Say curve K ed quadraticly by (from random sets of (x,y) in space) parabola A that concave up for x= a to b parabola B that concave down for x = b to c The two parabolas join smoothly at b with continuous first degree derivative. At b, C changes concavity, which suddenly causes the tangent vectors to be 'mirrored' against the curve. (Graphicly, instead of being always above parabola A, now the tangents are always below parabola B) This causes problem for me because in 3d, the Normal calculated from the Tangent suddenly become mirrored against the curve. (from the TNB coordinate's view, everything is now upside down after passing the change in concavity). Basicly I need an interpolation method that prevents the TNB system ==== Well, you sound like you want to suck Shaq off, and are jackin your load over getting payton and malone. I AM FROM hollywood unlike you hillbilly types. Don't shoot your load too soon. You got the porno on your webtv system I bet, so you can't even claim ignorance. But claim something, cause your usenet post just sucked rats ass. Perhaps you would like to suck Ronald Reagans cock ==== ermmm this post is very nice. only an artist can appreciate this kind of work. hahahahahaha > Well, you sound like you want to suck Shaq off, and are jackin your > load over getting payton and malone. I AM FROM hollywood unlike you > hillbilly types. Don't shoot your load too soon. You got the porno on your webtv > system I bet, so you can't even claim ignorance. But claim something, > cause your usenet post just sucked rats ass. Perhaps you would like to suck Ronald Reagans cock ==== > Well, you sound like you want to suck Shaq off, and are jackin your > load over getting payton and malone. I AM FROM hollywood unlike you > hillbilly types. Don't shoot your load too soon. You got the porno on your webtv > system I bet, so you can't even claim ignorance. But claim something, > cause your usenet post just sucked rats ass. Perhaps you would like to suck Ronald Reagans cock > Don't you love it when someone makes a post like this but leaves out all the attribution. AOL - still a bastion of idiocy. -- Go Yankees! Go Lakers!! Go Rangers! Go Giants!! Fire Fassle NOW!!! This is very upsetting to me (j/k, mostly) but I just cannot seem to really UNDERSTAND why a number with an irrational exponent exists. I believe it's true, because people much smarter than me say it is, but I don't see why. How can 2 ^ pi, for example, be a number? What does 2 ^ pi MEAN? 2 ^ 0.713 means the thousandth root of two ^ 713. But if the exponent is irrational, isn't it meaningless? It's been explained to me that since you can get closer and closer to the number without reaching it (2^3.14, 2^3.141, 2^3.1415...) it must exist since otherwise it would cause a hole in the rationals...but how can an utterly meaningless symbol like 2^pi be a hole? They see a hole, because it doesn't EXIST and they want it to exist. (Like I said, I DO believe them, I just don't Erin ==== Do you know calculus? Take a constant a, and imagine an object moving on the real line so that its position x(t) at time t obeys x'(t) = a x(t), with starting value x(0)=1. That is, the velocity is proportional to the coordinate. [Imagine a race-car driver starting at marker 1 meter going at speed 1 meter per second; when he reaches marker 10 meters, he should be going at speed 10 meters per second; when he reaches marker 20 meters, he should be goint at speed 20 meters per second; and so on. I used a=1 in this example.] Then the position at time t will be exp(at). That is the MEANING of exp(at). If we take the value a so that x(1) = 2, that is exp(a) = 2, we call that value a=ln(2). Using this value we have x(t) = exp(at) = 2^t for integer (or rational) t, so take this differential equation as the MEANING of 2^t for ALL values of t. If time pi makes sense, then 2^pi is the location of the race-car at that time. Does that help? If you don't know calculus, perhaps you will just have to wait. Every calculus text has a section on exponential growth and decay. As an aside: Sometimes people are confused by COMPLEX NUMBERS as exponents. Well, the same differential equation method works to explain them, too. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== For a given group G does there always exist a group H such that G=Aut(H)? ==== > For a given group G does there always exist a group H such that G=Aut(H)? Not always. For example there is no group H such that Aut(H) is a cyclic group of order 3. Alireza Abdollahi ==== > Not always. For example there is no group H such that Aut(H) is a > cyclic group of order 3. Can you prove this? ==== >> For a given group G does there always exist a group H such that G=Aut(H)? Not always. For example there is no group H such that Aut(H) is a > cyclic group of order 3. I guess if Aut(G) is of prime order p then G/Z(G) is of order p or 1, and so G is abelian. It is easy to deal with _finite_ abelian groups, but does the argument extend to infinite abelian groups? If an infinite abelian group has only a finite number of automorphisms is it necessarily finitely-generated (in which case it would be easily dealt with). I'm sure this is answered at length in Kaplansky's book. -- Timothy Murphy tel: +353-86-233 6090 ==== For a given group G does there always exist a group H such that > G=Aut(H)? >> Not always. For example there is no group H such that Aut(H) is a >> cyclic group of order 3. I guess if Aut(G) is of prime order p > then G/Z(G) is of order p or 1, and so G is abelian. It is easy to deal with _finite_ abelian groups, > but does the argument extend to infinite abelian groups? > If an infinite abelian group has only a finite number of automorphisms > is it necessarily finitely-generated > (in which case it would be easily dealt with). Most abelian groups have an obvious automorphism of order 2 :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > suppose H is a normal subgroup in G and K is normal in H. Of course > then K is a subgroup of G. > But is K normal in G? Proof or counterexample! This problem occured to me some time ago, I don't have time to figure > it out for myself these days so thanks for help! May be that I've overlooked something. But, as K is normal in H it follows that > forall kin K ==> kin H The only thing we gotta check now is whether > g^{-1}kg in G > but as this holds for every h in H by hypothesis and -- by the statement > above -- every k lies H it follows that K is also normal in G. You have to check that g^{-1}kg is in K, not G. The statement is not necessarily true. Consider S_4, the set of permutations of four elements. Let G = A_4, the alternating group of S_4 (i.e. the even permutations). Let H = {1, (12)(34), (13)(24), (14)(23)}, the Klein four-group. Then H is normal in S_4 and hence in A_4. Let K = {1, (12)(34)}. Then K is normal in H. However, K is not normal in A_4, since conjugating (12)(34) by (123) gives (13)(24) which is not in K. Rob ==== In Cantor's Diagonal Argument reals are listed by their decimal representations. However, if the digits in a decimal representation of a real are denoted d_1, d_2...d_n, then the set of numbers that comprise the subscripts of d are integers, and therefore are countably infinite. Assuming the validity of uncountablty infinite, a countably infinite set of subscript numbers is not comprehensive; there must exist decimal places greater than the nth place where n is an integer. As the value of a given real is completed in a countably infinite number of digits, any remaining digits must be zeros. If all reals have some point after which the remaining digits are zeros, diagonalization no longer works. ==== What argument? > In Cantor's Diagonal Argument reals are listed by their decimal > representations. However, if the digits in a decimal representation > of a real are denoted d_1, d_2...d_n, then the set of numbers that > comprise the subscripts of d are integers, and therefore are > countably infinite. OK > Assuming the validity of uncountablty infinite, ? > a countably infinite set of subscript numbers is not comprehensive; comprehensive? > there must exist decimal places greater than the nth place where n > is an integer. Hmmm. This sounds like the dreaded there are infinitely many integers do there are infinite integers argument. What do you mean here? That for each positive integer n there is a decimal place beyond the n-th? (yes there is, for instance, the (n+1)-th). or, that there is a decimal place beyond the n-th for all positive integers n? (no, the decimal expansion is a function from N to {0,..,9}) > As the value of a given real is completed in a > countably infinite number of digits, any remaining digits must be > zeros. Aha, you did mean the second after all! But there are no remaining digits. > If all reals have some point after which the remaining digits > are zeros, They don't. > diagonalization no longer works. ? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > In Cantor's Diagonal Argument reals are listed by their decimal > representations. However, if the digits in a decimal representation > of a real are denoted d_1, d_2...d_n, then the set of numbers that > comprise the subscripts of d are integers, and therefore are > countably infinite. Assuming the validity of uncountablty infinite, > a countably infinite set of subscript numbers is not comprehensive; > there must exist decimal places greater than the nth place where n > is an integer. As the value of a given real is completed in a > countably infinite number of digits, any remaining digits must be > zeros. If all reals have some point after which the remaining digits > are zeros, diagonalization no longer works. A decimal representation consists of a countable number of integers in the range 0..9. If you like you can think of a decimal representation as a function N->{0,1,2,...9} where N is the natural numbers. It's not clear what you mean when you say that a countably infinite set of subscript numbers is not comprehensive. It is true that for any natural number n there are decimal places past n. But each decimal place is indexed by a natural number. ==== >In Cantor's Diagonal Argument reals are listed by their decimal >representations. However, if the digits in a decimal representation >of a real are denoted d_1, d_2...d_n, then the set of numbers that >comprise the subscripts of d are integers, and therefore are >countably infinite. Assuming the validity of uncountablty infinite, >a countably infinite set of subscript numbers is not comprehensive; >there must exist decimal places greater than the nth place where n >is an integer. Uh, right. Just like if the idea of infinite sets is valid then every set must be infinite. Hmm. >As the value of a given real is completed in a >countably infinite number of digits, any remaining digits must be >zeros. If all reals have some point after which the remaining digits >are zeros, diagonalization no longer works. ************************ David C. Ullrich ==== > In Cantor's Diagonal Argument reals are listed by their decimal > representations. However, if the digits in a decimal representation > of a real are denoted d_1, d_2...d_n, then the set of numbers that > comprise the subscripts of d are integers, and therefore are > countably infinite. Assuming the validity of uncountablty infinite, > a countably infinite set of subscript numbers is not comprehensive; > there must exist decimal places greater than the nth place where n > is an integer. Your last statement is incorrect. There are uncountably many real numbers, but there also are uncountably many decimal digit sequences d_1, d_2, ..., where each such sequence is a countable collection of digits. Therefore, there are plenty of digit strings to go around, with no need for uncountably-long strings to represent the real numbers. >As the value of a given real is completed in a > countably infinite number of digits, any remaining digits must be > zeros. If all reals have some point after which the remaining digits > are zeros, diagonalization no longer works. Your premise doesn't hold. There are no remaining digits. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > In Cantor's Diagonal Argument reals are listed by their decimal > representations. However, if the digits in a decimal representation > of a real are denoted d_1, d_2...d_n, then the set of numbers that > comprise the subscripts of d are integers, and therefore are > countably infinite. I follow you up to here. > Assuming the validity of uncountablty infinite, > a countably infinite set of subscript numbers is not comprehensive; > there must exist decimal places greater than the nth place where n > is an integer. As the value of a given real is completed in a > countably infinite number of digits, any remaining digits must be > zeros. ... a bit fuzzy but... >If all reals have some point after which the remaining digits > are zeros, diagonalization no longer works. This point does not exist. Moreover, there are countably infinite sets such as the rationals for which the above statement isn't true either. ==== In Cantor's Diagonal Argument reals are listed by their decimal >representations. However, if the digits in a decimal representation >of a real are denoted d_1, d_2...d_n, then the set of numbers that >comprise the subscripts of d are integers, and therefore are >countably infinite. Assuming the validity of uncountablty infinite, >a countably infinite set of subscript numbers is not comprehensive; >there must exist decimal places greater than the nth place where n >is an integer. Uh, right. Just like if the idea of infinite sets is valid then every > set must be infinite. Hmm. Well, a number such as 1.0 _could_ be expressed in a finite number of places, but because of the validity of carrying out calculations past the zero (infinitely so), this number is thought to have an endless trail of zeros. I am asking: if uncountably infinite is valid, then in the same way we require a (countably) infinite number of decimal places even if the _value_ of a number can be expressed in a finite number of digits, why don't we require an uncountably infinite number of decimal places even though any number's _value_ will be expressed in a countably infinite amount of places? (allow calculations to continue as the new size permits....) dg >As the value of a given real is completed in a >countably infinite number of digits, any remaining digits must be >zeros. If all reals have some point after which the remaining digits >are zeros, diagonalization no longer works. ************************ David C. Ullrich ==== I have answered the following in response the OTHER place where you posted the same comment. To post the same comment twice under two different subject headings, presumably in the hope that one of the postings will remain unanswered and thus appear to be unanswerable, is very bad form. Since I have already given an answer to the other identical posting that you made, I won't bother here, except to point out that I have made an extra remark at the end. >>1. There exists two point masses moving towards each other. >>2. At time T they collide at point P. >>3. At time T two simultaneously paired opposite forces of equal magnitude occur >>at point P. >> This is a contact force. On a side note, in Special Relativity, all >> forces are contact forces. >>4. ________ >>Fill in the blanks. >> Point mass 1 exerts a force on point mass 2, and point mass 2 exerts the >> paired force on point mass 1. The motion of point mass 1 is affected by >> the force exerted on it by point mass 2. Since the force exerted on point >> mass 2 by point mass 1 is not on point mass 1, then it has no effect on >> the motion of point mass 1. While you continue to consider that there is >> no effect of two paired contact forces on a system because they balance, >> then you are not doing Newtonian Mechanics, because you are making an >> assumption which is contradictory to Newtonian Mechanics (specifically, >> it is contradictory to Newton's Second Law of Motion). >You still have it all wrong. >POINT mass A has mass Ma. >POINT mass B has mass Mb. >At time t, a collision occurs at POINT p. >So at time t we observe mass A and mass B existing simultaneously at >point P. An equivalent observation would be that there exists a SINGLE >POINT >mass C with mass Ma+Mb existing at POINT P at time t. >Also at time t, we have the paired force as predicted by Newtons 3rd >law. Given that there is a single POINT mass at point p, then the >superposition principle applies and the paired forces cancel each >other out as predicted. >Do you see the problem yet? >> Newton's Three Laws of Motion: >> 1. If a body has no forces acting on it, then it either remains >> stationary or it moves uniformly. >Empirical evidence implies otherwise. >> 2. The time-derivative of the momentum of a body is equal to the >> sum of the forces which are exerted on the body. >Contact forces can never exist as they are always cancelled out at the >point of contact. >> 3. Forces are paired in such a manner that the forces in a pair >> are equal in magnitude and opposite in direction. The two forces >> in a pair are caused by the same mechanism. The same body >> experiences one of the forces and exerts the other, so that if >> one force in a pair is exerted on body A by body B, then the >> other force in the pair is exerted on body B by body A. >The problem with Newtons laws is that they do not define what a body >is and isn't. Furthermore, Newtons laws do not prohibit the existence >of POINT >masses and - as i've clearly shown - fails to predict their behaviour >(unless logical fallacies constitute strong evidence in academic >circles). As I pointed out when I responded to your other ideration = 7.5058959(93) x 10^87 rad/s^2 > 139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K > 140) current density constant = 7.2263839(39) x 10^92 A/m^2 > 141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 > 142) energy density = 7.3778543(28) x 10^112 kg/m-s^ 2 > 143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr > 144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3