSH himself). > If you want to reply to this post, please remove the JSH from the subject > or I won't see it - a discussion about how to completely ignore JSH is > about the only conversation I want to read on this subject ... > Peter Webb I guess I'll try that. The problem is that not only do the JSH drop out of the replies a lot of times he will change it to SF or something else ;/ The only true way to fix this problem is for everyone to stop replying to him. Why not a special newsgroup just for him? All those that love Mr. Harris can visit there and discuss all the intricacies of is genius, his proofs on FLT, his factoring methods, etc... It would be much nicer than filling up this group with garbage? (which it surely is?) Not that a little garbage hurts but it gets kinda ridiculous. I count ~25 original posts by him from 5/13 to 6/4. This doesn't include all the replies which seems to grow exponentially. It wouldn't be so bad if he actually was asking for help but instead he berates us and claims we are in conspiracy against him. He then claims that he's wrong and apologies but continues to do the same thing. He doesn't want help he just wants attention. Surely there is no place for that in this newgroup(atleast on the scale of Mr. Harris). I guess I could make some complex kill file system that will scan the message body for anything remotely related to his posts and hope that takes care of it. Jon === > I killfiled JSH a long time ago. The problem is that (as someone else > pointed out) you still get the replies, which are almost as annoying. > The better strategy would be for people replying to Harris to add the > three letters JSH into the subject field. People completely uninterested > in... Have your butler screen the threads. === Subject: Re: Understanding the pathology >> On Wed, 07 Jun 2006 05:43:56 +1000, David Eather secret message would be typical of a psychopath. >> No, that's scizophrenia. >> A psychopath is someone who has no empathy with others, and is capable >> of committing murder without any particular reason or provocation. > Is an order to kill someone considered a valid reason or provocation? > Do assassins get off on a technicality simply because they are employed > for that very purpose? It's called the Nuremberg defense (I was just following orders)- it didn't work very well when it was first wheeled out. > Steve === Subject: Re: Understanding the pathology <5ydhg.382$Ye6.221@fe04.lga> <4485db05@dnews.tpgi.com.au> <44870331.3642154@news.usenetzone.com> <4487a006@dnews.tpgi.com.au> On Wed, 07 Jun 2006 05:43:56 +1000, David Eather secret message would be typical of a psychopath. >> No, that's scizophrenia. >> A psychopath is someone who has no empathy with others, and is capable >> of committing murder without any particular reason or provocation. > Is an order to kill someone considered a valid reason or provocation? > Do assassins get off on a technicality simply because they are employed > for that very purpose? > It's called the Nuremberg defense (I was just following orders)- it > didn't work very well when it was first wheeled out. Seems to work great today, although the people who use it also take pains to recapitulate the motives for their actions in a favourable light by way of excluding data that contradicts their a priori judgements. But anyhow I was asking whether assasins are exempt from the psychopath (or sociopath) label because their employers task them with killing, not whether they are protected by some bull concept like `unlimited liability'. Whether or not particular groups are given the 'legal right' to kill (steal, maim, whatever) is immaterial to the classification of their psychology. Steve === Subject: Re: Understanding the pathology >> On Wed, 07 Jun 2006 05:43:56 +1000, David Eather secret message would be typical of a psychopath. >> No, that's scizophrenia. >> A psychopath is someone who has no empathy with others, and is capable >> of committing murder without any particular reason or provocation. > Is an order to kill someone considered a valid reason or provocation? > Do assassins get off on a technicality simply because they are employed > for that very purpose? >> It's called the Nuremberg defense (I was just following orders)- it >> didn't work very well when it was first wheeled out. It doesn't work that well as a psyc. defense for many (not all) of the people who had to carry out the order. > Seems to work great today, although the people who use it also take > pains to recapitulate the motives for their actions in a favourable > light by way of excluding data that contradicts their a priori > judgements. > But anyhow I was asking whether assasins are exempt from the psychopath > (or sociopath) label because their employers task them with killing, > not whether they are protected by some bull concept like `unlimited > liability'. Whether or not particular groups are given the 'legal > right' to kill (steal, maim, whatever) is immaterial to the > classification of their psychology. There is a lot of research on the subject, some of the best (least stylized / most stark) comes from WW1. Considering the current state of the world, I think that it is not a debate that could be done on sci.crypt or sci.math without causing a lot of damage. If you wanted to look on your own you might start looking for phrases like the talking cure and the elephant or seen/saw the elephant if you find the right ones you should be dumped in a pile of information that will fill out most of the details if you dig. You could try searches for shell shock or post traumatic stress syndrome but they may actually miss exactly what you want. Then again I've been a little foggy in the brain for at least a few days - so I could be well off the mark. > Steve === Subject: Re: Understanding the pathology > On Wed, 07 Jun 2006 05:43:56 +1000, David Eather Things like believing the number plates on cars are giving you a >> secret message would be typical of a psychopath. > No, that's scizophrenia. Sorry. You are correct of course - i was just having a little break myself. > A psychopath is someone who has no empathy with others, and is capable > of committing murder without any particular reason or provocation. > John Savard > http://www.quadibloc.com/index.html > _________________________________________ > Usenet Zone Free Binaries Usenet Server > More than 140,000 groups > Unlimited download > http://www.usenetzone.com to open account === Subject: Re: Understanding the pathology >> On Wed, 07 Jun 2006 05:43:56 +1000, David Eather secret message would be typical of a psychopath. >> No, that's scizophrenia. >> A psychopath is someone who has no empathy with others, and is capable >> of committing murder without any particular reason or provocation. > Is an order to kill someone considered a valid reason or provocation? > Do assassins get off on a technicality simply because they are employed > for that very purpose? > Steve Doing things for money is considered to be sane. Assassination is naughty but sane. Andrew Swallow === Subject: Re: ? derive wave eqn with energy lost > ... > Of course energy goes somewhere. When describing the solution using > complex refractive index, no mention of where it went was included, > just that there was attenuation of the EM beam. It can go to heat, it > can just be light scattered out of the beam, or it can go into exciting > an atom or molecule, when could then fluoresce away, at the same or > different wavelength, some energy going into heat. >> Then we have heat transfer eqn: a*del(T) = div( b*grad(T) )+Q. >> Q = f(E) =? >> strange >> in two ways: 1) energy is then a vector, 2) energy is complex. >> In another book this EM energy is written as: 0.5*dot( J, E ) = >> 0.5*dot( >> sigma*E, E ), >> which is the same as Born's book. But my questions remain. What's wrong? >> .... > Energy is in different forms. In EM, you are showing resistive > heating. Looking into the Poynting vector. Energy flow does have a > direction. It is not complex. > I don't know what question remains? Or what is wrong? Seems that I forgot the dot operator is defined as: dot( a, b ) = conj( transpose(a) )*b = conj(a1)*b1+conj(a2)*b2+conj(a3)*b3 So, 0.5*dot( J, E ) = 0.5*dot( sigma*E, E ) = 0.5*sigma*sum( conj(E_i)*E_i, i = 1 to 3 ) = 0.5*sigma*dot( E, E ). Similarly, Q = 0.5*sigma*dot( E, conj(E) ) only need to be 0.5*sigma*dot( E, E ). But Born's notation is still strange. 0.5*sigma*vector(E)^2 = 0.5*sigma*( E_i^2*vector(i)+E_j^2*vector(j)+E_k^2*vector(k) ) This is a vector, and its magnitude square = 0.5*sigma*( E1^2+E2^2+E3^2 ) When E_i, E_j and E_k are complex, E_i^2 /= conj(E_i)*E_i. Why don't I get Q = 0.5*dot( J, E ) = 0.5*sigma*dot( E, E ), which is a real scalar? by Cheng Cosine Jun/08/2k6 NC === Subject: Re: ? derive wave eqn with energy lost > If you assume an exp(-i*w*t) time dependence, you can reduce the wave > equation to Helmholtz's equation, i.e., del^2(E)+k^2*E = 0 (homogeneous > equation). 'k' is the wavenumber, with value 2*pi*n/Lambda, where 'n' > is the index of refraction, 'Lambda' the wavelength in vacuum. To add > either absorption or gain, you make the refractive index complex. In 3D, Green's function of the above is: G(r) = exp( sqrt(-1)*k*r )/( 4*pi* r ). So vector(E_Helmholtz) = G(r)*unit directional vector. But a static charge in free space has e-field as vector(E) = 1/( 4*pi*r^2 )*unit directional vector. In steady state, vector(E_Helmholtz; k = 0) = G(r; k = 0)*unit directional vector = 1/( 4*pi* r )*unit directional vector. The powers of r are different, why? by Cheng Cosine Jun/08/2k6 NC === Subject: Re: ? derive wave eqn with energy lost <4xNhg.50504$Lg.21576@tornado.southeast.rr.com If you assume an exp(-i*w*t) time dependence, you can reduce the wave > equation to Helmholtz's equation, i.e., del^2(E)+k^2*E = 0 (homogeneous > equation). 'k' is the wavenumber, with value 2*pi*n/Lambda, where 'n' > is the index of refraction, 'Lambda' the wavelength in vacuum. To add > either absorption or gain, you make the refractive index complex. > In 3D, Green's function of the above is: G(r) = exp( sqrt(-1)*k*r )/( 4*pi* > r ). > So vector(E_Helmholtz) = G(r)*unit directional vector. > But a static charge in free space has e-field as vector(E) = 1/( > 4*pi*r^2 )*unit directional vector. > In steady state, vector(E_Helmholtz; k = 0) = G(r; k = 0)*unit directional > vector > = 1/( 4*pi* r )*unit directional vector. > The powers of r are different, why? > by Cheng Cosine > Jun/08/2k6 NC This is an interesting question. You might consider posting in at sci.physics or some science site. I'm only going to show you how you can figure it out. Take a look at an EM book, specifically the vector and scalar potentials, their general solutions in terms of current and charge, respectively, convolved with the Green's function. E is more than the gradient of the scalar potential, it has another term, the time derivative of the vector potential, while B is the curl of the vector potential. Look at it for DC, and AC sources. In addition, give some thought to a single fixed charge vs. a dipole, which is charge neutral, i.e., charge changes come in pairs. === Subject: Re: anybody got an IQ over 200 here? <4ecrl8F1efu6iU1@individual.net> <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> === Subject: Re: anybody got an IQ over 200 here? http://www.aceviper.net/aceviper_net/ace_intelligence/aceviper_famous_people _ iq_list/aceviper_famous_people_iq_list.html Question: What do Hillary Clinton, Madonna, and Adolf Hitler have in > common? Answer: The same IQ > Correction: Hitler's IQ was 1 point higher. > This list of I.Q.'s clearly shows that George W. Bush's I.Q. is on a > par with Abraham Lincoln's and Andrew Jackson's. It is, moreover, > clearly superior to George Washington's, John F. Kennedy's, and Ulysses > S. Grant's. But that was a list laundered by the White House of the current administration. :-) George W. Bush's IQ is comparable to Andy Warhol's except Bush never produced a 25 hour movie of the same scene in Manhattan. Bush has staged a 25 month theatre of the absurd in Iraq though. -- Bob. > tomcat === Subject: Re: anybody got an IQ over 200 here? That was a doddle compared to the neat trick of getting into the National Guard during the Vietnam War. Hum, Bush was the only person to get accepted to a National Guard unit from 1963 through 1973? Interesting. === Subject: Re: anybody got an IQ over 200 here? > That was a doddle compared to the neat trick of getting > into the National Guard during the Vietnam War. > Hum, > Bush was the only person to get accepted to a National > Guard unit from 1963 through 1973? Interesting. Were you there? === Subject: Re: anybody got an IQ over 200 here? David, Please don't muddle political correctness with logic or facts! You'll throw the PC crowd into a tailspin. > Like this one better? > http://www.truthorfiction.com/rumors/p/presidentialiq.htm > 147 Franklin D. Roosevelt (D) > 132 Harry Truman (D) > 122 Dwight D. Eisenhower (R) > 174 John F. Kennedy (D) > 126 Lyndon B. Johnson (D) > 155 Richard M. Nixon (R) > 121 Gerald Ford (R) > 175 James E. Carter (D) > 105 Ronald Reagan (R) > 099 George HW Bush (R) > 182 William J. Clinton (D) > 091 George W. Bush (R) > The non-partisan researchers who evaluated the twelve presidents Non-partisan. It was a panel of college professors, right? > President G. W. Bush was rated the lowest of all the Republicans with > an IQ of 91. The six Democrat presidents had IQs with an average of 156, Bet no one saw that coming. > Bush's IQs were challenged, but I don't recall whether the complaint > was that they were estimated too high or too low. Har. Neat trick, graduating from Yale with a below-average IQ. Or even meeting the entrance requirements. Dubya made a 1206 on his SAT (566v+640m, before they dumbed down the scale in 1994), which roughly translates to an IQ score of 123 (http://www.gnxp.com/MT2/archives/002360.html) to 129 (http://members.shaw.ca/delajara/Pre1974SAT.html). Clinton's supposed IQ of 182 is 5.5 standard deviations above the norm, meaning he would be expected to have easily made a perfect 1600 on his SAT score. He reputedly has an actual IQ closer to 135, which puts him somewhere around a 1320~1350 SAT score. === Subject: Re: anybody got an IQ over 200 here? attributable to historical personages: > Estimated how? By picking numbers out of a hat? The thoroughly discredited study by Cox (1926) used the number of lines devoted to the biography of each person in a standard reference book to estimate each person's IQ. This methodology is not replicable on current populations, and has been completely abandoned. See Shurkin, Joel N. (1992). Terman's Kids: The Groundbreaking Study of How the Gifted Grow Up. Boston: Little, Brown. for full details. > Of all the silly things I've seen written about IQ's, this list has to be one > of the silliest. Silly ideas about IQ, alas, are more commonplace than sound ideas. I'd appreciate comments from sci.math readers to what I have drafting about IQ tests for a forthcoming publication below: IQ scores are not a measure of anything. Psychologists have consistently conceived of IQ tests as ñmeasures.î For the first edition of his manual for the Stanford-Binet scale under the title The Measurement of Intelligence (a title used by more recent authors for their books on IQ testing) later making his title slightly less definite by changing it to (1937) Measuring Intelligence. Even the authors who have noted the Flynn effect, including James R. Flynn himself, have adopted this terminology and written under such titles as ñMassive IQ Gains in 14 Nations: What IQ Tests Really Measureî (Flynn 1987) or The Rising Curve: Long-Term Gains in IQ and Related Measures (Neisser 1998). All these titles assume that IQ tests ñmeasureî something, and some amateur writers on psychology specifically invite readers to analogize IQ scores to measurements on a yardstick (Shaine 2001). But professor of psychology Joel Michell points out that it took hundreds of years of scientific and philosophical development of the concept of measurement to establish the foundations of such simple measurements as length and weight, much more conjoint measurements such as density, and none of that work has been done for most psychological test scales (Michell 1999). Indeed, tests of mental ability are based on an untested theory of ñmeasurementî unrelated to any view of measurement used outside the field of psychology (Michell 1999). The best considered view is that IQ tests and other mental ability tests are not measures at all. Certainly they should not be analogized to rulers. An IQ score, despite the analogies appealed to by some authors, is not like a marking on a ruler related to an absolute scale but is on an ordinal scale (Mackintosh 1998, pp. 30-31). Psychologists, in a terminology that is largely confined to the field of psychology, speak of four kinds of scales that test scores can be reported on: nominal scales, ordinal scales, interval scales, and ratio scales. A nominal scale labels results according to distinct categories--the common example is distinguishing male and female test-takers. An ordinal scale sorts tests are ordinal scales. An interval scale has a measurement unit that is comparable in extent along all parts of the scale--many psychologists assume, without proof, that IQ scales are interval scales. A ratio scale, in psychological terminology, is a scale with a meaningful zero point such that any two scores can be compared by multiplication or division. The physical scales of length or of temperature are ratio scales, and one can meaningfully speak of an object being twice as long as another, or a temperature reading being twice as hot as another. IQ cannot be referred to an ñabsolute zero,î and thus is surely not a ratio scale. (It is very unfortunate that one large group of authors on gifted education refer to ñconventional IQsî as ñratio IQs,î which is a mistaken giftedness quote statements such as that ña profoundly gifted child of IQ 190 differs from his or her moderately gifted classmate of IQ 130 to the same degree that the latter differs from an intellectually handicapped child of IQ 70î (Gross 2000), such a statement is simply an incorrect inference caused by a misunderstanding of the mathematical nature of the IQ scale (Mackintosh 1998, pp. 30-31; Truch 1993, p. 35; Eysenck 1998, pp. 24-25; see also http://www.carleton.ca/~jlogan/PSYC2001/data.pdf for more discussion of IQ scores as scores from an ordinal scale). Alfred Binet warned against this error early on, writing, ñThis scale properly speaking does not permit the measure of the intelligence, [1] because intellectual qualities are not superposable, and therefore cannot be measured as linear surfaces are measured, but are on the contrary, a classification, a hierarchy among diverse intelligences;î (Binet 1905, English translation 1916). Lewis Terman (designer of the Stanford-Binet IQ test) was well aware of the arbitrary nature of IQ scoring and denied that IQ scores had interval characteristics. > The expression of a test result in terms of age norms is simple and > unambiguous, resting upon no statistical assumptions. A test so scaled does not > pretend to measure intelligence as a linear distance is measured by the equal > units of a foot-rule, but tells us merely that the ability of a given subject > corresponds to the average ability of children of such and such an age. This was > all that Binet claimed to accomplish, and one can well doubt whether the > voluminous output of psychometric literature since his day has enabled us to > accomplish more. We have accordingly chosen to retain this least pretentious of > units for the estimation of mental level. (Terman & Merrill 1937, p. 25) > Finally, the reader should understand that the mental levels we have added > beyond fifteen, where mental ability as measured by this scale practically > ceases to improve, are not true ñmental ages,î but purely arbitrary scores > designed to permit the computation of I.Q.'s for adult subjects of superior > ability. The magnitude of these units in score points has been adjusted in > such a way as to give approximately the same distribution of I.Q.'s for adult > subjects as is found for children. The adjustment assumes that in terms of an > absolute scale (if we had one), the distribution of brightness of adult > subjects would follow the same curve as the distribution for unselected > children of a given age. (Terman & Merrill 1937, p. 26) > In this connection it should be noted that mental ages above thirteen years > cease to have the same significance as at lower levels, since they are no > longer equivalent to the median performances of unselected populations of the > corresponding chronological ages. . . . Beyond fifteen, of course, mental > ages are artificial and are to be thought of as simply numerical scores. > (page 31) A very thorough critique of the approach that psychologists have taken in rewriting the standards for ñmeasurementî in science can be found in Michell (1999). An application of MichellÍs critique to IQ testing can be found in Barrett (2000). As usual, James Flynn expressed the matter very thoughtfully. He thinks, as is clear from his writings, that IQ tests are fairly reliable in sorting people into rank orders on the basis of a correlate of intelligence that is tapped by the tests. In other words, he agrees with most psychologists that the rank orderings produced by IQ tests are reasonable, and that they reflect a genuine rank ordering of real-world differences in intelligence. But Flynn points out that when you try to treat IQ scores as magnitude ñmeasuresî rather than as ordinal rankings, they may not correlate well with intelligence. In his elegant phrasing (Flynn 1987), > Imagine that we could not directly measure the population of cities but had > to take aerial photographs, which gave a pretty good estimate of area. In > 1952, ranking the major cities of New Zealand by area correlated almost > perfectly with ranking them by population, and in 1982, the same was true. > Yet anyone who found that the area of cities had doubled between 1952 and > 1982 would go far astray by assuming that the population had doubled. The > factors that intervene, such as central city decay, affluent creation of > suburbs, and more private transport, all of which can expand the city's area > without the help of increased population. REFERENCES Barrett, Paul (2000). Intelligence, Psychometrics, Iq, G, and Mental Abilities: Quantitative Methodology Dressed as Science. Psycoloquy, 11, No. 46. Received May 12, 2006 from http://psycprints.ecs.soton.ac.uk/archive/00000046/ Binet, Alfred. (1916). New methods for the diagnosis of the intellectual level of subnormals. In E. S. Kite (Trans.), The Development of Intelligence in Children. Vineland, NJ: Publications of the Training School at Vineland. (Originally published 1905 in L'Ann.8ee Psychologique, 12, 191-244.) (Available on the Web at Classics in the History of Psychology site) http://psychclassics.yorku.ca/Binet/binet1.htm Eysenck, Hans (1998). Intelligence: A New Look. New Brunswick: Transaction Publishers. Flynn, James R. (1984). The Mean IQ of Americans: Massive Gains 1932 to 1978. Psychological Bulletin vol. 95, pages 29-51. Gross, Miraca U. M. (2000) Exceptionally and Profoundly Gifted Students: An Underserved Population. Understanding Our Gifted, 12, (2) pp. 3-9, (Winter 2000). Available on the Web at the Hoagies Gifted Web site http://www.hoagiesgifted.org/underserved.htm Mackintosh, N. J. (1998). IQ and Human Intelligence. Oxford: Oxford University Press. Michell, Joel (1999). Measurement in Psychology: Critical History of a Methodological Concept. Cambridge: Cambridge University Press. Neisser, Ulric (Ed.) (1998). The Rising Curve: Long-Term Gains in IQ and Related Measures. Washington, DC: American Psychological Association. Shaine, Josh (2001). Those IQ Tests: Their Strengths and Weaknesses. Home EducatorÍs Family Times, 9 (April 2001). Retrieved May 8, 2006 from Spearman, Charles (1904). ñGeneral Intelligence,î Objectively Determined and Measured. American Journal of Psychology, 15, 201-293. (Available on the Web) http://psychclassics.yorku.ca/Spearman/ Terman, Lewis & Merrill, Maude (1937). Measuring Intelligence: A Guide to the Administration of the New Revised Stanford-Binet Tests of Intelligence. Boston: Houghton Mifflin. Truch, Steve (1993). The WISC-III(R) Companion: A Guide to Interpretation and Educational Intervention. Austin, TX: Pro-Ed. -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> <5CIhg.2788$o4.2126@newsread2.news.pas.earthlink.net> A rather lengthy scholarly essay, citing many references, but lacking some of the factual basis! > I'd appreciate comments from sci.math readers to what I have drafting > about IQ tests for a forthcoming publication below: > IQ scores are not a measure of anything. That is the silliest of all statements ever made on any score! While IQ scores may not be a valid or reliable measure of intelligence, it certainly is a well-defined score of test items measuring many basic factors in intelligence, such as vocabulary, quantitative reasoning, spatial visualization, logical deduction and inference, analogy, and many other facets of intelligence. The raw scores are normed according to a Gaussian scale to reflect the percentile standing of a subject's score relative to those of the general (adult) population. > The best > considered view is that IQ tests and other mental ability tests are not measures > at all. Certainly they should not be analogized to rulers. See the preceding paragraph. Each IQ test has its own normed ruler, by the normal probability distribution scale. > Psychologists, in a terminology that is largely > confined to the field of psychology, speak of four kinds of scales that test > scores can be reported on: > nominal scales, > ordinal scales, > interval scales, and > ratio scales. These scales are UNIVERSALLY understood -- the first three are standard measurement scales used in ALL statistical studies. > An ordinal scale sorts > tests are ordinal scales. That can be PROVEN to be false, based on the actual history of the construction of IQ scores! In the early days of Stanford Binet tests, IQ was based on a ratio scale (hence Quotient) for children. That was how Marilyn Vos Savant was rated to have an IQ of 228, based on her test score at 10 years of age, as comparable to that of a 22 year old. That is a RATIO scale. Later, IQ scores for adults were based on INTERVAL scales. > An interval scale has a measurement unit that is > comparable in extent along all parts of the scale--many psychologists > assume, without proof, that IQ scales are interval scales. It's not a matter of assumption. That is the way IQ scores ARE scored, in discretised version of the continuous scale of a normal random variable. Your references are rather dated and lacking the backing of those who construct and score IQ tests!! > It is very unfortunate that one large group of authors on > gifted education refer to conventional IQs as ratio IQs, which is a mistaken > usage in this context. Since Marilyn Vos Savant's IQ score has been sensationalized by the Guiness World Record, you should do a little research and can easily find documented evidence of exactly HOW her score was derived from the RATIO scale! > Lewis Terman (designer of the Stanford-Binet IQ test) was well aware of the > arbitrary nature of IQ scoring and denied that IQ scores had interval > characteristics. At what time frame? Stanford-Binet had RATIO scale for children, and NOW have interval scale in a Gaussian distribution with mean 100 and standard deviation between 15 and 16. > We have accordingly chosen to retain this least pretentious of > units for the estimation of mental level. (Terman & Merrill 1937, p. 25) Your material is OBSOLETE and has no semblance whatsoever to the Stanford-Binet scores or other IQ scores that are currently used! > (Terman & Merrill 1937, p. 26) to this obsolete reference. In later generations of IQ tests, such as Ron Hoeflin's Mega Test, his IQ scores were normed against the SAT scores -- thus making HIS scores invalid IQ scores, which are independent of one's scholastic Aptitude! -- Bob. === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> Dubya made a 1206 on his SAT (566v+640m, before they >> dumbed down the scale in 1994), which roughly translates to >> an IQ score of 123 (http://www.gnxp.com/MT2/archives/002360.html) >> to 129 (http://members.shaw.ca/delajara/Pre1974SAT.html). > Anyone who translates an IQ score from SAT scores is automatically > an ignoramus about what IQ scores are supposed to measure -- > intelligence, not APTITUDE -- which the SAT scores measure. So you're saying there is no correlation between the two? === Subject: Re: anybody got an IQ over 200 here? >> Anyone who translates an IQ score from SAT scores is automatically >> an ignoramus about what IQ scores are supposed to measure -- >> intelligence, not APTITUDE -- which the SAT scores measure. > So you're saying there is no correlation between the two? It sounds like David is asking an appropriate question about what psychologists call the naming fallacy. Here's what Julian Stanley and Kenneth Hopkins had to say about the matter in their authoritative book on educational testing: Hopkins, Kenneth D. & Stanley, Julian C. (1981). Educational and Psychological Measurement and Evaluation. Englewood Cliffs, NJ: Prentice Hall. Many parents are unaware that intelligence tests tend to measure primarily scholastic aptitude and that many other cognitive abilities that can be legitimately considered to reflect intelligence and special abilities are untapped. (page 363) In this, Stanley and Hopkins agree with Lewis Terman, the author of the Stanford-Binet scales, who by 1937 knew better than to claim that his tests tapped all aspects of intelligent human behavior. Discussing the relationship between IQ and scholastic achievement, Stanley and Hopkins make a very important point: Most authorities feel that current intelligence tests are more aptly described as 'scholastic aptitude' tests because they are so highly related to academic performance, although current use suggests that the term INTELLIGENCE TEST is going to be with us for some time. (He got that right.) This reservation is based NOT on the opinion that intelligence tests do not reflect intelligence but on the belief that there are other kinds of intelligence that are not reflected in current tests; the term INTELLIGENCE is too inclusive. (page 364, emphasis in original) It's really time to talk here about VALIDATION of IQ tests, scholastic aptitude tests, and other mental tests. It doesn't do to simply treat a test battery as a magical incantation that produces a unitless number with no agreed meaning. -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> I didn't see his reply, but the cited paragraph was by ME: >> Anyone who translates an IQ score from SAT scores is automatically >> an ignoramus about what IQ scores are supposed to measure -- >> intelligence, not APTITUDE -- which the SAT scores measure. > So you're saying there is no correlation between the two? Not supposed to. and in practice, the correlation should be LOW. > It sounds like David is asking an appropriate question about what psychologists > call the naming fallacy. Here's what Julian Stanley and Kenneth Hopkins had to > say about the matter in their authoritative book on educational testing: > Hopkins, Kenneth D. & Stanley, Julian C. (1981). Educational and > Psychological Measurement and Evaluation. Englewood Cliffs, NJ: Prentice > Hall. > Many parents are unaware that intelligence tests tend to measure primarily > scholastic aptitude and that many other cognitive abilities that can be > legitimately considered to reflect intelligence and special abilities are > untapped. (page 363) So much for the authoritative authors who fully qualify for my characterization of ignoramuses. First of all, it they have looked at the test items of a dozen of more legit IQ tests, they should have noticed that many of the problem types are NOT found in any scholastic APTITUDE tests, because they don't measure aptitude toward college learning, as the SAT tests, or LSAT toward studies in Law, or aptitude tests for Business. Second, there are IQ tests that are completely independent of LANGUAGE. They can be administered to children before they have acquired sufficient aptitude in reading OR to adults whose language is not English. Third, all the well-known aptitude tests are for college-bound or graduate-school bound candidates, for their admission considerations by the colleges/grad school to which they apply. A person's IQ is certainly NOT dependent on having reached that level of formal education. In fact, many of the most precocious kids are not ready for college, but they possess many cognitive, inferential, deductive, and spatial recognition skills than college students. > In this, Stanley and Hopkins agree with Lewis Terman, the author of the > Stanford-Binet scales, who by 1937 knew better than to claim that his tests > tapped all aspects of intelligent human behavior. That only shows how ignorant psychologists are about the subjects in which they write about. > Discussing the relationship between IQ and scholastic achievement, Stanley and > Hopkins make a very important point: > Most authorities feel that current intelligence tests are more aptly described > as 'scholastic aptitude' tests because they are so highly related to academic > performance, although current use suggests that the term INTELLIGENCE TEST is > going to be with us for some time. (He got that right.) He got that WRONG, in history, and at the present! > Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 -- Bob. === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> <4Vshg.11397$KB.1004@fed1read08> President G. W. Bush was rated the lowest of all the Republicans with >> an IQ of 91. >> Bush's IQs were challenged, but I don't recall whether the complaint >> was that they were estimated too high or too low. >> Har. Neat trick, graduating from Yale with a below-average IQ. >> Or even meeting the entrance requirements. > W met the entrance requirements. He had money and his father graduated from > Yale. I didn't realize those were the only criteria for acceptance into Yale. Of course, Kerry also got into Yale, and his SAT score was even lower than Bush's, so you might be right. === Subject: Re: anybody got an IQ over 200 here? >> President G. W. Bush was rated the lowest of all the Republicans with >> an IQ of 91. >> Bush's IQs were challenged, but I don't recall whether the complaint >> was that they were estimated too high or too low. >> Har. Neat trick, graduating from Yale with a below-average IQ. >> Or even meeting the entrance requirements. > W met the entrance requirements. He had money and his father graduated from > Yale. > I didn't realize those were the only criteria for acceptance into Yale. Those are not the only criteria. They are, however, adequate criteria. > Of course, Kerry also got into Yale, and his SAT score was even lower > than Bush's, so you might be right. === Subject: Re: anybody got an IQ over 200 here? <4jd382hqrqbt615oe83ctefs5d579hkqfe@4ax.com> <878xo9whwz.fsf@phiwumbda.org> The non-partisan researchers who evaluated the twelve presidents [...] >> Non-partisan. It was a panel of college professors, right? > There was no study. The list was made up and presented as if it was > the result of real research. It is instead just a lame hoax or a bad > joke. > So all the folks here wailing about liberal bias in the universities > or about how George W. Bush gamed the system by entering Yale with a > 91 IQ are getting their panties in a wad over nothing. There may be > liberal bias, but this list isn't evidence of it. George W. Bush may > be stupid, but this list doesn't prove it[1]. True. His SAT score is actual evidence of his intelligence (or lack thereof). === Subject: understanding PDE Problem statement, Let q_i denote the cartesian components of the heat flux vector, let u be the temperature and ket L be the heat supply per unit volume. Assume the heat flux vector is defined in terms of the temperature gradient by the generalized Fourier law: q_i = -K_ij * u,_j where u,_j means partial derivative of u wrt jth coordinate. The K_ij's are given functions of x. Given L: D-->R, f : T1-->R, h : T2-->R, find u : D-->R s.t. q_i,_i = L in D (Heat Equation) u = f on T1 -q_i *n_i = h on T2 I don't really understand how this is a heat equation. In my intro PDE book, heat equation looks like: u_t - k(u_xx + u_yy + u_zz) = 0 plus boundary/initial conditions. Can someone explain the difference? === Subject: Re: understanding PDE Problem statement, > Let q_i denote the cartesian components of the heat flux vector, let u be > the temperature and ket L be the heat supply per unit volume. Assume the > heat flux vector is defined in terms of the temperature gradient by the > generalized Fourier law: q_i = -K_ij * u,_j > where u,_j means partial derivative of u wrt jth coordinate. The K_ij's > are given functions of x. > Given L: D-->R, f : T1-->R, h : T2-->R, find u : D-->R s.t. > q_i,_i = L in D (Heat Equation) > u = f on T1 > -q_i *n_i = h on T2 > I don't really understand how this is a heat equation. In my intro PDE > book, heat equation looks like: > u_t - k(u_xx + u_yy + u_zz) = 0 plus boundary/initial conditions. > Can someone explain the difference? I don't completely understand the notation. It seems that the index is refering to the ith component but if this is true then there are probably limits on it. If this is some type of einstien notation for tensors then I can get close to the original heat equation but its off somewhat. I'll show you what I can do and maybe you can fill in the rest... it could be completely wrong. in vector notation if u = u(x,y,z) q = -K*grad(u) (i.e., in the original notation, u,_i means u_x = du/dx) then q_i,_i = L would be grad(q) = L but q = -K*grad(u). if K = k is constant then grad(q) = L ==> -k*laplacian(u) = L ==> -k*(u_xx + u_yy + u_zz) = L where did u_t go? I don't know. speration of variables might have been performeed on u to remove the time dependence. If thats the case then the above make sense. If not then I don't know but a similar method might make it work. Jon === Subject: Re: understanding PDE Problem statement, >> Let q_i denote the cartesian components of the heat flux vector, let u be >> the temperature and ket L be the heat supply per unit volume. Assume the >> heat flux vector is defined in terms of the temperature gradient by the >> generalized Fourier law: q_i = -K_ij * u,_j >> where u,_j means partial derivative of u wrt jth coordinate. The K_ij's >> are given functions of x. >> Given L: D-->R, f : T1-->R, h : T2-->R, find u : D-->R s.t. >> q_i,_i = L in D (Heat Equation) >> u = f on T1 >> -q_i *n_i = h on T2 >> I don't really understand how this is a heat equation. In my intro PDE >> book, heat equation looks like: >> u_t - k(u_xx + u_yy + u_zz) = 0 plus boundary/initial conditions. >> Can someone explain the difference? > I don't completely understand the notation. It seems that the index is > refering to the ith component but if this is true then there are probably > limits on it. If this is some type of einstien notation for tensors then I > can get close to the original heat equation but its off somewhat. > I'll show you what I can do and maybe you can fill in the rest... it could > be completely wrong. > in vector notation > if u = u(x,y,z) > q = -K*grad(u) > (i.e., in the original notation, u,_i means u_x = du/dx) > then > q_i,_i = L > would be > grad(q) = L > but q = -K*grad(u). if K = k is constant then > grad(q) = L ==> -k*laplacian(u) = L ==> -k*(u_xx + u_yy + u_zz) = L I guess I should say that q_i,_i = (grad(grad(-k*u)))_(i,i) but since one has repeated indicies one sums over them(some type of tensor notation or something). i.e. grad(grad(u)) = [u_xx u_xy u_xz] [u_yx u_yy u_yz] [u_zx u_zy u_zz]_(i,i) = u_xx if i = 1 u_yy if i = 2 u_zz if i = 3 but sense the index is repeat in the formula it is convention to sum over it. in essense one has trace(grad(grad(-Ku))) = L grad(u) is extendent to work on vectors and there is a certain convention to treat it it as a tensor when done so. i.e., given u:R^n->R then trace(grad(grad(u))) = laplacian(u). the reason for the notation with subscripts above is to get away from talking about specific variables such as x, y, z. (which is what tensors are all about). I'm pretty sure thats whats going on but who knows. Jon === Subject: Help with elliptic curve y^2 = p^3-34p^2+225p Hello all, I need to verify if the elliptic curve, y^2 = p(p-9)(p-25) only has five rational solutions p = 0,5,9,25,45, or if there are more? (Also, if this looks familiar or if you've come across this before as this was found in the context of a quintic equation.) --Titus === Subject: Re: Help with elliptic curve y^2 = p^3-34p^2+225p > Hello all, > I need to verify if the elliptic curve, > y^2 = p(p-9)(p-25) > only has five rational solutions p = 0,5,9,25,45, or if there are more? Best way to do that is to reduce it to a standard minimal model and then look it up. I get y^2 + xy + y = x^3+x^2-10x-10 === Subject: Re: Help with elliptic curve y^2 = p^3-34p^2+225p > Best way to do that is to reduce it to a standard minimal model and > then look it up. > I get > y^2 + xy + y = x^3+x^2-10x-10 I found my Cremona tables; looking it up, it is A15, (conductor 15) with a rank of zero and a torsion group of eight, so there are eight rational points on it. p=0, 9, and 25 come with one y value, p=5 and 45 come with two, so that makes seven, and the point at infinity eight. === Subject: Re: Help with elliptic curve y^2 = p^3-34p^2+225p > Best way to do that is to reduce it to a standard minimal model and > then look it up. > I get > y^2 + xy + y = x^3+x^2-10x-10 > I found my Cremona tables; looking it up, it is A15, (conductor 15) > with a rank of zero and > a torsion group of eight, so there are eight rational points on it. > p=0, 9, and 25 come with one y value, p=5 and 45 come with two, so that > makes seven, and the point at infinity eight. How very interesting that the curve in question is equivalent to, y^2 + xy + y = x^3+x^2-10x-10 I did mention that it appeared in the context of a quintic equation. The background is this: The general quintic can be reduced in radicals to the one-parameter forms, (z-5)(z^2+20)^2 + q1 = 0 (eq.1) (z-5)(z^2+15)^2 + q2 = 0 (eq.2) where the first is a version of the Brioschi quintic, and there are an infinite number of non-scaled rational q1 such that eq.1 is solvable in radicals. The second is by yours truly, and I conjectured that there was only a finite number of rational q2 such that eq.2 was solvable. In an email, Spearman was kind enough to confirm it by using Dummit's resolvent and Maple, though he didn't mention that it involved A15. I used my own resolvent and found the elliptic curve in the heading, and have an answer. === Subject: Re: Help with elliptic curve y^2 = p^3-34p^2+225p > I did mention that it appeared in the context of a quintic equation. > The background is this: The general quintic can be reduced in radicals > to the one-parameter forms, > (z-5)(z^2+20)^2 + q1 = 0 (eq.1) > (z-5)(z^2+15)^2 + q2 = 0 (eq.2) Another fun curve game to play is with genus zero curves; for instance if q2 is set to q2 = 128 (9t^2-40)/(t^2-5) then the discriminant is a square, and you get reduction to A5 at least. Of course, you probably knew that. :) === Subject: Re: Polynomial ring that aren't domains > 1. Assumed R is not integral (otherwise the question wouldnt be valid) >> Why that???? Let me tell you an example of how to reduce to the case >> where R is integral. >> You *assume* without loss of generality that R *is* integral, since you >> can pass to the quotient Ring R/P. >> In more detail, let i: R -> R[X] the natural injection and P a prime >> ideal of R. Then the natural injection i_P: R/P -> (R/P)[X] is naturally >> isomorphic to i (x)_R R/P (where (x)_R denotes the tensor product over >> R). Note that the closed immersions Spec(R/P) -> Spec(R) and >> Spec(R/P[X]) -> Spec(R[X]) commute with Spec(i) and Spec(i_P). This fact >> implies that the fibers of Spec(i) and Spec(i_P) over P are the same. >> One can say even more: they are homeomorphic. >> In my example you quote above take any ring S with a minimal prime ideal >> P with S/P=R (isomorphism). >> A similar mechanism works by tensoring with R_P (R localised at P) over >> R. What primes are filtered in R[X] by passing to R[X] (x)_R R? >> In addition it could be helpful to know nil(R[X])=nil(R)[X] (where >> nil(R) denotes the nil-radical of R). Note that if R has only one >> minimal prime ideal P then nil(R)=P. >> What do you think of these two mechanisms often used in commutative >> algebra or algebraic geometry to reduce to the case where R is integral >> or has just one minimal prime ideal, resp., and the prime ideal in >> question is the zero ideal or the nil-radical, resp. >> Note that your first question focuses on primes living above PR[X] !!! >> In case that I am missing something here, I would like you to tell me >> where I am off the road. >> J. > 2. Asked if phi^{-1}(P) / minspec(R[X])) = emptyset for some P in > minspec(R) > Well.. :) .. let me try to make it clear. Of course you can mod the > prime ideal and play with the functorial property of the spectrum. But > here you are thinking of the fibers of the spectral maps in general, > and I am talking of the fibers of a minimal point INTERSECTED with the > minimal spectrum. > Ok, maybe to make this clear. I will explain why it wouldnt work for > integral domain. I wanted to find an R and a P in Minspec R, by which > Spec(i)^(-1)(P) / minspec(R[X])) = emptyset > where i is the canonical injection i : R ---> R[x] > This is never true if R is an integral domain because the ONLY minimal > prime ideal in R and R[x] is (0), and clearly > (0) is in Spec(i)^(-1)(0) > Hopefully I clarified things :) > Jose Capco Now it is clear! :-) ... and Hagen solved the problem already in a general setting. Best wishes, J. === Subject: Re: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com> them, play with the equations for them. Trace out tangent lines, and >marvel over their properties--and I liked to rotate them into >hyperbolas. Just for fun. >>Rotate it all you will, you'll never make a parabola into a hyperbola. >>Parabola: y = ax^2 + bx + c >>Hyperbola: (x-a)(y-b) = c > Really? Are you sure? > ___JSH > The parabola is connected. The hyperbola is not connected. > There is no continuous transformation of the plane that turns > a parabola into a hyperbola. Even if you only consider one > of the components of the hyperbola, a rotation of the plane > will not yield a hyperbolic curve. > This can be done in the projective plane, if you push the > parabola through the line at infinity. The parabola is > tangent to that line, and if it is pushed in one direction > you get an ellipse; if you push in the opposite direction, > you get a hyperbola. > Dale. Try it. Rotate a parabola and see what you get. If it seems too abstract, post what you get. Like, rotate a parabola 45 degrees. Post the result. James Harris === Subject: Re: SF might work, bad luck? As a hobby when I was a kid I was fascinated with parabolas. I'd draw >them, play with the equations for them. Trace out tangent lines, and >marvel over their properties--and I liked to rotate them into >hyperbolas. Just for fun. >>Rotate it all you will, you'll never make a parabola into a hyperbola. >>Parabola: y = ax^2 + bx + c >>Hyperbola: (x-a)(y-b) = c > Really? Are you sure? ___JSH The parabola is connected. The hyperbola is not connected. > There is no continuous transformation of the plane that turns > a parabola into a hyperbola. Even if you only consider one > of the components of the hyperbola, a rotation of the plane > will not yield a hyperbolic curve. > This can be done in the projective plane, if you push the > parabola through the line at infinity. The parabola is > tangent to that line, and if it is pushed in one direction > you get an ellipse; if you push in the opposite direction, > you get a hyperbola. > Dale. > Try it. Rotate a parabola and see what you get. > If it seems too abstract, post what you get. > Like, rotate a parabola 45 degrees. > Post the result. I was a physoics major once. I changed majors (several times) before I got my degree, but I got far enough in the physics curriculum to take Theoretical Mechanics. In the introductory weeks of that course, we learned about coordinate transformation, including the mathematics needed to rotate standard Cartesian rectangualr coordinates. I would be surprised if your study of physics did not include a similar course. Summing up: A parabola rotated 45 degrees is still a parabola. === Subject: Re: SF might work, bad luck? >[...] >Try it. Rotate a parabola and see what you get. >If it seems too abstract, post what you get. >Like, rotate a parabola 45 degrees. >Post the result. Ok. If you rotate a parbola 45 degrees the result is a parabola. You never tire of finding _new_ ways to make a public fool of yourself. >James Harris ************************ David C. Ullrich === Subject: Re: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com> <4k1g82tjdtvlu8kaurs04u201h6o1k4t68@4ax.com[...] >Try it. Rotate a parabola and see what you get. >If it seems too abstract, post what you get. >Like, rotate a parabola 45 degrees. >Post the result. > Ok. If you rotate a parbola 45 degrees the result > is a parabola. > You never tire of finding _new_ ways to make a > public fool of yourself. Wait a minute - are you saying JSH was NOT the child prodigy he believes he was? I guess that explains a lot. >James Harris > ************************ > David C. Ullrich === Subject: Re: SF might work, bad luck? : Try it. Rotate a parabola and see what you get. : If it seems too abstract, post what you get. : Like, rotate a parabola 45 degrees. : Post the result. James, I think perhaps you don't know what a hyperbola is. It's a common mistake for my precalculus students to believe that a hyperbola is two parabolas pointing in opposite directions but this isn't true. Both are conic sections but they have different general equations. For a parabola only one of x and/or y is squared but for a hyperbola both are squared but one has a positive coefficient and one has a negative coefficient. There is no way to rotate one to the other. Just because a parabola turned sideways might LOOK LIKE a half of a hyperbola to you does not make it so. Justin === Subject: Re: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com> As a hobby when I was a kid I was fascinated with parabolas. I'd draw >them, play with the equations for them. Trace out tangent lines, and >marvel over their properties--and I liked to rotate them into >hyperbolas. Just for fun. >>Rotate it all you will, you'll never make a parabola into a hyperbola. >>Parabola: y = ax^2 + bx + c >>Hyperbola: (x-a)(y-b) = c > Really? Are you sure? The parabola is connected. The hyperbola is not connected. > There is no continuous transformation of the plane that turns > a parabola into a hyperbola. Even if you only consider one > of the components of the hyperbola, a rotation of the plane > will not yield a hyperbolic curve. > This can be done in the projective plane, if you push the > parabola through the line at infinity. The parabola is > tangent to that line, and if it is pushed in one direction > you get an ellipse; if you push in the opposite direction, > you get a hyperbola. > Dale. > Try it. Rotate a parabola and see what you get. > If it seems too abstract, post what you get. > Like, rotate a parabola 45 degrees. > Post the result. Consider Baby's First Parabola, given by y = x^2 [1] Let us rotate it 45 degress clockwise. One way to do this is by defining new axes x' and y', such that ( x ) = ( cos 45o -sin 45o ) ( x' ) ( y ) = ( sin 45o cos 45o ) ( y' ) [2] Which is expressed the other way round as ( x' ) = ( cos -45o -sin -45o ) ( x ) ( y' ) = ( sin -45o cos -45o ) ( y ) [3] So x' = x / sqrt(2) + y / sqrt(2) y' = -x / sqrt(2) + y / sqrt(2) [4] Now our rotated parabola is given by y' = x'^2 [2] Substituting [4] into [2]: (y - x) / sqrt(2) = ((x + y) / sqrt(2)) ^2 = 1/2 (x + y)^2 Multiply through by 2 sqrt(2) (y - x) = x^2 + 2xy + y^2 Putting into canonical form x^2 + 2xy + y^2 + sqrt(2)x - sqrt(2)y + 0 = 0 [5] Now, we look at our Big Book of Conic Sections and we see: In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section, and all conic sections arise in this way. If the equation is of the form ax^2 + 2hxy + by^2 +2gx + 2fy + c = 0 then: * if h^2 < ab, the equation represents an ellipse; o if we also have a = b and h = 0, the equation represents a circle; * if h^2 = ab, the equation represents a parabola; * if h^2 > ab, the equation represents a hyperbola; o if we also have a + b = 0, the equation represents a rectangular hyperbola. (wikipedia) Now what are our magic numbers in [5] ? Well, we have a = b = 1, and also h = 1. So h^2 = ab. So it's (still) a parabola. What were you expecting James? -- Larry Lard Replies to group please === Subject: Re: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com> <128c9fed6co0k3e@corp.supernews.com> I have a B.Sc. in physics. I taught myself calculus when I was a kid, >> having to go back and teach myself trigonometry when I kept bumping >> into all these things with sin and cos in them in the calculus >> texts. > Just becaues you have a B.S. in physics doesn't mean you are genius. I > know many phd's(even in mathematics) who don't know . Its easy to get > a piece of paper saying your a genius but it doesn't mean you are one. > BS = Bull > MBA = More Bull Accumulated > PhD = piled hip-deep... ITVIH, MS = More of the Same. --- Christopher Heckman === Subject: Re: SF might work, bad luck? <128ap1te73mhhad@corp.supernews.com If you truely realize what you are saying and are not just rebaiting the > hooks, I suggest you go take a couple of classes at your local community > college. > Start with the very easy stuff... even if it seems to easy so that you will > have a solid foundation. > Take pre-algebra, algebra, trig, then maybe even calculus if do good in > those(once class at a time though). > I have a B.Sc. in physics. I taught myself calculus when I was a kid, > having to go back and teach myself trigonometry when I kept bumping > into all these things with sin and cos in them in the calculus > texts. > Kind of funny in retrospect as when I was a kid I wondered what sin > had to do with math! > I taught myself geometry at Duke University as part of their Talent > Identification Program (TIP). > When I was in high school a math teacher came across me working on > partial differential equations and wanted to tutor me. History might > have changed if I had accepted, but I was already too independent. > As a hobby when I was a kid I was fascinated with parabolas. I'd draw > them, play with the equations for them. Trace out tangent lines, and > marvel over their properties--and I liked to rotate them into > hyperbolas. Just for fun. > One day in high school fascinated by a demonstration of gyroscopic > properties, I worked out the differential equations explaining why. > Yes there are other ways to do it but I used to be fascinated with > vectors and vector calculus. > After that I had a different perspective when I saw a spinning football > go straight. > I have forgotten more mathematics than most people ever learn. > I DID take abstract algebra at Duke after I finished up geometry, but > found it dry and boring, so I put the book down. Ironic. > You people talk so much you believe your own nasty press about me. > I am way off the scale with my mathematical knowledge, and have been > off the scale most of my life. All this and modesty, too! === Subject: Re: alternatives to logistic growth >>Hello group, >>There is a general solution for models of exponential >>growth, such that >>the size of the growing quantity at time t is >>S(t) = S(0) Exp[g t] >>where S(0) is the size of the quantity at time zero, >and >g is the >>instantaneous growth rate. >>There is no similar general solution for the logistic >>growth model. >> I don't know which logistic growth model you are >> referring to, but the differential equation >> dN/dt = r_0 * N * (1-N/K) >> (usually known as the _continuous_ logistic growth >model), >> can be solved analytically. >> For a solution, look at >> http://www.ento.vt.edu/~sharov/PopEcol/lec5/logist.html >> Maybe you mean the discrete model ? >>My question is: is there a model that is *like* the >>logistic growth >>model, in some sense, that does permit a general >>solution? If there is, >>then I would be pleased if someone could point it out >to >me. >> Best wishes >> Torsten. >have a model of 'competition' between two entities, >based on the >continuous logistic growth model, described by the >equations >dS1/dt = g1 S1 (1-((S1+ S2)/Smax)) >dS2/dt = g2 S2 (1-((S1+S2)/Smax)) >Then is there a general solution for {S1(t),S2(t)}? If >it helps, we can >assume that g2 = g1 + dg where dg->0. if you divide your two equations, you get dS1/dS2 = g1/g2 * S1/S2 with solution S1 = S1(t=0) * (S2/S2(t=0))^(g1/g2). You can insert this expression in the second equation to arrive at dS2/dt = g2 * S2 * (1-( S1(t=0) * (S2/S2(t=0))^(g1/g2) + S2) / Smax) To solve this by seperation of variables, you had to find an antiderivative for the function f(x) = 1/(g2*x*(1-( S1(t=0) * (x/S2(t=0))^(g1/g2) + x) / Smax) You could try to search in an integral table, but I doubt that this expression can be integrated analytically. Best wishes Torsten. === Subject: Re: surrogate factoring > A previous poster was right - having anything to do with JH's posts is > like rubbernecking a traffic accident. Sad but difficult to resist. > Anyway ... > The bit I don't get (well one of the bits) is what specifically JH > *thinks* is the point of surrogate factoring. I know it's total > nonsense but does anyone have a clue *why* he thinks it's worthwhile? I > guess I find the nature of the delusion somewhat puzzling and it's an > amusing diversion to work out what he actually thinks. > I really should get out more. The idea behind surrogate factoring is to factor a target number by using a surrogate factorization instead. It turns out it's trivially easy to prove that surrogate factoring has to work. If you consider f_1 f_2 = k + g_1 g_2 with 4 variables, you can get 4 linear equations from those four factors: f_1, f_2, g_1, and g_2 Explicitly then for your 4 variables you have these solutions using trivial mathematics. It's just the intersection of 4 lines. I figured out a way to use two of the lines, plus this variable I call k. You complete the square twice and you have this nifty solution, which trivially solves the factoring problem. Unfortunately, two sci.math posters have lied about what happens when you complete the square twice creating a gap. That gap has left the world in a quandary as the stock markets of the world tread water. When it's realized that I am right, the stock markets will correct, and I don't know, but I suspect the world will do its own justice against the two sci.math posters who lied. And they did ask for it. James Harris === Subject: Re: surrogate factoring [gjedwards@gmail.com] > A previous poster was right - having anything to do with JH's posts is > like rubbernecking a traffic accident. Sad but difficult to resist. > Anyway ... > The bit I don't get (well one of the bits) is what specifically JH > *thinks* is the point of surrogate factoring. I know it's total > nonsense but does anyone have a clue *why* he thinks it's worthwhile? I > guess I find the nature of the delusion somewhat puzzling and it's an > amusing diversion to work out what he actually thinks. > I really should get out more. Me too, but so long as we're both house-ridden today, and I've enjoyed more quality SF time than anyone other than James ... First, surrogate factoring doesn't really mean anything specific. There are literally dozens of distinct methods James has _called_ surrogate factoring over the past two years. While they've all been wrong (in the sense that none have been efficient factoring methods, and for some it was a miracle if they ever found a factor), total nonsense is either overstatement or understatement :-) The general idea is that you want to factor T, but factor some related integer S (the surrogate) instead and hope to relate the factors of S to T's factors. That's perfectly sensible so far as it goes. For example, many methods try to factor k*T instead, and that can even be useful by eyeball. Consider T=817. A tiny bit of thought shows it's not divisible by 2, 3, or 5, and then you might notice it's about a third of 2500. Indeed, 3 * T = 3 * 817 = 2451 = 2500 - 49 = 50^2 - 7^2 = (50+7)*(50-7) = 57*43 by inspection, so you _can_ know almost at once that 43 and 57/3 = 19 are factors of T, if only you think about 3*T instead of T. More, variations on that can be turned into straightforward algorithms that always find factors (Google for Fermat's method). Some of them don't even need division, which may be surprising at first sight. Alas, trial division is usually more efficient (if you suspected I put some care into picking the 817 example, you were right ;-)), and a _few_ of James's SF attempts have been inefficient variations on this theme, excruciatingly obfuscated ways of trying to find integers I and J such that I^2-J^2 = k*T, in which case gcd(I +- J, T) has a good chance of revealing a non-trivial factor of T. Most typical _last_ year were SF variants that derived a large (compared to T) S to factor, and while James didn't know how to do this, it actually was possible to find S of the required form efficiently so that S could be factored efficiently despite it being much larger than T (and wrt much larger, sometimes S grew like the 10th(!) power of T). Most criticism of those methods focused on a wrong thing, the supposed difficulty of factoring S. That wasn't the problem, but neither was it was obvious that wasn't the problem. James insisted he had proof that given the prime factorization of S, it must be the case that _some_ 2-integer factorization of S, S=f1*f2, revealed a factor of T after plugging f1 and f2 into some messy formula (which could change daily). That came in for more off-target criticism on the grounds that S may be expressible in a great many ways as the product of two integers. While that's true, it was also the case (although James didn't know how to do this either) that is was possible to find S of the required form such that S was both easy to factor and had few distinct prime divisors (the latter relates to how many ways there are to express S as the product of two integers). The real problem was that he had no proof in reality (although he claimed to on several occasions, in his charmingly cautious way ;-)), and it was typical that _no_ way of expressing S as f1*f2 revealed a factor of T. Undaunted, some of us went on to _search_ for an S of the required form such that some f1*f2 = S existed revealing a factor of T. All such algorithms of that nature performed worse (in number of attempts needed, and extremely worse in terms of time needed given the complexity of the operations and the large size of the integers) than picking integers I at random and seeing whether gcd(I, T) revealed a non-trivial factor of T. Those weren't really James's algorithms, though -- they were just enjoyable time-wasting exercises to see whether _anything_ could be salvaged. After that, IMO all subsequent SF attempts got worse. James got too frustated by working with integers, and started talking about rational factors instead. That was silly on the face of it, since given any integer T, _every_ non-zero rational r is a rational factor of T. Nevertheless, James went through increasingly elaborate piles of formulas showing how a rational factor of S could be transformed into a rational factor of T. But no transformation was necessary: every non-zero rational r is a rational factor of both S and T, for any S and T. His arguments for why this had to be a promising approach were just bizarre, and none of the resulting methods were actually better than this one: Given T, pick some non-zero rational r out of the air. If 1 < gcd(numerator(r), T) < T, stop, else try again. While you _could_ find rational factors of S to drive that, there was no point, since the set of all non-zero rationals constituted the search space regardless. That seemed to mark the start of what's still going on, but in much slower motion this year: 1. Take a pile of starting equations as formal identities. 2. Push them around endlessly: solve for some symbols in terms of others, multiply pairs from time to time, complete the square on occasion, and sometimes invoke the quadratic formula. Don't bother restricting yourself to operations that must yield integers, or even reals -- any transformation valid in the complex field is fair game. 3. Announce that the factoring problem is solved, and the end of civilization as we know it is at hand. 4. After time passes, insist that if nobody has found factors using it yet, it's because they're too stupid to understand the simple math, but be assured that other nameless people _are_ working on it and success is imminent. Or James is really the only person in the world smart enough to make it work, but he cares about humanity too much to risk factoring anything (but not _so_ much that it will bother him if someone else does). The excuses here change but are always a hoot :-) 5. Discover that the result of #2 is really just a useless respelling of the equations you started with in #1, and the only way to make them work efficiently is to know T's factorization before you start. This branches in two directions then: 5a. Add more independent variables to the same equations, and go back to #2. - or - 5b. Go back to #1. The _current_ round (although James may have announced a new method or two while I was typing this) is _almost_ at step #5 in the template above. Anyone who paid attention to Rick Decker's and my postings this time knows that #5 has been solidly demonstrated to be case, but last I saw James was still fighting that. I'm not sure he's ever gotten to #5 without discovering it for himself; until then, he's always got _some_ reason to believe it's more likely that others are lying than that he's just blinded himself with hope again. If nothing else, I hope I've convinced you that you didn't really waste your life by staying out of this :-) As to why he does it, I think he's told us the truth on occasion: he enjoys pushing formulas around, and it may be a bona fide compulsion for him. As to why he also has a demon compelling him to believe that the results of idly pushing formulas around must be a breakthrough, your guess may be better than mine. === Subject: Re: surrogate factoring A previous poster was right - having anything to do with JH's posts is > like rubbernecking a traffic accident. Sad but difficult to resist. > Anyway ... > The bit I don't get (well one of the bits) is what specifically JH > *thinks* is the point of surrogate factoring. I know it's total > nonsense but does anyone have a clue *why* he thinks it's worthwhile? I > guess I find the nature of the delusion somewhat puzzling and it's an > amusing diversion to work out what he actually thinks. > I really should get out more. > Me too, but so long as we're both house-ridden today, and I've enjoyed more > quality SF time than anyone other than James ... > First, surrogate factoring doesn't really mean anything specific. There > are literally dozens of distinct methods James has _called_ surrogate > factoring over the past two years. While they've all been wrong (in the > sense that none have been efficient factoring methods, and for some it was a > miracle if they ever found a factor), total nonsense is either > overstatement or understatement :-) > The general idea is that you want to factor T, but factor some related > integer S (the surrogate) instead and hope to relate the factors of S to > T's factors. That's perfectly sensible so far as it goes. For example, > many methods try to factor k*T instead, and that can even be useful by > eyeball. Consider T=817. A tiny bit of thought shows it's not divisible > by 2, 3, or 5, and then you might notice it's about a third of 2500. > Indeed, > 3 * T = > 3 * 817 = > 2451 = > 2500 - 49 = > 50^2 - 7^2 = > (50+7)*(50-7) = > 57*43 > by inspection, so you _can_ know almost at once that 43 and 57/3 = 19 are > factors of T, if only you think about 3*T instead of T. > More, variations on that can be turned into straightforward algorithms that > always find factors (Google for Fermat's method). Some of them don't even > need division, which may be surprising at first sight. > Alas, trial division is usually more efficient (if you suspected I put some > care into picking the 817 example, you were right ;-)), and a _few_ of > James's SF attempts have been inefficient variations on this theme, > excruciatingly obfuscated ways of trying to find integers I and J such that > I^2-J^2 = k*T, in which case gcd(I +- J, T) has a good chance of revealing a > non-trivial factor of T. > Most typical _last_ year were SF variants that derived a large (compared to > T) S to factor, and while James didn't know how to do this, it actually was > possible to find S of the required form efficiently so that S could be > factored efficiently despite it being much larger than T (and wrt much > larger, sometimes S grew like the 10th(!) power of T). Most criticism of > those methods focused on a wrong thing, the supposed difficulty of factoring > S. That wasn't the problem, but neither was it was obvious that wasn't the > problem. > James insisted he had proof that given the prime factorization of S, it must > be the case that _some_ 2-integer factorization of S, S=f1*f2, revealed a > factor of T after plugging f1 and f2 into some messy formula (which could > change daily). That came in for more off-target criticism on the grounds > that S may be expressible in a great many ways as the product of two > integers. While that's true, it was also the case (although James didn't > know how to do this either) that is was possible to find S of the required > form such that S was both easy to factor and had few distinct prime divisors > (the latter relates to how many ways there are to express S as the product > of two integers). > The real problem was that he had no proof in reality (although he claimed to > on several occasions, in his charmingly cautious way ;-)), and it was > typical that _no_ way of expressing S as f1*f2 revealed a factor of T. > Undaunted, some of us went on to _search_ for an S of the required form such > that some f1*f2 = S existed revealing a factor of T. All such algorithms of > that nature performed worse (in number of attempts needed, and extremely > worse in terms of time needed given the complexity of the operations and the > large size of the integers) than picking integers I at random and seeing > whether gcd(I, T) revealed a non-trivial factor of T. Those weren't really > James's algorithms, though -- they were just enjoyable time-wasting > exercises to see whether _anything_ could be salvaged. > After that, IMO all subsequent SF attempts got worse. James got too > frustated by working with integers, and started talking about rational > factors instead. That was silly on the face of it, since given any integer > T, _every_ non-zero rational r is a rational factor of T. Nevertheless, > James went through increasingly elaborate piles of formulas showing how a > rational factor of S could be transformed into a rational factor of T. But > no transformation was necessary: every non-zero rational r is a rational > factor of both S and T, for any S and T. His arguments for why this had to > be a promising approach were just bizarre, and none of the resulting > methods were actually better than this one: > Given T, pick some non-zero rational r out of the air. > If 1 < gcd(numerator(r), T) < T, stop, else try again. > While you _could_ find rational factors of S to drive that, there was no > point, since the set of all non-zero rationals constituted the search space > regardless. > That seemed to mark the start of what's still going on, but in much slower > motion this year: > 1. Take a pile of starting equations as formal identities. > 2. Push them around endlessly: solve for some symbols in terms > of others, multiply pairs from time to time, complete the > square on occasion, and sometimes invoke the quadratic formula. > Don't bother restricting yourself to operations that must yield > integers, or even reals -- any transformation valid in the > complex field is fair game. > 3. Announce that the factoring problem is solved, and the > end of civilization as we know it is at hand. > 4. After time passes, insist that if nobody has found factors > using it yet, it's because they're too stupid to understand > the simple math, but be assured that other nameless people > _are_ working on it and success is imminent. Or James is > really the only person in the world smart enough to make it > work, but he cares about humanity too much to risk factoring > anything (but not _so_ much that it will bother him if > someone else does). The excuses here change but are always > a hoot :-) > 5. Discover that the result of #2 is really just a useless > respelling of the equations you started with in #1, and the > only way to make them work efficiently is to know T's > factorization before you start. This branches in two > directions then: > 5a. Add more independent variables to the same equations, and > go back to #2. > - or - > 5b. Go back to #1. > The _current_ round (although James may have announced a new method or two > while I was typing this) is _almost_ at step #5 in the template above. > Anyone who paid attention to Rick Decker's and my postings this time knows > that #5 has been solidly demonstrated to be case, but last I saw James was > still fighting that. I'm not sure he's ever gotten to #5 without > discovering it for himself; until then, he's always got _some_ reason to > believe it's more likely that others are lying than that he's just blinded > himself with hope again. > If nothing else, I hope I've convinced you that you didn't really waste your > life by staying out of this :-) > As to why he does it, I think he's told us the truth on occasion: he enjoys > pushing formulas around, and it may be a bona fide compulsion for him. As > to why he also has a demon compelling him to believe that the results of > idly pushing formulas around must be a breakthrough, your guess may be > better than mine. === Subject: Re: surrogate factoring A previous poster was right - having anything to do with JH's posts is > like rubbernecking a traffic accident. Sad but difficult to resist. > Anyway ... > The bit I don't get (well one of the bits) is what specifically JH > *thinks* is the point of surrogate factoring. I know it's total > nonsense but does anyone have a clue *why* he thinks it's worthwhile? I > guess I find the nature of the delusion somewhat puzzling and it's an > amusing diversion to work out what he actually thinks. > I really should get out more. > Me too, but so long as we're both house-ridden today, and I've enjoyed more > quality SF time than anyone other than James ... > First, surrogate factoring doesn't really mean anything specific. There > are literally dozens of distinct methods James has _called_ surrogate > factoring over the past two years. While they've all been wrong (in the > sense that none have been efficient factoring methods, and for some it was a > miracle if they ever found a factor), total nonsense is either > overstatement or understatement :-) > The general idea is that you want to factor T, but factor some related > integer S (the surrogate) instead and hope to relate the factors of S to > T's factors. That's perfectly sensible so far as it goes. [...] As a digression, I'll repost what I considered a useful offshoot of this idea: Suppose you know T, and you know that T = p * q, where p and q are distinct (maybe odd) primes, and you want to find p and/or q. Now suppose you have an integer which factors the same way; i.e., as the product of two distinct (maybe odd) primes; thus: S = p' * q'. (I wanted to avoid S = r * s here.) Now, IF you can find a function f: Z -> Z such that f(p') = p and f(q') = q, without using the specific values of p and q, then the factors of S will map to the factors of T. This, of course, appears to be impossible, but seemed the only viable approach to anything which can properly be called surrogate factoring. You could also extend the idea by supposing you know how T factors, perhaps as T = p^2 * q * r^3, where p, q, and r are all distinct primes (possibly with p < q < r), construct a function f, using the fact that 15435 = 3^2 * 5 * 7^3, then calculate f(3), f(5), and f(7), to get p, q, and r. And if you can get _that_ to work, well, you're on your way to Milliways! --- Christopher Heckman === Subject: prove that B[a,b] is not separable. Can anybody show a hint to prove that B[a,b] is not separable? B[a,b] is set of bounded functions on [a,b]. === Subject: Re: prove that B[a,b] is not separable. > Can anybody show a hint to prove that B[a,b] is not separable? > B[a,b] is set of bounded functions on [a,b]. The problem is meaningless if you don't mention which topology you're working with. I'll assume that it is the one induced by the metric d(f,g) = sup |f - g|. Take _x_ in [a,b] and define the function f_x:[a,b] ---> R by f_x(x) = 1 and f_x(y) = 0 whenever _y_ is different from _x_. Then, if _x_ is different from _y_, d(f_x,f_y) = 1 and therefore the balls B(f_x,1/2) and B(f_y,1/2) have a non-empty intersection. Can you take it from here? Jose Carlos Santos === Subject: Re: prove that B[a,b] is not separable. On Thu, 08 Jun 2006 09:39:50 +0100, Jos.8e Carlos Santos >> Can anybody show a hint to prove that B[a,b] is not separable? >> B[a,b] is set of bounded functions on [a,b]. >The problem is meaningless if you don't mention which topology you're >working with. I'll assume that it is the one induced by the metric > d(f,g) = sup |f - g|. >Take _x_ in [a,b] and define the function f_x:[a,b] ---> R by f_x(x) = 1 >and f_x(y) = 0 whenever _y_ is different from _x_. Then, if _x_ is >different from _y_, d(f_x,f_y) = 1 and therefore the balls B(f_x,1/2) >and B(f_y,1/2) have a non-empty intersection. Typo: ... have an empty intersection. >Can you take it from here? >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: SF: Does this work? Such a simple answer to the problem, with such simple algebra. >It is remarkable beyond belief. >James Harris > I suggest you ask yourself two questions: > 1 Is there anything in your work that was not known to Euler, Fermat > or Gauss? > 2 Are you a better mathematician than Euler, Fermat or Gauss? > Having answered them, go back and check over your work thoroughly to > see if: > a) there are any mistakes in it, > b) it is faster than existing methods. > As you have said, there have been mistakes in some of the stuff that > you posted here. Given that you have made mistakes before you need to > check carefully that you have not made mistakes again. > I tried one of your previous factorising methods on my computer. It > found factors, but it was about 10,000 times slower than Fermat's > method. If a new method is to be useful, then it has to work faster > than existing methods. > rossum The math is trivial. The problem here is that sci.math posters have routinely lied about my math results. Given 4 linear equations with 4 variables, do you or do you not get explicit solutions? But if you look at claims by the posters Decker and Peters you get more solutions than are allowed by the 4 linear equations with 4 variables. Now then, if you believed in mathematics then there wouldn't be a debate. You'd know immediately that they have to be lying. The problem here is that many of you have not a clue what mathematics is. You think it's a social thing. But there are people who do know what mathematics is. So all Decker and Peters did was give them time to the world. James Harris === Subject: Re: SF: Does this work? >>The mathematics is beautiful, and easy to prove. >>Such a simple answer to the problem, with such simple algebra. >>It is remarkable beyond belief. >>James Harris >> I suggest you ask yourself two questions: >> 1 Is there anything in your work that was not known to Euler, Fermat >> or Gauss? >> 2 Are you a better mathematician than Euler, Fermat or Gauss? >> Having answered them, go back and check over your work thoroughly to >> see if: >> a) there are any mistakes in it, >> b) it is faster than existing methods. >> As you have said, there have been mistakes in some of the stuff that >> you posted here. Given that you have made mistakes before you need to >> check carefully that you have not made mistakes again. >> I tried one of your previous factorising methods on my computer. It >> found factors, but it was about 10,000 times slower than Fermat's >> method. If a new method is to be useful, then it has to work faster >> than existing methods. >> rossum >The math is trivial. So I take it that you agree that there is nothing in your work that was not known to Euler, Fermat or Gauss. You have not answered my second question: Are you a better mathematician than Euler, Fermat or Gauss? >The problem here is that sci.math posters have routinely lied about my >math results. In the past there have been mistakes in what you posted here, as you yourself have said. It is not surprising if others suspect that there might be mistakes in what you are posting now. One of the things you could (and should) do better at is checking your work more thoroughly before you post it here, rather than posting it unchecked. Do you like having to apologise for avoidable mistakes? >Given 4 linear equations with 4 variables, do you or do you not get >explicit solutions? You may or you may not. There can be zero, one or many solutions. Taking two equations in two variables for simplicity: x = y + 3; x = y - 3 has no solutions. x = y + 3; x = 2y + 2 has one solution (x=4, y=1). x = y + 3; 2x = 2y + 6 has many solutions. >But if you look at claims by the posters Decker and Peters you get more >solutions than are allowed by the 4 linear equations with 4 variables. It is possible for there to be many solutions to a system of four linear equations, as I showed with my simple two variable example above. Check how many solutions are there to: w + x + y + z = 2 2w + 2x + 2y + 2z = 4 3w + 3x + 3y + 3z = 6 4w + 4x + 4y + 4z = 8 Yes, this is a trivial example, but it does make the point: you cannot just assume that a single solution exists. In the case of your currently proposed factoring method the equations you are using do have a single solution, but that is not something you can just assume. You have to show firstly that a solution exists and secondly that it is unique. rossum >James Harris === Subject: Re: SF: Does this work? rossum a .8ecrit : >> The mathematics is beautiful, and easy to prove. >> Such a simple answer to the problem, with such simple algebra. >> It is remarkable beyond belief. >> James Harris > I suggest you ask yourself two questions: > 1 Is there anything in your work that was not known to Euler, Fermat > or Gauss? > 2 Are you a better mathematician than Euler, Fermat or Gauss? > Having answered them, go back and check over your work thoroughly to > see if: > a) there are any mistakes in it, > b) it is faster than existing methods. > As you have said, there have been mistakes in some of the stuff that > you posted here. Given that you have made mistakes before you need to > check carefully that you have not made mistakes again. > I tried one of your previous factorising methods on my computer. It > found factors, but it was about 10,000 times slower than Fermat's > method. If a new method is to be useful, then it has to work faster > than existing methods. > rossum >> The math is trivial. > So I take it that you agree that there is nothing in your work that > was not known to Euler, Fermat or Gauss. Wrong. I take a genius of the caliber of Gauss at least to see that any linear system of 4 equations in four unknowns has a unique solution. In fact, even Gauss had missed this trivial fact. > You have not answered my second question: Are you a better > mathematician than Euler, Fermat or Gauss? See above. His results (like the fact that integers are irrationals) speak for themselves. >> The problem here is that sci.math posters have routinely lied about my >> math results. > In the past there have been mistakes in what you posted here, as you > yourself have said. It is not surprising if others suspect that there > might be mistakes in what you are posting now. One of the things you > could (and should) do better at is checking your work more thoroughly > before you post it here, rather than posting it unchecked. Do you > like having to apologise for avoidable mistakes? >> Given 4 linear equations with 4 variables, do you or do you not get >> explicit solutions? > You may or you may not. This is not an answer. Have you stopped beating your wife yet? There can be zero, one or many solutions. > Taking two equations in two variables for simplicity: > x = y + 3; x = y - 3 has no solutions. > x = y + 3; x = 2y + 2 has one solution (x=4, y=1). > x = y + 3; 2x = 2y + 6 has many solutions. So what is *the* answer? >> But if you look at claims by the posters Decker and Peters you get more >> solutions than are allowed by the 4 linear equations with 4 variables. > It is possible for there to be many solutions to a system of four > linear equations, as I showed with my simple two variable example > above. Check how many solutions are there to: > w + x + y + z = 2 > 2w + 2x + 2y + 2z = 4 > 3w + 3x + 3y + 3z = 6 > 4w + 4x + 4y + 4z = 8 > Yes, this is a trivial example, but it does make the point: you cannot > just assume that a single solution exists. You are sadly mistaken too, I see (or maybe another liar). There is a unique solution to the above (w=x=y=z=1/2); any other solution must be a mistake or a false solution, as math dont lie. > In the case of your currently proposed factoring method the equations > you are using do have a single solution, See? You lie just for the fun of it, while you know perfectly well JSH is right. You will be one of the first to die before 2007. but that is not something you > can just assume. You have to show firstly that a solution exists and > secondly that it is unique. > rossum >> James Harris === Subject: Re: SF: Does this work? <4487f769$0$7764$7a628cd7@news.club-internet.fr w + x + y + z = 2 > 2w + 2x + 2y + 2z = 4 > 3w + 3x + 3y + 3z = 6 > 4w + 4x + 4y + 4z = 8 > Yes, this is a trivial example, but it does make the point: you cannot > just assume that a single solution exists. > You are sadly mistaken too, I see (or maybe another liar). There is a > unique solution to the above (w=x=y=z=1/2); any other solution must be a > mistake or a false solution, as math dont lie. Actually you're wrong. That system is singular, consider w=2, x=y=z=0 or any permutation of that, or w=1,x=1,y=z=0 or any permutation of that, etc, etc.... There is a solution but it isn't unique. Rule #1: Pass high school algebra before you insist on replying to the various trolls in usenet about the greater lessons of mathematics. JSH is just a jerk and people who consistantly reply to all his stupid threads he starts are jerks too. At some point you have to stop playing with the troll and just move on. Tom === Subject: Re: SF: Does this work? tomstdenis@gmail.com a .8ecrit : > w + x + y + z = 2 > 2w + 2x + 2y + 2z = 4 > 3w + 3x + 3y + 3z = 6 > 4w + 4x + 4y + 4z = 8 > Yes, this is a trivial example, but it does make the point: you cannot > just assume that a single solution exists. >> You are sadly mistaken too, I see (or maybe another liar). There is a >> unique solution to the above (w=x=y=z=1/2); any other solution must be a >> mistake or a false solution, as math dont lie. > Actually you're wrong. That system is singular, consider w=2, x=y=z=0 > or any permutation of that, or w=1,x=1,y=z=0 or any permutation of > that, etc, etc.... There is a solution but it isn't unique. As I said, you lie. How could (1,0,0,1) be a solution when the obvious solution is (1/2,1/2,1/2,1/2) and math says no other can exist? Singular system, indeed. You just invent words to try to obfuscate the simple truth. > Rule #1: Pass high school algebra before you insist on replying to the > various trolls in usenet about the greater lessons of mathematics. Rule #1bis : learn to recognize tongue-in-cheek replies, or at least look at the poster history. > JSH is just a jerk and people who consistantly reply to all his stupid > threads he starts are jerks too. At some point you have to stop > playing with the troll and just move on. No, no, all this is good clean fun. I am even beginning to see what JSH find in it > Tom === Subject: Re: SF: Does this work? <4487f769$0$7764$7a628cd7@news.club-internet.fr w + x + y + z = 2 > 2w + 2x + 2y + 2z = 4 > 3w + 3x + 3y + 3z = 6 > 4w + 4x + 4y + 4z = 8 Yes, this is a trivial example, but it does make the point: you cannot > just assume that a single solution exists. > You are sadly mistaken too, I see (or maybe another liar). There is a > unique solution to the above (w=x=y=z=1/2); any other solution must be a > mistake or a false solution, as math dont lie. > Actually you're wrong. That system is singular, consider w=2, x=y=z=0 > or any permutation of that, or w=1,x=1,y=z=0 or any permutation of > that, etc, etc.... There is a solution but it isn't unique. > Rule #1: Pass high school algebra before you insist on replying to the > various trolls in usenet about the greater lessons of mathematics. > JSH is just a jerk and people who consistantly reply to all his stupid > threads he starts are jerks too. At some point you have to stop > playing with the troll and just move on. ... else you might find yourself losing the ability to spot jocular posts, and that would never do. -- Larry Lard Replies to group please === Subject: Re: The list of all natural numbers don't exist HI Barb, > D.9asn't it give you the slightest pause that disproving Cantor is > canonical crank-fodder (followed closely by disproving Special > Relativity)? Did you ever experience cranks doing such a thing? ;-) Albrecht already has proven the no-answer in abundance ... or ad nauseam to use a more fitting term. As for the 'lists' that caused concern here: meanwhile I believe that Albrecht and his counter-crank WM deliberatedly keep using that term to avoid the much stronger notion of one-to-one mapping. Viele Gr.9f¤e Klaus === Subject: Re: The list of all natural numbers don't exist > ... > Your proof fails, because when you apply the diagonal argument to the > natural numbers you obviously get something that is not on the list. > (And I am talking about list, not about actual list, whatever that > may mean.) But you have not shown that what you obtain is indeed a > natural number. > Let's use unary notation for the natural numbers: > x = one > xx = two > xxx = three ... > Here is the diagonal formula I want to use: > Let n be the number of x's in the row. > Make sure the diagonal number has at least (n+1) x's. > Applying this rule to the list above we get: xxxx > xxxx = four > Are you claiming that four is not a natural number? No. I said that about the list of natural numbers, not about a finite sublist of the natural numbers. > It is easily proven that the diagonal number has > exactly one more x than some member of the list. More strict, it is easily proven that the diagonal number has one more x than the last member of the list. This works well with lists that have a last element, and the proof applies only to lists that have a last member. The argument does not go through with lists that do *not* have a last element. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: The list of all natural numbers don't exist > ... > Your proof fails, because when you apply the diagonal argument to the > natural numbers you obviously get something that is not on the list. > (And I am talking about list, not about actual list, whatever that > may mean.) But you have not shown that what you obtain is indeed a > natural number. > Let's use unary notation for the natural numbers: > x = one > xx = two > xxx = three > ... > Here is the diagonal formula I want to use: > Let n be the number of x's in the row. > Make sure the diagonal number has at least (n+1) x's. > Applying this rule to the list above we get: xxxx > xxxx = four > Are you claiming that four is not a natural number? > No. I said that about the list of natural numbers, not about a finite > sublist > of the natural numbers. The formula I described works on any list of natural numbers. > It is easily proven that the diagonal number has > exactly one more x than some member of the list. > More strict, it is easily proven that the diagonal number has > one more x than the last member of the list. This works well > with lists that have a last element, and the proof applies only to > lists that have a last member. The argument does not go through > with lists that do *not* have a last element. It is more accurate to say the diagonal number has one more x than the largest number in the list. The proof doesn't assume the list of natural numbers is in any particular order. Of course, the argument above also shows that any list of natural numbers has a largest number. I can provide other proofs in you want. Russell - 2 many 2 count === Subject: Re: The list of all natural numbers don't exist > That's why there must be a line in my sketch (in this special case > exactly at least two lines in sequence) with X but no 0 since the > X in the first column are infinite many and the 0 in every column > are only finite many. > Nonsense. Each column has only a finite number of 0 's. > However, there are an infinite number of columns, so the > number of 0 's is infinite. So we cannot conlcude that > there is a line with X' but no 0. Maybe the number of all 0's in all columns are infinite. I don't > argue about this. Do you disagree that an infinite number has at least one more than any > finite number? It is clear that any infinite number must be infinitely greater than > any finite number. So, I guess you can say that it has at least > one more. This does not say very much. > But enough for me. > [At this point you should stop and carefully define finite and > infinite numbers (one standard way is to use cardinalities, > equivalence > classes of sets under the bijection relation. Since any set can be > bijected > to itself, all sets have cardinalities.)] > If the X's in the first column are infinite many > True > and the 0's in any > column are finite many > True >and maybe all sequences of X's and 0's > starts at the same line downwards > Meaningless. > how could it be that there is no > line with X but no 0. > There are an infinite number of columns, and every line contains > a finite number of X's, so it is clear that no line contains > an X but no 0. > There are an infinite number of lines and every line contains a finite > number of X. Is there a column which contains an infinite number of > 0? No > If yes: This is an infinite index number. Irrelevent, the answer is no. > If no: how could be this although every line which contains X > contains 0 too? All this tells us is that there are an infinite number of 0's. (true). It does not tell us that there are an infinite number of 0's in at least one column (false). This is straight operator dyslexia. For all x there exists a y such that P(x,y) is true does *not* imply There exists a y such that for all x, P(x,y) is true. I this case x is the line, y is the column, and P(x,y) is there is an 0 in column y of line x. It is true for every x, there is a y, such that there is a 0 in column y of line x. This does not imply There exists a y such that for every x there is a 0 in column y of line x This second statement is trivially false. -William Hughes === Subject: Re: The list of all natural numbers don't exist >How so? Cantor's theorem merely says that |X| < |P(X)| for all X. The >proof shows that if f: X -> P(X) is any mapping, then there exists y in >P(X) that is not in the range of f. The proof does not even mention lists. >> The cardinality of the power set of any set X strictly exceeds the >> cardinality of X, said Tom listlessly. >> I have never questioned this part of Cantor's theorem. >> It obviously applies to any finite set. >> I question whether there exists a set that contains all natural numbers. >> The diagonal argument can be used to prove no set >> contains every natural number. > No, it can't. See below. >> (More precisely, no set that contains finite representations of >> natural numbers can contain every finite representation of >> a natural number.) > Hmm. Set of natural numbers is not finite? Novel idea. I never said the set of natural numbers is finite or infinite. I said no set can contain every finite representation of a natural number. I do agree that no finite set can contain every natural number. Since the diagonal argument (and other arguments) can be used to show that all sets are finite, one must question if there exists a set of all natural numbers. It is the assumption that there exists a set of all natural numbers that requires us to distinguish between finite and infinite sets. Finite sets are those sets that don't contradict the axioms of set theory. Infinite sets are those sets that would lead to contradictions if they were finite sets. Russell - C# give programmers greater opportunity to write bad code. === Subject: Re: The list of all natural numbers don't exist Discussion, linux) <7MCdnXban_MB_hrZnZ2dnUVZ_rSdnZ2d@comcast.com It is the assumption that there exists a set of all natural numbers > that requires us to distinguish between finite and infinite sets. > Finite sets are those sets that don't contradict the axioms > of set theory. Infinite sets are those sets that would lead > to contradictions if they were finite sets. Infinite sets (like N) don't contradict the axioms of set theory. Therefore, infinite sets are finite in the sense defined above. Brilliant reasoning that. I also like your definition of infinite. I want to try it out on something. Let's see. A number is prime if it is not non-trivially divisible. Composite numbers are those numbers that would lead to contradictions if they were prime numbers. It's like poetry, dude! That's way cooler than saying composite means not prime! -- Jesse F. Hughes I want to really eat myself, so then I'll be a coalgebra. -- Quincy P. Hughes, Age 3 1/2 === Subject: fourier series for rectified sine wave Why is the b1 coefficient non-zero for rectified half-wave but zero for rectified full wave? I uinderstand bk coefficients should be zero since a rectified full wave is an even function. But when doing the integration for b1, it works out to non-zero. Benn === Subject: Re: fourier series for rectified sine wave >Why is the b1 coefficient non-zero for rectified half-wave but zero for >rectified full wave? >I uinderstand bk coefficients should be zero since a rectified full >wave is an even function. >But when doing the integration for b1, it works out to non-zero. I'm not sure what a rectified full wave is, but if it's even, and assuming that bk is the sine coefficient, then bj is indeed 0. You did the integration wrong - hard to say what the error is since you don't show what you did. >Benn ************************ David C. Ullrich === Subject: Pedal Triangles I suspect the following fact: given a triangle ABC and a point O, the pedal triangle PQR, obtained projecting O on the sides of ABC, is similar to ABC if and only if O is the cirumcenter of ABC. What do you think about? Maury === Subject: Re: Pedal Triangles > I suspect the following fact: given a triangle ABC and a > point O, the pedal triangle PQR, obtained projecting > O on the sides of ABC, is similar to ABC if and only if > O is the cirumcenter of ABC. A very tempting conjecture, but alas not a correct one. Here is a simple counterexample: Consider the right triangle with vertices at (0,0), (0,20), (10,0). The circumcenter is at (5,10), and the corresponding pedal triangle is indeed similar to the original triangle, but it is not the only point with this property. Another is (8,4), also on the hypotenuse; the pedal triangle then has vertices (8,4), (8,0), (0,4), so it is also a right triangle with one leg twice the length of the other. Let's see if I can draw an ASCII picture: (0,20)* | | | | | | | (0,4)* *(8,4) | (0,0)*-------*-*(10,0) (8,0) -- John Garrideb Moorville, KS jgarrideb@springmail.com === Subject: Re: Pedal Triangles > I suspect the following fact: given a triangle ABC > and a > point O, the pedal triangle PQR, obtained > projecting > O on the sides of ABC, is similar to ABC if and > only if > O is the cirumcenter of ABC. > A very tempting conjecture, but alas not a correct > one. Here > is a simple counterexample: > Consider the right triangle with vertices at (0,0), > (0,20), (10,0). > The circumcenter is at (5,10), and the corresponding > pedal > triangle is indeed similar to the original triangle, > but it is not > the only point with this property. Another is (8,4), > also on > the hypotenuse; the pedal triangle then has vertices > (8,4), > (8,0), (0,4), so it is also a right triangle with one > leg twice > the length of the other. > Let's see if I can draw an ASCII picture: > (0,20)* > | > | > | > | > | > | > | > (0,4)* *(8,4) > | > (0,0)*-------*-*(10,0) > (8,0) > -- > John Garrideb > Moorville, KS > jgarrideb@springmail.com Ok! However, one can reinforce hypotheses in the following way. Given a triangle ABC and a point O, let PQR be the pedal triangle obtained projecting O respectively on BC, AC and AB. If the angle QPR is equal to BAC, and the angle PQR is equal to ABC, then O is the circumcenter. I think it's true. Anyway, exploring with a graphic software what happens when AB=AC, I find a surprise. There are three positions of O which make PQR similar to ABC. Besides, there infinitely many positions of O which make PQR isosceles: I couldn't identify, however, the locus which O describes in this way. It's a curve, but I don't know what type of curve. Maury === Subject: Re: Pedal Triangles > I suspect the following fact: given a triangle ABC > and a > point O, the pedal triangle PQR, obtained > projecting > O on the sides of ABC, is similar to ABC if and > only if > O is the cirumcenter of ABC. > A very tempting conjecture, but alas not a correct > one. Here > is a simple counterexample: > Consider the right triangle with vertices at (0,0), > (0,20), (10,0). > The circumcenter is at (5,10), and the corresponding > pedal > triangle is indeed similar to the original triangle, > but it is not > the only point with this property. Another is (8,4), > also on > the hypotenuse; the pedal triangle then has vertices > (8,4), > (8,0), (0,4), so it is also a right triangle with one > leg twice > the length of the other. > Let's see if I can draw an ASCII picture: > (0,20)* > | > | > | > | > | > | > | > (0,4)* *(8,4) > | > (0,0)*-------*-*(10,0) > (8,0) > -- > John Garrideb > Moorville, KS > jgarrideb@springmail.com Ok! However, one can reinforce hypotheses in the following way. Given a triangle ABC and a point O, let PQR be the pedal triangle obtained projecting O respectively on BC, AC and AB. If the angle QPR is equal to BAC, and the angle PQR is equal to ABC, then O is the circumcenter. I think it's true. Anyway, exploring with a graphic software what happens when AB=AC, I find a surprise. There are three positions of O which make PQR similar to ABC. Besides, there infinitely many positions of O which make PQR isosceles: I couldn't identify, however, the locus which O describes in this way. It's a curve, but I don't know what type of curve. Maury === Subject: Re: Pedal Triangles > > I suspect the following fact: given a triangle > ABC > and a > point O, the pedal triangle PQR, obtained > projecting > O on the sides of ABC, is similar to ABC if and > only if > O is the cirumcenter of ABC. > > A very tempting conjecture, but alas not a correct > one. Here > is a simple counterexample: > > Consider the right triangle with vertices at > (0,0), > (0,20), (10,0). > The circumcenter is at (5,10), and the > corresponding > pedal > triangle is indeed similar to the original > triangle, > but it is not > the only point with this property. Another is > (8,4), > also on > the hypotenuse; the pedal triangle then has > vertices > (8,4), > (8,0), (0,4), so it is also a right triangle with > one > leg twice > the length of the other. > > Let's see if I can draw an ASCII picture: > > (0,20)* > | > | > | > | > | > | > | > (0,4)* *(8,4) > | > (0,0)*-------*-*(10,0) > (8,0) > > -- > John Garrideb > Moorville, KS > jgarrideb@springmail.com > > Ok! However, one can reinforce hypotheses in the > following way. > Given a triangle ABC and a point O, let PQR be the > pedal > triangle obtained projecting O respectively on BC, AC > and AB. If the angle QPR is equal to BAC, and the > angle > PQR is equal to ABC, then O is the circumcenter. I was right. This is a simple consequence of the Sines Law. > I think it's true. Anyway, exploring with a graphic > software what happens when AB=AC, I find a surprise. > There are three positions of O which make PQR > similar > to ABC. Besides, there infinitely many positions of > which make PQR isosceles: I couldn't identify, > however, > the locus which O describes in this way. It's a > curve, > but I don't know what type of curve. Here methods of euclidean geometry are not sufficient, I think. Maybe an analytical study is needed. Maury > Maury === Subject: Anybody got an IQ over 20 here? ... if so, please confirm whether the US government (and mainstream media) are telling the truth about the events of September 11, 2001; or whether they are lying and covering up for the actual -zookumar- === Subject: Re: Anybody got an IQ over 20 here? > ... if so, please confirm whether the US government (and > mainstream media) are telling the truth about the events of September > 11, 2001; or whether they are lying and covering up for the actual > -zookumar- Danger and opportunity are confirmed; we face an over-ripened crisis. -uptimod The Chinese word for crisis is written by joining two ideograms together. When these ideograms, are presented separately they stand for two different ideas or concepts. The first ideogram stands for danger and the second ideogram stands for opportunity. DANGER-Danger of injury if we humans fail to understand. Danger of destruction if we humans act in ignorance. OPPORTUNITY-Opportunity for growth if we humans understand. Opportunity for survival if we humans act in knowledge. The life form that survives and evolves is the one that can grasp the opportunity and avoid the risk. This is the very process of life. It is the engine of evolution. http://futurepositive.synearth.net/stories/storyReader$53 === Subject: Series question Hello everybody, I encountered the following problem : let a_n and b_n are sequences of positive numbers such that b_n tends to infinity.Let SUM(a_n*b_n)n)----> 0? It is obvious that liminf b_n*Sum(a_k;k>n)=0. === Subject: Re: Series question On 8 Jun 2006 02:34:12 -0700, ManOfLight I encountered the following problem : >let a_n and b_n are sequences of positive numbers such that b_n tends >to infinity.Let SUM(a_n*b_n)b_n*Sum(a_k;k>n)----> 0? It is obvious that liminf b_n*Sum(a_k;k>n)=0. This is actually much simpler than I realized a second ago. If n is even let b_n = n^4, a_n = 1/n^6. If n is odd let b_n = n, a_n = 1/n^3. Then b_n -> infinity, sum b_n*a_n is finite, and if n is even then b_n*a_{n+1} = n^4/(n+1)^3, which tends to infinity. ************************ David C. Ullrich === Subject: Re: Series question On 8 Jun 2006 02:34:12 -0700, ManOfLight I encountered the following problem : >let a_n and b_n are sequences of positive numbers such that b_n tends >to infinity.Let SUM(a_n*b_n)b_n*Sum(a_k;k>n)----> 0? It is obvious that liminf b_n*Sum(a_k;k>n)=0. I think the answer is clearly no, although writing down an example explicitly might be a little tedious. The idea is this: There will be subsequence of n's with the property that b_n is much larger than b_{n+1} (although they're both very large), and then all the b_m for m > n+1 are even much much larger. This allows you to have b_n*a_{n+1} large... ************************ David C. Ullrich === Subject: Re: Anyone got any good (or bad) mathematical jokes? > What can I say, I'm bored. Complex analysis is all fun and games until somebody loses an i. === Subject: Re: Anyone got any good (or bad) mathematical jokes? > It doesn't work that way in USENET. I can only point you to go > elsewhere that are more appropriate for jokes; but your request > for someone to go away simply because he pointed out your > anti-netiquette cluelessness is neither appropriate nor effective. > No its quite simple, what I'm saying don't respond to this post if you > can't do anything other than be insulting. It's pointing out to you that there are JOKE newsgroups for jokes and humor, mathematical or otherwise. sci.math is simply NOT a newsgroup designed for mathematical jokes. -- Bob. === Subject: Re: Anyone got any good (or bad) mathematical jokes? >A bunch of Polish scientists decided to flee their repressive government by > hijacking an airliner and forcing the pilot to fly them to a western > half-plane. Are you positive? Phil -- The man who is always worrying about whether or not his soul would be damned generally has a soul that isn't worth a damn. -- Oliver Wendell Holmes, Sr. (1809-1894), American physician and writer === Subject: [NT] How many (minimum) arithmetic progressions descompose a set? Def: Let be S an arbitrary non-empty set. We say that S is n-{arithmetic descomposable} if there are A_1, ...., A_n, with all A_i arithmetic progressions. My question is if we could determine a good bound for n (obviusly n = card(S) ;) Respectively with geometrically descomposable Xan. === Subject: Re: [NT] How many (minimum) arithmetic progressions descompose a set? > Def: Let be S an arbitrary non-empty set. > We say that S is n-{arithmetic decomposable} if there are A_1, ...., > A_n, with all A_i arithmetic progressions. Rather that guess how to rewrite this, would you clarify it? > My question is if we could determine a good bound for n (obviously n = > card(S) ;) What if S is infinite? > Respectively with geometrically decomposable Nonsense respectively grows exponentially instead of arithmetically. === Subject: Re: PDE solution help >I have hurt myself trying to solve the following PDE the last few days > and need help. The PDE is a heat transfer problem that boils down to. > d2T/dr2 + (2/r)dT/dr - AdT/dt +B/r = 0 > A,B = const. > Bounday conditions > T(r=0,t) ~= To > T(r=a,t) = Tm > Any help pointing me in the right direction would be a big help. > TIA > Zine Use seperation of variables in product and sum form Y_rr + 2/r*Y_r - AY_t + B/r = 0 r*Y_rr + B + 2Y_r = r*A*Y_t Y(r,t) = R(r)*T(t) Y_rr = T*R_rr Y_r = T*R_r Y_t = R*T_t r*T*R_rr + B + 2*T*R_r = r*A*R*T_t if B = 0 then (r*R_rr + 2*R_r)/(r*A*R) = A*T_t/T T_t = C*T/A => T(t) = c*exp(C*t/A) and r*R_rr + 2*R_r = r*R*C R(r) = 1/r*(a*sinh(sqrt(C)*r) + b*cosh(sqrt(C)*r)) so we can try a solution of the form Y(r,t) = R(r)*T(t) - 1/2*B*(a1*r + a2 + a3/r) (actually one might want to try R(r)*T(r) - B*sum(a_k*r^k,k=-oo..oo) but should simplify whats above) after trying you will have to set a1 = 1 which gives the solution to the pde as Y(r,t) = c*1/r*(a*sinh(sqrt(C)*r) + b*cosh(sqrt(C)*r))*exp(C*t/A) - 1/2*B*(r + a2 + a3/r) Hope that makes sense, Jon === Subject: Re: PDE solution help >I have hurt myself trying to solve the following PDE the >last few days >and need help. The PDE is a heat transfer problem that >boils down to. >d2T/dr2 + (2/r)dT/dr - AdT/dt +B/r = 0 >A,B = const. >Bounday conditions >T(r=0,t) ~= To >T(r=a,t) = Tm >Any help pointing me in the right direction would be a >big help. >TIA >Zine your equation can be written as A*dT/dt = 1/r^2 * d/dr ( r^2 * dT/dr) + B/r. This is the heat conduction equation in a sphere with a singularity at r=0 (the B/r - term). So the solution will be unbounded there and I doubt that it is well-posed in the way it stands. What do you want to model by the B/r - term ? Best wishes Torsten. === Subject: Re: PDE solution help > your equation can be written as > A*dT/dt = 1/r^2 * d/dr ( r^2 * dT/dr) + B/r. > This is the heat conduction equation in a sphere with > a singularity at r=0 (the B/r - term). > So the solution will be unbounded there and I doubt > that it is well-posed in the way it stands. Well, if the solution behaves approximately as T=a+b*r around r=0, then plugging in and requiring the right hand side to be finite, we get the condition 2b+B=0, which can be solved for b. Furthermore, the boundary condition gives T0=a. So it seems that a well-behaved solution might exist. -Michael. === Subject: Re: PDE solution help >> your equation can be written as >> A*dT/dt = 1/r^2 * d/dr ( r^2 * dT/dr) + B/r. >> This is the heat conduction equation in a sphere with >> a singularity at r=0 (the B/r - term). >> So the solution will be unbounded there and I doubt >> that it is well-posed in the way it stands. >Well, if the solution behaves approximately as T=a+b*r >around r=0, then >plugging in and requiring the right hand side to be >finite, we get the >condition 2b+B=0, which can be solved for b. >Furthermore, the boundary >condition gives T0=a. >So it seems that a well-behaved solution might exist. >-Michael. Yes, you are right. If a stationary solution to the problem above exists, it can be shown that it is of the form T(r) = -B/2*r + constant if T is assumed bounded at r=0. Because physically, it seems reasonable to expect a stationary solution to exist if Dirichlet boundary conditions are imposed, there is only one free parameter in the solution. This seems to indicate that it's impossible to impose both T(r=0) and T(r=a). Maybe T(r=a)=T_a and |T(r=0)| < oo is a better choice ... Best wishes Torsten. === Subject: Re: JSH: SF: Finally, surrogate factoring On Wed, 07 Jun 2006 10:51:53 -0400, Rick Decker On Tue, 06 Jun 2006 20:54:03 -0400, Rick Decker [...] >>What I don't understand is how anything other than that outcome could be >>_hoped_ for here. No amount of rearranging and cross-substituting the >>initial equations (whatever they may be) is going to yield new information, >>and there's never a step that even requires the quantities to be integers >>(as opposed to, e.g., arbitrary complex numbers). How can someone imagine >>that insight into integer factorization could result from this insight-less >>symbol-pushing? >I think that what we interpret as obfuscation on James' part is actually >a consequence of the fact that his understanding is extremely shallow. >This is, I think, the reason that he thinks his prime counting >function is truly new and innovative--he really is incapable of even >the slightest bit of abstraction that to all mathematicians is as >natural as breathing. >> Huh. Ya think? >You missed my point. Until recently I thought that it was James' >narcisism that motivated his jaw-dropping refusal to see that >there was nothing new in his prime-counting function. Now, though, >I have come to the view that he's lacking an important faculty >that all mathematicians have. It's pitiful, really. I don't think I missed that point - I thought that the point you make was obvious. Sorry. >> Both you guys must be new here. >Hrrumph. Now you're playing the newbie card. I think I'll >contact the OSU administration. Oh my gosh. Ok, never mind. You're absolutely right about > ... > So completing the square w.r.t y first yields > (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T > Completing the square w.r.t. z first yields > (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T [jstevh@msn.com] >> The only problem I have with that is that you have too many solutions >> for y. >> There is one explicit solution for y given by solving the 4 linear >> equations. >> That solution is >> y = (7g_1 - 3g_2 + 5f_1 - 3f_2)/4 Woo hoo! Correct! >> which what you give covers, but you could have gotten that by working >> backwards FROM the explicit solution, He could have, but he didn't. Neither did I. >> and there is one problem. >> If T has only two prime factors p_1 and p_2, there are 8 possible >> values for y for a given f_1 and f_2, which represent g_1 = T, p_1, >> p_2, or 1, and the negatives, and g_2 = 1, p_2, p_1, or 1, and the >> negatives. >> But your solution has more than that because it gives >> y = (5f_1 - 3f_2 + 21g_1 - g_2)/4 >> as a solution as well. [Rick Decker] > No. However, it would be interesting to see how you got this. > Ah. Perhaps you were working from h_1 * h_2 = 21 * g_1 * g_2, > like this: Are you psychic or what? I had no idea how he came up with the thing containing 21g_1, and never would have guessed he was just pulling it out of his butt :-) > Let h_1 and h_2 be chosen so that h_1 * h_2 = 21 * T > h_1 + h_2 = 10*y + 42*z + 19*f_1 - 3*f_2 [1] > h_2 - h_1 = 4*y - 5*f_1 + 3*f_2 [2] I think you meant to write h_1 - h_2 on the LHS of [2]. > Then we can write > (10*y + 42*z + 19*f_1 - 3*f_2)^2 = (4*y - 5*f_1 + 3*f_2)^2 + 84*T > in the form > (h_1 + h_2)^2 = (h_1 - h_2)^2 + 4 * h_1 * h_2 This part would be clearer with the correction above. > Then, from [1] and [2] we solve for y to get > y = (5*f_1 - 3*f_2 + h_1 - h_2) / 4 [3] While this conclusion _needs_ the correction above. > Then, since h_1 * h_2 = 21 * T = 21 * g_1 * g_2 we may as well > pick h_1 = 21 * g_1 and h_2 = g_2 so [3] becomes > y = (5*f_1 - 3*f_2 + 21*g_1 - g_2) / 4 > Right? Yes, you are psychic! > If that was your reasoning, it's wrong. You can't pick any old > values for h_1 and h_2. Watch: > Solving [1] and [2] for h_1 and h_2 we get > h_1 = 7(y + 3 * z + f_1) > h_2 = 3(y + 7 * z + 4 * f_1 - f_2) That also needs the correction above ;-) > But from your original four linear equations we can derive > g_1 = y + 3 * z + f_1 > g_2 = y + 7 * z + 4 * f_1 - f_2 > in other words, we are forced to choose > h_1 = 7 * g_1 > h_2 = 3 * g_2 > and not your h_1 = 21 * g_1 and h_2 = g_2. And to force the conclusion, in that case [3] becomes y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 But let's give James something else to worry about :-) Take (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T expand it, use the quadratic equation to solve for y, and then substitute to get rid of z and T: z = -(3*f_1 - f_2 + g_1 - g_2)/4 T = g_1*g_2 The result is: y = (5*f_1 - 3*f_2 + 5*g_1 - 5*g_2 +/- 2*(g_1 + g_2))/4 Pick + and you get the result James wants: y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 Pick - and it's different: y = (5*f_1 - 3*f_2 + 3*g_1 - 7*g_2)/4 Woo hoo! Centuries of mathematics down the tubes again, or can James spot the bogosity? Hint #1: this isn't an algebraic error; you really do get that result for y. Hint #2: you get the same two results for y if you do the same thing but starting from (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T instead. > It's why I decided after thinking that this method MUST lead to a >> surrogate factoring solution as the only way the algebra can avoid the >> contradiction is to use a surrogate factorization. >> You claim otherwise with your post. > No. That was your (incorrect) deduction, as I show above. The suspense is killing me. >> Your claims mean that 4 linear equations can be wrong. >> I wonder if you just lied. > You just can't resist, can you? Are you naturally boorish, or do > you have to work at it? I strongly suspect that bit of gratuitous assholishness was deliberate. God only knows why, but James got it into his head that he needs to _provoke_ people into replying when he thinks they know something he wants to find out. That's just his despicable way of trying to goad you into doing his work for him. It's especially idiotic in this case, since if he had any memory he'd recall that you typically respond much better to polite requests than to his stupid baiting tactics. But, in this case, I'm afraid what he'll take away is ha! it worked again, without a shadow of a clue that it was neither necessary nor helpful to behave like an ass. === Subject: Re: JSH: SF: Finally, surrogate factoring > [Rick Decker] y = (5f_1 - 3f_2 + 21g_1 - g_2)/4 >as a solution as well. > [Rick Decker] >>No. However, it would be interesting to see how you got this. I had no idea how he came up with the thing > containing 21g_1, and never would have guessed he was just pulling it out of > his butt :-) Surely you're not surprised. >>Let h_1 and h_2 be chosen so that h_1 * h_2 = 21 * T >> h_1 + h_2 = 10*y + 42*z + 19*f_1 - 3*f_2 [1] >> h_2 - h_1 = 4*y - 5*f_1 + 3*f_2 [2] > I think you meant to write h_1 - h_2 on the LHS of [2]. Indeed I did. >>Then we can write >> (10*y + 42*z + 19*f_1 - 3*f_2)^2 = (4*y - 5*f_1 + 3*f_2)^2 + 84*T >>in the form >> (h_1 + h_2)^2 = (h_1 - h_2)^2 + 4 * h_1 * h_2 > This part would be clearer with the correction above. Yes. >>Then, from [1] and [2] we solve for y to get >> y = (5*f_1 - 3*f_2 + h_1 - h_2) / 4 [3] > While this conclusion _needs_ the correction above. Yes, yes. >>Then, since h_1 * h_2 = 21 * T = 21 * g_1 * g_2 we may as well >>pick h_1 = 21 * g_1 and h_2 = g_2 so [3] becomes >> y = (5*f_1 - 3*f_2 + 21*g_1 - g_2) / 4 >>Right? > Yes, you are psychic! I knew you'd say that. >>If that was your reasoning, it's wrong. You can't pick any old >>values for h_1 and h_2. Watch: >>Solving [1] and [2] for h_1 and h_2 we get >> h_1 = 7(y + 3 * z + f_1) >> h_2 = 3(y + 7 * z + 4 * f_1 - f_2) > That also needs the correction above ;-) (Grr). Yes, yes, yes! >>But from your original four linear equations we can derive >> g_1 = y + 3 * z + f_1 >> g_2 = y + 7 * z + 4 * f_1 - f_2 >>in other words, we are forced to choose >> h_1 = 7 * g_1 >> h_2 = 3 * g_2 >>and not your h_1 = 21 * g_1 and h_2 = g_2. > And to force the conclusion, in that case [3] becomes > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > But let's give James something else to worry about :-) Take > (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T > expand it, use the quadratic equation to solve for y, and then substitute to > get rid of z and T: > z = -(3*f_1 - f_2 + g_1 - g_2)/4 > T = g_1*g_2 > The result is: > y = (5*f_1 - 3*f_2 + 5*g_1 - 5*g_2 +/- 2*(g_1 + g_2))/4 > Pick + and you get the result James wants: > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > Pick - and it's different: > y = (5*f_1 - 3*f_2 + 3*g_1 - 7*g_2)/4 > Woo hoo! Centuries of mathematics down the tubes again, or can James spot > the bogosity? Hint #1: this isn't an algebraic error; you really do get > that result for y. Hint #2: you get the same two results for y if you do > the same thing but starting from > (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T > instead. Hehe. I predict that this section (cute, BTW) will generate no response. >>>You just can't resist, can you? Are you naturally boorish, or do >>you have to work at it? > I strongly suspect that bit of gratuitous assholishness was deliberate. God > only knows why, but James got it into his head that he needs to _provoke_ > people into replying when he thinks they know something he wants to find > out. That's just his despicable way of trying to goad you into doing his > work for him. It's especially idiotic in this case, since if he had any > memory he'd recall that you typically respond much better to polite requests > than to his stupid baiting tactics. > But, in this case, I'm afraid what he'll take away is ha! it worked again, > without a shadow of a clue that it was neither necessary nor helpful to > behave like an ass. Sadly, I predict you're right again. Rick === Subject: Re: JSH: SF: Finally, surrogate factoring <5N6dnYHVYq4DBBjZnZ2dnUVZ_rWdnZ2d@hamilton.edu> But your solution has more than that because it gives y = (5f_1 - 3f_2 + 21g_1 - g_2)/4 as a solution as well. > [Rick Decker] >>No. However, it would be interesting to see how you got this. > to anyone who hasn't been following closely Are you psychic or what? > Yes, and I knew you'd ask that. > I had no idea how he came up with the thing > containing 21g_1, and never would have guessed he was just pulling it out of > his butt :-) > Surely you're not surprised. >>Let h_1 and h_2 be chosen so that h_1 * h_2 = 21 * T >> h_1 + h_2 = 10*y + 42*z + 19*f_1 - 3*f_2 [1] >> h_2 - h_1 = 4*y - 5*f_1 + 3*f_2 [2] > I think you meant to write h_1 - h_2 on the LHS of [2]. > Indeed I did. >>Then we can write >> (10*y + 42*z + 19*f_1 - 3*f_2)^2 = (4*y - 5*f_1 + 3*f_2)^2 + 84*T >>in the form >> (h_1 + h_2)^2 = (h_1 - h_2)^2 + 4 * h_1 * h_2 > This part would be clearer with the correction above. > Yes. >>Then, from [1] and [2] we solve for y to get >> y = (5*f_1 - 3*f_2 + h_1 - h_2) / 4 [3] > While this conclusion _needs_ the correction above. > Yes, yes. >>Then, since h_1 * h_2 = 21 * T = 21 * g_1 * g_2 we may as well >>pick h_1 = 21 * g_1 and h_2 = g_2 so [3] becomes >> y = (5*f_1 - 3*f_2 + 21*g_1 - g_2) / 4 >>Right? > Yes, you are psychic! > I knew you'd say that. >>If that was your reasoning, it's wrong. You can't pick any old >>values for h_1 and h_2. Watch: >>Solving [1] and [2] for h_1 and h_2 we get >> h_1 = 7(y + 3 * z + f_1) >> h_2 = 3(y + 7 * z + 4 * f_1 - f_2) > That also needs the correction above ;-) > (Grr). Yes, yes, yes! >>But from your original four linear equations we can derive >> g_1 = y + 3 * z + f_1 >> g_2 = y + 7 * z + 4 * f_1 - f_2 >>in other words, we are forced to choose >> h_1 = 7 * g_1 >> h_2 = 3 * g_2 >>and not your h_1 = 21 * g_1 and h_2 = g_2. > And to force the conclusion, in that case [3] becomes > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > But let's give James something else to worry about :-) Take > (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T > expand it, use the quadratic equation to solve for y, and then substitute to > get rid of z and T: > z = -(3*f_1 - f_2 + g_1 - g_2)/4 > T = g_1*g_2 > The result is: > y = (5*f_1 - 3*f_2 + 5*g_1 - 5*g_2 +/- 2*(g_1 + g_2))/4 > Pick + and you get the result James wants: > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > Pick - and it's different: > y = (5*f_1 - 3*f_2 + 3*g_1 - 7*g_2)/4 > Woo hoo! Centuries of mathematics down the tubes again, or can James spot > the bogosity? Hint #1: this isn't an algebraic error; you really do get > that result for y. Hint #2: you get the same two results for y if you do > the same thing but starting from > (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T > instead. > Hehe. I predict that this section (cute, BTW) will generate no response. >>>You just can't resist, can you? Are you naturally boorish, or do >>you have to work at it? > I strongly suspect that bit of gratuitous assholishness was deliberate. God > only knows why, but James got it into his head that he needs to _provoke_ > people into replying when he thinks they know something he wants to find > out. That's just his despicable way of trying to goad you into doing his > work for him. It's especially idiotic in this case, since if he had any > memory he'd recall that you typically respond much better to polite requests > than to his stupid baiting tactics. > But, in this case, I'm afraid what he'll take away is ha! it worked again, > without a shadow of a clue that it was neither necessary nor helpful to > behave like an ass. > Sadly, I predict you're right again. > Rick You're both lying. I am going to warn you. Neither of you will live to see 2007 because of this lie becaue there will be angry people who will kill you, as there is so much money at stake. Billions will be lost. It's not like you can reverse it now either. You killed yourselves. It might have seemed like a small lie to both of you, but your names will live in infamy, while you will not live at all. James Harris === Subject: Re: JSH: SF: Finally, surrogate factoring [...] > James Harris I encourage you to read this: http://www.nimh.nih.gov/publicat/NIMHocd.pdf David Bernier === Subject: Re: JSH: SF: Finally, surrogate factoring >> [Rick Decker] >> [jstevh@msn.com] >> But your solution has more than that because it gives >>y = (5f_1 - 3f_2 + 21g_1 - g_2)/4 >>as a solution as well. >> [Rick Decker] >No. However, it would be interesting to see how you got this. >> > to anyone who hasn't been following closely> Are you psychic or what? >> Yes, and I knew you'd ask that. >> I had no idea how he came up with the thing >> containing 21g_1, and never would have guessed he was just pulling it >> out of >> his butt :-) >> Surely you're not surprised. >Let h_1 and h_2 be chosen so that h_1 * h_2 = 21 * T h_1 + h_2 = 10*y + 42*z + 19*f_1 - 3*f_2 [1] > h_2 - h_1 = 4*y - 5*f_1 + 3*f_2 [2] >> I think you meant to write h_1 - h_2 on the LHS of [2]. >> Indeed I did. >Then we can write (10*y + 42*z + 19*f_1 - 3*f_2)^2 = (4*y - 5*f_1 + 3*f_2)^2 + 84*T in the form (h_1 + h_2)^2 = (h_1 - h_2)^2 + 4 * h_1 * h_2 >> This part would be clearer with the correction above. >> Yes. >Then, from [1] and [2] we solve for y to get y = (5*f_1 - 3*f_2 + h_1 - h_2) / 4 [3] >> While this conclusion _needs_ the correction above. >> Yes, yes. >Then, since h_1 * h_2 = 21 * T = 21 * g_1 * g_2 we may as well >pick h_1 = 21 * g_1 and h_2 = g_2 so [3] becomes y = (5*f_1 - 3*f_2 + 21*g_1 - g_2) / 4 Right? >> Yes, you are psychic! >> I knew you'd say that. >If that was your reasoning, it's wrong. You can't pick any old >values for h_1 and h_2. Watch: Solving [1] and [2] for h_1 and h_2 we get h_1 = 7(y + 3 * z + f_1) > h_2 = 3(y + 7 * z + 4 * f_1 - f_2) >> That also needs the correction above ;-) >> (Grr). Yes, yes, yes! >But from your original four linear equations we can derive g_1 = y + 3 * z + f_1 > g_2 = y + 7 * z + 4 * f_1 - f_2 in other words, we are forced to choose h_1 = 7 * g_1 > h_2 = 3 * g_2 and not your h_1 = 21 * g_1 and h_2 = g_2. >> And to force the conclusion, in that case [3] becomes >> y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 >> But let's give James something else to worry about :-) Take >> (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T >> expand it, use the quadratic equation to solve for y, and then >> substitute to >> get rid of z and T: >> z = -(3*f_1 - f_2 + g_1 - g_2)/4 >> T = g_1*g_2 >> The result is: >> y = (5*f_1 - 3*f_2 + 5*g_1 - 5*g_2 +/- 2*(g_1 + g_2))/4 >> Pick + and you get the result James wants: >> y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 >> Pick - and it's different: >> y = (5*f_1 - 3*f_2 + 3*g_1 - 7*g_2)/4 >> Woo hoo! Centuries of mathematics down the tubes again, or can James >> spot >> the bogosity? Hint #1: this isn't an algebraic error; you really do >> get >> that result for y. Hint #2: you get the same two results for y if you >> do >> the same thing but starting from >> (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T >> instead. >> Hehe. I predict that this section (cute, BTW) will generate no response. >I wonder if you just lied. >You just can't resist, can you? Are you naturally boorish, or do >you have to work at it? >> I strongly suspect that bit of gratuitous assholishness was deliberate. >> God >> only knows why, but James got it into his head that he needs to >> _provoke_ >> people into replying when he thinks they know something he wants to >> find >> out. That's just his despicable way of trying to goad you into doing >> his >> work for him. It's especially idiotic in this case, since if he had >> any >> memory he'd recall that you typically respond much better to polite >> requests >> than to his stupid baiting tactics. >> But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful to >> behave like an ass. >> Sadly, I predict you're right again. >> Rick > You're both lying. > I am going to warn you. > Neither of you will live to see 2007 because of this lie becaue there > will be angry people who will kill you, as there is so much money at > stake. > Billions will be lost. > It's not like you can reverse it now either. > You killed yourselves. > It might have seemed like a small lie to both of you, but your names > will live in infamy, while you will not live at all. > James Harris James, you're the one lying. When you say someone's lying, you really mean you don't understand what they're talking about. Sorry to bust your bubble, but you're not the genius you think you are. Dave === Subject: Re: JSH: SF: Finally, surrogate factoring > [Rick Decker] >> > [Rick Decker] >> No. However, it would be interesting to see how you got this. >> > to anyone who hasn't been following closely Are you psychic or what? >> Yes, and I knew you'd ask that. > I had no idea how he came up with the thing > containing 21g_1, and never would have guessed he was just pulling it out of > his butt :-) >> Surely you're not surprised. >> Let h_1 and h_2 be chosen so that h_1 * h_2 = 21 * T >> h_1 + h_2 = 10*y + 42*z + 19*f_1 - 3*f_2 [1] >> h_2 - h_1 = 4*y - 5*f_1 + 3*f_2 [2] > I think you meant to write h_1 - h_2 on the LHS of [2]. >> Indeed I did. >> Then we can write >> (10*y + 42*z + 19*f_1 - 3*f_2)^2 = (4*y - 5*f_1 + 3*f_2)^2 + 84*T >> in the form >> (h_1 + h_2)^2 = (h_1 - h_2)^2 + 4 * h_1 * h_2 > This part would be clearer with the correction above. >> Yes. >> Then, from [1] and [2] we solve for y to get >> y = (5*f_1 - 3*f_2 + h_1 - h_2) / 4 [3] > While this conclusion _needs_ the correction above. >> Yes, yes. >> Then, since h_1 * h_2 = 21 * T = 21 * g_1 * g_2 we may as well >> pick h_1 = 21 * g_1 and h_2 = g_2 so [3] becomes >> y = (5*f_1 - 3*f_2 + 21*g_1 - g_2) / 4 >> Right? > Yes, you are psychic! >> I knew you'd say that. >> If that was your reasoning, it's wrong. You can't pick any old >> values for h_1 and h_2. Watch: >> Solving [1] and [2] for h_1 and h_2 we get >> h_1 = 7(y + 3 * z + f_1) >> h_2 = 3(y + 7 * z + 4 * f_1 - f_2) > That also needs the correction above ;-) >> (Grr). Yes, yes, yes! >> But from your original four linear equations we can derive >> g_1 = y + 3 * z + f_1 >> g_2 = y + 7 * z + 4 * f_1 - f_2 >> in other words, we are forced to choose >> h_1 = 7 * g_1 >> h_2 = 3 * g_2 >> and not your h_1 = 21 * g_1 and h_2 = g_2. > And to force the conclusion, in that case [3] becomes > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > But let's give James something else to worry about :-) Take > (42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T > expand it, use the quadratic equation to solve for y, and then substitute to > get rid of z and T: > z = -(3*f_1 - f_2 + g_1 - g_2)/4 > T = g_1*g_2 > The result is: > y = (5*f_1 - 3*f_2 + 5*g_1 - 5*g_2 +/- 2*(g_1 + g_2))/4 > Pick + and you get the result James wants: > y = (5*f_1 - 3*f_2 + 7*g_1 - 3*g_2)/4 > Pick - and it's different: > y = (5*f_1 - 3*f_2 + 3*g_1 - 7*g_2)/4 > Woo hoo! Centuries of mathematics down the tubes again, or can James spot > the bogosity? Hint #1: this isn't an algebraic error; you really do get > that result for y. Hint #2: you get the same two results for y if you do > the same thing but starting from > (2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T > instead. >> Hehe. I predict that this section (cute, BTW) will generate no response. >> > You just can't resist, can you? Are you naturally boorish, or do >> you have to work at it? > I strongly suspect that bit of gratuitous assholishness was deliberate. God > only knows why, but James got it into his head that he needs to _provoke_ > people into replying when he thinks they know something he wants to find > out. That's just his despicable way of trying to goad you into doing his > work for him. It's especially idiotic in this case, since if he had any > memory he'd recall that you typically respond much better to polite requests > than to his stupid baiting tactics. > But, in this case, I'm afraid what he'll take away is ha! it worked again, > without a shadow of a clue that it was neither necessary nor helpful to > behave like an ass. >> Sadly, I predict you're right again. >> Rick > You're both lying. > I am going to warn you. > Neither of you will live to see 2007 because of this lie becaue there > will be angry people who will kill you, as there is so much money at > stake. > Billions will be lost. > It's not like you can reverse it now either. > You killed yourselves. > It might have seemed like a small lie to both of you, but your names > will live in infamy, while you will not live at all. > James Harris priceless === Subject: Re: JSH: SF: Finally, surrogate factoring >> [...] >You're both lying. >I am going to warn you. >Neither of you will live to see 2007 because of this lie becaue there >will be angry people who will kill you, as there is so much money at >stake. >Billions will be lost. >It's not like you can reverse it now either. >You killed yourselves. >It might have seemed like a small lie to both of you, but your names >will live in infamy, while you will not live at all. Jesus. The next time you feel like whining about people calling you crazy, think back to this post you just made. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: SF: Finally, surrogate factoring <5N6dnYHVYq4DBBjZnZ2dnUVZ_rWdnZ2d@hamilton.edu> I am going to warn you. > Neither of you will live to see 2007 Is that a threat?! -- Larry Lard Replies to group please === Subject: Re: JSH: SF: Finally, surrogate factoring Larry Lard a .8ecrit : >> You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 > Is that a threat?! Not really. Cassandre does not make threats. Nor does Nostradamus === Subject: Re: JSH: SF: Finally, surrogate factoring > But, in this case, I'm afraid what he'll take away is ha! it worked again, > without a shadow of a clue that it was neither necessary nor helpful to > behave like an ass. >> Sadly, I predict you're right again. > You're both lying. > I am going to warn you. > Neither of you will live to see 2007 because of this lie becaue there > will be angry people who will kill you, as there is so much money at > stake. > Billions will be lost. > It's not like you can reverse it now either. > You killed yourselves. > It might have seemed like a small lie to both of you, but your names > will live in infamy, while you will not live at all. The poor guys! Now you really scared them. :-) Is it just me, or this post will go down in history as one of the finest quote it completely. Jose Carlos Santos === Subject: Re: JSH: SF: Finally, surrogate factoring >> But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful to >> behave like an ass. > Sadly, I predict you're right again. >> You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. > The poor guys! Now you really scared them. :-) > Is it just me, or this post will go down in history as one of the finest > quote it completely. Smells like a forgery to me. Rick === Subject: Re: JSH: SF: Finally, surrogate factoring > >> But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful to >> behave like an ass. > Sadly, I predict you're right again. > You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. > > > The poor guys! Now you really scared them. :-) > > Is it just me, or this post will go down in history as one of the finest > quote it completely. > Smells like a forgery to me. > Rick He's pullin' your leg. -- Paul Sperry Columbia, SC (USA) === Subject: Re: JSH: SF: Finally, surrogate factoring >> But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful to >> behave like an ass. > Sadly, I predict you're right again. > You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. > The poor guys! Now you really scared them. :-) > Is it just me, or this post will go down in history as one of the finest > quote it completely. > Smells like a forgery to me. I smell a conspiracy (or a JSH setup). First we see a posting that everyone agrees must be a forgery, even though it looks like a genuine JSH posting (except for its rational acceptance of reality). Then we see some especially outrageous JSH warning/threat/accusation postings (but consistent with JSH posting history). James now has a means of plausible deniability - I didn't write that - There have been many forgeries of me lately - etc - that are better than his attempts to scrub the record by deleting postings from the Google Goup archive. === Subject: Re: JSH: SF: Finally, surrogate factoring <5N6dnYHVYq4DBBjZnZ2dnUVZ_rWdnZ2d@hamilton.edu> <4eq5ffF1f6067U2@individual.net> > But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful > to >> behave like an ass. > Sadly, I predict you're right again. > You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. > The poor guys! Now you really scared them. :-) Is it just me, or this post will go down in history as one of the finest > quote it completely. Smells like a forgery to me. > I smell a conspiracy (or a JSH setup). First we see a posting that everyone > agrees must be a forgery, even though it looks like a genuine JSH posting > (except for its rational acceptance of reality). Then we see some > especially outrageous JSH warning/threat/accusation postings (but consistent > with JSH posting history). James now has a means of plausible deniability - > I didn't write that - There have been many forgeries of me lately - > etc - that are better than his attempts to scrub the record by deleting > postings from the Google Goup archive. I'm not sure why everyone had to agree that it was a forgery. It stated as much at the end of the post. The person who posted it had done so before, and signs the end of the posts with something like /not jsh or something like that. === Subject: Re: JSH: SF: Finally, surrogate factoring > But, in this case, I'm afraid what he'll take away is ha! it > worked again, > without a shadow of a clue that it was neither necessary nor > helpful to > behave like an ass. > Sadly, I predict you're right again. > You're both lying. > I am going to warn you. > Neither of you will live to see 2007 because of this lie becaue there > will be angry people who will kill you, as there is so much money at > stake. > Billions will be lost. > It's not like you can reverse it now either. > You killed yourselves. > It might have seemed like a small lie to both of you, but your names > will live in infamy, while you will not live at all. >> The poor guys! Now you really scared them. :-) >> Is it just me, or this post will go down in history as one of the finest >> quote it completely. > Smells like a forgery to me. > Rick signed James Harris and with e-mail adress: jstevh@msn.com in addition to the one containing the passage: It might have seemed like a small lie to both of you, but your names The headers of all three match for NNTP-Posting-Host and the posting-account values: (Google Groups)-posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY If there are generally-accepted JSH forgery/ies, it might be fun to take a look at the headers, especially the Google posting-account value. NNTP-Posting-Host = my_IP_address. The IP adress can change over time. I tend to believe that the real JSH has been posting through Google frequently, so I'd also look at the posting-account This brings up a hypothetical: would some committed JSH trickster bother to enter JSH's computer, log on to his Google account, and make a fake post? If so, what might be the trickster's motive? David Bernier === Subject: Re: JSH: SF: Finally, surrogate factoring >>But, in this case, I'm afraid what he'll take away is ha! it >>worked again, >>without a shadow of a clue that it was neither necessary nor >>helpful to >>behave like an ass. > >Sadly, I predict you're right again. >You're both lying. >>I am going to warn you. >>Neither of you will live to see 2007 because of this lie becaue there >>will be angry people who will kill you, as there is so much money at >>stake. >>Billions will be lost. >>It's not like you can reverse it now either. >>You killed yourselves. >>It might have seemed like a small lie to both of you, but your names >>will live in infamy, while you will not live at all. >The poor guys! Now you really scared them. :-) >Is it just me, or this post will go down in history as one of the finest >quote it completely. >>Smells like a forgery to me. >>Rick > signed James Harris and with e-mail adress: > jstevh@msn.com in addition to the one containing > the passage: > It might have seemed like a small lie to both of you, but your names > The headers of all three match for NNTP-Posting-Host > and the posting-account values: > (Google Groups)-posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY > If there are generally-accepted JSH forgery/ies, it might be > fun to take a look at the headers, especially > the Google posting-account value. > NNTP-Posting-Host = my_IP_address. The IP adress can > change over time. > I tend to believe that the real JSH has been posting through > Google frequently, so I'd also look at the posting-account > This brings up a hypothetical: would some committed JSH > trickster bother to enter JSH's computer, log on to his > Google account, and make a fake post? If so, what > might be the trickster's motive? James made last night. It appears that this one was the real thing after all. I haven't noticed any angry mobs bearing torches and pitchforks yet, but I'm keeping watch. Actually, now that I think of it, it's hard to imagine what their motivation would be. I mean, if Tim Peters and I had concocted a plot to supress James' miracle factoring result, just where would those missing billions of dollars have come from? Angry cryptographers? I can see it now: Rivest, Shamir, and Adleman leading a black-ops gang from NSA in an assault on the secret lab of the arch-factorers Peters and Decker. I'd better tell Tim to make sure we have enough piranhas in the moat. I love sci.math. As I've said before, you couldn't pay for entertainment like this. Rick === Subject: Re: JSH: SF: Finally, surrogate factoring <5N6dnYHVYq4DBBjZnZ2dnUVZ_rWdnZ2d@hamilton.edu> <4eq5ffF1f6067U2@individual.net> But, in this case, I'm afraid what he'll take away is ha! it worked >> again, >> without a shadow of a clue that it was neither necessary nor helpful to >> behave like an ass. > Sadly, I predict you're right again. > You're both lying. >> I am going to warn you. >> Neither of you will live to see 2007 because of this lie becaue there >> will be angry people who will kill you, as there is so much money at >> stake. >> Billions will be lost. >> It's not like you can reverse it now either. >> You killed yourselves. >> It might have seemed like a small lie to both of you, but your names >> will live in infamy, while you will not live at all. > The poor guys! Now you really scared them. :-) > Is it just me, or this post will go down in history as one of the finest > quote it completely. > Smells like a forgery to me. You're being charitable again - you know it doesn't help :) Headers all match the last post in this thread by James (the one starting The only problem I have with that is that you have too many solutions for y.). If it IS a forgery, it's the best we've seen yet. -- Larry Lard Replies to group please === Subject: Ulam's Conjecture - proof Yesterday my friend Przemek Drochomirecki asked about status of Ulam's Conjecture. http://mathworld.wolfram.com/UlamsConjecture.html I've decided to present my proof on the group. I'm not sure if it's correct and I'm waiting for your oppinion. Blazej Podsiadlo ---------------------------------------------------------------------------- ----------------------------------------------------- Ulam's Conjecture Let graph G have n points vi and graph H have n points ui, where n>=3. Then if for each i, the subgraphs Gi=G-vi and Hi=H-ui are isomorphic, then the graphs G and H are isomorphic. ---------------------------------------------------------------------------- ---------------------------------------------------- Proof: Each graph could be represented as Adjacency Matrix. (http://mathworld.wolfram.com/AdjacencyMatrix.html) Having some graph G: for ex: n=4 A B C D ----------- A | 1 0 0 1 B | 1 0 1 0 C | 0 0 0 1 D | 1 1 0 0 we can add some modifications: 1. label all edges, but each one shall receive a different label: in our case it would be: A B C D ----------- A | a 0 0 b B | c 0 d 0 C | 0 0 0 e D | f g 0 0 2. extend to fully connected graph A B C D ----------- A | a 1 1 b B | c 1 d 1 C | 1 1 1 e D | f g 1 1 3. label all new edges, but each one shall receive a different label: A B C D ----------- A | a h i b B | c j d k C | l m n e D | f g o p // above representation can be converted to Adjacency Matrix, be lab function: lab: label->{0,1} Let's do for all subgraphs of G and H. If we know that: Gi ~ Hi => there is some function fi: {v1,v2, ..., v(i-1), v(i+1), ..., vn} -> {u1,u2, ..., u(i-1), u(i+1), ..., un} in other words there is some map between vertices in G and H. Our modification have to be made in some specific way: edges (vj, vk) and (fi(vj), fi(vk)) will be labeled by the same label. in out example: A B C D ----------- A | 1 0 0 1 B | 1 0 1 0 C | 0 0 0 1 D | 1 1 0 0 firstly lets do it for G-{A} A B C D ----------- A | 1 0 0 1 B | 1 j d k C | 0 m n e D | 1 g o p then for G-{B} A B C D ----------- A | a 0 i b B | 1 j d k C | l m n e D | f g o p and for G-{C} A B C D ----------- A | a h i b B | c j d k C | l m n e D | f g o p in this way we've just received LABELED form of graph G *// :) by the way; it's enough to do it for 3 subgraphs :) Let's assume that we've made it for both G and H. in result G and H have the same set of labels. ---------------------------------------------------------------------- Let's talk for a moment about Adjacency Matrix. Each vertice in graph G is represented as raw and column in Adjacency Matrix; if we change of order of columns and rows simultaneously, they will be still representing the same graph. in our ex: A B C D ----------- A | a h i b B | c j d k C | l m n e D | f g o p and D B C A ----------- D | p h i f B | k j d c C | e m n l A | b g o a represent the same graph. // labeled graph has n! of different forms; (sic!) So graph G: 1 2 ... i ... n ---------------- 1| X 2| X .| X i|XXXXXXXXXXXXXXX .| X n| X can be represented as: 1 2 . i-1 i+1 . n i --------------------- 1 | X 2 | X . | X i-1| Gi X i+1| X . | X n | X i |XXXXXXXXXXXXXXXXXX to do it, we are changing in places (i, i+1), then (i+1, i+2) and so on ..., (n-1, n); !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!! Very important is that: set of values in row/column is unchangeable!!!! so if labels x, y and z are in the same row/column before transformation they will be there after!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!! --------------------------------------------- so let's take the everything together: 1. from assumption Gi ~ Hi and fi: Gi->Hi 2. G and H are labeled and they are in form: 1 2 . i-1 i+1 . n i --------------------- 1 | X 2 | X . | X i-1| Hi X i+1| X . | X n | X i |XXXXXXXXXXXXXXXXXX let's extend function fi to fi': fi'=fi u {i->i} then fi'(G) has form: 1 2 . i-1 i+1 . n i --------------------- 1 | Y 2 | Y . | Y i-1| fi(Gi)=Hi Y i+1| Y . | Y n | Y i |YYYYYYYYYYYYYYYYYY fi(Gi)=Hi means that in this range {1, .., n}{i} fi(Gi) and Hi have the same values/labels (in sense of indexes H(1,1) = G(1,1)) At this moment we can say that: if values on X/Y positions in G and H are the same then G ~ H to show it, let's get some raw from G: 1 2 . i-1 i i+1 . n --------------------- j | x y z let's assume that in this raw are values {x, y, z}, y is on i-position, so in Gi they are only {x, z}. what does it give us? in fi'(G) 1 2 . i-1 i+1 . n i --------------------- fi(j) | Y this unknown value/label Y have to be y, because: !! set of values in row/column is unchangeable !!! even more: Gi ~ Hi => fi (j) = j' so rows j and j' has the same values in range {1..n}{i}. in H let's assume that in this raw are values {x, y, z}, y is on i-position, so in Hi thay are only {x, z}. so fi'(G)[j', i] = H[j', i] that means fi'(G) = H for all positions expect (i,i), but we know that G and H have the same set of labels, and all expect this last one are used, so there is only one value and it's the same in both G and H representation. so G ~ H and fi' is the isomorphism function between them. === Subject: Re: Does the Calculus rest on Euclid? On Wed, 07 Jun 2006 13:44:23 -0400, Hatto von Aquitanien >I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >have never felt satisfactorily convinced that the transition from the >Riemann sum approximation to a smooth curve is logically founded upon >axioms I have assumed at the outset. These axioms are those of formal >logic and those of Euclid. Well, those are not the axioms that are used in the standard approach to a rigorous treatment of calculus. > When using a geometric argument to justify the >transition from the chord-length approximation to a continuous curve - for >example in the typical proof of the arclength theorem - there seems to be >an unacknowledged step of faith. >Is this a subject which has been discussed in mathematics? ************************ David C. Ullrich === Subject: Re: Does the Calculus rest on Euclid? > On Wed, 07 Jun 2006 13:44:23 -0400, Hatto von Aquitanien >>I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >>have never felt satisfactorily convinced that the transition from the >>Riemann sum approximation to a smooth curve is logically founded upon >>axioms I have assumed at the outset. These axioms are those of formal >>logic and those of Euclid. > Well, those are not the axioms that are used in the > standard approach to a rigorous treatment of calculus. I've checked several so-called advanced treatments of analysis, and to be quite honest, they don't seem to state their assumptions very clearly. I will concede that there is no need to appeal to Euclid in the calculus of single variable functions. Nonetheless, I do believe there is an assumption of continuity which is geometric in nature, and it is at the foundation of all our mathematical reasoning. implication, those of axiomatic set theory which is isomorphic to symbolic logic. In differential geometry it is not infrequently stated that every n-dimensional non-Euclidian space is described in terms of an n+m Euclidian embedding space. Where m is a positive integer such that m >= 1. Now, that usually appears in the context of discussing physics, so there may be a certain amount of poetic license taking place. For example, I do not know if it would be more correct to say the embedding space is Lorentzian in the case of general relativity. Nonetheless, I believe when all is said and done, we cannot reason about any of these ideas without appeal to our innate Euclidian space. The developments I have seen for the real numbers beginning with Peano's axioms seem to take a step of faith either explicitly through the continuum hypothesis, or implicitly when they jump off the firm ground of the rational numbers using the demonstrably irrational numbers as a justification for the existence of a continuum of real numbers. -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? > I've checked several so-called advanced treatments of analysis, and to be > quite honest, they don't seem to state their assumptions very clearly. I > will concede that there is no need to appeal to Euclid in the calculus of > single variable functions. Nonetheless, I do believe there is an > assumption of continuity which is geometric in nature, and it is at the > foundation of all our mathematical reasoning. You haven't looked very carefully, then. The standard axioms for the real numbers for the purposes of analysis are that they form a complete, ordered field. The completeness axiom (which can be put forward in many equivalent ways---Cauchy sequences converge, or bounded monotone sequences converge, or the least upper bound axiom, or Dedekind's axiom about cuts) is what gives you the continuity of the real line. Read Dedekind's Continuity and irrational numbers for a very nice treatment. There are lots of standard real analysis texts that deal with this quite explicitly. I've used Gaughan's Introduction to Analysis in my courses several times and I know he goes through this. > implication, those of axiomatic set theory which is isomorphic to symbolic > logic. ...plus the axioms for the real numbers, which I mentioned above. > The > developments I have seen for the real numbers beginning with Peano's axioms > seem to take a step of faith either explicitly through the continuum > hypothesis, or implicitly when they jump off the firm ground of the > rational numbers using the demonstrably irrational numbers as a > justification for the existence of a continuum of real numbers. The continuum hypothesis is never taken as an axiom in an elementary treatment of the real numbers. The demonstration of the existence of irrational numbers shows the need for axioms beyond those of an ordered field to deal with the real numbers. === Subject: Re: Does the Calculus rest on Euclid? > The > developments I have seen for the real numbers beginning with Peano's axioms > seem to take a step of faith either explicitly through the continuum > hypothesis, or implicitly when they jump off the firm ground of the > rational numbers using the demonstrably irrational numbers as a > justification for the existence of a continuum of real numbers. Have you ever seen the development of the reals from the rationals based on either Dedekind cuts or Cauchy sequences? Either development gives a satisfactorily concrete model for a continuum of real numbers having all the expected properties . === Subject: Re: Does the Calculus rest on Euclid? > I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I > have never felt satisfactorily convinced that the transition from the > Riemann sum approximation to a smooth curve is logically founded upon > axioms I have assumed at the outset. These axioms are those of formal > logic and those of Euclid. When using a geometric argument to justify the > transition from the chord-length approximation to a continuous curve - for > example in the typical proof of the arclength theorem - there seems to be > an unacknowledged step of faith. I think a Euclidean approach to calculus would be possible, and to some extent you find Newton doing it. To do it right you need to nail down the ordering properties which Euclid assumed, and you need to add an axiom something like this: (Euclidean convexity) Any convex collection of points on a line is either the interval between two points, a ray extending from one point, or the entire line. By a convex collection is meant a set of points such that any point without ever defining the field of real numbers and sticking to the language of Euclid, you could do calculus--albeit from a peculiar point of view. By the way, it would be interesting to hear an argument that Euclidean convexity could have been proven by Euclid from his assumptions; one hears that Eudoxus basically constructed the positive reals, but it seems to be there are problem with that view. If it's actually all in Euclid, Euclidean convexity (is there an actual name for this axiom?) should be provable. === Subject: Re: Does the Calculus rest on Euclid? > I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I > have never felt satisfactorily convinced that the transition from the > Riemann sum approximation to a smooth curve is logically founded upon > axioms I have assumed at the outset. These axioms are those of formal > logic and those of Euclid. When using a geometric argument to justify the > transition from the chord-length approximation to a continuous curve - for > example in the typical proof of the arclength theorem - there seems to be > an unacknowledged step of faith. > Is this a subject which has been discussed in mathematics? Yes, to great depth, over many years. There was some unease about the logical underpinnings of the calculus from the beginning, and significant progress in justifying the work by Cauchy, Weierstrass and others in the mid-19th century. You have to go beyond Euclid, you need a theory of the real numbers to begin with, but it has all been explored, debated, and settled to our satisfaction. You may not find the details in your average calculus textbook, because faculty have learned through hard experience that 99.99% of all students are uninterested, but it's all out there, somewhere. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Does the Calculus rest on Euclid? >> I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >> have never felt satisfactorily convinced that the transition from the >> Riemann sum approximation to a smooth curve is logically founded upon >> axioms I have assumed at the outset. These axioms are those of formal >> logic and those of Euclid. When using a geometric argument to justify >> the transition from the chord-length approximation to a continuous curve >> - for example in the typical proof of the arclength theorem - there seems >> to be an unacknowledged step of faith. >> Is this a subject which has been discussed in mathematics? > Yes, to great depth, over many years. There was some unease about > the logical underpinnings of the calculus from the beginning, and > significant progress in justifying the work by Cauchy, Weierstrass > and others in the mid-19th century. You have to go beyond Euclid, > you need a theory of the real numbers to begin with, but it has > all been explored, debated, and settled to our satisfaction. You > may not find the details in your average calculus textbook, > because faculty have learned through hard experience that 99.99% > of all students are uninterested, but it's all out there, somewhere. Glancing over Weyl's _The Continuum;/A Critical Examination of the Foundation of Analysis/_ has convinced me that I do not have time to delve too deeply into this matter. My conviction is that we innately believe it is meaningful to transcend the concept of discrete partitions as these partitions become infinitesimal, but the concept of continuity cannot be constructed from arguments which begin with the whole numbers. -- Nil conscire sibi === Subject: Re: Does the Calculus rest on Euclid? <2sudncAQi9yiVRrZnZ2dnUVZ_vmdnZ2d@speakeasy.net> I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >> have never felt satisfactorily convinced that the transition from the >> Riemann sum approximation to a smooth curve is logically founded upon >> axioms I have assumed at the outset. These axioms are those of formal >> logic and those of Euclid. When using a geometric argument to justify >> the transition from the chord-length approximation to a continuous curve >> - for example in the typical proof of the arclength theorem - there seems >> to be an unacknowledged step of faith. >> Is this a subject which has been discussed in mathematics? > Yes, to great depth, over many years. There was some unease about > the logical underpinnings of the calculus from the beginning, and > significant progress in justifying the work by Cauchy, Weierstrass > and others in the mid-19th century. You have to go beyond Euclid, > you need a theory of the real numbers to begin with, but it has > all been explored, debated, and settled to our satisfaction. You > may not find the details in your average calculus textbook, > because faculty have learned through hard experience that 99.99% > of all students are uninterested, but it's all out there, somewhere. > Glancing over Weyl's _The Continuum;/A Critical Examination of the > Foundation of Analysis/_ has convinced me that I do not have time to delve > too deeply into this matter. My conviction is that we innately believe it > is meaningful to transcend the concept of discrete partitions as these > partitions become infinitesimal, but the concept of continuity cannot be > constructed from arguments which begin with the whole numbers. > -- > Nil conscire sibi Huh? You can't be bothered to look too deeply into the arguments behind something so you assume your intuition is right and the mathematical arguments are wrong? -- mike. === Subject: Re: Does the Calculus rest on Euclid? On Thu, 08 Jun 2006 02:47:58 -0400, Hatto von Aquitanien > I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I > have never felt satisfactorily convinced that the transition from the > Riemann sum approximation to a smooth curve is logically founded upon > axioms I have assumed at the outset. These axioms are those of formal > logic and those of Euclid. When using a geometric argument to justify > the transition from the chord-length approximation to a continuous curve > - for example in the typical proof of the arclength theorem - there seems > to be an unacknowledged step of faith. > > Is this a subject which has been discussed in mathematics? >> Yes, to great depth, over many years. There was some unease about >> the logical underpinnings of the calculus from the beginning, and >> significant progress in justifying the work by Cauchy, Weierstrass >> and others in the mid-19th century. You have to go beyond Euclid, >> you need a theory of the real numbers to begin with, but it has >> all been explored, debated, and settled to our satisfaction. You >> may not find the details in your average calculus textbook, >> because faculty have learned through hard experience that 99.99% >> of all students are uninterested, but it's all out there, somewhere. >Glancing over Weyl's _The Continuum;/A Critical Examination of the >Foundation of Analysis/_ has convinced me that I do not have time to delve >too deeply into this matter. My conviction is that we innately believe it >is meaningful to transcend the concept of discrete partitions as these >partitions become infinitesimal, but the concept of continuity cannot be >constructed from arguments which begin with the whole numbers. You're entitled to your conviction. But it's incorrect. Alas it does take some time to explain exactly _how_ we get from the integers to the continuous. But it doesn't require innate belief, nothing but simple logic. ************************ David C. Ullrich === Subject: Re: Does the Calculus rest on Euclid? > I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I > have never felt satisfactorily convinced that the transition from the > Riemann sum approximation to a smooth curve is logically founded upon > axioms I have assumed at the outset. These axioms are those of formal > logic and those of Euclid. When using a geometric argument to justify the > transition from the chord-length approximation to a continuous curve - for > example in the typical proof of the arclength theorem - there seems to be > an unacknowledged step of faith. No geometry is required to produce the fundemental theorem of calculus. It can all be done analytically. Geometry is eye candy for the intuition. Bob Kolker === Subject: Re: Does the Calculus rest on Euclid? >> I suspect the answer to this may be 'Yes.', 'No.', and 'It depends.'. I >> have never felt satisfactorily convinced that the transition from the >> Riemann sum approximation to a smooth curve is logically founded upon >> axioms I have assumed at the outset. These axioms are those of formal >> logic and those of Euclid. When using a geometric argument to justify the >> transition from the chord-length approximation to a continuous curve - for >> example in the typical proof of the arclength theorem - there seems to be >> an unacknowledged step of faith. >No geometry is required to produce the fundemental theorem of calculus. >It can all be done analytically. Quite right, as long as it (the fundamental theorem) is stated in purely analytic terms: the derivative of a function at a point is a certain limit (if that limit exists), the integral of a function on an interval is a certain limit (if that limit exists), and then a certain relation holds between (anti)differentiation and integration. All analytic, and very nice it is. But, to say that: >Geometry is eye candy for the intuition. misses some important points (which may not be points that HvA means to make; he may be a just target for your scorn, for all we can tell so far). There *is* an elementary theory of area for plane polygonal regions, which can be developed from Euclidian axioms; when those axioms are appropriately added to (to justify limiting processes that were known, and used, but inadequately justified), that theory can be extended to an analytic theory of area for more general plane regions, in particular, the regions that the standard (bad) first-year calculus curriculum pretends to use to define int_a^b f(x)dx for (piecewise) continuous f. There is a similar theory of tangency that can be developed without limits for plane algebraic curves (we could argue about without, I suppose) and then extended by Euclidian geometry and analysis to more general plane curves. Both these analytic-geometric theories can then be *interpreted* in terms of analytic integration and differentiation, and at that point the Fundamental Theorem of Calculus actually says something interesting *about area and tangency*: not eye-candy, but a good, filling meal. The situation that HvA mentions explicitly, of arc length, is rather different. Lee Rudolph === Subject: Re: Does the Calculus rest on Euclid? <4eor9aF1env59U1@individual.net> In , on 06/07/2006 at 04:59 PM, lrudolph@panix.com (Lee Rudolph) said: >The situation that HvA mentions explicitly, of arc length, is rather >different. The axioms and postulates that Euclid gave are not adequate to prove all of the geometrical theorems in The Elements. If you add the missing axioms then you can develop Real Analysis geometrically. That includes defining arc length for simple curves. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: exam results Hi all, I recently got the results back from my US Navy advancement exam. I scored in the 92nd percentile, correctly answered 56.5% of the questions (200 question exam), and scored 64.28 points on a point scale that ranges from 20 - 80. My (completely frivolous) question is: how many questions did the guy who scored 80 get right? :-D Best wishes, Allen === Subject: Re: exam results > Hi all, > I recently got the results back from my US Navy advancement exam. > I scored in the 92nd > percentile, correctly answered 56.5% of the questions (200 question > exam), and scored 64.28 points on a point scale that ranges from 20 - > 80. > My (completely frivolous) question is: how many questions did the > guy who scored 80 get right? :-D Insufficient information. However, we can conclude a few things. The point scale (20-80) suggests that the Navy is using a scheme similar to that of the College Board, which uses 200-800. It evidently indicates that the Navy scores are intended to have a mean of 50 and a standard deviation of 10. To check this hypothesis, let's compute what percentage of the scores should lie below 64.28 on this scale. Mathematica says: In[1]:= < === Subject: Re: exam results I recently got the results back from my US Navy advancement exam. > I scored in the 92nd > percentile, correctly answered 56.5% of the questions (200 question > exam), and scored 64.28 points on a point scale that ranges from 20 - > 80. > My (completely frivolous) question is: how many questions did the > guy who scored 80 get right? :-D > Insufficient information. However, we can conclude a few things. The > point scale (20-80) suggests that the Navy is using a scheme similar to > that of the College Board, which uses 200-800. It evidently indicates > that the Navy scores are intended to have a mean of 50 and a standard > deviation of 10. > To check this hypothesis, let's compute what percentage of the scores > should lie below 64.28 on this scale. Mathematica says: > In[1]:= < In[2]:= CDF[NormalDistribution[50,10],64.28] > Out[2]= 0.923354 > which agrees nicely with your claim that this score is in the 92nd > percentile. We can also say that a score of 80 is 3 standard deviations > above the mean: > In[3]:= CDF[NormalDistribution[50,10],80.0] > Out[3]= 0.99865 > which is around the 99.8th percentile. But we have only a single data > point concerning the actual distribution of questions answered correctly > (56.5% right is in 92nd percentile) and this is insufficient to determine > either the mean or the standard deviation, even if we assume the raw > scores are normally distributed. Allen === Subject: Re: exam results > Hi all, > I recently got the results back from my US Navy advancement exam. > I scored in the 92nd > percentile, correctly answered 56.5% of the questions (200 question > exam), and scored 64.28 points on a point scale that ranges from 20 - > 80. > My (completely frivolous) question is: how many questions did the > guy who scored 80 get right? :-D > Best wishes, > Allen I'm not sure if this is the best way to do it, but, here we go. 200(.565)=113 questions correct 113/64.28=x/80 x= 140.6 (so about 141 questions) Dave === Subject: Re: exam results <7tUhg.40676$fG3.27810@dukeread09 Hi all, > I recently got the results back from my US Navy advancement exam. > I scored in the 92nd > percentile, correctly answered 56.5% of the questions (200 question > exam), and scored 64.28 points on a point scale that ranges from 20 - > 80. > My (completely frivolous) question is: how many questions did the > guy who scored 80 get right? :-D > I'm not sure if this is the best way to do it, but, here we go. > 200(.565)=113 questions correct > 113/64.28=x/80 > x= 140.6 (so about 141 questions) Hi Dave, Stats is not my specialty, but that doesn't look quite right. I'm fairly certain that the point scale (20 - 80) is based on some kind of curve and so it seems (to me anyway) that the scores would need to be normalized first (or some missing step to that effect) not just compared in a linear fashion. Allen === Subject: Re: Atheism - Defined Expires: 28 days On 27 May 2006 22:26:52 -0000, Daniel Joseph Min >-----BEGIN PGP SIGNED MESSAGE----- >Not that we need a dictionary to confirm what we already >know, but Atheism is properly defined as a belief, e.g., >quoting from Webster's New World Dictionary & Thesaurus: > Atheism - > the *belief* that there is no God, or denial that > God or gods exist. [end quote, emphasis added] >Thus Atheism is properly a (ir)religious belief, and so >adherents and proponents of Atheism are equally as much >believers in their godless religion of Atheism as other >believers in other contrastingly less-godless religions >are likewise believers their respective religions; some >devout believers, some less devout in their convictions. >Of course other more Atheist-biased dictionaries define >Atheism as scientific fact, intellectual superiority >and the like. But we non-Atheists know that Atheists are >every bit as religious in their beliefs as anybody else >in the world is religious in his or her (non-Atheistic) >beliefs. By the strict definition, Atheism *is* a belief. >But don't expect Atheists to admit to this any time soon. >For example, Atheists devoutly believe In the beginning, >Atheists created the heaven and the earth--in a big bang. >Atheists devoutly believe their sacred gospel 'According >to Charles Darwin', etc. They hold these beliefs sacred. 'Atheist' is a label used to describe perfectly normal people by a bunch of deluded misfits, whose ego wont let them accept they are not going to live forever. The whole idea of giving a name to a 'non-thing' when the 'thing' itself is also a non-thing, epitomises human stupidity at its worst. HW. www.users.bigpond.com/hewn/index.htm Appropriate message snipping is considerate and painless. === Subject: The secret of series I have done some researches regarding the polynomial sequences which is available in the following website link, A formulla is derived which is applicable to a great variety of functions indipendent of the law behind the sequence, I would be glad if you help me go further... http://www.mathhelpforum.com/math-help/showthread.php?p=11525 Mubashir === Subject: Re: The secret of series Am 08.06.2006 14:33 schrieb Mubashir: > I have done some researches regarding the polynomial sequences which is available in the following website link, A formulla is derived which is applicable to a great variety of functions indipendent of the law behind the sequence, I would be glad if you help me go further... > http://www.mathhelpforum.com/math-help/showthread.php?p=11525 > Mubashir Mubashir - for you - in the bernoulli-en-paper I deal with pascal-matrices and sums of powers. http://www.uni-kassel.de/fg_pur/helms/math/bernoulli/bernoulli_en.pdf some additional information are here (concerning bernoulli - and the related stirling-numbers): http://www.uni-kassel.de/fg_pur/helms/math/bernoulli/stirling2.pdf an introduction to the pascal-matrix as matrix-exponential of an amazing simple parameter is in http://www.uni-kassel.de/fg_pur/helms/math/pascalmatrix/PascalDreieckMatrixL o g.pdf Comments welcome Gottfried Helms === Subject: how to prove this I am learning linear algebra, and can not figure it out. A: m*n matrix, B: n*s matrix, prove that r( A * B ) >= r( A )+ r( B ) - n, where r is the abbr of rank. === Subject: Statistics questions If I roll three six-sided dice (3d6) and only take the median roll (not the mean or average), what kind of bell curve or distribution odds would it give me for the result 1 to 6? If I roll two 20-sided dice (2d20) and only look at the higher result of the two, what is the average result I will get? What about 3d20 and only look at the highest die? 4d20 etc. up to 16d20...? It's for a RPG system called Donjon: http://en.wikipedia.org/wiki/Donjon_%28role-playing_game%29 Sonja www.kisa.ca === Subject: Re: Statistics questions > If I roll two 20-sided dice (2d20) and only look at the higher result > of the two, what is the average result I will get? What about 3d20 and > only look at the highest die? 4d20 etc. up to 16d20...? Asumming M dice with N sides (and higher means the highest number shown, so if two or more dice are both the highest it is stil valid) each you get: Average = (sum of ( i* (i^M-(i-1)^M)) (for i= 1 to N)) / N^M Using this formula you get (from 2 to 16) 13,825 15,4875 16,4833375 17,14584375 17,61787797 17,97086977 18,24450271 18,46258734 18,6402762 18,7876713 18,91176665 19,01755823 19,10871003 19,18797026 19,2574407 With the other question I am thinking about it. And don't ask for a proof. Georg === Subject: Re: Statistics questions Sonja Elen Kisa nous a r.8ecemment amicalement signifi.8e : > If I roll three six-sided dice (3d6) and only take the median roll > (not the mean or average), what kind of bell curve or distribution > odds would it give me for the result 1 to 6? You can just count the 216 possibilities. You can also distinguish 7 cases for computing p(a) witj a in [1,6] : dice 1 dice 2 dice 3 =a =a ==> (a-1)(7-a)/216 =a =a any ==> 6 /216 =a >a <=a ==> a (6-a)/216 >a =a <=a ==> a (6-a)/216 >a (a-1)(6-a)/216 =a ==> (a-1)(7-a)/216 a =a ==> (a-1)(6-a)/216 Which gives p(a) = (-3a^2 + 21a -10)/108 and : p(1) = 8/108 = 0,074 p(2) = 20/108 = 0,185 p(3) = 52/108 = 0,241 p(4) = 52/108 = 0,241 p(5) = 20/108 = 0,185 p(6) = 8/108 = 0,074 > If I roll two 20-sided dice (2d20) and only look at the higher result > of the two, what is the average result I will get? What about 3d20 and > only look at the highest die? 4d20 etc. up to 16d20...? If I call p(n,a) the probability of having the highest die in nD20 being <= a, we have p(n,a) = (a/20)^n So, the probabilty q(n,a) of having the highest die being =a is : q(n,1) = 1/20^n q(n,a) = p(n,a) - p(n,a-1) for a > 1 So : q(n,a) = (a^n - (a-1)^n)/20^n The average result you are looking for is : e(n) = sum_a [a(a^n - (a-1)^n)/20^n] = 20 - sum_{i=1, 19}(i^n)/20^n Application : n=1 (1D20) average result : 20 - sum_{i=1, 19}(i)/20 = 10.5 (obvious) n=2 (2D20) average result : 20 - sum_{i=1, 19}(i^2)/20^2 = 13,825 n=3 (3D20) average result : 20 - sum_{i=1, 19}(i^3)/20^3 = 15,487 4D20 ==> 16,48 5D20 ==> 17,15 6D20 ==> 17,62 7D20 ==> 17,97 8D20 ==> 18,24 9D20 ==> 18,46 10D20 ==> 18,64 11D20 ==> 18,79 12D20 ==> 18,91 13D20 ==> 19,02 14D20 ==> 19,11 15D20 ==> 19,19 16D20 ==> 19,26 -- Patrick === Subject: Re: Statistics questions > If I roll three six-sided dice (3d6) and only take the median roll (not > the mean or average), what kind of bell curve or distribution odds > would it give me for the result 1 to 6? There are 216 possible rolls, and by symmetry the number of 1s and 6s are the same, as are 2s and 5s, and 3s and 4s, so we only need to consider those. How to get a 1 (or 6)? you need to roll at least two 1s on the three dice to have the median be 1. There are 16 ways this can happen (3 ways to get two 1s, wih 6 choices for the other die in each, minus 2 for the triple-counting of 1-1-1). There are 16 ways to get a 1(6) How to get 2 (or 5)? you need (i) exactly one 1 and at least one 2, or (ii)no 1s and at least two 2s. (i) 3 ways for a 1, 9 ways for each to have at least one 2 and no more 1s. total: 27 (ii) 3 ways for two 2s, each with 5 choices for the other die, less 2 for the triple-counted 2-2-2. total: 13. There are 40 ways to get a 2(5) 216 - 16 - 16 - 40 - 40=104 remaining die rolls. By symmetry, 52 for each of 3 and 4. So, the probability of getting each number is (rounded to 3 decimal places): .074, .185, .241, .241, .185, .074 > If I roll two 20-sided dice (2d20) and only look at the higher result > of the two, what is the average result I will get? What about 3d20 and > only look at the highest die? 4d20 etc. up to 16d20...? You are looking for a number where the probability of rolling at least one number higher is 50%. let p=probability one die is less than or equal to n=n/20. For M dice, we want the 1-prob(all dice <=n)=.5 Thus, we have: 1-p^M=.5, so p^M=.5 and p=0.5^(1/M) thus n=20*(0.5^(1/M))=20/2^(1/M) For 2 dice, n=20/sqrt(2)=14.1..., so 14 is the most likely outcome For 3 dice, n=20/2^(1/3)=15.87, so 16 is the most likely outcome To determine whether to round up or down, find the probability of all dice less than the bracketing numbers. Whichever is closer to 50% is what you most likely want. You can find the probabilities for up to 16 dice on your own. === Subject: How to calculate (estimate) sigma from lot data I have a stastical question. I have a large number of lots measured. For each lot the mean and standard deviation and the number of devices measured are stored. Now I want to have the standard deviation and the mean of all the measurements combined. Can I do that? Example: Lot1 500dev, Mean 1.105, Sigma 0.001 Lot2 400dev, Mean 1.107, Sigma 0.002 Lot3 800dev, Mean 1.104, Sigma 0.001 ... So can I calculate (or estimate) the mean and the standard deviation for all 1700 devices combined? Assumption would be that normal distribution is valid for each lot and for the means as well. Any help is very much appreciated. Georg === Subject: Re: How to calculate (estimate) sigma from lot data > I have a stastical question. I have a large number of lots measured. > For each lot the mean and standard deviation and the number of devices > measured are stored. Now I want to have the standard deviation and the > mean of all the measurements combined. Can I do that? > Example: > Lot1 500dev, Mean 1.105, Sigma 0.001 > Lot2 400dev, Mean 1.107, Sigma 0.002 > Lot3 800dev, Mean 1.104, Sigma 0.001 > ... > So can I calculate (or estimate) the mean and the standard deviation > for all 1700 devices combined? If you have the lot sizes for each lot, one can calculate a weighted (or grouped) mean and a weighted (or grouped) standard deviation for the whole collection of lots by counting each mean or mean squared in your formulae as many times as its lot size. Your lot size sigmas are not needed in this process, only the lot means. If Lot k has mean m_k and size n_k with sum n_k = n then the mean for all lots combined will be approximately M = sum(m_k * n_k) / n and the combined (sample) standard deviation will be approximately S = sqrt[sum((m_k-M)^2 * n_k)/(n-1)] > Assumption would be that normal distribution is valid for each lot and > for the means as well. > Any help is very much appreciated. > Georg === Subject: Re: Convergence problem [ about the place where I probably made a mistake ] Han de Bruijn === Subject: Fourier transform I have tried lot to find the fourier transform of e(-5 cos x).Can somebody help me === Subject: Re: The Numbers Game I'm very astonished by the magic property of nine. I did a lot of magic games with my students in which nine digit was applied. For example: games A: think of a number (the player must think for a 2 digits number and he does not reveal this one). Then he is invited to multiply to 9, then to 11, then to 9 again then to 11 again. How anyone can discover the beginning number in a few second? Simply reading the last two digits of the revealed last result! Yeah! If I think 56 (and nobody know what I have thought) I will get these following hidden results: 56x9 = 504x11 = 5544x9 = 49896 If the player finally will multiply 49896x11 it will reach the following result: 548856. Nice, isn't it? Someone knows other funning games in which 9 is the magic number? === Subject: Re: The Numbers Game <14777047.1149779315155.JavaMail.jakarta@nitrogen.mathforum.org I'm very astonished by the magic property of nine. I did a lot of magic games with my students in which nine digit was applied. Students in elementary high school? This kind of trick is as old as sin, for anyone who knows any algebra. > For example: > games A: think of a number (the player must think for a 2 digits number and he does not reveal this one). Let the two-digit number be X. > Then he is invited to multiply to 9, then to 11, then to 9 again then to 11 again. That would yield 99X times 99 which is X (100-1)^2 = (10,000 - 200 + 1) X. So, the number is 9800*X + X. But 9800 times ANY X always equal a number whose last TWO digits are 00. So, the last two digits must be X. > How anyone can discover the beginning number in a few second? > Simply reading the last two digits of the revealed last result! As you can see from the algebra, there is no need for X to be 2 digits. any number from 0 to 99 will work. So. you actually short-changed the range of values for which the trick works. > Someone knows other funning games in which 9 is the magic number? Yes. -- Bob. === Subject: Re: bayesians versus frequentists Bill Taylor says... I have another very basic comment about frequentism. Frequentism assumes *objective* probabilities. As far as we know, the only (things like neutron decay, absorption of radiation by atoms, etc.) When we compile statistics about some macroscopic system, for example, the recovery rate of patients with such and such a condition given such and such a treatment, the variability does *not* reflect objective probability. It reflects our ignorance about the relevant details about each patient (and also our practical inability to solve the quantum we get from such a study does not tell us anything about objective probabilities that we didn't know already. It's plausible that the recovery isn't probabilistic at all in any meaningful sense. Each patient will either recover or not, based on the details of his particular situation. But we don't know those details, or else don't know how to take them into account. It's hard to see how a *frequentist* model makes any sense here. If there are any objective probabilities, it seems that have a pretty good grasp of the probabilities (given by QED, or QCD, or whatever). So if it is only appropriate to apply probability theory when probabilities are objective, then probability has no -- Daryl McCullough Ithaca, NY === Subject: Re: bayesians versus frequentists Bill Taylor says... >> Bayesian would say it makes sense to talk about the probability that >> there is life on Mars, but the frequentist would see that as nonsense. >This is an excellent example. It's a good example of why frequentism is useless in the real world. Suppose you are thinking of launching a mission to Mars to bring back soil samples for the purpose of looking for evidence of life. Now, people don't just launch billion-dollar missions to check out random facts. Maybe there is an alien restaurant hidden on the dark side of the moon. Maybe one of the undiscovered moons of Jupiter is a perfect cube. Maybe Elvis is alive and living in a secret underwater headquarters off the cost of Venezuela. We have to make some kind of decision about what to investigate based on the cost of the investigation, the likelihood of finding something interesting, and the value of that thing if we do find it. But what is the notion of likelihood that we use? I claim it is Bayesian, not frequentist. It's a combination of subjective notions of likelihood, continually adjusted by accumulated evidence. >Someone noted that this question could be embedded into a wider >question like how many Mars-like planets have life on them? >But this is very much NOT the same question; >when we ask the Mars question, we are very definitely thinking >of our Mars, not any other, and ultimately want a reply >concerning that. Objective statistics has little to say here. And for similar reasons, objective statistics have little to say about *anything* in the real world. The statistics by themselves don't tell us what's a sensible thing to do, they don't tell us what is likely to happen. The statistics are only useful if augmented with some guesses or assumptions that go beyond what the statistics say. >OC Bayesians admit this, and defend themselves with the comment that >there is *always* going to be subjectivity in statistics (or anything >else), and that at least they put it all up front, in the subjective >prior, where everyone can see and criticise it. (Though what such, >criticisms could possibly carry in terms of intellectual weight >they never specify). IMHO this justification is nonsense, like >a child painter saying look at the lovely mess I've made, it's ART. The frequentist is making the same sorts of messes. The mess is not *caused* by Bayesianism, it is inherent in trying to apply mathematics to the real world. The epistemological situation we are in in the real world is that not only do we do not know *what* the probabilities are, we don't even know *whether* things are probabilistic (in the frequentist sense) at all. It's possible that things are completely deterministic, and the appearance of randomness is due to complexity, lack of knowledge of the forces at work, and lack of knowledge of initial conditions. (Of course, quantum mechanics makes determinism very implausible). So we don't know enough to know whether frequentism has *any* application in the real world. The frequentist arguments against Bayesianism are actually arguments against using probability theory, period. >> What paradigm is more popular among economists and other applied >> scientists? >Orthodox; though this is largely because Bayesians have been very slow >in developing stats packages that *automatically* require the >subjective input of the users. There have been some efforts >in this direction; but it also means users are eventually going >to wonder WHY am I being asked all these silly things? >It may turn out counter-productive, in the very long run! The reason they are silly is because in many situations, the prior probabilities have negligible impact on the posterior probabilities. >BTW - I *do* think that Bayesian ideas have a good place! >Namely, in subjective circumstances. I want to know when frequentist ideas ever have a good place. It seems to me that it is only for simplicity; the frequentist keeps track of fewer numbers. So computations using frequentism might be simpler. And, as you say, the frequentist doesn't need to fill out annoying questionnaires. So from my point of view, frequentism seems like an ad hoc methodology that attempts to get good results with less hassle and computation. It's a heuristic, rather than anything principled. Which is fine, as long as you realize that it's a heuristic. >But if there is to be any suggestion of public concerns, >especially e.g. of medical/pharmaceutical concerns, and other >dangerous contexts, then a subjective approach is totally >irresponsible and hateful. I don't see how frequentism is any less dangerous or irresponsible in such cases. Yes, if you give somebody a powerful drug, the drug itself might kill the patient. If you *don't* give the drug, then the disease might kill him. Whether you are a frequentist or a Bayesian does not change this dilemma. How is frequentism a less dangerous approach? Here's an example Suppose that there are two competing theories about some medical condition, and these theories give different conclusions about what the proper treatment should be. How do you decide which treatment to give? The frequentist approach doesn't have an answer. It can only give odds if we *know* the probabilities, and only know the probabilities if we know which theory is correct. You can decide to wait until one theory or the other is proven correct beyond reasonable doubt (which I claim is a subjective criterion, by the way), but your patient may be dead by then. Alternatively, you can just pick whichever theory has the most impressive supporting evidence, and go with that one. But what if there is only a slight advantage for one side over another? Let's add some more details: Suppose that theory A implies that the patient's odds will be made slightly worse by doing procedure P. Suppose that theory B implies that the patient will almost surely recover if you do procedure P (and will almost surely die if you don't). Suppose that theory A has slightly better evidence in its favor than theory B. Which theory do you go with? Do you perform the procedure, or not? The frequentist has no answer in this case. He can just say: If we do procedure P, you'll either live or die, and we don't know which. The Bayesian has an answer in this case: If there is only slightly better reasons for preferring A over B, then do the procedure. He computes a certain posterior probability that A is true, and a certain posterior probability that B is true, and then uses these probabilities to compute the probability that the patient will recover given procedure P, and the probability that he will recover if he isn't given procedure P. The odds are better with procedure P. Frequentism cannot do this sort of weighing of probabilities, because theories being true is *not* a probabilistic event in the sense of frequentism. >It is true, we can never eliminate the subjective element from >public concerns completely, nor from science or math. >But we ought to be seeking ways to DETECT and >MINIMIZE the subjective element. NOT building it into the system!!! It seems to me that Bayesianism does a *better* job at this than frequentism. The way I would put it is this: if we are trying to reason in the presence of uncertainty, then we should try to eliminate or reduce the uncertainty. Subjectivity is *inevitable* when there is uncertainty. We should try to eliminate the *need* for subjectivity. Eliminating subjectivity without reducing uncertainty doesn't make sense to me. -- Daryl McCullough Ithaca, NY === Subject: Re: bayesians versus frequentists > If is meaningful to ask what is the probability that Bush will be > re-elected in the 2008 election. A meaningful and operational > value of p can be assigned to that future event, and it's likely > that everyone will have a slightly different value. That is the > crux of SUBJECTIVE probability. That is entirely correct. But it all depends on the conditional: IF it is meaningful to ask.... I don't think it IS meaningful, except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot be spoken of (IMHO) in a discussion between two or more people. The utter failure of Bayesians to extend their methods to multi-person or (better still) public conclusions, in spite of decades of failed attempts, is highly significant. > Jimmie (L.J.) Savage uses the example of a 10 repeated > trials of asking a subject if a piece of music was written by > Hyden or Mozart. If the outcome was 9 successes out of > 10, a frequentist would assign a value of 0.9 to the > probability that the subject can distinguish the style of the composers. That is NOT what frequentist statistics is about, nor would a frequentist ever conclude any such simplistic nonsense. The frequentist postulates merely that there IS a probability of picking the right composer, but that he doesn't know what it is, but can get some decent clues as to what it is by examining the experimental results. > How about the probability that Team A will beat Team B in the game > this weekend? Exactly the same, as you imply. > Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. EXACTLY SO! Got it in one. This is a crucial argument against the rationality of subjectivism. Bayesians make a HUUUGE song and dance about the fact that a subjectivist is obliged to make his probabilities coherent, so that a Dutch book of bets cannot be taken out against him. i.e. one cannot be SURE to win money off him. Well and good. But as bookies know, one can make money FOR SURE against a combination of subjectivists, merely by playing them off against each other. IMHO, this embarrassing fact is a crucial criterion showing the imbecility of using subjective probabilities in public, or even 2-person matters. For one chap by himself, fine. But he MUST keep it to himself and not talk to anyone else about it, or doom lies in wait. > Out of ignorance rather than any foundational or coherent argument. Charming. I expect I know more about it than you do. > Your ignorance about Bayesian statistics is much deeper than I ever > thought any anti-Bayesian would be. Charming. I'm sure we are all stunned by the force of your debating style. > That explains why the rich history and foundation of Bayesian > statistics in Europe (UK and Italy) and the USA has not filtered > down to the isolation of the continent of Aboriginies and > criminals sent exile by the UK. :-) How ironic of someone who speaks of others' ignorance, to confuse Australia and New Zealand. But then I expect you are an American, so such ignorance is explicable, if hardly admirable. ------------------------------------------------------------------------ Bill Taylor W.Taylor@math.canterbury.ac.nz ------------------------------------------------------------------------ USA: The only country in the western world where the right to superstitious ignorance is constitutionally enshrined. ------------------------------------------------------------------------ === Subject: Re: bayesians versus frequentists > ...Exactly the same, as you imply. > Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > EXACTLY SO! Got it in one. This is a crucial argument against > the rationality of subjectivism. Bayesians make a HUUUGE > song and dance about the fact that a subjectivist is obliged > to make his probabilities coherent, so that a Dutch book > of bets cannot be taken out against him. > i.e. one cannot be SURE to win money off him. > Well and good. But as bookies know, one can make money FOR SURE > against a combination of subjectivists, merely by playing them off > against each other. IMHO, this embarrassing fact is a crucial > criterion showing the imbecility of using subjective probabilities > in public, or even 2-person matters. For one chap by himself, fine. > But he MUST keep it to himself and not talk to anyone else about it, > or doom lies in wait... This is a dubious argument, if I'm reading it correctly. In the Bayesian view (or indeed any reasonable view, in the context of gambling) each individual bettor would be wise to take care that he is not vulnerable to Dutch Book considerations --- for example, being willing to take the Sharks versus the Jets even-up while giving odds taking the Jets versus the Sharks. Bill Taylor's argument seems to depend on the notion that while the individual bettor may be able to pass the Dutch Book test, most any given pool of bettors will _collectively_ fail the test. But I don't see how this is so devastating to the whole idea of Bayesian subjectivism. The key Bayesian concept is that as new evidence accumulates, your probability estimates and mine will tend to converge in the long run. If you and I have different betting preferences, but we want to become partners for some reason (that is, we want our bets to be _collectively_ rational) then we're pooling our evidential and theoretical resources as well as our financial resources. So it will behoove us to update our probabilities accordingly. This is by no means an insurmountable problem for the Bayesian gambler, as BT seems to be suggesting. According to the Bayesian model, while our respective betting systems may be difficult to square with each other in the short run, the differences will steadily diminish in the long run, assuming that we're not impervious to learning from experience. An entertaining and informative web site on Bayesian thinking is the late Richard Jeffrey's: http://www.princeton.edu/~bayesway/ Jeffrey's specialty was what he called probability kinematics --- the art of updating one's probabilities in the light of new evidence. He thought of this as an extension of classical logic. Larry T. === Subject: Re: bayesians versus frequentists > If is meaningful to ask what is the probability that Bush will be > re-elected in the 2008 election. A meaningful and operational > value of p can be assigned to that future event, and it's likely > that everyone will have a slightly different value. That is the > crux of SUBJECTIVE probability. > That is entirely correct. But it all depends on the conditional: > IF it is meaningful to ask.... I don't think it IS meaningful, > except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot > be spoken of (IMHO) in a discussion between two or more people. I don't follow your illogic. > The utter failure of Bayesians to extend their methods to multi-person > or (better still) public conclusions, in spite of decades > of failed attempts, is highly significant. > Jimmie (L.J.) Savage uses the example of a 10 repeated > trials of asking a subject if a piece of music was written by > Hyden or Mozart. If the outcome was 9 successes out of > 10, a frequentist would assign a value of 0.9 to the > probability that the subject can distinguish the style of the composers. > That is NOT what frequentist statistics is about, > nor would a frequentist ever conclude any such simplistic nonsense. What other simplistic nonsense can frequentist offer? > The frequentist postulates merely that there IS a probability of > picking the right composer, but that he doesn't know what it is, > but can get some decent clues as to what it is by examining > the experimental results. > How about the probability that Team A will beat Team B in the game > this weekend? > Exactly the same, as you imply. Sentence unclear. Everyone has a different probability on that event. > Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > EXACTLY SO! Got it in one. This is a crucial argument against > the rationality of subjectivism. Bayesians make a HUUUGE > song and dance about the fact that a subjectivist is obliged > to make his probabilities coherent, so that a Dutch book > of bets cannot be taken out against him. > i.e. one cannot be SURE to win money off him. Of course you can't. I don't think you understand anything about gambling and subjective probability, or Dutch book, for that matter. > Well and good. But as bookies know, one can make money FOR SURE > against a combination of subjectivists, merely by playing them off > against each other. That's how ALL gambling houses work, regardless whether the bettor has subjective probability or just wanted to bet. IMHO, this embarrassing fact is a crucial > criterion showing the imbecility of using subjective probabilities > in public, or even 2-person matters. For one chap by himself, fine. > But he MUST keep it to himself and not talk to anyone else about it, > or doom lies in wait. You are rather naive in game theory and gambling. > Out of ignorance rather than any foundational or coherent argument. > Charming. I expect I know more about it than you do. Really? How do you know so much about Bayesian statistics and probabilities? I learned mine from L.J.Savage (in person) and from learn your Bayesian ideas? Ronald Fisher? :[) > Your ignorance about Bayesian statistics is much deeper than I ever > thought any anti-Bayesian would be. > Charming. I'm sure we are all stunned by the force of your debating > style. It's inference based on what you exhibited in your pseudo arguments. > That explains why the rich history and foundation of Bayesian > statistics in Europe (UK and Italy) and the USA has not filtered > down to the isolation of the continent of Aboriginies and > criminals sent exile by the UK. :-) > How ironic of someone who speaks of others' ignorance, to confuse > Australia and New Zealand. But then I expect you are an American, > so such ignorance is explicable, if hardly admirable. I was speaking of the CONTINENTS. I was in New Zealand AND Australia earlier this year, and previous years. -- Bob. === Subject: Re: bayesians versus frequentists >> If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >> That is entirely correct. But it all depends on the conditional: >> IF it is meaningful to ask.... I don't think it IS meaningful, >> except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >> be spoken of (IMHO) in a discussion between two or more people. >I don't follow your illogic. An interesting point in this *particular* exchange is that unless the U.S. Constitution is amended again, and very quickly, Bush *cannot* legally be (re)elected (to the Presidency) ever again, and in particular not in 2008. So it is to be hoped that not *too* many of the everyone above (limiting quantification to USAn citizens of voting age, perhaps; and perhaps excluding those posting to Usenet in the heat of the moment) will have a value for p that is even slightly different from 0. Unless, of course, they know something I don't... Lee Rudolph === Subject: Re: bayesians versus frequentists If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >> That is entirely correct. But it all depends on the conditional: >> IF it is meaningful to ask.... I don't think it IS meaningful, >> except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >> be spoken of (IMHO) in a discussion between two or more people. >I don't follow your illogic. > An interesting point in this *particular* exchange is that unless > the U.S. Constitution is amended again, and very quickly, Bush > *cannot* legally be (re)elected (to the Presidency) ever again, > and in particular not in 2008. I stand corrected if this is Dubya's 2nd term already. That shows how much attention I pay to the presidency of the USA. :-) In that case, the subjective probability would be 0 for most, and some non-zero number by people like me; OR some more creative Constitution scholars who consider Bill Clinton's chance to hold a THIRD term to be possible -- because he is eligible to run for Vice President, and he (or Bush) could work out a deal for the elected president to resign, so that the VP would become Prez, for the 3rd time. :-) > So it is to be hoped that not > *too* many of the everyone above (limiting quantification to > USAn citizens of voting age, perhaps; and perhaps excluding those > posting to Usenet in the heat of the moment) will have a value for > p that is even slightly different from 0. Unless, of course, they > know something I don't... See the above. :-) -- Bob. === Subject: Re: bayesians versus frequentists Suppose a frequentist goes to his doctor, and his doctor says: You have a terminal disease. The chance of you surviving as long as 6 months is 40%. Does the frequentist say this is nonsense and ignore it? >> If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >> That is entirely correct. But it all depends on the conditional: >> IF it is meaningful to ask.... I don't think it IS meaningful, >> except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >> be spoken of (IMHO) in a discussion between two or more people. I don't follow your illogic. > An interesting point in this *particular* exchange is that unless > the U.S. Constitution is amended again, and very quickly, Bush > *cannot* legally be (re)elected (to the Presidency) ever again, > and in particular not in 2008. > I stand corrected if this is Dubya's 2nd term already. That shows > how much attention I pay to the presidency of the USA. :-) > In that case, the subjective probability would be 0 for most, and > some non-zero number by people like me; OR some more > creative Constitution scholars who consider Bill Clinton's > chance to hold a THIRD term to be possible -- because he > is eligible to run for Vice President, and he (or Bush) could > work out a deal for the elected president to resign, so that > the VP would become Prez, for the 3rd time. :-) > So it is to be hoped that not > *too* many of the everyone above (limiting quantification to > USAn citizens of voting age, perhaps; and perhaps excluding those > posting to Usenet in the heat of the moment) will have a value for > p that is even slightly different from 0. Unless, of course, they > know something I don't... > See the above. :-) > -- Bob. === Subject: Re: bayesians versus frequentists Bill Taylor says... >> If is meaningful to ask what is the probability that Bush will be >> re-elected in the 2008 election. A meaningful and operational >> value of p can be assigned to that future event, and it's likely >> that everyone will have a slightly different value. That is the >> crux of SUBJECTIVE probability. >That is entirely correct. But it all depends on the conditional: >IF it is meaningful to ask.... I don't think it IS meaningful, >except as a purely subjective, ONE-PERSON-ONLY, matter. It cannot >be spoken of (IMHO) in a discussion between two or more people. I don't understand what you are talking about. It can certainly be spoken of. In talking about events in the real world, we are *always* in the situation where we don't have enough knowledge to actually make decisions. Bayesianism is just a disciplined way of dealing with it. Frequentism has no empirical content until you make some *guess* about what the probabilities are. Now, you might say that it's not a guess that a coin has a 50/50 probability; you can toss it 1000 times and check it out. But you can't *really* check it out, because the fact that a coin lands heads 500 times out of 1000 doesn't imply that it was a fair coin. Of course, what you can do is argue as follows: If the probabilities were significantly different from 50/50, then the probability of getting 500 heads out of 1000 would be very, very tiny. But that kind of careful reasoning is what Bayesianism is all about. At any given time, there are many, many physical theories (infinitely many, actually) that are consistent with what we have observed so far. If you have to make a decision about what to do next, which theory do you go with? The right thing to do, according to one theory, may not be the right thing, according to another theory. In practice, what you have to do is decide that some theories are less plausible than others, so you dismiss them, at least until some future date where there is much more evidence in their favor. But if you are doing that, you're doing Bayesianism, but in a not very systematic way. You're combining a priori preferences for theories (based on plausibility) with more mathematical, objective information based on statistics. >> Las Vegas sports gambling makes bets on THEIR >> subjective odds on sporting events. >EXACTLY SO! Got it in one. This is a crucial argument against >the rationality of subjectivism. Bayesians make a HUUUGE >song and dance about the fact that a subjectivist is obliged >to make his probabilities coherent, so that a Dutch book >of bets cannot be taken out against him. >i.e. one cannot be SURE to win money off him. >Well and good. But as bookies know, one can make money FOR SURE >against a combination of subjectivists, merely by playing them off >against each other. IMHO, this embarrassing fact is a crucial >criterion showing the imbecility of using subjective probabilities >in public, or even 2-person matters. That's an argument against *gambling*, period. Believing that there *is* an objective probability, but we just don't know what it is, doesn't do any good. To *use* probability, you have to take the plunge and assume some probabilities. -- Daryl McCullough Ithaca, NY === Subject: Re: bayesians versus frequentists > How about the probability that Team A will beat Team B in the game > this weekend? That's no more nonsense than any other probability > assessment. Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. Odds makers do no such thing. They do not have to know anything about the teams, their records, or even what sport they are playing, so in that way they have no subjective odds. They merely add a sufficient point spread to ensure that the the same number of bettors (actually, dollars) for each side offset each other. This way, regardless of the outcome of the game, the money given by the losers will be used to pay the winners, minus the bookkee's cut. So when Las Vegas says Team A is a 3.5 point favorite over Team B, it is not making a subjective statement about probabilities. In fact, NO probability (subjective or otherwise) is involved here at all. What this statement actually means is: the aggregate number of dollars bet on Team A winning by more than 3 points is approximately the same as the aggregate number of dollars bet on Team A NOT winning by more than 3 points. Although those making the bets may be engaging in subjective probabilities, those running the betting, such as bookkees, casinos, etc., deal only with hard math. === Subject: Re: bayesians versus frequentists When a bet is first introduced, only subjective odds can be used. Later, If you put a 3.5 point spread and signifcantly more money is spent on one side than the other, the spread will be dynamically changed to a new value in an attempt to balance future wagers... This is actually a much harder thing to do without risk exposure than for bets where you were giving odds in the first place (If you placed 11-10 odds on a team winning, and the bets are 15-12 you can change the odds to 15-12. If you had pittsburgh giving 3.5 points, and the bets were 15-12 for pittsburgh at this spread, its not obvious if you need to change the spread to 4.5, 5.5 or what to have future bets balance). > How about the probability that Team A will beat Team B in the game > this weekend? That's no more nonsense than any other probability > assessment. Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > Odds makers do no such thing. They do not have to know anything about > the teams, their records, or even what sport they are playing, so in > that way they have no subjective odds. They merely add a sufficient > point spread to ensure that the the same number of bettors (actually, > dollars) for each side offset each other. This way, regardless of the > outcome of the game, the money given by the losers will be used to pay > the winners, minus the bookkee's cut. > So when Las Vegas says Team A is a 3.5 point favorite over Team B, it > is not making a subjective statement about probabilities. In fact, NO > probability (subjective or otherwise) is involved here at all. What > this statement actually means is: the aggregate number of dollars bet > on Team A winning by more than 3 points is approximately the same as > the aggregate number of dollars bet on Team A NOT winning by more than > 3 points. > Although those making the bets may be engaging in subjective > probabilities, those running the betting, such as bookkees, casinos, > etc., deal only with hard math. === Subject: Re: bayesians versus frequentists some bookies, however, are not fully balanced in their bets and carry risk exposure they feel is in their favor. > How about the probability that Team A will beat Team B in the game > this weekend? That's no more nonsense than any other probability > assessment. Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > Odds makers do no such thing. They do not have to know anything about > the teams, their records, or even what sport they are playing, so in > that way they have no subjective odds. They merely add a sufficient > point spread to ensure that the the same number of bettors (actually, > dollars) for each side offset each other. This way, regardless of the > outcome of the game, the money given by the losers will be used to pay > the winners, minus the bookkee's cut. > So when Las Vegas says Team A is a 3.5 point favorite over Team B, it > is not making a subjective statement about probabilities. In fact, NO > probability (subjective or otherwise) is involved here at all. What > this statement actually means is: the aggregate number of dollars bet > on Team A winning by more than 3 points is approximately the same as > the aggregate number of dollars bet on Team A NOT winning by more than > 3 points. > Although those making the bets may be engaging in subjective > probabilities, those running the betting, such as bookkees, casinos, > etc., deal only with hard math. === Subject: Re: bayesians versus frequentists > some bookies, however, are not fully balanced in their bets and carry > risk exposure they feel is in their favor. This is true, particularly of some family run businesses. Of course, the risk is being taken by the bookie who thinks he knows more than the average bettor. If he is correct, then the times he's wrong will be more than paid for by the times he's right. If he is not correct, of course, the opposite is true. If a bookie truly believes he knows more than the average joe (since it is the average joe in aggregate who determines the spread), he could certainly just place his own bet. Of course, anyone who places a bet presumably thinks they know more than the average person...otherwise, he wouldn't be taking that bet, would he? === Subject: Re: bayesians versus frequentists And now (in case you didn't already know) the FDA are publishing draft guidelines for Bayesian clinical trials. http://www.fda.gov/cdrh/meetings/072706-bayesian.html Duncan === Subject: Re: bayesians versus frequentists > sci.math. > Is it true that most professors in the stats department consider > themselves to be > in one camp or the other camp? >>Bayesian would say it makes sense to talk about the probability that there is life on Mars, but the frequentist would see that as nonsense. > Isn't this a pointless distinction because a frequentist can still > allow that, in principle, one can take statistics across a large sample > of intergalactic planets and talk about the probabilty of finding that > mars is among the subset of those with life? A Bayesian would think about it in those terms, but not a frequentist. A frequentist would be happy enough with the probability of life existing on a planet that was to be randomly selected, but not a specific planet. Bayesian and frequentist probabilities are not the same, although they can be equal. Duncan === Subject: Re: bayesians versus frequentists so, according to your examples, a person who is charting stocks is a bayesian because he is making the leap that a stock's historical price trends will indicate future performance. 2 different chartists will chart in different ways do to their different assumptions even though both might rationally dervie the conclusions of their differing assumptions. so chartists are bayesians in action? > I am going to go one step farther here. > First, let me give a common situation in reliability analysis. Lets say > you are building a rocket booster. Your rocket booster contains an > engine which has been used in a number of previous rockets. How do you > get an estimate for the reliability of this rocket booster. A simple > minded approach is to say > A. this engine has failed in X out of Y previous launches, so its > reliability R= (Y-X)/Y > booster is not exactly replicating any previous experiment so you have > no information. > Now we can take this one step father, suppose instead of blindly > considering all previous launches as equivalent, the statististian has > brilliantly identified all factors a1,a2,..., an that might effect the > probability of failure and also brilliantly picked the form of the > model that relates these covariates to the probability (perhaps one of > those logodds models you see in discrete choice analysis). Even with > this model, unless the range of the covariates from previous data > actually includes the values from the given experiment, you still have > no relevent information about the current reliability. > Unless you make a leap. > The leap is that, previous results are applicable to the current > situation. > But there is still more. > Suppose you had 100 previous rocket launches (from the old rockets) and > there was 1engine failure. > Now you have used the new rocket 10 times with no engine failures. How > do you combine these results? Is the previous experience (from slightly > different operating conditions) given equal weight? Do you pick > weights? Once you have to combine these results (even if you CHOOSE to > give them equal weights or CHOOSE to give the first set of data 0 > weight) you are doing bayesian statistics. Just bayesian folks are more > self conscious about the fact that they are combining two sets of data, > with different uncertainty distributions, based on a subjective belief > pertaining to the question: how applicable was the old data? > There is one other thing to note. While bayes rule is often applied to > a situation where you have a prior distribution, and new evidence, it > is equally applicable to the situation where you have two different > distributions that you are trying to combine. There is no need to give > them the interpretations of prior subjective belief and Evidence. > Instead they can just be evidence 1 and evidence 2. > For instance. Suppose, you are making ball bearings and want to know > the mean diameters of the ball bearings. You have two different > experimental set ups with different lasers to take the measurement. > Both experiments include some measurment error. In fact one laser's > error tends to be worse on the smaller ball bearings and the other > tends to be worse on the larger ones (do to the different wavelengths > of the laser beams). So both give you a distribution for the mean, > rather than an exact answer. You use bayes's rule to combine the two. > Bayes rule is essentially: > Given that BOTH results are correct (that is the uncertainty > distributions adequitely account for the correct answers), what would > you deduce? > This is similar to one person saying that there are between 7000-8000 > jelly beans in a jar, and someone else saying there are between > 7500-9000 jelly beans. IF both people are correct (that is, there > uncertainty distributions are adequite) then there are between > 7500-8000 jelly beans. You are just combining data in a logical manner. > Summing up, whenever you are making a leap that go as follows: > a. You have some phenomenon that you are interested in measuring > statistics for (maybe the average income in a town, or the probability > that someone in particular goes default on a loan, or the probability > that their is life on MARS, etc.) > b. We have data that is probably relevent to this problem but the > situation is not 100% identical. The source of the data is irrelevent. > In the case on life on MARS, we could have used a sample of 1000 other > planets, or I could have down a computer simulation of planet > formation, or I could have just thought about it and decided that I > think the answer is 1/1000, I am willing to give 10-1 odds on a bet > that it is at least 1/10000, I am willing to give 100-1 odds on a bet > that it is at least 1/1000000. etc. > c. we have a second set of data that also pertains to problem. This > might be a real measurement (I asked 1000 people in this town their > income, and I checked their bank statements and audited their books to > verify this [and there still may be an error if they are sufficently > crooked]). This may be a second guess (my friend estimated the > probability distribution that there is life on MARS). > d. we want to combine these data source. So, we use Bayes's rule... === Subject: Re: bayesians versus frequentists This isn't exactly the paradym, but a similar intelectual leap is occuring when you are updating your model. The stock price modelers are given a time series of data on a particular stock price: p(1),p(2),p(3),p(4),.... and perhaps on some other variables (which could be stock prices, inflation rates, etc.) x(1),x(2),.... y(1),y(2),.... z(1),z(2),... They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) This model is an attempt to find the best function to predict p(n) from past data. There will be two errors in your answers: 1. When you build a model, you usually Select a functional form for the function F and the set of variables that are used. This historical function will not have perfect results even on past data. (For instance if you did a linear regression, not all the data points will be exactly on the line). Call this model fitting error. 2. There is also the fundamental assumption that you have included every possible covariate and the correct functional form (and that that form doesn't change with time). This is a knowledge error. For instance it may turn out that the price of steel varies with the price of computers. This, of course, wouldn't have been in your model before computers were invented... So on top of the error you get in 1, there is additional error pertaining to the question: is the formula you are using really a law of the universe or did it just happen to work in the past. This additional uncertainity should be taken into account, and is based on a subjective assessment. Now, given this model, the bayesian paradigm is one way you would update your model given additional data if you wanted to give the old model and the new model their own uncertainity distributions before combining them. Often people implicity give the old data equal weight to the new data and just refit the model using all the data. But you could beleive that the more recent data is more important and do something different... === Subject: Re: bayesians versus frequentists the queer thing here is that stock prices can be self-fullfilling prophecies. If a large fraction of chartists make predictions based on some function F, their buying and selling will drive stock price movements based on their function F. If enough chartists use that function F, then F actually becomes actually descriptive of stock price behavior whether it had be previously or not! > This isn't exactly the paradym, but a similar intelectual leap is > occuring when you are updating your model. > The stock price modelers are given a time series of data on a > particular stock price: > p(1),p(2),p(3),p(4),.... > and perhaps on some other variables (which could be stock prices, > inflation rates, etc.) > x(1),x(2),.... > y(1),y(2),.... > z(1),z(2),... > They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) > This model is an attempt to find the best function to predict p(n) from > past data. There will be two errors in your answers: > 1. When you build a model, you usually Select a functional form for the > function F and the set of variables that are used. This historical > function will not have perfect results even on past data. (For instance > if you did a linear regression, not all the data points will be exactly > on the line). Call this model fitting error. > 2. There is also the fundamental assumption that you have included > every possible covariate and the correct functional form (and that that > form doesn't change with time). This is a knowledge error. For instance > it may turn out that the price of steel varies with the price of > computers. This, of course, wouldn't have been in your model before > computers were invented... > So on top of the error you get in 1, there is additional error > pertaining to the question: is the formula you are using really a law > of the universe or did it just happen to work in the past. This > additional uncertainity should be taken into account, and is based on a > subjective assessment. > Now, given this model, the bayesian paradigm is one way you would > update your model given additional data if you wanted to give the old > model and the new model their own uncertainity distributions before > combining them. Often people implicity give the old data equal weight > to the new data and just refit the model using all the data. But you > could beleive that the more recent data is more important and do > something different... === Subject: Re: bayesians versus frequentists Yeah thats why there is a big difference between economics where discovering patterns almost always changes the fundamental economic relationship (feedback) and other sciences where the observer has limited effect on the system observed (yes I understand that at the microscopic level there is an effect...). Hence there will always be much greater knowledge error in economics/finance. Once you learned how to make money from something getting misspriced, your buying or selling the thing gets rid of the mispricing from a basic supply and depand mechanism .... There is an old joke about someone walking down the street with his friend, a finance professor. The guy sees a $20 bill on the ground as they walked by and tells the professor. The professor says, no there couldn't be a $20. If there was someone would have already picked it up. (This is a commentary on the efficient market hypothesis which argues that arbitrage opportunitities can't exist for more than a split second since someone would identify it and make money while re-balancing the market.). > the queer thing here is that stock prices can be self-fullfilling > prophecies. If a large fraction of chartists make predictions based on > some function F, their buying and selling will drive stock price > movements based on their function F. If enough chartists use that > function F, then F actually becomes actually descriptive of stock price > behavior whether it had be previously or not! > This isn't exactly the paradym, but a similar intelectual leap is > occuring when you are updating your model. > The stock price modelers are given a time series of data on a > particular stock price: > p(1),p(2),p(3),p(4),.... > and perhaps on some other variables (which could be stock prices, > inflation rates, etc.) > x(1),x(2),.... > y(1),y(2),.... > z(1),z(2),... > They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) > This model is an attempt to find the best function to predict p(n) from > past data. There will be two errors in your answers: > 1. When you build a model, you usually Select a functional form for the > function F and the set of variables that are used. This historical > function will not have perfect results even on past data. (For instance > if you did a linear regression, not all the data points will be exactly > on the line). Call this model fitting error. > 2. There is also the fundamental assumption that you have included > every possible covariate and the correct functional form (and that that > form doesn't change with time). This is a knowledge error. For instance > it may turn out that the price of steel varies with the price of > computers. This, of course, wouldn't have been in your model before > computers were invented... > So on top of the error you get in 1, there is additional error > pertaining to the question: is the formula you are using really a law > of the universe or did it just happen to work in the past. This > additional uncertainity should be taken into account, and is based on a > subjective assessment. > Now, given this model, the bayesian paradigm is one way you would > update your model given additional data if you wanted to give the old > model and the new model their own uncertainity distributions before > combining them. Often people implicity give the old data equal weight > to the new data and just refit the model using all the data. But you > could beleive that the more recent data is more important and do > something different... === Subject: Re: bayesians versus frequentists mispricing is a nebulous concept that sounds good on paper but cannot be determined objectively in practice. When is cisco or wal-mart mispriced? Can anybody accurately assess mispricing? Mispricing is an immeasurable concept. > Yeah thats why there is a big difference between economics where > discovering patterns almost always changes the fundamental economic > relationship (feedback) and other sciences where the observer has > limited effect on the system observed (yes I understand that at the > microscopic level there is an effect...). Hence there will always be > much greater knowledge error in economics/finance. Once you learned how > to make money from something getting misspriced, your buying or selling > the thing gets rid of the mispricing from a basic supply and depand > mechanism .... > There is an old joke about someone walking down the street with his > friend, a finance professor. The guy sees a $20 bill on the ground as > they walked by and tells the professor. The professor says, no there > couldn't be a $20. If there was someone would have already picked it > up. > (This is a commentary on the efficient market hypothesis which argues > that arbitrage opportunitities can't exist for more than a split second > since someone would identify it and make money while re-balancing the > market.). > the queer thing here is that stock prices can be self-fullfilling > prophecies. If a large fraction of chartists make predictions based on > some function F, their buying and selling will drive stock price > movements based on their function F. If enough chartists use that > function F, then F actually becomes actually descriptive of stock price > behavior whether it had be previously or not! > This isn't exactly the paradym, but a similar intelectual leap is > occuring when you are updating your model. The stock price modelers are given a time series of data on a > particular stock price: > p(1),p(2),p(3),p(4),.... > and perhaps on some other variables (which could be stock prices, > inflation rates, etc.) > x(1),x(2),.... > y(1),y(2),.... > z(1),z(2),... They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) This model is an attempt to find the best function to predict p(n) from > past data. There will be two errors in your answers: > 1. When you build a model, you usually Select a functional form for the > function F and the set of variables that are used. This historical > function will not have perfect results even on past data. (For instance > if you did a linear regression, not all the data points will be exactly > on the line). Call this model fitting error. > 2. There is also the fundamental assumption that you have included > every possible covariate and the correct functional form (and that that > form doesn't change with time). This is a knowledge error. For instance > it may turn out that the price of steel varies with the price of > computers. This, of course, wouldn't have been in your model before > computers were invented... So on top of the error you get in 1, there is additional error > pertaining to the question: is the formula you are using really a law > of the universe or did it just happen to work in the past. This > additional uncertainity should be taken into account, and is based on a > subjective assessment. Now, given this model, the bayesian paradigm is one way you would > update your model given additional data if you wanted to give the old > model and the new model their own uncertainity distributions before > combining them. Often people implicity give the old data equal weight > to the new data and just refit the model using all the data. But you > could beleive that the more recent data is more important and do > something different... === Subject: Re: bayesians versus frequentists > mispricing is a nebulous concept that sounds good on paper but cannot > be determined objectively in practice. When is cisco or wal-mart > mispriced? Can anybody accurately assess mispricing? Mispricing is an > immeasurable concept. >> Yeah thats why there is a big difference between economics where >> discovering patterns almost always changes the fundamental economic >> relationship (feedback) and other sciences where the observer has >> limited effect on the system observed (yes I understand that at the >> microscopic level there is an effect...). Hence there will always be >> much greater knowledge error in economics/finance. Once you learned how >> to make money from something getting misspriced, your buying or selling >> the thing gets rid of the mispricing from a basic supply and depand >> mechanism .... >> There is an old joke about someone walking down the street with his >> friend, a finance professor. The guy sees a $20 bill on the ground as >> they walked by and tells the professor. The professor says, no there >> couldn't be a $20. If there was someone would have already picked it >> up. >> (This is a commentary on the efficient market hypothesis which argues >> that arbitrage opportunitities can't exist for more than a split second >> since someone would identify it and make money while re-balancing the >> market.). > the queer thing here is that stock prices can be self-fullfilling > prophecies. If a large fraction of chartists make predictions based on > some function F, their buying and selling will drive stock price > movements based on their function F. If enough chartists use that > function F, then F actually becomes actually descriptive of stock price > behavior whether it had be previously or not! >> This isn't exactly the paradym, but a similar intelectual leap is >> occuring when you are updating your model. >> The stock price modelers are given a time series of data on a >> particular stock price: >> p(1),p(2),p(3),p(4),.... >> and perhaps on some other variables (which could be stock prices, >> inflation rates, etc.) >> x(1),x(2),.... >> y(1),y(2),.... >> z(1),z(2),... >> They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) >> This model is an attempt to find the best function to predict p(n) from >> past data. There will be two errors in your answers: >> 1. When you build a model, you usually Select a functional form for the >> function F and the set of variables that are used. This historical >> function will not have perfect results even on past data. (For instance >> if you did a linear regression, not all the data points will be exactly >> on the line). Call this model fitting error. >> 2. There is also the fundamental assumption that you have included >> every possible covariate and the correct functional form (and that that >> form doesn't change with time). This is a knowledge error. For instance >> it may turn out that the price of steel varies with the price of >> computers. This, of course, wouldn't have been in your model before >> computers were invented... >> So on top of the error you get in 1, there is additional error >> pertaining to the question: is the formula you are using really a law >> of the universe or did it just happen to work in the past. This >> additional uncertainity should be taken into account, and is based on a >> subjective assessment. >> Now, given this model, the bayesian paradigm is one way you would >> update your model given additional data if you wanted to give the old >> model and the new model their own uncertainity distributions before >> combining them. Often people implicity give the old data equal weight >> to the new data and just refit the model using all the data. But you >> could beleive that the more recent data is more important and do >> something different... My bank offers me a higher interest rate for a certificate of deposit of 2 or more years than for a savings account. [Actually, the savings account rate doesn't seem too bad because there is no minimal balance to keep for any given month, to avoid lower rates.] The bank advertises certificates of deposit at higher rates, the longer the term. I'm generally skeptical of locking in money for two or more years, given that (1) the difference in rates is not that big. (2) I figure that interest rates for savings accounts could very well be higher than at present in two years anyway. In the US, interest rates are, I think, very much dependent on the target rates of the Fed., which says it seeks to avoid too much inflation. Part of inflation can be explained perhaps by higher crude oil prices which result in increased costs being passed on to buyers of transportation services, materials made from fossil fuels, etc; perhaps the increased transportation costs affect the cost of oranges from Florida only months after the increase in crude oil costs. I sometimes look at the NYMEX light sweet oil futures contracts for the next delivey month, and subsequent months. One question I have about stock investors who outperform is: how much is due to luck? A few years ago, there were reports on television about a group of retired women of a certain age who read about stocks, did their research, and had outperformed. Subsequently, say a year later, I read that their portfolio hadn't outperformed in the last year or two; I'm not sure how much attention that development got in the media. One other thing I look at is how interest rates vary by country. David Bernier === Subject: Re: bayesians versus frequentists Luck is the most important factor in picking stocks. It is easy to underperform, but with optimal skill, you cannot expect to consistently overperform. Most of the true experts I know, myself included, simply buy index funds. Thos who want something for free in the stock market eventually lose all their money. > mispricing is a nebulous concept that sounds good on paper but cannot > be determined objectively in practice. When is cisco or wal-mart > mispriced? Can anybody accurately assess mispricing? Mispricing is an > immeasurable concept. >> Yeah thats why there is a big difference between economics where >> discovering patterns almost always changes the fundamental economic >> relationship (feedback) and other sciences where the observer has >> limited effect on the system observed (yes I understand that at the >> microscopic level there is an effect...). Hence there will always be >> much greater knowledge error in economics/finance. Once you learned how >> to make money from something getting misspriced, your buying or selling >> the thing gets rid of the mispricing from a basic supply and depand >> mechanism .... >> There is an old joke about someone walking down the street with his >> friend, a finance professor. The guy sees a $20 bill on the ground as >> they walked by and tells the professor. The professor says, no there >> couldn't be a $20. If there was someone would have already picked it >> up. >> (This is a commentary on the efficient market hypothesis which argues >> that arbitrage opportunitities can't exist for more than a split second >> since someone would identify it and make money while re-balancing the >> market.). > the queer thing here is that stock prices can be self-fullfilling > prophecies. If a large fraction of chartists make predictions based on > some function F, their buying and selling will drive stock price > movements based on their function F. If enough chartists use that > function F, then F actually becomes actually descriptive of stock price > behavior whether it had be previously or not! > This isn't exactly the paradym, but a similar intelectual leap is >> occuring when you are updating your model. >> The stock price modelers are given a time series of data on a >> particular stock price: >> p(1),p(2),p(3),p(4),.... >> and perhaps on some other variables (which could be stock prices, >> inflation rates, etc.) >> x(1),x(2),.... >> y(1),y(2),.... >> z(1),z(2),... >> They build a model: p(n)=F(p(1),...,p(n-1),x(1),...,x(n-1), y(1),.....) >> This model is an attempt to find the best function to predict p(n) from >> past data. There will be two errors in your answers: >> 1. When you build a model, you usually Select a functional form for the >> function F and the set of variables that are used. This historical >> function will not have perfect results even on past data. (For instance >> if you did a linear regression, not all the data points will be exactly >> on the line). Call this model fitting error. >> 2. There is also the fundamental assumption that you have included >> every possible covariate and the correct functional form (and that that >> form doesn't change with time). This is a knowledge error. For instance >> it may turn out that the price of steel varies with the price of >> computers. This, of course, wouldn't have been in your model before >> computers were invented... >> So on top of the error you get in 1, there is additional error >> pertaining to the question: is the formula you are using really a law >> of the universe or did it just happen to work in the past. This >> additional uncertainity should be taken into account, and is based on a >> subjective assessment. >> Now, given this model, the bayesian paradigm is one way you would >> update your model given additional data if you wanted to give the old >> model and the new model their own uncertainity distributions before >> combining them. Often people implicity give the old data equal weight >> to the new data and just refit the model using all the data. But you >> could beleive that the more recent data is more important and do >> something different... > My bank offers me a higher interest rate for a certificate of deposit > of 2 or more years than for a savings account. [Actually, the > savings account rate doesn't seem too bad because there is no > minimal balance to keep for any given month, to avoid > lower rates.] > The bank advertises certificates of deposit at higher rates, the > longer the term. I'm generally skeptical of locking in > money for two or more years, given that > (1) the difference in rates is not that big. > (2) I figure that interest rates for savings accounts > could very well be higher than at present in > two years anyway. > In the US, interest rates are, I think, very much dependent on > the target rates of the Fed., which says it seeks to avoid > too much inflation. Part of inflation can be explained perhaps by > higher crude oil prices which result in increased costs > being passed on to buyers of transportation services, > materials made from fossil fuels, etc; perhaps the increased > transportation costs affect the cost of oranges from > Florida only months after the increase in crude oil > costs. > I sometimes look at the NYMEX light sweet oil futures > contracts for the next delivey month, and subsequent > months. > One question I have about stock investors who > outperform is: how much is due to luck? > A few years ago, there were reports on television about a group of > retired women of a certain age who read about stocks, > did their research, and had outperformed. > Subsequently, say a year later, I read that their > portfolio hadn't outperformed in the last year or > two; I'm not sure how much attention that development > got in the media. > One other thing I look at is how interest rates vary by country. > David Bernier === Subject: Re: bayesians versus frequentists > As a result, most applied Bayesians are restricted to the use of conjugate priors because those are the only mathematically tractible priors that is informative -- though they hardly represent any Bayesian's real beliefs. The other kinds are the un-informative priors or diffuse priors that reduces the strongest element of Bayesian inference to the wimpy use of NO subjective info as input, thus defeating the most attractive part of the Bayesian system of statistical inference. Criticism is cheaper than back-seat driving unless you can do better. What priors do you suggest that economists use then? > Can somebody explain to us the argument between bayesians > and frequentists? > David Petry has made an excellent brief summary. > I will just emphasize a particular point. > Bayesian is a quick term for subjective probabilist, > or perhaps more aptly, subjective-probability statistician. > The key is that, at heart and root, they conceive that probability > is *inherently* subjective, and that thus any study based on it, > such as statistics, inherits that subjectivity, and thus is, > in some sense, objectively meaningless. Of course they go on > to say that a proper, *sensible* subject, will always change > their subjective beliefs about the state of the world in certain > ways, according to Bayesian probabilities. But this is already > a severe wedge in their ground concept that > one can believe as one wants. > The first half of the paragraph is a reasonable summary. The > second half if pure nonsense. > First, one must admit that there are meaningful PROBABILITIES > that can only be made subjectively because the frequentist > notion doesn't exist, because it is a single-occurrence event! > If is meaningful to ask what is the probability that Bush will be > re-elected in the 2008 election. A meaningful and operational > value of p can be assigned to that future event, and it's likely > that everyone will have a slightly different value. That is the > crux of SUBJECTIVE probability. > Beyond that the event that have operational frequentist > methods and interpretations can be argued to be flawed for > lack of use of known facts that can only be incorporated > subjectively. > Jimmie (L.J.) Savage uses the example of a 10 repeated > trials of asking a subject if a piece of music was written by > Hyden or Mozart. If the outcome was 9 successes out of > 10, a frequentist would assign a value of 0.9 to the > probability that the subject can distinguish the style of > the composers. > A Bayesian would view the same evidence differently, > depending on the subject. A Baysesian, as any reasonable > person, would put a stronger belief on the same outcome > had the subject been a musicologist, over another subject > who is a drunk picked up at the corner of a street who > happened to have made some lucky guesses. > Why should the two identical outcomes be view the same way? > Bayesian would say it makes sense to talk about the probability that > there is life on Mars, but the frequentist would see that as nonsense. > How about the probability that Team A will beat Team B in the game > this weekend? That's no more nonsense than any other probability > assessment. Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > This is an excellent example. Someone noted that this question could > be embedded into a wider question like how many Mars-like planets > have life on them? But this is very much NOT the same question; > when we ask the Mars question, we are very definitely thinking > of our Mars, not any other, and ultimately want a reply > concerning that. Objective statistics has little to say here. > That's because what you call objective statistics cannot make any > frequentist interpretation out of all those events in which it makes > perfect sense to speak of probability assessments. > It will be no surprise after the above, to learn that I am quite set > against Bayesianism. > Out of ignorance rather than any foundational or coherent argument. > There have been three attempts in the 20th C > to build subjectivity into the system, in some science-like study. > IMHO such attempts are anathema, and ought to be resisted at all costs, > being effectively a surrender to the forces of chaos and darkness. > What did you consider to be the three attempts? > OC Bayesians admit this, and defend themselves with the comment that > there is *always* going to be subjectivity in statistics (or anything > else), and that at least they put it all up front, in the subjective > prior, where everyone can see and criticise it. (Though what such, > criticisms could possibly carry in terms of intellectual weight > they never specify). IMHO this justification is nonsense, like > a child painter saying look at the lovely mess I've made, it's ART. > Non sequitur. > Incidentally, the other two domains where subjectivity has been built > into the system are Copenhagenism in QM, and Intuitionism in math. > So THAT's what you mean by your three attempts? > Your ignorance about Bayesian statistics is much deeper than I ever > thought any anti-Bayesian would be. Most of them know SOMETHING > about how Bayesian statistics work to criticize (sometimes justifiably) > certain parts of the practice. > In your case, you know NOTHING about Bayesian statistics or its > foundation (Try L.J.Savage's book The Foundations of Statistics > for a starter). > Who is winning the argument? > No-one can ever win such an argument; > The rational person who is DEEPLY familiar with BOTH Bayesian > statistics AND the frequentist/classical statistics will always pick > Bayesian as the winner, on rational and logical grounds, without > exception. > What paradigm is more popular among economists and other applied > scientists? > Why economists? As the saying goes, if you line them end to end > they would point to all directions. As far as Bayesians go, ALL > economists who claim to be Bayesians are pseudo-Bayesians of > the worst kind (see the weakest link below). > Orthodox; though this is largely because Bayesians have been very slow > in developing stats packages that *automatically* require the > subjective input of the users. > THAT is the weakest link in the practice of Bayesian statistical > inference because of the difficulty of assessing one's prior > distribution > as well as the mathematical intractibility of nearly all realistic > priors. > As a result, most applied Bayesians are restricted to the use of > conjugate priors because those are the only mathematically > tractible priors that is informative -- though they hardly represent > any Bayesian's real beliefs. The other kinds are the un-informative > priors or diffuse priors that reduces the strongest element of > Bayesian inference to the wimpy use of NO subjective info as > input, thus defeating the most attractive part of the Bayesian > system of statistical inference. > BTW - I *do* think that Bayesian ideas have a good place! > Namely, in subjective circumstances. If a theoretician/user > decides he wants to do some respectable stats about his OWN beliefs, > with no thought of publicity or public use, then Bayesianism > could well be the way to go! (Though it NEED not.) > Why shouldn't Bayesianism be the way to go for PUBLIC use? > Any public action by anyone is always controversial -- take > Bush's policies. Your attitude seems to be a cop out, yielding > to the lowest common denominator of the unthinking non-Bayesians. > But if there is to be any suggestion of public concerns, > especially e.g. of medical/pharmaceutical concerns, and other > dangerous contexts, then a subjective approach is totally > irresponsible and hateful. It is true, we can never eliminate > the subjective element from public concerns completely, nor from > science or math. But we ought to be seeking ways to DETECT and > MINIMIZE the subjective element. NOT building it into the system!!! > Your paragraph is self-defeating and self-contradictory. Your last > sentence is self-destruct. Why should one MINIMIZE the subjective > elements of the most competent to make those decisions in > medical/pharmaceutical matters? > Would you like to go to a hospital for treatment and rely on the > decision by an automated computer, rather than by a competent > medical doctor in the specialty area? > ------------------------------------------------------------------ > Bill Taylor W.Taylor@math.canterbury.ac.nz > ------------------------------------------------------------------ > That explains why the rich history and foundation of Bayesian > statistics in Europe (UK and Italy) and the USA has not filtered > down to the isolation of the continent of Aboriginies and > criminals sent exile by the UK. :-) > -- Bob. === Subject: Re: bayesians versus frequentists > As a result, most applied Bayesians are restricted to the use of conjugate priors because those are the only mathematically tractible priors that is informative -- though they hardly represent any Bayesian's real beliefs. The other kinds are the un-informative priors or diffuse priors that reduces the strongest element of Bayesian inference to the wimpy use of NO subjective info as input, thus defeating the most attractive part of the Bayesian system of statistical inference. > Criticism is cheaper than back-seat driving unless you can do better. > What priors do you suggest that economists use then? The own prior if they know HOW to assess it AND use it. Failing to do so, they should just practice whatever else they (mal)practice without giving Bayesian statistics a bad name. -- Bob. > Can somebody explain to us the argument between bayesians > and frequentists? David Petry has made an excellent brief summary. > I will just emphasize a particular point. Bayesian is a quick term for subjective probabilist, > or perhaps more aptly, subjective-probability statistician. > The key is that, at heart and root, they conceive that probability > is *inherently* subjective, and that thus any study based on it, > such as statistics, inherits that subjectivity, and thus is, > in some sense, objectively meaningless. Of course they go on > to say that a proper, *sensible* subject, will always change > their subjective beliefs about the state of the world in certain > ways, according to Bayesian probabilities. But this is already > a severe wedge in their ground concept that > one can believe as one wants. > The first half of the paragraph is a reasonable summary. The > second half if pure nonsense. > First, one must admit that there are meaningful PROBABILITIES > that can only be made subjectively because the frequentist > notion doesn't exist, because it is a single-occurrence event! > If is meaningful to ask what is the probability that Bush will be > re-elected in the 2008 election. A meaningful and operational > value of p can be assigned to that future event, and it's likely > that everyone will have a slightly different value. That is the > crux of SUBJECTIVE probability. > Beyond that the event that have operational frequentist > methods and interpretations can be argued to be flawed for > lack of use of known facts that can only be incorporated > subjectively. > Jimmie (L.J.) Savage uses the example of a 10 repeated > trials of asking a subject if a piece of music was written by > Hyden or Mozart. If the outcome was 9 successes out of > 10, a frequentist would assign a value of 0.9 to the > probability that the subject can distinguish the style of > the composers. > A Bayesian would view the same evidence differently, > depending on the subject. A Baysesian, as any reasonable > person, would put a stronger belief on the same outcome > had the subject been a musicologist, over another subject > who is a drunk picked up at the corner of a street who > happened to have made some lucky guesses. > Why should the two identical outcomes be view the same way? Bayesian would say it makes sense to talk about the probability that > there is life on Mars, but the frequentist would see that as nonsense. > How about the probability that Team A will beat Team B in the game > this weekend? That's no more nonsense than any other probability > assessment. Las Vegas sports gambling makes bets on THEIR > subjective odds on sporting events. > This is an excellent example. Someone noted that this question could > be embedded into a wider question like how many Mars-like planets > have life on them? But this is very much NOT the same question; > when we ask the Mars question, we are very definitely thinking > of our Mars, not any other, and ultimately want a reply > concerning that. Objective statistics has little to say here. > That's because what you call objective statistics cannot make any > frequentist interpretation out of all those events in which it makes > perfect sense to speak of probability assessments. It will be no surprise after the above, to learn that I am quite set > against Bayesianism. > Out of ignorance rather than any foundational or coherent argument. > There have been three attempts in the 20th C > to build subjectivity into the system, in some science-like study. > IMHO such attempts are anathema, and ought to be resisted at all costs, > being effectively a surrender to the forces of chaos and darkness. > What did you consider to be the three attempts? OC Bayesians admit this, and defend themselves with the comment that > there is *always* going to be subjectivity in statistics (or anything > else), and that at least they put it all up front, in the subjective > prior, where everyone can see and criticise it. (Though what such, > criticisms could possibly carry in terms of intellectual weight > they never specify). IMHO this justification is nonsense, like > a child painter saying look at the lovely mess I've made, it's ART. > Non sequitur. Incidentally, the other two domains where subjectivity has been built > into the system are Copenhagenism in QM, and Intuitionism in math. > So THAT's what you mean by your three attempts? > Your ignorance about Bayesian statistics is much deeper than I ever > thought any anti-Bayesian would be. Most of them know SOMETHING > about how Bayesian statistics work to criticize (sometimes justifiably) > certain parts of the practice. > In your case, you know NOTHING about Bayesian statistics or its > foundation (Try L.J.Savage's book The Foundations of Statistics > for a starter). > Who is winning the argument? No-one can ever win such an argument; > The rational person who is DEEPLY familiar with BOTH Bayesian > statistics AND the frequentist/classical statistics will always pick > Bayesian as the winner, on rational and logical grounds, without > exception. > What paradigm is more popular among economists and other applied > scientists? > Why economists? As the saying goes, if you line them end to end > they would point to all directions. As far as Bayesians go, ALL > economists who claim to be Bayesians are pseudo-Bayesians of > the worst kind (see the weakest link below). Orthodox; though this is largely because Bayesians have been very slow > in developing stats packages that *automatically* require the > subjective input of the users. > THAT is the weakest link in the practice of Bayesian statistical > inference because of the difficulty of assessing one's prior > distribution > as well as the mathematical intractibility of nearly all realistic > priors. > As a result, most applied Bayesians are restricted to the use of > conjugate priors because those are the only mathematically > tractible priors that is informative -- though they hardly represent > any Bayesian's real beliefs. The other kinds are the un-informative > priors or diffuse priors that reduces the strongest element of > Bayesian inference to the wimpy use of NO subjective info as > input, thus defeating the most attractive part of the Bayesian > system of statistical inference. > BTW - I *do* think that Bayesian ideas have a good place! > Namely, in subjective circumstances. If a theoretician/user > decides he wants to do some respectable stats about his OWN beliefs, > with no thought of publicity or public use, then Bayesianism > could well be the way to go! (Though it NEED not.) > Why shouldn't Bayesianism be the way to go for PUBLIC use? > Any public action by anyone is always controversial -- take > Bush's policies. Your attitude seems to be a cop out, yielding > to the lowest common denominator of the unthinking non-Bayesians. > But if there is to be any suggestion of public concerns, > especially e.g. of medical/pharmaceutical concerns, and other > dangerous contexts, then a subjective approach is totally > irresponsible and hateful. It is true, we can never eliminate > the subjective element from public concerns completely, nor from > science or math. But we ought to be seeking ways to DETECT and > MINIMIZE the subjective element. NOT building it into the system!!! > Your paragraph is self-defeating and self-contradictory. Your last > sentence is self-destruct. Why should one MINIMIZE the subjective > elements of the most competent to make those decisions in > medical/pharmaceutical matters? > Would you like to go to a hospital for treatment and rely on the > decision by an automated computer, rather than by a competent > medical doctor in the specialty area? > ------------------------------------------------------------------ > Bill Taylor W.Taylor@math.canterbury.ac.nz > ------------------------------------------------------------------ > That explains why the rich history and foundation of Bayesian > statistics in Europe (UK and Italy) and the USA has not filtered > down to the isolation of the continent of Aboriginies and > criminals sent exile by the UK. :-) > > -- Bob. === Subject: Wrong Reason for the Right Answer; was (Re: Funny wrong proofs) > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. > 2) Proof of the Cayley-Hamilton theorem: in order to prove that a matrix > M is a zero of its characteristic polynomial, all you have to do is to > notice that this polynomila is det(x.Id - M) and that it is obvious that > det(M.Id - M) = det(0) = 0. > Jose Carlos Santos Wrong proofs yield wrong answers (most of the time). Your examples fit my revised subject better, IMO. In statistics, there is a well-known example of the pool variance of the difference between two sample means, as in a T-test, is given by [(n-1) u^2 + (m-1)v^2]/(n+m-2) where u^2 and v^2 are the unbiased estimates of the variance of x1bar and x2bar respectively. A correct result will be obtained by those students who take this as the correct algebraic formula: a/b + c/d = (a+b)/(c+d), using the numerator of u^2 as a, and the numerator of v^2 as c, and (c+d) = sum of the denominators = (n-1)+(m-1) = n+m-2. -- Bob. === Subject: Re: Funny wrong proofs > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. > 2) Proof of the Cayley-Hamilton theorem: in order to prove that a matrix > M is a zero of its characteristic polynomial, all you have to do is to > notice that this polynomila is det(x.Id - M) and that it is obvious that > det(M.Id - M) = det(0) = 0. > Jose Carlos Santos This is actually a valid proof, but still rather amusing. A problem on a calculus exam: Let F(x) = [integral from t=1 to t=2 of sin(t^2) dt] . Compute the derivative F'(x). Solution given by one student: Break up the integral as a sum: F(x) = [integral from t=1 to t=x of sin(t^2) dt] + [integral from t=x to t=2 of sin(t^2) dt] Switch the limits of integration on the second integral: F(x) = [integral from t=1 to t=x of sin(t^2) dt] - [integral from t=2 to t=x of sin(t^2) dt] Differentiate using the 2nd fundamental theorem of calculus: F'(x) = sin(x^2) - sin(x^2) = 0. === Subject: Re: Funny wrong proofs Discussion, linux) > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. Wow. I love this one. Lots of students would do something like: (a^2 - b^2)/(a + b) = a^2/a - b^2/b = a - b. That's obvious. But dividing the minus sign by the plus sign: that's insight, baby! -- Jesse F. Hughes I think the burden is on those people who think he didn't have weapons of mass destruction to tell the world where they are. -- White House spokesman Ari Fleischer === Subject: Re: Funny wrong proofs > Hi all: > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. It's not exactly what you're looking for, by years ago I saw a comical gaffe in a book written by the one of the best-known mathematicians in the game. The fellow sets out to prove a certain theorem by induction, proves it for n=1, and then moves on to the next theorem! === Subject: Re: Funny wrong proofs >> I would like to have examples of funny wrong proofs. Here are two >> examples. They are real, in the sense that they were made by >> math students. > It's not exactly what you're looking for, by years ago I saw a comical > gaffe in a book written by the one of the best-known mathematicians in > the game. The fellow sets out to prove a certain theorem by induction, > proves it for n=1, and then moves on to the next theorem! That reminds me that in Yosida's Functional Analysis (sixth edition), the author proves that a certain subset A of a metric space M has a non-empty interior by proving that it is homeomorphic to another subset B of M whose interior is non-empty. Jose Carlos Santos === Subject: Re: Funny wrong proofs On Thu, 08 Jun 2006 09:23:51 +0100, Jos.8e Carlos Santos > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. >> It's not exactly what you're looking for, by years ago I saw a comical >> gaffe in a book written by the one of the best-known mathematicians in >> the game. The fellow sets out to prove a certain theorem by induction, >> proves it for n=1, and then moves on to the next theorem! >That reminds me that in Yosida's Functional Analysis (sixth edition), >the author proves that a certain subset A of a metric space M has a >non-empty interior by proving that it is homeomorphic to another subset >B of M whose interior is non-empty. Hmm. Was this a general metric space? Of course the sketch above is invalid in general, but for example it's valid in R^n, by Invariance of Domain. Which raises an obvious question... hmm, no it's not valid even in a Hilbert space (for example consider one-sided l^2, the unit ball and the image of the unit ball under a right shift.) >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Funny wrong proofs > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b Fantastic corollary: x is the unknown. tLet a = -b = x/2. Thus x = 0/0. > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. === Subject: Re: Funny wrong proofs > I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b Fantastic corollary: x is the unknown. Let a = -b = x/2. Thus x = 0/0. which = 0, right? > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. === Subject: Re: Funny wrong proofs <128fkf0c7glff88@corp.supernews.com I would like to have examples of funny wrong proofs. Here are two > examples. They are real, in the sense that they were made by > math students. > 1) Proof of the fact that, in a field, you have > (a^2 - b^2)/(a + b) = a - b > Fantastic corollary: x is the unknown. > Let a = -b = x/2. Thus x = 0/0. > which = 0, right? No, you belittle it's magnitude. Recall tx = a/b when a = bx Now as ta = 0 = b, tx = 1 = 2 = 3 = ... because t0 = 0*1 = 0*2 = 0*3 = ... Indeed tx = sqr 2 = pi because t0 = 0(sqr 2) = 0.pi There you see, we don't know what x is, it can be anything. However we do notice that x is a thing for it can't be two places at once, say for example at pi and at sqr 2. Oh no no no, that is not possible, that pi = x = sqr 2, that a mere computably algebraic number dare rise unto transcendental eminence. > Proof: (a^2 - b^2)/(a + b) = a^2/a + (-/+) + b^2/b = a - b. === Subject: Re: Internal Bisectors of a Triangle > the Italian Mathematician O. Chisini proved in > his > work Sulla costruzione di un triangolo date le > tre > bisettrici, Periodico di Mat., (4), 1, 1921, > 43-51 > and > 108-121 that the construction of a triangle, > given > its three internal bisectors, can't be made only > with > ruler > and compass. > > By internal bisectors do you mean the > perpendicular > bisectors of the > triangle's sides? If so, I'm a little surprised > that > you can't > construct a triangle having given bisectors using > ruler and compass. It > looks to me like you can do the calculations with > no > more than > addition, subtraction, multiplication and division > of > lengths, which > should mean you can do it with ruler and compass, > no? > > Surely this contruction should be possible. But I > don't > know your reasons: what calculations do you refer? > I don't think you can mean the angle bisectors > either, as a > ruler-and-compass construction is possible in that > case (by > coincidence, see > se_frm/thread/9634317e4e965288). > > > Maybe I have made a mistake, or maybe you mean > something else. > > Sorry, I have made a howler. In the problem studied > by > Chisini is totally different: to construct a > triangle > give the lengths of its three angle bisectors. I > think > the existence is not ensured, but maybe the > uniqueness > is. What do you think about? > I have now two more elementary questions: > (I) given three distinct straight lines > concurrent > in a > point I, is there a triangle having these lines > as > internal bisectors? > (II) if such a triangle exists, is it > essentially > unique > (that is every other triangle which solves the > problem > can be obtained by the first by a homothety with > center > O)? > Maury > > Many other problems like that studied by Chisini can > be > put. E.g., one can assign the lenghts of the three > perpendicular bisectors (the segmnents which join > the > circumcenter to the midpoints of each side). Do you > know > a bibliography about them? > Maury I found the wonderful page http://www.cut-the-knot.org/triangle/index.shtml where many constructions are listed. I found here a simple proof of the impossibility of the construction of a triangle given the lengths of its bisectors. However, I didn't yet understand what are the answer to my questions. (I) Given three arbitrary lenghts, is there a triangle whose angle bisectors have the given lengths? (II) Is it unique? My Best Regsrds, Maury === Subject: Re: Internal Bisectors of a Triangle > Maury Barbato nous a r.8ecemment amicalement signifi.8e : > I have > now two more elementary questions: (I) given three > distinct straight > lines concurrent in a > point I, is there a triangle having these lines as > internal bisectors? > Yes, as soon as the angles made by the three lines > are all greater than > pi/2 > In fact, you just have to notice the following point > Given a triangle ABC and the intersection I of > of angles bissectors : > angle AIB = pi/2 + angle ACI > angle BIC = pi/2 + angle BAI > angle CIA = pi/2 + angle CBI > (II) if such a triangle exists, is it essentially > unique > (that is every other triangle which solves the > problem > can be obtained by the first by a homothety with > center > O)? > Yes, if you accept also homothety with n.8egative k > -- > Patrick my reply to matt271829-news@yahoo.co.uk, the problem studied by Chisini is quite different (and not so trivial, surely!). However, thank you very very much for your help. Maury === Subject: Re: Standard Deviation of PSIA {...} >That's why God commanded us to put to death a race whore who has sex >with a nigger, > God commanded no such thing. But God did command us to do away with > abominations, and you certainly qualify as an abomination. More to the point, God commanded John to seperate himself from the likes of us -- and he's been strangely reluctant go obey this particular command from God. I think he may need our help: anyone want to set up an Exile John Knight poll to give him a bit of friendly encouragement? -- cary === Subject: Re: Standard Deviation of PSIA >Mamzers are not normal people. How would you know, nincompoop? >Since jews like jew schlochley like to claim that jews are descendants >of niggers, I've never heard him make such a claim. There are some Jewish blacks. But not a particularly large number. But no one knows what nincompoops like you are descended from, including you. >That's why the wall that Wall Street was named after was built--to keep >the beasts out. http://en.wikipedia.org/wiki/Wall_Street That's why God commanded us to put to death a race whore who has sex >with a nigger, God commanded no such thing. But God did command us to do away with abominations, and you certainly qualify as an abomination. lojbab === Subject: Re: Another Reason Why Collatz is Unprovable <447ffe1b$0$11352$3b214f66@aconews.univie.ac.at #> My proof on arXiv.org rules out a proof by contradiction because it > #> rules out any proof. It is a completely logical proof. No one on usenet > #> has found any flaws in the proof. The only things people have been > #> doing are claiming that my definition of random is not rigorous > #> (because it does not specify a formal language, which is really > #> irrelevant in the context of my proof) and giving straw-man arguments > #> which prove false statements and claiming that my proof uses the same > #> type of arguments. If you have faith in logic, then you should have > #> faith in my proof. > # My proof never claims that you have to treat every integer > # individually. My proof says: let's pretend that we have a proof of > # Collatz with L bits. It then shows that there is a specific n for which > # any proof that Collatz halts at one with input n requires at least L+1 > # least L+1 bits, so we have a contradiction, as our pretend proof only > # has L bits. Therefore, Collatz is unprovable. > In your paper, you do not discuss at all the axiom system that > your are using. All your statements on proofs are absolute > statements, that do not even touch the question of the underlying > axiom systems in the slightest way. > Suppose, that I use as axiom system PA plus the axiom that the > Collatz algorithm always reaches 1. > In this axiom system, there is a very short and very trivial > proof that the Collatz algorithm always reaches 1. > This horribly collides with your absolute statement that > Collatz is unprovable. Using this line of reasoning, you can make 2+2=5 an axiom and then prove 2+2=5, contradicting the well-established fact that 2+2=5 is unprovable, i.e., that it is impossible to prove that 2+2=5. If you think mathematics is some kind of game, then this is perfectly OK, but if you believe like me that mathematics is more than just a game, then your argument is very weak. The only axiom needed for my proof to work is the axiom that in order to prove that T^k(n)=1, it is necessary to specify the formula for T^k(n) in the proof. I don't know anyone who doesn't accept this axiom as obviously true. Craig > --Gerhard > ___________________________________________________________ > Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Re: Another Reason Why Collatz is Unprovable <447ffffe$0$11352$3b214f66@aconews.univie.ac.at> It collides with Theorem 2 of your paper. > There is no contradiction with Theorem 2 in my paper. If n is a number > of the form 2^k, then it is true that n will cause Collatz to halt at > one. But still, how can one know that the number n is of the form 2^k? > One has to prove this. And proving this requires specification of the > parity vector of n, the vector of k zeroes, just as Theorem 2 predicts. > So there is no contradiction with Theorem 2. > It would also be sufficient, to show, > 1) that not only numbers of the form 2^k would eventually stop at 1 > but also all numbers of the from (2^k-1)/3 > 2) and by induction all numbers of the form (2^k-1)/3*2^j > 3) and, that by math induction from (2^k0-1)/3 would follow, > that this is also true for (2^k0-1)/3*2^k1 > and > 4) that all the numbers created by 3) would fully cover the > positive integer numbers. > 3) is evident; 4) is the open problem. 4) is definitely not true because it only includes numbers with parity vectors with only one 1. But if you were to follow this general idea and include a wider array of formulas in 3), then you could arrive at a parameterized family of all numbers which cause the Collatz algorithm to converge. However, then 4) would be equivalent to proving that the set of all numbers which cause Collatz to converge is equal to the set of all numbers. And of course, we're back at our original problem to prove the Collatz conjecture. So such a strategy that you have outlined unfortunately doesn't produce any new information that is useful for proving the Collatz conjecture. And this is really the heart of the matter as to why Collatz is unprovable. Many of the posters who have argued against my proof have argued based on the flawed assumption that the general idea that you have outlined and proposed has a chance of working. Craig > why this should be systematically impossible. > Gottfried Helms === Subject: Re: Another Reason Why Collatz is Unprovable <447ffffe$0$11352$3b214f66@aconews.univie.ac.at # #> PROPOSITION. T^(k)(2^n) is eventually 1, if n is is a nonnegative > #> integer. > # #> Proof. Mathematical Induction. > # #> (a) If n = 0, then 2^n = 1, so we can take k = 0. > #> (b) Assume that T^(k)(2^N) is eventually 1; we now prove that > #> T^(k)(2^(N+1)) is eventually 1. > #> Note that T(2^(N+1)) = 2^N, and that applying T a finite number of > #> times will make > #> T^(k)(2^N) = 1. If we let K = k+1, then > #> T^(k+1)(2^(N+1)) = T^(k)[T(2^(N+1))] = 1, > #> so the proposition is also true for N+1. > #> (c) Since we've shown that P(0) is true and P(N) implies P(N+1), the > #> Principle of Mathematical Induction proves that P(n) is true for all > #> integers n >= 0. QED. > # # > # I agree that your proposition and proof are correct. > # However, it doesn't matter. > It actually should matter a lot to you... > It collides with Theorem 2 of your paper. > There is no contradiction with Theorem 2 in my paper. If n is a number > of the form 2^k, then it is true that n will cause Collatz to halt at > one. But still, how can one know that the number n is of the form 2^k? > One has to prove this. And proving this requires specification of the > parity vector of n, the vector of k zeroes, just as Theorem 2 predicts. > So there is no contradiction with Theorem 2. Another way of explaining why Theorem 2 is not contradicted by the proof of Proginoskes' proposition is to point out the obvious, that the proof by induction of the proposition actually does specify the parity vector of 2^k by showing that at each step of the Collatz algorithm, the algorithm divides by two, so the parity vector is of k zeroes. This way of understanding Theorem 2 is a little different from the way I was explaining it before, and I think it's a better way of understanding it. Before, I was interpretting the word specify to mean that no variables can be used in the specification. In such a case, Theorem 2 could not possibly apply to parameterized families like 2^k, but only plain numbers. In this other way of interpretting Theorem 2, it's a much more natural way to counter Gerhard's argument that the fact that one can prove that 2^k causes Collatz to converge contradicts Theorem 2. Of course, both interpretations of Theorem 2 are still valid. Craig > --Gerhard > ___________________________________________________________ > Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Cool puzzles Speaking of IQs, from another newsgroup: http://analogik.com/multimedia_samorost.asp If you manage to solve it, give the sequel a try: http://www.samorost2.net/ This one's a toughie: Took me some serious head busting to finish it. Best online game/puzzles I found so far. Wonderful and imaginative animation. Flash and fast line recommended. -- Ioannis === Subject: Cosines Suppose you have two quadratic irrationals a, b. Is is possible that Cos[2 * pi * a] = b? I suspect the answer is no. If this is the case, could somebody please provide a reference to a proof? === Subject: Re: Cosines > Suppose you have two quadratic irrationals a, b. Is is possible that > Cos[2 * pi * a] = b? I suspect the answer is no. If this is the case, > could somebody please provide a reference to a proof? http://en.wikipedia.org/wiki/Gelfond-Schneider_theorem exp(2*Pi*a*i) is a value of (-1)^(2*a), so if a is algebraic irrational, then exp(2*Pi*a*i) is transcendental, and therefore its real part cos(2*Pi*a) is transcendental. === Subject: Fractions I know this is an easy question to answer but I need some help with fractions. The question is 6 & 2/3 * 1/4 ... so I convert 6 2/3 into an improper fraction which becomes 20/3 now, do I assume 1/4 is the equivelant to 1 + 1/4 or do I convert it to it's reciprocal? === Subject: Re: Fractions days. My association with the Department is that of an alumnus. >I know this is an easy question to answer but I need some help with >fractions. >The question is 6 & 2/3 * 1/4 >... so I convert 6 2/3 into an improper fraction which becomes 20/3 >now, do I assume 1/4 is the equivelant to 1 + 1/4 Why would you assume that? >or do I convert it to >it's reciprocal? Why would you do that? Yes, six and two thirds is the same as 20/3. And how do you multiply 20/3 by 1/4? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu ===