mm-2909 ==== Subject: JSH: Finally simple disproof I've been working a bit at getting a disproof showing the problem with Galois Theory that removes all possibility for disagreement that isn't just clearly bogus grasping at failed ideas. Here it is. Again starting with the equations given by Rick Decker of Hamilton College, In the ring of algebraic integers, consider 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2) and 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) where the a's are defined by a^2 - (x - 1)a + 7(x^2 + x) = 0 where Decker chose this example because when x=1 mod 7, the middle term has 7 as a factor which he I guess thought was a counterexample to my ideas because at x=0, the a's are 0 and 1, indicating that 7 divides through only one of the expressions, except for the special cases. The first step in the disproof is to focus on that last quadratic and write it with the focus on x, which gives a non-monic polynomial that becomes monic if 'a' has 7 as a factor: 7x^2 + (7-a)x + a^2 + a = 0 Now choose an integer value for a, like a=3, so that you have a non-monic polynomial irreducible over Q, and therefore, you have that x cannot be an algebraic integer. But because 3 is coprime to 7 I know that the other solution for 'a' MUST have 7 as a factor, right? But now I just go back to a^2 - (x - 1)a + 7(x^2 + x) = 0 and note that for any solution for x that resulted from my previous equation, where again, I know that x cannot be an algebraic integer there is now NO way that x can be a ratio of algebraic integers with a denominator wiht only SOME factors in common with 7, and that quadratic have a root that has 7 as a factor. It's a mathematical impossibility. So there is a direct contradiction with the result from the ring of algebraic integers, as with one integer 'a' such that a^2 + a is coprime to 7, it MUST be the case that the other 'a' has 7 as a factor, while that is impossible in the ring of algebraic integers, while there is no meaningful way for that 'a' to be like some kind of fraction i.e. a ratio of algebraic integers where the numerator and denominator are actually coprime. However, it CAN be written as a ratio of algebraic integers where the numerator and denominator are coprime in the ring of algebraic integers, which directly shows the apparent contradiction, which is resolved by recognizing the ring of algebraic integers is incomplete. Denial of the mathematics does not change anything. Bogus math ideas that DO NOT WORK are not worth fighting for. I hope that some of you choose to be mathematicians versus people who just claim to be mathematicians. Each day that you let the false mathematical ideas reign supreme is another day you take away time with the correct and more powerful ideas from humanity. You are attacking the future of humanity, and fighting against correct and powerful mathematical ideas to hold on to bogus ones that DO NOT ACTUALLY WORK, which can be proven wrong with trivial algebra!!! Why do it? James Harris. ==== Subject: Re: JSH: Finally simple disproof > I've been working a bit at getting a disproof showing the problem with > Galois Theory that removes all possibility for disagreement that isn't > just clearly bogus grasping at failed ideas. > Here it is. What is this, the fifth in 24 hours? > Again starting with the equations given by Rick Decker of Hamilton > College, In the ring of algebraic integers, consider > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0 > where Decker chose this example because when x=1 mod 7, the middle term > has 7 as a factor which he I guess thought was a counterexample to my > ideas because at x=0, the a's are 0 and 1, indicating that 7 divides > through only one of the expressions, except for the special cases. > The first step in the disproof is to focus on that last quadratic and > write it with the focus on x, which gives a non-monic polynomial that > becomes monic if 'a' has 7 as a factor: > 7x^2 + (7-a)x + a^2 + a = 0 > Now choose an integer value for a, like a=3, so that you have a > non-monic polynomial irreducible over Q, and therefore, you have that x > cannot be an algebraic integer. > But because 3 is coprime to 7 I know that the other solution for 'a' > MUST have 7 as a factor, right? > But now I just go back to > a^2 - (x - 1)a + 7(x^2 + x) = 0 > and note that for any solution for x that resulted from my previous > equation, where again, I know that x cannot be an algebraic integer > there is now NO way that x can be a ratio of algebraic integers with a > denominator wiht only SOME factors in common with 7, and that quadratic > have a root that has 7 as a factor. Why? This is not clear at all? Are you still using, if a doesn't divide b and a doesn't divide c, then a doesn't divide (b+c)? -William Hughes ==== Subject: Re: JSH: Finally simple disproof >> I've been working a bit at getting a disproof showing the problem with >> Galois Theory that removes all possibility for disagreement that isn't >> just clearly bogus grasping at failed ideas. >> Here it is. > What is this, the fifth in 24 hours? A fifth almost certainly is at the heart of this. Remember, Sunday is JSH's favorite drinking day. No doubt he followed his brilliant epistle with a relaxing evening of drunken crying and singing to the walls (his other favored weekend pursuits). -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: JSH: Finally simple disproof > I've been working a bit at getting a disproof showing the problem with > Galois Theory that removes all possibility for disagreement that isn't > just clearly bogus grasping at failed ideas. > Here it is. > What is this, the fifth in 24 hours? So? This exposition leaves no room for arguing, which is why I guess you started with social crap. Who cares how long it takes me to explain in a way that no one can refute, once I achieve the goal? See below. The goal is achieved. > Again starting with the equations given by Rick Decker of Hamilton > College, In the ring of algebraic integers, consider > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0 > where Decker chose this example because when x=1 mod 7, the middle term > has 7 as a factor which he I guess thought was a counterexample to my > ideas because at x=0, the a's are 0 and 1, indicating that 7 divides > through only one of the expressions, except for the special cases. > The first step in the disproof is to focus on that last quadratic and > write it with the focus on x, which gives a non-monic polynomial that > becomes monic if 'a' has 7 as a factor: > 7x^2 + (7-a)x + a^2 + a = 0 > Now choose an integer value for a, like a=3, so that you have a > non-monic polynomial irreducible over Q, and therefore, you have that x > cannot be an algebraic integer. > But because 3 is coprime to 7 I know that the other solution for 'a' > MUST have 7 as a factor, right? > But now I just go back to > a^2 - (x - 1)a + 7(x^2 + x) = 0 > and note that for any solution for x that resulted from my previous > equation, where again, I know that x cannot be an algebraic integer > there is now NO way that x can be a ratio of algebraic integers with a > denominator wiht only SOME factors in common with 7, and that quadratic > have a root that has 7 as a factor. > Why? This is not clear at all? Are you still using, if a doesn't > divide b and a doesn't divide c, then a doesn't divide (b+c)? > -William Hughes Choose an 'a' that is an integer, such that a^2 + a is coprime to 7, for instance, let one of the a's equal 3. Then 7x^2 + (7-a)x + a^2 + a = 0 will be irreducible over Q, proving that x cannot be an algebraic integer. Now let x = d/c, where d and c are pairwise coprime and algebraic integers, and make the substitution into a^2 - (x - 1)a + 7(x^2 + x) = 0 which gives a^2 - (d/c - 1)a + 7(d^2/c^2 + d/c) = 0 which is a^2 - (d/c - 1)a + (7(d^2 + cd))/c^2 = 0 and if c is not coprime to 7, then neither solution for 'a' can have 7 as a factor, but given the integer 'a' previously chosen, it MUST be the case that 7 is a factor of the other one, and the contradiction is shown. You see, if c has just some factors in common with 7 while being coprime to others then those factors will divide off from 7, leaving them unavailable for the a's, but with one of the a's coprime to 7, like if it's 3, the other of the a's MUST have 7 itself as a factor. Trivial it's so easy. You see, c must be a unit, but provably cannot be a unit in the ring of algebraic integers, proving quite simply that that ring is incomplete. There is no room left for argument. The ring of algebraic integers has been shown to be incomplete, disproving the standard usage of Galois Theory, and showing the theory of ideals is flawed. It's so simple that it's hard to imagine what would convince you if that doesn't. James Harris ==== Subject: Re: JSH: Finally simple disproof >[...] >It's so simple that it's hard to imagine what would convince you if >that doesn't. Curiously, this reminded me of something that happened once when I was doing some work with a colleague. He asserted something one day, I said I didn't see why it was so. He explained, I still didn't get it. He got exasperated and said that if I didn't follow his last explanation there was nothing more he could say about it. The next day he came in with a revised claim - that one I could follow. >James Harris David C. Ullrich ==== Subject: Re: JSH: Finally simple disproof On Mon, 06 Feb 2006 05:21:39 -0600, David C. Ullrich >>[...] >>It's so simple that it's hard to imagine what would convince you if >>that doesn't. >Curiously, this reminded me of something that happened once >when I was doing some work with a colleague. He asserted >something one day, I said I didn't see why it was so. He >explained, I still didn't get it. He got exasperated and >said that if I didn't follow his last explanation there >was nothing more he could say about it. >The next day he came in with a revised claim - that one >I could follow. >>James Harris > >David C. Ullrich Your department should hire Harris, then you could have such fruitful discussions every day. Of course, the subject would need be limited to polynomials of low order, and this might lose its charm before the first day is done. ==== Subject: Re: JSH: Finally simple disproof > I've been working a bit at getting a disproof showing the problem with > Galois Theory that removes all possibility for disagreement that isn't > just clearly bogus grasping at failed ideas. > > Here it is. > What is this, the fifth in 24 hours? > So? This exposition leaves no room for arguing, which is why I guess > you started with social crap. > Who cares how long it takes me to explain in a way that no one can > refute, once I achieve the goal? > See below. The goal is achieved. See below. The goal is not achieved. > > Again starting with the equations given by Rick Decker of Hamilton > College, In the ring of algebraic integers, consider > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > > and > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > where the a's are defined by > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > where Decker chose this example because when x=1 mod 7, the middle term > has 7 as a factor which he I guess thought was a counterexample to my > ideas because at x=0, the a's are 0 and 1, indicating that 7 divides > through only one of the expressions, except for the special cases. > > The first step in the disproof is to focus on that last quadratic and > write it with the focus on x, which gives a non-monic polynomial that > becomes monic if 'a' has 7 as a factor: > > 7x^2 + (7-a)x + a^2 + a = 0 > > Now choose an integer value for a, like a=3, so that you have a > non-monic polynomial irreducible over Q, and therefore, you have that x > cannot be an algebraic integer. > > But because 3 is coprime to 7 I know that the other solution for 'a' > MUST have 7 as a factor, right? > > But now I just go back to > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > and note that for any solution for x that resulted from my previous > equation, where again, I know that x cannot be an algebraic integer > there is now NO way that x can be a ratio of algebraic integers with a > denominator wiht only SOME factors in common with 7, and that quadratic > have a root that has 7 as a factor. > Why? This is not clear at all? Are you still using, if a doesn't > divide b and a doesn't divide c, then a doesn't divide (b+c)? > -William Hughes > Choose an 'a' that is an integer, such that a^2 + a is coprime to 7, > for instance, let one of the a's equal 3. Then > 7x^2 + (7-a)x + a^2 + a = 0 > will be irreducible over Q, proving that x cannot be an algebraic > integer. And at this point we know that x is not an algebraic integer, and a^2 - (x - 1)a + 7(x^2 + x) = 0 has two solutions, one of them is a=3, call the other a_2. Note we do not know if a_2 is an algebraic integer. So 3a_2 = 7(x^2 + x), but as x is not an algebraic integer this tells us nothing about whether a_2 is divisible by 7. > Now let x = d/c, where d and c are pairwise coprime and algebraic > integers, and make the substitution into > a^2 - (x - 1)a + 7(x^2 + x) = 0 > which gives > a^2 - (d/c - 1)a + 7(d^2/c^2 + d/c) = 0 > which is > a^2 - (d/c - 1)a + (7(d^2 + cd))/c^2 = 0 > and if c is not coprime to 7, then neither solution for 'a' can have 7 > as a factor, but given the integer 'a' previously chosen, it MUST be > the case that 7 is a factor of the other one, Nope, you know nothing about the divisibility of the other factor (you do not even know if it is an algebraic integer). >and the contradiction is > shown. and even if you did you have shown no contradiction. You are still relying on if a doesn't divide b and a doesn't divide c then a doesn't divide (b+c). -William Hughes ==== Subject: Re: Measuring time passage If the circular nature of clocks bothers you, just think about the universal cover whenever you are thinking about timespans longer than 12 hours. Peace out ==== Subject: Applications of integration and differentiation Could you give me some real world applications of integration and differentiation? I know that, by using differentiation, you can find how fast the slope changes of a quadratic, allowing you to find stuff like acceleration and jerk, but what are the applications for integration? To me at least, finding the area under a curve seems ==== Subject: Re: Applications of integration and differentiation > Could you give me some real world applications of integration and > differentiation? I know that, by using differentiation, you can find > how fast the slope changes of a quadratic, allowing you to find stuff > like acceleration and jerk, but what are the applications for > integration? To me at least, finding the area under a curve seems In real life, it doesn't matter what the opinions of the electorate are. Reality is reality. Imagine the results of a poll asking whether Newton's first law is true or not. I wouldn't be surprised to see such a question asked. Try reading beyond the first paragraph of your book. --Ron Bruck ==== Subject: Re: Applications of integration and differentiation On 5 Feb 2006 16:03:48 -0800, Protoman differentiation? I know that, by using differentiation, you can find >how fast the slope changes of a quadratic, allowing you to find stuff >like acceleration and jerk, but what are the applications for >integration? To me at least, finding the area under a curve seems Before the days of GPS, consider the problem of inertial navigation. You have an accelerometer and a fast computer on board. So the accelerometer gives a real time readout of the acceleration Vector at any time t. You know your initial position and velocity, so two integrations of acceleration give you your position vector. If your computer is fast enough to do numerical integration in real time, you have your position at time t. It must have worked pretty well because they apparently were able to hit things with ballistic missiles even before GPS. ==== Subject: Re: Applications of integration and differentiation > Could you give me some real world applications of integration and > differentiation? I know that, by using differentiation, you can find > how fast the slope changes of a quadratic, allowing you to find stuff > like acceleration and jerk, but what are the applications for > integration? To me at least, finding the area under a curve seems Calculus is used a lot in economics. If you think about it, the national debt is the integral of the annual deficit; and the deficit is the derivative of the debt. 90% of the public doesn't understand the difference. ==== Subject: Re: Applications of integration and differentiation differentiation? I know that, by using differentiation, you can find > how fast the slope changes of a quadratic, allowing you to find stuff > like acceleration and jerk, but what are the applications for > integration? To me at least, finding the area under a curve seems > Calculus is used a lot in economics. If you think about it, the national > debt is the integral of the annual deficit; and the deficit is the > derivative of the debt. 90% of the public doesn't understand the > difference. That's cool. ==== Subject: Re: Applications of integration and differentiation > Could you give me some real world applications of integration and > differentiation? It is useful for keeping mathematicians employed as instructors of the subject. There are other uses for the calculus but this is the best, by far. Rich ==== Subject: Re: Applications of integration and differentiation > To me at least, finding the area under a curve seems If I recall correctly, historically the need in integration arised from the need to estimate the volumes ( or at least this is one of the primary motivations). E.g, how much much wine can this barrel contain. Probably the whole issue became important with the rise of commerce, especially at the age of Renaissance. It's interesting you find the differentiation is clearly useful, and integration is useless, because --- again, historically --- the idea of differentation appears (in the works of Fermat) much later than Integration. For me, the idea of integration is also intuitively more obvious than the idea of differentation and more practical as well. (ok, so you've find the slope or acceleration rate, and what are going to do with it. At least with computing the volume, I know if I've been cheated etc...) Yet most of the Calculus textbooks (with a notable exception of T. Apostol) introduce differentation first, and then - integration. Getting back to finding the area under the curve, which is very useful is working with probabilities. The area under the curve (normalized to 1) is a 'total probability', the area under the part of the curve f(x) for a < x < b is the probability of event x such that a < x < b. You have to do lots of integration when you do probabilities and statistics. In fact, *both* integration and differentation arised from the practical needs. The theoretical justification came much much later. ==== Subject: Re: Applications of integration and differentiation So, could you tell me how to find the volume of a coke can given the dimensions? ==== Subject: Re: Applications of integration and differentiation http://tutorial.math.lamar.edu/AllBrowsers/2413/VolumeWithCylinder.asp ==== Subject: Re: Applications of integration and differentiation Could anyone show me here how to do that? ==== Subject: Re: Applications of integration and differentiation > Could anyone show me here how to do that? Do what? You have removed all reference to the problem. You need to learn how to do replies properly, so that your on-line conversations can be understood. RGV ==== Subject: Re: JSH: Simple key > Now that is also the definition of evens. But it doesn't change it > from being a rule that for evens 2 is a factor. > Your example that any rational number must be the quotient of two > integers is more like just a definition in that it doesn't quite make > sense as a rule. >> Exactly what happens with the concept of algebraic integer! > In contrast, the rule is that to be an algebraic integer a number must > be a root of a monic polynomial with integer coefficients, which > importantly DOES NOT FOLLOW from simply defining algebraic integers to > be roots of monic polynomials with integer coefficients. >> roots of monic polynomials with integer coefficients, it DOES NOT FOLLOW >> that an algebraic integer must be a root of a monic polynomial with >> integer coefficients. I would not have believed it if I hadn't seen it >> myself. :-) > It takes a proof that any algebraic integer must be the root of some > monic polynomial with integer coefficients. Even if algebraic integer means root of some monic polynomial with integer coefficients? > The example is to show how an exclusionary rule can cause apparent > contradictions. >> They are apparent for you alone. > I doubt it. > So 2 is coprime to 6 in a ring made up only of evens because 3 is > excluded from that ring because it's not even. >> Define coprime in the ring of even numbers. > Not sharing factors within the ring. > That is, 2 and 6 do not share a factor in the ring of evens. > Understand? Yes. Now please give me an example of a textbook in which coprimeness is defined like that for general commutative rings with or without a unit. After all, if you are going to persuade the world that the Mathematics that is being taught is flawed, you'll have to present proofs of your assertions. > BUT with 7x^2 + (7-a)x + a^2 + a = 0 if 'a' does not have 7 as a factor, then you are forced into having a > non-monic polynomial. At least one solution of that polynomial MUST be out of the ring of > algebraic integers. >> Really? Which root of the polynomial 2x - 2 does not belong to the ring >> of algebraic integers? > The polynomial is non-monic, so you know that at least one of its > solutions is outside of the ring of algebraic integers, if 'a' does not > have 7 as a factor. >> The polynomial 2x - 2 is non-monic. Can you please tell me which of its >> roots is not an algebraic integer? > You have a factor of 2, which isn't applicable to the polynomial I was > talking about. I was *not* talking about your specific polynomial. However *you* stated that a non-monic polynomial must have a root which is not an algebraic integer. Or did you change your mind (again)? > Your comment is like I should explain every tiny little detail as if > you can't look at > 7x^2 + (7-a)x + a^2 + a = 0 > and see that it does NOT have 2 as a factor. I never claimed otherwise. > That's basic. > For example, with > 2x^2 + 3x + 1 = (2x + 1)(x + 1) > notice that one root of x is -1/2 which is not an algebraic integer. >> Again, can you please tell me which of the roots of the polynomial >> 2x - 2 is not an algebraic integer? > This is getting tedious. >> I agree! > You're being deliberately obtuse. That's funny! I thought that you had taken the decision of not making comments such as this one. > I guess you just want to be aggravating. Or this one. > So childish. And much less this one. Jose Carlos Santos ==== Subject: Re: JSH: Simple key <44kbqsF2d84dU1@individual.net> <44kjh1F2knidU1@individual.net> <44lp5bF2q7m5U1@individual.net> <44n6ebF31mgcU1@individual.net> <44niooF33s1rU1@individual.net Now that is also the definition of evens. But it doesn't change it > from being a rule that for evens 2 is a factor. Your example that any rational number must be the quotient of two > integers is more like just a definition in that it doesn't quite make > sense as a rule. >> Exactly what happens with the concept of algebraic integer! > In contrast, the rule is that to be an algebraic integer a number must > be a root of a monic polynomial with integer coefficients, which > importantly DOES NOT FOLLOW from simply defining algebraic integers to > be roots of monic polynomials with integer coefficients. >> roots of monic polynomials with integer coefficients, it DOES NOT FOLLOW >> that an algebraic integer must be a root of a monic polynomial with >> integer coefficients. I would not have believed it if I hadn't seen it >> myself. :-) > It takes a proof that any algebraic integer must be the root of some > monic polynomial with integer coefficients. > Even if algebraic integer means root of some monic polynomial with > integer coefficients? Yup. I'll take an example from your playbook, and define Santos numbers as roots of non monic polynomials, so 2x + 2 = 0 is, by that definition, a Santos number. But x=1 is also the root of polynomials that are NOT monic, so the definition doesn't create the rule that a Santos number must be the root of only non monic polynomials. You're way past tiresome, as you just keep arguing about dumb stuff. The pure math that comes from the ideas I've shot down has been impractical, with mathematicians saying maybe, someday it might be useful, but I say, it has never been practical, because it's wrong (with some questions now in my mind about the use of group theory in physics). Yet, I give proof, and you people fight the proof. I get proof published and you fight publication. I come back with examples showing integers behaving as predicted by my research, and you try to ignore that, so I use an example one of you came up with, to finally crush any semblance of a mathematical objection, and you still fight the mathematical truth. What DO you believe in? James Harris ==== Subject: Re: JSH: Simple key > I'll take an example from your playbook, and define Santos numbers as > roots of non monic polynomials, so > 2x + 2 = 0 > is, by that definition, a Santos number. Wow! I never dreamed that I would have my name attached to a concept! There's a small problem however. You say first that a Santos number is a root of a non monic polynomial and then you say that 2x + 2 is a Santos number. But 2x + 2 = 0 is not a number, it's a polynomial equation! I guess that what you meant was that *the roots* of 2x + 2 = 0 are Santos numbers. > But x=1 is also the root of polynomials that are NOT monic, so the > definition doesn't create the rule that a Santos number must be the > root of only non monic polynomials. I fully agree with you. However, it's *you* who said that at least one root of a non-monic polynomial must be out of the ring of algebraic integers. Or have changed your opinion on this? > You're way past tiresome, as you just keep arguing about dumb stuff. Look who's talking. :-) > Yet, I give proof, and you people fight the proof. It's not a proof. > I get proof published and you fight publication. And... ? Should we accept as sacred anything that has ever been published? Do you follow that rule? Besides, you never told us what explanation you got from the editor of the Southwest Journal of Pure and Applied Mathematics for deleting was published by mistake. But if I am wrong, tell us what he told you. Jose Carlos Santos ==== Subject: Re: JSH: Simple key <44kbqsF2d84dU1@individual.net> <44kjh1F2knidU1@individual.net> <44lp5bF2q7m5U1@individual.net> <44n6ebF31mgcU1@individual.net> <44niooF33s1rU1@individual.net> <44oj82F32m4dU1@individual.net I'll take an example from your playbook, and define Santos numbers as > roots of non monic polynomials, so > 2x + 2 = 0 > is, by that definition, a Santos number. > Wow! I never dreamed that I would have my name attached to a concept! > There's a small problem however. You say first that a Santos number is > a root of a non monic polynomial and then you say that 2x + 2 is a > Santos number. But 2x + 2 = 0 is not a number, it's a polynomial > equation! I guess that what you meant was that *the roots* of 2x + 2 = 0 > are Santos numbers. > But x=1 is also the root of polynomials that are NOT monic, so the > definition doesn't create the rule that a Santos number must be the > root of only non monic polynomials. > I fully agree with you. However, it's *you* who said that at least one > root of a non-monic polynomial must be out of the ring of algebraic > integers. Or have changed your opinion on this? > You're way past tiresome, as you just keep arguing about dumb stuff. > Look who's talking. :-) > Yet, I give proof, and you people fight the proof. > It's not a proof. > I get proof published and you fight publication. > And... ? Should we accept as sacred anything that has ever been > published? Do you follow that rule? > Besides, you never told us what explanation you got from the editor > of the Southwest Journal of Pure and Applied Mathematics for deleting > was published by mistake. There was more to it than merely being published by mistake. JSH was fully aware that there were issues unresolved when he sent that paper. It was clearly a case of fraud. Although the editor may have been wrong yanking the paper, it was certainly justified. > But if I am wrong, tell us what he told you. Don't hold your breath. > Jose Carlos Santos ==== Subject: Re: Well-defined operation??? Harmonic Mean. Arithmetic Mean <24442197.1138995715120.JavaMail.jakarta@nitrogen.mathforum.org Sorry Sir, I cannot understand what you are talking about. > You say that the rational mean is not a valid arithmetical operation because > it produces different results for the same decimal values: a/b, 2a/2b, 3a/3b, ... I did not mention the word decimal. This has nothing to do with any particular system of numerals. > At the same time, all of you (world-wide) state that the arithmetic mean, the harmonic mean and in general > the well-known and widely accepted weighted mean certainly are well-defined operations in the set of rational numbers. > No doubts, the weighted mean has been world-wide accepted as a well-defined operation in the set of rationals. > Then according to you the following weighted mean of the rational values a1/b1, a2/b2,...an/bn constitutes a well > defined operation: > Weighted Mean = (b1/(b1+b2+...+bn)) (a1/b1) + (b2/(b1+b2+...+bn)) (a2/b2) + ... + (bn/(b1+b2+...+bn)) (an/bn) > The weighting factors are: (b1/(b1+b2+...+bn)), (b2/(b1+b2+...+bn)), (bn/(b1+b2+...+bn)) These are only well-defined if you specify that your rational numbers are expressed in lowest terms. > The Cartesian-decimal system states that, say, 1/2 = 2/4 =3/6,... However, I am sure we are not in front of the Holy Office. I am not in front of any office, holy or not. Are you denying that 1/2 = 2/4? If so, your problem goes back a lot farther than Descartes or the decimal system. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Well-defined operation??? Harmonic Mean. Arithmetic Mean >I did not mention the word decimal. This has nothing >to do with any >particular >system of numerals. It has certainly something to do with the system of numerals adopted by Cartesian system. The decimal system led the way for replacing, say, 3/2 by the result [CapitalEth]in decimal notation-- of dividing 3 by 2, that is :1.5, which is a decimal value and the system of quantity adopted by Cartesian System: just the decimal value coming out from divisions. Of course, this is the main reason to consider the rational number 3/2 as the set: 3/2=6/4=9/6=... I understand that such rational-number conception could probably have been reached without the decimal-notation practice, however it impelled that with so much force. The Cartesian system adopted such rational-number conception which is the cause for the modern words: Well-defined and Bad-defined in the set of rational numbers. My point is that 3/2 has a form which has a very different meaning from that of 6/4. Such form has no significance for Cartesian system, but, what about any other system based on the Rational Mean? The new proposed rational-number concept --based on the Rational Mean-- asserts that 3/2 and 6/4 have the same magnitude but also have different forms, and such forms play a crucial role in a new system of quantity which clearly differs from the Cartesian system. I hope you have find some time to check the new higher-order arithmetical algorithms based on the Rational Mean which appear at: http://mipagina.cantv.net/arithmetic/rmdef.htm Based on all the evidences there are no precedents for such trivial algorithms even though mathematicians of past times have at hand the basic tool (The rational mean) for handling them so easily. As you can see there the form of the rational number plays the main role. This means that we donÍt need the _geometrical_ artifice which I use to call: Cartesian-decimal-fluxions system (along with hundred of years it was required for its creation) to develop all those ńsuperbî algorithms that we learned under the supremacy and the dominion of cartesian imperialism. All this means that we can handle Number just by agency of Number itself. I recall now the following outstanding phrase from the American philosopher Charles Sanders Peirce, who when dealing with the Rational Mean (he didnÍt gave it any name) said: (Collected papers, Hardvard University Press, 1933, Vol. IV, art. 681, pag. 580) ń.83It is because [of] this form of relation of rational consequence that numbers are of such stupendous importance in reasoning. But the highest and last lesson which the numbers whisper in our ears is that of the supremacy of the forms of relation for which their tawdry outside is the mere shell of the casket.83î Domingo G.97mez Mor.92n mipagina.cantv.net/arithmetic ==== Subject: Re: Standard Deviation of PSIA <0001HW.C00741FA00070AF4F0284550@news.giganews.com> <9pa8u15g5crtdk9vhvd78g8sno5vj1l11c@4ax.com> The politicians in DC quietly implemented their deniggerizer plan, and guess what happened? The percentage of niggers dropped from 83% to 60%. Guess what ELSE happened???!!! The murder rate PLUNGED by more than half, from 80.6 to only 35.8 (a tidy 54% drop which SAVED 266 lives PER YEAR, in ONE city, year after year). Our politicos finally figured out--without ever telling anybody else--that deniggerizing by only 1% reduces the murder RATE by 2.3 per 100k. Why does DC get to deniggerize while they try to niggerize the rest of us? ANOTHER point you forgot--every single one of these niggerized cities have ONE thing in common--the world's HIGHEST murder rates, BAR NONE [US Statistical Abstract, Crime Rates by Type--selected large cities, Table No. 290]: >Ten Cities of 100,000 or More with Highest Percentage of Blacks or African Americans, 2000 >City Percent >Gary, Ind. 84.0% WITH A MURDER RATE OF 86 >Detroit, Mich. 81.6 WITH A MURDER RATE OF 57 >Birmingham, Ala. 73.5 WITH A MURDER RATE OF 47 >Jackson, Miss. 70.6 WITH A MURDER RATE OF ?? >New Orleans, La. 67.3 WITH A MURDER RATE OF 86 >Baltimore, Md. 64.3 WITH A MURDER RATE OF 41 >Atlanta, Ga. 61.4 WITH A MURDER RATE OF 59 >Memphis, Tenn. 61.4 WITH A MURDER RATE OF 32 >Washington, DC 60.0 WITH A MURDER RATE OF 36 >Richmond, Va. 57.2 WITH A MURDER RATE OF ?? Not a single one of these cities has a murder RATE lower than 160 TIMES higher than nigger-free Singapore, or North Dakota, or Saudi Arabia, or Qatar, where the murder rates hover around 0.2. The LOWEST rate of the above cities is STILL TEN TIMES HIGHER than a multitude of nigger-free US cities, like Mesa, Arizona; Omaha, Ne; Arlington, Tx; and Colorado Spring, Co, which I'm sure is the ultimate objective of District of Columbians. I can't wait http://blackexile.com ==== Subject: Re: Standard Deviation of PSIA >The politicians in DC who are mostly black >quietly implemented their deniggerizer plan, What was this plan? Something like create a lot of new jobs and cheap housing in Prince Georges County MD? And the DC politicians implemented this plan how? >and guess what happened? The nincompoop pulled some more noisome nonsense out of its strange orifice - that is what happened. >The percentage of niggers dropped from 83% to 60%. No. >Guess what ELSE happened???!!! The murder rate PLUNGED by more than >half, from 80.6 to only 35.8 (a tidy 54% drop which SAVED 266 lives >PER YEAR, in ONE city, year after year). The DC murder rate hit its peak during drug wars in the 1990s, when the percentage of blacks was between 60-66% (1990 66%, 2000 60%). >Why does DC get to deniggerize while they try to niggerize the rest of >us? They don't. >ANOTHER point you forgot--every single one of these niggerized cities >have ONE thing in common--the world's HIGHEST murder rates, BAR NONE >[US Statistical Abstract, Crime Rates by Type--selected large cities, >Table No. 290]: >>Ten Cities of 100,000 or More with Highest Percentage of Blacks or African Americans, 2000 >>City Percent >>Gary, Ind. 84.0% WITH A MURDER RATE OF 86 >>Detroit, Mich. 81.6 WITH A MURDER RATE OF 57 >>Birmingham, Ala. 73.5 WITH A MURDER RATE OF 47 >>Jackson, Miss. 70.6 WITH A MURDER RATE OF ?? >>New Orleans, La. 67.3 WITH A MURDER RATE OF 86 >>Baltimore, Md. 64.3 WITH A MURDER RATE OF 41 >>Atlanta, Ga. 61.4 WITH A MURDER RATE OF 59 >>Memphis, Tenn. 61.4 WITH A MURDER RATE OF 32 >>Washington, DC 60.0 WITH A MURDER RATE OF 36 >>Richmond, Va. 57.2 WITH A MURDER RATE OF ?? >Not a single one of these cities has a murder RATE lower than I notice that you don't have a number for two of them. >160 TIMES higher than nigger-free Singapore, Feel free to move to Singapore. We won't miss you. >or North Dakota, North Dakota isn't a city. >or Saudi Arabia, or Qatar, Nor are they. But you are welcome to go there. >where the murder rates hover around 0.2. >The LOWEST rate of the above cities is STILL TEN TIMES HIGHER than a >multitude of nigger-free US cities, >like Mesa, Arizona; 3% black, 20% Hispanic Omaha, Ne; 8.3% black, 5% Hispanic >Arlington, Tx; 14% black, 19% Hispanic and Colorado Spring, Co 6.5% black, 11% Hispanic >which I'm sure is the ultimate objective of District of Columbians. I rather doubt it. lojbab ==== Subject: Re: a question about cohomology of groups ok ! are you learning the book volume 4 in the Graduate Texts in Mathematics series from Springer ? you have passed chapter six at least :-). Glad to meet you ==== Subject: Re: a question about cohomology of groups >ok ! >are you learning the book volume 4 in the Graduate Texts in >Mathematics >series from Springer ? you have passed chapter six at least :-). >Glad to meet you To whom is this post addressed? Derek Holt. ==== Subject: Re: a question about cohomology of groups $$$$$$$$ to m...@mimosa.csv.warwick.ac.uk $$$$$$$$$$$ Mr. Derek Holt are you learning the book volume 4 in the Graduate Texts in Mathematics series from Springer ? you have passed chapter six at least :-). Glad to meet you ==== Subject: Re: a question about cohomology of groups 9;................... ok i have find it ! ------from Mr. nonono ==== Subject: Re: a question about cohomology of groups days. My association with the Department is that of an alumnus. NOT EVERYONE READS NEWS THROUGH GOOGLE. There is nothing left to look, in my interface. Please learn some basic nettiquette and QUOTE the message you are replying to. link at the bottom; click on More Options at the TOP of the message you are replying to, and then click on the Reply link that appears then (at the top). This will provide a QUOTE of the message you are replying to, indicating who you are replying to. Just like my message here includes the following: which shows I am replying to you and to the words I am quoting. -- ' ==== Subject: Re: a question about cohomology of groups a list of the newsgroups I subscribe to. Further left, and I'm at the desktop. ==== Subject: Re: a question about cohomology of groups your advice has been accpeted ! ==== Subject: Re: a question about cohomology of groups >your advice has been accpeted ! Whose advice would that be? Derek Holt. ==== Subject: Re: a question about cohomology of groups of course to Mr.Arturo Magidin ==== Subject: Re: a question about cohomology of groups days. My association with the Department is that of an alumnus. >of course to Mr.Arturo Magidin Of course to me what? Learn to provide context, or stop posting. -- ' ==== Subject: Re: Linear System of Equations > I have a great problem. I have a set of 3n-3 linear equations in 2k-1 > variables and I would like to prove that (0,0,...,0) is the only > solution in Z/2Z. The equations are: > $sum_{j=0}^{i-3}{C_{i-1}^{j} x_{i+j+1}}+ix_{2i-1}=0$ pour $3leq ileq > n-1$, > $sum_{j=0}^{i-1}{C_{i-1}^{j} x_{i+j}}=0$ pour $1leq i leq n$, > $x_i+x_{2n-i}=0$ pour $1leq ileq n-1$ et $x_k=0$. > I have check that (0,0,....,0) is the only solution for n<300 with Something wrong here. What is k as in 2 k - 1 variables and x_k = 0? Won't matters depend on the coefficients? E.g., if n = 4, and C_0^0 = C_3^3 = 0, then there's a solution with x_1 = 17, x_7 = -17, all other variables zero (I've ignored x_k = 0 as I don't know what k is). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) ==== Subject: Extremely basic topology notation question: S' ? Sorry to bother the group with such a trivial question, but I am having no luck finding what this denotes, and good freaking luck single-quotes! S' Is it the interior of S? Or the closure? Or the exterior? I know it's one of those things, it's just ridiculous to have such arbitrary notation for something which shows up maybe once per 10,000 pages of literature, for something so rare something like int(S) would be so much more appropriate! ==== Subject: Re: Extremely basic topology notation question: S' ? > Sorry to bother the group with such a trivial question, but I am > having no luck finding what this denotes, and good freaking luck > single-quotes! > S' > Is it the interior of S? Or the closure? Or the exterior? I > know it's one of those things, it's just ridiculous to have such > arbitrary notation for something which shows up maybe once per 10,000 > pages of literature, for something so rare something like int(S) > would be so much more appropriate! Probably the derived set. http://en.wikipedia.org/wiki/Derived_set_%28mathematics%29 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== Subject: Re: Extremely basic topology notation question: S' ? > Sorry to bother the group with such a trivial question, but I am > having no luck finding what this denotes, and good freaking luck > single-quotes! > S' > Is it the interior of S? Or the closure? Or the exterior? I > know it's one of those things, it's just ridiculous to have such > arbitrary notation for something which shows up maybe once per 10,000 > pages of literature, for something so rare something like int(S) > would be so much more appropriate! You provide zero context and you're complaining? ==== Subject: Re: Extremely basic topology notation question: S' ? > Sorry to bother the group with such a trivial question, but I am > having no luck finding what this denotes, and good freaking luck > single-quotes! > S' > Is it the interior of S? Or the closure? Or the exterior? I > know it's one of those things, it's just ridiculous to have such > arbitrary notation for something which shows up maybe once per 10,000 > pages of literature, for something so rare something like int(S) > would be so much more appropriate! I know that in Munkres' Topology book, the prime notation denotes the set of limit points of any subspace. Perhaps this is what the notation means in your situation as well. Mike ==== Subject: Re: Extremely basic topology notation question: S' ? >> Sorry to bother the group with such a trivial question, but I am >>having no luck finding what this denotes, and good freaking luck >>single-quotes! >> S' >> Is it the interior of S? Or the closure? Or the exterior? I >>know it's one of those things, it's just ridiculous to have such >>arbitrary notation for something which shows up maybe once per 10,000 >>pages of literature, for something so rare something like int(S) >>would be so much more appropriate! >> >I know that in Munkres' Topology book, the prime notation denotes the >set of limit points of any subspace. Perhaps this is what the notation >means in your situation as well. Yes, I would think it meant the set of limit points of S. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan ==== Subject: Re: Extremely basic topology notation question: S' ? I've never seen it. Could it have been S^1 (the circle)? ==== Subject: Re: JSH: Better explanation? days. My association with the Department is that of an alumnus. >[Tim Peters, lampoons Cauchy and Lam.8e's reactions upon being told > that their puported proofs of FLT were wrong] >[Arturo Magidin] >> You should check out the corresponding chapter in Edwards Fermat's >> Last Theorem: A Genetic Introduction to Number Theory. >Ah, good, so Edwards _was_ there :-) >I've noted the book for reference, because I would indeed enjoy it. Alas, >the local libraries don't have it, and US$80 is a bit steep. A bit late for this year, but come September keep an eye out for Springer's annual Yellow Sale. Edwards's book has been offered a few times over the past few years, if memory serves, and the discounts are often quite good. [.snip.] >> for several more weeks. He also made some noises about how he was >> coming to the notions of Kummer's ideal numbers (in so far as he could >> glimpse them from the brief description Liouville had given) and that >> he was sure he would be able to connect his research and final results >> to Kummer's. But by the end of the summer he stopped writing on the >> topic; he never acknowledged defeat, he simply switched to publishing >> about mathematical astronomy. >So, when we complete the tree showing the great mathematicians and their >intellectual ancestors, should there be a dotted line from Harris back to >Cauchy -- or does JSH stand alone ;-)? Well, while changing the subject when stumped is something we have observed in the latter case, you'll note that Cauchy never came back to the subject. Once he left it behind he was truly done with it. -- ' ==== Subject: Re: JSH: Better explanation? [Tim Peters, lampoons Cauchy and Lam.8e's reactions upon being told that their puported proofs of FLT were wrong] [Arturo Magidin] > You should check out the corresponding chapter in Edwards Fermat's > Last Theorem: A Genetic Introduction to Number Theory. Ah, good, so Edwards _was_ there :-) I've noted the book for reference, because I would indeed enjoy it. Alas, the local libraries don't have it, and US$80 is a bit steep. > Lame sketched his argument; Lame made a point of noting that he > had the idea after a casual conversation with Liouville. Immediately > after Lame finished, Liouville took the floor; he declined all credit > to himself, and pointed out that many others had had the idea of > factoring diophantine equations over the complex numbers,mentioning > explicitly Euler, Lagrange, Gauss, Cauchy, and above all Jacobi; in > fact, Lagrange had used the exact same factorization as Lame, > explicitly, before. He then immediately pointed out that the proposed > argument had a rather large gap in it, namely, unique factorization > (Edwards notes that the fact that Liuoville immediately noted the gap > meant that he was aware of the gap in Euler's proof of the n=3 case, > although apparently no writer had yet to criticize the argument in > print; and there was a very good chance, in view of how quickly his > point was accepted, that several people were already well aware of the > role of unique factorization). > Cauchy took the floor immediately after [Liouville], and not only > said there was a very good chance that Lame's argument would work, he > essentially said that he (Cauchy) had pretty much done the same thing > a year earlier, presented it to the Academy, but hadn't bothered to > flesh it out. > Lame acknowledged the gap, though he was certain he would be able to > bridge it. While I wasn't here for this one either ;-), about 8 years earlier Lam.8e had the first proof for the n=7 FLT case, so his opinion carried (or should have carried) some weight. According to ... In 1832 Dirichlet published a proof of Fermat's Last Theorem for n=14. Of course he had been attempting to prove the n=7 case but had proved a weaker result. The n=7 case was finally solved by Lam.8e in 1839. It showed why Dirichlet had so much difficulty, for although Dirichlet's n=14 proof used similar (but computationally much harder) arguments to the earlier cases, Lam.8e had to introduce some completely new methods. Lam.8e's proof is exceedingly hard and makes it look as though progress with Fermat's Last Theorem to larger n would be almost impossible without some radically new thinking. > ... > Almost three months after the original announcement, in late May 1847, > Liouville read into the proceedings a letter from Kummer; in that > letter Kummer told Liouville that his questioning of the method was > dead on the money, and then pointed Liouville to his (Kummer's) memoir > proving that unique factorization fails in some of the cyclotomic > cases, and in fact specifically in some cases for which Lame was > claiming his 'lemmas' proved it held. Although the memoir was already > three years old, it had been published in an obscure journal, so the > fact that it was not known is understandable. Just noting that the reference above says Kummer included an off-print of his 1844 paper with the letter. Helpful fellow! I wonder whether he'd hang out on Usenet ... > ... > Lame simply fell silent at this revelation. Beats ranting -- good for him! > for several more weeks. He also made some noises about how he was > coming to the notions of Kummer's ideal numbers (in so far as he could > glimpse them from the brief description Liouville had given) and that > he was sure he would be able to connect his research and final results > to Kummer's. But by the end of the summer he stopped writing on the > topic; he never acknowledged defeat, he simply switched to publishing > about mathematical astronomy. So, when we complete the tree showing the great mathematicians and their intellectual ancestors, should there be a dotted line from Harris back to Cauchy -- or does JSH stand alone ;-)? ==== Subject: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? There is a variation of random chess that is basically the same as the standard chess, except that the back row is shuffled randomly with a digital device, so that there are many more opening positions that the regular opening. (i put random in quotes, because there is no such a thing in reality) This makes it much more difficult to memorize the opening moves, and thus places more emphasis on the skill of the player, and not just rote memorization. The question i have is probably a basic one, but how do you compute the number of variations in this case? That is, the number of different ways you can order 2 rooks, 2 knights, 2 bishops, and one king and one queen. I know what the number is already, i just wanted to know how you would calculate this with permutations, combinatorics and such. Slick ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? > There is a variation of random chess that > is basically the same as the standard chess, > except that the back row is shuffled randomly > with a digital device, so that there are many > more opening positions that the regular opening. > (i put random in quotes, because there is no > such a thing in reality) > This makes it much more difficult to memorize > the opening moves, and thus places more > emphasis on the skill of the player, and not > just rote memorization. > The question i have is probably a basic > one, but how do you compute the number > of variations in this case? That is, the > number of different ways you can order > 2 rooks, 2 knights, 2 bishops, and one > king and one queen. I know what the > number is already, Is your known number =48? Is one still required to observe the rules of chess - such as castling of king needs minimum distance of two square between a rook and the king and same colored bishops always stand on opposite colored squares. >i just wanted to know > how you would calculate this with permutations, > combinatorics and such. > Slick -- Respectfully, Mohan Pawar MIO Instruments LLC ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? is basically the same as the standard chess, > except that the back row is shuffled randomly > with a digital device, so that there are many > more opening positions that the regular opening. > (i put random in quotes, because there is no > such a thing in reality) > This makes it much more difficult to memorize > the opening moves, and thus places more > emphasis on the skill of the player, and not > just rote memorization. > The question i have is probably a basic > one, but how do you compute the number > of variations in this case? That is, the > number of different ways you can order > 2 rooks, 2 knights, 2 bishops, and one > king and one queen. I know what the > number is already, > Is your known number =48? 960 > Is one still required to observe the rules of chess - such as castling of > king needs minimum distance of two square between a rook and the king and > same colored bishops always stand on opposite colored squares. http://www.dwheeler.com/essays/Fischer_Random_Chess.html ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? is basically the same as the standard chess, > except that the back row is shuffled randomly > with a digital device, so that there are many > more opening positions that the regular opening. > (i put random in quotes, because there is no > such a thing in reality) > This makes it much more difficult to memorize > the opening moves, and thus places more > emphasis on the skill of the player, and not > just rote memorization. > The question i have is probably a basic > one, but how do you compute the number > of variations in this case? That is, the > number of different ways you can order > 2 rooks, 2 knights, 2 bishops, and one > king and one queen. I know what the > number is already, > Is your known number =48? > Is one still required to observe the rules of chess - such as castling of > king needs minimum distance of two square between a rook and the king and > same colored bishops always stand on opposite colored squares. link: http://www.dwheeler.com/essays/Fischer_Random_Chess.html which explains things nicely. --- Christopher Heckman ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? There is a variation of random chess that > is basically the same as the standard chess, > except that the back row is shuffled randomly > with a digital device, so that there are many > more opening positions that the regular opening. > (i put random in quotes, because there is no > such a thing in reality) > > This makes it much more difficult to memorize > the opening moves, and thus places more > emphasis on the skill of the player, and not > just rote memorization. > > The question i have is probably a basic > one, but how do you compute the number > of variations in this case? That is, the > number of different ways you can order > 2 rooks, 2 knights, 2 bishops, and one > king and one queen. I know what the > number is already, > Is your known number =48? > Is one still required to observe the rules of chess - such as castling of > king needs minimum distance of two square between a rook and the king and > same colored bishops always stand on opposite colored squares. > link: > http://www.dwheeler.com/essays/Fischer_Random_Chess.html > which explains things nicely. It's also possible use this procedure to see why there are exactly 960 possible initial positions. Each bishop can take one of 4 positions, the Queen one of 6, and the two knights can have 5 or 4 possible positions, respectively. This means that there are 4*4*6*5*4 = 1920 possible positions if the two knights were different in some way. However, the two knights are indistinguishable during play; if they were swapped, there would be no difference. This means that the number of distinguishable positions is half of 1920, or 1920/2 = 960 possible distinguishable positions. S. ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? > There is a variation of random chess that >is basically the same as the standard chess, >except that the back row is shuffled randomly >with a digital device, so that there are many >more opening positions that the regular opening. >(i put random in quotes, because there is no >such a thing in reality) [...] > The question i have is probably a basic >one, but how do you compute the number >of variations in this case? That is, the >number of different ways you can order >2 rooks, 2 knights, 2 bishops, and one >king and one queen. I know what the >number is already, i just wanted to know >how you would calculate this with permutations, >combinatorics and such. The king must be between the rooks. Therefore, the king and rooks together can be placed in (8 3) = 56 ways. Out of these 56 ways, there are (4 3) = 4 ways to place them solely on light squares and similarly (4 3) = 4 ways to place them solely on dark squares. The remaining five squares can be broken down into five cases: I. 3 dark, 2 light II. 2 dark, 3 light III. 4 dark, 1 light <=> KRR all on light IV. 1 dark, 4 light <=> KRR all on dark Since the bishops must be placed on different colored squares, in 4 + 4 = 8 cases (III and IV.) we can place them in 4 different ways, and the remaining 56 - 8 = 48 cases (I. and II.) we can place them in 3 * 2 = 6 ways. The queen can be freely placed on the of the remaining squares in (3 1 ) = 3 ways. Thus the total number of ways to place the pieces is: 8 * 4 * 3 + 48 * 6 * 3 = 96 + 864 = 960 ==== Subject: Re: COMPUTE VARIATIONS OF FISCHER RANDOM CHESS????? is basically the same as the standard chess, >except that the back row is shuffled randomly >with a digital device, so that there are many >more opening positions that the regular opening. >(i put random in quotes, because there is no >such a thing in reality) > [...] > The question i have is probably a basic >one, but how do you compute the number >of variations in this case? That is, the >number of different ways you can order >2 rooks, 2 knights, 2 bishops, and one >king and one queen. I know what the >number is already, i just wanted to know >how you would calculate this with permutations, >combinatorics and such. > The king must be between the rooks. Therefore, the king and rooks > together can be placed in (8 3) = 56 ways. How did you get (8 3)=56? Out of these 56 ways, there > are (4 3) = 4 ways to place them solely on light squares and similarly > (4 3) = 4 ways to place them solely on dark squares. > The remaining five squares can be broken down into five cases: > I. 3 dark, 2 light > II. 2 dark, 3 light > III. 4 dark, 1 light <=> KRR all on light > IV. 1 dark, 4 light <=> KRR all on dark > Since the bishops must be placed on different colored squares, in 4 + > 4 = 8 cases (III and IV.) we can place them in 4 different ways, and > the remaining 56 - 8 = 48 cases (I. and II.) we can place them in 3 * > 2 = 6 ways. > The queen can be freely placed on the of the remaining squares in > (3 1 ) = 3 ways. > Thus the total number of ways to place the pieces is: > 8 * 4 * 3 + 48 * 6 * 3 = 96 + 864 = 960 Ok, you got the same answer as: http://www.dwheeler.com/essays/Fischer_Random_Chess.html But they calculated it in an easier fashion by saving the last three open squares for the Rook-King-Rook sequence. There is also a simplified version of Fischer Random called Chess_18, which has the Rooks and King in the standard positions, and shuffles the remaining 2 bishops, 2 knights, and 1 queen: 3 (black or white for first bishop) * 2(opposite color for remaining bishop) * 3 (first knight) * 2 (remaining knight) = 36 But because the knights are identical, it's 36 / 2 = 18 positions. S ==== Subject: Re: expansion for polylog(x,-1) for x<0 <15099752.1139141519016.JavaMail.jakarta@nitrogen.mathforum.org> So what about for x<0, i.e. is there a series or product which converges to this wildly oscillating function on the negative real axis? ==== Subject: Re: expansion for polylog(x,-1) for x<0 <15099752.1139141519016.JavaMail.jakarta@nitrogen.mathforum.org> Your post and Google helped! Searching under dirichlet eta quickly led to http://www.geocities.com/hjsmithh/Numbers/Eta.html where I read For negative x < -4, let y = 1-x, then eta(x) = 2 * (2*Pi)^y * sin(Pi*x/2) * gam(y) * eta(y) * (1-2^y)/(1-2^x). This is pretty much what I wanted. ==== Subject: Re: expansion for polylog(x,-1) for x<0 > Your post and Google helped! > Searching under dirichlet eta quickly led to > http://www.geocities.com/hjsmithh/Numbers/Eta.html > where I read > For negative x < -4, let y = 1-x, then > eta(x) = 2 * (2*Pi)^y * sin(Pi*x/2) * gam(y) * eta(y) > * (1-2^y)/(1-2^x). > This is pretty much what I wanted. This result follows from the fact that zeta(1 - z) = 2 * (2pi)^(-z) * cos(z pi / 2) * gamma(z) * zeta(z) ..which also reflects the nature of the trivial zeros of the zeta function. Kyle ==== Subject: differential equations with random coefficients a limited series of differential equations with constant coefficients is easy to be solved, but, how to solve differential equations whose coefficients are random ==== Subject: Angled projectile Could you give me the formula for a projectle launched at an angle? ==== Subject: Finding info about a projectile How do I find what the maxium height is, how long it takes to get there, the max displacement, and the time to reach max displacement if the projectile is launched at an ANGLE? I know how when its straight PS. Here's the question: Find the max height and displacement, and the time for each, when a ball is thrown at a 45 degree angle with a initial acceleration of 16ft/sec. ==== Subject: Re: Finding info about a projectile > How do I find what the maxium height is, how long it takes to get > there, the max displacement, and the time to reach max displacement if > the projectile is launched at an ANGLE? I know how when its straight > PS. Here's the question: > Find the max height and displacement, and the time for each, when a > ball is thrown at a 45 degree angle with a initial acceleration of > 16ft/sec. Still looking for someone to do your homework for you? ==== Subject: Re: Finding info about a projectile On 5 Feb 2006 21:02:49 -0800, Protoman there, the max displacement, and the time to reach max displacement if >the projectile is launched at an ANGLE? I know how when its straight >PS. Here's the question: >Find the max height and displacement, and the time for each, when a >ball is thrown at a 45 degree angle with a initial acceleration of >16ft/sec. Feed the following words (or other similar search words) to Google and follow any/all of the first several links returned: equations projectile motion BTW, there is something wrong with the part of your question that reads initial acceleration of 16ft/sec. A velocity can have units of ft/sec, an acceleration can have units of ft/sec^2. ==== Subject: Re: Finding info about a projectile reads initial acceleration of 16ft/sec. A velocity can have units > of ft/sec, an acceleration can have units of ft/sec^2. Oh, that was a typo. ==== Subject: Re: Finding info about a projectile > How do I find what the maxium height is, how long it takes to get > there, the max displacement, and the time to reach max displacement if > the projectile is launched at an ANGLE? I know how when its straight > PS. Here's the question: > Find the max height and displacement, and the time for each, when a > ball is thrown at a 45 degree angle with a initial acceleration of > 16ft/sec. Well you have two accelerations g and part of 16 ft/sec(hint), and one opposes the other (-)sign you have two coordinates x and y, get two equations one for x then one for y and you have it ==== Subject: dimension of (x+y+z) such that x,y,z are subspaces of V is the following true: dim(x+y+z)=dim(x)+dim(y)+dim(z)-dim(x^y)-dim(x^z)-dim(y^z)-dim(x^y^z) (^ means intersection) If not what would be a counter example? if it is true could anyone give me a clue to start proving it? Ardi ==== Subject: Re: dimension of (x+y+z) such that x,y,z are subspaces of V > is the following true: > dim(x+y+z)=dim(x)+dim(y)+dim(z)-dim(x^y)-dim(x^z)-dim(y^z)-dim(x^y^z) > (^ means intersection) > If not what would be a counter example? if it is true could anyone > give me a clue to start proving it? If x = y = z then your right side is negative, which seems unlikely. Clue: draw a Venn diagram, n = 3. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) ==== Subject: Re: dimension of (x+y+z) such that x,y,z are subspaces of V Arturo you were right I made a mistake on the sign where you mentioned. Your hint should get me going with the proof tough. ==== Subject: Re: dimension of (x+y+z) such that x,y,z are subspaces of V days. My association with the Department is that of an alumnus. >is the following true: >dim(x+y+z)=dim(x)+dim(y)+dim(z)-dim(x^y)-dim(x^z)-dim(y^z)-dim(x^y^z) >(^ means intersection) Not quite. >If not what would be a counter example? Take V to be 4 dimensional, generated by v1, v2, v3, v4. Let x = , y = , and z= Than dim(x)=dim(y)=dim(z) = 2 dim(x^y) = dim(x^z) = dim(y^z) = 1 dim(x^y^z) = 1 dim(x+y+z) = dim(V) = 4. Your left hand side is 4, but your right hand side is 2+2+2-1-1-1-1 = 2. Of course, it is possible you meant to write >dim(x+y+z)=dim(x)+dim(y)+dim(z)-dim(x^y)-dim(x^z)-dim(y^z)+dim(x^y^z) ^ | instead. If so, it is true for finite dimensional spaces. For infinite dimensional spaces you would need to rewrite it so that there is no subtraction (there being no subtraction of cardinals). Probably best as: dim(x+y+z) + dim(x^y) + dim(x^z) + dim(y^z) = dim(x)+dim(y)+dim(z)+dim(x^y^z) > if it is true could anyone >give me a clue to start proving it? Start with a basis for x^y^z. Extend it in three ways to bases for x^y, x^z, and y^z. Then extend those to bases of x, y, and z. -- ' ==== Subject: Re: Interesting Problem >Protoman escribiÖ: >The minimal non trivial solution is (x, y) = (1151, 120). To find that you >need to develop sqrt(92) as continued fraction. Then is easy to see that >x(n) + sqrt(92)y(n) = (1151 + sqrt(92)*120)^n >for x(n) and y(n) and explcit formulaes also. I have worked through some of the details of the continued fraction solution of the given Pell equation. None of the results are new, I believe that they are all mentioned in other threads on this topic. The continued fraction for sqrt(92) is ________________ [9,1,1,2,4,2,1,1,18] The best approximations come from the convergents preceding the partial quotient of 18. For these approximants, we have 1 | p - sqrt(92) q | <= --- 18q Therefore, 2 2 | p - 19 q | = | p - sqrt(92) q | | p + sqrt(92) q | 1 1 <= --- ( 2 | sqrt(92) q | + --- ) 18q 18q sqrt(92) 1 2 = -------- + ( --- ) 9 18q sqrt(92) 1 <= -------- + --- 9 324 < 1.07 Thus, since 19 is not a perfect square and everything is an integer, 2 2 | p - 19 q | = 1 Recursion for p and q --------------------- To derive a recursive relation for p and q, we note that 2231 x + 1320 [18,1,1,2,4,2,1,1,x] = ------------- 120 x + 71 So if we define two sequences of integers { p_n } and { q_n } where p = 2231 p + 1320 q [1a] n n-1 n-1 q = 120 p + 71 q [1b] n n-1 n-1 then we have that p_{n-1} p_n [18,1,1,2,4,2,1,1,-------] = --- q_{n-1} q_n Recursion for p_n ----------------- Multiply [1b] by 1320 to get 1320 q = 1320(120 p + 71 q ) n-1 n-2 n-2 and use [1a] to write 1320 q_{n-1} = p_n - 2231 p_{n-1}. Thus, p - 2231 p = 1320*120 p + 71 (p - 2231 p ) n n-1 n-2 n-1 n-2 p = 2302 p - p [2a] n n-1 n-2 Recursion for q_n ----------------- Multiply [1a] by 120 to get 120 p = 120(2231 p + 1320 q ) n-1 n-2 n-2 and use [1b] to write 120 p_{n-1} = q_n - 71 q_{n-1}. Thus, q - 71 q = 2231(q - 71 q ) + 120*1320 q n n-1 n-1 n-2 n-2 q = 2302 q - q [2b] n n-1 n-2 Solutions of x^2 - 92 y^2 = 1 ----------------------------- We can write the continued fraction ________________ [9,1,1,2,4,2,1,1,18] as ________________ [-9,0,18,1,1,2,4,2,1,1] The convergents preceding 18 are 1/0 and 1151/120 and both of these satisfy x^2 - 92 y^2 = 1. So start with (p_0,q_0) = (1,0) and (p_1,q_1) = (1151,120) and use [2] to generate more. The next four solutions are (2649601,276240), (6099380351,635904360), (14040770918401,1463851560480), (32321848554778751,3369785656320600). Rob Johnson take out the trash before replying ==== Subject: Re: First homology group question.. > I have a question about homology and would be grateful for any response. If > f : X ----> Y is a covering map, is f_* : H_1(X) -----> H_1(Y) injective? > James Take the wedge (= one-point union = union with one point of each space identified) of two circles: Y = S^1 v S^1. The homology of Y is: n 0 1 H_n(Y) Z Z(+)Z and the Euler characteristic chi(Y) is -1. There are many connected covering spaces of Y, in particular at least one 3-fold cover (one of the standard exercises in this is to construct a 3-fold connected covering space of Y, for which the only self-map covering the identitity of Y is itself the identity). The cover, X, must have three separate 0-cells, and six separate 1-cells. Thus, since it's connected, it must have the following homology: n 0 1 H_n(Z) Z Z(+)Z(+)Z(+)Z which must be the case in order for the Euler characteristic of -3 to agree with the alternating sum of the Betti numbers (ranks of the homology groups). X is in fact homotopy-equivalent to the wedge of four circles. There is no monomorphism from H_1(X) to H_1(Y), since such a monomorphism would lead (after tensoring with the rationals Q) to a 4-dimensional subspace Q^4 within Q^2, or a 4x2 matrix over the rationals having rank 4. In particular, the homomorphism H_1(X) ---> H_1(Y) cannot be a monomorphism. Dale. ==== Subject: Re: First homology group question.. > I have a question about homology and would be grateful for any response. If > f : X ----> Y is a covering map, is f_* : H_1(X) -----> H_1(Y) injective? My first inclination was to say no, what about the projective plane but that's H_2. However: no, what about disconnected spaces?. With any space Y you can take its disjoint union with itself, call that X, and take f to be the map identifying two corresponding points. Clearly this never induces an injective map on any homology group, unless that group is trivial. On the other hand, I think it follows from the Hurewicz isomorphism theorem (that H_1 is naturally the abelianization of pi_1 for a connected space) that when X and Y are connected this is true, and follows from what Mr. Santos was saying about covering spaces. Actually, the same argument shows that if X and Y are connected and n-connected, then the induced map on H_{n + 1} is injective (even an isomorphism!), for any n. -- Ryan Reich ryan.reich@gmail.com ==== Subject: Re: First homology group question.. >> I have a question about homology and would be grateful for any response. >> If >> f : X ----> Y is a covering map, is f_* : H_1(X) -----> H_1(Y) >> injective? > My first inclination was to say no, what about the projective plane > but that's H_2. However: no, what about disconnected spaces?. With > any space Y you can take its disjoint union with itself, call that X, > and take f to be the map identifying two corresponding points. Clearly > this never induces an injective map on any homology group, unless that > group is trivial. Why does this never induce an injective map on any homology group (unless that group is trivial) ? > On the other hand, I think it follows from the Hurewicz isomorphism > theorem (that H_1 is naturally the abelianization of pi_1 for a > connected space) that when X and Y are connected this is true, and > follows from what Mr. Santos was saying about covering spaces. > Actually, the same argument shows that if X and Y are connected and > n-connected, then the induced map on H_{n + 1} is injective (even an > isomorphism!), for any n. > -- > Ryan Reich > ryan.reich@gmail.com ==== Subject: Re: First homology group question.. I have a question about homology and would be grateful for any response. >> If >> f : X ----> Y is a covering map, is f_* : H_1(X) -----> H_1(Y) >> injective? > My first inclination was to say no, what about the projective plane > but that's H_2. However: no, what about disconnected spaces?. With > any space Y you can take its disjoint union with itself, call that X, > and take f to be the map identifying two corresponding points. Clearly > this never induces an injective map on any homology group, unless that > group is trivial. > Why does this never induce an injective map on any homology group (unless > that group is trivial) ? Since the homology of the union of two copies of Y is the direct sum of the homology of Y itself, and the map induced by identifying corresponding points is the map which takes a pair (a,b) in the direct sum to a + b in the homology of Y; in particular, this vanishes when b = -a. However, as Mr. Hall has observed, I'm wrong anyway about the following paragraph. For the curious, the reason I'm wrong is that the Hurewicz isomorphism, though natural, still isn't good enough to turn an injection of homotopy groups into an injection of homology groups. That is, we're given an injective map of groups f: G -> H and we divide each by its commutator subgroup to get a map of their abelianizations. The problem is that the crucial intersection of [H, H] with G doesn't have to be [G, G], and likely won't be. Certainly everything in [G, G] is also in [H, H] (it has to be; that's how you get an induced map between the abelianizations at all), but there could be a great deal more in [H, H] (coming from the commutators of elements that aren't in G) and therefore H/[H, H] is a lot smaller relative to G/[G, G] than H was to G; more precisely, anything in the intersection of [H, H] with G, and not in [G, G], is killed by f in the quotient. -- Ryan Reich ryan.reich@gmail.com ==== Subject: Re: surface normal formula- Correct, a normal can be inward or outward and any length greater than 0. A gradient is one specific normal. A unit normal is used a lot and that's just a unit length normal. It must have length 1, but a normal in general can have any length greater than 0. If f is differentiable at a point and the gradient of f at at that point is not 0, then the gradient of f at that point is normal to the level curve through that point. ==== Subject: Re: surface normal formula- Correct, a normal can be inward or outward and any length greater than 0. A gradient is one specific normal. A unit normal is used a lot and that's just a unit length normal. It must have length 1, but a normal in general can have any length greater than 0. If f is differentiable at a point and the gradient of f at at that point is not 0, then the gradient of f at that point is normal to the level curve through that point. ==== Subject: Re: surface normal formula- >> I am learning the method of characteristics for solving first order >>pdes and I am trying to remember the formula for a surface normal to >>the surface F = F(x1,x2,x3). I know that for 3d surfaces, the surface >>normal to F = F(x1, x2) is given by: >>( d/dx1 F, d/dx2 F, -1) = ( Fx1, Fx2, -1) >>Does this formula extend to higher dimensions? I.e. Is the surface >>normal to F = F(x1,x2,x3) given by >>( Fx1, Fx2, Fx3, -1)? >Yes: >In 3D you give the explicit function z = f(x, y) to represent the surface. >Now one has dz = df/dx . dx + df/dy . dy or as a dot product of vectors, >(df/dx, df/dy, -1).(dx, dy, dz) = 0 >Exactly the same formula and reasoning hold - with due changes, of >course - in any number of dimensions. >Mr Rob Johnson says in his earlier reply (see below) essentially the >same in terms of implicit functions. In your problem the implicit >function is G(x, y, z) = f(x, y) - z = 0. >Strictly speaking, the gradient lies along the normal to a >(hyper)surface - it is a vector, it is not the normal itself. I have been thinking about this and I am confused by what you say here. The normal to a surface is just a vector that is perpindicular to the surface at a given point. A normal does not need to be a unit vector nor does it need to point inward or outward, unless specified. Why is the gradient not the normal itself? >BTW, one could better write x4 = F(x1, x2, x3) - then one sees better >what F actually means. >The normal to F(x1,x2,x3) = 0, is grad F = (F_x1,F_x2,F_x3). Rob Johnson take out the trash before replying ==== Subject: Re: surface normal formula-- >I am learning the method of characteristics for solving first order >pdes and I am trying to remember the formula for a surface normal to >the surface F = F(x1,x2,x3). I know that for 3d surfaces, the surface >normal to F = F(x1, x2) is given by: > >( d/dx1 F, d/dx2 F, -1) = ( Fx1, Fx2, -1) > >Does this formula extend to higher dimensions? I.e. Is the surface >normal to F = F(x1,x2,x3) given by > >( Fx1, Fx2, Fx3, -1)? > >>Yes: >>In 3D you give the explicit function z = f(x, y) to represent the surface. >>Now one has dz = df/dx . dx + df/dy . dy or as a dot product of vectors, >>(df/dx, df/dy, -1).(dx, dy, dz) = 0 >>Exactly the same formula and reasoning hold - with due changes, of >>course - in any number of dimensions. >>Mr Rob Johnson says in his earlier reply (see below) essentially the >>same in terms of implicit functions. In your problem the implicit >>function is G(x, y, z) = f(x, y) - z = 0. >>Strictly speaking, the gradient lies along the normal to a >>(hyper)surface - it is a vector, it is not the normal itself. >I have been thinking about this and I am confused by what you say >here. The normal to a surface is just a vector that is perpindicular >to the surface at a given point. A normal does not need to be a unit >vector nor does it need to point inward or outward, unless specified. >Why is the gradient not the normal itself? No wonder that one gets confused by the terminology. Often the adjective normal is used without substantive, but rather as a substantive itself. This word is then no longer unequivocal, and indeed has different meanings depending on the context. I did a little research into the concepts of normal line, normal vector, unit normal vector. First, my printed encyclopedia (Grote Winkler Prins Encyclopedie - 7th printing - 1972) defines (in the case of three dimensions) the normal line at a certain point P of a surface as the line through P perpendicular to the tangent plane at P. Apparently the context is here classical differential geometry with no special emphasis on the vector concept. Second, Google >normal line yields at the top of about 249,000 hits MathWorld takes care to give several different meanings of the word normal - see http://mathworld.wolfram.com/Normal.html , a kind of desambiguation page; BTW, Mathworld's own item normal line is missing from this page. Third, in Wikipedia one gets after some trial and error at http://en.wikipedia.org/wiki/Surface_normal , where the word normal is consistently used with the meaning of normal vector. The geometrical term normal line is unknown at Wikipedia. I guess that with normal one means very often normal vector because in practice normal vectors are more important than the normal line itself. >>BTW, one could better write x4 = F(x1, x2, x3) - then one sees better >>what F actually means. >>The normal to F(x1,x2,x3) = 0, is grad F = (F_x1,F_x2,F_x3). >Rob Johnson >take out the trash before replying ==== Subject: Re: surface normal formula-- >I am learning the method of characteristics for solving first order >pdes and I am trying to remember the formula for a surface normal to >the surface F = F(x1,x2,x3). I know that for 3d surfaces, the surface >normal to F = F(x1, x2) is given by: ( d/dx1 F, d/dx2 F, -1) = ( Fx1, Fx2, -1) Does this formula extend to higher dimensions? I.e. Is the surface >normal to F = F(x1,x2,x3) given by ( Fx1, Fx2, Fx3, -1)? >Yes: >>In 3D you give the explicit function z = f(x, y) to represent the >surface. >>Now one has dz = df/dx . dx + df/dy . dy or as a dot product of vectors, >>(df/dx, df/dy, -1).(dx, dy, dz) = 0 >>Exactly the same formula and reasoning hold - with due changes, of >>course - in any number of dimensions. >>Mr Rob Johnson says in his earlier reply (see below) essentially the >>same in terms of implicit functions. In your problem the implicit >>function is G(x, y, z) = f(x, y) - z = 0. >>Strictly speaking, the gradient lies along the normal to a >>(hyper)surface - it is a vector, it is not the normal itself. >I have been thinking about this and I am confused by what you say >here. The normal to a surface is just a vector that is perpindicular >to the surface at a given point. A normal does not need to be a unit >vector nor does it need to point inward or outward, unless specified. >Why is the gradient not the normal itself? >No wonder that one gets confused by the terminology. >Often the adjective normal is used without substantive, but rather as >a substantive itself. This word is then no longer unequivocal, and >indeed has different meanings depending on the context. >I did a little research into the concepts of normal line, normal vector, >unit normal vector. [extensive research snipped] >I guess that with normal one means very often normal vector because >in practice normal vectors are more important than the normal line itself. Yes, when I see normal in the context of differential geometry, I usually think of normal vector instead of normal line, but I can see how an unqualified reference to one could be interpreted as the other. The line has an aesthetic appeal as there is only one normal line, but there is a whole family of normal vectors. Furthermore, the vectors bear no reference back to the point on the surface, whereas the normal line at least intersects the surface at the point of normalcy. Rob Johnson take out the trash before replying ==== Subject: Re: >I am learning the method of characteristics for solving first order >pdes and I am trying to remember the formula for a surface normal to >the surface F = F(x1,x2,x3). I know that for 3d surfaces, the surface >normal to F = F(x1, x2) is given by: ( d/dx1 F, d/dx2 F, -1) = ( Fx1, Fx2, -1) Does this formula extend to higher dimensions? I.e. Is the surface >normal to F = F(x1,x2,x3) given by ( Fx1, Fx2, Fx3, -1)? >Yes: >>In 3D you give the explicit function z = f(x, y) to represent the >surface. >>Now one has dz = df/dx . dx + df/dy . dy or as a dot product of vectors, >>(df/dx, df/dy, -1).(dx, dy, dz) = 0 >>Exactly the same formula and reasoning hold - with due changes, of >>course - in any number of dimensions. >>Mr Rob Johnson says in his earlier reply (see below) essentially the >>same in terms of implicit functions. In your problem the implicit >>function is G(x, y, z) = f(x, y) - z = 0. >>Strictly speaking, the gradient lies along the normal to a >>(hyper)surface - it is a vector, it is not the normal itself. >I have been thinking about this and I am confused by what you say >here. The normal to a surface is just a vector that is perpindicular >to the surface at a given point. A normal does not need to be a unit >vector nor does it need to point inward or outward, unless specified. >Why is the gradient not the normal itself? >No wonder that one gets confused by the terminology. >Often the adjective normal is used without substantive, but rather as >a substantive itself. This word is then no longer unequivocal, and >indeed has different meanings depending on the context. >I did a little research into the concepts of normal line, normal vector, >unit normal vector. [extensive research snipped] >I guess that with normal one means very often normal vector because >in practice normal vectors are more important than the normal line itself. Yes, when I see normal in the context of differential geometry, I usually think of normal vector instead of normal line, but I can see how an unqualified reference to one could be interpreted as the other. The line has an aesthetic appeal as there is only one normal line, but there is a whole family of normal vectors. Furthermore, the vectors bear no reference back to the point on the surface, whereas the normal line at least intersects the surface at the point of normalcy. Rob Johnson take out the trash before replying ==== Subject: Re: Fermat's actual proof of FLT? Discussion, linux) <878xsp1oqe.fsf@phiwumbda.org> I<- was very clear. Look, I don't want to argue, but it was *me* that wasn't clear. My words were opaque as mud. Not yours. Mine. -- I arrest anybody I think needs arresting, Mr. Carter, and I'm not in the habit of explaining why. There's a law about that --- You're in Dodge, Mr. Carter. -- Gunsmoke radio show / John Ashcroft ==== Subject: Re: poker Stephen, I have thought of some way out. Please comment on this See, the random variable is uniformly distributed in [0,1]. So the mean value is 0.5 and mean deviation from mean is 0.2 So now we have following values to be considered for their strategies values. 0 0.3 0.5 0.7 1 Now A will bet an additional unit in two cases, (1)if his random variable is between 0.3 and 0.5, in this case he would try to bluff, so that opponent may loose (2)if his random variable is greater thn 0.5 in which he his somewhat confident that his number is greater than the opponent's one And A will put down when his number is less than 0.3 Stephen, can we start off like this?Please comment on this ==== Subject: Re: poker <33459497.1139208934955.JavaMail.jakarta@nitrogen.mathforum.org Stephen, I have thought of some way out. Please comment on this > See, the random variable is uniformly distributed in [0,1]. So the mean value is 0.5 > and mean deviation from mean is 0.2 > So now we have following values to be considered for their strategies values. > 0 0.3 0.5 0.7 1 > Now A will bet an additional unit in two cases, > (1)if his random variable is between 0.3 and 0.5, in this case he would try to bluff, so that opponent may loose > (2)if his random variable is greater thn 0.5 in which he his somewhat confident that his number is greater than the opponent's one > And A will put down when his number is less than 0.3 > Stephen, can we start off like this?Please comment on this If I understand you correctly, you are saying that A should bet if his number is greater than 0.3. This is a special case of the A bets if his number is greater than some cutoff strategy which is the first thing I tried too. In general, if A's strategy is to bet if his number is greater than k, I make it that B should call if his number is greater than (3*k + 1)/4, and A's overall expectation is then -k^2/8 + k/4 - 1/8. Perhaps surprisingly, this is always negative, except if k = 1 (A never calls), when it is zero. In particular, if k = 0.3 then A's expectation is -49/800. A strategy of this type can therefore never be optimal - we know of at least one which gives a positive expectation for A (A calls if his number is less than 0.1 or greater than 0.7). ==== Subject: Re: poker <33459497.1139208934955.JavaMail.jakarta@nitrogen.mathforum.org A strategy of this type can therefore never be optimal - we know of at > least one which gives a positive expectation for A (A calls if his > number is less than 0.1 or greater than 0.7). I too meant A bets if... !! ==== Subject: Re: poker >> The two players each ante 1 unit to a pot. Then each player receives a >random number uniformly distributed between 0 and 1. Each player knows >the value of his number but not the value of his opponent's number. The >first player is then given an opportunity to bet one additional unit. If >the first player doesn't bet there is a showdown and the player with the >highest number collects the antes. If the first player bets the second >player may call by matching the bet or drop out (giving the antes to the >first player). If the second player calls there is again a showdown and >the player with the highest number collects the pot (consisting of 4 >units, the bets and the antes). If both players follow their optimal >strategy what is the value of the game? In other words if they play >(optimally) a large number of games how much is the first player >expected to win (or lose if the value is negative) per game? >Following the suggestion in some of the other replies, I have worked >through the whole thing again under the assumption that A's optimal >strategy is to bet if and only if his number is less than some >predetermined number c or greater than some other predetermined number >d. Rather astonishingly, since my calculations cover about six pages, I >arrived at the same answer as others have given - namely that the best >choice is c = 0.1, d = 0.7, when the value of the game for A is 0.1. >If A follows this strategy, I make it that in order to peg his losses >to the best case of -0.1, B should call if his number is greater than >0.7, should fold if his number is less than 0.1, and between these >values can do either. >I did the calculations under the assumption that A and B both know each >other's optimal strategies and behave accordingly. (This must be the >case since if we can work it out then so can they.) >At the risk of repeating what has already been said, I am still very >unclear about several points: >1. How do we know that A's optimal strategy is to bet if his number is >less than some number c or greater than some number d? The most >obvious strategy (my first attempt, assume that A will bet if his >number is greater than some predetermined cutoff) is not optimal, so >what confidence do we have that this new one - the second most >obvious - has any better chance of being optimal? >2. How do we know that A cannot profit by deviating from this supposed >optimal strategy under the assumption that B will continue playing as >if A was following it? The more I think about this the more confused I >get... it's a kind of double-triple-quadruple bluff thing spiralling on >forever. >3. How do we know that A's optimal strategy will be to follow the same >rule for every game (if a series of games is to be played)? The optimal strategies for A and B should satisfy the basic optimality conditions: 1) if A follows his optimal strategy, the expected payoff to A is always at least v, no matter what B does. 2) if B follows his optimal strategy, the expected payoff to A is always at most v, no matter what A does. Unless I've made an error, these conditions are both satisfied for the claimed optimal strategies, with v = 1/10. Thus if A follows his claimed optimal strategy and B bets with probability g(t) when A bets and he is dealt t, the expected payoff is int_0^{1/10} ds (int_0^s dt (1 + g(t)) + int_s^1 dt (1 - 3 g(t))) + int_{1/10}^{7/10} ds (int_0^s dt - int_s^1 dt) + int_{7/10}^1 ds (int_0^s dt (1 + g(t)) + int_s^1 dt (1 - 3 g(t))) = 7/25 + int_0^{1/10} dt (2/5 - 4 t) g(t) + int_{7/10}^{1} dt (14/5 - 4 t) g(t) Since 2/5 - 4 t > 0 for 0 < t < 1/10 and 14/5 - 4 t < 0 for 7/10 < t < 1, this is minimized if g(t) = 0 for t < 1/10 and g(t) = 1 for t > 7/10, which makes the expected payoff 1/10. Next, suppose B follows his claimed optimal strategry and A bets with probability f(s) when he is dealt s. The expected payoff is int_0^{2/5} ds (int_0^s dt +int_s^{2/5} dt (-1 + 2 f(s)) + int_{2/5}^1 dt (-1 - f(s))) + int_{2/5}^1 ds (int_0^{2/5} dt + int_{2/5}^s dt (1 + f(s)) + int_s^1 dt (-1 - f(s)) = int_0^{2/5} ds (1/5 - 2 s) f(s) + int_{2/5}^1 ds (-7/5 + 2 s) f(s) This is maximized when f(s) = 1 for 0 < s < 1/10, 0 for 1/10 < s < 7/10, and 1 for 7/10 < s < 1, which makes the expected payoff 1/10. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: poker P(a,b,c) = (a + 1 - c) b + int_a^c ds int_0^1 dt X(s,t) + 2 int_0^a int_b^1 dt X(s,t) + 2 int_c^1 int_b^1 dt X(s,t) = 2 b^2 + (c-a-1) b + (a-c)(a+c-1) if b < a or (3 a + c - 1) b - a + c - a^2 - c^2 if a < b < c or 2 b^2 + (3a - 3c - 1) b - (a-c)(a+1-c) if c < b Robert can you please explain properly, i am a bit confused by the way of representing (1)int_0^1 (2)int_0^a and how did you determine that all the summation terms? ==== Subject: Re: poker <43E5AA6E.7030209@netscape.net> The two players each ante 1 unit to a pot. Then each player receives a >>random number uniformly distributed between 0 and 1. Each player knows >>the value of his number but not the value of his opponent's number. The >>first player is then given an opportunity to bet one additional unit. If >>the first player doesn't bet there is a showdown and the player with the >>highest number collects the antes. If the first player bets the second >>player may call by matching the bet or drop out (giving the antes to the >>first player). If the second player calls there is again a showdown and >>the player with the highest number collects the pot (consisting of 4 >>units, the bets and the antes). If both players follow their optimal >>strategy what is the value of the game? In other words if they play >>(optimally) a large number of games how much is the first player >>expected to win (or lose if the value is negative) per game? >Let's call the players A and B rather than first and second. B has no reason to bluff, so his optimal strategy will be to call >if his number is greater than some b in [0,1], otherwise >drop out. A, on the other hand, might want to bluff >with a low number, in the hope that B will drop out. So let's >say A calls if his number is less than a or greater than c, where >Sorry, I meant bets. >0 <= a <= c <= 1. >>Is it clear that the set where A bets is an interval? How do we know >>that a randomized strategy for A is not optimal? >In this case, especially knowing B's optimal strategy, it should be >easy. But I think there's a general principal here, that I don't know >the details of. Essentially, A may as well use the random number he was >dealt to do any randomizing that is necessary. > It may well be that A can just as well use his hand to do > whatever randomizing is needed. I don't see how it follows > that the optimal strategy for A is a fixed interval. On second thought, maybe it doesn't in general, although it does in this example. Somewhat more generally, consider a two-person zero-sum game where A and B are dealt independent random variables S and T with uniform distribution in [0,1], and each makes a choice from two alternatives (1 and 2). The payoff to A if A chooses i and B chooses j is a_{ij} if S > T, b_{ij} otherwise. In the given example, a_{11} = a_{12} = a_{21} = b_{21} = 1, b_{11} = b_{12} = -1, a_{22} = 2, b_{22} = -2. I seem to need a_{11} - a_{12} - b_{11} + b_{12} = 0 (which of course is true here). Suppose B knows A is using a mixed strategy giving probability f(s) of choosing 2 when S = s, where f is a measurable function on [0,1] with 0 <= f <= 1. Consider B's choice when T = t. If he chooses j, the expected payoff will be P_j(f,t) = int_0^t ds b_{1j} + f(s) (b_{2j} - b_{1j}) + int_t^1 ds a_{1j} + f(s) (a_{2j} - a_{1j}) If P_1(f,t) < P_2(f,t), then B would choose 1 in this situation, while if P_1(f,t) > P_2(f,t) he would choose 2. Note that (a.e.) d/dt P_j(f,t) = f(t)(b_{2j} - b_{1j} - a_{2j} + a_{1j}) + b_{1j} - a_{1j} Consider changing f to f + g on an interval J = (c,d), leaving it unchanged outside the interval. If t < c, P_j(f+g,t) - P_j(f,t) = int_J ds g(s) (a_{2j} - a_{1j}) while if t > d, P_j(f+g,t) - P_j(f,t) = int_J ds g(s) (b_{2j} - b_{1j}). Thus by making int_J ds g(s) = 0, we can ensure that P_j is unchanged outside the interval J. Now if P_1(f,t) <> P_2(f,t) on the interval J, A would like to increase whichever of these is less, and that can be done by moving all the area under the graph of f over the interval J as much to the left or right as possible, depending on the sign of b_{2j} - b_{1j} - a_{2j} + a_{1j}. Thus we can say that for an optimal f, on any interval where P_1(f,t) <> P_2(f,t) we will have f(t) in {0,1} (a.e.). However, there may also be a set of positive measure where P_1(f,t) = P_2(f,t). Now (a.e.) we have d/dt (P_1(f,t) - P_2(f,t)) = C f(t) where C is a constant, using my assumption a_{11} - a_{12} - b_{11} + b_{12} = 0. So almost everywhere on the set where P_1(f,t) = P_2(f,t) we must have f(t) = 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: poker >><12015809.1139031629539.JavaMail.jakarta@nitrogen.mathforum.org>, >The two players each ante 1 unit to a pot. Then each player receives a >random number uniformly distributed between 0 and 1. Each player knows >the value of his number but not the value of his opponent's number. The >first player is then given an opportunity to bet one additional unit. If >the first player doesn't bet there is a showdown and the player with the >highest number collects the antes. If the first player bets the second >player may call by matching the bet or drop out (giving the antes to the >first player). If the second player calls there is again a showdown and >the player with the highest number collects the pot (consisting of 4 >units, the bets and the antes). If both players follow their optimal >strategy what is the value of the game? In other words if they play >(optimally) a large number of games how much is the first player >expected to win (or lose if the value is negative) per game? >>Let's call the players A and B rather than first and second. >>B has no reason to bluff, so his optimal strategy will be to call >>if his number is greater than some b in [0,1], otherwise >>drop out. A, on the other hand, might want to bluff >>with a low number, in the hope that B will drop out. So let's >>say A calls if his number is less than a or greater than c, where >>Sorry, I meant bets. >>0 <= a <= c <= 1. >Is it clear that the set where A bets is an interval? How do we know >that a randomized strategy for A is not optimal? >>In this case, especially knowing B's optimal strategy, it should be >>easy. But I think there's a general principal here, that I don't know >>the details of. Essentially, A may as well use the random number he was >>dealt to do any randomizing that is necessary. >> It may well be that A can just as well use his hand to do >> whatever randomizing is needed. I don't see how it follows >> that the optimal strategy for A is a fixed interval. >On second thought, maybe it doesn't in general, although it does in >this example. >Somewhat more generally, consider a two-person zero-sum game where >A and B are dealt independent random variables S and T with uniform >distribution >in [0,1], and each makes a choice from two alternatives (1 and 2). The >payoff to A >if A chooses i and B chooses j is a_{ij} if S > T, b_{ij} otherwise. >In the given >example, >a_{11} = a_{12} = a_{21} = b_{21} = 1, b_{11} = b_{12} = -1, a_{22} = >2, b_{22} = -2. >I seem to need a_{11} - a_{12} - b_{11} + b_{12} = 0 (which of course >is true here). This is not such an arbitrary condition: it is true because B only has a choice to make in case A bets. If we also wanted to give B a choice when A starts with a pass, and B is supposed to know whether A has passed or not, then B actually has four possible choices (bet/bet, pass/bet, bet/pass and pass/pass, where the first member of each pair refers to the situation when A has passed and the other to the situation when A has bet). I haven't checked, but I wouldn't be surprised if it turned out here too that there are optimal strategies that don't involve randomization. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Maximal 4 Color Mapping Re: Adjacency is the Primary Dependent Variable over that of Coloring Re: one paragraph proof of 4-Color-Mapping <43E39704.5090809@dtgnet.com> <43E46113.4050402@dtgnet.com> <43E6479A.3040806@dtgnet.com [...] > So let us talk about Maximal coloring in the plane. > Let us apply Kepler Packing to density of coloring. > So what is the Maximal 4 Color Mapping? I would guess it is the plane > tiled in triangles such as this: > ^ > / / > / / > /2 / 3 > / / 4 > /---------------------- > 1 > ------------------------- > So a triangle with its 3 sides as strips forming 3 countries and > numbered 1,2,3 and the landlocked country as 4. > Can there be a more dense 4 Color Mapping. The trouble with my above is > that adjacent triangles, although requiring 4 colors for each triangle > cell, can we increase the density of coloring between countries inside > one triangle cell adjacent to countries inside another triangle cell. > So is that the Maximal density of 4 Coloring? You haven't said anything about what the density of 4 Coloring is. For the moment, it's as useless a phrase as noon blue apples. You could be asking for the largest number of countries in a map that requires 4 colors; this is infinite, because you can always add more countries around this particular map. > Chris you chewed on this problem longer than I have, what does current > mathematics say about this. I have no idea what concept you're trying to convey here. --- Christopher Heckman ==== Subject: Re: Adjacency is the Primary Dependent Variable over that of Coloring Re: one paragraph proof of 4-Color-Mapping <43E39704.5090809@dtgnet.com> <43E46113.4050402@dtgnet.com [...] > The two properties I have in mind are not quite the two you > mentioned. Here are I hope clearer versions of them (everything > is on the plane, n always an integer >= 1): > M(n) - a map has this property if it has n colors and has the fewest > possible number of countries I assume this means the Map requires n colors, and of every map requiring n colors, this map is the smallest. > A(n) - a map has this property if all its countries are mutually > adjacent. Property A(n) does not depend on n. (Maybe you mean the Map has n countries, and all are mutually adjacent ?) > I want to show that these two are equivalent for all n - (they are both > mutually true for n = 1, 2, 3, but that doesn't mean except in a > trivial sense that they're 'equivalent' ). Then, for n = 5, > a minimal map with 5 colors will have 5 mutually adjacent > countries (again on the plane) which contradicts Mobius. > I'm not sure this equivalence has been shown for n=4, but it > SEEMS like it ought to be relatively easy to show for n = 1,2,3. > NB: when I say things like 'I want to show' or 'IF I could show', I'm > pretty sure that, even if my speculations turn out to be true, > it will not be me who does the showing. > To resume - is it possible, are there examples, in mathematics, of > proving an equivalence 'on n' (M(n) <=> A(n) in this > case) to be true for all n even if one side or the other (possibly > both) fails to be true if let's say n > 5? Yes; such statements are called vacuuously true. For instance, your statement M(6) => A is vacuuously true, because there is no planar map that requires 6 colors.* Hence, M(6) will be False for every map, and False implies anything is True. --- Christopher Heckman * A 5CT was extracted out of Kempe's failed attempt to prove the 4CT, and this proof is recognized as being correct; it does not depend on a computer. ==== Subject: Re: Adjacency is the Primary Dependent Variable over that of Coloring Re: one paragraph proof of 4-Color-Mapping <43E39704.5090809@dtgnet.com> <43E46113.4050402@dtgnet.com [...] > The two properties I have in mind are not quite the two you > mentioned. Here are I hope clearer versions of them (everything > is on the plane, n always an integer >= 1): > M(n) - a map has this property if it has n colors and has the fewest > possible number of countries > I assume this means the Map requires n colors, and of every map > requiring n colors, this map is the smallest. Yes, it is the smallest in the sense of having the fewest countries. > A(n) - a map has this property if all its countries are mutually > adjacent. I messed this one up. I meant what you say below, IOW, A(n) - map has this property if it has n countries and they are mutually adjacent. > Property A(n) does not depend on n. (Maybe you mean the Map has n > countries, and all are mutually adjacent ?) > I want to show that these two are equivalent for all n - (they are both > mutually true for n = 1, 2, 3, but that doesn't mean except in a > trivial sense that they're 'equivalent' ). Then, for n = 5, > a minimal map with 5 colors will have 5 mutually adjacent > countries (again on the plane) which contradicts Mobius. > I'm not sure this equivalence has been shown for n=4, but it > SEEMS like it ought to be relatively easy to show for n = 1,2,3. > I > NB: when I say things like 'I want to show' or 'IF I could show', I'm > pretty sure that, even if my speculations turn out to be true, > it will not be me who does the showing. > To resume - is it possible, are there examples, in mathematics, of > proving an equivalence 'on n' (M(n) <=> A(n) in this > case) to be true for all n even if one side or the other (possibly > both) fails to be true if let's say n > 5? > Yes; such statements are called vacuuously true. For instance, your > statement > M(6) => A is vacuuously true, because there is no planar map that > requires 6 colors.* Hence, M(6) will be False for every map, and False > implies anything is True. > --- Christopher Heckman What I'm looking for is the possibility that the equivalence M(n) <=> A(n) can be proven non-vacuously for all n, even if, as is the case, one or the other side fails. (that is, there is no country with 5 maps with all countries mutually adjacent, in fact; there is no planar map that requires 5 colors). Is that logically impossible? Ken ==== Subject: Re: Adjacency is the Primary Dependent Variable over that of Coloring Re: one paragraph proof of 4-Color-Mapping <43E39704.5090809@dtgnet.com> <43E46113.4050402@dtgnet.com > [...] > > The two properties I have in mind are not quite the two you > mentioned. Here are I hope clearer versions of them (everything > is on the plane, n always an integer >= 1): > > M(n) - a map has this property if it has n colors and has the fewest > possible number of countries > I assume this means the Map requires n colors, and of every map > requiring n colors, this map is the smallest. > Yes, it is the smallest in the sense of having the fewest countries. > A(n) - a map has this property if all its countries are mutually > adjacent. > I messed this one up. I meant what you say below, IOW, > A(n) - map has this property if it has n countries and they > are mutually adjacent. > Property A(n) does not depend on n. (Maybe you mean the Map has n > countries, and all are mutually adjacent ?) > I want to show that these two are equivalent for all n - (they are both > mutually true for n = 1, 2, 3, but that doesn't mean except in a > trivial sense that they're 'equivalent' ). Then, for n = 5, > a minimal map with 5 colors will have 5 mutually adjacent > countries (again on the plane) which contradicts Mobius. > > I'm not sure this equivalence has been shown for n=4, but it > SEEMS like it ought to be relatively easy to show for n = 1,2,3. > I > NB: when I say things like 'I want to show' or 'IF I could show', I'm > pretty sure that, even if my speculations turn out to be true, > it will not be me who does the showing. > > To resume - is it possible, are there examples, in mathematics, of > proving an equivalence 'on n' (M(n) <=> A(n) in this > case) to be true for all n even if one side or the other (possibly > both) fails to be true if let's say n > 5? > Yes; such statements are called vacuuously true. For instance, your > statement > M(6) => A is vacuuously true, because there is no planar map that > requires 6 colors.* Hence, M(6) will be False for every map, and False > implies anything is True. > What I'm looking for is the possibility that the equivalence > M(n) <=> A(n) can be proven non-vacuously for all n, even if, > as is the case, one or the other side fails. (that is, there is no > country with 5 maps with all countries mutually adjacent, in fact; > there is no planar map that requires 5 colors). If one side or the other fails, then the result _is_ vacuuous, by definition. --- Christopher Heckman ==== Subject: Re: one paragraph proof of 4-Color-Mapping <6134905.1139086473819.JavaMail.jakarta@nitrogen.mathforum.org Consider a graph with five vertices A,B,C,D,E who adjacent pairs are > {A,B} > {B,C} > {C,D} > {D,E} > {E,A} > Observe that > -- the graph is not 3-colorable. > -- the graph DOES NOT contain 3 pairwise adjacent vertices. > Back to the drawing board, A.Pu. > But the graph IS 3-colorable as follows; > A = 1; B = 2; C = 1; D = 2; E = 3 > [...] Larry's first observation should probably have been the graph requires 3 colors. (And why didn't I catch this the last time around?) --- Christopher Heckman ==== Subject: Re: ? adjoint and Fredholm theorem > To allow A*x = b having solutions, Fredholm thm says the sufficient and >necessary condition is that b is orthogonal to null space of adjoint of A. Yes in finite dimensions, no in infinite dimensions. In a Hilbert space, the orthogonal complement of the null space of the adjoint of A is the closure of the range of A. So it's necessary, but not sufficient unless A has closed range. >eigenvectors (or eigen functions) of A. No, that's not what the spectral theorem says. First of all, the spectral theorem only applies if A is a normal operator. Second, you only get a series expansion in eigenvectors in the case of discrete spectrum. > It seems that there could be some >relations between the two theorems. More explicitly, could one express >the null space of adjoint(A) by eigen vectors of A? Are there more >relations? If A is a normal operator, A and its adjoint have the same null space. The nonzero members of the null space are the eigenvectors for eigenvalue 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: combinatorics . > How many three digit positive integers are odd and do not contain the digit > 5? I answered this in your other thread. --- Christopher Heckman ==== Subject: Re: combinatorics > How many three digit positive integers are odd and do not contain the digit > 5? > x = d1d2d3, for d_i in {1, 3, 7, 9} > For a d_i, there are 4! orderings, and the digits can be ordered 3!. So > 4!*3! = 288 is the answer. > How come this formula m!*n! does not seem to work for the case a digit can > take on 3 values and 2 digits. Because the formula m!*n! isn't correct. (It gave you a correct answer above, though, by coincidence.) You have 8 choices for d_1 (all but 0 and 5), 9 choices for d_2, and 4 choices for d_3. The answer is 8 * 9 * 4, which also happens to be 288. > There should be 2!*3! = 12 integers, but I > see only 9: > d1d2, for d_i in {0, 1, 2} > 00 > 01 > 02 > 10 > 11 > 12 > 20 > 21 > 22 You're basically solving different problems here. In the first one, you want choose 3 digits without choosing 5s and making sure the number is odd. In the second, you only want to choose 2 digits, with no extra conditions. Apples and oranges. --- Christopher Heckman ==== Subject: Re: Proving the P=NP problem > i am in no way aligning myself with AP -- but i was hoping someone > could briefly explain the practical applications of a proof that P=NP > or P!=P or point to a relevant reference that does. I completely > believe it has practical implications, don't get me wrong, i would just > like to know more about the details. Cook proved that the matching problem was in P, back in 1970. P has become the ruler for what problems can be solved efficiently. It's basically a definition. --- Christopher Heckman ==== Subject: Re: Maximal and Minimal density of Coloring in 4 Color Mapping Theorem > [CCH clipping unneeded stuff throughout] > So let us talk about Maximal coloring in the plane. > Let us apply Kepler Packing to density of coloring. > ^ > / / > / / > /2 / 3 > / / 4 > /---------------------- > 1 > ------------------------- > I need to define density of coloring. And so I define it as the > frequency of using all 4 colors. [...] If all four colors are used the same > amount means it is the highest density. > So, I need to find out from Chris Heckman if that intuition is correct. > What is the literature on the density of coloring as I described and > defined above? The number of colors need to color a map is called the chromatic number. If you want a map which needs 4 colors, and those colors have to be used evenly, then you can just divide up a bunch of islands the way you've done above (with 4 countries). If you'd like a map which is all one landmass, you can just glue the sides together: ^ / / / / /2 / 3 / / 4 /---------------------- 1 -------------------------| 2 ----------------------/ 4 / / 1 /3 / / / / / V > And is there a quick and easy proof that what I described is the > Maximal Density. Well, in the maps above, you need each color 1/4 of the time, and you can't get a more equitable distribution of colors than this. > For you see, a proof of 4 Color Mapping is easily begot from Minimal > Coloring If you're talking about greedy coloring (defined below), then no, it won't work; greedy coloring can require a large number of colors. Greedy coloring: The object is to color each country with a natural number. (1) Start with an empty map. (2) Choose a country. (3) Color that country with the smallest natural number which is not a color of any of its neighbors. (4) Repeat (2) and (3) until all countries have been colored. > And an observation comment: Notice that if the above is the most dense > coloring that those triangles are what I would call Discrete. They are > like isolated packets of 4 colors all disjoint from one another because > they cannot require 4 colors outside their individual cell. So we get > into Discrete Mathematics if that is the maximal density. The Discrete in Discrete Mathematics is not what you think. It's the part of mathematics that deals with finite sets and the structures you can build from them. --- Christopher Heckman ==== Subject: Re: Maximal and Minimal density of Coloring in 4 Color Mapping Theorem The number of colors need to color a map is called the chromatic number. (snipped) The Discrete in Discrete Mathematics is not what you think. It's the part of mathematics that deals with finite sets and the structures you can build from them. I want a stronger concept than chromatic number. I want a density of coloring defining density as to frequency of colors where the highest density is an even distribution of colors of X/4 for large number X countries. So if you have 16 countries and all 4 colors were used 4 times is the highest density. Question: has anyone done a concept like this on color mapping? Has anyone made a conjecture that triangles tiling the plane such as this one: ^ / / / / /2 / 3 / / 4 /---------------------- 1 -------------------------| forms the highest density of coloring. Has anyone noted before that this highest density causes packets throughout the plane. Call them packets or cells of 4 adjacent countries. A proof of that fact seems to me to be an alternate proof of the 4 Color Mapping theorem because a 5th color would be a more dense coloring than those triangles. Much like a Kepler Packing cell with 13 kissing points would be a denser packing. So does the literature of mathematics have anything resembling what I call density of color and does the literature have that conjecture I made of the most dense coloring in the plane? Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Subject: Re: Maximal and Minimal density of Coloring in 4 Color Mapping Theorem > The number of colors need to color a map is called the chromatic > number. > (snipped) > The Discrete in Discrete Mathematics is not what you think. It's > the part of mathematics that deals with finite sets and the structures > you can build from them. > I want a stronger concept than chromatic number. I want a density of > coloring defining density as to frequency of colors where the highest > density is an even distribution of colors of X/4 for large number X > countries. So if you have 16 countries and all 4 colors were used 4 > times is the highest density. > Question: has anyone done a concept like this on color mapping? It would be called a uniform coloring or maybe an equitable coloring. The only reference I've seen to a uniform coloring is its opposite (a non-uniform coloring, in Planar Ramsey Numbers). > Has anyone made a conjecture that triangles tiling the plane such as > this one: > ^ > / / > / / > /2 / 3 > / / 4 > /---------------------- > 1 > -------------------------| > forms the highest density of coloring. > Has anyone noted before that this highest density causes packets > throughout the plane. Call them packets or cells of 4 adjacent > countries. I mentioned before that there is no higher density than these triangles, but there could be more. > A proof of that fact seems to me to be an alternate proof of the 4 > Color Mapping theorem because a 5th color would be a more dense > coloring than those triangles. How, exactly? --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT <25094907.1139173401165.JavaMail.jakarta@nitrogen.mathforum.org [...] > I'm pretty sure I've even constructed a map where it never happens that four > countries are all adjacent to all the other three but at the same time that you > must use four colors. [...] AP doesn't believe an analogies. I've tried the much-simpler example of a pie chart with 5 slices, which needs 3 colors, but no 3 countries are adjacent. If you count the ocean around it as another region, you have a map where no 4 countries are adjacent, but you need 4 colors. > You make a false assumption and prove the theorem off of that. Even if what > you said were true, do you think you'd be the first person ever to take a theorem > and say This theorem is true, therefore this other obvious (only because I make > a false assumption) corollary to the theorem is also true! I've also used this argument, with Occam's razor. --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT > [...] > Here are the two basic facts that AP is ignoring: > You seem to be ignoring the fact the 4CT applies only to planar graphs. [...] To AP, every map is planar, and I tend to follow along. Occasionally, when I post, I refer to non-planar maps. --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT Hi Christopher - > > noting the effort, you've put into this discussion, I want > to add one remark, that at least to one reader you were able > to open eyes ;-) > > Some years ago I had a similar discussion with some mathematician, > and I was in the position as AP: I could not understand, why > he denied my no-5-adjacent idea as a proof. Well, we had no > hassle about that, and I gave it a chance to possibly come out > to be understandable later. The current discussion solved that > inherent problem in my understanding. What I take now with me, > is that if the non-5-adjacent/inside-triange-outside-triangle > argument should be the core of the proof, then someone must prove > that any configuration can be reduced to a representation of > such (nested) non-5-adjacent-sub-configurations. > Is it that/something like, what A&H did with their computer- > proof? > It's like it. A&H's proof is direct, and they showed that there is a > fixed list of 1476 graphs such that: if you have a planar graph, then > it contains a graph on that list. (This is the unavoidable > configuration list.) AP's graphs do not form a list of unavoidable > configurations. > Every maximal planar graph contains a wheel graph of order six or less (This is equivalent to the statement that some vertex has degree <= 5.) > Is this graph one of the 1475 A&H graphs? The ones where the hub vertex has degree 1 -- 4. The wheels whose hub vertex is 5 isn't one of them. > If you initially consider > only maximal planar graphs; then you have only to consider the wheel > graph of order 6. Yes, because Kempe showed how to handle the other cases. Kempe thought he handled the case where the wheel had order 6 (a hub vertex of degree 5), but he was wrong. A&H's 1475 graphs all contain a vertex of degree 5. > The decsion to study only maximal planar graphs is > justified by the observation that any planar graphs can be maximalized > by the addition of a sufficient number of edges. That, and because if you add edges and get a coloring, that is also a coloring of the original graph. --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT >[...] > Here are the two basic facts that AP is ignoring: > (1) If you can color every triangulated map with 4 colors, that does > not mean you can color any map with 4 colors. > What is a 'triangulated' map? Every region is a triangle, and you are coloring regions. (Note that if you are coloring the VERTICES, you CAN assume that every region is a triangle.) > BTW, the error in ignoring (2) is called arguing from ignorance, > which is basically, I can't imagine any other possibility, so my > possibility must be true. The same error is present, in the same form, > in the following proof: > What is frustrating is; > I can't imagine any other possibility, therefore, I can't evaluate > the other possibilities! Well, then you don't have a proof then, and you need to do some more thinking. > BTW, a list of countries and a list of which countries are adjacent to > which is all that is needed in order to color a map. The following > scenario should make it clear, even to AP: > The leaders of the various countries got together and decided they > would like to have high-speed roads (highways) connecting the capital > cities of their countries. To reduce the cost and eliminate routing > conflicts, it was decided that two capitals would have a highway > between them only if those two countries were adjacent (but not at a > point), and that the route between the two capitals should pass through > their common boundary. > I imagine that the citizens of Denver and Phoenix might be disappointed > with this plan? [...] Colorado and Arizona aren't adjacent, because they only have a single point in common. To be adjacent, the regions have to have a line segment in common. --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT Also, from my first post I made an error. I said AP, check this. Your argument supposes that for 5 colors to be needed, 5 countries must all be adjacent to each other. But that's all it does... it supposes, it assumes that is true. It does not prove that's the only case. Maybe you're right that it's the only case, well in fact you probably are because we know FCT is true. But you have not proven that it's the only case. In reality, to say that for 5 colors to be needed, 5 countries must all be adjacent is silly since we know that FCT is already prove true. There is no case where 5 colors are needed. 5 countries can not be adjacent so it is not a possible case of when 5 colors would be needed. But, if we were to look at the FCT, not knowing if it were true or not, trying to prove it. We could use the Jordan Curve Theorem to help. We would know that at most four countries can all be adjacent, one to another. That eliminates an infinite number of cases where 5, 6, 7, and more countries are all adjacent and thus we would not need to think about any thing of that sort. But there would still be hundreds of cases, actual possible configurations, we would need to look at where at most two, three, and four countries were all adjacent to each other. We would need to show that, in all of those cases of possible configurations, never do we have an instance where five colors are needed. ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT <4718916.1139205769608.JavaMail.jakarta@nitrogen.mathforum.org Also, from my first post I made an error. > I said > AP, check this. Your argument supposes that for 5 colors to be needed, 5 > countries must all be adjacent to each other. But that's all it does... it supposes, > it assumes that is true. It does not prove that's the only case. Maybe you're right > that it's the only case, well in fact you probably are because we know FCT is > true. But you have not proven that it's the only case. > In reality, to say that for 5 colors to be needed, 5 countries must all be adjacent > is silly since we know that FCT is already prove true. [...] But if you're proving the 4CT, you can't assume the 4CT is true! --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT Which I explained in the same post you quoted. ==== Subject: Re: (Alpha-Beta) Triangle I wonder if my triangles are neither for this century nor for any I wonder if shall I present my coming equations in non-mathematical categories I wonder if many will appreciate your absolute silence better than your opinions Note: Non of my messages reflect my employer attitudes Bassam King Karzeddin Al-Hussein Bin Talal University JORDAN ==== Subject: Affine geometry: line-less axiomization? Hi all, Triggered by a problem in a math magazine, I've been looking into affine geometry a bit. The axiomization that is usually given is the following: 1. Two distinct points determine exactly one line. 2. Parallel postulate Given a point and a line not containing that point, there is exactly one line through that point that does not intersect the first line. 3. There are at least four points, no three collinear (i.e., on the same line). While trying to formalize this a bit more, I wondered whether it is possible to leave out the concept of 'line' in the axioms, and only use the concept of collinearity. It seems it is, but I'm not sure whether I'm correct here. My question: do the axioms below reflect the essence of affine geometry, and if not, where am I going wrong? And also: If this is possible, then this approach must be known already, but I couldn't find anything out there on the web. Pointers are appreciated. * * * p, q, r, etc. are points, drawn from a finite set of points. The notation pqr informally means p, q, and r are collinear or there is a line through p, q, and r. Obviously this is a symmetrical concept: pqr == qpr pqr == rpq The first part of axiom 1 says that (1a) ppq for any p and q; or informally: there is at least one line through p and q. (Note that the restriction p /= q is not necessary, since we obviously also want ppp to hold.) The other way around, (1b) p /= q / pqr / pqs / pqt => rst for any p, q, r, s, and t. (I don't really have a solid argument yet why this is equivalent to p and q determine at most one line, but I'm pretty sure it is.) With these axioms we can already prove (at least, it seems I have proven) that, for any p /= q and r /= s, exactly one of the following three is true: pq and rs don't intersect (i.e., are parallel) pq and rs intersect in exactly one point pq and rs are the same line Axiom 2 can be translated as (2) =/= rsu>> for any p, q, and r. Informally, given a line pq and a point r, there is a point u such that every line rs either intersects pq, or passes through u, but not both. (This is the simplest translation I could find which seems to be equivalent to axiom 2. Again, no solid argument yet.) And obviously axiom 3 translates to (3) * * * Groetjes, <>< Marnix ==== Subject: Re: An exact 1-D limit challenge - 17 I tried with Mathematica, but I don't know if it's correct : In[1]:=$Version Out[1]=5.1 for Microsoft Windows (January 28, 2005) In[2]:=f[z_]:=z-z^3/Sqrt[16*Pi]*MeijerG[{{1/4,1/2},{}},{{0},{-1/2,-3/4}},z^4 ] In[3]:=Series[f[z]//FullSimplify[#,z>0]&,{z,Infinity,2}]//Normal Series::esss:Essential singularity encountered [...] Out[3]=Gamma[3/4]/Sqrt[Pi] In[4]:=%//N Out[4]=0.691367 v.a. ==== Subject: Re: An exact 1-D limit challenge - 17 VA> Out[3]=Gamma[3/4]/Sqrt[Pi] 100% right. Actually, many years ago, I reported to WRI a set of bugs related to such stuff. And WRI took my data seriously. As you maybe observed, Bhuvanesh Bhatt, a WRI's QA Engineer tells us that now the things run smoothly at this point, and we obtain the desired answer instantly. In[1]:= Limit[z-z^3/Sqrt[16*Pi]*MeijerG[{{1/4,1/2},{}},{{0}, {-1/2,-3/4}}, z^4], z->Infinity] 3 Gamma[-] 4 Out[1]= -------- Sqrt[Pi] Sanity check produces a result fitting your exact answer ideally In[2] := Limit[z - z^3/Sqrt[16*Pi]*MeijerG[{{1/4,1/2},{}},{{0}, {-1/2, -3/4}}, z^4], z -> 100.] Out[2] = 0.691367 As for Maple, none of its version can calculate this limit limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z=infinity); -------------------- (2005) Maple 10.00 ----------------- limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) -------------------- (2002) Maple 8 --------------------- limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) -------------------- (2001) Maple 7 --------------------- limit(z-1/4/Pi^(1/2)*z^3*MeijerG([[1/4, 1/2], []],[[0], [-1/2, -3/4]],z^4),z = infinity) -------------------- (2000) Maple 6 --------------------- limit(z-1/4/Pi^(1/2)*z^3*MeijerG([1/4, 1/2], [],[0], [-1/2, -3/4],z^4),z=infinity); # different syntax -------------------- (1997) Maple V Rel 5 --------------- N/A: not implemented -------------------- (1995) Maple V Rel 4 --------------- N/A: not implemented -------------------- (1994) Maple V Rel 3 --------------- N/A: not implemented ---------------------------------------------------------- Best luck to you, VB http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ==== Subject: Re: Cantorian pseudomathematics |> I don't see how one can practically speaking prevent oneself, |> as one is doing constructive mathematics, from dealing |> sometimes with statements of the form X->Y where it turns |> out afterward that (sometimes) X is not constructively valid, |> but is nonconstructively provable. |> We had a discussion of statements of the form X->Y where X |> is false, and one of my points there was that we need to be |> able to proceed in a way that doesn't require us to declare |> things to be nonsensical retroactively. If I can't tell, just |> by looking at X, whether it's okay to use it as a premise, |> even just hypothetically, then there's something wrong with |> your notion of okay. | |But we can tell, just by looking at X, whether it is a constructively |meaningful statement. Sometimes it is not so clear when the statement |is made in informal natural language, though. I think that when you just look at a statement and decide whether it is constructively meaningful you do it in a naive way. Going with your gut on these things is a bad idea. You need to spend more time dealing with criteria that are explicit enough that someone can tell whether a statement qualifies or not without being you. |When you write we had a discussion..., are you talking about fifteen |years ago? No, I didn't mean you and I; I meant a recent discussion of implication where someone had another one of these naive gut reactions to the effect that if X is false, X->Y should be regarded as meaningless rather than true. | I vaguely recall such a discussion, but the problem was |with the interpretation of X->Y. Usually we say that X->Y is equivalent |to I have a way to convert a proof of X into a proof of Y, in which |case X must be a constructively meaningful statement (i.e. it must make |predictions about the results of computational experiments). But it |might also be interpreted as not(X) OR Y, in which case, not(X) must be |constructively meaningful. For X->Y to be meaningful X has to be meaningful, but this just begs the question of what it takes for X to be meaningful. My point here is that one has to distinguish between X being meaningless and X being unacceptable for some other reason, like its being false. Errett Bishop's constructive analysis textbook in at least one edition contains a mistake regarding AC. He made a mistake which led him to think it was true. So if there is some way that one can just magically intuit its unacceptability from a constructive point of view by just looking at it, for the first time, it is at least a method that had escaped his attention. No, AC is something that is eventually realized to be not constructively acceptable *AFTER* some consequences have been deduced from it. If your claim about it was correct, one would have to decide retroactively to invalidate the deductions of the form AC->X that had been made, once it was discovered to imply the law of excluded middle and the rest. |> |> And also it's not clear that every |> |> constructive theorem in the sense he and I were considering |> |> is constructive in the sense you and I are considering. |> | |> |I would say not. |> So, I was pointing out the contrast between your idea and |> the one I more often consider. |> In the context of explaining why it's not clear to me that |> a lot of Cantorian mathematics will never have a |> constructive counterpart, in your sense, the point is (1) it |> always has a constructive counterpart in a broad sense, and | |Proving theorems like AC->X is not doing constructive mathematics, even |in a broad sense. So now you're ready to explain what core, essential aspect of the concept of constructivity it violates? No. Of course not. You derive too much enjoyment from glibly tossing off such claims for stopping to examine the principles in question to be attractive to you. Constructivity is typically explained in terms of the meaning assigned to existential quantifiers. In particular, each existence claim should be accompanied by a construction, where it is often left rather vague as to what should be counted as a construction. One case, however, is more crucial than most people who've heard of constructivism realize, and that's the case of integers (and consequently other finite combinatorial constructions). A constructive proof that an integer exists must (by the nature of the concept of constructive) give a way of computing it (in decimal form, say). A second case that is less crucial but generally fits people's intuitions about what constructive should mean is that a construction of a set should give us a way to express it as {x:P(x)} for some explicit property P. So for an axiom system we have two natural criteria for constructivity, known as the numeric existence property and the set existence property. It has the numeric existence property if there is a procedure for converting a proof of a statement of the form (En)Q(n) where n ranges over the integers into a proof of Q(n) for a specific n given in decimal form. It has the set existence property if there is a procedure for converting a proof of a statement of the form (ES)Q(S) where S ranges over sets (of some kind) into a proof of a statement of the form Q({x:R(x)}) for some predicate R. There are two fundamental axioms famous for creating problems with the numeric and set existence principles. Most mathematically trained people are aware of proofs using the axiom of choice of the existence of things like a Vitali set (with one representative from each member of R/Q) where one doesn't expect to be able to give an explicit example. This is associated with ZFC's failure of the set existence principle. Now a lot of people don't realize this can be patched over by going to ZF+V=L, where V=L implies AC. But set theorists usually seem to prefer axioms that contradict V=L so it doesn't make much difference. The second and more serious issue is with the way the law of excluded middle destroys the numeric existence principle. Not every constructivist is happy with every formal system satisfying the numeric and set existence properties. I don't know of any additional requirement, however, for which there is a consensus that it is a necessity for a system to be a constructive system, and its proofs constructive proofs. Among people who want proofs to be constructive, I suppose there's often interest in their being predicative as well. It strikes me as somewhat arbitrary and severe to say that this is some kind of absolute requirement. Then there's the issue of computational meaning. Some famous judge said of pornography that he couldn't define it precisely but could recognize it when he saw it. If one wants to be able to say something more satisfying than that about mathematics that is obscenely lacking in computational meaning, then one has further work to do. You seem not to believe me when I say it, but there's a simple, straightforward way of applying the mathematical theories must make predictions rule that tells us to dump the same two axioms, and nothing else. Take computational predictions to mean only the most obvious kind of thing, namely, statements of the form (n)P(n) where n ranges over the integers and P is a primitive recursive predicate. Then say that an axiom system should be no more complicated than necessary to entail all conclusions of that form that it does. By the work of Goedel on AC, we can show that removing it has no effect on which arithmetic statements we can prove, let alone the kind of computational prediction I'm considering here. Likewise there are some basic results on the law of excluded middle that show it also doesn't give us any more predictions than we had to begin with. I don't know whether it counts as a coincidence or not, but these are the same two axioms as were nonconstructive in the obvious way. The rest of the axioms in ZFC do give new predictions. The axiom of replacement is often written in a form implying the law of excluded middle, but it can be written not to. Without AC or LEM this leaves us with a system called IZF that satisfies the numeric and set existence properties. I guess it's not very well liked for practical use, but in a theoretical way it nicely matches ZFC. One can prove the existence of the same recursive functions in each for instance. IZF can prove the consistency of IZF with the axiom of replacement removed, which IZF with the axiom of replacement removed can't do (by Goedel's second incompleteness theorem). Since a consistency statement is equivalent to one of the form (n)P(n) this counts as some kind of additional prediction. IZF with the axiom of replacment removed can prove the consistency of IZF with the power set axiom and the axiom of replacement both removed. IZF with both the power set axiom and the axiom of replacement removed can prove the consistency of IZF with the axiom of infinity removed. If the axiom of infinity is replaced with its negation, we get a system essentially the same as elementary arithmetic. There's a model of it consisting only of the sets that can be built up in finitely many steps from the empty set. The only way to get the full computational content is to keep all the axioms in IZF. Presumably there's some way to simplify the system by replacing axioms, but none of the axioms can be just tossed out. |> (2) I don't see where, within mathematics seen from a |> constructive point of view, it is supposed to be that it |> departs from having computational meaning. We could consider |> cases where someone was investigation a question of the form |> X->Y and eventually found that X is false, that there were |> no examples of the kind of thing they were investigating, and |> things like this, but as long as it's decided after the fact, |> and there isn't a criterion we can use in advance to see that |> the question lacked content, I don't think it counts. | |Actually, I can't really follow that paragraph. But the question of |whether a statement has computational meaning is independent of whether |it has been proven. I'm talking in that paragraph about your apparent wish to draw a line between computationally meaningful and not computationally meaningful. I can relatively easily see how you could decide that certain bits of nonconstructive mathematics should be left on the not computationally meaningful side of the line. But then you appear to want to draw your line so that it passes somewhere *inside* of what I describe as constructive mathematics. The criteria I described above give us a principled way to designate certain kinds of mathematics as nonconstructive. Is there a principled way to designate some of the mathematics that counts as constructive by those criteria, and say that, nevertheless, it lacks computational meaning? As far as I can tell, you just go on your own gut reaction, without any principle that someone not sharing your intuitions could apply in a consistent way. It's hard to see how one could in a principled way dispose of all the Cantorian mathematics you have such a problem with (Cantor's theorem as usually understood, for pity's sake) without being rather draconian. It's hard to see how one could in a principled way accept the applied math you are friendly toward without accepting just about everything. |> |> Some results like Cantor's theorem are perfectly natural for |> |> a constructive mathematician like Bishop. |> | |> |You might say that, but the constructive version of Cantor's theorem is |> |not the same theorem as the classical version. |> I don't know in what relevant sense you consider them not |> the same. | |I know we've discussed this a few times before. The result that is not |disputed by either constructivists or classical mathematicians is that |if we are given a well defined list of well defined real numbers (so |that every digit of every number can be computed), then the diagonal |method gives us a new number not on that list. The classical |mathematicians claim, essentially, that the argument is still valid |when the list and the numbers on the list may not be well defined. You're equivocating between computable, which is a property of certain objects, and definedness which is really a property of the description of the object that needs to be satisfied before one can ask whether the object described by the description has some property or not. You describe constructivists as if they automatically assume that reals are computable. They don't. The argument makes no use of any such assumption. we can, moreover, understand what the classical mathematicians have in mind when they talk about real numbers. They are thinking of things that we can represent with sets of rationals. Given two sets of rationals L and R, with L downward closed and R upward closed, where (*) For some epsilon>0, for every l in L and r in R, we have r-l>epsilon. leads to a contradiction, we have a classical real. These are not undefined, although its nonconstructive to say that these are the same as the honest-to-goodness reals. We can use the same (Cantorian) reasoning to take a sequence of classical reals and produce a new classical real different from the terms of the sequence. Keith Ramsay ==== Subject: Re: Cantorian pseudomathematics Discussion, linux) > You seem not to believe me when I say it, but there's a > simple, straightforward way of applying the mathematical > theories must make predictions rule that tells us to dump > the same two axioms, and nothing else. Take computational > predictions to mean only the most obvious kind of thing, > namely, statements of the form (n)P(n) where n ranges over > the integers and P is a primitive recursive predicate. > Then say that an axiom system should be no more complicated > than necessary to entail all conclusions of that form that > it does. Sorry, Keith, but I don't understand what you mean by that last sentence. Can you say that again, maybe a bit slower and with smaller words? -- I don't want to wine and dine and date you once or twice. I want to hold you now. I just want to spend the night. You tell me a better plan. Baby, I'm not a patient man. -- Jimmy Lafave, the romantic troubadour. ==== Subject: Re: Cantorian pseudomathematics >>http://hdebruijn.soo.dto.tudelft.nl/jaar2005/winter.pdf > By the way, what you are doing is not new. Gaussian smoothing > is a well-known, useful and rigorously-analyzed technique > http://www.cee.hw.ac.uk/hipr/html/gsmooth.html my paper that I haven't encountered anywhere else: subsection Sampling theorem. The original Shannon theorem is about a rectangle function in the Fourier domain and a comb of 'sinc' functions in the 'time' domain. Mine is about a Gaussian in the Fourier domain and a comb of Gaussians in the 'time' domain. The original theorem by Shannon is _exact_, while mine is approximate. Because, unlike the rectangle function, a Gaussian isn't bandwidth limited, theoretically. I don't claim anything, but ... Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics > Something I've wondered about, though, is: > Is it possible to construct axiomatic systems using fuzzy logic in > such a way that we get theorems that are somewhat true within the > system? Isn't that the subject of: http://omega.math.albany.edu:8008/JaynesBook.html First two chapters? Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics <952d9$43ccb12d$82a1e2b0$19955@news2.tudelft.nl> <3424c$43ce146e$82a1e2b0$25410@news1.tudelft.nl> ... > [Obligatory there is no universe in ZF text snipped.] > Infinite sets are equivalent. > Dude, t, = -t. I look at equations and want to integrate both sides even when the function is not separable. f(x, y) = g(x) h(y) = g(x) f(1/y) So, y is non-zero and all that, where division by zero is verboten in terms of the vertical asymptote and horizontal rule, so anyways with something along the lines of: f(x, y) = xy / (x^2 + y^2) that is not separable. It's reciprocal 1/f is (x^2 + y+2) / xy, say, its inverse f^-1 is non-existent because it's not 1-to-1, even though some notion of a principle inverse is considerable. So then I wonder about the quadrature, basically in terms of z = xy, what that is and what it's supposed to mean. For something along the lines of non-separable linear first order equation: y' + y = x dy/dx + y = x dy + y dx = x dx dy + ydx = 1 + C Then, it's not OK to integrate both sides,to just toss up an integral bar on either side, even though each side has a differential, because h(y) and g(x) are not separable. 1 + Int ydx = 1 + C That there Int y dx has, with dx in terms of dy being dy/dx dx by the chain rule, again, damme. In finding an integrating factor mu = e^ Int 1 dx = e^x, then mu y = Int mu x dx + C e^x y = Int xe^x dx + C integrate y = e^-x (Int xe^x dx + C) That appears to be something along the lines of: y = e^-x ((-1)^2 incompletegamma(2, -oo)) = e^-x ((-1)^-2 incompleteGamma(2, 0)) although that's integrated definitely over [0,oo). So, y = incompletegamma(2, -oo)/ e^x = incompleteGamma(2, 0)/ e^x, and, incompletegamma(2, oo) = incompleteGamma(2, 0). That's only because the constant mu in e^(mu x) is equal to one, i.e. it's not so for, uh, non-unitary values. Well, that's simply a solution to y' = x - y, except defined in terms of two different special functions, and assuming they're defined and so forth or not caring. I got those from Gradshteyn and Ryzhik, says incompletegamma(x, y) is Int_0^y e^-t t^x-1 dt. As above I am looking at 1 and 3 from section 3.381 on page 342 of 6'th ed, incompletegamma from 8.35, incompleteGamma from 8.35, the incomplete Gamma function being along the lines of Gamma(x, y) = Int_0^oo e^-t t^x-1 dt. So, that makes sense, they just drop out as they are equal to each other. They just have different parameterizations in the variables which happen to be equal, so incompletegamma(2, oo) = incompleteGamma(2, 0) = Int_0^oo t / e^t dt. So, y = e^-x Int_0^oo t / e^t dt + C So, Int xe^x dx = Int_0^oo te^(-t) dt where y is a solution. That might seem only where e^t = e^(-t), i.e. t = 0, t = -t, as I see from unconditioning separability, in integration. There t is not jsut a constant, it's zero, although I also think about it in terms of the oo vis-a-vis +-oo thing(s). It appears that quadrature is not what I am wondering about, but actually that is what I'm wondering about. Basically I'm trying to find forms of differential equations that are in general soluble, similarly to how e^x's derivative, with respect to x, is itself, but different. Simple, different. Ross ==== Subject: Re: Cantorian pseudomathematics <43087$43e32c1e$82a1e2b0$1563@news1.tudelft.nl> <7b71c$43e37a9b$82a1e2b0$7242@news2.tudelft.nl> <43e37b44$0$19705$8fcfb975@news.wanadoo.fr> <23eau15cuv8ekjpm1t46348g7snjp83vnq@no.spam> statement seems to be impossible in many cases. How can you for > example state how true the Goldbach Conjecture is? There are > infinitely many cases that need to be checked. Actually, it is very easy to say how true the Goldbach conjecture is. Starting with the heuristic that the probability that a given number of size 'k' is prime is 1/log(k), we can fuzzily get a formula for the probability that for a given even number N, both 'N' and 'N-k' are prime, and from there we can get an approximate formula for the probability that a given even number of size 'N' is equal to the sum of two primes. And then if we do a little more work (for example, looking at the factorization of 'N') we can get a formula that experiments show to be very very accurate as a predictor of the number of ways in which a given even number can be the sum of two primes. Furthermore, we can verify the conjecture for all even numbers up to some very large number (it's been done!), and then calculate a probability the there exists a counterexample for some larger number. Doing all these calculations suggests that the probability that there exists a counterexample to Goldbach's conjecture is on the order of exp( - log(N)^2 ), where the conjecture has been verified for all even numbers less than 'N'. The conclusion: our best estimate is that Goldbach's conjecture is true with a probability of about 1 - 1/10^2000. ==== Subject: New mathematics/physical sciences positions at http://jobs.phds.org, February 06, 2006 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Research Professor, Environmental, Embry-Riddle Aeronautical University: Academic Careers Online, Luxembourg. Environmental research takes flight at Embry-Riddle in Luxembourg. 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Essential Functions Manage the MARC, including identifying and implementing MARC services, recruiting and supervising tutors and other staff, managing budget and setting program goals, managing instructional resources and promoting the services of the MARC to... http://jobs.phds.org/jobs/academiccareers/listing_2006_02_01_4 ==== Subject: How many types of data graphs do exist? repository web sites? How many types of data graphs do exist? Any repository web sites? I am trying to analyze trillion and trillions numbers of data and to get some idea how large data is usually analyzed? My data is in an array which is 1500 x1600 multiplied by 18 layers. ( 1500x1600X18) Each number is in the range 1 to 20 million. I've found a few data graphs at http://www.math.yorku.ca/SCS/Gallery/noframes.html Any other websites or depositaries? Guy (Gnh888) ==== Subject: Z_p and the p-adic numbers Q_p One construction of the field Q_p of p-adic numbers uses Q, the p-adic norm |.|_p, and equivalence classes of Cauchy sequences (of elements of Q): Section 1.2 of Dubravka Ban's Introduction to Representation Theory of p-adic Groups and Langlands Program, which I got here: http://www.math.siu.edu/ban/Math522.pdf/ . [ A second definition of Q_p in Section 1.2 starts with Z_p defined as an inverse limit, and then Q_p is the field of fractions of Z_p .] But sticking to the first definition of Q_p as equivalence classes of Cauchy sequences, the ring Z_p is then defined as Z_p = {x in Q_p such that |x|_p <=1}. (*) So in this case Z_p is defined non-algebraically from Q_p. The smallest subring of Q_p containing the identity is Z (not Z_p). I'm trying to define Z_p algebraically (from Q_p), without using the p-adic norm. Perhaps Z_p is contained in any unital subring of Q_p R having the property that any x in Q_p is a quotient of elements of R ... David Bernier ==== Subject: Re: Z_p and the p-adic numbers Q_p [snipped some discussion] > I'm trying to define Z_p algebraically (from Q_p), without using the > p-adic norm. Perhaps Z_p is contained in any unital subring of Q_p > R having the property that any x in Q_p is a quotient of elements of R ... > David Bernier Would you be happy with the following? Assume p>2, and x is in Q_p. Then x is an element of Z_p, if and only if the element S(x)=1+px is a square of a non-zero element of Q_p. If p=2, then you can use S(x)=1+8x instead. The proofs do involve simple analytic methods (like Newton's approximation in creating a sequence of more and more accurate square roots). However, the result gives a purely algebraic description of of Q_p must keep Z_p fixed as a set. Consequently such an automorphism must be continuous, and hence actually equal to the identity mapping. Jyrki ==== Subject: Re: Z_p and the p-adic numbers Q_p [...] > Would you be happy with the following? Assume p>2, and x is in Q_p. > Then x is an element of Z_p, if and only if the element > S(x)=1+px > is a square of a non-zero element of Q_p. If p=2, then you can use > S(x)=1+8x [...] So for p=5, 3/5 is in Z_5 ? (because 1 + 5*(3/5) = 4 = 2^2). Yet, |3/5|_5 = 5 > 1 , => 3/5 not in Z_5, so maybe I'm missing something... David Bernier ==== Subject: Sum of two *nearly* bivariate normal distributed random variables let (X,Y) be a bivariate normal distributed random vector. Can anyone compute the distribution function of Z := max(x_0, X) + Y, (x_0 a real number)? I am afraid there is no explicit solution. However, on the one hand one can calculate the distribution of X + Y (x_0 = - infty). It is, in hopefully self-explanatory notation, normal distributed with mean mu_x + mu_y and variance sigma_x^2 + sigma_y^2 + 2 sigma_{xy}. On the other hand, one can compute the distribution function of max(x_0, X) alone. I get Phi_{max(x_0, X)} = Phi_X (X) * 1_{[x_0, infty)} (X) where Phi is the respective distribution function and 1_M the characteristic function of the set M. But what about Z? ==== Subject: Re: Sum of two *nearly* bivariate normal distributed random variables >let (X,Y) be a bivariate normal distributed random vector. >Can anyone compute the distribution function of >Z := max(x_0, X) + Y, >(x_0 a real number)? Consider the cdf F(z) = P(Z <= z) = P((X,Y) in G(z)) where G(z) = {(x,y): y <= z - x_0, x + y <= z}. With f(x,y) the joint density for (X,Y), we then have F(z) = int_{-infty}^{z - x_0} dy int_{-infty}^{z - y} dx f(x,y) and the density is the derivative of that: f_Z(z) = int_{-infty}^{z-x_0} f(z-y,y) dy + int_{-infty}^{x_0} f(x,z-x_0) dx which should be no problem to evaluate as a rather complicated expression in closed form (of course containing the error function). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: product space of two triangulable spaces are triangulable Let X and Y be two triangulable spaces. Show that X x Y is triangulable. ==== Subject: Re: product space of two triangulable spaces are triangulable > Let X and Y be two triangulable spaces. Show that X x Y is > triangulable. It comes down to triangulating the product of two simplexes. I don't recall the details, but it involves the shuffle product. The details ought to be findable in any decent algebraically text, or at least a reference. ==== Subject: product space of two triangulable spaces are triangulable Let X and Y be two triangulable spaces. Show that X x Y is triangulable. ==== Subject: product space of two triangulable spaces are triangulable Let X and Y be two triangulable spaces. Show that X x Y is triangulable. ==== Subject: Re: Are there known ways to solve g(3*x/(x+1), y) = d/dy g(x,y) - 2/y*g(x,y) ? > I am looking for ways to solve the given > functional equation g:R^2 -> R continuously > differentiable for x and y . > We may consider the following process: > when x => 3*x/(x+1) , g => (d/dy - 2/y ) o g > Wait for your help , > Alain. I would say: change variables appropriately, then you have to solve C(x+1,y) = (d/dy) C(x,y) For this, define C(x,y) any way you like for 0 <= x < 1, then the other values are multiple derivatives (or integrals) of this. [Of course, for each x, the function C(x,y) should have infinitely many integrals and derivatives.] ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse >Take A and B for integers <> 0 with gcd(A,B)=1. >A solution in integers X_0,Y_0 to the equation >AX+BY=1 can be calculated by means of the >Extended Euclidean Algorithm (EEA). >Hereby you get >AX_0+BY_0=1 >AX_0 - 1 = -BY_0 >AX_0 - 1 = 0 mod B >AX_0 = 1 mod B >-> X_0 is multiplicative inverse to A mod B >BY_0 - 1 = -AX_0 >BY_0 - 1 = 0 mod A >BY_0 = 1 mod A >-> Y_0 is multiplicative inverse to B mod A. >Question: >Is EEA's X_0 already reduced modulo B ? >(Is abs(X_0) < abs(B) ?) >Is EEA's Y_0 already reduced modulo A ? >(Is abs(Y_0) < abs(A) ?) They should be. If you use the algorithm that is described at , then they are. Rob Johnson take out the trash before replying ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse > They should be. If you use the algorithm that is described at > , then they > are. I use this algorithm. Also I see that in the example presented on that web-page, the solutions x=-4, y=3 to mx + ny = gcd(m,n); m = 17, n = 23 are reduced-multiplicative-inversions to 17 mod 23 / 23 mod 17. But unfortunately I cannot see why this is the case for all m,n in Z. Am I overlooking something trivial here? Sam ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse >> They should be. If you use the algorithm that is described at >> , then they >> are. >I use this algorithm. >Also I see that in the example presented on that web-page, the >solutions x=-4, y=3 to mx + ny = gcd(m,n); m = 17, n = 23 >are reduced-multiplicative-inversions to 17 mod 23 / 23 mod 17. >But unfortunately I cannot see why this is the case for all >m,n in Z. >Am I overlooking something trivial here? I am not sure what your question is. Are you uncertain that the solutions to mx + ny = gcd(m,n) are reduced or that they are inverses? Let me address each of these questions. They are reduced because the each step in the algorithm produces continued fraction convergents for m/n. The last step is the reduced fraction m/n; the previous step is the solution. The numbers increase in each step, so the solutions must be less than the m and n. If gcd(m,n) = 1, then the equation mx + ny = 1 is just a rephrasing of being inverses. If gcd(m,n) is not 1, there is no inverse of m mod n or of n mod m. For example, there is no inverse of 6 mod 15 since gcd(6,15) = 3. Suppose x is the inverse of 6 mod 15, then 6x + 15y = 1 and we have that 3(2x + 5y) = 1, which is impossible since 3 does not divide 1. Does this answer your question, or am I missing your point? Rob Johnson take out the trash before replying ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse > I am not sure what your question is. Are you uncertain that the > solutions to mx + ny = gcd(m,n) are reduced or that they are > inverses? Let me address each of these questions. > AX_0+BY_0=1 > AX_0 - 1 = -BY_0 > AX_0 - 1 = 0 mod B > AX_0 = 1 mod B > -> X_0 is multiplicative inverse to A mod B > BY_0 - 1 = -AX_0 > BY_0 - 1 = 0 mod A > BY_0 = 1 mod A > -> Y_0 is multiplicative inverse to B mod A. So it is clear that they are inverse. I didn't see that they are reduced in all cases. [By the way: I was not uncertain: Rob Johnson already told me that they are reduced, which is nearly as good as a proof. ;-) I just did not see why.] > They are reduced because the each step in the algorithm produces > continued fraction convergents for m/n. Aaargh!!!!! This is so painful! Yes of course! I am so stupid! > The last step is the reduced > fraction m/n; the previous step is the solution. The numbers increase > in each step, so the solutions must be less than the m and n. > If gcd(m,n) = 1, then the equation mx + ny = 1 is just a rephrasing > of being inverses. If gcd(m,n) is not 1, there is no inverse of m > mod n or of n mod m. For example, there is no inverse of 6 mod 15 > since gcd(6,15) = 3. Suppose x is the inverse of 6 mod 15, then > 6x + 15y = 1 and we have that 3(2x + 5y) = 1, which is impossible > since 3 does not divide 1. > Does this answer your question, or am I missing your point? It does answer my question. You did not miss my point. Sam ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse I haven't proved it yet but the answer seems to be yes. Let me show an example so that other readers will know what the question is about. We are given two positive coprime intergers X and Y, with X>Y. For example X=103 and Y=61. 1 0 103 0 1 61 1 -1 42 -1 2 19 3 -5 4 -13 22 3 16 -27 1 The third row, and the remaining ones, are linear combintations L-nM where L and M are the previous two rows, and n>0. When we get 1 in the right column, we can write (in this example) 16*103 - 27*61 = 1 which shows that 27 is the inverse of 61 mod 103 and 16 is the inverse of 103 mod 61. ==== Subject: Re: Extended Euclidean Algorithm/multiplicative inverse > I haven't proved it yet but the answer seems to be > yes. > Let me show an example so that other readers will > know what the question is about. We are given two > positive coprime intergers X and Y, with X>Y. For > example X=103 and Y=61. > 1 0 103 > 0 1 61 > 1 -1 42 > -1 2 19 > 3 -5 4 > -13 22 3 > 16 -27 1 > The third row, and the remaining ones, are linear > combintations L-nM where L and M are the previous two > rows, and n>0. When we get 1 in the right column, we > can > write (in this example) > 16*103 - 27*61 = 1 > which shows that 27 is the inverse of 61 mod 103 and > 16 is the inverse of 103 mod 61. Er, I meant -27 is the inverse of 61 mod 103. The conclusion we seek is (in this example) |-27|<=|61|/2. See also Rob Johnson's answer, which seems to have escaped from the response-tree :) ==== Subject: summary of {1/n} - woody ==== Subject: Re: summary of {1/n} I'm guessing that by summary, you mean sum or summation. However, I can't guess whether you mean the sum from n=1 to infinity or from n=1 to some finite N. Do you want the exact formula, which is not as useful for numerical purposes, or an asymptotic expansion, which is not exact, but better for numerical purposes. The complete series diverges, that is oo --- 1 > - = oo --- n n=1 The exact sum is N --- 1 Gamma'(N) > - = gamma + --------- --- n Gamma(N) n=1 where gamma is the Euler-Mascheroni constant and Gamma is the Gamma function The asymptotic expansion is N --- 1 1 1 1 1 1 1 > - ~ log(N) + gamma + -- - ----- + ------ - ------ + ------ - ------- + ... --- n 2N 12N^2 120N^4 252N^6 240N^8 132N^10 n=1 See . Rob Johnson take out the trash before replying ==== Subject: Re: summary of {1/n} days. My association with the Department is that of an alumnus. The infinite sum or series (which I suspect is what you mean by summary; summary means something else) 1 + (1/2) + (1/3) + ... + (1/n) + ... diverges. It has no solution. -- ' ==== Subject: Re: summary of {1/n} summary; summary means something else) > 1 + (1/2) + (1/3) + ... + (1/n) + ... > diverges. It has no solution. ==== Subject: Re: summary of {1/n} >> The infinite sum or series (which I suspect is what you mean by >> summary; summary means something else) >> 1 + (1/2) + (1/3) + ... + (1/n) + ... >> diverges. It has no solution. Arturo, you are amazing. I never would have figured out what he meant. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. ==== Subject: Re: summary of {1/n} days. My association with the Department is that of an alumnus. > The infinite sum or series (which I suspect is what you mean by > summary; summary means something else) > 1 + (1/2) + (1/3) + ... + (1/n) + ... diverges. It has no solution. >Arturo, you are amazing. I never would have figured out what he meant. It helps not being a native speaker. I could make an intelligent guess as to what summary might mean; I could easily have been wrong, of course. (Much easier when the mistakes go from English to Spanish, for me, at least. The three most typical mistakes I see that way are: aplicar for application as in school or job applications, which should be 'solicitudes'; libreria for library, and biblioteca for bookstore, which should be reversed; and perversely, 'lima' for 'lime' and 'limon' for 'lemon', when they should also be reversed: 'lima'=lemon and 'limon' = lime. Go figure) -- ' ==== Subject: Re: summary of {1/n} Here's a way to see that it diverges: First term = 1 Next three terms = 1/2 + 1/3 + 1/4 > 1 Then collect the next eight or so terms and they add to more than 1 With an infinite series, you never run out of terms! ==== Subject: Re: JSH: Even shorter, disproof ... > I pick x=3. That gives a = 1 + sqrt(-83). > > Working it out, hopefully I got the algebra right, I get > > x = (-6 + sqrt(-83) +/- sqrt(2221 - 96sqrt(83))/14 > > That should be > > x = (-6 + sqrt(-83) +/- sqrt(2221 - 96sqrt(-83)))/14 Note the fix. > > (increment the Oops counter). > No need, as I just left off a parenthesis on the end which doesn't > change anything. The error is not just a left off parenthesis... > I pick x=3. That gives a = 1 + sqrt(-83). > Working it out, hopefully I got the algebra right, I get > x = (-6 + sqrt(-83) +/- sqrt(2221 - 96sqrt(83)))/14 Fix it. > and I can multiply top and bottom by > (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83)) > use x=3, and simplify to get > 1 = (-54 + 2sqrt(-83)/((-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83))) This is wrong. When you do the calculations you will find that: (-6 + sqrt(-83) + sqrt(2221 - 96.sqrt(-83))/14 = 3 and (-6 + sqrt(-83) - sqrt(2221 - 96.sqrt(-83))/14 = (-27 + sqrt(-83)/7. (You may notice that sqrt(2221 - 96.sqrt(-83)) = 48 + sqrt(-83)...) > (-6 + sqrt(-83) -/+ sqrt(2221 - 96sqrt(83))) = -54 + 2sqrt(-83) Wrong. See above. -- 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: JSH: Even shorter, disproof >> a^2 - (x - 1)a + 7(x^2 + x) = 0 >> And we have many many cases where neither root is divisible >> by 7 nor coprime to 7. >> E.g. Take x=2 >> We get >> a^2 - a + 42 = 0 >> (note the middle term does not have 7 as a factor) >> The roots are >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > ----------------------------------------------------------------- >> It was noted in a long-distant post by Keith Ramsay that >> a_1 = (1 + sqrt(-167))/2 is such that >> a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and >> 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). >> Therefore a_1^11 and 7^11 share an algebraic integer factor, >> namely >> 44555 - 222*a_1. >> Therefore a_1 and 7 share the algebraic integer factor >> (44555 - 222*a_1)^(1/11). >> Note that 44555 is divisible by 7, but a_1 is not (in the >> algebraic integers). Therefore this common algebraic integer >> factor is not divisible by 7. >It's not divisible by 7 in the ring of algebraic integers. Yes - that's what I said. Neither a_1 nor a_2 is divisible by 7, and neither is coprime to 7. Since I gave an explicit common factor, this is not a case where you depend on Dedekind's theorem (on ideals) to show that a common factor exists. Therefore your claim that you have found a problem with the theory of ideals does not apply. >You keep going in circles around that central point. >My counter would be that one solution of the 11 generated by that >radical expression as it is taking the 11th root, does have 7 as a >factor. All you need to consider here is a_1. Since it does not have 7 as a factor and it is not coprime to 7, and since a_1 * a_2 is divisible by 7, both a_1 and a_2 are not coprime to 7 and not divisible by 7. >But rather than argue that point let me show you the simple way to see >a contradiction if I'm wrong, as I just figured it out. >> Which proves very explicitly that Harris is wrong. The algebraic >> integers are NOT incomplete. >> Marcus >They are, as this simple algebra shows. >I have from before >a^2 - (x - 1)a + 7(x^2 + x) = 0 >from which I can easily get >7x^2 + (7-a)x + a^2 + a = 0 >and this time I'll choose an integer value for a, like a=3, so that I >have a non-monic polynomial irreducible over Q for x. >But because 3 is coprime to 7 I know that the other solution for 'a' >MUST have 7 as a factor, right? No - you are now considering this as a polynomial in x. You chose a value for a, and then you solve for x. You have to decide one at a time which variables you are treating as constants and which as 'unknowns'. Here you are treating x as the unknown, and a as a constant. >But now I just go back to >a^2 - (x - 1)a + 7(x^2 + x) = 0 >and note that for any solution for x, remember x is NOT an algebraic >integer, where x as a ratio of algebraic integers has factors in common >with 7, while being coprime to some factors of 7, it is NOT possible >that the a's have 7 as a factor, which is a direct contradiction with >one of the a's being 3, and thus forcing the other of the a's to have 7 >as a factor. No - you're just confusing yourself on this. You either choose x and solve for a, or you choose a and solve for x. You don't get to do both at the same time. >Interestingly enough, there is no meaningful way to consider the second >a which is NOT an algebraic integer to be a fraction, so the ring of >algebraic integers is easily shown to be incomplete as I've said. >Done arguing yet? >Why fight for ideas that can so easily be shown to be mathematically >incorrect? >What do you think, you get some prize for using bogus math? The shoe is on the other foot here. Pick an x, and solve for a. That is what William Hughes did. I then showed that the solutions for a are not divisible by 7 and not coprime to 7, without appealing to Dedekind's theorem. It's just arithmetic. Therefore Dedekind's theorem is not the problem. Your own math is the problem. Marcus James Harris ==== Subject: Re: JSH: Even shorter, disproof >> a^2 - (x - 1)a + 7(x^2 + x) = 0 >> And we have many many cases where neither root is divisible >> by 7 nor coprime to 7. >> E.g. Take x=2 >> We get >> a^2 - a + 42 = 0 >> (note the middle term does not have 7 as a factor) >> The roots are >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > ----------------------------------------------------------------- >> It was noted in a long-distant post by Keith Ramsay that >> a_1 = (1 + sqrt(-167))/2 is such that >> a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and >> 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). >> Therefore a_1^11 and 7^11 share an algebraic integer factor, >> namely >> 44555 - 222*a_1. >> Therefore a_1 and 7 share the algebraic integer factor >> (44555 - 222*a_1)^(1/11). >> Note that 44555 is divisible by 7, but a_1 is not (in the >> algebraic integers). Therefore this common algebraic integer >> factor is not divisible by 7. >It's not divisible by 7 in the ring of algebraic integers. > Yes - that's what I said. Neither a_1 nor a_2 is divisible > by 7, and neither is coprime to 7. Since I gave an explicit > common factor, this is not a case where you depend on > Dedekind's theorem (on ideals) to show that a common factor > exists. Therefore your claim that you have found a problem > with the theory of ideals does not apply. But I've proven that the ring of algebraic integers is incomplete. The belief that it is complete relies on a false claim of proof using the theory of ideals made by Dedekind, brining the theory of ideals into question. >You keep going in circles around that central point. >My counter would be that one solution of the 11 generated by that >radical expression as it is taking the 11th root, does have 7 as a >factor. > All you need to consider here is a_1. Since it does not have 7 > as a factor and it is not coprime to 7, and since a_1 * a_2 is > divisible by 7, both a_1 and a_2 are not coprime to 7 and not > divisible by 7. That's the case IN THE RING OF ALGEBRAIC INTEGERS where I can show that you get a contradiction with that ring, as I did in reply to you, quite simply. You can do the same thing with 2 and 6 in the ring of evens, where one person can claim that 2 is not coprime to 6, while the other keeps claiming it is, while they remain in the ring of evens. You cannot disprove my claims by relying on results in the ring of algebraic integers, any more than someone could prove that 2 is coprime to 6 by remaining in the ring of evens. It just doesn't fly. >But rather than argue that point let me show you the simple way to see >a contradiction if I'm wrong, as I just figured it out. >> Which proves very explicitly that Harris is wrong. The algebraic >> integers are NOT incomplete. >> Marcus >They are, as this simple algebra shows. >I have from before >a^2 - (x - 1)a + 7(x^2 + x) = 0 >from which I can easily get >7x^2 + (7-a)x + a^2 + a = 0 >and this time I'll choose an integer value for a, like a=3, so that I >have a non-monic polynomial irreducible over Q for x. >But because 3 is coprime to 7 I know that the other solution for 'a' >MUST have 7 as a factor, right? > No - you are now considering this as a polynomial in x. You > chose a value for a, and then you solve for x. You have to > decide one at a time which variables you are treating as constants > and which as 'unknowns'. Here you are treating x as the unknown, > and a as a constant. That's not mathematics! Obviously if one of the choices for 'a' is 3, then there is a resultant x, which must fulfil the conditions of all the equations. >But now I just go back to >a^2 - (x - 1)a + 7(x^2 + x) = 0 >and note that for any solution for x, remember x is NOT an algebraic >integer, where x as a ratio of algebraic integers has factors in common >with 7, while being coprime to some factors of 7, it is NOT possible >that the a's have 7 as a factor, which is a direct contradiction with >one of the a's being 3, and thus forcing the other of the a's to have 7 >as a factor. > No - you're just confusing yourself on this. You either choose > x and solve for a, or you choose a and solve for x. You don't get > to do both at the same time. Um, do you think that if you do one you don't do the other? That is, given a^2 - (x - 1)a + 7(x^2 + x) = 0 and a choice for 'a', you are claiming that you can't use that solution with 7x^2 + (7-a)x + a^2 + a = 0 the same exact equation, just written in a slightly different way? With the second and a choice for one of the a's that is an integer and coprime to 7, where a^2 + a is coprime to 7, solutions for x must fit the same equation. damn equation! If math people are going to go this far then I wonder how there can be any hope. You throw algebra away as if it were trash. James Harris ==== Subject: Re: JSH: Even shorter, disproof >> a^2 - (x - 1)a + 7(x^2 + x) = 0 >> And we have many many cases where neither root is divisible >> by 7 nor coprime to 7. >> E.g. Take x=2 >> We get >> a^2 - a + 42 = 0 >> (note the middle term does not have 7 as a factor) >> The roots are >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > ----------------------------------------------------------------- >> It was noted in a long-distant post by Keith Ramsay that >> a_1 = (1 + sqrt(-167))/2 is such that >> a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and >> 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). >> Therefore a_1^11 and 7^11 share an algebraic integer factor, >> namely >> 44555 - 222*a_1. >> Therefore a_1 and 7 share the algebraic integer factor >> (44555 - 222*a_1)^(1/11). >> Note that 44555 is divisible by 7, but a_1 is not (in the >> algebraic integers). Therefore this common algebraic integer >> factor is not divisible by 7. >It's not divisible by 7 in the ring of algebraic integers. > Yes - that's what I said. Neither a_1 nor a_2 is divisible > by 7, and neither is coprime to 7. Since I gave an explicit > common factor, this is not a case where you depend on > Dedekind's theorem (on ideals) to show that a common factor > exists. Therefore your claim that you have found a problem > with the theory of ideals does not apply. > But I've proven that the ring of algebraic integers is incomplete. > The belief that it is complete relies on a false claim of proof using > the theory of ideals made by Dedekind, brining the theory of ideals > into question. >You keep going in circles around that central point. >My counter would be that one solution of the 11 generated by that >radical expression as it is taking the 11th root, does have 7 as a >factor. > All you need to consider here is a_1. Since it does not have 7 > as a factor and it is not coprime to 7, and since a_1 * a_2 is > divisible by 7, both a_1 and a_2 are not coprime to 7 and not > divisible by 7. > That's the case IN THE RING OF ALGEBRAIC INTEGERS where I can show that > you get a contradiction with that ring, as I did in reply to you, quite > simply. > You can do the same thing with 2 and 6 in the ring of evens, where one > person can claim that 2 is not coprime to 6, while the other keeps > claiming it is, while they remain in the ring of evens. > You cannot disprove my claims by relying on results in the ring of > algebraic integers, any more than someone could prove that 2 is coprime > to 6 by remaining in the ring of evens. > It just doesn't fly. >But rather than argue that point let me show you the simple way to see >a contradiction if I'm wrong, as I just figured it out. >> Which proves very explicitly that Harris is wrong. The algebraic >> integers are NOT incomplete. >> Marcus >They are, as this simple algebra shows. >I have from before >a^2 - (x - 1)a + 7(x^2 + x) = 0 >from which I can easily get >7x^2 + (7-a)x + a^2 + a = 0 >and this time I'll choose an integer value for a, like a=3, so that I >have a non-monic polynomial irreducible over Q for x. >But because 3 is coprime to 7 I know that the other solution for 'a' >MUST have 7 as a factor, right? > No - you are now considering this as a polynomial in x. You > chose a value for a, and then you solve for x. You have to > decide one at a time which variables you are treating as constants > and which as 'unknowns'. Here you are treating x as the unknown, > and a as a constant. > That's not mathematics! Obviously if one of the choices for 'a' is 3, > then there is a resultant x, which must fulfil the conditions of all > the equations. >But now I just go back to >a^2 - (x - 1)a + 7(x^2 + x) = 0 >and note that for any solution for x, remember x is NOT an algebraic >integer, where x as a ratio of algebraic integers has factors in common >with 7, while being coprime to some factors of 7, it is NOT possible >that the a's have 7 as a factor, which is a direct contradiction with >one of the a's being 3, and thus forcing the other of the a's to have 7 >as a factor. > No - you're just confusing yourself on this. You either choose > x and solve for a, or you choose a and solve for x. You don't get > to do both at the same time. > Um, do you think that if you do one you don't do the other? > That is, given > a^2 - (x - 1)a + 7(x^2 + x) = 0 > and a choice for 'a', you are claiming that you can't use that solution > with > 7x^2 + (7-a)x + a^2 + a = 0 > the same exact equation, just written in a slightly different way? > With the second and a choice for one of the a's that is an integer and > coprime to 7, where a^2 + a is coprime to 7, solutions for x must fit > the same equation. > damn equation! > If math people are going to go this far then I wonder how there can be > any hope. > You throw algebra away as if it were trash. You know, the bottom line on this is the following. You say that in the algebraic integers, one root of a^2 - a + 42 = 0 *should* have 7 as a factor, while the other is coprime to 7. Yet you know, and we all know, that a theorem of Dedekind and a theorem in Galois theory both say that that is not true. So you concluded that Dedekind's theorem is false and Galois theory is wrong. You also concluded that the algebraic integers are incomplete in the sense that they do not permit a factorization of the kind that you want. But I showed explicitly, without any appeal to either Dedekind's theorem or Galois theory, that neither root is coprime to 7 and neither root is divisible by 7. Both roots have algebraic integer factors in common with 7. Factors of 7 are distributed across both roots. You say this cannot happen in the algebraic integers. But I showed it can. You can verify it by just carrying out the arithmetic. It is just a matter of brute calculation, just as true as 2 + 2 = 4. Therefore what you claimed is wrong. Therefore you have an error which you have not yet recognized. Getting mad at me for pointing this out is rather unproductive, don't you think? Shouldn't you be trying to find the mistake in your reasoning? Marcus > James Harris ==== Subject: Re: JSH: Even shorter, disproof > >> a^2 - (x - 1)a + 7(x^2 + x) = 0 > >> And we have many many cases where neither root is divisible >> by 7 nor coprime to 7. > >> E.g. Take x=2 > >> We get > >> a^2 - a + 42 = 0 > >> (note the middle term does not have 7 as a factor) > >> The roots are > >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > > ----------------------------------------------------------------- > >> It was noted in a long-distant post by Keith Ramsay that > >> a_1 = (1 + sqrt(-167))/2 is such that > >> a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and > >> 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). > >> Therefore a_1^11 and 7^11 share an algebraic integer factor, >> namely > >> 44555 - 222*a_1. > >> Therefore a_1 and 7 share the algebraic integer factor > >> (44555 - 222*a_1)^(1/11). > >> Note that 44555 is divisible by 7, but a_1 is not (in the >> algebraic integers). Therefore this common algebraic integer >> factor is not divisible by 7. > >It's not divisible by 7 in the ring of algebraic integers. > > Yes - that's what I said. Neither a_1 nor a_2 is divisible > by 7, and neither is coprime to 7. Since I gave an explicit > common factor, this is not a case where you depend on > Dedekind's theorem (on ideals) to show that a common factor > exists. Therefore your claim that you have found a problem > with the theory of ideals does not apply. > But I've proven that the ring of algebraic integers is incomplete. > The belief that it is complete relies on a false claim of proof using > the theory of ideals made by Dedekind, brining the theory of ideals > into question. >You keep going in circles around that central point. > >My counter would be that one solution of the 11 generated by that >radical expression as it is taking the 11th root, does have 7 as a >factor. > > All you need to consider here is a_1. Since it does not have 7 > as a factor and it is not coprime to 7, and since a_1 * a_2 is > divisible by 7, both a_1 and a_2 are not coprime to 7 and not > divisible by 7. > That's the case IN THE RING OF ALGEBRAIC INTEGERS where I can show that > you get a contradiction with that ring, as I did in reply to you, quite > simply. > You can do the same thing with 2 and 6 in the ring of evens, where one > person can claim that 2 is not coprime to 6, while the other keeps > claiming it is, while they remain in the ring of evens. > You cannot disprove my claims by relying on results in the ring of > algebraic integers, any more than someone could prove that 2 is coprime > to 6 by remaining in the ring of evens. > It just doesn't fly. >But rather than argue that point let me show you the simple way to see >a contradiction if I'm wrong, as I just figured it out. > >> Which proves very explicitly that Harris is wrong. The algebraic >> integers are NOT incomplete. > >> Marcus > >They are, as this simple algebra shows. > >I have from before > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > >from which I can easily get > >7x^2 + (7-a)x + a^2 + a = 0 > >and this time I'll choose an integer value for a, like a=3, so that I >have a non-monic polynomial irreducible over Q for x. > >But because 3 is coprime to 7 I know that the other solution for 'a' >MUST have 7 as a factor, right? > > No - you are now considering this as a polynomial in x. You > chose a value for a, and then you solve for x. You have to > decide one at a time which variables you are treating as constants > and which as 'unknowns'. Here you are treating x as the unknown, > and a as a constant. > That's not mathematics! Obviously if one of the choices for 'a' is 3, > then there is a resultant x, which must fulfil the conditions of all > the equations. > >But now I just go back to > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > >and note that for any solution for x, remember x is NOT an algebraic >integer, where x as a ratio of algebraic integers has factors in common >with 7, while being coprime to some factors of 7, it is NOT possible >that the a's have 7 as a factor, which is a direct contradiction with >one of the a's being 3, and thus forcing the other of the a's to have 7 >as a factor. > > No - you're just confusing yourself on this. You either choose > x and solve for a, or you choose a and solve for x. You don't get > to do both at the same time. > Um, do you think that if you do one you don't do the other? > That is, given > a^2 - (x - 1)a + 7(x^2 + x) = 0 > and a choice for 'a', you are claiming that you can't use that solution > with > 7x^2 + (7-a)x + a^2 + a = 0 > the same exact equation, just written in a slightly different way? > With the second and a choice for one of the a's that is an integer and > coprime to 7, where a^2 + a is coprime to 7, solutions for x must fit > the same equation. > damn equation! > If math people are going to go this far then I wonder how there can be > any hope. > You throw algebra away as if it were trash. > You know, the bottom line on this is the following. You say > that in the algebraic integers, one root of > a^2 - a + 42 = 0 > *should* have 7 as a factor, while the other is coprime to 7. And that's just NOT true, as it's possible to prove that they do not. I can then show a contradiction with the ring of algebraic integers, which is equivalent to showing a contradiction with someone in the ring of evens claiming that 2 is coprime to 6. The analogy should not escape you. > Yet you know, and we all know, that a theorem of Dedekind > and a theorem in Galois theory both say that that is not true. So > you concluded that Dedekind's theorem is false and Galois theory > is wrong. You also concluded that the algebraic integers are > incomplete in the sense that they do not permit a factorization > of the kind that you want. That's not true. I prove that you get the appearance of a contradiction if you assume that the ring of algebraic integers is complete. That proof is so trivial, you just ignored it in your reply, though at least you didn't delete it out! By picking an 'a' that is an *integer* and coprime to 7, like 3, I can easily show a contradiction with standard teaching, and I see you are just blissfully trying to ignore the proof. I have a question for you if you wish to maintain that you can only use those equations once, how do you suppose that knowing 3 is one of the a's, you would get the other one? > But I showed explicitly, without any appeal to either Dedekind's > theorem or Galois theory, that neither root is coprime to 7 and > neither root is divisible by 7. Both roots have algebraic integer > factors in common with 7. Factors of 7 are distributed across > both roots. That is true IN THE RING OF ALGEBRAIC INTEGERS so around and around we go. Just like if I were arguing with someone claiming that 2 is coprime to 6 who stayed in the ring of evens, repeating over and over again that 2 is coprime to 6 as they can prove--but needing me to keep reminding that the ring is evens!!! It's exactly the same thing. > You say this cannot happen in the algebraic integers. But I No I DO NOT!! Why make a false claim? The very point I make is that it's true in the ring of algebraic integers proving that ring is incomplete!!! Can you hear yourself? Take away your personal feelings. Forget about me. This is not personal. What's correct is what's important. Focus on the math, and not the social stuff. > showed it can. You can verify it by just carrying out the arithmetic. > It is just a matter of brute calculation, just as true as 2 + 2 = 4. > Therefore what you claimed is wrong. Therefore you have an > error which you have not yet recognized. Getting mad at me > for pointing this out is rather unproductive, don't you think? > Shouldn't you be trying to find the mistake in your reasoning? > Marcus I'm not mad at you. I am a bit exasperated yes, but look at what you're doing! In one reply you claim I can't use the same equation twice, making up some bizarre non-math rule. Here you claim that I make a claim that I do not, relying on proof of something being true in the ring of algebraic integers when I repeatedly explain why that's specious, and you end by just declaring that I am wrong. Where in your training as a math student did they teach you to make up things that are not true, deny mathematical proof, and ultimately just state something that you should realize is mathematically false--because the proof is in front of you and EASY--but instead you just go with your own opinion as if mathematics were trash? Is mathematics of no real importance to you? Is it just junk with no meaning or great purpose? Do you see math as garbage? If not, then why lie? I am serious. If you actually value mathematics, why lie about it? James Harris ==== Subject: Re: JSH: Even shorter, disproof > >> a^2 - (x - 1)a + 7(x^2 + x) = 0 > >> And we have many many cases where neither root is divisible >> by 7 nor coprime to 7. > >> E.g. Take x=2 > >> We get > >> a^2 - a + 42 = 0 > >> (note the middle term does not have 7 as a factor) > >> The roots are > >> a_1 = (1+sqrt(-167))/2 and a_2=(1+sqrt(-167))/2 > > ----------------------------------------------------------------- > >> It was noted in a long-distant post by Keith Ramsay that > >> a_1 = (1 + sqrt(-167))/2 is such that > >> a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and > >> 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). > >> Therefore a_1^11 and 7^11 share an algebraic integer factor, >> namely > >> 44555 - 222*a_1. > >> Therefore a_1 and 7 share the algebraic integer factor > >> (44555 - 222*a_1)^(1/11). > >> Note that 44555 is divisible by 7, but a_1 is not (in the >> algebraic integers). Therefore this common algebraic integer >> factor is not divisible by 7. > >It's not divisible by 7 in the ring of algebraic integers. > > Yes - that's what I said. Neither a_1 nor a_2 is divisible > by 7, and neither is coprime to 7. Since I gave an explicit > common factor, this is not a case where you depend on > Dedekind's theorem (on ideals) to show that a common factor > exists. Therefore your claim that you have found a problem > with the theory of ideals does not apply. > > But I've proven that the ring of algebraic integers is incomplete. > > The belief that it is complete relies on a false claim of proof using > the theory of ideals made by Dedekind, brining the theory of ideals > into question. > >You keep going in circles around that central point. > >My counter would be that one solution of the 11 generated by that >radical expression as it is taking the 11th root, does have 7 as a >factor. > > All you need to consider here is a_1. Since it does not have 7 > as a factor and it is not coprime to 7, and since a_1 * a_2 is > divisible by 7, both a_1 and a_2 are not coprime to 7 and not > divisible by 7. > > That's the case IN THE RING OF ALGEBRAIC INTEGERS where I can show that > you get a contradiction with that ring, as I did in reply to you, quite > simply. > > You can do the same thing with 2 and 6 in the ring of evens, where one > person can claim that 2 is not coprime to 6, while the other keeps > claiming it is, while they remain in the ring of evens. > > You cannot disprove my claims by relying on results in the ring of > algebraic integers, any more than someone could prove that 2 is coprime > to 6 by remaining in the ring of evens. > > It just doesn't fly. > >But rather than argue that point let me show you the simple way to see >a contradiction if I'm wrong, as I just figured it out. > >> Which proves very explicitly that Harris is wrong. The algebraic >> integers are NOT incomplete. > >> Marcus > >They are, as this simple algebra shows. > >I have from before > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > >from which I can easily get > >7x^2 + (7-a)x + a^2 + a = 0 > >and this time I'll choose an integer value for a, like a=3, so that I >have a non-monic polynomial irreducible over Q for x. > >But because 3 is coprime to 7 I know that the other solution for 'a' >MUST have 7 as a factor, right? > > No - you are now considering this as a polynomial in x. You > chose a value for a, and then you solve for x. You have to > decide one at a time which variables you are treating as constants > and which as 'unknowns'. Here you are treating x as the unknown, > and a as a constant. > > That's not mathematics! Obviously if one of the choices for 'a' is 3, > then there is a resultant x, which must fulfil the conditions of all > the equations. > >But now I just go back to > >a^2 - (x - 1)a + 7(x^2 + x) = 0 > >and note that for any solution for x, remember x is NOT an algebraic >integer, where x as a ratio of algebraic integers has factors in common >with 7, while being coprime to some factors of 7, it is NOT possible >that the a's have 7 as a factor, which is a direct contradiction with >one of the a's being 3, and thus forcing the other of the a's to have 7 >as a factor. > > No - you're just confusing yourself on this. You either choose > x and solve for a, or you choose a and solve for x. You don't get > to do both at the same time. > > Um, do you think that if you do one you don't do the other? > > That is, given > > a^2 - (x - 1)a + 7(x^2 + x) = 0 > > and a choice for 'a', you are claiming that you can't use that solution > with > > 7x^2 + (7-a)x + a^2 + a = 0 > > the same exact equation, just written in a slightly different way? > > With the second and a choice for one of the a's that is an integer and > coprime to 7, where a^2 + a is coprime to 7, solutions for x must fit > the same equation. > > damn equation! > > If math people are going to go this far then I wonder how there can be > any hope. > > You throw algebra away as if it were trash. > You know, the bottom line on this is the following. You say > that in the algebraic integers, one root of > a^2 - a + 42 = 0 > *should* have 7 as a factor, while the other is coprime to 7. > And that's just NOT true, as it's possible to prove that they do not. > I can then show a contradiction with the ring of algebraic integers, > which is equivalent to showing a contradiction with someone in the ring > of evens claiming that 2 is coprime to 6. > The analogy should not escape you. > Yet you know, and we all know, that a theorem of Dedekind > and a theorem in Galois theory both say that that is not true. So > you concluded that Dedekind's theorem is false and Galois theory > is wrong. You also concluded that the algebraic integers are > incomplete in the sense that they do not permit a factorization > of the kind that you want. > That's not true. I prove that you get the appearance of a > contradiction if you assume that the ring of algebraic integers is > complete. > That proof is so trivial, you just ignored it in your reply, though at > least you didn't delete it out! > By picking an 'a' that is an *integer* and coprime to 7, like 3, I can > easily show a contradiction with standard teaching, and I see you are > just blissfully trying to ignore the proof. > I have a question for you if you wish to maintain that you can only use > those equations once, how do you suppose that knowing 3 is one of the > a's, you would get the other one? > But I showed explicitly, without any appeal to either Dedekind's > theorem or Galois theory, that neither root is coprime to 7 and > neither root is divisible by 7. Both roots have algebraic integer > factors in common with 7. Factors of 7 are distributed across > both roots. > That is true IN THE RING OF ALGEBRAIC INTEGERS so around and around we > go. > Just like if I were arguing with someone claiming that 2 is coprime to > 6 who stayed in the ring of evens, repeating over and over again that 2 > is coprime to 6 as they can prove--but needing me to keep reminding > that the ring is evens!!! > It's exactly the same thing. > You say this cannot happen in the algebraic integers. But I > No I DO NOT!! Why make a false claim? > The very point I make is that it's true in the ring of algebraic > integers proving that ring is incomplete!!! > Can you hear yourself? > Take away your personal feelings. Forget about me. > This is not personal. What's correct is what's important. > Focus on the math, and not the social stuff. > showed it can. You can verify it by just carrying out the arithmetic. > It is just a matter of brute calculation, just as true as 2 + 2 = 4. > Therefore what you claimed is wrong. Therefore you have an > error which you have not yet recognized. Getting mad at me > for pointing this out is rather unproductive, don't you think? > Shouldn't you be trying to find the mistake in your reasoning? > Marcus > I'm not mad at you. I am a bit exasperated yes, but look at what > you're doing! > In one reply you claim I can't use the same equation twice, making up > some bizarre non-math rule. > Here you claim that I make a claim that I do not, relying on proof of > something being true in the ring of algebraic integers when I > repeatedly explain why that's specious, and you end by just declaring > that I am wrong. > Where in your training as a math student did they teach you to make up > things that are not true, deny mathematical proof, and ultimately just > state something that you should realize is mathematically > false--because the proof is in front of you and EASY--but instead you > just go with your own opinion as if mathematics were trash? > Is mathematics of no real importance to you? Is it just junk with no > meaning or great purpose? > Do you see math as garbage? > If not, then why lie? I am serious. If you actually value > mathematics, why lie about it? It would help if you actually stated a theorem. My impression is that you think that one of the roots should be divisible by 7, while the other one should be coprime to 7. You appear to know that this is not true in the algebraic integers, but for some reason you think it should be. Your belief that one root should be divisible by 7 in the algebraic integers either is true or it isn't true. If it isn't true, then you agree with Dedekind and Galois and you do not have anything that contradicts either of them. Both of their results apply in the ring of algebraic integers. They do not apply to some other rings. If you believe that one root MUST BE divisible by 7 in the algebraic integers, then you are wrong. I showed that without any recourse to using Dedekind's theorem or Galois theory. It is just calculation. If you believe that some other ring is better than the ring of algebraic integers because it does have the property that you want, who cares? Dedekind and Galois probably do not apply in that ring. You have not found a problem with their theorems or with other theorems. Here's the summary. Your belief that one root should be divisible by 7 is meaningless. Should be is not a mathematical statement. It's just an expression of your preference. But the math doesn't care what you prefer. Should be is not the same as Has to be, unless you actually have a proof that it has to be. But you don't. In the algebraic integers, you CAN factor 7 out of the product of the two polynomial factors without factoring 7 itself out of either one. I showed that. But it is not done in the way you _prefer_. Too bad, Harris. Your _preference_ is not real math and happens not to be the way things work. Should be in math is like being close in playing darts. It doesn't count. Marcus > James Harris ==== Subject: Re: JSH: Even shorter, disproof > It was noted in a long-distant post by Keith Ramsay that > a_1 = (1 + sqrt(-167))/2 is such that > a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and > 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). > Therefore a_1^11 and 7^11 share an algebraic integer factor, > namely > 44555 - 222*a_1. > Therefore a_1 and 7 share the algebraic integer factor > (44555 - 222*a_1)^(1/11). > Note that 44555 is divisible by 7, but a_1 is not (in the > algebraic integers). Therefore this common algebraic integer > factor is not divisible by 7. This can't be quite right, as (44555 - 222*a_1)*(44555 + 222*a_1) can't be a rational integer. (a_1)^11 = (44444 - 111*sqtr(-167)) * (-27781 - 2017*sqrt(-167))/2 which is indeed (44555 - 222*a_1) * (-12882 - 2017*a_1) but 7^11 = (44444 - 111*sqrt(-167))*(44444 + 111*sqrt(-167)) = (44555 - 222*a_1) * (44333 + 222*a_1) So (44555 - 222*a_1) remains a common factor of 7^11 and (a_1)^11. -- David Hartley ==== Subject: Re: JSH: Even shorter, disproof - Hide quoted text - - Show quoted text - > It was noted in a long-distant post by Keith Ramsay that > a_1 = (1 + sqrt(-167))/2 is such that > a_1^11 = (44555 - 222*a_1)*(-12882 - 2017*a_1) and > 7^11 = (44555 - 222*a_1)*(44555 + 222*a_1). > Therefore a_1^11 and 7^11 share an algebraic integer factor, > namely > 44555 - 222*a_1. > Therefore a_1 and 7 share the algebraic integer factor > (44555 - 222*a_1)^(1/11). > Note that 44555 is divisible by 7, but a_1 is not (in the > algebraic integers). Therefore this common algebraic integer > factor is not divisible by 7. > This can't be quite right, as (44555 - 222*a_1)*(44555 + 222*a_1) can't > be a rational integer. > (a_1)^11 = (44444 - 111*sqtr(-167)) * (-27781 - 2017*sqrt(-167))/2 > which is indeed (44555 - 222*a_1) * (-12882 - 2017*a_1) > but 7^11 = (44444 - 111*sqrt(-167))*(44444 + 111*sqrt(-167)) > = (44555 - 222*a_1) * (44333 + 222*a_1) > So (44555 - 222*a_1) remains a common factor of 7^11 and (a_1)^11. David, You're right, a misprint. Fortunately the important factor was correct. Marcus -- David Hartley ==== Subject: Re: JSH: Even shorter, disproof days. My association with the Department is that of an alumnus. >I guess that I shall soon become one of those who no longer respond to > In contrast, the rule is that to be an algebraic integer a number > must be a root of a monic polynomial with integer coefficients, which > importantly DOES NOT FOLLOW from simply defining algebraic integers > to be roots of monic polynomials with integer coefficients. >I think that this will become one of those JSH's immortal sentences. I think I see here what is going on there: there is a or two word missing. Algebraic integers are defined as roots of monic polynomials algebraic results, that an algebraic integer must be the root of a monic IRREDUCIBLE polynomial with integer coefficients. This particular restriction is not immediately contained in the definition, and some people had, at some time in the not-so-distant past, unwilling to grant the latter. -- ' ==== Subject: Re: JSH: Even shorter, disproof >> I guess that I shall soon become one of those who no longer respond to >> In contrast, the rule is that to be an algebraic integer a number >> must be a root of a monic polynomial with integer coefficients, which >> importantly DOES NOT FOLLOW from simply defining algebraic integers >> to be roots of monic polynomials with integer coefficients. >> I think that this will become one of those JSH's immortal sentences. > I think I see here what is going on there: there is a or two word > missing. Algebraic integers are defined as roots of monic polynomials > algebraic results, that an algebraic integer must be the root of a > monic IRREDUCIBLE polynomial with integer coefficients. So that's where the problem is! I had a different conjecture: that the problem was that, given a monic polynomial whose coefficients are algebraic integers, its roots must also be algebraic integers. True, but not obvious. > This > particular restriction is not immediately contained in the definition, > and some people had, at some time in the not-so-distant past, > unwilling to grant the latter. I wonder who you might be talking about... :-) Jose Carlos Santos ==== Subject: Re: JSH: Even shorter, disproof I get the impression that you're new here - in any case, I don't recall seeing you responding to JSH until recently. You should know that many people have been down the path that you're on, taking JSH seriously, patiently and exhaustively detailing the errors in his mathematical efforts, trying to explain them to him in the clearest possible terms. Most of these people no longer respond to JSH. They eventually tired of being rewarded with insults for attempting to educate him. You see, if you disagree with JSH, and in particular if you do it a very reasoned way, he will accuse you of being a liar. It does you credit that you persevere, but I predict that you'll soon discover how fruitless is this effort. Gib ==== Subject: Re: JSH: Even shorter, disproof > I get the impression that you're new here - in any case, I don't recall > seeing you responding to JSH until recently. You should know that many > people have been down the path that you're on, taking JSH seriously, > patiently and exhaustively detailing the errors in his mathematical > efforts, trying to explain them to him in the clearest possible terms. > Most of these people no longer respond to JSH. They eventually tired of > being rewarded with insults for attempting to educate him. You see, if > you disagree with JSH, and in particular if you do it a very reasoned > way, he will accuse you of being a liar. I guess that I shall soon become one of those who no longer respond to In contrast, the rule is that to be an algebraic integer a number must be a root of a monic polynomial with integer coefficients, which importantly DOES NOT FOLLOW from simply defining algebraic integers to be roots of monic polynomials with integer coefficients. I think that this will become one of those JSH's immortal sentences. Jose Carlos Santos ==== Subject: Re: JSH: Even shorter, disproof > What do you think, you get some prize for using bogus math? It's obvious you don't care about math; you just care about some award that you'll never get, aside from World's Biggest Crank. Dave ==== Subject: Linear diophantine word problem suggestions days. My association with the Department is that of an alumnus. I am looking for a word problem or two that devolves into solving a linear diophantine equation (i.e., an equation of the form ax+by = c). I know one, which I mention below, but I can't seem to find others that do not rely on tricks (e.g., problems about horses and geese in a farm that can be solved by 'counting legs' and similar). The problem I know is: At a small town, a movie theater charges $5 admission to men, $2 to women, and 10 cents to children. One afternoon, the owner goes to check and finds that there are $100 in the box office. When he goes in, he can count in the darkened theater 100 heads, but cannot tell how many are men, how many women, and how many children. How many of each are in the theater? The problem has a unique solution in positive integers, which can be found after eliminating one variable from the two obvious equations, solving the resulting linear diophantine equation, and finding the only solution where both variables are nonnegative. You can either post or e-mail me any suggestions you might have by way -- ' ==== Subject: Re: Linear diophantine word problem suggestions >I am looking for a word problem or two that devolves into solving a >linear diophantine equation (i.e., an equation of the form >ax+by = c). >I know one, which I mention below, but I can't seem to find others >that do not rely on tricks (e.g., problems about horses and geese in a >farm that can be solved by 'counting legs' and similar). >The problem I know is: > At a small town, a movie theater charges $5 admission to men, $2 to > women, and 10 cents to children. One afternoon, the owner goes to > check and finds that there are $100 in the box office. When he goes > in, he can count in the darkened theater 100 heads, but cannot tell > how many are men, how many women, and how many children. How many > of each are in the theater? >The problem has a unique solution in positive integers, which can be >found after eliminating one variable from the two obvious equations, >solving the resulting linear diophantine equation, and finding the >only solution where both variables are nonnegative. >You can either post or e-mail me any suggestions you might have by way The Sylvester/Frobenius postage stamp problem: given an unlimited supply of postage stamps of certain denominations, find the largest integer number of cents you can't make using these stamps. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: CSP 5.6 : make your Cutting stock patterns in less than a second!!! CSP has these features : makes an optimum cutting pattern in just a second!! imports your SAP and ETABS results for best perfomance in making the best pattern from your and exports to MS Word,MS Excel and HTML for publishing your results in high standards save more and more ... For more inormation check up www.iranciviltools.com ==== Subject: Re: about volume under geometrical transformations? > Why? Do you have a proof? > Hi all, > > I am wondering about the following: > > given a set B={x | x satisfies a constraint}, > this > gives a region of > all possible vector x in R^p space. And let's say > is closed. > > Now I want to do a transformation, C={y | y=A*x > for > all x in set B}, > > so now C is a new closed region. > > What is the relation between the volume of B and > the > volume of the new > C? > > -------------------------------------------- > > Suppose B={x | x'*F*x <= c}, where ' denotes > the > matrix > transposition. > > I obtain C by transformation: C={y | y=A*x for > all x > in set B}, > > Want to know the metric of the set C, > > for example, if C is a one dimensional set -- an > interval, I want to > know its length, its min, and its max, etc. > > if C is a two dimension set, I want to know its > area; > > if C is more than 2D, I want to know its > volume... > > is there a way that I can compute all these > things? Vol(C) = |det(A)|*vol(B) - MO Look for Jacobian in advanced calculus and differential geometry textbooks. In these texts you will find formulas for converting volume elements in integrals from one set of parameters to another. The Jacobian provides the link. - MO ==== Subject: Re: about volume under geometrical transformations? >> Hi all, >> I am wondering about the following: >> given a set B={x | x satisfies a constraint}, this >> gives a region of >> all possible vector x in R^p space. And let's say B >> is closed. >> Now I want to do a transformation, C={y | y=A*x for >> all x in set B}, >> so now C is a new closed region. >> What is the relation between the volume of B and the >> volume of the new >> C? >> -------------------------------------------- >> Suppose B={x | x'*F*x <= c}, where ' denotes the >> matrix >> transposition. >> I obtain C by transformation: C={y | y=A*x for all x >> in set B}, >> Want to know the metric of the set C, >> for example, if C is a one dimensional set -- an >> interval, I want to >> know its length, its min, and its max, etc. >> if C is a two dimension set, I want to know its area; >> if C is more than 2D, I want to know its volume... >> is there a way that I can compute all these things? >> Vol(C) = |det(A)|*vol(B) >Why? Do you have a proof? The proof is based on the following three facts/rules which define the determinant: 1. The determinant is a multilinear function on the columns (or rows) of a matrix. 2. Swapping two columns (or rows) of a matrix negates its determinant. 3. The determinant of the identity matrix is 1. Here is a summarization of the proof: Suppose M is a matrix whose columns (or rows) are the edge vectors of a parallelopiped P. to orthogonalize the edge vectors of P which preserve its volume do not alter the determinant of M. That is, altering an edge vector so that the altitude on the corresponding side is not changed will alter neither the volume nor the determinant. When the edges of P are parallel to the axes, M is a permuted diagonal matrix. In this case, the volume of P is the product of the lengths of its edges and the absolute value of the determinant of M is the product of its elements. Therefore, the determinant of P is the absolute value of the determinant of M. The sign of the determinant of M depends on whether the permutation of the diagonal mentioned above is even or odd, but this is not important here. Rob Johnson take out the trash before replying ==== Subject: Coefficient Correlation Let f(x, y) = 2, 0Let f(x, y) = 2, 0I have figured out the conidtional means to be (1 + x)/2 0 In , on 02/05/2006 at 06:23 PM, rouben@pc18.math.umbc.edu (Rouben Rostamian) said: >If you think too much about it, it is a bit disconcerting, Not at all. (x + 4x^2) refers to Lambda x (x + 4x^2), i.e., the function mapping x->(x + 4x^2). It's one of many cases in Mathematics where an abbreviation is used that requires reading between the lines. -- 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: Re: A question of axiom of choice >Theorem. Every infinite set has a countable subset. >The proof usually looks like the following: A is infinite. Pick a1 in >A, A-{a1} is not empty. So we can pick a2 in A-{a1}, then A-{a1,a2} is >not empty, and from which we can pick a3. Continue this process... >Where have we implicitly used Axiom of Choice? >How to write a proof rigorously using axiom of choice explicitly? This proof does not use the full axiom of choice, but uses the principle of dependent choices. The easy way to see this is to consider the relation R between finite subsets of A, x R y if x is a subset of y and y has exactly one element more than x. Since for every x there exists a y with x R y, there exists a sequence such that for all n, x_n R x_{n+1}. This last step is the principle of dependent choices. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: axiom of choice and equivalent statements >> Prove the following statements are equivalent: >> (1) axiom of choice; >> (2) for any sets A and B, it holds either A <= B or B <= A. (X <= Y >> means that there is an injection from X to Y) >> (3) Zorn's lemma >> (4) Well-ordering Principle : every set can be well-ordered. >> (5) Every binary relation can be single-valued: Given a binary relation >> R, there exists a function F subset R such that dom F = dom R >> (6) A is a nonempty set, then A^a is nonempty for any ordinal number a. >> I know that (1)(3)(4) are equivalent and have a proof of (3)=>(2). >The simplest solution to your problem is to get from your school library a >copy of Equivalents of the Axiom of Choice. First edition is Rubin & >Rubin. Second addition added someone else whose name I can't recall. The second edition is largely a rewriting and expansion by the same authors; it is self-contained. There is a book by J. E. Rubin and Paul Howard, _Consequences of the Axiom of Choice_. > Norm -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: axiom of choice and equivalent statements >Prove the following statements are equivalent: >(1) axiom of choice; >(2) for any sets A and B, it holds either A <= B or B <= A. (X <= Y >means that there is an injection from X to Y) >(3) Zorn's lemma >(4) Well-ordering Principle : every set can be well-ordered. >(5) Every binary relation can be single-valued: Given a binary relation >R, there exists a function F subset R such that dom F = dom R >(6) A is a nonempty set, then A^a is nonempty for any ordinal number a. >I know that (1)(3)(4) are equivalent and have a proof of (3)=>(2). What is the precise meaning of A^a? I am unaware of a version of (6) which is true with choice and false without. Proofs of (2) => 4 typically use Hartogs' function. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Cofinite topology on the reals Hello there, Have a question about the cofinite topology on the reals. i know if a subset is open then it is a nhd of each of its points (and the nhd -neigbhourhood, depends on the actual topology) Im not sure how i should think about this, under this topology as when i have the usual topology on the reals, the interval (0,1) there exists an epsilon >0 s.t for all pts in (0,1) there exists a ball of radius epsilon which is contained in the interval for example (back to the cofinite topolology) say i have the open set X=R/{0} now for any pt in this set, i can find open sets which are contained in (0,1) it just doesnt feel as natural as with the usual topology, so im wondering how should i think about if a subset is open then it is a nhd of each of its points in this topology? ==== Subject: Re: Cofinite topology on the reals > Have a question about the cofinite topology on the reals. > i know if a subset is open then it is a nhd of each of its points (and the > nhd -neigbhourhood, depends on the actual topology) > Im not sure how i should think about this, under this topology > as when i have the usual topology on the reals, the interval (0,1) there > exists an epsilon >0 s.t for all pts in (0,1) there exists a ball of radius > epsilon which is contained in the interval This is false. You can't use the same epsilon for every point. > for example (back to the cofinite topolology) say i have the open set > X=R/{0} > now for any pt in this set, i can find open sets which are contained in > (0,1) Not a non-empty one! A subset of (0,1) is missing infinitely many points of R, so if non-empty it can't be open in this topology. > it just doesnt feel as natural as with the usual topology, so im > wondering how should i think about if a subset is open then it is a nhd of > each of its points in this topology? I think the problem is with your understanding of the word neighborhood. A neighborhood of a point is just an open set that contains it. That's all. I think you must have in mind some definition using open balls, but this has nothing to do with the cofinite topology, which is defined only in terms of cardinality, ignoring all the order and metric properties of R. When considering a topology radically different from the standard one, don't expect specific properties of the neighborhoods you're used to to carry over. ==== Subject: Re: Cofinite topology on the reals > Hello there, > Have a question about the cofinite topology on the reals. > i know if a subset is open then it is a nhd of each of its points (and the > nhd -neigbhourhood, depends on the actual topology) > Im not sure how i should think about this, under this topology > as when i have the usual topology on the reals, the interval (0,1) there > exists an epsilon >0 s.t for all pts in (0,1) there exists a ball of radius > epsilon which is contained in the interval > for example (back to the cofinite topolology) say i have the open set > X=R/{0} > now for any pt in this set, i can find open sets which are contained in > (0,1) it just doesnt feel as natural as with the usual topology, so im > wondering how should i think about if a subset is open then it is a nhd of > each of its points in this topology? This definition of open is not useful in general topological spaces - in fact it is circular. A neighbourhood of a point x in a topological space X is any subset N of X such that there exists an open set U with x in U subset N. Neighbourhood is defined in terms of open, rather than the other way round. If U is any open set and x in U then U is a neighbourhood of x and so trivially U contains a neighbourhood of each of its points (viz. U itself). ==== Subject: Graph Theory Help on Proofs Could someone help me improve or make these proofs stronger for the question: Let G be a graph with average degree d. Then there exists a subgraph of G with minimum degree at least d/2. proof: If the minimum degree of G is at least d/2, one is done. Otherwise, choosing a vertex v in V(G) with deg(v) < d/2, the induced subgraph G-v has average degree at least d. Continuing with this procedure if necessary, will result in the desired subgraph. question: Let G be a graph with average degree d. Then there exists a bipartite subgraph of average degree at least d/2. proof: If the minimum of G is at least d/2, then we are done. Otherwise, choosing a random vertex in G, the bipartite subgraph G-v has average degree at least d. Continue with process until result is achieved. ==== Subject: Re: Graph Theory Help on Proofs >Could someone help me improve or make these proofs stronger for the >question: >Let G be a graph with average degree d. Then there exists a subgraph of >G with minimum degree at least d/2. >proof: >If the minimum degree of G is at least d/2, one is done. Otherwise, >choosing a vertex v in V(G) with deg(v) < d/2, the induced subgraph G-v >has average degree at least d. Continuing with this procedure if >necessary, will result in the desired subgraph. This is more or less correct. But to make the induction work properly, you need to prove the stronger statement: Let G be a graph with average degree at least d. Then there exists a subgraph of G with minimum degree at least d/2. Also, you need to justify your claim that G-v has average degree at least d, >question: >Let G be a graph with average degree d. Then there exists a bipartite >subgraph of average degree at least d/2. >proof: >If the minimum of G is at least d/2, then we are done. Otherwise, >choosing a random vertex in G, the bipartite subgraph G-v has average >degree at least d. Continue with process until result is achieved. This proof makes no sense at all. Firstly, why do say we are done is the minimal degree is at least d/2. Where does your bipartite subgraph come from? Secondly, G-v need not be bipartite, so the bipartite subgraph G-v makes no sense. Derek Holt. ==== Subject: Re: Future Prospects for Computer-Assisted Mathematics > While it is true that nano 3d-chips and Chou-Esaki SOI quantum >effects may beat Moore's law, it is far from forgone that Moore's law >has more than a few years to go. > One thing I haven't seen symbolics tackle is chemistry. I mean >Organic Chemistry is just as much a memorisation tangle excercise to >undergrads as calc II (semester 2/4). The Symbolic Synthesis world >hasn't taken off as I'd've expected. I have seen a posting which indicated that Organic Chemistry, at least as the course usually goes, can be learned by learning four basic principles and using them. > The other thing is assymptotic induction. I mean when you have some >numerical stuff that you can solved, it would be cool if the computer >took the extra step and tried to find the closed form, or at least >give you some good guesses to choose from. Stuff like control theory >and partial differential equations needs to be represented more >naturally, less cryptically, so folks can really grab it by the >handles and do something with it. I mean we need innovations like >Einstein Notation for tensors (repeated indices summed), which lets >us get working without needless formalities. For crying out loud, >anything in math that seems like it was written by lawyers, just junk it. The use of mathematical notation made a big difference. Modern mathematical notation for numbers dates to the late 16th century. For functions, Euler is usually given credit. For other things, it was the middle of the 19th century before anything was done, with the exception of geometry as done by Euclid. The idea of symbols which can be used to represent anything, be they nouns, verbs, adjective, sentences, points, numbers, etc., is simple when done generally, and much more difficult when done gradually. This belongs with beginning reading. However, the esoteric words might look odd, but most are not, and are derived from logical application. However, vocabulary is brought in very gradually. > You have to have people in applications specifying the software. >Letting the theoreticals in there too much never helps innovate much. >The folks out there with real life problems are the ones who will >come up with the best ideas. It is unfortunate, but while any sufficiently large computer-language combination can emulate any other, much of the capabilities of the early computers have been made difficult because the designers could not see the need for them. This applies to hardware as well as software. > Symbolics offers a great opportunity to upgrade education. It is not used until middle school, and only in a restricted manner, at the present time. The general concepts, and full mathematical logic, have been taught at the elementary school level; they belong there. However, I despair of even high school teachers being able to learn how to use a formal language. What's >now college level should become mandatory on high school, because the >rote skills aren't needed. (although, for crying out loud, if my bill >is 2.76 and I give the clerk $3.01, I sure wish they wouldn't stare >blankly like I hit them with a brick) This is not so important. While I am good at arithmetic, and would do such a thing as to give a cashier $1.13 for an 88 cent bill, there are many good mathematicians who are poor at arithmetic. Being able to do arithmetic is a useful skill, but adds nothing to understanding. Educators are the dumbest people >in our society (teachers have the lowest SAT scores) and they will >drag their heels in doing this. The educationists who set up the current system, started out with the idea that being with one's age group is more important than learning. Also, they came up with the idea that one can only learn concepts by being gradually brought to that point, and expected to intuit them. This is utter nonsense, promulgated by those who cannot understand the importance of structure and concepts. (I mean factory workers have to be >retaught the trig they supposedly learned in high school just to use a >stupid robot; then what good were their freaking teachers in the first >place) We need factory workers who understand ODEs and PDEs. But a >century ago they did the same thing: >Diane Ravitz 2000 Left Back S&S 0-684-84417-6 >p30 The two most influential educators in the 1890s were Charles W >Eliot, president of Harvard >p31 Eliot urged educators to shorten the grammar school course by >eliminating redundant work in arithmetic and grammar while introducing ^^^^^^^^^ >natural sciences, such as botany, zoology, and geology, as well as >physics, algebra, geometry and foreign languages Redundant work in anything should be eliminated for all except those with weak mentalities. The others are bored. There is ONE reasonable treatment for those; to quote a famous admiral, Damn the torpedoes! Full speed ahead! >p62 According to one popular saying, it didn't matter what children >studied as long as they didn't like it; doing unpleasant things was >supposed to train the will How is this going to produce people who want to learn? ------------ > I once dated a teacher who was a business major - she chose >teaching because it gave her the max time off to work on her >tan. Another one was a social studies teacher and took a test to teach >math because it paid more. No wonder students fear math. Te darn >curricullum is written by freaking lawyers. Gosh, my prep school >couldn't mathc public school benefits though it paid more, so it hired >grad students, and doggonnit, those grad students were more interested Those graduate students, and not just in math, understand the subjects far more than the teachers. Because they understand, they do more than teach memorization and routine; they teach why, not just how. The schools are designed not to even attempt to teach why. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Hadamard product Hi! I'm facing a cute problem about matrix multiplications. I hope I'll be clear :) Let me denote . the matrix multiplication (row/column), and * the hadamard product of two matrices (element per element). I have a coefficient W = e^(-j 2 PI / N) and with this I construct a matrix S constituted by elements of the type W^(i * j). So, S will be (columns and rows are 0-indexed) S={{W^(0*0), W^(0*1), ..., W^(0 * (N-1)}, {W^(1*0), W^(1*1), ..., W^(1*(N-1)}, ... {W^((N-1)*0), W^((N-1)*1), ..., W^((N-1)*(N-1)}} This means that S is symmetric. If I'm wrong just tell me :) Now, I have a vector x = {X_0, X_1, ..., X_(N-1)} and what I want is having a matrix with each row being the product of X_i with W^(i*j), better explained this way: R={{X_0 * W^(0*0), X_1 * W^(0*1), ..., X_(N-1) * W^(0 * (N-1)}, {X_0 * W^(1*0), X_1 * W^(1*1), ..., X_(N-1) * W^(1*(N-1)}, ... {X_0 * W^((N-1)*0), X_1 * W^((N-1)*1), ..., X_(N-1) * W^((N-1)*(N-1)}} The matrix R can be so obtained from an N*N matrix (here I indicate each row) X := {x, x, ... x} with the hadamard product R = S*X What I'm trying to produce, is converting the hadamard product to a normal row/column product, so obtaining a matrix T such that R = S*X = T*X I find it interesting, but I'm not so into these matters... anyway, it's not necessary that these matrices are exactly defined that way. They are free of being transposed, the only real constraint is having as the result the coefficients (under whatever form): X_0 * W^(0 * 0) X_1 * W^(0 * 1) ... X_(N-1) * W^(0 * (N-1)) X_0 * W^(1 * 0) X_1 * W^(1 * 1) ... X_(N-1) * W^(1 * (N-1)) ... X_0 * W^((N-1) * 0) X_1 * W^((N-1) * 1) ... X_(N-1) * W^((N-1) * (N-1)) -- Sensei Part of the inhumanity of the computer is that, once it is competently programmed and working smoothly, it is completely honest. (Isaac Asimov) ==== Subject: Re: Hadamard product On 2006-02-06 19:56:16 +0100, Sensei said: Sorry, correction: > What I'm trying to produce, is converting the hadamard product to a > normal row/column product, so obtaining a matrix T such that > R = S*X = T*X The last line should obviously R = S*X = T.X -- Sensei Part of the inhumanity of the computer is that, once it is competently programmed and working smoothly, it is completely honest. (Isaac Asimov) ==== Subject: Re: I have just rediscovered Fermat's proof of the Last Theorem > I think that dividing the problem into many cases is > actually a trap. For n = 4, maybe Fermat would like > to present a new method. As you know, this method did > not operate in general case. Many people like me > misunderstood that proving for n being an odd prime > is easier than for any integer n greater than or > equal 3. This made the discovery being lasted more > than three hundred years. However, many new methods > are created from it. I'd like to follow Your idea: I used to mention in my previous post, that for n=3 there are several methods of proper proof. There are also some for n=5, but less then for n=3 and so on for some optional big prime number we'll come to may be only one proper proof procedure ? But for sure even if Your integer n next to infinity but divisible by 3 so proof for p=3 is covering Your proof for this n ... See: let some x^n + y^n =z^n.....(1) where x;y;z;n>=3 integers but for n = p1p2 once a^(p1p2)= (a^p1)^p2 or (a^p2)^p1 (x^p2)^p1 + (y^p2)^p1 = (z^p2)^p1 .................(2) where once X^p1 + Y^p1 # Z^p1 was already proved for such p1=3 etc. so also for such numbers as X=x^p2; Y=y^p2; Z=z^p2 then once equation (2) with these numbers is false so it can be rewritten also as: (x^p1)^p2 + (y^p1)^p2 # (z^p1)^p2 what means, that for extremely big X;Y;Z we'd extend value p2 also to infinity and so on Q.E.D to FLT if it was only proved for p1 = 3 * * * You agree ? I am still skeptical ... With compliments Ro-bin ==== Subject: Re: I have just rediscovered Fermat's proof of the Last Theorem <13166219.1139254553875.JavaMail.jakarta@nitrogen.mathforum.org> See: let some x^n + y^n =z^n.....(1) where x;y;z;n>=3 integers but for n = p1p2 once a^(p1p2)= (a^p1)^p2 or (a^p2)^p1 (x^p2)^p1 + (y^p2)^p1 = (z^p2)^p1 .................(2) where once X^p1 + Y^p1 # Z^p1 was already proved for such p1=3 etc. so also for such numbers as X=x^p2; Y=y^p2; Z=z^p2 then once equation (2) with these numbers is false so it can be rewritten also as: (x^p1)^p2 + (y^p1)^p2 # (z^p1)^p2 what means, that for extremely big X;Y;Z we'd extend value p2 also to infinity and so on Q.E.D to FLT if it was only proved for p1 = 3 ********************** I was my understanding that no math would be required for this usergroup. ==== Subject: Complex Exponent Ok, I can't seem to solve this one: Let r be a real number. Let z = a+bi be a complex number. Write r^z as x+yi that is, r raised to the power of z would be at point x,y in the complex plain, x,y reals of course. What is the general formula for x and y give r and z? ==== Subject: Re: Complex Exponent > Ok, I can't seem to solve this one: > Let r be a real number. Assumed to be >= 0? > Let z = a+bi be a complex number. > Write r^z as x+yi > that is, r raised to the power of z would be at point x,y in the > complex plain, x,y reals of course. What is the general formula for x > and y give r and z? That's what you're supposed to find, of course. Let me see if I can point you in the right direction. First, use the rule that a^(b+c) = (a^b)*(a^c) to write r^z = r^(x+iy) = (r^x)*(r^iy) The first term is a real raised to a real power. If r is nonnegative, that's a real value. The second term is a real raised to an imaginary power. You should have seen the formula that tells you how to expand e^(iy) into real and imaginary parts: How do you use that to expand r^(iy)? Hint: r = e^(ln r). If r can take on negative values, there are some extra complications in the above. - Randy ==== Subject: Constructing the natural numbers from the reals -- a formal proof Following up on my previous postings, here is a formal proof constructing the natural numbers from the reals (field axioms, etc.) http://www.dcproof.com/ConstructN.html Starting with the field, order and completeness axioms for the real numbers R, I show that there exists unique subset N with properties corresponding Peano's Axioms. Finally, I have the Archimedean Axiom in terms of N as I have constructed it. The above proof was written with the aid of my DC Proof software, a PC-based proof checker. Dan Download my DC Proof software at http://www.dcproof.com ==== Subject: Re: Constructing the natural numbers from the reals -- a formal proof > Following up on my previous postings, here is a formal proof constructing > the natural numbers from the reals (field axioms, etc.) > http://www.dcproof.com/ConstructN.html > Starting with the field, order and completeness axioms for the real numbers > R, I show that there exists unique subset N with properties corresponding > Peano's Axioms. Finally, I have the Archimedean Axiom in terms of N as I > have constructed it. > The above proof was written with the aid of my DC Proof software, a PC-based > proof checker. > Dan > Download my DC Proof software at http://www.dcproof.com But, is it N = {0, 1, 2, 3, ...}, or is it N0 + SA? The second statement is somewhat more succinct. Hey cool, I believe you, except you misspelled Archimedean. I think that's cool that you write your own symbolic proof checker. I'm for that. I can't say that I've actually verified that. So, how do you get the reals from the naturals? Spinoza suggests the naturals are a continuum, and while a finite proof in PA is a finite proof in PA, that theory would still appear to be, as they say, incomplete, except along the lines of Spinoza's continuum as introduced by Dean Buckner some months ago, leading to a variety of metatheoretical statements expressible and thus manipulable, tractable, in terms of symbolic logic. The reals are qualifiedly a field, in their normal, or standard, statement. Ross F. ==== Subject: basic topology question Hi there, have a question ive been working on X is a subset of R^2, where X is defined as X={(x,y)| x=1/2^n, n=1,2........., 0=/ Hi there, > have a question ive been working on > X is a subset of R^2, where X is defined as X={(x,y)| x=1/2^n, > n=1,2........., 0=/ X is a topological space with the topology induced from the usual topology > on R^2 > So i see the graph as vertical lines (height 1) getting closer to 0 and > getting grouped together more > I want to prove that an open or closed subset containing (0,0) MUST contain > (0,1) > i dont really follow why this is the case. (i do see that any open set > around (0,0) must contain infintely many of those vertical straight lines, > but i dont see how that is related to (0,1)) > Any help would be appreciated I guess you read the problem wrong, and what you need to prove is that an subset that is open *and* closed containing (0,0) must contain (0,1). Am I right? ==== Subject: Re: basic topology question >> Hi there, >> have a question ive been working on >> X is a subset of R^2, where X is defined as X={(x,y)| x=1/2^n, >> n=1,2........., 0=/> X is a topological space with the topology induced from the usual >> topology >> on R^2 >> So i see the graph as vertical lines (height 1) getting closer to 0 and >> getting grouped together more >> I want to prove that an open or closed subset containing (0,0) MUST >> contain >> (0,1) >> i dont really follow why this is the case. (i do see that any open set >> around (0,0) must contain infintely many of those vertical straight >> lines, >> but i dont see how that is related to (0,1)) >> Any help would be appreciated > I guess you read the problem wrong, and what you need to prove is that > an subset that is open *and* closed containing (0,0) must contain > (0,1). Am I right? yeah, do you mind telling me the difference, (i thought the whole set and the empty set are ALWAYS open and closed.. (and connected components have this property)) ==== Subject: Re: basic topology question Hi there, >> have a question ive been working on >> X is a subset of R^2, where X is defined as X={(x,y)| x=1/2^n, >> n=1,2........., 0=/> X is a topological space with the topology induced from the usual >> topology >> on R^2 >> So i see the graph as vertical lines (height 1) getting closer to 0 and >> getting grouped together more >> I want to prove that an open or closed subset containing (0,0) MUST >> contain >> (0,1) >> i dont really follow why this is the case. (i do see that any open set >> around (0,0) must contain infintely many of those vertical straight >> lines, >> but i dont see how that is related to (0,1)) >> Any help would be appreciated > I guess you read the problem wrong, and what you need to prove is that > an subset that is open *and* closed containing (0,0) must contain > (0,1). Am I right? > yeah, > do you mind telling me the difference, The difference between OR and AND? > (i thought the whole set and the empty set are ALWAYS open and closed.. (and > connected components have this property)) Yes, the whole set and the empty set are always open and closed, and both of them satisfy that if they contain (0,0) they also contain (0,1). The note about connected components is also true, and in fact what the exercise is asking is proof that (0,0) and (0,1) are in the same connected component. The proof could go something like this: - Assume S is an open and close set that contains (0,0). - Show that there must be an N such that n>N implies that the line (1/2^n,y) intersects S. - Show that S actually contains all those lines. - Show that (1,0) is in S. ==== Subject: Re: What Software to Type Math In? I use Scientific Notebook by MacKichan, which no one has yet mentioned here. I used it to take notes throughout Calculus without ever having to resort to pencil and paper. It provides a friendly end for creating Latex files. It depends on your purpose though: if you're looking for submit things for publication, it's probably not ideal, beause it throws in a bunch of stuff which makes it obvious it was created in Scientific Notebook, although that could probably be tweaked. However, if your purpose is to submit a hardcopy paper to someone, Scientific Notebook, works great, imho; you can print directly from the product or output the work as HTML. Here are notes I took throughout Calculus II: http://www.sundrystuff.net/CalculusII/index.htm if you want to see the sort of stuff one can do with it. > What software should I type a math paper in? I have heard microsoft word has something called equation, but I doubt if it would be adequate. I have also heard about Latex, but I think that is a language and not a software product. > Does anybody have any suggestions or favorite programs? > Bijan ==== Subject: Re: What Software to Type Math In? > Did LaTex put something in everyone's coffee? You folks remind me of > religious converts who constantly have to convince themselves they like > their new religion. You protest and protect far too much. Use what > you like and I'll get on with my life never having had a problem with > the equation editor I have. > Fair enough. > My experience. > I have used equation editor (and the somewhat > nicer extention MathType) with mixed results. I find equation editor > very cumbersome to use in technical papers, especially for anything > but simple equations. Furthermore the fact that word is so bad at > producing > technical papers (poor or no equation numbering and references, > what you see is what we want, and don't you dare try to change it , > no easy way to have several people work on one document...) > makes producing technical papers in word problematic. > On the other hand I have found no good solution for power point. > Equations built with equation editor are OK if simple enough, > but far from wonderful. For more complicated equations I typeset > using e.g. Latex and import as bimap (yech!). Two solutions for PowerPoint. The best is to drop it altogether. Use a Macintosh, and Apple's software, called Keynote (part of iWork). Since Mac OS X is essentially Display PDF, it's trivial to embed PDF's in it. And a program like iTeXit allows you to typeset fragments and paste them as PDF's. Alternatively, if you MUST use Windows, use a little program called TeX4PPT, found at Microsoft uses XML as its underlying graphics language, and TeX4PPT converts the TeX'd mathematics into XML. The only problem I've had with it has been when Microsloth changed the specs on PowerPoint. But I think the current version of TeX4PPT is OK for current PowerPoints. The slides class for LaTeX, suggested by someone else, is getting a little long in the tooth. I prefer a package called pdftalk. I think this works on any platform. But Keynote is the best solution. Integration of other formats into it, and fancy tricks available, is much, much better than anything else out there. --Ron Bruck ==== Subject: Re: What Software to Type Math In? <43e514b6$0$3810$ba620e4c@news.skynet.be> <6geau1lcjme96gd7foegs9rfvsabar9k7c@no.spam On the other hand I have found no good solution > for power point. Equations built with equation > editor are OK if simple enough, but far from > wonderful. For more complicated equations I > typeset using e.g. Latex and import as bimap (yech!). Why not use the Latex slide style form? I've given perhaps a dozen talks using this since about 1992, when I started using Latex, although I've probably given even more talks that used nothing more than handwritten transparencies (and I'm not even counting those before 1992, when the only thing I used were handwritten transparencies). Dave L. Renfro ==== Subject: Re: What Software to Type Math In? <43e514b6$0$3810$ba620e4c@news.skynet.be> <6geau1lcjme96gd7foegs9rfvsabar9k7c@no.spam> Please try with MathMagic equation editor. They have free trials for both Mac(OS X & Classic) & Windows. It's got better equation quality, more like TeX equation look. > On the other hand I have found no good solution > for power point. Equations built with equation > editor are OK if simple enough, but far from > wonderful. For more complicated equations I > typeset using e.g. Latex and import as bimap (yech!). ==== Subject: Re: What Software to Type Math In? >> I think folks are missing the point: Why should I change when what I >> use works just fine? Telling me You'll like it! doesn't help at all. >I think you will find that all professional mathematical journals >prefer submissions in Latex, and many accept nothing else. >But, if what you are using is fine for you, why ask about anything else? Again, this is for typesetting. Getting out class notes, and producing easy drafts, cannot be done this way. WYSIWYG, and direct placing the material on the screen, are needed for that. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: What Software to Type Math In? : Again, this is for typesetting. Getting out class notes, and : producing easy drafts, cannot be done this way. WYSIWYG, and : direct placing the material on the screen, are needed for that. I'm not entirely sure why this would be the case. I can spit out a document in TeX far faster than I can boot up microsoft word and deal with all the menu driven crap. This is true no matter what type of document I'm writing. The only thing faster and more convenient is handwritten notes and these of course have their own failing. Justin ==== Subject: Re: What Software to Type Math In? >> I think folks are missing the point: Why should I change when what I >> use works just fine? Telling me You'll like it! doesn't help at all. >I think you will find that all professional mathematical journals >prefer submissions in Latex, and many accept nothing else. >But, if what you are using is fine for you, why ask about anything else? > Again, this is for typesetting. Getting out class notes, and > producing easy drafts, cannot be done this way. WYSIWYG, and > direct placing the material on the screen, are needed for that. I agree that LaTeX has more typing overhead than plain TeX, which is why I prefer the latter for short notes and special formats. But why is WYSIWYG necessary? Do you have some examples available? (no attempt to heckle, I am really curious) Marc ==== Subject: Re: What Software to Type Math In? <43e514b6$0$3810$ba620e4c@news.skynet.be> <040220062056437056%anniel@nym.alias.net.invalid> I think folks are missing the point: Why should I change when what I >> use works just fine? Telling me You'll like it! doesn't help at all. >I think you will find that all professional mathematical journals >prefer submissions in Latex, and many accept nothing else. >But, if what you are using is fine for you, why ask about anything else? > Again, this is for typesetting. Getting out class notes, and > producing easy drafts, cannot be done this way. I don't know if you are pushing for the use of WORD, or whether you have some other tool in mind. However, if it is WORD that you are referring to, I must say that my experiences are the precise opposite of those you describe. Class notes, exams, etc., were always horrible to prepare in WORD, and a snap in LaTeX. Statements like ...let i be the annual interest rate ... were a headache in WORD. Despite all my attempts, I was never able to stop the program from converting the i to an I,which could only be undone manually by backspacing (not deleting) and re-typing. Anything with formulas was a source of high blood pressure. Even when writing co-authored papers, I would produce my suggestions and corrections in LaTeX and let my co-authors go ahead and re-type them in WORD (since that is what the journal wanted). The only times I found WORD easier and cleaner than LaTeX was when producing tables and inserting graphics. R.G. Vickson Adjunct Professor, University of Waterloo > WYSIWYG, and > direct placing the material on the screen, are needed for that. > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: What Software to Type Math In? > I think folks are missing the point: Why should I change when what I > use works just fine? Telling me You'll like it! doesn't help at all. >>I think you will find that all professional mathematical journals >>prefer submissions in Latex, and many accept nothing else. >>But, if what you are using is fine for you, why ask about anything else? >> Again, this is for typesetting. Getting out class notes, and >> producing easy drafts, cannot be done this way. >I don't know if you are pushing for the use of WORD, or whether you >have some other tool in mind. However, if it is WORD that you are >referring to, I must say that my experiences are the precise opposite >of those you describe. Class notes, exams, etc., were always horrible >to prepare in WORD, and a snap in LaTeX. Statements like ...let i be >the annual interest rate ... were a headache in WORD. Despite all my >attempts, I was never able to stop the program from converting the i to >an I,which could only be undone manually by backspacing (not deleting) >and re-typing. Anything with formulas was a source of high blood >pressure. Even when writing co-authored papers, I would produce my >suggestions and corrections in LaTeX and let my co-authors go ahead and >re-type them in WORD (since that is what the journal wanted). The only >times I found WORD easier and cleaner than LaTeX was when producing >tables and inserting graphics. I am definitely NOT pushing for WORD, or in fact any editor which does not place the font/style information in with the text. It should be possible to take a segment of the text and treat it as an entity without all the messy material which is in a hidden section of the binary file. >R.G. Vickson >Adjunct Professor, University of Waterloo >> WYSIWYG, and >> direct placing the material on the screen, are needed for that. -- -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: What Software to Type Math In? <43e514b6$0$3810$ba620e4c@news.skynet.be> <040220062056437056%anniel@nym.alias.net.invalid> In my very first encounter with Word, there were problems because my colleagues were using a different version of Word. This obviously won't be a problem for someone typing and printing only his own documents. ==== Subject: Re: What Software to Type Math In? >: Did LaTex put something in everyone's coffee? You folks remind me of >: religious converts who constantly have to convince themselves they like >: their new religion. You protest and protect far too much. Use what >: you like and I'll get on with my life never having had a problem with >: the equation editor I have. >The difference being that religions are all relative whereas with >typesetting there are some tangible measurements. >Latex is more powerful and more flexible than any other typesetting method >out there. That's not to say that others won't fit the bill in small >applications or if your needs are fairly limited, but if you need the >maximum amount of kick it's Latex or bust, basically. This is why it's >grown to be the default in mathematics, and this is what the original >poster was asking anyway. The problem is that Latex, and the others, are typeSETTIMG methods, and this is always complicated. Very often, one wants typing methods which do not necessarily produce good looking output, but produce readable output quickly, and without using horrible notational strings. Typing 100 characters when 10 will do is not a good idea. I frequently type mathematics without writing it first, and TeX, which I have used, and its derivative Latex, which I have not, just require too much window-dressing. Another advantage of at typewriter rather than a typesetter is that the author has easy control of line breaks. Also, fixed width fonts are necessary for easy communication. This goes completely against the typesetting mentality. >If Microsoft Word, for example, has worked for you then that's fine, but I >wouldn't recommend it because I have no idea of the needs and goals of the >original poster and by recommending Latex I am 99.99% sure he will be >satisfied. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ==== Subject: Re: What Software to Type Math In? <6geau1lcjme96gd7foegs9rfvsabar9k7c@no.spam> In , on 02/06/2006 >The problem is that Latex, and the others, are typeSETTIMG methods, >and this is always complicated. Very often, one wants typing methods >which do not necessarily produce good looking output, but produce >readable output quickly, and without using horrible notational >strings. Typing 100 characters when 10 will do is not a good idea. That's a false dichotomy. There are editors that will generate the proper Tex or LaTex when you want them to. >Another advantage of at typewriter rather than a typesetter is that >the author has easy control of line breaks. That's true only if you never revise your text. If you do revise your text then you have to manually add and delete line breaks, which is not my idea of easy. >Also, fixed width fonts are necessary for easy communication. So are variable width fonts. >This goes completely against the typesetting mentality. There's nothing in any document preparation software that I have ever used that prevents me from using fixed pitch fonts when I want. I've never heard of a typesetter that didn't understand the use of fixed pitch fonts for, e.g., tables. Now, a professional typesetter[1] might disagree with you as to what font to use when, but the typesetter mentality is to use the appropriate fonts for the text at hand, including the use of fixed pitch. In , on 02/06/2006 >Again, this is for typesetting. Getting out class notes, and >producing easy drafts, cannot be done this way. Of course they can be, and have. >WYSIWYG, and direct placing the material on the screen, are >needed for that. Such tools are available for markup languages. [1] I'm not talking about low bidder shlock print shops that aren't willing to pay for professional typesetters. Such places will mangle your test in more serious ways than using the wrong font. -- 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: Re: What Software to Type Math In? >: Did LaTex put something in everyone's coffee? You folks remind me of >: religious converts who constantly have to convince themselves they like >: their new religion. You protest and protect far too much. Use what >: you like and I'll get on with my life never having had a problem with >: the equation editor I have. >The difference being that religions are all relative whereas with >typesetting there are some tangible measurements. >Latex is more powerful and more flexible than any other typesetting method >out there. That's not to say that others won't fit the bill in small >applications or if your needs are fairly limited, but if you need the >maximum amount of kick it's Latex or bust, basically. This is why it's >grown to be the default in mathematics, and this is what the original >poster was asking anyway. > The problem is that Latex, and the others, are typeSETTIMG > methods, and this is always complicated. Very often, one > wants typing methods which do not necessarily produce good > looking output, but produce readable output quickly, and > without using horrible notational strings. Typing 100 > characters when 10 will do is not a good idea. I frequently > type mathematics without writing it first, and TeX, which I > have used, and its derivative Latex, which I have not, just > require too much window-dressing. I wonder, what an alternative typing method should be and which horrible notational strings you have met. Alternating between keyboard and mouse-driven menues does not strike me as particular timesaving alternative. Of course, I do not know what you use now; you may well feel that none of the currently available systems meets your needs. The more fundamental point is, to what extent you can take advantage of the division of labour in the process of typesetting your document. If there is somebody else available who can take care of the typesetting aspect of your document, you may as well say I just want to type it quickly, the _final_appearance_ of the document is not _my_ business. And of course this makes sense. The same applies, if the final appearance does not matter as such, e.g. in a personal not for yourself. (even in this case, I would feel _very_ uncomfortable with a format that is not plain text and where readability depends on the existence of a particular version of a particular software) If instead, the final appearance matters and can not be delegated to somebody else, one may as well start right away with the real thing. As far as I could see, this is to be the situation of the OP. > Another advantage of at typewriter rather than a typesetter > is that the author has easy control of line breaks. Also, > fixed width fonts are necessary for easy communication. > This goes completely against the typesetting mentality. Depending on the final format it might also go completely against the idea of readability. Nobody suggested that you typeset your e-mail. In fact, easy documentation depends on plain text format. This is one of the reasons for using TeX and LaTeX, even if all the fine points of typesetting are ignored. But of course, I have no interest in missionising anybody. You will know much better than I, what you want. Marc ==== Subject: Re: JSH: Headlines around the world > Let's just say for the sake of argument that mathematicians just came > out and acknowledged that through accidents of history and subtlety of > some difficult concepts an erroneous group of techniques became > dominant in number theory. > That news would make headlines around the world. In a few math journals, maybe. No one else would care. Try an experiment: Stop a few people on the street and ask who Andrew Wiles is. You'll get blank stares. Explain that he's the guy who proved Fermat's Last Theorem. You'll get more blank stares. Then try to explain what Fermat's Last Theorem is. By that time they'll be walking away, and if you persist, they'll tell you they don't have time to listen to such silly gibberish. Face, it, James: Even *if* you succeeded in making one of the major breakthroughs you so desperately crave, no one except the Mathematical Establishment you despise would notice or care. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: JSH: Headlines around the world Discussion, linux) >> Let's just say for the sake of argument that mathematicians just came >> out and acknowledged that through accidents of history and subtlety of >> some difficult concepts an erroneous group of techniques became >> dominant in number theory. >> That news would make headlines around the world. > In a few math journals, maybe. No one else would care. > Try an experiment: Stop a few people on the street and ask who Andrew > Wiles is. You'll get blank stares. Explain that he's the guy who proved > Fermat's Last Theorem. You'll get more blank stares. Then try to explain > what Fermat's Last Theorem is. By that time they'll be walking away, > and if you persist, they'll tell you they don't have time to listen to > such silly gibberish. When Andrew Wiles proved FLT, it was reported in the general press. A retraction of not only that result but a hundred years of number theory would also be reported in the general press. (Mind, this is a hypothetical.) -- Memoirists like Frey and Augusten Burroughs belong to the long list of those who should never have stopped using drugs. The drugs might have made Frey more interesting, or they might have killed him. Either way, American literature would have benefited. --John Dolan, www.exile.ru ==== Subject: Re: JSH: Headlines around the world >> Let's just say for the sake of argument that mathematicians just came >> out and acknowledged that through accidents of history and subtlety of >> some difficult concepts an erroneous group of techniques became >> dominant in number theory. >> That news would make headlines around the world. > In a few math journals, maybe. No one else would care. > Try an experiment: Stop a few people on the street and ask who Andrew > Wiles is. You'll get blank stares. Explain that he's the guy who proved > Fermat's Last Theorem. You'll get more blank stares. Then try to explain > what Fermat's Last Theorem is. By that time they'll be walking away, > and if you persist, they'll tell you they don't have time to listen to > such silly gibberish. > When Andrew Wiles proved FLT, it was reported in the general press. Yes, in fact that is the only time I can ever remember any math news ever making the general press. ==== Subject: Re: JSH: Headlines around the world >> Try an experiment: Stop a few people on the street and ask who Andrew >> Wiles is. You'll get blank stares. Explain that he's the guy who proved >> Fermat's Last Theorem. You'll get more blank stares. Then try to explain >> what Fermat's Last Theorem is. By that time they'll be walking away, >> and if you persist, they'll tell you they don't have time to listen to >> such silly gibberish. > When Andrew Wiles proved FLT, it was reported in the general press. I wonder how many people actually saw that news item, and of those who did, how many paid any attention or would remember it now? Not many, I'd guess. I probably pay more attention to that sort of thing than many other non-mathematicians in the general public, having been interested in FLT since first learning of it in junior school in the late '60s. Yet I missed the news when it came out and didn't find out FLT had been proven until a few years later. > A retraction of not only that result but a hundred years of number > theory would also be reported in the general press. Possibly, though I doubt it would gather much attention. > (Mind, this is a hypothetical.) As is any discussion of the results of JSH's accomplishments. :-) -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: JSH: Headlines around the world > there is > over 10,000 years of human civilization There is?? -- Larry Lard Replies to group please ==== Subject: Re: JSH: Headlines around the world >>Nothing else to do on a Sunday afternoon, why _not_ fight >>the future? >So, you're not a football fan either, then? Didn't realize it was Super Sunday until Monday morning when I got a breaking headline email from CNN saying that some team won... Luckily I was born in this country, can't be deported. David C. Ullrich ==== Subject: Re: JSH: Headlines around the world Nothing else to do on a Sunday afternoon, why _not_ fight >>the future? >So, you're not a football fan either, then? > Didn't realize it was Super Sunday until Monday morning > when I got a breaking headline email from CNN saying > that some team won... > Luckily I was born in this country, can't be deported. Obviously you are not one of the advertising fans either. I was fast-forwarding the VCR this morning, looking at beer ads. - Randy ==== Subject: Re: JSH: Headlines around the world > >>Nothing else to do on a Sunday afternoon, why _not_ fight >>the future? > >So, you're not a football fan either, then? > Didn't realize it was Super Sunday until Monday morning > when I got a breaking headline email from CNN saying > that some team won... > Luckily I was born in this country, can't be deported. > Obviously you are not one of the advertising fans either. > I was fast-forwarding the VCR this morning, looking at > beer ads. They're on the web, but most of the sites seem to be too busy this morning. However, this one still works: http://www.godaddy.com/gdshop/superbowl05/landing.asp?se=%2B ==== Subject: Re: JSH: Headlines around the world >>0_ >> `. ___ >> / __>0 >> / / |/' / >> / / ` ,`'--. >> / /(___________)_ >> |/ //.-. .-. >> 0 // :@ ___ @: / >> ( o ^(___)^ o ) 0 >> _______/ / I'm back! I'm back! >> / '._______.'--. >> /| |<_____> | >> __|<_____>____/|__ >> ____<_____>_______/ >> |<_____> | >> |<_____> | >> :<_____>____: >> / <_____> /| >> / <_____> / | >> /___________/ | >> | | _|__ >> | | ---||_ >> | |L/|/ | | [__] >> | |||| | / >>jhs | | / >> |___________|/ > Now that is the best piece of ascii art I have seen in quite a while. > R.G. Vickson Agreed - except for the typo... ==== Subject: Re: Headlines around the world > Let's just say for the sake of argument that mathematicians just came > out and acknowledged that through accidents of history and subtlety of > some difficult concepts an erroneous group of techniques became > dominant in number theory. > That news would make headlines around the world. > Now though, there is quiet, so let's look at the other way this can > work out, as mathematicians can instead try to ignore the result. > Then, as history shows, the result will emerge eventually as there is > over 10,000 years of human civilization where these kinds of battles > have played out, and the side opposing the truth, has always lost. > If the story emerges within a couple of years then clearly > mathematicians will have been gambling on it not coming out within > their careers, and will have lost that gamble. > Why is it a gamble? > Well these things have happened before. In the physics field recently > there was the emergence of quantum mechanic and relativity, where > adjustments had to be made with dramatic consequences for the entire > world. > The physicists absorbed the impact, fought a bit, yes, but quickly came > on the side of truth. > If mathematicians do not, then they will still lose, but they also get > a much darker story, and those mathematicians who are in powerful > positions when the story comes out, will probably take the worst of it. > Social castigation. Their pictures in the papers. Reporters hounding > them with hard questions. > And it won't end during their lifetimes. > These stories are huge on a scale that's hard to comprehend. > They keep coming but people never see them coming. The stories get > bigger with time, but while you're living in it, it can seem > unimaginable. > Simple self-preservation would make some of you sing like canaries at > this point if you had any sense of what is going to come, while > otherwise you have to hold on to the belief that you can play the odds, > play for time, and hope that the story stays buried long enough for you > to have a long career in mathematics, retire, and die with no one ever > knowing the truth. > Let the future handle it, you may think. > But you will not get that time. I'll make sure you don't get that > time. > You will not get to grow old and die with the world thinking you're > something you're not. > The story will come out before then and instead you'll be castigated by > world society. > I'll paint you for what you are--a dangerous element in society > fighting against the foundations of society and technological > progress--and remind that if people like you ever succeed then our > Progress as a species, comes to a screeching halt. > What if physicists had tried and succeeded at what some of you may now > think you can succeed at? What if they'd shut down Einstein's work, > ignored his papers? What if they'd fought quantum mechanics > tooth-and-nail? > What if they'd blocked the knowledge? > Well, I wouldn't be typing this up on this computer as computers > wouldn't be here. Or maybe we'd have some kind of clunky mechanical > computer, but would we have had the transistor, and the technological > revolution? > Or might we have physicists fighting to explain odd behavior within the > framework of the old knowledge, vigorously attacking cranks and > crackpots who attempted to push through the ideas of quantum > mechanics? > You will not succeed in blocking the knowledge. But if you attempt to > do so, when you are broken, you lose so much. > So why bother? > Why not just tell the truth now? > Why fight the future? > James Harris I seriously doubt the average person is going to care about something like this. Most people probably haven't even heard of number theory, let alone know what it is. Dave ==== Subject: Re: creating game theory trees I have created game trees using the graphics part of the statistical package R that is available for free at: http://www.r-project.org/. It takes a little effort to learn how to use R and its graphics facilities, but once you are familiar with it, it is relatively easy to create game trees in R. You can create PDF output, and then import the PDF files for example into Latex. Once you have created a few templates it is easy to modify them to create new trees. ==== Subject: Re: creating game theory trees > Does anyone here know a good way of creating game trees (decision > trees) with Microsoft Word? Or, does anyone know a good > shareware/freeware program to do it, if MS Word is simply deficient on > that score? I've been trying to do it by hand using the Draw feature > and it's just been lousy. use the index/contents then export it as text then use global replace to turn it into 1 1.1 1.2 2 2.1 2.1.1 Use Excel if You have it. ? Which Cat do you want to skin? Microsoft -- The world is flat it's pi that's round! There is only one number. ==== Subject: Re: creating game theory trees trees) with Microsoft Word? Or, does anyone know a good > shareware/freeware program to do it, if MS Word is simply deficient on > that score? I've been trying to do it by hand using the Draw feature > and it's just been lousy. > use the index/contents then export it as text > then use global replace to turn it into > 1.1 > 1.2 > 2.1 > 2.1.1 > Use Excel if You have it. Could you please spell this out a little further? I'm not at all clear as to how this lets me draw/create a tree. ==== Subject: Re: creating game theory trees > Does anyone here know a good way of creating game trees (decision > trees) with Microsoft Word? Or, does anyone know a good > shareware/freeware program to do it, if MS Word is simply deficient on > that score? I've been trying to do it by hand using the Draw feature > and it's just been lousy. Do you want to draw them or generate them? If you want to generate game trees, you're better off using a programming language (Visual Basic, C, Pascal, etc.). --- Christopher Heckman ==== Subject: Cramer-Rao lower bound In statistics, we aim at approximating the actual value t_0 of some parameter using some estimators. The Cramer-Rao theorem then states that a lower bound for the variance of any unbiased estimator is given by the reciprocal of the Fischer information. I now have a stupid question : what about taking the (stupid) estimator being constantly equal to t_0 ? It is clearly unbiased and has zero variance. Doesn't this contradict Cramer-Rao theorem ? Most probably I don't place things in the proper setting, but I really feel very confused. Herve ==== Subject: Re: Cramer-Rao lower bound >In statistics, we aim at approximating the actual value t_0 of some >parameter using some estimators. The Cramer-Rao theorem then states >that a lower bound for the variance of any unbiased estimator is given >by the reciprocal of the Fischer information. I now have a stupid >question : what about taking the (stupid) estimator being constantly >equal to t_0 ? It is clearly unbiased and has zero variance. Doesn't >this contradict Cramer-Rao theorem ? Most probably I don't place things >in the proper setting, but I really feel very confused. This would not be an unbiased estimator. An unbiased estimator is a function of a random variable (with distribution depending on a parameter), such that the expected value of the estimator is equal to the parameter for all possible values of the parameter. The Cramer-Rao inequality, since it involves derivatives with respect to the parameter, assumes the parameter can take values in an open interval. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: the historical misconception of the 4-Color Mapping Problem > Before I departed from sci.math a week ago, I had a lot of loose > strings to tie. One of them was the 4 Color Mapping theorem. > I conceded there was a valid 4 Color Mapping theorem, and also my own 2 > Color Mapping theorem. > The error spot, or the spot in which it was easy to make a error is > that 4 Color Mapping is not a geometry problem but strictly that of > Topology. This is a mistake and error not of me but of the entire rest > of the math community for never really clearing up that error. Once the > error is cleared up, then there is a simple one page proof of the 4 > Color Mapping which ignores borderlines and is strictly in topology and > not geometry, as well as the 2 Color Mapping that includes borderlines > and is in geometry and not topology. > So I have the rest of the mathematics community to blame for never > really making this 4 Color Mapping clear and is the reason why so many > mathematicians in the past history have abandoned working on 4 Color > Mapping is because it was not a clear problem. > The Jordan Curve theorem, JCT, proves both the 4-Color Mapping which > ignores borderlines and resides purely in topology theory and JCT > proves the 2-Color Mapping which resides in geometry and includes > borderlines. > The 4 Color Mapping is simply a statement that when you have 4 > countries that fill 1 continent and that continent is surrounded by > ocean water (5th color is blue for water) that it is impossible to have > all 4 countries own ocean beachfront and that at least one of those > countries is landlocked and thus colored blue. That is ridiculous! What are you really trying to say? And the Jordan Curve > theorem easily proves this problem because the countries have to be > closed figures which raises a contradiction. > The 2 Color Mapping is not topology but geometry and the Jordan Curve > theorem applies also. Only 2 colors are necessary and sufficient to > color all maps in the plane where the borderlines are color blue and > the interiors are white. Proof: JCT. > So, yes, I conceded my mistake of thinking that 4-Color Mapping was > fakery. Because I did not realize until 2005 that it was all in > topology and not geometry. Yet the rest of the mathematics community > needs to stand up and admit their mistake in that they never really > made 4Color Mapping a clear and concise problem of mathematics by > indicating it was not a geometry problem but a topology problem and > that its cousin-- 2 Color Mapping where borderlines are not ignored is > in the realm of geometry and not topology. > So I conceded my mistake, yet the rest of the mathematics community has > not conceded their mistake of a fuzzy and unclear 4 Color Mapping. They > simply needed to state it was topology and that it is connected to 2 > Color Mapping with the Jordan Curve Theorem as the engine of both > proofs. > Considering how awfully arrogant are the people of the mathematics > community, I suspect they will never concede or recognize their big and > awful mistakes about 4 Color Mapping. > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies ==== Subject: Help with parametric curves? Suppose I had any random spatial function. For ex, y(x)=x^2. If I wanted to write this as a function of time instead of space (how would you say that mathematically, make y and x both functions of a parameter t) -- can you just say, ok, let x=t and y=t^2? I mean, that turns it into a function of a variable which I have labeled 't'...is that really a function of time, though? If not, how would I go about this? It seems like this is something that should be really simple and definitely should be contained in my calculus book in the section on parametric curves...but it's not. -n. ==== Subject: Re: Help with parametric curves? >Suppose I had any random spatial function. For ex, y(x)=x^2. If I >wanted to write this as a function of time instead of space (how would >you say that mathematically, make y and x both functions of a >parameter t) -- can you just say, ok, let x=t and y=t^2? I mean, that >turns it into a function of a variable which I have labeled 't'...is >that really a function of time, though? Yes. If x depends on the time t in a certain way, say x = g(t), then y = g(t)^2. Thus you get a parametric curve x = g(t), y = g(t)^2, which describes a point moving on the parabola y = x^2 in a certain way. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Help with parametric curves? > Suppose I had any random spatial function. For ex, y(x)=x^2. If I > wanted to write this as a function of time instead of space (how would > you say that mathematically, make y and x both functions of a > parameter t) -- can you just say, ok, let x=t and y=t^2? I mean, that > turns it into a function of a variable which I have labeled 't'...is > that really a function of time, though? If not, how would I go about > this? How would you go about what? > It seems like this is something that should be really simple > and definitely should be contained in my calculus book in the section > on parametric curves...but it's not. Maybe because the notion of time is entirely irrelevant and does not really exist in the realm of mathematics. Time is either an intrinsic property of the Universe, or an abstraction in our minds that works to model, to visualize some property of the Universe. That our notion of time has aspects that can be directly modeled using mathematics, that's an entirely different aspect. But why would a function (which is a set) *have* to have something to do with time? The notion of time in parametric curves is convenient for the learning process. It's quite easy to visualize for many people if we picture the parameter as being time, and then the curve is nothing else than the trajectory that a point traverses as time moves. But when you think about it, introducing time makes things a little ambiguous -- what's the speed at which the point is moving? Do points moving at different speeds represent the same curve? Why? Or why not? Now, if you're using mathematics to model and solve a physics problem (or whatever other science), then you can choose to use time as the basis for a parametric representation of something -- in parametric representations, the parameter is entirely arbitrary; so you're free to decide that you want to use time as your parameter -- provided, of course, that it does represent what you want to represent. HTH, Carlos -- ==== Subject: Re: LDPC <43e14e5a$0$64361$892e7fe2@authen.yellow.readfreenews.net> though closed solutions were found for SOME of them, there have been tremendous amount of work in their construction, giving them fine mathematical structures based upon techniques such as finite geometry analysis, their applications etc etc which can be explored/shared. ==== Subject: Eigenvalue Need help to solve this problem Let A be an nxn matrix over field K and k_1 , k_2 , ... , k_r be eigenvalues of A . Let f(x) be a polynomial over field K. Show that f(k_1) , f(k_2) , ... , f(k_r) are ==== Subject: Re: Eigenvalue > Need help to solve this problem Let A be an nxn matrix over field K > and k_1 , k_2 , ... , k_r be eigenvalues of A . Let f(x) be a > polynomial over field K. Show that f(k_1) , f(k_2) , ... , f(k_r) are Showing that they are eigenvalues of f(A) is easy---you really should be able to do it, just by using the definitions. I'm not sure if the wording of the problem implies that these are the _only_ eigenvalues of f(A); and if so, this might be harder. R.G. Vickson ==== Subject: Re: Eigenvalue > Need help to solve this problem Let A be an nxn matrix over field K > and k_1 , k_2 , ... , k_r be eigenvalues of A . Let f(x) be a > polynomial over field K. Show that f(k_1) , f(k_2) , ... , f(k_r) are > Showing that they are eigenvalues of f(A) is easy---you really should > be able to do it, just by using the definitions. I'm not sure if the > wording of the problem implies that these are the _only_ eigenvalues of > f(A); and if so, this might be harder. Harder, nothing, it might be impossible, e.g., if f(A) is the zero matrix and A isn't. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) ==== Subject: Re: JSH: Decker example, end of an era <87ek2i1966.fsf@phiwumbda.org > So Galois Theory as a way to determine where factors go is refuted, and > it is proven that the ring of algebraic integers is incomplete[...] > James, I know you're a busy guy and all, but I am still waiting for a > definition of complete here. What does it mean when you say that > the ring of algebraic integers is not complete? Which of Dedekind's > theorems is thereby refuted? And how does this contradict Galois > theory? > Some of us are a bit slow here and could use even simpler > explanations. > Galois Theory has been used to make claims about factors and how they > distribute. > When it comes down to particulars though claims relevant to this > discussion generally are based on whether or not a factor is a factor > in the ring of algebraic integers. > With the Decker example you can see how 7 as a factor of one of the a's > works for quite a few cases, except when x = 1 mod 7 (I've been saying > when x=1 but that's the correct requirement) and when the equation > defining the a's is irreducible over Q, as then neither of the a's can > have 7 as a factor in the ring of algebraic integers. > Completeness or lack of it is best shown with an example, where my > favorite is to give 2 and 6 considering evens a ring, and note that 2 > is coprime to 6 IN THAT RING because 3 is not even. > Similarly, numbers are excluded from the ring of algebraic integers > because they are not roots of monic polynomials with integer > coefficients, which can lead to apparent contradictions, where you can > appear to prove two different and contradictory things. > Back to my 2 and 6 example, it's like being able to prove that 2 is not > coprime to 6, but then someone coming back at you that it IS coprime to > 6, in the ring of evens. > You have a contradiction unless you understand that 3 is excluded so > you are both right. > I can prove 7 as a factor in a way I call algebraically which is > equivalent to being in the ring of objects, which I defined. > Dedekind claimed to have a proof depending on the theory of ideals > which would it turns out prove that the ring of algebraic integers is > complete, and cannot have the problem I outline, like with evens where > 2 and 6 are coprime. > Here completenes means that you couldn't algebraically contradict with > a coprimeness result in the ring of algebraic integers if it were > complete. So that coprimeness in that ring means coprimeness in > general, versus like with evens, coprimeness in the ring, merely > meaning you may have an element that is not in the ring, like 3 is not > even so 2 is coprime to 6. I've left most of this in because it is to my knowledge the most complete... um... well, at least the *longest* explanation we've ever had about the 'completeness' (or lack of it) of the algebraic integers. If I follow the analogy correctly, we have: E (ring of even integers) is not Harris-complete, because E subset Z, and there exist a, b in E such that a and b are coprime in E but not in Z - because a/gcd(a,b) and/or b/gcd(a,b) are not in E, but only in Z ~magic analogy link~ A (ring of algebraic integers) is not complete, because A subset A' (the algebraic numbers), and there exist a, b in A such that a and b are coprime in A but not in A' - because a/gcd(a,b) and/or b/gcd(a,b) are not in A, but only in A' I don't know that any of James' latest set of manipulations actually produce such an a and b - I'm working on the premise that his 'like's are accurate. But then he talks about the common factor being 7, which obviously is in A, so there's something wrong there straight off. -- Larry Lard Replies to group please ==== Subject: Re: JSH: Decker example, end of an era >> This post marks the end of an era in the world of mathematics. >> When will the rest of the world notice that? >> Jose Carlos Santos > A couple of days? I'm not sure, but it's likely to be quick. The > problem with my previous examples were things like very complicated > expressions and the ability of people to find areas where they could I > guess think they saw reason for doubt. If you honestly believe anyone outside sci.math will even *notice* this, much less accept it as correct, then you are even more pathetically deluded than usual. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: limit of multivariables >> How do you find the limit of (x^3-y^3)/(x^2+y^2) as >> (x,y)->(0,0)? >P.S. If anyone could post a simple proof using rectangular >coordinates I'd apprechiate it. ^_^ x^2 <= x^2+y^2 so |x^3/(x^2+y^2)| <= |x| = (x^2)^(1/2) <= (x^2+y^2)^(1/2). Similarly |y^3/(x^2+y^2)| <= (x^2+y^2)^(1/2), so | f(x,y) | <= 2 (x^2+y^2)^(1/2). That is, if (x,y) is within delta of (0,0), then | f(x,y) | is no larger than 2 delta. So we have proved that lim_{(x,y) to (0,0)} f(x,y) = 0 . dave ==== Subject: Re: Well Ordering the Reals Pure Box Theory: >> Let's say I have a box. This box holds red boxes. Now I claim that >> since the first box contains only red boxes, the first box itself must >> be red. Brian Chandler >> I think your box theory would be much more convincing with large >> instead of red. Everyone (including Tony) can see that a shed of >> green apples isn't necessarily green*. But obviously, only a large box >> can contain a large box. > Trying to use color as a property in this analogy is David's attempt to make me > look stupid and obscure the situation. It's typical false QD accusation, used > in desperation. See, Brian, Tony thinks any set of finite naturals must be finite itself. My analogy was that any box containing red apples must be red itself. Needless to say, Tony does not like this logical analogy. ==== Subject: Re: Well Ordering the Reals David R Tribble said: >> Uh, no. I was describing your binary infinite naturals as >> p-adics. >> Despite the fact that they are distinctly NOT the p-adics? What is >> the point of that obfuscation? Virgil said: >> HOW are they different form p-adics? You keep claiming a difference that >> is not immediately obvious. > It's not immediately obvious that there are limit points which define digits at > infinite positions? Do the T-riffics have ellipses to the left or in the > middle, or both. Where are the ellipses in the p-adics? Define immediately > obvious, and then explain how 1:000...000 and ....1111 look the same to you. Well, ...111 is an infinite number because it has only one end, and does not have a last digit at the other end. Your T-riffic 0:111...111 has two ends, yet you claim it somehow also has an infinite number of digits, which is a contradiction in terms. ==== Subject: Re: Well Ordering the Reals David R Tribble said: >> But you are still willing to say oo+1/2 > oo, right? Is oo+1/2 < oo+1? >> Under some circumstances. Like I said, there are two ways to view the number >> circle. Virgil said: >> No one else is viewing a number circle, as such a representation >> destroys the order property of the standard reals. > http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html Apparently you did not notice that oo+1=oo in R*. So adding oo to the real number line does not change its behavior; it still does not act like a finite number.