mm-619 -=== Subject: resolving Will's misunderstanding WILLS 1st POST >it's input (recurse as desired), so the whole thing is silly. It would >be like writing a Windows emulator and running it on Windows instead of >Mac/Linux. > Exactly, but how else are you going to perform Barbs modified diagonal function? > G(j) = UTM(j,j) mod 2 + 1, Huh? G(j) is UTM modified. You don't even need the Godel number of UTM to encode it, just modify UTM to read in j, replicate it, run UTM on it (which results in UTM(j,j) ), then mod it and add 1. *************** #1 ************** > Assuming UTM is only total functions. Here's the problem. You are *adding* an assumption. > Also, assemble and compile mean the same thing, its not machine language, particularly > when it uses labels assembly is just compiled. You can't ASSUME a language in a > proof of contradiction, you have to show all possible languages will result in the > contradiction. Raw TMs don't have labels accessible outside of that single function, > its a set of states not a list. If the proof works in a language that is equivalent to the instructions on a TM, it will work for ALL languages that are equivalent to the instructions on a TM. There may be some languages that would generate a contradiction, but that indicates a limitation on the language, not the proof. ************ #2 ************** WILLS 2nd POST > you have some proof otherwise. The only rebuttal was diagonalisation over the > set, which is invalid considering diagonalisation will only work on a function > calling paradigm of algorithms, not more elementary systems. Where did you get this idea? The diagonalization argument works with any method of uniquely identifying a function. Godel numbering comes to mind. Then you *emulate* the function based on its number. ****************** #3 ****************** > You've all tossed away WHAT functions do, and are entrenched with your > theory about WHO the functions are, giving them names and twisting them. The halting problem has no reference to functions. Functions are irrelevent to the problem. Try talking about the halting problem without using the word function, it may help you see what we are talking about. There are only machines with an input and an output. ******************* #4 ****************** > My theory is so trivially simple, I don't see how 1,000 so called computer maths > theorists here are so blind about it. This statement should be a clue to you. > TMS dont have names, you can't link one state in one TM to another state in > another TM and say THIS TM CALLED THAT TM. YOU CANT DO THAT! We never said you did. However, you can think of the Godel number of a TM as its name, and you can tell a TM to emulate another TM's processing. I don't know why you don't understand this. #1 Here Will is saying to make the Ôbroken' function we modify the existing UTM. I've posted the proof that this is not a valid modification several times now. #2 Will says : If the proof works in a language that is equivalent to the instructions on a TM, it will work for ALL languages that are equivalent to the instructions on a TM This is like saying, every contradiction has an analogy in every language. This is clearly wrong. In functional programming there are many subtle paradoxes of reference due the to consise expressive power of the language. As languages get more primitive, the amount of code to perform the same calculation is much larger. Will seems to think that because a TM has a reference, its godel number that the UTM can emulate it with, that functions can call one another in a nested fashion exactly like Pascal. Say TM-333 adds 2 to its argument. Say TM-444 outputs number 7. Does UTM 0000000000033300000444 output 9? No it doesn't. Only the UTM can emulate the function, functions can't call one another. Making up a contradiction using G = lamba x (mod(F(x, x))) is NOT POSSIBLE with a TM. It might be possible depending how easy F is to modify, but its not always possible, F might be defined as mod2(G(x)), G is not free in F. Do these situations arise with RAW TMs? Not at all. #3 Will says : The diagonalization argument works with any method of uniquely identifying a function. Godel numbering comes to mind. Then you *emulate* the function based on its number. This comes just after saying #1, G(j) is UTM modified. You don't even need the Godel number of UTM to encode it, just modify UTM to read in j, replicate it, run UTM on it (which results in UTM(j,j) ), then mod it and add 1. So which is it Will? Can you stick to some method of accessing a function and derive results based on that one method? #4 Rebuttal Will says : Functions are irrelevent to the problem. ... There are only machines with an input and an output. Try looking up a defn of a function. Herc -- oo ____|mn / /_/ / _ - FREE THE TRUeMAN - / K-9/ /_/ - Join www.chatty.net - /____/_____ - Webmasters join www.BannerX.net - === Subject: Re: resolving Will's misunderstanding > WILLS 1st POST >it's input (recurse as desired), so the whole thing is silly. It would >be like writing a Windows emulator and running it on Windows instead of >Mac/Linux. >Exactly, but how else are you going to perform Barbs modified diagonal function? >G(j) = UTM(j,j) mod 2 + 1, > Huh? G(j) is UTM modified. You don't even need the Godel number of UTM > to encode it, just modify UTM to read in j, replicate it, run UTM on it > (which results in UTM(j,j) ), then mod it and add 1. > *************** #1 ************** >Assuming UTM is only total functions. > Here's the problem. You are *adding* an assumption. >Also, assemble and compile mean the same thing, its not machine language, particularly >when it uses labels assembly is just compiled. You can't ASSUME a language in a >proof of contradiction, you have to show all possible languages will result in the >contradiction. Raw TMs don't have labels accessible outside of that single function, >its a set of states not a list. > If the proof works in a language that is equivalent to the instructions > on a TM, it will work for ALL languages that are equivalent to the > instructions on a TM. There may be some languages that would generate a > contradiction, but that indicates a limitation on the language, not the > proof. > ************ #2 ************** > WILLS 2nd POST >you have some proof otherwise. The only rebuttal was diagonalisation over the >set, which is invalid considering diagonalisation will only work on a function >calling paradigm of algorithms, not more elementary systems. > Where did you get this idea? The diagonalization argument works with > any method of uniquely identifying a function. Godel numbering comes to > mind. Then you *emulate* the function based on its number. > ****************** #3 ****************** >You've all tossed away WHAT functions do, and are entrenched with your >theory about WHO the functions are, giving them names and twisting them. > The halting problem has no reference to functions. Functions are > irrelevent to the problem. Try talking about the halting problem > without using the word function, it may help you see what we are > talking about. There are only machines with an input and an output. > ******************* #4 ****************** >My theory is so trivially simple, I don't see how 1,000 so called computer maths >theorists here are so blind about it. > This statement should be a clue to you. >TMS dont have names, you can't link one state in one TM to another state in >another TM and say THIS TM CALLED THAT TM. YOU CANT DO THAT! > We never said you did. However, you can think of the Godel number of a > TM as its name, and you can tell a TM to emulate another TM's > processing. I don't know why you don't understand this. > #1 > Here Will is saying to make the Ôbroken' function we modify the existing UTM. > I've posted the proof that this is not a valid modification several times now. I will be looking at your latest rebuttal in the HALTING DISPROOF thread. > #2 > Will says : > If the proof works in a language that is equivalent to the instructions > on a TM, it will work for ALL languages that are equivalent to the > instructions on a TM > This is like saying, every contradiction has an analogy in every language. This > is clearly wrong. In functional programming there are many subtle paradoxes > of reference due the to consise expressive power of the language. As languages > get more primitive, the amount of code to perform the same calculation is much larger. > Will seems to think that because a TM has a reference, its godel number that the > UTM can emulate it with, that functions can call one another in a nested fashion exactly > like Pascal. Say TM-333 adds 2 to its argument. Say TM-444 outputs number 7. > Does > UTM > 0000000000033300000444 > output 9? No it doesn't. Only the UTM can emulate the function, functions can't > call one another. The UTM is not he only TM that can emulate another. Start with the UTM then append code/states/whatever to further modify what the emulated TM would have produced. You can make use of the existence of the UTM. Now, I do not believe that TM's can call each other. I do believe that you can work with the UTM to create a larger TM that will alternate between one or more TMs, possibly using the output of a TM as its input on the next emulation cycle. Think of it this way, a TM can have a subroutine in much the way that BASIC does, by going into a seperate loop of code. It has no functions, but it can have pieces of code that act like functions. There are no function calls, but there is an internal structure that resembles called functions. > Making up a contradiction using G = lamba x (mod(F(x, x))) is NOT POSSIBLE > with a TM. It might be possible depending how easy F is to modify, but its > not always possible, F might be defined as mod2(G(x)), G is not free in F. > Do these situations arise with RAW TMs? Not at all. Let G be the states g1, g2, g3, g4, ... gm, UTM states, gm+1, gm+2, .., gh Let F be the states f1, f2, f3, f4, ... fn, UTM states, fn+1, fn+2, .., fk G has Godel number G1, F has Godel number F1. What if G lays down F1 on the tape followed by x, x, then enters the UTM states, then (rather than halting) performs lambda? What if F is similarly set up to lay G1 followed by x on the tape, enter the UTM, then perform mod2? This appears to do what you said is impossible. > #3 > Will says : > The diagonalization argument works with > any method of uniquely identifying a function. Godel numbering comes to > mind. Then you *emulate* the function based on its number. > This comes just after saying #1, > G(j) is UTM modified. You don't even need the Godel number of UTM > to encode it, just modify UTM to read in j, replicate it, run UTM on it > (which results in UTM(j,j) ), then mod it and add 1. > So which is it Will? Can you stick to some method of accessing a function > and derive results based on that one method? I concede point 3. A TM can be thought of as a function in a program (note: I'm thinking of a valid C function as an example). I prefer to think of them as programs rather than as functions because a mathematical function has different connotations of behavior from a programming function. Also, I generally associate the word function with the mathematical term used in other branches of mathematics. > #4 Rebuttal > Will says : > Functions are irrelevent to the problem. ... There are only machines with an input and an output. > Try looking up a defn of a function. The issue (having looked it up) is that we are dealing with partially recursive functions, not necessarily total recursive functions. Note that this differs from functions in other contexts where a function is a map from one set to another, regardless how of the mapping is achieved. Now the real question is whether the definitions I have found correspond with what you are thinking of. You appear to want to discuss only the primitive recursive functions, but don't like the idea that there are total functions which are not primitive recursive. prompt for me to brush up on the topic. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Help Needed Understanding Article > > You are trying to take credit for my accomplishment [...] When you said that I was following Type Theory, you were saying that they have what I did. But they don't have anything that I did. They do not synthesize programs and they don't use the 5 principles (questions) that I asked. There is no comparison. I solved the problem of Program Synthesis and they didn't. So saying that I followed them is totally false. If I followed them, then why are you unable to use Type Theory to accomplish what I demonstrate? This is the basic false assertion that you are making. > What good is a system that has not been *proven* to solve the problem? > Such a system may serve a practical purpose (but only if someone made a > computer program out of it), but scientifically it is worthless imho. No offense, but to say that a practical solution to a problem has no scientific value is to not understand the purpose of science. The ultimate goal of science is to solve practical problems (such as building nuclear weapons to solve the problem of over-population.) Actually, the theoretical implications of my axiomatic system are no less than amazing. First of all, how do you think Mathematicians feel about axiomatizing an area of study? As I said, the rules of inference are well-known theorems from the Theory of Computation. Did you look at them? Do you know how the system works? Here is what I say (section III). Here is the first rule, the NOT rule: The procedures for creating the new programs follow. P represents the given program that solves the wff p. [unused references to Q omitted] 1. NOT: Change each write (true) to write (false) and vice-versa. Example: if P write (true) => if P write (false) write (false) write (true) Now, we have a program P that solves predicate p, and the NOT rule creates a program that solves ~p. The procedure applied to program p changes each WRITE (TRUE) to WRITE (FALSE) and vice-versa. The general form of such program P, as shown, is: IF P WRITE (TRUE) / WRITE (FALSE), where P in the program represents any program code. This procedure changes that to: IF P WRITE (FALSE) / WRITE (TRUE). That is the result of applying that procedure. Now, assuming that you read the description of this 4-command programming language, you can see that the new program does in fact solve ~p. The code says that: if p is true then it halts NO (i.e., false). Otherwise it continues and halts YES (i.e., true.) So if p is true then it returns false, and otherwise it returns true. Do you doubt that the new program solves ~p? Similar simple logic applies to each of the 8 rules of inference. They are simple procedures for constructing computer programs, one step at a time. They are formal and the results are represented by Predicate Calculus wffs, as I described. > For example, where is your theorem prover described? I gave a detailed description here of these algorithms, including: 1. The theorem is FAC(I,J). It is line 13 in proof 1 of 2 proofs. This is a predicate calculus wff that is true iff input I is a factor of input J. We EXPRESS the factor of relation. The system attempts to REPRESENT it. 2. Note DEF 5: FAC(a,b) , (eA)MUL(A,a,b) in section IV. The theorem FAC(I,J) in of the form FAC(a,b) in the definition. a is I and b is J. So let a = I and b = J, and form the equivalent wff, which is (eA)MUL(A,I,J). This is line 12 in the final proof. Is that clear? The theorem at this point is FAC(I,J). There is a DEF which has FAC(a,b) as one of the two equivalent wffs. FAC(I,J), the theorem, is an instance of the wff FAC(a,b), where a is I and b is J. So we can substitute I for a and J for b, to create FAC(I,J) from FAC(a,b), and the DEF is FAC(I,J) , (eA)MUL(A,I,J). So we can change FAC(I,J) to (eA)MUL(A,I,J). It is straightforward substitution of variables in wffs. We substitute any terms for the variables a, b, . . . in wff P(a, b, . . .) to apply a DEF of the form P(a, b, . . .) , Q. P(a, b, . . .) represents any wff of that form and by substituting for the variables a, b, . . . we can create specific wffs with terms instead of the variables a, b, . . . Substitution like this is common in logic. Then I continue: 3. Similarly, apply definitions DEF 4, 3, 1 and 2 to produce lines 11, 10, 9 and 8 in the final proof. 4. Line 8 is (eA)~LT(J,A)^(eB)MUL(A,I,B)^EQ(B,J). Note the QUIT Rule (section III): P(x)H -> (eA)P(A). Line 8 is of the form (eA)P(A) which is the conclusion in this rule, where set P(a) is (eA)~LT(J,A)^MUL(A,I,a)^EQ(a,J), i.e., B -> a. Reverse this rule (as described in paragraph 10 at the end of the paper), so that (eA)P(A) -> P(x)H, substituting x for a in P(a) to produce (eA)~LT(J,A)^MUL(A,I,x)^EQ(x,J)H, which is line 7. I am simply describing more instances of substitution in order to apply the rules. It is merely substitution and modus ponens, like any axiomatic system. Except, rather than creating TRUE wffs, I create wffs that that are solved by computer programs that are created in parallel. > In your future research section you > write: What would [...] the Program Calculus do if asked to solve each > of the above wffs ? Well, if you have such a nice automatic system, > why didn't you just *try*? I hadn't programmed it at that point, actually. But you have no Program Synthesis system at all, neither programmed nor even on paper. And what happened to my questions about where Type Theory solves the programming problems and uses the principles that I use, since you said that I am just following Type Theory? I don't think any of what I do is in Type Theory. And besides, how can they claim to have what I have when I show how to synthesize programs and they don't? There is no comparison. You can't make any claims about what Type Theory does because it doesn't produce Program Synthesis in the first place. It is irrelevant as to what it does, because it is not a Program Synthesis system anyway. And, I am still waiting for you to show me how my 2 example programs and 5 principles (questions) are part of Type Theory. Why don't you either answer the 5 questions or just admit that Type Theory does not use the principles that I use, and does not produce the results that I do? Personally, I think it's terrible when people don't give other people credit for what they accomplish. I am glad to give people credit for their ideas. But unfortunately, there are many people who, for whatever reason, seem to resent admitting that someone who is not famous or in a prestigious position has succeeded at what others have not. It definitely is a psychological problem, I firmly believe. When you say that I copied a system that doesn't even accomplish what I do, it makes no technical sense. It is just a desperate attempt to keep credit away from me. I'm sorry, but that is what I see. And you'll never be able to justify your assertions that attempt to deny what I have done. BTW What do you think of the famous stories, The Emperor Has No Clothes or The Wizard of Oz or Alice in Wonderland? The really perceptive people in this world know that the masses are led around by BS artists. I think that it is just people being too lazy to think for themselves, so they identify with the famous people and just assume that whatever they do is the best and, conversely, an unknown person couldn't possibly do anything as well. But they are very wrong. Charlie Volkstorf Cambridge, MA > groente > -- Sander === Subject: Re: Help Needed Understanding Article >What good is a system that has not been *proven* to solve the problem? >Such a system may serve a practical purpose (but only if someone made a >computer program out of it), but scientifically it is worthless imho. > No offense, but to say that a practical solution to a problem has no > scientific value is to not understand the purpose of science. The > ultimate goal of science is to solve practical problems (such as > building nuclear weapons to solve the problem of over-population.) Yes, solving practical problems is one of the ultimate goals of science. But it goes from theory to practice, and not the other way around, as you are trying to do. Other important goals of science are knowledge and understanding. In theoretical computer science proofs play an important role in both: proofs tell you *that* something is true (knowledge) and *why* it is true (understanding). > Actually, the theoretical implications of my axiomatic system are no > less than amazing. There are no theoretical implications of your axiomatic system because there is no theory. All you have is examples, and some not explicitly stated unproved conjectures. > Here is what I say (section III). Here is the first > rule, the NOT rule: Yes, I understand how you translate proofs to programs. >For example, where is your theorem prover described? > I gave a detailed description here of these algorithms, including: [snip explanation of definitions] > Substitution like this is common in logic. Indeed. Apparently you didn't understand my question. > I am simply describing more instances of substitution in order to > apply the rules. It is merely substitution and modus ponens, like any > axiomatic system. No, you have no modus ponens rule in your rules of inference. And you substitution. Do you understand your system yourself? So the question remains, how do you *automatically* select the rule of inference that you're going to use? >In your future research section you >write: What would [...] the Program Calculus do if asked to solve each >of the above wffs ? Well, if you have such a nice automatic system, >why didn't you just *try*? > I hadn't programmed it at that point, actually. But you have no > Program Synthesis system at all, neither programmed nor even on paper. Have you programmed it already, then? If you did, may I see this program? If not, since creating a prototype implementation in a language like Haskell would take about 2 days tops, may I presume you don't have this automatic system at all? > And what happened to my questions about where Type Theory solves the > programming problems and uses the principles that I use, since you > said that I am just following Type Theory? I hate repeating arguments. Besides, most of your criticism of type theory applies to your own system even more than it does to type theory. > Personally, I think it's terrible when people don't give other people > credit for what they accomplish. If you had accomplished anything I would happily give you credit for it. > When you say that I copied a system that doesn't even accomplish what > I do, it makes no technical sense. I didn't say you copied it, I said you follow the same approach. Type theory is just that: a *theory* (or actually a family of theories). And theoretical justification is exactly what lacks in your system. > It is just a desperate attempt to > keep credit away from me. I'm sorry, but that is what I see. You see it plain wrong. Actually, I tried to help you by pointing out points. Fine, that's your right as author. Do what you want. But if you looks like you want this --, it requires much more work. It might even prove to be an interesting achievement, but in its current form it surely isn't. make it better. For example, I, and others, have pointed out to you that how you can fix them. It is a one hour job to fix the fonts, but you haven't done it yet. And if you don't even want to spend one more hour on the paper, how can we expect that you'll work on the other necessary improvements? > I think that it is just people being too lazy to think > for themselves, No, I think it is just *you* being too lazy to prove your own assertions. groente -- Sander === Subject: Re: Help Needed Understanding Article > Yes, solving practical problems is one of the ultimate goals of science. > But it goes from theory to practice, and not the other way around, as > you are trying to do. Oh heavens no. IMHO essentially all creativity in science (and perhaps all of cognition) is going from the particular (practice) to the general (theory.) We start with our 5 senses. We observe physical processes. We formalize it. And we find patterns, which become rules. Then the practical application is the routine substituting of different particular (constant, literal) values and applying the rule to that. In my case, I have taken particular programming tasks, abstracted them into 8 universal Rules of Inference, and show how to apply these rules to create a variety of programs, from Number Theory to Database Query Processing to Theory of Computation theorems. > There are no theoretical implications of your axiomatic system because > there is no theory. All you have is examples, and some not explicitly > stated unproved conjectures. That's not true. What am I lacking? I showed you how my Rules of Inference are valid. I also showed how they are well-known theorems from the Theory of Computation. Tell me what else you need. I'll tell you (again) just one of the many theoretical implications of my axiomatic system. One of the 4 axioms that are used to generate theorems from the Theory of Computation is TRUE(x). Do you know what it means? It means that the universal set is recursively enumerable. And who else said that but not in those words, but in a much more complex, less formal way? Peano's Axioms. My system shows me that Peano's 5 axioms are equivalent to TRUE(x). And from that we can see why a recursive set is recursively enumerable, and even consider more general models of computing than Turing et. al do: a base of computing in which the universal set is not recursively enumerable, and where a recursive set is not always recursively enumerable. (And there's lots more, as well.) To me that is a great achievement, to simplify a well-known part of Mathematics and to generalize (in a constructive way) an existing theory. And then to axiomatize the Theory of Computation. Honestly, anyone who doesn't see the value of that doesn't know about the history of Mathematics and the role axiomatic systems have played. Personally, whenever I see a paper that talks about axiomatizing an area of study, I eagerly read what they are proposing. Are you aware of the rich history of axiomatic systems in Mathematics? > No, you have no modus ponens rule in your rules of inference. And you > substitution. Do you understand your system yourself? The Rules are applied by substituting particular wffs for the variables in the Rules. Then the implication that the Rule states is made, i.e. Modus Ponens. For example, the NOT Rule again: P => ~P. Now, suppose we have axiom LT(I,J) (less than is recursive.) We substitute LT(I,J) for P [SUBSTITUTION] and infer that ~LT(I,J) [MODUS PONENS.] That is, we conclude that not less than is a recursive relation, while at the same time constructing a program that solves wff ~LT(I,J). > So the question remains, how do you *automatically* select the rule of > inference that you're going to use? By following the procedure that I described 2 or 3 messages ago and gave more details to in my previous message (that you snipped above). > Have you programmed it already, then? If you did, may I see this > program? If not, since creating a prototype implementation in a language > like Haskell would take about 2 days tops, may I presume you don't have > this automatic system at all? I don't know Haskell and I don't know what you are talking about taking 2 days. All I can present in these posts is a description of my system, just as others do. You are right (as I've said repeatedly) that it would help tremendously if I had it running on the internet. I would love to (and plan to) do that. But sometimes I spend a lot of time doing other things, like e.g. posting to Google Groups! In the meantime I have a paper describing my system, and I am comparing it to other systems. I will certainly announce when I get it on the internet - believe me (or not.) > I hate repeating arguments. Besides, most of your criticism of type > theory applies to your own system even more than it does to type theory. You never answered the questions and it is not a criticism. It is a response to your saying that I follow Type Theory. Then where does Type Theory solve the 2 programming tasks and use the 5 principles that I listed? Why don't you either substantiate your statement or admit that it's not true? > If you had accomplished anything I would happily give you credit for it. Did I show how to synthesize programs? What is wrong with my paper or posts? I responded to every objection and question that you had. > Type > theory is just that: a *theory* (or actually a family of theories). And > theoretical justification is exactly what lacks in your system. What is wrong with the explanation (informal proof) that I gave that demonstrates that my Rules of Inference are valid? > You see it plain wrong. Actually, I tried to help you by pointing out > points. That's not true. Where did I do that? I did the opposite. I said every question. > make it better. For example, I, and others, have pointed out to you that > how you can fix them. It is a one hour job to fix the fonts, but you > haven't done it yet. And if you don't even want to spend one more hour > on the paper, how can we expect that you'll work on the other necessary > improvements? That's not true. You don't know that I can fix the fonts in one hour. It took me weeks just to write the original paper. Now I have to first. In the meantime, I think that people can still read it, no? I'll gladly send an MS Word version to anyone who wants it. > I think it is just *you* being too lazy to prove your own assertions. That is a lie. I have written more messages, and longer messages, than almost anyone in this forum. You are saying the exact opposite of reality. You are making statements without regard to the truth. :( Charlie Volkstorf Cambridge, MA > groente > -- Sander === Subject: Least Square Error problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4I1cuD18192; I am trying to compute a solution using the least- square- error principle to an overdetermined system of equations. The equation is as below: Error = summation (y[n] summation (ai * x[n-i] ) ).... in Ôai Ô term Ôi' is subscript where Ôn' ranges from 1 to 300 and Ôi' ranges from 1 to 18. The outside summation is for `n' (1,2,3,4,299,300) whereas the inside summation is for Ôi' (1,2,3,418) I am interested in calculating `ai ` ( i is subscript). I have y[n] and x[n] available but not able to evaluate that _expression. I would be obliged if anybody could help me in solving/evaluating the above equation so that I can find ai's. Is it possible to solve it using Maple, Matlab, Mathematica etc. or to convert it to Ax=B matrix form. I am interseted in writing a C program for the above _expression. Rahul Parthasarthy === Subject: simple question My guess was 6^5 which is 7767...The question is what are the odds of rolling 5 of a kind if you are rolling 5 dice (six sided obviously). Please enlighten me! ps-there is over 6G on stake currently... === Subject: Re: simple question > My guess was 6^5 which is 7767...The question is > what are the odds of rolling 5 of a kind if you > are rolling 5 dice (six sided obviously). Please > enlighten me! > ps-there is over 6G on stake currently... No. What you have calculated is the odds of rolling a specific 5 of a kind e.g. 5 sixes. But if 5 ones, twos etc would be acceptable then you must allow for them. One route is to say that there are 6 solutions out of 7767 throws. The other is the route that * used and say that the first die may be anything and the other 4 must be the same hence 6^4. Fortunately, these two numbers are the same. But it is easy to make mistakes and if there are two ways to look at the problem, then it is good to use both and see if the answers agree. If they don't, then one, the other, or both are wrong. If you start to calculate the odds of other combinations then things will get a bit more complicated since they can be achieved in more ways. Try the odds of four of a kind (four dice the same and one different) and then try full house (3 of one and 2 of another). Try all hands and see if the answers add up to 1 (a good test). Se.87n O'Leathl.97bhair === Subject: Re: simple question first dice sets the match number Prob is 1 second dice must match first, prob is 1/6 third etc, 1/6 fourth etc, 1/6 fifth etc, 1/6 so probability is (1/6)^4 or 0.0772% or 1/1296 ps-better get the money back > My guess was 6^5 which is 7767...The question is > what are the odds of rolling 5 of a kind if you > are rolling 5 dice (six sided obviously). Please > enlighten me! > ps-there is over 6G on stake currently... === Subject: Re: simple question > first dice sets the match number Prob is 1 > second dice must match first, prob is 1/6 > third etc, 1/6 > fourth etc, 1/6 > fifth etc, 1/6 > so probability is (1/6)^4 or 0.0772% or 1/1296 > ps-better get the money back rolling 5 of a kind! >My guess was 6^5 which is 7767...The question is >what are the odds of rolling 5 of a kind if you >are rolling 5 dice (six sided obviously). Please >enlighten me! >ps-there is over 6G on stake currently... === Subject: Re: TEARING DOWN ABU-GHARIB PRISON IS NOT EVEN MEAN > Joseph Hertzlinger > And that has what to do with Iraq? Except for the fact that the US > and Britain occupying Iraq has become the best recruiting tool al > Qaida ever had. >The main recruiting tool was the US retreat from Somalia. > How about the retreat from Beirut in the 1980s? That was another brick in the foundation. OTOH, we wanted to oppose one enemy at a time. -- http://hertzlinger.blogspot.com === Subject: Re: co-ord free description of 3-D rotations >Hi. >I am stuck on deriving what is evidently a very common description of a >rotation operator R: R^3 --> R^3 which has a vector e describing the >axis of rotation and an angle theta describing the amount, reckoned >positive by the right hand rule. >v = R(u) R rotates vector u into vector v. >Here is the problem statement [Simmonds A Brief on Tensor Analysis p.42]: >By resolving v into components along the 3 mutually perp vectors >{ e, e x u, e x (e x u) } show that >v = u + sin(theta) (e x u) + (1 - cos(theta)) (e x (e x u) ) > = (cos(theta)) u + (1 - cos(theta)) (e.u) e + sin(theta) (e x u) >I can get the 2nd expression from the first, but I can't derive the first. >I just get lost in a jungle of algebra... >--Jeff The first equation looks very wrong. It says v = u, the original vector, plus functions of theta. Unit vector e and u form a plane so exu = u sin alpha, and so does ex(exu). So it says v = u +u sinTh*sinAl + u(1-cosTh)*sinAl or v = u + u(Th +Th^2/2)*SinAl using small angle approximations for sin/cos theta. For positive Theta v just gets larger and larger than u. I'm pretty sure the first term should be u dot e or u cos Al, not just u. ÔNuff time spent. Mr. Dual Space (If you have something to say, write an equation. If you have nothing to say, write an essay). === Subject: Re: co-ord free description of 3-D rotations >Hi. >I am stuck on deriving what is evidently a very common description of a >rotation operator R: R^3 --> R^3 which has a vector e describing the >axis of rotation and an angle theta describing the amount, reckoned >positive by the right hand rule. >v = R(u) R rotates vector u into vector v. >Here is the problem statement [Simmonds A Brief on Tensor Analysis p.42]: >By resolving v into components along the 3 mutually perp vectors >{ e, e x u, e x (e x u) } show that >v = u + sin(theta) (e x u) + (1 - cos(theta)) (e x (e x u) ) > = (cos(theta)) u + (1 - cos(theta)) (e.u) e + sin(theta) (e x u) >I can get the 2nd expression from the first, but I can't derive the first. >I just get lost in a jungle of algebra... >--Jeff >The first equation looks very wrong. It says v = u, the original >vector, plus functions of theta. >Unit vector e and u form a plane so exu = u sin alpha, and so does >ex(exu). >So it says > v = u +u sinTh*sinAl + u(1-cosTh)*sinAl > v = u + u(Th +Th^2/2)*SinAl >using small angle approximations for sin/cos theta. >For positive Theta v just gets larger and larger than u. >I'm pretty sure the first term should be u dot e or u cos Al, not just >u. >'Nuff time spent. >Mr. Dual Space >(If you have something to say, write an equation. > If you have nothing to say, write an essay). The equation is all wrong. Before rotation we have u resolved in a single plane to v = e(e*u) + exu where * = Ôdot There are only two sides. To make it 3 is to overspecify! With rotation, only the second side exu rotates: v = e(e*u) + exu(cos theta) + ex(exu)sin theta This is an abomination. Better to state the components: v = (e*u, |exu|cos theta, |exu|sin theta) 1 - cos theta, or versine theta, is never seen in rotations. Mr. Dual Space (If you have something to say, write an equation. If you have nothing to say, write an essay). === Subject: Re: help with derivative > My calculus is so rusty that I need help with this: > (d/dx) [(F(x))^G(x)] > What is it? _______________________________________ In addition to the previous posts (which use the fact that M = e^(ln M) if M > 0), you can also use logarithmic differentiation. First define y = (F(x))^G(x). Take ln of both sides of this equation giving ln y = ln[(F(x))^G(x)] = G(x) ln(F(x)). * Now differentiate both sides wrt (with respect to) x (1/y) (dy/dx) = (d/dx) [G(x) ln(F(x))] (**The left side was found using implicit differentiation -- a special version of the chain rule.) The right side needs to be expanded using the product rule (1/y) (dy/dx) = G'(x) ln(F(x)) + (F'(x)/F(x))G(x) Now isolate the derivative: multiply both sides by y and replace y with its defining expression to obtain the derivative as a function of x (dy/dx) = [F(x))^G(x)] * [G'(x) ln(F(x)) + (F'(x)/F(x))G(x)]. Additional algebraic simplification can be done to give forms more closely resembling solutions in previous posts. Suggestion: Try this process on some simpler exercises, e.g., y = x^x, y = x^(sin x), etc. You might even try y = 2^x. If you're interested, I would be willing to post solutions. ____________________________________________________ *A property of logarithmic functions used here is ln (B^a) = a * ln B. **Recall the chain rule for finding the derivative of a composite function (a function of a function): (d/dx) f(g(x)) = f Ô (g(x)) * g Ô(x). Also recall that (d/dx) ln x = 1/x Combining these (let f = ln in the chain rule) gives (d/dx) ln(g(x)) = [ 1/(g(x)) ] * g Ô(x). = g Ô(x) / (g(x)) and because we know y is a function of x just let y = g(x) in the chain rule; then (d/dx) ln(y) = [ 1/y ] * y Ô = [ 1/y ] * (dy/dx). _______________________________________ === Subject: Re: help with derivative > My calculus is so rusty that I need help with this: > (d/dx) [(F(x))^G(x)] > What is it? F(x)^G(x) is defined to be e^{G(x)*log F(x)} (d/dx)e^{G(x)*log F(x)} = e^{G(x)*log F(x)} * (G(x)*(1/F(x))*F'(x) + G'(x)*log F(x) . -- G.C. === Subject: Re: JSH published electronically OK, then. your mission is to assist HSJ in publishing his proof; what ever in Hell is the meaning of definition. in other word, Doh! > fyi, there is no comedy going on here, except possibly from your end. > i just decided to exploit a bit some of your apparent inconsistencies > - that's all. > Enjoy the last word, although I won't see it. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/HowTheNation.PDF http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate02. html === Subject: Loves Mathematics and Physics? Pledges Please Ladies and Gentlemen: Do you have a lot of money? Here is a decent place you might want to give some money to: http://www.ima.umn.edu Read the first line, it shows: We're out of money and almost out of room, but videos of the talks should eventually become available on the web It's a sad case. === Subject: Re: Loves Mathematics and Physics? Pledges Please > Ladies and Gentlemen: > Do you have a lot of money? Here is a decent place you might want to > give some money to: > http://www.ima.umn.edu > Read the first line, it shows: > We're out of money and almost out of room, but videos of the talks > should eventually become available on the web > It's a sad case. The IMA was founded by and receives major support from the National Science Foundation Division of Mathematical Sciences to carry out a crucial interdisciplinary mission. 1) Either they are absolutely rolling in folding green, or 2) Somebody pulled the plug on their revealed bull. Probability and Statistics in Complex Systems: Genomics, Networks, and Financial Engineering, 1) March to their own tune and thrive, or 2) March to their own tune and perish. No! Wait for it... HETEROSKEDASTICITY! The worse they fail, the more funding they need to succeed. They should change their name to Project Head Start. This single slum bunny sump now gets the same annual budget (within a factor of ~2) as the entire NSF. But Project Head Start is better! It has unlimited potential, as opposed to the niggardly net outputs of the NSF. But Project Head Start cannot achieve its first success in 35 years unless it receives even more funding. Uncle Al DEMANDS that Brown v. Topeka Board of Education be enforced. Admit Whites and watch those test scores zoom up toward... normal. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Loves Mathematics and Physics? Pledges Please > Head Start is better! It has unlimited potential, as opposed to the > niggardly net outputs of the NSF. Contrast the niggardly outputs of NSF with the niggerly outputs of Head Start. Hee. Hee. Sorry. I just could not resist the pun. Bob Kolker === Subject: Re: Loves Mathematics and Physics? Pledges Please > Ladies and Gentlemen: > Do you have a lot of money? Here is a decent place you might want to > give some money to: > http://www.ima.umn.edu > Read the first line, it shows: > We're out of money and almost out of room, but videos of the talks > should eventually become available on the web > It's a sad case. Correct links are: http://www.ima.edu/categories/ http://www.math.ucr.edu/home/baez/ Sorry. === Subject: Re: Loves Mathematics and Physics? Pledges Please > Correct links are: > http://www.ima.edu/categories/ > http://www.math.ucr.edu/home/baez/ > Sorry. The first link does not work, at least for me. Is the second link related? Socks === Subject: Re: Loves Mathematics and Physics? Pledges Please > The first link does not work, at least for me. Hmm. That's strange! Mine works just fine. > Is the second link related? > Socks Yes. Category theory is his speciality - last I heard. === Subject: Any tips how to train on epsilon-delta proofs? I've recently (after a returning from a very long detour into algebra) come to the realization that I am woefully lacking in skill at epsilon-delta type proofs. Being that these are of of critical importance, I am going to devote some substantial time to mastering them. I want to be good enough that when I read a book on (insert anything continuous here), I can guess the proofs (or at least the basic details, like what to set delta to) before reading them. The problem is of course that as far as I know there is no epsilon-delta-ology course. What would you recommend I do? I was thinking maybe start back at elementary calculus and just start doing all the delta-epsilon proofs there by bare hand, then go through analysis/complex analysis/wherever else they show up and repeating the process there. Is there any easier way? Respectfully, SA === Subject: Re: Any tips how to train on epsilon-delta proofs? | | I've recently (after a returning from a very long detour into |algebra) come to the realization that I am woefully lacking in skill |at epsilon-delta type proofs. Being that these are of of critical |importance, I am going to devote some substantial time to mastering |them. I want to be good enough that when I read a book on (insert |anything continuous here), I can guess the proofs (or at least the |basic details, like what to set delta to) before reading them. | The problem is of course that as far as I know there is no |epsilon-delta-ology course. What would you recommend I do? I was |thinking maybe start back at elementary calculus and just start doing |all the delta-epsilon proofs there by bare hand, then go through |analysis/complex analysis/wherever else they show up and repeating the |process there. Is there any easier way? just in case you'd like the opinion of someone who's interested in almost every part of mathematics (geometry, algebra, topology, logic, combinatorics, physics, computer science, etc.) but who finds that the fuzzily-defined subject of analysis usually amounts to studying all the same things that other mathematicians study except with an emphasis on trying to make it all seem as boring as possible: personally i find that the best way to try to understand epsilon-delta-ology is by thinking in terms of trying to construct desired skolem functions by composing (or otherwise combining) previously given skolem functions in intuitively sensible ways. a skolem function is the function that's implicitly given to you by a statement of the form for all x there exists a y such that...; think of x as the input of the skolem function and y as the output. technically it requires the axiom of choice to get from a for all... there exists... statement to a skolem function, but even if you don't subscribe to the axiom of choice you can still use the method i'm talking about here as a discovery method. so for example, suppose it's your job to prove a conclusion of the form for all x there exists a z such that.... and you examine the hypothesises that you've been given, and you find that one of the objects that you've been handed obeys a definition that says for all x there exists a q such that..., and another one obeys a definition that says for all x there exists a y such that..., and yet another one obeys a definition that says for all q there exists a z such that.... then the x|->y skolem function is obviously completely useless, but you should check whether composing the x|->q one with the q|->z one solves your problem for you. of course there's another level of problem that requires a level of cleverness even beyond this, but some mathematicians manage to make a career without rising above it. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Any tips how to train on epsilon-delta proofs? > .... I've recently (after a returning from a very long detour into > algebra) come to the realization that I am woefully lacking in skill > at epsilon-delta type proofs. Being that these are of of critical > importance, I am going to devote some substantial time to mastering > them.... What would you recommend I do? .... First, don't feel ashamed. I think this is one of the two most intrinsically subtle concepts in undergraduate mathematics. (The other is statistical confidence intervals and tests of hypotheses, which have a backwards-and-forwards logical structure like epsilon-delta proofs.) It took some of the best mathematical minds in the world a couple of centuries to get these things right, so it's perfectly reasonable for you to take a few weeks or months. Remember even Newton and Euler didn't have it completely clear. Now, how to do it. As you know, the idea is to set a small target (using epsilon) and then aim to hit it (using a good enough delta). The neatly polished presentations in text-books simply present you with a good delta and then show that it works, just like a conjuring trick. Reading a whole lot of such finished products may only lead to despair. You need to know what happens on scrap paper beforehand to make the tricks work. I'll use the very simple example f(x) = 3x + 1, and prove that f is continuous at 2. We must show that (for every epsilon > 0)(there exists delta > 0) (|x - 2| < delta) => (|f(x) - f(2)| < epsilon). [Now, everything inside these square brackets is done on scrap paper and officially secret. The logical statement above starts with epsilon, but the discovery process starts with delta and calculates what it has to do. |x - 2| < delta => - delta < x - 2 < delta => 2 - delta < x < 2 + delta => 3(2 - delta) + 1 < 3x + 1 < 3(2 + delta) + 1 (which can be a more troublesome step with other functions) => 7 - 3(delta) < f(x) < 7 + 3(delta) => - 3(delta) < f(x) - 7 < 3(delta) => |f(x) - 7| < 3(delta). What we want to get is |f(x) - 7| < epsilon, so we need to make sure that 3(delta) =< epsilon. The obvious simple choice is delta = epsilon/3, so let's choose that. With more difficult functions, solving for delta in terms of epsilon may require fiddling about with inequalities. Now take a fresh piece of paper, smile innocently, and write the following story.] For each positive epsilon, choose delta = epsilon/3. Then |x - 2| < delta => [then write out the above calculation, but always using epsilon/3 instead of delta, so that you end up with:] => |f(x) - 7| < epsilon just like they do in the books. The last step is to throw away the scrap paper, conceal everything you ever have to explain the idea to someone else, take pity on them and let them into the secret. ;-) Ken Pledger. === Subject: Re: Any tips how to train on epsilon-delta proofs? Tim Gowers has a section on how to solve any basic epsilon delta proof using a simple Ôformula' www.dpmms.cam.ac.uk/~wtg10/autoanalysis.html once you get the idea, they become quite simple. === Subject: Re: Any tips how to train on epsilon-delta proofs? > I've recently (after a returning from a very long detour into > algebra) come to the realization that I am woefully lacking in skill > at epsilon-delta type proofs. Being that these are of of critical > importance, I am going to devote some substantial time to mastering > them. I want to be good enough that when I read a book on (insert > anything continuous here), I can guess the proofs (or at least the > basic details, like what to set delta to) before reading them. I'm surprised by your enthusiasm for these epsilon-delta proofs. Of all the math I got in school and in college, this was one of the lows for me; mathematical bureaucracy at its worst. I understand that thorough epsilon-delta proofs are required for the mathematical rigor of the rest of the calculus story: limits, derivatives, integrals etc. But why pester students with these unattractive chores ? There are other fields in maths where the laborious proceedings are spared from the students, so that they can start using the handsome results right away. -- __________ ~ALCATEL/~~~~Walter Baeck ~~BELL~/~~~~~Alcatel Belgium ~~~~~~/~~~~~~DSL Microelectronics Design ~~~~~/~~~~~~~E-mail : walter.baeck@alcatel.be ~~~~/~~~~~~~~Phone : +32 3 240 73 86 === Subject: Re: Any tips how to train on epsilon-delta proofs? > I understand that thorough epsilon-delta proofs are required for > the mathematical rigor of the rest of the calculus story: > limits, derivatives, integrals etc. But why pester students with > these unattractive chores ? There are other fields in maths where > the laborious proceedings are spared from the students, so that > they can start using the handsome results right away. Without epsilon-delta there isn't much in the way of calculus (I havent studied the so-called infinitesimal approach but I can't imagine it being much other than a disguised version of epsilon-delta), having kids do calculus without that stuff would be like doing group theory without the group axioms, topology without the definition of a topology, etc... Without epsilon-delta, if a calculus book just thrust the various derivative formulae at you with no proof, it wouldnt be math, it would be religion, it would be astrology and mysticism, it would be the numerology calculus... Kids would notice that 1-1+1-1+1-1+...=(1-1)+(1-1)+(1-1)+...=0=1-(1-1)-(1-1)...=1 Without epsilon-delta (or in this case epsilon-M) the above would be just as logically valid as any power series Besides, calculus is already a horrific journey into the unbearable world of the engineer/physicist. Epsilon-delta is basically its sole beauty, deprive it of that and you may as well just force kids to memorize rote tables of periodic elements. === Subject: Re: Any tips how to train on epsilon-delta proofs? > I understand that thorough epsilon-delta proofs are required for > the mathematical rigor of the rest of the calculus story: The goal of a mathematical lesson is to make you revisit what was found by your predecessors, not only to learn it by heart like a data-greedy historian, nor apply it without knowing what you are doing (like a robotic engineer ... oops, i said it). > limits, derivatives, integrals etc. But why pester students with > these unattractive chores ? There are other fields in maths where > the laborious proceedings are spared from the students, so that > they can start using the handsome results right away. You said it all ... handsome ... all that shines isn't made of gold ;) I myself would never apply a theorem if I am not able to prove it from scratch, that's just a matter of being honest with oneself. === Subject: Re: Any tips how to train on epsilon-delta proofs? > I'm surprised by your enthusiasm for these epsilon-delta proofs. > Of all the math I got in school and in college, this was one of > the lows for me; mathematical bureaucracy at its worst. > I understand that thorough epsilon-delta proofs are required for > the mathematical rigor of the rest of the calculus story: > limits, derivatives, integrals etc. But why pester students with > these unattractive chores ? There are other fields in maths where > the laborious proceedings are spared from the students, so that > they can start using the handsome results right away. Certainly there are many areas of mathematics where annoying detail is required and for some people the rigorous study of continuity in metric spaces (epsilon-delta proofs) is one of them. However, without this one is not likely to understand continuity in more general settings and this lack of topology will hurt in many places where one would not expect the notion of continuity to be important. Even in abstract algebra (Zariski topology) and number theory (p-adic topology) this is a fundamental concept. Yes there are many handsome results that do not use topological ideas but if you want to be a mathematician you'll have to come to terms with topology or else be very limited :-). === Subject: Re: Any tips how to train on epsilon-delta proofs? > I've recently (after a returning from a very long detour into > algebra) come to the realization that I am woefully lacking in skill > at epsilon-delta type proofs. Being that these are of of critical > importance, I am going to devote some substantial time to mastering > them. I want to be good enough that when I read a book on (insert > anything continuous here), I can guess the proofs (or at least the > basic details, like what to set delta to) before reading them. > The problem is of course that as far as I know there is no > epsilon-delta-ology course. What would you recommend I do? I was > thinking maybe start back at elementary calculus and just start doing > all the delta-epsilon proofs there by bare hand, then go through > analysis/complex analysis/wherever else they show up and repeating the > process there. Is there any easier way? Forget a calculus book. You need to move forward, not backward, to undergrad real analysis. Get hold of Principles of Mathematical Analysis by Rudin, start at page one, and work slowly and carefully. I think this would be a tall order for self-study unless you're very motivated and disciplined. Taking the class would be much better. === Subject: Re: Any tips how to train on epsilon-delta proofs? > I've recently (after a returning from a very long detour into > algebra) come to the realization that I am woefully lacking in skill > at epsilon-delta type proofs. Being that these are of of critical > importance, I am going to devote some substantial time to mastering > them. I want to be good enough that when I read a book on (insert > anything continuous here), I can guess the proofs (or at least the > basic details, like what to set delta to) before reading them. > The problem is of course that as far as I know there is no > epsilon-delta-ology course. What would you recommend I do? I was > thinking maybe start back at elementary calculus and just start doing > all the delta-epsilon proofs there by bare hand, then go through > analysis/complex analysis/wherever else they show up and repeating the > process there. Is there any easier way? > Respectfully, > SA You would be well advised to pursue Rudin's Principles of Mathematical Analysis. If this is too advanced you might try Knopp's Infinite Sequences and Series, a Dover edition. In the former case do the exercises, and in the latter reconstruct the proofs, since there are no exercises. Going back and reviewing elementary calculus would be a waste of time; the real ideas are not in making your delta-estimates sharp (as in the proof that showing that x^2 is a continuous function of a real variable), but knowing when and where a sufficiently small neighborhood of a point actually exists. Best. === Subject: Re: Ubiquitous Naturals, Infinity In lieu of contradiction, it appears the concept of a model of ubiquitous ordinals is accepted. Such a conception offers multiple representation of sets, and thus a route about the bulwark of the powerset result, perhaps extending it to demand ubiquitous ordinals. It also offers a framework for having the considerations of infinity as zero and negative one, and why or why not, and/or both or neither. Thus, in one stroke, Cantor's and Burali's paradoxes can be rationalized away, as all paradoxes should be. With the powerset result being the sole justification for and against a variety of arguments, the availability of a model where it is not assumed opens vast vistas for mathematical development. All in all, it's just another brick in the wall. Ross F. === Subject: Re: Ubiquitous Naturals, Infinity > In lieu of contradiction, it appears the concept of a model of > ubiquitous ordinals is accepted. Your Fields medal will follow shortly. Ross, have you been smoking out of JSH's pipe? === Subject: Pulling hair out here, I really think this textbook is ßawed.... OK, this book defines L_s (the Language of sets) as a language consisting of: 1. Parentheses 2. Connectives (not and if..then) 3. Quantifier (upside down A, for all) 4. Variables v0,v1,v2,... 5. Equality (the binary relation, that is) 6. The binary relation element of The book is asking me to write, in this language, the informal statement There is an empty set in L_s Yeah, yeah.. trivial.. Exists x such that for all v1 not (v1 element of x) One problem... exists and such that are not in the above language! Considering the above language does not possess constants (x) either, I am fairly convinced this problem is impossible, since even if by some amazing trick we managed to conjure the empty set out of just the quantifier etc., what would it possibly be?? Without being a constant? Now, one solution might be for all v1 not (v1 element of v0). But here we are basically cheating and using v0 as a constant... surely this is not allowed?? This stuff is going to make me go bald before my time! (Book btw is Bilaniuk's A Problem Course in Mathematical Logic available online off the links on Wilf's homepage) === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... > OK, this book defines L_s (the Language of sets) as a language > consisting of: > 1. Parentheses > 2. Connectives (not and if..then) > 3. Quantifier (upside down A, for all) > 4. Variables v0,v1,v2,... > 5. Equality (the binary relation, that is) > 6. The binary relation element of > The book is asking me to write, in this language, the informal > statement There is an empty set in L_s ... > Now, one solution might be for all v1 not (v1 element of v0). ... Almost for all v1 not (v1 element of v0) says that v0 is empty. You now need to say that such a v0 exists: there exists v0 such that for all v1 not (v1 element of v0) which in your primitives is: not for all v0 not for all v1 not (v1 element of v0) -- G.C. === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... >OK, this book defines L_s (the Language of sets) as a language >consisting of: >1. Parentheses >2. Connectives (not and if..then) >3. Quantifier (upside down A, for all) >4. Variables v0,v1,v2,... >5. Equality (the binary relation, that is) >6. The binary relation element of >The book is asking me to write, in this language, the informal >statement There is an empty set in L_s >Yeah, yeah.. trivial.. Exists x such that for all v1 not (v1 element >of x) >One problem... exists and such that are not in the above language! A formula like Ex P(x) translates into English as there exists x such that P(x). In many formal systems Ex is not a primitive, it is an abbreviation for not Ax not. >Considering the above language does not possess constants (x) >either, Not that there's any reason x could not be used as a constant, but it's usually not - usually x is a variable. You have v0 and v1 here... >I am fairly convinced this problem is impossible, since even >if by some amazing trick we managed to conjure the empty set out of >just the quantifier etc., what would it possibly be?? Without being a >constant? >Now, one solution might be for all v1 not (v1 element of v0). No, that doesn't mean what you were asked to translate. for all v1 not (v1 element of v0) says v0 is empty. You were not asked to translate v0 is empty, you were asked to translate there exists an empty set. > But >here we are basically cheating and using v0 as a constant... surely >this is not allowed?? >This stuff is going to make me go bald before my time! >(Book btw is Bilaniuk's A Problem Course in Mathematical Logic >available online off the links on Wilf's homepage) ************************ David C. Ullrich === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... Quine discusses, probably in his Methods of Logic, using the single connective neither nor in lieu of the usual bunch. John. -- John T Lowry, PhD Flight Physics 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 jlowry100@earthlink.net > OK, this book defines L_s (the Language of sets) as a language > consisting of: > 1. Parentheses > 2. Connectives (not and if..then) > 3. Quantifier (upside down A, for all) > 4. Variables v0,v1,v2,... > 5. Equality (the binary relation, that is) > 6. The binary relation element of > The book is asking me to write, in this language, the informal > statement There is an empty set in L_s > Yeah, yeah.. trivial.. Exists x such that for all v1 not (v1 element > of x) > One problem... exists and such that are not in the above language! > Considering the above language does not possess constants (x) > either, I am fairly convinced this problem is impossible, since even > if by some amazing trick we managed to conjure the empty set out of > just the quantifier etc., what would it possibly be?? Without being a > constant? > Now, one solution might be for all v1 not (v1 element of v0). But > here we are basically cheating and using v0 as a constant... surely > this is not allowed?? > This stuff is going to make me go bald before my time! > (Book btw is Bilaniuk's A Problem Course in Mathematical Logic > available online off the links on Wilf's homepage) === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... not for all x not P(x) is equivalent to there exists x such that P(x) === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... > OK, this book defines L_s (the Language of sets) as a language > consisting of: > 1. Parentheses > 2. Connectives (not and if..then) > 3. Quantifier (upside down A, for all) > 4. Variables v0,v1,v2,... > 5. Equality (the binary relation, that is) > 6. The binary relation element of > The book is asking me to write, in this language, the informal > statement There is an empty set in L_s > Yeah, yeah.. trivial.. Exists x such that for all v1 not (v1 element > of x) > One problem... exists and such that are not in the above language! some x with for all y, y not in x can be rendered not for all x, not for all y, y not in x or not (for all x, not (for all y, not (y in x))) (Ex)Fx is usually defined as ~(Ax)~Fx (~ is Ônot') which has intended meaning or intension of some x with Fx or exists x such that Fx The semantical equivalence of some x with Fx and not for all x, not Fx is intuitive exercise. > Considering the above language does not possess constants (x) > either, I am fairly convinced this problem is impossible, since even > if by some amazing trick we managed to conjure the empty set out of > just the quantifier etc., what would it possibly be?? Without being a > constant? Yes, languages without constants require much circumlocution. Virtual constants can be constructed with abbreviation schemata. (Ex)Fx for ~(Ax)~Fx is simple example of abbreviation schemata. By the end of a text, the accumulation of nested abbreviations can reach up to a single line representing millions of symbols. === Subject: Re: Pulling hair out here, I really think this textbook is ßawed.... >OK, this book defines L_s (the Language of sets) as a language >consisting of: >1. Parentheses >2. Connectives (not and if..then) >3. Quantifier (upside down A, for all) >4. Variables v0,v1,v2,... >5. Equality (the binary relation, that is) >6. The binary relation element of >The book is asking me to write, in this language, the informal >statement There is an empty set in L_s >Yeah, yeah.. trivial.. Exists x such that for all v1 not (v1 element >of x) >One problem... exists and such that are not in the above language! >Considering the above language does not possess constants (x) >either, I am fairly convinced this problem is impossible, since even >if by some amazing trick we managed to conjure the empty set out of >just the quantifier etc., what would it possibly be?? Without being a >constant? >Now, one solution might be for all v1 not (v1 element of v0). But >here we are basically cheating and using v0 as a constant... surely >this is not allowed?? Just translate there exists to not for all not: not (for all v0 not (for all v1 not (v1 element of v0))) -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Boolean functions' invariant properties. > Why you use 1/4 for negation? Your assumption that I made an excentric choice of notation is wrong. There could be a rendering anomaly in your computer. Your other criticisms may seem unfounded if you read my post at leisure. === Subject: Re: non-Schauder bases >Does there exist any example of a complete locally-convex >Hausdorff-space which has a topological basis >with unbounded coefficient functionals? > >Examples of complete locally-convex Hausdorff-spaces with >topological bases with non-continuous coefficient functionals >seem to exist, but I can't find any references. >My first question was what _is_ a topological basis, if not >a Schauder basis. I found >http://etd.lib.nsysu.edu.tw/ETD-db/ETD-search/getfile?URN= etd-0709103-115 608&filename=etd-0709103-115608.pdf >where topological basis is defined, and then Schauder basis is >defined to be a topological basis with continuous coefficient >functionals. >Yes, that is the definition I had in mind: >a sequence e_n is a topological basis if every vector x in the space >has a unique series representation x = x_1*e_1 + x_2*e_2 + ... >which converges in the topology of the space. >And Schauder basis is usually defined as a topological basis where >the coefficients depend continuously on x. But there are locally convex >spaces on which not every bounded map is continuous. > ??? I thought that boundedness was equivalent to continuity in > general TVS's, not even locally convex. I am pretty sure, that this is wrong. However, I can't think of a counterexample right now. > Let's see, A is bounded if for every neighborhood N of 0 there > exists r > 0 with A contained in rN. And T is bounded if TA is > bounded for every bounded A. > Now... ok, I'm stuck here. But let's add local convexity. > Say V and W are locally convex, with topologies defined > by the familes of seminorms S_V and S_W. To make the > notation simpler below, let's assume that S_V and > S_W are closed under finite sums. Which of > the following is wrong? > (i) The set A in V is bounded if and only if every seminorm > in S_V is bounded on A (and similarly for W.) I believe this one (i.e. I just proved it in my head) > (ii) T: V -> W (linear) is bounded if and only if for every > rho in S_W there exists rho' in S_V such that > rho(Tx) <= c rho'(x) for all x in V. I can't see this one. Obviously, if the inequality holds, then T is bounded. How do you propose to do the other direction? > (iii) T is continuous if and only if it satisfies the same condition. This one is true (i.e. I found this result in a book) I looked in Functional Analysis by K. Yosida, and Theorem I.7.3 reads: Let X be bornologic. If a linear operator on X into a locally convex linear topological space Y maps every bounded set into a bounded set, then T is continuous. Yosida calls a space bornologic if every balanced, convex and absorbing set is a neighbourhood of 0. Here is the condensed proof of that theorem: Let p be the Minkowski functional of some convex, balanced neighbourhood U of 0 in Y. Define the seminorm q via q(x) = p(T x). This seminorm is bounded on any bounded set B (because T B gets absorbed by U ). -> The seminorm q is continuous because X is bornologic. <- Therefore, T^(-1) U is a neighbourhood of 0 in X. QED I consider Yosidas book as circumstancial evidence that one needs an extra condition on the first space (your V) for the implication T bounded => T continuous. I hope, I find some time later today for more work on this. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: non-Schauder bases >Does there exist any example of a complete locally-convex >Hausdorff-space which has a topological basis >with unbounded coefficient functionals? > >Examples of complete locally-convex Hausdorff-spaces with >topological bases with non-continuous coefficient functionals >seem to exist, but I can't find any references. >My first question was what _is_ a topological basis, if not >a Schauder basis. I found >http://etd.lib.nsysu.edu.tw/ETD-db/ETD-search/getfile?URN= etd-0709103-11 5608&filename=etd-0709103-115608.pdf >where topological basis is defined, and then Schauder basis is >defined to be a topological basis with continuous coefficient >functionals. >Yes, that is the definition I had in mind: >a sequence e_n is a topological basis if every vector x in the space >has a unique series representation x = x_1*e_1 + x_2*e_2 + ... >which converges in the topology of the space. >And Schauder basis is usually defined as a topological basis where >the coefficients depend continuously on x. But there are locally convex >spaces on which not every bounded map is continuous. > ??? I thought that boundedness was equivalent to continuity in > general TVS's, not even locally convex. >I am pretty sure, that this is wrong. However, I can't think >of a counterexample right now. > Let's see, A is bounded if for every neighborhood N of 0 there > exists r > 0 with A contained in rN. And T is bounded if TA is > bounded for every bounded A. > Now... ok, I'm stuck here. But let's add local convexity. > Say V and W are locally convex, with topologies defined > by the familes of seminorms S_V and S_W. To make the > notation simpler below, let's assume that S_V and > S_W are closed under finite sums. Which of > the following is wrong? > (i) The set A in V is bounded if and only if every seminorm > in S_V is bounded on A (and similarly for W.) >I believe this one (i.e. I just proved it in my head) > (ii) T: V -> W (linear) is bounded if and only if >[*] for every > rho in S_W there exists rho' in S_V such that > rho(Tx) <= c rho'(x) for all x in V. >I can't see this one. Obviously, if the inequality holds, then >T is bounded. How do you propose to do the other direction? Well if you can't do it in your head it's no doubt false. Um [pause for a few minutes] I thought it was clear but I don't see how the proof goes right now, sorry. > (iii) T is continuous if and only if it satisfies the same condition. >This one is true (i.e. I found this result in a book) >I looked in Functional Analysis by K. Yosida, and Theorem I.7.3 >reads: >Let X be bornologic. If a linear operator on X into a locally >convex linear topological space Y maps every bounded set into a >bounded set, then T is continuous. >Yosida calls a space bornologic if every balanced, convex and >absorbing set is a neighbourhood of 0. >Here is the condensed proof of that theorem: >Let p be the Minkowski functional of some convex, balanced >neighbourhood U of 0 in Y. Define the seminorm q via >q(x) = p(T x). >This seminorm is bounded on any bounded set B (because T B gets >absorbed by U ). >-> The seminorm q is continuous because X is bornologic. <- >Therefore, T^(-1) U is a neighbourhood of 0 in X. >QED >I consider Yosidas book as circumstancial evidence that one needs >an extra condition on the first space (your V) for the implication >T bounded => T continuous. I hope, I find some time later >today for more work on this. >HTH, >Michael. ************************ David C. Ullrich === Subject: Re: non-Schauder bases ... >I looked in Functional Analysis by K. Yosida, and Theorem I.7.3 >reads: >Let X be bornologic. If a linear operator on X into a locally >convex linear topological space Y maps every bounded set into a >bounded set, then T is continuous. >Yosida calls a space bornologic if every balanced, convex and >absorbing set is a neighbourhood of 0. ... >I consider Yosidas book as circumstancial evidence that one needs >an extra condition on the first space (your V) for the implication >T bounded => T continuous. I hope, I find some time later >today for more work on this. A locally convex space X is bornologic if and only if every linear map from X into any other locally convex space Y, which maps bounded sets into bounded sets, is continuous. (see e.g. Koethe I, 28.2(3)) === Subject: Re: Why is b*exp(-|t|^a) a characteristic function? Karl === Subject: Re: can you solve this simple integral equation? >To summarize, if we broaden the OP's problem to read: find h so that >Integral dy f(y) h(a x + y) = b * h(x) (all x) >for some constants 0< a < 1 and b > 0 (where f is the Gauss density with >variance s), then >(*) h(x) = exp(c*x^2) >is a solution provided a*b^2 = 1. Here c = (1-a^2)/2s. The condition I get for this is ab = 1. Moreover, for b = 1/a^n and c = (1-a^2)/(2s), it seems there are solutions h(x) = const H_{n-1}(i sqrt(c) x) exp(c x^2) where H_j is the j'th Hermite polynomial. Thus for suitable values of the constant, the first few cases are b=1/a, h = exp(c x^2) b=1/a^2, h = x exp(c x^2) b=1/a^3, h = (x^2 + 1/(2 c)) exp(c x^2) b=1/a^4, h = (x^3 + 3x/(2 c)) exp(c x^2) b=1/a^5, h = (x^4 + 3x^2/c + 3/(4 c^2)) exp(c x^2) b=1/a^6, h = (x^5 + 5x^3/c + 15x/(4 c^2)) exp(c x^2) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: can you solve this simple integral equation? >To summarize, if we broaden the OP's problem to read: find h so that >Integral dy f(y) h(a x + y) = b * h(x) (all x) >for some constants 0< a < 1 and b > 0 (where f is the Gauss density with >variance s), then >(*) h(x) = exp(c*x^2) >is a solution provided a*b^2 = 1. Here c = (1-a^2)/2s. > The condition I get for this is ab = 1. > Moreover, for b = 1/a^n and c = (1-a^2)/(2s), it seems there are solutions > h(x) = const H_{n-1}(i sqrt(c) x) exp(c x^2) > where H_j is the j'th Hermite polynomial. For a proof of this, use the generating function G(t,x) = exp(-t^2 + 2 t x) = sum_{k=0}^infinity t^k/k! H_k(x) Thus h_k(x) = H_k(i sqrt(c) x) exp(c x^2) has the generating function U(t,x) = exp(-t^2 + 2it sqrt(c) x + c x^2) Now if c = (1-a^2)/(2 s), int_{-infinity}^infinity dy f(y) U(t,a x + y) = 1/a U(t/a, x) and taking the series of both sides with respect to t gives int_{-infinity}^infinity dy f(y) h_k(a x + y) = a^(-1-k) h_k(x) Does this make sense for non-integer k (using the analytic extension of the Hermite polynomials, called HermiteH by Maple)? I don't know if the integral converges - does anyone know asymptotics of HermiteH for imaginary arguments? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: can you solve this simple integral equation? RESTATEMENT of problem I'm sorry I was not particular enough regarding the parameter range I am especially interested in, so let me re-state it precisely: We want to solve h(x) = q * integral( f(y)*h(ax - y)dy ) where f(y) = exp(-y^2/(2*s^2)) /sqrt(2 pi s^2) is the Gaussian distribution, 0 < a < 1 s is some fixed real number, and 0 < q < 1. I know that h(x) = exp(-r x) is a solution (for some special r) when a=1. I am looking for the extensions of this solution when 0 I'm sorry I was not particular enough regarding the parameter range I > am especially interested in, so let me re-state it precisely: > We want to solve > h(x) = q * integral( f(y)*h(ax - y)dy ) > where f(y) = exp(-y^2/(2*s^2)) /sqrt(2 pi s^2) is the Gaussian > distribution, > 0 < a < 1 > s is some fixed real number, and > 0 < q < 1. > I know that h(x) = exp(-r x) is a solution (for some special r) when > a=1. > I am looking for the extensions of this solution when 0= 0} a^{2k} X_k, which is readily seen to be almost surely (absolutely) convergent. Let F_n denote the sigma-field generated by X_n, X_{n+1}, .... Then (*) (with q=1) may be interpreted as saying that (**) E[h(ax+Z) | F_1] = h(x+Z(T)), where T is the shift operator on sequences: T(X_0,X_1,...)=(X_1,X_2,...), so that Z(T) = sum_{k >= 0} a^{2k} X_{k+1}. By induction, (***) E[h((a^n)*x+Z | F_n] = h(x+Z(T^n)). Assuming h to be continuous and suitably integrable, the left side of (***) converges to E[h(Z) | F_infinity] = E[h(Z)] as n goes to infinity, because the tail sigma-field F_infinity is trivial by Kolmogorov's 0-1 law. On the other hand, the distribution of h(x+Z(T^n)) does not depend on n. It follows that h(x+Z) = E[h(Z)] almost surely, for each real x. This is more than enough to imply that h is constant. When 0 Probabalistically, some of the runners will keep running forever. This can only occur > when the sticker stated HALT. So no matter what stickers Doctor Health decides > to give the Australians, his diagnosis can never be perfect. Although I understand that there exist certain circumstances where this sort of language is appropriate, I'm not sure that this is one of them. Hence, using the words probabalistically and never in the same argument, with the second depending on the first, seems oxymoronic to me. Ôcid Ôooh === Subject: Re: All new halting proof oo ____|mn / /_/ / _ - FREE THE TRUeMAN - / K-9/ /_/ - Join www.chatty.net - /____/_____ - Webmasters join www.BannerX.net - > Probabalistically, some of the runners will keep running forever. This can only occur > when the sticker stated HALT. So no matter what stickers Doctor Health decides > to give the Australians, his diagnosis can never be perfect. > Although I understand that there exist certain circumstances where > this sort of language is appropriate, I'm not sure that this is one of > them. Hence, using the words probabalistically and never in the > same argument, with the second depending on the first, seems > oxymoronic to me. With Robert Decastella doing the rounds on Australian TV commercials our prospects of a marathon runner in for the long run has never looked better. All I need is one Australian to be a RUN type person and never give up. Doing a single test or introspection on the runner won't guarantee he will last forever, a large sample comes close to guaranteeing a long time runner, which would show Dr Healths diagnosis is ßawed. Looking at your own sticker once and deciding to run forever is a weak assertion, he might join the chocolate club next week. Herc === Subject: Re: All new halting proof >EARTH 234249 > >Nanotechnology has prevailed, the human race is strong and healthy >and live as immortals. > >A popular sport is the marathon club. Each day millions of runners >get up in the morning, run their hearts out, have a break, run again >during the day, all day every day. As immortals they apparently run >forever, but occasionaly some will quit from the club for one reason >or another. Every marathon runner is given a top with his unique >number printed on it, maybe one day to feature on the 6:02 ptpm >sports break broadcast on www.YeOldeCoffeeShoppe.com. > >Doctor Health was concerned about the runners giving up, so he devised >a medical test to determine which runners will keep running forever and >which runners will stop. After he examines the runners he places one of 2 >stickers on their gernsey HALT or RUN, so each runner knows if they would >last forever doing marathon runs. This was great, it reduced the stress of >long term members leaving and everyone was happy. The test was perfect, >every runner with RUN on their gernsey was still in the marathon club, only >people diagnosed with HALT ever left. > >Doctor Health obtained a lot of notoriety having his perfect diagnosis record, >then one day an Australian joined the marathon club, from down under... > >Being new, the Australian was given the task of washing the gernseys every >Saturdaymk737. He wasn't happy with this, so he decided if the last gernsey >he pulled out from the wash has a RUN sticker on it, he would give up marathon >running. If the sticker was a HALT, he would keep running. Weeks went by >and for a while every last gernsey he pulled from the wash was HALT so he >endured the washing duties. Then one week, he was extra tired from a large >marathone that day, he pulled out the last gernsey anxious to see what he should >do. To his suprise the gernsey was #47546846, HIS OWN GERNSEY!! > >Does the Australian marathon runner keep running the next week or give up? > >He does the opposite of what's on his gernsey, which is HALT. Next week >he pulls a gernsey that says RUN and he promptly obeys his what his >gernsey said. Doctor Health said nothing about *when* he would halt. > > > >How does this affect Doctor Health's diagnosis rate? > >It doesn't, since Doctor Health's diagnosis will be validated when the >Australian finally draws some OTHER jersey labelled RUN. > >Now,modify this so that the Australian would check only one time and >then abide by what it said no matter what, and drew his own. This would >be an example of where the contradiction lies. >That would simplify the problem, but would be an odd thing to happen to >get his own gernsey 1st go. I might draw it out a bit more...... remember he can >be taken off washing duty when the next newbie runner joins the club, so >your assumption his gernsey has HALT on it is not valid. >The key is that the usual presentation of the halting problem >corresponds to what happens when he draws his own gernsey *once*, and >obeys its opposite for his infinite life. Your example above is close, >but differs in details that make it a different problem. Specifically, >you only describe his behavior for a while, and give no statements about >how long he will be on wash duty. If you specified he came off wash >duty at the same time he drew his own last, then we would be back to the >case I described. > If his gernsey says RUN, then he will HALT. Once you resign that's it! > The doctors diagnosis is proven ßawed. > If the gernsey says HALT, then he will keep running. Atleast up until next > week. He might leave wash duty at any time so he can keep running. > The only way Doctor Healths record is left intact is if he gives a HALT sticker, > and the Australian happens to halt afterwards. Fair interpretation but the > runner just pledging to always run won't guarantee that he will, so a single > wash day won't solve the problem of always KNOWING what he will do. > To seal Doctor Healths track record into mediocrity, we need an ongoing supply > of Australians joining the marathon club in a possion distribution. Most get a fair turn > at the washing, they all make the same pledge and a lot of them will decide their > running carreer based on their own gernsey sticker. > Probabalistically, some of the runners will keep running forever. This can only occur > when the sticker stated HALT. So no matter what stickers Doctor Health decides > to give the Australians, his diagnosis can never be perfect. Let's remove probability. The runner, when he quits washing gernseys, takes responsibility for washing his own, and therefor sees what it is each week (the only one he owns) and does the opposite. Regardless, at this point you are moving further and further away from an accurate analogy to the halting problem. It is stated with insane levels of precision precisely to avoid all of this. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: analysis 100 > > Would a less provocative nom-de-plume > provide you with comfort of less ßack? > > um....good idea~ > but Are you sure ? > > Only of death and taxes and I'm not so sure about > death but I am sure about when I have to pay taxes. > um....what's taxes? Huh? You no pay taxes? Taxes is money citizens must pay to government so government can do all that nice and nasty stuff government does. > you are William Shakespeare ? No, I'm too old for my spear to shake. > maybe, hot-girl will eliminate soon. > you will meet my new name. === Subject: Re: analysis 100 > let 0 < a < 1 > show that there is c, r such that n(a^n) <= c(r^n) > for all positive integers n. > (c is real number , a > n(a^n) <= c(r^n) > n <= c*(r/a)^n > > let c = n > since r/a > 1 , very trivial~ > > n <= c(1.00000000000000000000000000000000000000000000000000000001) ^n > right ? > > No, what happened at > n+1 <= c(1.00000000000000000000000000000000000000000000000000000001) ^(n+1) > no, i mean that in this case, let c = n+1 thus, trivial No, c is independent of n. === Subject: Re: what are the elements in a topological space? Ju > G2 = U {]a_i,b_i[ x ]c_i,d_i[; a_i<=b_i, c_i<=d_i, i in I} > where I can be taken such that I is at most countable (prove it). Now notice > that those balls are not balls for the euclidian metric (such balls would be > not that easy to write down, they are open discs) but the product metric and > the euclidian metric are equivalent so they lead to the same topology. > (intuitively you can fill an open disc with open rectangles, and you can > fill an open rectangle with open discs) More than intuitively, the metric is d((a,b),(c,d)) = Max ( |c-a| , |d-b| ) That way squares are balls -:) === Subject: Re: Proofs involving power sets >Hence, using Lemma 2, x is an element of P(A) intersect P(B). > Lemmas, unlike orangas, aren't for free, you have to prove them. > To be honest, I didn't think you wanted me to show my proof of the > lemmas. I proved the lemmas to myself here and then thought you would take > them as a given. Since you are helping me, I didn't think you would want to > have to read any more of my writing than necessary. I will provide proof for > lemmas in the future. Up to you, however they're short and good opportunity to shine and show. > The cancellation law for sets: > A/B = A/C, A/B = A/C ==> B = C > drawing some Venn diagrams may be useful > to give additional insight and meaning > Hint: B = A/B / (A/B)-A > state and prove some applicable lemmas > I will do my best to work on these. I need to prove to myself some > lemmas regarding operations dealing with equality of sets. Once I do that > figuring out the cancellation law will be much easier. Ok, see you later in the next thread. === Subject: Re: HALTING DISPROOF In sci.logic, |-|erc : > Some minor errors with the # of parameters of the UTM, it always > needs 2 to run, and with the tape counter sweeping left to right all > the function applications must be left associative. > > > > UTM > > 000009990000000860000050000 > > > This is UTM(UTM(86,5)) X > = TM-86(5) > >= UTM(999, 86) 5 >= UTM(86,5) ** >= TM-86(5) > > ** assume the emulated UTM has an ability to parse the right of the tape > (the adjacent 5) into the function it is emulating. > > must be why they call is parse, it scans past the function. :) > > > > An ordinary UTM definition with 2 parameters works fine. > > UTM > 0009990009990000999000 > > = UTM(UTM(999,999)) X > = UTM(999,999) > = UTM(999) > = halt > > >= UTM(999, 999) 999 >= UTM(999, 999) >= UTM(999) >= halt > > The 1st 999 in each line is reducing to a UTM as long as it has a parameter. > A modified UTM, mUTM(86) = UTM(86, 86) is not a halting function on > all inputs, even if UTM is! Anyone see that? > A modified UTM is not a UTM. It is a modified UTM. However, said > modified UTM can be encoded and fed back into the original UTM. > NO, only if non halting functions are allowed to be introduced!!!! > the parameter doubling code sets up a potential infinite loop. > Basically, you've got three problems. Which one(s) have you solved? > [1] The enumeration of all demonstrably halting TM's. > actually that's easy, but the functions demonstrated will all > have 0 parameters. > tan(0.1) halts > tan(0.2) halts > tan(0.3) halts The requirement of demonstrably halting does not require a nil-parameter Turing machine. It merely requires that one can prove that the Turing machine will halt on all (finite) inputs. [*] > you can just never say tan halts! Enumerate all TMs with a blank tape and > keep testing each one. > [2] The enumeration of all halting TM's. > That's impossible. If you can enumerate them, then you > can input a given TM and run through the enumeration until > you find or don't find that TM, hence you could program Halt, > option [3]. Yes. > [3] The determination of a procedure for whether a given TM halts > on a given input. > That's impossible, that's Halt. Yes. > [4] An enumeration of a subset of halting TMs that is equivalent > in computational power to the entire set. > Though I haven't solved it, I've shown that such a set is consistent > despite diagonalisation, as diagonalisation is dependant on the type > of computing model. Basically, if one has a set L of 1-parameter functions, each of which is guaranteed to halt, then one can specify a function f which implements L_i(i). That function, however, needs a UTM in order to implement it. Since UTM is a 2-parameter function one gets into some interesting logistical issues, although not insuperable, but I'd have to do some work. > HENCE, we can have a complete ordering of functions, and reals to boot. The reals are already ordered. Also, how does L become a set of all reals? > Herc [*] one assumes the tape is infinite but only a finite number of symbols are non-blank, and that these symbols are located within a finite area of the tape, near where the machine's read/write head starts. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > The arguments in > Panu's paper are not very pleasing, but I'm trying to fully understand > his point of view, and how he attacks the elegant proofs in AIT. Raatikainen doesn't attack any proofs. Why not comment on something actually stated or argued by Raatikainen? So far you have quoted one statement of his, and claimed that on the contrary Chaitin shows that there *is* a direct dependence between the complexity of a theory and its power to prove theorems. So what is this dependence? === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > The arguments in > Panu's paper are not very pleasing, but I'm trying to fully understand > his point of view, and how he attacks the elegant proofs in AIT. > Raatikainen doesn't attack any proofs. Why not comment on something > actually stated or argued by Raatikainen? So far you have quoted one > statement of his, and claimed that on the contrary Chaitin shows that > there *is* a direct dependence between the complexity of a theory and > its power to prove theorems. So what is this dependence? Does he not? Okay, I think you constantly believe that I engage in foul rhetoric by making others say things they have not said, while I think it is Raatikainen whom engages in foul rhetoric. Anyway, let me then quote from the glorious paper, and you tell me what this means: ABSTRACT. The aim of this paper is to comprehensively question the validity of the standard way of interpreting Chaitin's famous incompleteness theorem, which says that for every formalized theory of arithmetic there is a finite constant c such that the theory in question cannot prove any particular number to have Kolmogorov complexity larger than c. The received interpretation of theorem claims that the limiting constant is determined by the complexity of the theory itself, which is assumed to be good measure of the strength of the theory. I exhibit certain strong counterexamples and establish conclusively that the received view is false. Moreover, I show that the limiting constants provided by the theorem do not in any way reßect the power of formalized theories, but that the values of these constants are actually determined by the chosen coding of Turing machines, and are thus quite accidental. KEY WORDS: algorithmic information theory, incompleteness, Kolmogorov complexity This abstract claims that Theorem LB in AIT is wrong (incompleteness of formal axiomatic systems). Now, I must ask you: did you read the algorithmic information theory monograph? If so, how can you not see that this abstract claims that the theorem is wrong? Don't tell me that he is just wrestling with an interpretation, he is not. The guy is openly saying that I show that the limiting constants provided by the theorem do not in any way reßect the power of formalized theories, but that the values of these constants are actually determined by the chosen coding of Turing machines, and are thus quite accidental which means that Theorem LB is wrong! That's what he tries to establish through his great knowledge of kolmogorov complexity and recursive function theory in the following pages.. Raatikainen says interpreting, because he doesn't know what he is talking about. He must think that mathematical truth looks different in the morning and in the evening! Let's continue discussing this, and I think I will eventually tell you what O(f(x)) really means. BTW, I have actually quoted two statements of his, and the latter about reasons for digits of Omega plainly shows his incompetence and it doesn't need further arguments on my part to show what a rigorous philosophy he is wielding. It's interesting though, a page of mathematics can put a philosopher into eternal trouble. -- Eray Ozkural === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. I am sitting on the sideline and do not understand: ... > Raatikainen doesn't attack any proofs. Why not comment on something > actually stated or argued by Raatikainen? ... > ABSTRACT. The aim of this paper is to comprehensively question the > validity of the standard way of interpreting Chaitin's famous > incompleteness theorem, which says that for every formalized theory of > arithmetic there is a finite constant c such that the theory in > question cannot prove any particular number to have Kolmogorov > complexity larger than c. Where is the attack on the proof? As far as I see, there is an attack on the *interpretation*. > This abstract claims that Theorem LB in AIT is wrong (incompleteness > of formal axiomatic systems). Not as far as I can see. It only claims that the standard *interpretation* is wrong. > Don't tell me that he is just wrestling with an interpretation, he is > not. The guy is openly saying that I show that the limiting constants > provided by the theorem do not in any way reßect the power of > formalized theories, but that the values of these constants are > actually determined by the chosen coding of Turing machines, and are > thus quite accidental which means that Theorem LB is wrong! Eh? Does Chaitkin claim in LB that the constants are independent of the Turing machines chosen? If so, where? I think that is the actual question from Torkel Franzen. -- 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: Panu Raatikainen's review of two of Chaitin's books. > The received interpretation of theorem > claims that the limiting constant is determined by the complexity of > the theory itself, which is assumed to be good measure of the strength > of the theory. If this really is the received interpretation, then it is false. However I don't actually know if Chaitin ever said exactly that, or even implied it. It certainly can be fixed by adding a bit of fine print. (e.g. for a fixed encoding of TMs, or in some asymptotic sense). The point is that such statements *without* the fine print can be very misleading. They can make a theorem appear to say something that it does not. For example, one could come to the false conclusion that the following statement implies that the theorem is wrong. > The guy is openly saying that I show that the limiting constants > provided by the theorem do not in any way reßect the power of > formalized theories, but that the values of these constants are > actually determined by the chosen coding of Turing machines, and are > thus quite accidental which means that Theorem LB is wrong! It means nothing of the kind! Read the theorem, and read Raatikainen's statement *very* *carefully*, the precise wording matters. They are not incompatible (which shouldn't be surprising since they are both true). If you claim this is false, show me what's wrong with the proof that has appeared in this list. Don't bother attacking the credentials of the proof's author, that isn't a valid argument in mathematics. Perhaps I would not agree with the puff phrase in any way, since even though the rest of the sentence is true, I could think of a way in which it does reßect the power of formalized theories (e.g. if the coding is set in advance, or asymptotically, but the rest of the sentence makes clear that that isn't what is meant). Ralph Hartley === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. <87ekpkqoru.fsf@phiwumbda.org> 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUps o1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+ QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+ NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+ CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7C PkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F 7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+ YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/ 8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+ JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/ yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/ PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuh ey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) > The received interpretation of theorem > claims that the limiting constant is determined by the complexity of > the theory itself, which is assumed to be good measure of the strength > of the theory. > If this really is the received interpretation, then it is false. However I > don't actually know if Chaitin ever said exactly that, or even implied it. > It certainly can be fixed by adding a bit of fine print. (e.g. for a fixed > encoding of TMs, or in some asymptotic sense). That doesn't really fix it, does it? As Torkel and others have said, there are very complex theories in which each axiom is of the form n+1 = 1+n. Evidently, Chaitin's complexity is not a good measure of the strength of the theory, even if we fix our encoding. The problem is deeper than just that the complexity depends on encoding choice. -- Come on people!!! The US just blew up a lot of people in Iraq, don't you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving Fermat's last theorem === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > Perhaps I would not agree with the puff phrase in any way, since even > though the rest of the sentence is true, I could think of a way in which it > does reßect the power of formalized theories (e.g. if the coding is set > in advance, or asymptotically, but the rest of the sentence makes clear > that that isn't what is meant). What reßection do you have in mind here? === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > The arguments in > Panu's paper are not very pleasing, but I'm trying to fully understand > his point of view, and how he attacks the elegant proofs in AIT. > Raatikainen doesn't attack any proofs. Why not comment on something > actually stated or argued by Raatikainen? So far you have quoted one > statement of his, and claimed that on the contrary Chaitin shows that > there *is* a direct dependence between the complexity of a theory and > its power to prove theorems. So what is this dependence? >Does he not? Okay, I think you constantly believe that I engage in >foul rhetoric by making others say things they have not said, while I >think it is Raatikainen whom engages in foul rhetoric. Anyway, let me >then quote from the glorious paper, and you tell me what this means: >ABSTRACT. The aim of this paper is to comprehensively question the >validity of the standard way of interpreting Chaitin's famous >incompleteness theorem, which says that for every formalized theory of >arithmetic there is a finite constant c such that the theory in >question cannot prove any particular number to have Kolmogorov >complexity larger than c. The received interpretation of theorem >claims that the limiting constant is determined by the complexity of >the theory itself, which is assumed to be good measure of the strength >of the theory. I exhibit certain strong counterexamples and establish >conclusively that the received view is false. Moreover, I show that >the limiting constants provided by the theorem do not in any way >reßect the power of formalized theories, but that the values of these >constants are actually determined by the chosen coding of Turing >machines, and are thus quite accidental. KEY WORDS: algorithmic >information theory, incompleteness, Kolmogorov complexity >This abstract claims that Theorem LB in AIT is wrong (incompleteness >of formal axiomatic systems). This is incredible. The abstract doesn't say any such thing. It says that it's going to question the _interpretation_ of the theorem in the first sentence. Then it gives a rough statement of the theorem, then says more about the _interpretation_ of it. >Now, I must ask you: did you read the >algorithmic information theory monograph? Did you read the abstract that you just posted? Do you see a place where he states that it's not true that for every formalized theory of arithmetic there exists c such that the theory cannot prove that any particular number has Kolmogorov complexity greater than c? >If so, how can you not see >that this abstract claims that the theorem is wrong? >Don't tell me that he is just wrestling with an interpretation, he is >not. The guy is openly saying that I show that the limiting constants >provided by the theorem do not in any way reßect the power of >formalized theories, but that the values of these constants are >actually determined by the chosen coding of Turing machines, and are >thus quite accidental which means that Theorem LB is wrong! Huh????? How does the (_obvious_ _fact_) that the value of the constants depends on the chosen coding of Turing machines show that the theorem is wrong? >That's >what he tries to establish through his great knowledge of kolmogorov >complexity and recursive function theory in the following pages.. >Raatikainen says interpreting, because he doesn't know what he is >talking about. He must think that mathematical truth looks different >in the morning and in the evening! >Let's continue discussing this, and I think I will eventually tell you >what O(f(x)) really means. Better yet, give us an absolute definition of Kolomogorov complexity, that's independent of the choice of coding our Turing machines. This is really quite amusing: Many years ago I heard some stuff about Chaitin this and that on sci.math, all of which was obviously wrong, because it seemed to assume that K complexity was an absolute sort of thing, independent of the way we defined our Turing machines. I was greatly relieved one day, when I looked at something that Chaitin had actually written, to see that he understood this - he acknowledged that of course the actual value of Omega depended on exactly what version of LISP he was using, so it wasn't really absolute in the sense that it was clear to me it couldn't be (but had seemingly been claimed to be by a few people.) >BTW, I have actually quoted two statements of his, and the latter >about reasons for digits of Omega plainly shows his incompetence and >it doesn't need further arguments on my part to show what a rigorous >philosophy he is wielding. It's interesting though, a page of >mathematics can put a philosopher into eternal trouble. ************************ David C. Ullrich === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > This is incredible. The abstract doesn't say any such thing. It says > that it's going to question the _interpretation_ of the theorem > in the first sentence. Then it gives a rough statement of the > theorem, then says more about the _interpretation_ of it. Sure. It's interpretation all the way down, about the correctness of Chaitin's claim, which is part of what is being said in the theorem, and the interpretation is ultimately wrong. >Now, I must ask you: did you read the >algorithmic information theory monograph? > Did you read the abstract that you just posted? Do you see > a place where he states that it's not true that for every formalized > theory of arithmetic there exists c such that the theory cannot > prove that any particular number has Kolmogorov complexity > greater than c? That is *not* Theorem LB! You are reiterating Shen's statement and it's not exactly the same thing. [Which perhaps I should not have given, it seems] Theorem LB of AIT says If H(s)>=n is a theorem only if it is true, then it is a theorem only if n <= H(axioms) + O(1). But that doesn't satisfy Raatikainen or yourself. You have to trip over the O(1) there don't you? But wait, that's not the O(1) you are tripped over. You are tripped over the following, even worse, O(1): | K_A(x) - K_B(x) | in O(1) where A and B are universal computers. Are not you? LOL, now I think I know why Chaitin didn't even bother to respond to that brilliant criticism of his theorems. Maybe so should I have! BTW, Kolmogorov complexity *is* universal, because it's concerned only with asymptotic behavior. The reason is given above. And it's the same concern as in time complexity analysis, in which we give bounds for asymptotic behavior... I'm a little tired of these elementary (but fundamental) facts which are usually given in the first few pages on Kolmogorov complexity texts... so forgive me for skipping the rest of your post... === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > Does he not? Okay, I think you constantly believe that I engage in > foul rhetoric by making others say things they have not said, while I > think it is Raatikainen whom engages in foul rhetoric. Forget about rhetoric. Just explain in what sense there is a direct dependence between the complexity of a theory and its power to prove theorems. === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > Does he not? Okay, I think you constantly believe that I engage in > foul rhetoric by making others say things they have not said, while I > think it is Raatikainen whom engages in foul rhetoric. > Forget about rhetoric. Just explain in what sense there is a direct > dependence between the complexity of a theory and its power to prove > theorems. Why are you avoiding answering my comments about Raatikainen's wonderful paper and its abstract? Does it say that Theorem LB, an elementary and therefore basic result in Kolmogorov complexity, is wrong or not? I am telling you it does. But you are failing to answer me, which keeps me wondering... What's more, you are acting like an inquisitor. I remind that you don't get to decide what gets asked and what gets answered. I promised I'll tell you, in a language that anybody can comprehend, just what the relation is according to Chaitin and other researchers in Kolmogorov complexity, which does *not* depend on silly indices. I'll come around to this later, of course: Kolmogorov complexity is universal, and it doesn't depend on any indexing whatsoever. Read Shen's lectures, you can learn a thing or two. Enumerability matters, indices not. If you're so curious, for the moment, be content with the following observation: First we have to formalize power to prove theorems, and one formalization of this is, apparently, how many particular halting problems can be solved by a machine. This is exactly what is being established by Chaitin's deep results. Not too hard to understand is it? Just like your insistence oh, but provable in what system? as if you have been talking to a non-mathematician, here we have a question of formalism, power in what sense? Chaitin provides one sense for us, and unless bearded apes of logic learn enough mathematics to come up with a better one, it is the best standard around. [*] That is not all as you know. You gave a trivial example of a formal axiomatic system, of your liking, that you think has increasing complexity but a constant power to prove theorems, a conclusion which perhaps you arrived through your intuition. But I must digress here, I cannot really keep serious, because your intuition about ZF set theory, etc. is no match for formalism! Nobody has an intuition that good, perhaps except people who found these new results in metamathematics! Now, having established that your intuition is not some magical source of truth, let me continue. I have thought, without sketching on paper, in what sense the power to prove theorems increases in the exemplary FAS that you described. I imagined an algorithm that does such, but when I later examined AIT, I saw that Chaitin was giving the proofs in more concise form, by directly computing the complexity of p.rec. functions. He uses simple facts like subadditivity of information and complexity of programs that tell their size, etc. So, before I give you the really easy solution that even Raatikainen can understand, I suggest you to look at the following theorems' proofs, Theorem LB, Theorem X, Theorem A, Theorem B, Theorem C, Corollary C in AIT. The one we've been talking about is Corollary C, and it's not difficult at all. Try to understand these, and we'll talk further. The idea is there. -- Eray Ozkural [*] And if the universe is discrete, it is the best possible standard of such. === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > Does he not? Okay, I think you constantly believe that I engage in > foul rhetoric by making others say things they have not said, while I > think it is Raatikainen whom engages in foul rhetoric. > Forget about rhetoric. Just explain in what sense there is a direct >dependence between the complexity of a theory and its power to prove >theorems. Yes, it seems to me that there is something missing from Chaitin's program of using complexity to understand incompleteness. He wants to say that some theorems are hard, and they can only be proved by a powerful theory. But how do you quantify hardness or power? Computational complexity alone doesn't do it, because you can concoct extremely complex theories that only prove a subset of the sentences of the form x = x. The only way that I can think of for Chaitin's connection between complexity and power is to make it true by definition: Letting K(T) = Kolmogorov complexity (the size of the smallest Turing machine that can generate the theorems of T), we define (in the case T is a subset of True Arithmetic, and S is a true sentence in the language of arithmetic) hardness(S) = min { K(T) | T is a subset of True Arithmetic, and T proves S } power(T) = max { hardness(S) | T proves S } These definitions have the consequence that more powerful theories can prove harder theorems, but I don't think they capture the intuitive notion of hardness very well. In particular, this definition of hardness makes it impossible to have a simple sentence that is hard to prove, because every true sentence S can be trivially proved by the theory { S }, whose power is at most K(S). -- Daryl McCullough Ithaca, NY === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > The only way that I can think of for Chaitin's connection between > complexity and power is to make it true by definition: Letting > K(T) = Kolmogorov complexity (the size of the smallest Turing > machine that can generate the theorems of T), we define > (in the case T is a subset of True Arithmetic, and S is a > true sentence in the language of arithmetic) > hardness(S) = min { K(T) | T is a subset of True Arithmetic, > and T proves S } > power(T) = max { hardness(S) | T proves S } > These definitions have the consequence that more powerful theories > can prove harder theorems, but I don't think they capture the > intuitive notion of hardness very well. In particular, this > definition of hardness makes it impossible to have a simple > sentence that is hard to prove, because every true sentence S > can be trivially proved by the theory { S }, whose power > is at most K(S). Here's a likely looking definition. Given T and a statement S, such that T+S is consistent, the information content of S relative to T (let's write it I(S|T)) is the length of the shortest set of axioms A such that S is a theorem of T+A and ~S is not. This definition is not limited to true arithmetic (whatever that is). If S is a theorem of T, then then I(S|T)=0. You can always prove S by adding S as an axiom, so I(S|T) <= |S|. You might want to call any statement for which I(S|T) = |S| a random fact. There is no algorithm that takes T and S and computes I(S|T) (or even tells if it is zero or not). This definition has a problem! The problem is that it depends critically on the language in which axioms are written. We would like something that only depends on logical relationships, and not on syntax. One might replace the length of the axioms with their Kolmogorov complexity, but that doesn't really help. It just replaces one dependency with another. Actually, I'm pretty sure I've seen a definition like this before. I's been a long time since I studied this stuff. Ralph Hartley === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > Does he not? Okay, I think you constantly believe that I engage in > foul rhetoric by making others say things they have not said, while I > think it is Raatikainen whom engages in foul rhetoric. > Forget about rhetoric. Just explain in what sense there is a direct >dependence between the complexity of a theory and its power to prove >theorems. > Yes, it seems to me that there is something missing from Chaitin's > program of using complexity to understand incompleteness. He wants > to say that some theorems are hard, and they can only be proved > by a powerful theory. But how do you quantify hardness or power? > Computational complexity alone doesn't do it, because you can concoct > extremely complex theories that only prove a subset of the sentences > of the form x = x. > The only way that I can think of for Chaitin's connection between > complexity and power is to make it true by definition: Letting > K(T) = Kolmogorov complexity (the size of the smallest Turing > machine that can generate the theorems of T), we define > (in the case T is a subset of True Arithmetic, and S is a > true sentence in the language of arithmetic) > hardness(S) = min { K(T) | T is a subset of True Arithmetic, > and T proves S } > power(T) = max { hardness(S) | T proves S } > These definitions have the consequence that more powerful theories > can prove harder theorems, but I don't think they capture the > intuitive notion of hardness very well. In particular, this > definition of hardness makes it impossible to have a simple > sentence that is hard to prove, because every true sentence S > can be trivially proved by the theory { S }, whose power > is at most K(S). > -- > Daryl McCullough > Ithaca, NY Do you know what this statement * by Ed Fredkin means? So, it's hard to separate program from data. Simple CA's (with simple programs) can generate the great complexity. *In Kolmogorv - Solomonoff complexity, it's obviously not just the size of the program, it's the size of the package of both program and data.* What Turing and Church taught us is that a complex program can be emulated by a simple program with data. Stephen === Subject: Re: Intertwining maps and Peter-Weyl Theorem question > I guess my quesiton is : Do we use the same definition for equivalence > when we > talk about unitary representations rather than any old representations? > Shouldn't > we say something also about inner products? The answers are yes an no respectively. Jose Carlos === Subject: Re: Intertwining maps and Peter-Weyl Theorem question > I guess my quesiton is : Do we use the same definition for equivalence > when we > talk about unitary representations rather than any old representations? > Shouldn't > we say something also about inner products? > The answers are yes an no respectively. > Jose Carlos Would you please scroll down to Definition 12 and explain to me what the discrepancy is here? We might not be communicating the same thing to each other. http://www.amsta.leeds.ac.uk/~kisilv/courses/sp-reprf.html What do you think? === Subject: Re: Intertwining maps and Peter-Weyl Theorem question > I guess my quesiton is : Do we use the same definition for equivalence > when we > talk about unitary representations rather than any old representations? > Shouldn't > we say something also about inner products? > The answers are yes an no respectively. > Would you please scroll down to Definition 12 and explain to me what the > discrepancy is here? We might not be communicating the same thing to each > other. > http://www.amsta.leeds.ac.uk/~kisilv/courses/sp-reprf.html > What do you think? I think that you were talking about *equivalent* representations, whereas Vladimir Kisil's notes define *unitary equivalent* representations. Besides, nothing is gained with this new definition: two irreducible unitary representations are equivalent if and only if they are unitary equivalent. Jose Carlos Santos === Subject: Re: Intertwining maps and Peter-Weyl Theorem question > I guess my quesiton is : Do we use the same definition for equivalence > when we > talk about unitary representations rather than any old representations? > Shouldn't > we say something also about inner products? > > The answers are yes an no respectively. > Would you please scroll down to Definition 12 and explain to me what the > discrepancy is here? We might not be communicating the same thing to each > other. > http://www.amsta.leeds.ac.uk/~kisilv/courses/sp-reprf.html > What do you think? > I think that you were talking about *equivalent* representations, > whereas Vladimir Kisil's notes define *unitary equivalent* > representations. Besides, nothing is gained with this new definition: > two irreducible unitary representations are equivalent if and only if > they are unitary equivalent. Why is it true that if two irreducible unitary representations are equivalent, then they are unitary equivalent? I don't see this... If f : G --> GL(V) and h : G --> GL(W) are unitary representations and T : V --> W is such that T is invertible, a homomorphism, and Tf = hT, then you are saying that T is unitary? Why? > Jose Carlos Santos === Subject: Re: Intertwining maps and Peter-Weyl Theorem question > I think that you were talking about *equivalent* representations, > whereas Vladimir Kisil's notes define *unitary equivalent* > representations. Besides, nothing is gained with this new definition: > two irreducible unitary representations are equivalent if and only if > they are unitary equivalent. > Why is it true that if two irreducible unitary representations are > equivalent, then they are unitary equivalent? > I don't see this... > If f : G --> GL(V) and h : G --> GL(W) are unitary representations and T : > V --> W is such that T is invertible, a homomorphism, and Tf = hT, > then you are saying that T is unitary? Why? No, I most certainly *do not* say this. What I said is that if such a T exists then f and g are unitary equivalent. That's quite easy, at least when V and W are finite-dimensional. Let <.,.> denote the inner products in V and W. Define a new inner product B in V by B(v,w) = . Then you have two isomorphisms from V onto V*, namely v |-> <.,v> and v |-> B(.,v). If you compose the inverse of the second one with the first one, you'll get an automorphism of V which, by Schur's lemma, must be a multiple of the identity. But this means that, for some constant k tou always have = k. This constant must be positve, bacause the inner products take positive values when v = w. Let r be the square root of 1/k and put T' = t.T. Then = = r^2. = . Jose Carlos Santos === Subject: Weighted Linear Least Squares Problem I was hoping that someone would be able to point me in the right direction regarding a linear least squares problem (weighted). The following height differences have been observed between four stations below: Change Distance between station 0 1 61.478 10000 1 2 16.994 15000 2 3 -25.051 9000 3 0 -53.437 18000 0 2 78.465 20000 Station 0 has a known height of 214.880 metres above sea level. The standard deviation in metres of an observed height difference is 0.005 * sqrt distance. I would like to to use least squares to compute the height of stations 1, 2 and 3. The layout is as below : 1------- 2 | / | | / | | / | |_/______| 0 3 ( Station 0 is linked to Station 2 however I couldnt draw this well using ascii) These are the steps that I have taken: 1. Functional model - Fx = L (describes the relationship between the observed heights and the unknown paramters to be estimated ) 2. Assigned provisional values to parameters x0 = x In order to linearise the functional model in the form Ax = b + v where A is the design matrix, b is the (observed minus computed vector) and v is a vector containing the residuals. b, (the observed computed vector) is computed from observations and the Functional Model, evaluated with provisional values of parameters. Using L1- L5, we get: F1(x) = l1 = -x0 + x1 = l1 : L2 -x1 + x2 = l2 L3 - x2 + x3 = l3 L4 x0 x3 = l4 F5(x) = l5 -x0 + x2 = l5 These are my provisional values. I have calculated A and x however, I am having difficulty working out the vector b, which contains the observed minus computed values in the form: [ L1 - f1(xo) ]. I have calculated that the first entry in the vector b(1) is 0. ( 61.478 - 61.478) The other entries for b I have not been able to calculate. Any suggestions? Markorio P.S Please note that this is a similar problem to one at www.imm.dtu.dk/~aa/adjustment.pdf (Example 6, page 14) === Subject: Re: Separation property of non-blocks in transitive group actions > Hi there, > This is an exercise from Dixon and Mortimer's Permutation Groups that > I'm badly stuck with. It should be relatively easy, based only on the > definition of a block and of a transitive action - but somehow, I can't > make it work. > Let $G$ be a group acting transitively on a set $Omega$ containing at > least two points. Show that a non-empty $Delta subset Omega$ is > emph{not} a block iff for every pair of distinct points $alpha, beta > in Omega$ there exist $x in G$ such that only one of them is in > $Delta^x$. Show that if $G$ is finite then this separation property can > be strengthened to $alpha in Delta^x, beta notin Delta^x$. > This does not seem right to me. Suppose Delta is the union of two blocks, > Delta_1 and Delta_2 from the same block system, but is not a block > itself. Now choose alpha, beta in Delta_1. Then, for any x in G, alpha^x > and beta^x are either both in Delta or neither is. I would expect the > separation property to be equivalent to Delta being a union of blocks > from a block system. > As a specific example, let G be the wreath product of cyclic groups of > orders 2 and 3 acting on 6 points, with generators (1,2), (1,3,5)(2,4,6) > (so |G| = 24), let Delta = {1,2,3,4} and let alpha=1, beta=2. So the only if holds, but the if fails. Would this be worth writing about to Dixon or Mortimer, in case there's ever a 2nd edition? (It's exercise 1.5.5 in the 1996 edition.) Maybe they'd even consider setting up an online list of errata, as more and more authors are doing these days. -- Jim Heckman === Subject: Re: Is Matrix inversion O(N^3)? > It is not so unusual to use O() notation even when the > real problem may be in the range where N isn't so large. > > This comes up often in comparing sort algorithms, but > also is important in matrix operations. > Indeed. Even worse. I have seen algorithms abused because their O() > performance was bad. Nevertheless, they sometimes are good enough > for small N, and frequently even better. (And here small may even be > some reasonably large number...) And it gets even more fun when you throw quick sort into the mix, and have to specify whether you were talking worst-case (O(n^2)) or expected (O(n.log(n))) behaviour. Phil -- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: Re: Is Matrix inversion O(N^3)? ... > Indeed. Even worse. I have seen algorithms abused because their O() > performance was bad. Nevertheless, they sometimes are good enough > for small N, and frequently even better. (And here small may even be > some reasonably large number...) > And it gets even more fun when you throw quick sort into the mix, > and have to specify whether you were talking worst-case (O(n^2)) > or expected (O(n.log(n))) behaviour. O, yes. I got abused because in a routine to calculate singular values I used an off-the-cuff sorting routine to sort them. Granted, the routine used was O(n^2) (insertion sort, or somesuch, I have to research the literature for nomenclature), but it was only five lines of code, while the main calculation was O(n^3). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: A group of order 32 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IAY6h31953; I have come across a group of order 32 and would like to know if this can be decomposed to some sort of a direct product, or if someone can throw some light on its properties, etc. The generating elements, all of order 2, are {C,H,V,Q,P} with all pairs of these commutating except for the below two cases 1) HQ = QHC 2) PV = VPC What group is this??? === Subject: Re: A group of order 32 > I have come across a group of order 32 and would like to know if this can be decomposed to some sort of a direct product, or if someone can throw some light on its properties, etc. > > The generating elements, all of order 2, are {C,H,V,Q,P} with all pairs of these commutating except for the below two cases > 1) HQ = QHC > 2) PV = VPC > What group is this??? Well, clearly the element C is central, and it isn't too hard to see that in fact, the centre of your group is , which is also its commutator subgroup. (Hence, your group is solvable.) So we've got a short exact sequence 0 -> -> G -> V -> 0, where G is your 32 element group and V is the abelian group of vectors in (Z/2)^4 under addition. Next, note that any normal subgroup of G is either trivial, or else it contains C. Therefore, G cannot be expressed as a (non-trivial) direct product. === Subject: Re: A group of order 32 > I have come across a group of order 32 and would like to know if this can be decomposed to some sort of a direct product, or if someone can throw some light on its properties, etc. > The generating elements, all of order 2, are {C,H,V,Q,P} with all pairs of these commutating except for the below two cases > 1) HQ = QHC > 2) PV = VPC >What group is this??? It's an extraspecial group of plus type and is isomorphic to SmallGroup(32,49). It does not decompose as a direct product, but it is a central product (= direct product with amalgamated centre) of two dihedral groups of order 8. (That is clear from the presentation - the two dihedral groups are and with amalgamated. Derek Holt. === Subject: Re: dispute > whose roots are (Wow!!) the same as the original equation. Looks like > an advanced game of Ring around the Rosey to me. I think you may be right, even if it is a way of looking at the equation, it's not a way of finding the roots really....especially not in an algebraic form :-) === Subject: Re: ZFC-independent results in arithmetic > And one last question if I may: the Aa in Aatu is pronounced like > a long A or like an O? (I'm thinking of Aarhus and such). Finnish is by and large phonetic, doubled vowels are double length. The habit of mutating non-ASCII vowels into pairs of ASCII vowels causes Finnish TV sports commentators, upon seeing names rendered as Raeikkoenen or Sillanpaeae, for example, to swallow their tongues. Phil -- 1st bug in MS win2k source code found after 20 minutes: scanline.cpp 2nd and 3rd bug found after 10 more minutes: gethost.c Both non-exploitable. (The 2nd/3rd ones might be, depending on the CRTL) === Subject: Re: ZFC-independent results in arithmetic > But this brings > me back to my original question: can there be a specific diophantine > equation that is *proven* to be independent of ZFC? That A is independent of ZFC implies that ZFC is consistent, for any A, so in considering questions of this kind we factor out ZFC is consistent. We can prove in elementary arithmetic (in fact using only finitary reasoning) a statement of the form if ZFC is consistent, the equation P=0 has no solution although this is unprovable in ZFC, for a specific equation P=0. === Subject: Re: ZFC-independent results in arithmetic > But this brings > me back to my original question: can there be a specific diophantine > equation that is *proven* to be independent of ZFC? > That A is independent of ZFC implies that ZFC is consistent, for any > A, so in considering questions of this kind we factor out ZFC is > consistent. We can prove in elementary arithmetic (in fact using only > finitary reasoning) a statement of the form if ZFC is consistent, the > equation P=0 has no solution although this is unprovable in ZFC, for a > specific equation P=0. Got it. In the case of PA, we don't need the if PA is consistent part, since PA *is* consistent assuming ZFC, right? Only we don't have (I suppose) a natural larger-than-ZFC system in which Con(ZFC) is a theorem. Alan === Subject: Re: Units of a Ring === Subject: Re: Units of a Ring >Assume for all x in ring R, x or 1-x is unit. >Now if 1-x and 1-y are units, why is 1 - x - y a unit? >Note, such a ring would then be local. >if, IN ADDITION TO THE HYPOTHESIS 1-x and 1-y are units, we also have >that either x or y are not units, then yes, 1-x-y will be a unit. Oh. When for all x in ring R, x or 1-x is unit, then yes. R local ring; non units closed under addition If 1 - x - y not unit and wlog x not unit, then 1 - y not unit, contradicts premise. Otherwise no. 1 = 1-0 and -1 = 1-2 are units in Z; 0 not unit; 1 - 0 - -1 = 2 not unit -- For every x in R, x is a unit, or 1-x is a unit, and not both. >Is F_2^n example? char F_2^n = 2, 2x /= 1, (0) maximal ideal > F_2^2 isn't >F_p^n not example. char F_p^n is odd >No field with more than two elements can be an example of a ring in >which the condition Ôx is a unit or 1-x is a unit, but not both' >holds. For if the field has more than one 2 elements, then let x be >any element other than 0 and 1. Then x is a unit, and 1-x is not >equal to 0, hence it is also a unit. Ok, fields are hardly a consideration. ---- === Subject: Re: Units of a Ring === Subject: Re: Units of a Ring > Assume [#]: for all x in ring R, x or 1-x is unit. > Now if 1-x and 1-y are units, why is 1 - x - y a unit? > It's been noted that the condition [#] is equivalent to > R is a local ring for if R is local with maximal ideal M > QUESTION: What can be said of a commutative ring for which > satisfies: For all x in R, x or 1-x is a unit, but not both. >William's hypothesis is A below; Arturo's is B (and R local). >As I show below, for local rings R these conditions are all >equivalent to the statement that R has residue field R/M = Z/2. >I.e. said simply, the arithmetic of nonunits and units in R >behaves like the arithmetic of even and odd integers, resp., Don't follow most of your proofs, yet you do show the structure. Includes my example Z/2^n. Z/2^n / (2) = Z/2 as [Z/2^n : (2)] = 2 >THEOREM Among the following statements A,B,C in a ring R > A: u,v units => u+v-1 unit > B: u unit => 1-u nonunit > C: u,v units => u+v nonunit >holds A => B <=> C. All are equivalent if R is local. A ==> (B <=> C) C ==> B. If u, 1-u units: 1 = u + 1-u nonunit which cannot be A,B ==> C. If u,v units: -u,-v units; -u-v-1 unit u+v = 1 - (-u-v-1) nonunit >For R local, C => A: u+v nonunit => u+v-1 unit by [#]. QED >COROLLARY A local ring R with maximal ideal I satisfies one >of the equivalent statements A,B,C if and only if R/I = Z/2 If R local ring with maximal ideal I, then R-I = set of units; I = set of non units [R:I] = 2 ==> C. If u,v units: -v unit; u,v not in I u+I = -v+I; u+v+I = I; u+v in I; u+v nonunit C ==> [R:I] = 2. u,v not in I ==> u+I = v+I if u,v not in I: u,v units; -v unit u-v nonunit; u-v in I; u+I = v+I [R:I] = 2 iff R/I = Z/2 ---- === Subject: Re: Smallest enclosing circle method proposed The method I proposed is faulty, as shown by a counterexample furnished by B. Gaertner of Zurich, because it depended upon the smallest circle (except in trivial cases) having at least two points on the convex hull. But that restriction is improper. My lemma using the power of a point with respect to a circle seems to be OK, however. John. -- John T Lowry 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 jlowry100@earthlink.net > I now have a four-page pdf file discussing the details of my proposed > way of finding the smallest circular disk covering a (finite) set of > points in the plane. If you'd like a copy, let me know by e-mail. > (Repetitive new post for those interested in the problem who might have > gotten fatigued by the older recent one!) > -- > John T Lowry, PhD > Flight Physics > 5217 Old Spicewood Springs Rd, #312 > Austin, Texas 78731 > (512) 231-9391 > jlowry100@earthlink.net === Subject: Re: Smallest enclosing circle method proposed Errr, I mean it depended on their being two CONSECUTIVE points on the CH contained in the smallest circle enclosing. J -- John T Lowry 5217 Old Spicewood Springs Rd, #312 Austin, Texas 78731 (512) 231-9391 jlowry100@earthlink.net > The method I proposed is faulty, as shown by a counterexample furnished > by B. Gaertner of Zurich, because it depended upon the smallest circle > (except in trivial cases) having at least two points on the convex hull. > But that restriction is improper. > My lemma using the power of a point with respect to a circle seems to > be OK, however. > John. > -- > John T Lowry > 5217 Old Spicewood Springs Rd, #312 > Austin, Texas 78731 > (512) 231-9391 > jlowry100@earthlink.net > I now have a four-page pdf file discussing the details of my proposed > way of finding the smallest circular disk covering a (finite) set of > points in the plane. If you'd like a copy, let me know by e-mail. > (Repetitive new post for those interested in the problem who might > have > gotten fatigued by the older recent one!) > -- > John T Lowry, PhD > Flight Physics > 5217 Old Spicewood Springs Rd, #312 > Austin, Texas 78731 > (512) 231-9391 > jlowry100@earthlink.net === Subject: Re: Peano's space-filling curve > No, you are wrong. Do you want to send us your proof? > No point just now. See my post replying to David Ullrich. > When I've read Peano's proof I'll do a bit more heavy > thinking and then let you know :-) > John OK. I did the heavy thinking and found a few answers. First I realise that Peano's curve doesn't do what some reports I read said it does - viz. pass through every point in the square. There exists a countable infinity of points that are isolated, like integers and rationals in the space of real numbers. If I understand it correctly you mathies call such sets sparse? Because the Lebesque measure of the set of all such points not on Peano's curve will be zero therefore, the set of points crossed by the Peano curve will have the same L-measure as the set of points in the boundary square holding the construct. At which point ran into a slight problem with the logic. Peano's curve is placed in a square on the Cartesian plane. The traditional way to measure such a square is to use the Cartesian product of orthogonal vectors. Taking the square with unit vectors as sides we need all possible real number pairs (x1, x2) where both sets x1 and x2 comprise all points in the continuum 0---->1. Peano's curve lacks some (x1,x2) as I have already pointed out. Helpppp!!!!!! John -- John Morgan Although the masters make the rules Of the wise men and the fools I got nothing, ma To live up to - Bob Dylan === Subject: Re: Peano's space-filling curve > > No, you are wrong. Do you want to send us your proof? > No point just now. See my post replying to David Ullrich. > When I've read Peano's proof I'll do a bit more heavy > thinking and then let you know :-) > John >OK. I did the heavy thinking and found a few answers. >First I realise that Peano's curve doesn't do what some >reports I read said it does - viz. pass through every point >in the square. Don't know how you came to realize that, but it's not true. The curve _does_ pass through every point in the square. >There exists a countable infinity of points >that are isolated, like integers and rationals in the space >of real numbers. Yes, there are such sets. So? >If I understand it correctly you mathies >call such sets sparse? Because the Lebesque measure of the >set of all such points not on Peano's curve will be zero The set of all points (in the square) not on the curve is _empty_. (So yes it does have measure zero, but...) >therefore, the set of points crossed by the Peano curve will >have the same L-measure as the set of points in the boundary >square holding the construct. >At which point ran into a slight problem with the logic. >Peano's curve is placed in a square on the Cartesian plane. >The traditional way to measure such a square is to use the >Cartesian product of orthogonal vectors. Taking the square >with unit vectors as sides we need all possible real number >pairs (x1, x2) where both sets x1 and x2 comprise all points >in the continuum 0---->1. Peano's curve lacks some (x1,x2) >as I have already pointed out. You've pointed this out, but you haven't given any reason why it's true (which is understandable, because it's false.) >Helpppp!!!!!! >John ************************ David C. Ullrich === Subject: Re: Peano's space-filling curve > OK. I did the heavy thinking and found a few answers. > First I realise that Peano's curve doesn't do what some > reports I read said it does - viz. pass through every point > in the square. There exists a countable infinity of points > that are isolated, like integers and rationals in the space > of real numbers. If I understand it correctly you mathies > call such sets sparse? Because the Lebesque measure of the > set of all such points not on Peano's curve will be zero > therefore, the set of points crossed by the Peano curve will > have the same L-measure as the set of points in the boundary > square holding the construct. Incorrect. The Peano curve passes through every point of the square. Some points more than once. The Peano curve is a continuous function from the interval [0,1] to the plane. The function is continuous. It is defined on the compact set [0,1]. Therefore its range is compact. If it passes through a dense set of points (as you can see it does by considering the corners of the squares in the approximations), then in fact is passes through all points. Let's rotate so the square we map to is [0,1] x [0,1]... Then if you are given the base 3 expansions of two reals x,y in [0,1], you can construct the base 9 expansion of a number t in [0,1] such that the Peano map sends t to point (x,y). > At which point ran into a slight problem with the logic. > Peano's curve is placed in a square on the Cartesian plane. > The traditional way to measure such a square is to use the > Cartesian product of orthogonal vectors. Taking the square > with unit vectors as sides we need all possible real number > pairs (x1, x2) where both sets x1 and x2 comprise all points > in the continuum 0---->1. Peano's curve lacks some (x1,x2) > as I have already pointed out. Helpppp!!!!!! > John > -- > John Morgan > Although the masters make the rules > Of the wise men and the fools > I got nothing, ma > To live up to - Bob Dylan -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Peano's space-filling curve >OK. I did the heavy thinking and found a few answers. >First I realise that Peano's curve doesn't do what some >reports I read said it does - viz. pass through every point >in the square. But it does; you'd better do some heavier thinking and find some better answers. >There exists a countable infinity of points >that are isolated, like integers and rationals in the space >of real numbers. It is not entirely clear what you are trying to say here (whatever it is, of course, if it's in support of the erronious claim above, it is either wrong in its own right, or about to be misused; that is, if it's not entirely meaningless). One possible interpretation of part of the sentence is that the image of the closed unit interval by the Peano map is a subset of the closed unit square whose complement in the closed unit square is countably infinite. This statement, however, is false: for the Peano map is continuous, therefore the image of the closed unit interval by the Peano map is compact; yet the complement of a countably infinite subset of the closed unit square cannot be compact. >If I understand it correctly you mathies >call such sets sparse? Because the Lebesque measure of the >set of all such points not on Peano's curve will be zero >therefore, the set of points crossed by the Peano curve will >have the same L-measure as the set of points in the boundary >square holding the construct. >At which point ran into a slight problem with the logic. >Peano's curve is placed in a square on the Cartesian plane. >The traditional way to measure such a square is to use the >Cartesian product of orthogonal vectors. Taking the square >with unit vectors as sides we need all possible real number >pairs (x1, x2) where both sets x1 and x2 comprise all points >in the continuum 0---->1. Peano's curve lacks some (x1,x2) >as I have already pointed out. Helpppp!!!!!! Before you can be given any useful help, you will have to explain clearly why you believe that Peano's curve lacks some (x1,x2): an unelaborated reference to heavy thinking taking place off-stage gives us no _point d'appui_. Lee Rudolph === Subject: Re: Peano's space-filling curve Transfinites and covering by counting: A fifth type exists and comes from Siegel disk Julia theory and number theory. An irrational rotation function like ( countable functions ) f1(n)=Mod[Pi*n,1] or f2(n)=Mod[Sqrt[2]*n,1] A circle is: g[n]=Cos(2*Pi*f1(n))+I*Sin(2*Pi*f1(n)) both functions are said to cover the interval [0,1) ( although in experience the Pi based function works less well in Siegel disks and irrational rotations) By plotting {f1(n),f2(n)} you get a space filling curve ( in practice you can't tell it from a Sierpinski space fill). but if you look at the numbers for the points they are only close together and with 200 digits expansions in Mathematica you see that the are never rational ( all rational points are actually excluded). But Dr. Edgar loves to point out that the set of rational numbers has measure zero ( and the points here come very close to the rational points). As far as I know this space fill isn't in the texts.... All the space fills are countable and use the fact that they draw lines ( L Systems) between points as coverings of uncountable numbers of points, so they are of a totally different type.. but when these functions are taken as chaos game IFS that do plot only individual points , then the same problem of coverage exists between space fills and irrational rotations: how can countably many points cover uncountably many points? It is a Cantor transfinite problem and the answer is that using a counting of points { 1,2,3,...,n} you can never cover a transfinite number of points making up a tiling area of the plane. So if transfinites do exist in the Cantor sense, it does not matter how long we run a program that defines a space filling curve, we can never cover all the transfinite points as we are limited to a counted number of points. Thus, we are left with using L Systems that draw lines? We still have countablely many lines to cover an area of uncountable many lines, so we still come up short, unless we can come up with a better definition of a space fill. Such a one is the box covering type L system... with such a set of finite boxes it is possible to cover space since each box itself contains a transfinite number of points of the order of the space to be filled. To conclude : to actually cover a space with a counting operation , the type of cover has to be of the same Cantor transfinite types as the area being covered: Point sets cover point sets lines cover lines areas cover areas volumes cover volumes I hate to do this... people always get distrubed by such reasonable arguments based on definitions and axiomatics. > Major types of space fills seem to be > 1) fractal tiles / Pisot affines ... fractiles ( 2,3,5... L systems and IFS) > 2) Peano-Gosper (eg. ßow snake L systems) > 3) Hilbert-McClure ( L systems and IFS) > 4) Sierpinski square types ( 4,9, 16, 25...etc IFS and L systems) > As for proof, continuity and technical aspects of the mathematics > involved , only Gosper seems to grasp that these are all > doing the same sort of operation in dividing space in a tiling way > that involves coving a 2d space. > If any one had references or links on the overall theory of tiling space > fills that covers all the kinds, I would appreciate it. > If other kinds are known I'd like to know as well. > [Please do not mail me a copy of your followup] > Nice post, Roger. === Subject: Re: Peano's space-filling curve > It is not entirely clear what you are trying to say here > (whatever it is, of course, if it's in support of the erronious > claim above, it is either wrong in its own right, or about to > be misused; that is, if it's not entirely meaningless). One > possible interpretation of part of the sentence is that > the image of the closed unit interval by the Peano map > is a subset of the closed unit square whose complement in > the closed unit square is countably infinite. This statement, > however, is false: for the Peano map is continuous, therefore > the image of the closed unit interval by the Peano map is compact; > yet the complement of a countably infinite subset of the closed > unit square cannot be compact. An appropriate mispelling of that adjective up there. The rest of this message is a reply to the original poster. The construction of space-filling curves is not hard if you take it step by step. Step 1. There is a continuous surjection from C^2 to I, where C is the Cantor set and I is the closed unit interval. Step 2. C^n = C, for every n from 1 to aleph_zero (inclusive), where = means is homeomorphic to. For Step 1, let f(x+y) = (x+y)/2. You already know that f is continuous. To show that it's surjective, work in base 3. Note that the formula (x+y)/2 defines a surjection from {0,2}^2 to {0,1,2}. Step 2 is as easy as aleph_zero + aleph_zero = aleph_zero or aleph_zero * aleph_zero = aleph_zero. With Steps 1 and 2 you get a continuous surjection from C to I^n. In a way, that's even more impressive than a space-filling curve, since C has dimension zero. To get a real space-filling curve takes one more step: Step 3. Any continuous real-valued function defined on C can be extended to a continuous function on I. I guess this is the trickiest step. You don't need any fancy extension theorems. Just extend the function in the most obvious way (linear interpolation) and use a simple epsilon-delta argument to show that the extension is continuous. -- They may be morons, but they're fine mathematicians. It's their one talent. - Hudson Hastings, The Big Night === Subject: Re: Peano's space-filling curve <0405181430430.28702-100000@gandalf.math.ukans.edu> > Step 1. There is a continuous surjection from C^2 to I, where C is the > Cantor set and I is the closed unit interval. > Step 2. C^n = C, for every n from 1 to aleph_zero (inclusive), where > = means is homeomorphic to. > For Step 1, let f(x+y) = (x+y)/2. ^^^^^^ Oops, sorry, I mean f(x,y). > -- > They may be morons, but they're fine mathematicians. It's their one > talent. - Hudson Hastings, The Big Night === Subject: Re: Peano's space-filling curve > erronious >An appropriate mispelling of that adjective up there. Just a tip o' the hat to that version of Russell's paradox phrased in terms of adjectives that do or don't describe themselves. Since you've kind enough (in the rest of your message) to describe for the original poster a construction of space-filling curves using the Cantor set C, I'll be mean enough to try to blow his mind with another neat fact about C, Euclidian space, and dimension. Let L and L' be a skew pair of lines in Euclidian 3-space E_3. Let C be a Cantor set on L, C' a Cantor set on L'. There exists a collection of line segments I_x in E_3 (indexed by x in, say, an interval), each with one endpoint on C and one endpoint on C', such that if you let G be the collection of connected components of the union of all the I_x's, and let X be the topological space E_3/G (that is, the quotient space of E_3 by the equivalence relation such that p is equivalent to q if either p and q both belong to the same element of G, or p = q does not belong to any element of G), then X is infinite-dimensional. This is sort of a dual surprise to the surprise the original poster feels when confronted with the claim that the continuous image of a 1-dimensional space can be n-dimensional for n>1. Or maybe it's the same surprise. I don't know. Lee Rudolph === Subject: Re: Peano's space-filling curve >[...] > an unelaborated reference to heavy thinking >taking place off-stage gives us no _point d'appui_. And not only that, but it's hard to tell where to begin. >Lee Rudolph ************************ David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > .... > deduction theorem says that if Sigma (a set of logical premisses) > suffices to prove that A->B (If A then B), then Sigma union {A} > suffices to prove B.... > This thread puzzles me, as I'd always thought the Deduction >Theorem was the converse of that. I for one simply didn't read the question very carefully, assuming that he was asking about what he said he was asking about... >If I'm mistaken, then what do you >call the following much deeper theorem? >if Sigma union {A} entails B, then Sigma entails (A implies B) > (Herbrand, 1930) > Ken Pledger. ************************ David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > .... > deduction theorem says that if Sigma (a set of logical premisses) > suffices to prove that A->B (If A then B), then Sigma union {A} > suffices to prove B.... > This thread puzzles me, as I'd always thought the Deduction >Theorem was the converse of that. > I for one simply didn't read the question very carefully, assuming > that he was asking about what he said he was asking about... So that's why you didn't recognize Modus Ponens? Yeah, that happens to me too. If someone asks me a question in a funny way, sometimes I suddenly think that 2+2=5. Charlie Volkstorf Cambridge, MA > ************************ > David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > if Sigma union {A} entails B, then Sigma entails (A implies B) What do you mean by entail. As far as I know, entailment is the relationship of (logical) consequence. The deduction theorem does not deal with the consequence. It is a statement about the existence of deriviations in a calculus. So, the other answers seem to be more correct. Thomas Kaeuß === Subject: Re: Can the deduction theorem be used recursively? > > > .... > deduction theorem says that if Sigma (a set of logical premisses) > suffices to prove that A->B (If A then B), then Sigma union {A} > suffices to prove B.... > > This thread puzzles me, as I'd always thought the Deduction > Theorem was the converse of that. If I'm mistaken, then what do you > call the following much deeper theorem? > > if Sigma union {A} entails B, then Sigma entails (A implies B) > > (Herbrand, 1930) > Cor! You're right mate, and the other four answerers didn't spot it! I've seen it presented this way and as an iff statement. Ôcid Ôooh === Subject: Re: Can the deduction theorem be used recursively? >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. Deduction Theorem twice? That's Modus Ponens. Gad! Did you skip Logic 101? Honestly, all the times you've used the word stupid and you don't know one of the most fundamental Rules of Inference? (Some axiomatic systems use only 2 rules, Modus Ponens and Substitution.) Charlie Volkstorf Cambridge, MA > ************************ > David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. >Deduction Theorem twice? That's Modus Ponens. Good point - I wasn't readind very carefully, just assumed he was asking about the Deduction Theorem since that's what he said he was asking about. Strictly speaking it's not _just_ Modus Ponens either; MP is the rule S |- A -> B S |- A ________ S |- B, while here we're talking about S |- A -> B _________ S, A |- B Which of course follows from MP via a few trivial rules (typically derived rules). Yes, it's certainly closer to being just MP than it is to being the Deduction Theorem. >Gad! Did you skip >Logic 101? >Honestly, all the times you've used the word stupid and you don't >know one of the most fundamental Rules of Inference? Giggle. What would be stupid would be if I wasted a lot of space insisting that we _were_ talking about the Deduction Theorem, after it was pointed out that that's not so. Going on and on about how I was the first person in history to _realize_ that the above is really a question about the Deduction Theorem... Giggle. >(Some axiomatic >systems use only 2 rules, Modus Ponens and Substitution.) Giggle. Of course this is so. It's hard to see what your point in pointing this out could be, since many formal systems use only the _one_ rule MP... >Charlie Volkstorf >Cambridge, MA > ************************ > David C. Ullrich ************************ David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. >Deduction Theorem twice? That's Modus Ponens. > Good point - I wasn't readind very carefully, just assumed > he was asking about the Deduction Theorem since that's > what he said he was asking about. So if he was asking about the Deduction Theorem, then the inference does in fact require the Deduction Theorem twice. Hmmm . . . More interesting logic from David Ullrich. (That's 2.) >Gad! Did you skip Logic 101? > What would be stupid would be if I wasted a lot of space > insisting that we _were_ talking about the Deduction Theorem, > after it was pointed out that that's not so. Going on and on about > how I was the first person in history to _realize_ that the above > is really a question about the Deduction Theorem... So if someone points out something that they learned in Logic 101, they are claiming to be the first person in history to realize it? Very interesting! (That's 3. Bang!) And it's NOT the Deduction Theorem, dammit! >(Some axiomatic >systems use only 2 rules, Modus Ponens and Substitution.) > It's hard to see what your point > in pointing this out could be, since many formal systems use > only the _one_ rule MP... Without substitution (in any of its manifestations!), you'll have a hard time making much use of variables. Charlie Volkstorf Cambridge, MA > ************************ > David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > Some axiomatic systems use only 2 rules, Modus Ponens and > Substitution. > Giggle. Of course this is so. It's hard to see what your point > in pointing this out could be, since many formal systems use > only the _one_ rule MP... Indeed... In Frege's Begriffsschrift, logic itself was organized as an axiomatic system, with explicitly defined rules of inference, and over the next fifty years various axiomatizations were suggested. Whitehead and Russell's axioms for propositional logic are these: (A1) |- (P v P) -> P (A2) |- Q -> (P v Q) (A3) |- (P v Q) -> (Q v P) (A4) |- (P v (Q v R)) -> (Q v (P v R)) (A5) |- (Q -> R) -> ((P v Q) -> (P v R)) There are two rules of inference, modus ponens (from |- A and |- [A->B] to infer |- B), and a rule of substitution. The latter permits the replacement of each occurrence of a specific atomic sentence in an axiom by another sentence. Thus we may replace ÔQ' in (A2) by Ô(Q -> R)' to obtain, |- (Q -> R) -> (P v (Q -> R)). A proof is is a sequence of statements, each of them either an axiom or obtained from earlier sentences by one of the rules of inference. In 1927 von Neumann suggested (though with Hilbert in mind rather than Whitehead and Russell) that a rule of substitution could be dispensed with if axiom schemata were used instead of axioms. [...] This practice is now widely adopted. (R. I. G. Hughes, On First-Order Logic, 1993) F. === Subject: Re: Can the deduction theorem be used recursively? >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. > Deduction Theorem twice? That's Modus Ponens. Gad! Did you skip > Logic 101? > Honestly, all the times you've used the word stupid and you don't > know one of the most fundamental Rules of Inference? (Some axiomatic > systems use only 2 rules, Modus Ponens and Substitution.) Uh, the deduction theorem isn't a rule of inference in the sense that modus ponens and substitution are. It's a theorem about deductive systems. Some call it a derived rule, but this is ßawed terminology since you in fact need some proof or model-theoretic machinery to prove the deduction theorem. If you want to get really precise, the deduction theorem says that there is a deduction of S |- A -> B if and only if there is a deduction of S u {A} |- B. See, it's a statement about the existence of deductions. This really is Logic 200 stuff, where you get into doing some metatheory. Did you skip that class? Ôcid Ôooh === Subject: Re: Can the deduction theorem be used recursively? >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. > > Deduction Theorem twice? That's Modus Ponens. Gad! Did you skip > Logic 101? > > Honestly, all the times you've used the word stupid and you don't > know one of the most fundamental Rules of Inference? (Some axiomatic > systems use only 2 rules, Modus Ponens and Substitution.) > > Uh, the deduction theorem isn't a rule of inference in the sense that > modus ponens and substitution are. It's a theorem about deductive > systems. I didn't say it was. The given derivation has nothing to do with the Deduction Theorem per se. (And yes, the Deduction Theorem is a Theorem - LOL. Duh!) It was an example of using Modus Ponens twice, which David Ullrich didn't recognize as such. Some friendly advice: If you spend time trying to defend an idiot, people are liable to not know which of you is which. :) > Some call it a derived rule, but this is ßawed terminology > since you in fact need some proof or model-theoretic machinery to > prove the deduction theorem. It can be used as a rule and it is derived, so what the heck? As far as that goes, the real issue is that, in the final analysis, there is little or no difference between axioms, rules and theorems (and axiom schemata.) For example, an axiom is a rule with no antecedent, and a theorem can often be used as a rule. It is really begging us to see the more general sense of assertions and how they relate to each other. > If you want to get really precise, the > deduction theorem says that there is a deduction of S |- A -> B if and > only if there is a deduction of S u {A} |- B. See, it's a statement > about the existence of deductions. > This really is Logic 200 stuff, where you get into doing some > metatheory. Did you skip that class? Well, if you want to talk about metamathematics, how's about: How does the Deduction Theorem relate to Godel's 1st. Incompleteness Theorem? (I recently posted here comments relating to the principle involved, although I didn't mention the Deduction Theorem's involvement.) If you read my comments concerning my Program Synthesis system, you will see how this whole issue is formalized by my axiomatization of programming. But I won't give you any more hints right now. ;) Charlie Volkstorf Cambridge, MA > Ôcid Ôooh === Subject: Re: Can the deduction theorem be used recursively? >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > > Yes. This has nothing to do with using the Deduction Theorem > recursively, whatever that means, it just requires using the > Deduction Theorem _twice_. > > Deduction Theorem twice? That's Modus Ponens. Gad! Did you skip > Logic 101? > > Honestly, all the times you've used the word stupid and you don't > know one of the most fundamental Rules of Inference? (Some axiomatic > systems use only 2 rules, Modus Ponens and Substitution.) > > > Uh, the deduction theorem isn't a rule of inference in the sense that > modus ponens and substitution are. It's a theorem about deductive > systems. > I didn't say it was. The given derivation has nothing to do with the > Deduction Theorem per se. (And yes, the Deduction Theorem is a > Theorem - LOL. Duh!) It was an example of using Modus Ponens twice, > which David Ullrich didn't recognize as such. I see what you're saying, but it seems fishy to me. I'll cite the relevant quotation to make my objection clear: >It seems to me then that if Sigma suffices to prove A->(B->C) >then Sigma union {A,B} suffices to prove C > In this context, if Sigma suffices to prove a A->(B->C), , it means that there is a deduction of A->(B->C) from Sigma. Similarly for the other sentence. So, if we parse it in this (correct) way, we see that the conditional quoted is a statement about the existence of deductions. You seem to be parsing the statement in a way that implies that modus ponens applies *to* deductions. > Some friendly advice: If you spend time trying to defend an idiot, > people are liable to not know which of you is which. :) *yawn*--you bore me. (too easy, had to move on) > Some call it a derived rule, but this is ßawed terminology > since you in fact need some proof or model-theoretic machinery to > prove the deduction theorem. > It can be used as a rule and it is derived, so what the heck? It's obviously not derived, since you can't deduce it using your object language. >As far > as that goes, the real issue is that, in the final analysis, there is > little or no difference between axioms, rules and theorems (and axiom > schemata.) For example, an axiom is a rule with no antecedent, and a > theorem can often be used as a rule. It is really begging us to see > the more general sense of assertions and how they relate to each > other. Fair enough, but there are functional reasons why the rules of inference and axioms are considered seperately. In particular, the rules of inference apply to _every_ instance of a FOL, whereas some axioms do not. The axioms that do apply in every case are in fact tautologies/FO validities, so they can be proved using nothing but the rules of inference. Now this really is Logic 101 stuff. > If you want to get really precise, the > deduction theorem says that there is a deduction of S |- A -> B if and > only if there is a deduction of S u {A} |- B. See, it's a statement > about the existence of deductions. > > This really is Logic 200 stuff, where you get into doing some > metatheory. Did you skip that class? > Well, if you want to talk about metamathematics, how's about: How does > the Deduction Theorem relate to Godel's 1st. Incompleteness Theorem? > (I recently posted here comments relating to the principle involved, > although I didn't mention the Deduction Theorem's involvement.) If > you read my comments concerning my Program Synthesis system, Oh no, not this clap trap again. >you will > see how this whole issue is formalized by my axiomatization of > programming. But I won't give you any more hints right now. ;) *whew* Ôcid Ôooh === Subject: Re: Can the deduction theorem be used recursively? > In this context, if Sigma suffices to prove a A->(B->C), , it means > that there is a deduction of A->(B->C) from Sigma. Similarly for the > other sentence. So, if we parse it in this (correct) way, we see that > the conditional quoted is a statement about the existence of > deductions. You seem to be parsing the statement in a way that > implies that modus ponens applies *to* deductions. Modus ponens causes certain deductions to exist, such as the above ones. > Some call it a derived rule, but this is ßawed terminology > since you in fact need some proof or model-theoretic machinery to > prove the deduction theorem. > > It can be used as a rule and it is derived, so what the heck? > It's obviously not derived, since you can't deduce it using your > object language. It is still derived in the general sense of being constructed. But I am not worried about ones choice of terminology, so long as it is consistent. > Fair enough, but there are functional reasons why the rules of > inference and axioms are considered seperately. In particular, the > rules of inference apply to _every_ instance of a FOL, whereas some > axioms do not. The axioms that do apply in every case are in fact > tautologies/FO validities, so they can be proved using nothing but the > rules of inference. Now this really is Logic 101 stuff. There are differences and distinctions can be made, of course, especially in certain limited contexts. I am just suggesting that at a certain level of abstraction there is no distinction. (So the task is to determine the lowest such level of abstraction.) > Well, if you want to talk about metamathematics, how's about: How does > the Deduction Theorem relate to Godel's 1st. Incompleteness Theorem? > (I recently posted here comments relating to the principle involved, > although I didn't mention the Deduction Theorem's involvement.) If > you read my comments concerning my Program Synthesis system, > Oh no, not this clap trap again. Relating the Deduction Theorem to Godel's 1st. Incompleteness Theorem is outside your area of expertise, or you think that it is not worthwhile? I am referring to truth versus provability, and principles that unite a number of results of logic (just as my Rules of Inference apply to a large number of domains, including mathematical programming, database programming and the Theory of Computation.) > you will > see how this whole issue is formalized by my axiomatization of > programming. But I won't give you any more hints right now. ;) > *whew* > Ôcid Ôooh It's not that hard, really. Think in terms of a really formal, programmable system. Do you have much experience programming? Charlie Volkstorf Cambridge, MA PS No offense, but if your comments are going to be heavy on sarcasm and light on technical analysis, then I can see why you might relate well to David. Oh my gosh I never noticed that. What's truly amazing about this insight of yours is that nobody else has noticed it, even though it's so glaringly obvious once it's pointed out. More to say about this, but I need to sign off now - I feel a stack overßow coming on... - David C. Ullrich === Subject: hello, math gurus! welcome to my website: www.easyworm.com ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Inverting Laplace Transforms I have a hopefully interesting question here: To give a simple example, it is trivial to show that the laplace transform of e^{at} is frac{1}{s-a}, for s > a. It's then possible to use the inversion formulae to recover e^{at}. However, what about the case where I wish to invert frac{1}{s-a} where now s < a? So my question is, if I have a transformed function *but* it is only well defined for a particular non-empty set of s in S, is it possible to extend the inversion into the region where the transform is not defined? Any references, in particular, would be greatly appreciated. Tronc === Subject: Characters of S^1 Define a character of S^1 as a continuous group homomorphism from S^1 to S^1. All unitary irreducible representations of S^1 are Characters of S^1 because an irreducible representation of an abelian group is one-dimensional, and if the representation is unitary, then the representation maps to S^1. However, if we have a character of S^1, does this define a unitary irreducible representation of S^1? Why? Moshe === Subject: Re: Characters of S^1 It it helps, you can show that the only irreducible 1-d reps of S^1 are the maps z-> z^n for n in Z. > Define a character of S^1 as a continuous group homomorphism from S^1 to > S^1. > All unitary irreducible representations of S^1 are Characters of S^1 because > an irreducible representation of an abelian group is one-dimensional, and if > the representation is unitary, then the representation maps to S^1. > However, if we have a character of S^1, does this define a unitary > irreducible representation of S^1? Why? > Moshe === Subject: Re: Characters of S^1 > Define a character of S^1 as a continuous group homomorphism from S^1 to > S^1. > All unitary irreducible representations of S^1 are Characters of S^1 > because an irreducible representation of an abelian group is > one-dimensional, and if the representation is unitary, then the > representation maps to S^1. > However, if we have a character of S^1, does this define a unitary > irreducible representation of S^1? Why? If f:G -> S^1 is a group homomorphism then we have a 1-dimensional representation of G on C given by (g,z) |-> phi(g) z. This is unitary wrt the Hermitan product = z w* (w* complex conjugate of w). It is in the nature of the beast that one-dimensional representations are irreducible. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: L^2(S^1) question For every f in L^2(S^1), we can expand f as a Fourier series f = summation (from -infinity to infninty) of e_n where e_n = e^{2pi inx}, n in Z and is the n-th Fourier coefficient. Here, the series does not converge pointwise, but the partial sums converge to f in the L^2 norm. My question is : When is the convergence pointwise and why? Moshe === Subject: Re: L^2(S^1) question >For every f in L^2(S^1), we can expand f as a Fourier series f = summation >(from -infinity to infninty) of e_n >where e_n = e^{2pi inx}, n in Z and is the n-th Fourier coefficient. >Here, the series does not converge pointwise, but the partial sums converge >to f in the L^2 norm. The series does converge almost everywhere, although this is a very deep theorem. >My question is : When is the convergence pointwise and why? There are many books devoted to giving partial answers to this question. >Moshe ************************ David C. Ullrich === Subject: Re: L^2(S^1) question >For every f in L^2(S^1), we can expand f as a Fourier series f = summation >(from -infinity to infninty) of e_n >where e_n = e^{2pi inx}, n in Z and is the n-th Fourier coefficient. >Here, the series does not converge pointwise, but the partial sums converge >to f in the L^2 norm. > The series does converge almost everywhere, although this is a very > deep theorem. >My question is : When is the convergence pointwise and why? > There are many books devoted to giving partial answers to this > question. ... and, among these, the one I suggest is Korner's book on Fourier Analysis. Jose Carlos Santos === Subject: Re: L^2(S^1) question >For every f in L^2(S^1), we can expand f as a Fourier series f = summation >(from -infinity to infninty) of e_n >where e_n = e^{2pi inx}, n in Z and is the n-th Fourier coefficient. >Here, the series does not converge pointwise, but the partial sums converge >to f in the L^2 norm. > The series does converge almost everywhere, although this is a very > deep theorem. >My question is : When is the convergence pointwise and why? > There are many books devoted to giving partial answers to this > question. >... and, among these, the one I suggest is Korner's book on Fourier >Analysis. ??? _If_ that's the book I'm thinking of then it would be one of the very best books in the universe if the question were why would one care about Fourier series, but far from the best for the current question. I'd look at something like Katznelson Introduction to Harmonic Analysis, maybe. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: L^2(S^1) question > There are many books devoted to giving partial answers to this > question. >... and, among these, the one I suggest is Korner's book on Fourier >Analysis. > ??? _If_ that's the book I'm thinking of then it would be one of the > very best books in the universe if the question were why would > one care about Fourier series, but far from the best for the current > question. I'd look at something like Katznelson Introduction to > Harmonic Analysis, maybe. Oops, I guess you're right. Jose Carlos Santos === Subject: Re: L^2(S^1) question > For every f in L^2(S^1), we can expand f as a Fourier series f = summation > (from -infinity to infninty) of e_n > where e_n = e^{2pi inx}, n in Z and is the n-th Fourier > coefficient. > Here, the series does not converge pointwise, but the partial sums > converge to f in the L^2 norm. > My question is : When is the convergence pointwise and why? It's a very big question! There are many well-known sufficient criteria for convergence to be pointwise everywhere, for instance that f be of bounded variation. (Continuity is not sufficent!) Carleson proved in the 1960s that convergence holds at almost all points. This was regarded as a major result, and still there is no simple proof known. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: L^2(S^1) question > For every f in L^2(S^1), we can expand f as a Fourier series f = summation > (from -infinity to infninty) of e_n > where e_n = e^{2pi inx}, n in Z and is the n-th Fourier > coefficient. > Here, the series does not converge pointwise, but the partial sums > converge to f in the L^2 norm. > My question is : When is the convergence pointwise and why? >It's a very big question! >There are many well-known sufficient criteria for convergence >to be pointwise everywhere, for instance that f be of bounded variation. >(Continuity is not sufficent!) If we want the series to converge _to f_ at every point then bounded variation is not quite sufficient either... >Carleson proved in the 1960s >that convergence holds at almost all points. This was regarded as a major >result, and still there is no simple proof known. ************************ David C. Ullrich === Subject: Re: Usenet dangerous to researchers > paper and sent it to a peer reviewed electronic math journal. > I sent that paper late July of last year, and waited these nine months > for word. > The word from the chief editor was that it passed peer review, and was > to be published. > Then someone posted that information on sci.math, and they started > lashing out at the math journal talking about how bad the journal was > and THEN started talking about emailing the editors about my paper. > So I worked for months and waited for MONTHS, and it took them a few > days to get my work yanked: > http://rattler.cameron.edu/swjpam/vol2-03.html > It looked to me like the consensus of opinion here (sci.physics, sci.math) > was that your paper should not have been yanked, but that a rebuttal or > correction should have followed in a later edition of the same journal. > It looks like the journal editors are trying to cover up the fact that > your paper was ever published there. This website now says: Advanced Polynomial Factorization by James Harris Withdrawn Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: The Grothendieck Group Hello. Let M be a commutative monoid, and let F be the free abelian group with basis indexed by the elements of M. Denote the generator corresponding to x by [x]. In Lang's Algebra, the Grothendieck group K(M) is defined as F/B, where B is the subgroup of F generated by all elements [x+y]-[x]-[y] with x,y in M. And now to my question: If f:M-->A is a monoidhomomorphism, and A is an abelian group, f factors through K(M). Precisely f = g * h, where h:M-->K(M) is the map x-->[x], and where g is a map K(M)-->A. We see that g must satisfy g([x]+B) for all x in M. Why is it well defined to define g in this way? I guess it should be fairly obvious... -- Michael Knudsen === Subject: Re: The Grothendieck Group > Hello. > Let M be a commutative monoid, and let F be the free abelian group with > basis indexed by the elements of M. Denote the generator corresponding to > x by [x]. In Lang's Algebra, the Grothendieck group K(M) is defined as > F/B, where B is the subgroup of F generated by all elements [x+y]-[x]-[y] > with x,y in M. At this stage it might be better to introduce yet a further piece of notation, say (x) for the coset [x] + B :-). > And now to my question: If f:M-->A is a monoidhomomorphism, and A is an > abelian group, f factors through K(M). Precisely f = g * h, where > h:M-->K(M) is the map x-->[x], and where g is a map K(M)-->A. We see that > g must satisfy g([x]+B) for all x in M. Why is it well defined to define g > in this way? I guess it should be fairly obvious... It is :-) One defines first G:F -> A by G([x]) = f(x) (unproblematic, F is free) and notes that B is contained in ker(G) (need G([x+y] - [x] - [y]) = 0) so that G factors through the projection F -> F/B. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Peano's space-filling curve > Perhaps somebody on this newsgroup can enlighten a non-mathy (but > one who has rather more than a passing interest in topics > mathematical) whether the curve construct described by Peano that > claims to fill a square, does actually do so. I'm going to go out on a limb here, and assume something that may not be true. That is, I'm assuming that you and Peano might have different notions about what space-filling means. Do you mean: A line can be squiggled enough until it turns into the whole plane? That may be true, but I haven't seen it yet, and I will guess that it is false. Do you mean: A line can be squiggled enough until it is arbitrarily close to any point in the plane? That is true, and is the sense meant by everything I've read about space-filling curves. Hope this helps. Ted Shoemaker === Subject: Re: Peano's space-filling curve > Perhaps somebody on this newsgroup can enlighten a non-mathy (but > one who has rather more than a passing interest in topics > mathematical) whether the curve construct described by Peano that > claims to fill a square, does actually do so. > I'm going to go out on a limb here, and assume something that may not > be true. > That is, I'm assuming that you and Peano might have different notions > about what space-filling means. > Do you mean: > A line can be squiggled enough until it turns into the whole plane? > That may be true, but I haven't seen it yet, and I will guess that it > is false. I don't know what squiggled enough means, but what is true is that there is a continuous surjection from the unit interval onto the unit square. > Do you mean: > A line can be squiggled enough until it is arbitrarily close to any > point in the plane? > That is true, and is the sense meant by everything I've read about > space-filling curves. Having an image that is dense in the square is not sufficient for a surjection. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Peano's space-filling curve > Perhaps somebody on this newsgroup can enlighten a non-mathy (but > one who has rather more than a passing interest in topics > mathematical) whether the curve construct described by Peano that > claims to fill a square, does actually do so. > I'm going to go out on a limb here, and assume something that may not > be true. > That is, I'm assuming that you and Peano might have different notions > about what space-filling means. > Do you mean: > A line can be squiggled enough until it turns into the whole plane? > That may be true, but I haven't seen it yet, and I will guess that it > is false. >I don't know what squiggled enough means, but what is true is that >there is a continuous surjection from the unit interval onto the unit >square. > Do you mean: > A line can be squiggled enough until it is arbitrarily close to any > point in the plane? > That is true, and is the sense meant by everything I've read about > space-filling curves. >Having an image that is dense in the square is not sufficient for a >surjection. Actually for a continuous map from [0,1] to the square it _is_ sufficient (so that if one believes that the image is dense one has to believe the map is surjective.) ************************ David C. Ullrich === Subject: Re: Peano's space-filling curve >Having an image that is dense in the square is not sufficient for a >surjection. > Actually for a continuous map from [0,1] to the square it _is_ > sufficient (so that if one believes that the image is dense one > has to believe the map is surjective.) Right, but I meant it's not sufficient in the absence of continuity. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Peano's space-filling curve > Perhaps somebody on this newsgroup can enlighten a non-mathy (but > one who has rather more than a passing interest in topics > mathematical) whether the curve construct described by Peano that > claims to fill a square, does actually do so. >I'm going to go out on a limb here, and assume something that may not >be true. >That is, I'm assuming that you and Peano might have different notions >about what space-filling means. >Do you mean: >A line can be squiggled enough until it turns into the whole plane? >That may be true, but I haven't seen it yet, and I will guess that it >is false. Well you guessed wrong - that's exactly what's meant when people talk about space-filling curves. To be precise: These curves we're talking about are continuous functions f: [0,1] -> [0,1]x[0,1] such that for every (x,y) in the square [0,1]x[0,1] there exists t in [0,1] such that f(t) = (x,y). >Do you mean: >A line can be squiggled enough until it is arbitrarily close to any >point in the plane? >That is true, and is the sense meant by everything I've read about >space-filling curves. No, you've misunderstood various things you've read about those curves. Probably the misunderstanding is because the curve is typically constructed as the limit of a sequence of simple curves. The simple curves eventually come very close to every point; it follows from that that the limit, the actual space-filling curve, actually hits every point. (It follows from that is a slight exaggeration; it follows from that, given the way the things are constructed.) >Hope this helps. >Ted Shoemaker ************************ David C. Ullrich === Subject: Re: Peano's space-filling curve > One possible point that _might_ > help here: you've seen pictures of those polygonal curves > that are used in the construction, right? The Peano curve > is not the union of those curves, it's the _limit_ of those > curves. I have always felt uncomfortable with the standard Peano curve for this same reason. I have only read about it in recreational maths books, and not any in depth texts, and those books just say the Peano curve is the limiting curve of this process, of which they give the first few iterations. With the blancmange curve or the Koch snowßake you can intuitively see that there is a limit, but with the Peano curve I was never totally convinced. With ordinary limits you would have to prove convergence beforehand, and to me the construction of the Peano curve was a bit too weird to accept that immediately. To overcome this, I reasoned that you could build a union map of these curves by mapping each interval [1/2^(2n), 1/2^(2n+1) ) to the nth Peano iteration, and the interval [1/2^(2n+1), 1/2^(2n+2) ) to a line connecting the end point of one iteration with the starting point of the next. This sort of avoids the whole limit curve idea I had trouble with, and achieves the same aim, the only difference being that this new curve is self-intersecting. Jaap === Subject: Re: Peano's space-filling curve > One possible point that _might_ > help here: you've seen pictures of those polygonal curves > that are used in the construction, right? The Peano curve > is not the union of those curves, it's the _limit_ of those > curves. >I have always felt uncomfortable with the standard Peano curve for >this same reason. I have only read about it in recreational maths >books, and not any in depth texts, and those books just say the Peano >curve is the limiting curve of this process, of which they give the >first few iterations. With the blancmange curve or the Koch snowßake >you can intuitively see that there is a limit, >but with the Peano Maybe it is possible to visualize the limiting curve in 3-space, i.e. by looking at the Graph of the map [0,1] -> [0,1]x[0,1]. Or would that just be a disappointment? It is hard to visualize this (self intersecting) curve from its image (just a filled rectangle). Maybe the graph looks more interesting? The reason I wonder, is because the Cantor map [0,1] -> [0,1] (the devil's staircase) can be visualized via its graph, and it has very strange properties, too. Just an idea... Thomas >curve I was never totally convinced. With ordinary limits you would >have to prove convergence beforehand, and to me the construction of >the Peano curve was a bit too weird to accept that immediately. >To overcome this, I reasoned that you could build a union map of these >curves by mapping each interval [1/2^(2n), 1/2^(2n+1) ) to the nth >Peano iteration, and the interval [1/2^(2n+1), 1/2^(2n+2) ) to a line >connecting the end point of one iteration with the starting point of >the next. >This sort of avoids the whole limit curve idea I had trouble with, and >achieves the same aim, the only difference being that this new curve >is self-intersecting. >Jaap === Subject: Re: Peano's space-filling curve > One possible point that _might_ > help here: you've seen pictures of those polygonal curves > that are used in the construction, right? The Peano curve > is not the union of those curves, it's the _limit_ of those > curves. >I have always felt uncomfortable with the standard Peano curve for >this same reason. I have only read about it in recreational maths >books, and not any in depth texts, and those books just say the Peano >curve is the limiting curve of this process, of which they give the >first few iterations. With the blancmange curve or the Koch snowßake >you can intuitively see that there is a limit, but with the Peano >curve I was never totally convinced. With ordinary limits you would >have to prove convergence beforehand, and to me the construction of >the Peano curve was a bit too weird to accept that immediately. The fact that the limit exists is _not_ supposed to be obvious! It needs to be proved. It's not hard to prove, but it might be tricky to try to give the proof in the sort of language one sees in recreational math books. _Do_ you know _exactly_ what the two terms Cauchy sequence and uniform convergence (or uniform limit) mean? If yes I'll try to post a hint as to why the construction works - if not I don't see much point to trying, there are simply so many preliminaries about what continuity is and how it works required that it's not going to be possible to give the proof in a short space. >To overcome this, I reasoned that you could build a union map of these >curves by mapping each interval [1/2^(2n), 1/2^(2n+1) ) to the nth >Peano iteration, and the interval [1/2^(2n+1), 1/2^(2n+2) ) to a line >connecting the end point of one iteration with the starting point of >the next. >This sort of avoids the whole limit curve idea I had trouble with, and >achieves the same aim, the only difference being that this new curve >is self-intersecting. No, this does _not_ achieve the same thing. There's a technicality about how the curve you've defined has domain (0,1] instead of [0,1], but the important difference is that the curve you've defined is _not_ a space-filling curve! There are many points in the square that it misses, in fact it misses most of them, in the sense of measure theory. (Also note that the Peano curve _is_ self-intersecting, so one difference you cite is no difference at all.) >Jaap ************************ David C. Ullrich === Subject: Re: Haar measure on positive reals with mult. > There seems to be few places in the literature where the > simple calculation of Haar measure for continuously > differentiable groups is pointed out. One place, which includes the Haar measure on additive and multiplicative groups of R, C, p-adic numbers, adeles, general linear groups, special linear groups, etc., is Weil's book, Basic Number Theory. There is more information on other groups in Weil's other books and papers. -- Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions and * comments do not reßect in any way on MIT. Also, I am nowhere near Boston. === Subject: can somebody help me with this topology problem? I need to solve this problem and I have the idea but I don't know how to write it: let (R^2,g) be a topological space where g is the standard topology on R^2. (R is the set of real numbers.) I have to prove that R^2(G1 U G2) never becomes a finite set. G1 and G2 are nonempty disjunctiv sets, and they are elements of g. that is they are open sets in R^2. solution in words. if we let a point P be in G1 and a point Q in G2 then we can always make a linesegment PQ and find a point C that neither is in G1 nor in G2. But since the points G1 and G2 are arbitrary and since there are always an infinite number of points in an open set and since the points don't lie in a straight line, then you will always have an infinite number of points C. does the argument sound fine? How can I write it with opensets? or how can I prove that there are always an infinite number of points in a open set? === Subject: Re: can somebody help me with this topology problem? > does the argument sound fine? > How can I write it with opensets? or how can I prove that there are always > an infinite number of points in a open set? Well, in R^2,g, you use the fact that every open set contains an open x-ball for some sufficiently small x. Then you use the fact that the x-ball always contains an infinite number of points. === Subject: Re: can somebody help me with this topology problem? > I have to prove that R^2(G1 U G2) never becomes a finite set. > G1 and G2 are nonempty disjunctiv sets, and they are elements of g. that is > they are open sets in R^2. [Snip] > does the argument sound fine? Not to me ... how can you assert the existence of that point which does not belong to G1 nor G2 ? The other parts of your reasoning are quite inaccurate too. What you may want to use here is the fact that if R^2(G1UG2) is finite, then G1UG2 is path-connected (whence connected), and open in R^2. As such, G1 and G2 being open sets of G1UG2 -for the induced topology-, either G1 or G2 must be empty (which is false). Contradiction. -- Julien Santini === Subject: Re: can somebody help me with this topology problem? It should also be possible to show that in the linesegment PQ there will always be a point C that neither is in G1 nor G2. That is obvious, I just don't know how to write it. can you help me also with this? === Subject: Re: can somebody help me with this topology problem? >It should also be possible to show that in the linesegment PQ there will >always be a point C that neither is in G1 nor G2. That is obvious, I just >don't know how to write it. >can you help me also with this? Hint: Consider the point in the closure of the intersection of one of your sets with the line segment that is closest to the endpoint not in that set. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: can somebody help me with this topology problem? > It should also be possible to show that in the linesegment PQ there will > always be a point C that neither is in G1 nor G2. Yes, precisely because G1UG2 is connected. But what will you do with that point ? Nothing tells you immediately that you will be able to produce infinintely many point outside G1UG2 using that method... === Subject: Re: can somebody help me with this topology problem? > It should also be possible to show that in the linesegment PQ there will > always be a point C that neither is in G1 nor G2. > Yes, precisely because G1UG2 is connected. But what will you do with that > point ? Nothing tells you immediately that you will be able to produce > infinintely many point outside G1UG2 using that method... But if I really want to show that C exists how can I do it without using the path-connection stuff? ( I haven't learned that yet) === Subject: Re: can somebody help me with this topology problem? > But if I really want to show that C exists how can I do it without using the > path-connection stuff? ( I haven't learned that yet) Look at Robert Israel's post .. it says it all; in passing, that method is useful when you want to show that, for instance, the connected subsets of R are its intervals. === Subject: Re: can somebody help me with this topology problem? > What you may want to use here is the fact that if R^2(G1UG2) is finite, > then G1UG2 is path-connected (whence connected), and open in R^2. How can you proof your last statement? I mean how R^2(G1UG2) being finite implies that? what do you mean with path-connected? === Subject: Re: can somebody help me with this topology problem? > What you may want to use here is the fact that if R^2(G1UG2) is finite, > then G1UG2 is path-connected (whence connected), and open in R^2. > How can you proof your last statement? I mean how R^2(G1UG2) being finite > implies that? > what do you mean with path-connected? Path-connected means that for any two points a and b in G1UG2, there exists a continuous map: f: [0,1] -> G1UG2 such that f(0)=a and f(1)=b. Showing G1UG2 is path - connected is trivial (due to the finiteness of its complement in R^2). Path-connectedness implies connectedness (for instance you may use the notion of connected component to show that). Now G1UG2 is connected, and open in R^2 (because a finite subset of R^2 is always closed ...). Whence G1 and G2, which are open in R^2, are also open in G1UG2 (induced topology). But G1UG2 is connected, so that would require either G1 or G2 to be empty (since they're both open and disjoint). === Subject: Subgroups of simple groups Can every group be realized as the subgroup of some simple group? Tyler Smith University of Illinois Urbana-Champaign === Subject: Re: Subgroups of simple groups graduate school days. My association with the Math Department is only that of an alumnus. Cc: >Can every group be realized as the subgroup of some simple group? Yes for a finite group: every group can be realized as a subgroup of S_n for n=|G|, and S_n can be identified with the simple group A_{n+2}. It is also true for torsion-free groups, since they can be embedded in groups with only two conjugacy classes. I have to think a bit for more general infinte groups. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Subgroups of simple groups > Cc: >Can every group be realized as the subgroup of some simple group? > Yes for a finite group: every group can be realized as a subgroup of > S_n for n=|G|, and S_n can be identified with the simple group > A_{n+2}. > It is also true for torsion-free groups, since they can be embedded in > groups with only two conjugacy classes. I have to think a bit for more > general infinte groups. Well, for groups with a *countable* infinity of elements, the group can be embedded in the symmetric group on a countable infinity of elements, which can be embedded in the alternating group on a countable infinity of elements by duplicating each element, which, if I'm not mistaken, is simple. === Subject: Re: Subgroups of simple groups days. My association with the Department is that of an alumnus. >Well, for groups with a *countable* infinity of elements, the group >can be embedded in the symmetric group on a countable infinity of >elements, which can be embedded in the alternating group on a >countable infinity of elements by duplicating each element, which, if >I'm not mistaken, is simple. Hmm.. What is the infinite alternating group? The group of all permutations on an infinite set, such that all permutations of finite support are even? In any case, the permutations of finite support form a nontrivial normal subgroup, don't they? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: What is a function? There is a work by Gottlob Frege entitled Was ist eine funktion?. I understand, there, pitifully, has not been a translation, thus my question. My recent conclusion is that the term function is central to mathematics. It is underlying (all) formal calculi (A. Church) and is the whole secret (L. Wittgenstein). Please, will you be kind enough as to point me in the direction of research which will allow me to earn the way of thinking of a function such that this particular way will be isomorphic to all function representations in any (scientific) discourse. Andrew === Subject: Re: What is a function? > There is a work by Gottlob Frege entitled Was ist eine funktion?. I > understand, there, pitifully, has not been a translation, thus my > question. It's been translated into English, but I don't know if it is still in print. Was ist eine Funktion?, in Festschrift Ludwig Boltzmann gewidmet zum sechzigsten Geburtstage, 20. Februar 1904, S. Meyer (ed.), Leipzig: Barth, 1904, pp. 656-666. Translated as What is a Function? by P. Geach in Translations from the Philosophical Writings of Gottlob Frege, P. Geach and M. Black (eds. and trans.), Oxford: Blackwell, third edition, 1980. > My recent conclusion is that the term function is central to > mathematics. It is underlying (all) formal calculi (A. Church) and > is the whole secret (L. Wittgenstein). > Please, will you be kind enough as to point me in the direction of > research which will allow me to earn the way of thinking of a function > such that this particular way will be isomorphic to all function > representations in any (scientific) discourse. > Andrew -- G.C. === Subject: Re: What is a function? > There is a work by Gottlob Frege entitled Was ist eine funktion?. I > understand, there, pitifully, has not been a translation, thus my > question. > My recent conclusion is that the term function is central to > mathematics. It is underlying (all) formal calculi (A. Church) and > is the whole secret (L. Wittgenstein). > Please, will you be kind enough as to point me in the direction of > research which will allow me to earn the way of thinking of a function > such that this particular way will be isomorphic to all function > representations in any (scientific) discourse. > Andrew en.wikipedia.org/wiki/Function perhaps? === Subject: Re: What is a function? > Please, will you be kind enough as to point me in the direction of > research which will allow me to earn the way of thinking of a function > such that this particular way will be isomorphic to all function > representations in any (scientific) discourse. It look difficult. The concept of function in mathematics is the result of a very long process of maturation. Natural selection works well in the world of ideas. Consequently the current definition of function is far from the concept of function in common sense or in other sciences. Mathematic definition : A function is a collection of 3 sets 1. The domain of the function (those elements who have an image) 2. The destination of the function (those elements who may be images) 3. A set describing the relations between the elements of the two set with the rule 4. Every elements of the domain have one and only one image. (Note that this is not symetrical, element of the destination set may be image or not, they also may be the image of many elements) This definition is purerly descriptive and statical. Note that nothing in this say why the image are in relation with the elements in the domain. This can be the result of a process like in physics, of a formula (x -> sin(x+1)), but it is not an obligation. So this definition can apply to many situations and is very ßexible. This is typical of the way of thinking in mathematics. In sciences, one want explanation of how the worlds works with causes and effects. In mathematics, the ideal is to see what we have, to forget the origin of the problem, and to concentrate on the study of facts. This way, mathematical solutions So, mathematical functions are usefull to interpret functions in other areas, but it is more general. This very general definition of function leads to problems informaticians describe very well. There are too much mathematical functions when we look at function we can really compute. So they invent concepts like constructive function, recursive function, etc... to limit the study to function with good properties. === Subject: Re: What is a function? > Mathematic definition : > A function is a collection of 3 sets > 1. The domain of the function (those elements who have an image) > 2. The destination of the function (those elements who may be images) > 3. A set describing the relations between the elements of the two set > with the rule > 4. Every elements of the domain have one and only one image. > (Note that this is not symetrical, element of the destination set may be > image or not, they also may be the image of many elements) I expected that somebody would post a definition like the above. The delta function is not a function by the above definition. I think a set of functions can be extended in some sense, much like the rational numbers are extended to form the reals, to include the delta function. I don't know of a name for such an extension. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: What is a function? > Mathematic definition : > A function is a collection of 3 sets ... >I expected that somebody would post a definition like the above. >The delta function is not a function by the above definition. Very true. It isn't a function (of a real variable) in the usual sense. >I think a set of functions can be extended in some sense, much like >the rational numbers are extended to form the reals, to include the >delta function. I don't know of a name for such an extension. The Dirac delta function can be understood as a measure or as a distribution. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: What is a function? > Mathematic definition : > A function is a collection of 3 sets > 1. The domain of the function (those elements who have an image) > 2. The destination of the function (those elements who may be images) > 3. A set describing the relations between the elements of the two set > with the rule > 4. Every elements of the domain have one and only one image. > (Note that this is not symetrical, element of the destination set may be > image or not, they also may be the image of many elements) > I expected that somebody would post a definition like the above. > The delta function is not a function by the above definition. Yup. Most mathematicians do indeed not think that the delta function is a function. > I think a set of functions can be extended in some sense, much like > the rational numbers are extended to form the reals, to include the > delta function. I don't know of a name for such an extension. I have seen functional used. -- 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: What is a function? > There is a work by Gottlob Frege entitled Was ist eine funktion?. I > understand, there, pitifully, has not been a translation, thus my > question. > My recent conclusion is that the term function is central to > mathematics. It is underlying (all) formal calculi (A. Church) and > is the whole secret (L. Wittgenstein). Almost anything on Lambda Calculus will consider a function as the rule of calculation. That is most congruent with the view of computation and algorithms. Then there is the view of function emphasized in category theory. Bob Kolker > Please, will you be kind enough as to point me in the direction of > research which will allow me to earn the way of thinking of a function > such that this particular way will be isomorphic to all function > representations in any (scientific) discourse. > Andrew === Subject: Re: The Liar Paradox is merely an ill-formed statement : There is no such thing as recursion. Some people may think there is, : but they are victims of a popular misconception. Damn, you're good. -- --- The history of our nation has demonstrated that separate is seldom, if ever, equal. === Subject: Re: The Liar Paradox is merely an ill-formed statement > : There is no such thing as recursion. Some people may think there is, > : but they are victims of a popular misconception. > Damn, you're good. Kind of a random question: if you have an inductive proof in a constructive logic, does that always correspond to a recursive program in a functional language? And is the other direction true, too, for programs that have well-founded recursion? -- Neel Krishnaswami neelk@cs.cmu.edu === Subject: rank of a random matrix R is a random m x 2m matrix with entries identically, independently chosen from some continuous distribution, say N(0,1), in the real domain. R' is a m x m matrix formed by removing any m columns from the matrix R. I was wondering what is the probability that R' is of full rank. === Subject: Re: rank of a random matrix >R is a random m x 2m matrix with entries identically, independently >chosen from >some continuous distribution, say N(0,1), in the real domain. >R' is a m x m matrix formed by removing any m columns from the matrix R. >I was wondering what is the probability that R' is of full rank. The probability is 1. Each m x m determinant is a nonconstant polynomial function of the matrix entries. For any nonconstant polynomial f(x_1,...,x_N), if X_1,...,X_N are random variables with a distribution absolutely continuous wrt Lebesgue measure on R^N, Prob[f(X_1,...,X_N) = 0] = 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: rank of a random matrix >R is a random m x 2m matrix with entries identically, independently >chosen from >some continuous distribution, say N(0,1), in the real domain. >R' is a m x m matrix formed by removing any m columns from the matrix R. >I was wondering what is the probability that R' is of full rank. >The probability is 1. Each m x m determinant is a nonconstant >polynomial function of the matrix entries. For any nonconstant >polynomial f(x_1,...,x_N), if X_1,...,X_N are random variables with >a distribution absolutely continuous wrt Lebesgue measure on R^N, >Prob[f(X_1,...,X_N) = 0] = 0. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 Mathematician: Think of a number between 1 and 100. Guy: OK, I have one. Mathematician: Is it rational? Guy: Yes. Mathematician: How unlikely! --Lynn === Subject: interesting, but lost Re: More Coin Tossing by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4IKKif25490; . My probability is so rusty I cannot properly understand the dispute between you and mensanator, but my crude intuition would be that the two events (a total number of heads, and a run of so many heads) would be, yes, partly connected and not independent. Could anybody indulge me with a childishly simplified numerical version of this disagreement, perhaps involving a few throws of one 6-die, or five throws of a coin, or something numerically very simple indeed? My apologies if someone did just this later in the thread - I only read 8 or 9 responses into this discussion. Mark Griffith >As others (e.g., G.A. Edgar) have pointed out, knowing >something about the longest run gives you information about the total>number of tosses. > >About the _expected_ total number of tosses. >More than that - about the entire distribution of the random variable. I.e., just what I said. >-- >Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: interesting, but lost Re: More Coin Tossing Perhaps this is what you want. Flip a fair coin twice. The event that ßip one is a head is independent of the event that ßip two is a head; the probability of each of these events is 1/2. Now suppose I tell you that at least one of the ßips is a head. Conditionally, the outcomes are no longer independent or fair. For now I have told you that any of the three equally likely following outcomes occurred: HT TH HH Hence, the conditional probability that a particular ßip is a head is 2/3, whereas the conditional probability that both are heads is 1/3 != 2/3 * 2/3. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Conclusion when infinity is possible. Say u is a measure, set G contains set A, u(A) is finite, and one can conclude that u(G-A)The entire point is to factor some number M by factoring an easy >number T, off M. I'm unfamiliar with this mathematical concept of off. Given two numbers a and b, how does one determine if a is off b? === Subject: Re: Factorizaton idea, revisited >The entire point is to factor some number M by factoring an easy >number T, off M. > I'm unfamiliar with this mathematical concept of off. He means, differing from M by some small offset. === Subject: Newton's mass versus the standard kilogram A recent translation of Sir Isaac Newton's The Principia: Mathematical Principles of Natural Philosophy, originally published in 1686,5 puts the intellectual challenge into perspective. With the following words, Newton opened the monumental treatise that laid the foundation for modern scientific and mathematical thought: Quantity of matter is a measure of matter that arises from its density and volume jointly. Thus, Newton articulated, for the first time in scientific history, a mathematical relationship between density, volume and quantity of matter (which he later called mass.) Still the concept of mass was confusing until quite recently, because no one clearly understood what he meant by Ôdensity and volume jointly': Still no one understands; but it has become a moot point, and hardly worth arguing about. This was finally cleared up by the metric commitee's minting of the standard kilogram of the archives, and declaring it to be the standard against which all other masses - quantities of matter - would henceforth be compared; using balance scales. === Subject: Re: Newton's mass versus the standard kilogram Aren't you the same idiot who recently claimed that Newton invented mass? Or are you a different idiot? > A recent translation of Sir Isaac Newton's The Principia: > Mathematical Principles of Natural Philosophy, originally published in > 1686,5 puts the intellectual challenge into perspective. With the > following words, Newton opened the monumental treatise that laid the > foundation for modern scientific and mathematical thought: > Quantity of matter is a measure of matter that arises from its > density and volume jointly. > Thus, Newton articulated, for the first time in scientific history, a > mathematical relationship between density, volume and quantity of > matter (which he later called mass.) > Still the concept of mass was confusing until quite recently, because > no one clearly understood what he meant by Ôdensity and volume > jointly': Still no one understands; but it has become a moot point, > and hardly worth arguing about. > This was finally cleared up by the metric commitee's minting of the > standard kilogram of the archives, and declaring it to be the standard > against which all other masses - quantities of matter - would > henceforth be compared; using balance scales. === Subject: Re: Newton's mass versus the standard kilogram > A recent translation of Sir Isaac Newton's The Principia: > Mathematical Principles of Natural Philosophy, originally published in > 1686,5 puts the intellectual challenge into perspective. With the > following words, Newton opened the monumental treatise that laid the > foundation for modern scientific and mathematical thought: > Quantity of matter is a measure of matter that arises from its > density and volume jointly. Idiot - what is the citation? Is it Motte and Cajori or Cohen and Whitman? What page, Dumb Donny Head? We all know your lbirary card was confiscated and destroyed for abuse of sub-literacy. any progress has been made in the intervening 318 years? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Newton's mass versus the standard kilogram > A recent translation of Sir Isaac Newton's The Principia: > Mathematical Principles of Natural Philosophy, originally published in > 1686,5 puts the intellectual challenge into perspective. With the > following words, Newton opened the monumental treatise that laid the > foundation for modern scientific and mathematical thought: > Quantity of matter is a measure of matter that arises from its > density and volume jointly. > Idiot - what is the citation? Is it Motte and Cajori or Cohen and > Whitman? What page, Dumb Donny Head? We all know your lbirary > card was confiscated and destroyed for abuse of sub-literacy. > any progress has been made in the intervening 318 years? > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > Quis custodiet ipsos custodes? The Net! Looks the Shead used: http://tlc.ousd.k12.ca.us/~acody/densitymisc.html Which paraphrased from 6. Newton, Isaac, The Principia: Mathematical Principles of Natural Philosophy, Translated by I. Bernard Cohen and Anne Whitman, Preceded by A Guide to Newton's Principia by I. Bernard Cohen, University of California Press, Berkeley, 1999, pp. 86 -95 and pp. 403 - 404. ^^ === Subject: Re: Newton's mass versus the standard kilogram > > A recent translation of Sir Isaac Newton's The Principia: > Mathematical Principles of Natural Philosophy, originally published in > 1686,5 puts the intellectual challenge into perspective. With the > following words, Newton opened the monumental treatise that laid the > foundation for modern scientific and mathematical thought: > > Quantity of matter is a measure of matter that arises from its > density and volume jointly. > Idiot - what is the citation? Is it Motte and Cajori or Cohen and > Whitman? What page, Dumb Donny Head? We all know your library > card was confiscated and destroyed for abuse of sub-literacy. > any progress has been made in the intervening 318 years? > Looks the Shead used: > http://tlc.ousd.k12.ca.us/~acody/densitymisc.html > Which paraphrased from > 6. Newton, Isaac, The Principia: Mathematical > Principles of Natural Philosophy, Translated by > I. Bernard Cohen and Anne Whitman, Preceded > by A Guide to Newton's Principia by I. Bernard > Cohen, University of California Press, Berkeley, > 1999, pp. 86 -95 and pp. 403 - 404. > ^^ That is a public school site. Dumb Donny Head is feigning scholarship from an 8th grader curriculum summary written by a diversity-humped female PhD in Science Education consulting? Perhaps, we should not be surprised that young (and not-so-young) learners of the 21st century, finds [sic] these ideas challenging. [Note the grammar error.] The whole point of the quote was to show the Newton's difficulty in grappling with the then poorly characterized concept of mass. Newton didn't have decimal points, either. They had been introduced to Engalnd about 40 years earlier and were still distrusted. It's a moot point in any case. Strongly curved spacetimes have no easily defined volumes (e.g., what is the diameter of a black hole's event horison? It's inarguably infinite on the inside and enirely finite on the outside if you derive it from circumference, which is also bogus). A relativistic universe has four independent distances - luminosity (inverse square), angular diameter, parallax, and proper motion. No two of them need agree to maintain consistency. 4-mass (more properly, 4-momentum) is easily defined and manipulated. Footnote 4: Oakland Unified School District, Science Content Standards, Oakland Board of Education, 1997, Eighth Grade Science Standards. The Oakland School board was the initiating proponent of Ebonics, attmepting to drop the California State median somewhere closer to Oakland's mean. The only way you're gonna see a White kid in an Oakland public school is if Mayor Jerry Brown reproduces. You'd have to import from San Francisco across the Oakland Bay Bridge - and keep the poor bastard under armed guard, and guard the guards or nobody keeps their lunch money. Dumb Donny Head missed the real goodie somewhat further down, Furthermore, I mean this quantity whenever I use the term body or mass in the following pages. It can always be known from a body's weight, for - by making very accurate measurements with pendulums - I have found it to be proportional to the weight, as will be shown below. That is Newton's statement of the Equivalence Principle, the critical founding postulate of General Relativity. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Newton's mass versus the standard kilogram > A recent translation of Sir Isaac Newton's The Principia: > Mathematical Principles of Natural Philosophy, originally published in > 1686,5 puts the intellectual challenge into perspective. With the > following words, Newton opened the monumental treatise that laid the > foundation for modern scientific and mathematical thought: > Quantity of matter is a measure of matter that arises from its > density and volume jointly. Not unlike today's mathematical description http://scienceworld.wolfram.com/physics/Mass.html m = Integral [ rho dV ] where rho equals density. And we go even further by assigning mass as a fundamental quantity in our measurement system. http://physics.nist.gov./cuu/Units/units.html === Subject: pair of winding cables question If two cables in mid-air strike each other's cross-sections, there may be a moment of deßection, but in the real world the general rule seems to be that they will always find themselves wrapping around each other. If I know the radius of each cable (its usual thickness, no variation from expansion from heating or shrinkage from compression), and the relative velocities (x, y, z per unit time) of the actual physical mid-point of each cable, and the distance from that midpoint where the two cables first collide, is there a formula available to me to predict the speed with which the cables will wrap themselves? I will assume that in those cases where the two cables strike each other away from each other's mid-points, at least one end will end up hanging free of the other. The most important variable I want to find out, is that of Ôt' for time. If it makes life any simpler, assume that both cables are straight lines when they first come into contact. If I can ignore factors relating to friction or compression, and if I can assume that the cables are equally highly pliant, what other variables do I need to answer my question? === Subject: Re: pair of winding cables question |If I can ignore factors relating to friction or compression, and if I can |assume that the cables are equally highly pliant, what other variables do |I need to answer my question? For instance, if I know the speed of the mid-points through space, is there a way of predicting the speed of the end-points as they wrap around each other, starting at the point of initial contact? Are the speeds constant? Or do the speeds of the end-points decrease as they wrap around each other, approaching some speed representing an average of the speeds they had before their first contact? === Subject: Re: pair of winding cables question ||If I can ignore factors relating to friction or compression, and if I can ||assume that the cables are equally highly pliant, what other variables do ||I need to answer my question? | |For instance, if I know the speed of the mid-points through space, is |there a way of predicting the speed of the end-points as they wrap |around each other, starting at the point of initial contact? | |Are the speeds constant? Or do the speeds of the end-points decrease |as they wrap around each other, approaching some speed representing an |average of the speeds they had before their first contact? Finally, I know that striking a string in some particular point between its two ends can initiate a vibration (described in cycles per unit time) that continues until it is damped, and however the damping process process works, it is bound to conclude when the lengths of the cable have been run through and wrapped, and a free end (if there is one) extends beyond the wrap. === Subject: Re: pair of winding cables question |I will assume that in those cases where the two cables strike each other |away from each other's mid-points, at least one end will end up hanging |free of the other. The most important variable I want to find out, is |that of Ôt' for time. If it makes life any simpler, assume that both |cables are straight lines when they first come into contact. I also would like to assume that the cables are shaped like segments of straight lines, that when they make contact, the angles of their initial contact when measured are found to be less than 90 degrees but greater than 0 degrees. === Subject: Re: smallest disk covering a set of points > I don't think your analogy holds. For example, constructing a tangent to > a circle at a given point is an example of a practical problem that > involves an infinite point set (the circle). Similar examples abound. > The problem being considered in this thread remains practical and > solvable in many situations where the point set is infinite, provided it > can be finitely described (a union of finitely many simple geometric > figures, perhaps). Or a closed-form description of a sequence of points; for instance, the sequence {Pk}, where Pk = (1-2^-k, 0), k = 0,1, ..... Carlos -- === Subject: Re: smallest disk covering a set of points >And I'm in comp.programming. However the existence proof must be >trivial. For any bounded set, express the points in polar >coordinates. Sort on magnitude. The circle at the origin with a >radius equal to the largest magnitude must suffice. > That's what I meant by every bounded point set clearly has one. > The slightly less trivial question I was answering was that a *minimum* > covering disk exists. The existance of /any/ covering disk is proof of existance of a /minimum/ covering disk. What further proof is required? -- Corey Murtagh The Electric Monk Quidquid latine dictum sit, altum viditur! === Subject: Re: smallest disk covering a set of points 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > That's what I meant by every bounded point set clearly has one. > The slightly less trivial question I was answering was that a *minimum* > covering disk exists. > The existance of /any/ covering disk is proof of existance of a > /minimum/ covering disk. What further proof is required? Well, the existence of a covering disk does not prove the existence of a maximum covering disk, since there never exists a maximum. So, some proof that is capable of distinguishing minima from maxima is required. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: smallest disk covering a set of points <%gbpc.73721$Qe.846289@wagner.videotron.net> > That's what I meant by every bounded point set clearly has one. > The slightly less trivial question I was answering was that a *minimum* > covering disk exists. > The existence of /any/ covering disk is proof of existence of a > /minimum/ covering disk. What further proof is required? Your first statement is only vacuously true (since just about *any* premise could be considered (vacuous) proof of a true conclusion, and David Eppstein has shown the conclusion to be true). However, your proof is not satisfying, because you haven't told us why it holds in the case of minimal covering disk while obviously *not* holding in the case of minimal integer or minimal circle enclosing but not containing all points in the set. I.e., what is the essential difference between > The existence of /any/ covering disk is proof of existence of a > /minimum/ covering disk and | The existence of /any/ integer is proof of existence of a | /minimum/ integer The second statement is ridiculously untrue. So why should the first statement, syntactically equivalent, be considered a rigorous proof of itself? [FWIW, when I originally posed the question, I was not sure whether there might exist sets of points not coverable by a disk of radius 1, but coverable by any disk of radius 1+epsilon. Apparently this is ruled out by something called compactness, about which I do not know but into which I will try to remember to look.] HTH, -Arthur === Subject: Re: smallest disk covering a set of points >Is there a fast algorithm for finding the smallest disk that covers a set >of points in the plane? > Turns out that's not necessary... apart from the stuff I got from other > posters to the thread, there's an excellent page about the problem here: > http://www.cs.wustl.edu/~pless/506/l11.html I have implemented the algorithm described in these notes. It might be of use if people want to compare methods, or get stuck on the geometry. http://visula.org/calum/cover.cpp There is a small error in the notes: step (1) of MinDisk should probably read |P| < 3, since if we have three points, we don't necessarily want to go through all of them. This method is fast: 3.5s for 100M points! Calum === Subject: Re: smallest disk covering a set of points Calum coughed up the following: > Is there a fast algorithm for finding the smallest disk that > covers a set of points in the plane? > Turns out that's not necessary... apart from the stuff I got from > other posters to the thread, there's an excellent page about the > problem here: > http://www.cs.wustl.edu/~pless/506/l11.html > I have implemented the algorithm described in these notes. It might > be > of use if people want to compare methods, or get stuck on the > geometry. > http://visula.org/calum/cover.cpp > There is a small error in the notes: step (1) of MinDisk should > probably read |P| < 3, since if we have three points, we don't > necessarily want > to go through all of them. > This method is fast: 3.5s for 100M points! > Calum I'd love to implement this in java / javascript / ßash / whatever, in a web page, along with an interface for folks to plop points down and test the resulting circle. It would be a great way to see this sucker at work. I'm too busy :( :( :( :( -- While using is ok, actually /writing/ free software is a disingenuous activity. You can afford to write software for free only because of someone else somewhere actually paying for it. Just say no. === Subject: Re: smallest disk covering a set of points >Is there a fast algorithm for finding the smallest disk that covers a set >of points in the plane? >(Don't they teach you harvard weenies about convex hulls?) > not in the Art History department they don't, no. >This is what I would do: >Step 1: Find the minimal set of points which generates the convex hull >of the set of points. > Turns out that's not necessary... apart from the stuff I got from other > posters to the thread, there's an excellent page about the problem here: > http://www.cs.wustl.edu/~pless/506/l11.html > which reduces it to a problem that is linear in n. Does it? The algorithm given at the end is actually cubic. What have I missed? Is the probability of calling MinDiskWith1Pt and MinDiskWith2Pts sufficiently low that these cubic cases occur sufficiently rarely to give a linear average case? This seems likely, intuitively, the probability of calling MinDiskWith1Pt decreases (linearly??) with i. Can the cubic worst case ever occur? If so, we should try to devise an algorithm that does not possess this worst case. Calum === Subject: AIM Research only for Ph. D.'s? AIM Executive Director and ARCC Director Mathematics research is only for those who have advanced degrees.... I'm very puzzled that you only allow PH. D.'s to apply for fellowships. There are a very large number of people who have made contribution to Mathematics who had no such degree. They continue to do so even today. I'm really sorry to find such a narrow scope to an organization that seems to have lofty ideas about Mathematics. Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: AIM Research only for Ph. D.'s? > AIM Executive Director and > ARCC Director > Mathematics research is only for those who have advanced degrees.... > I'm very puzzled that you only allow PH. D.'s to apply for fellowships. > There are a very large number of people > who have made contribution to Mathematics who had no such degree. > They continue to do so even today. > I'm really sorry to find such a narrow scope to an > organization that seems to have lofty ideas about Mathematics. > Respectfully, Roger L. Bagula > tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : > URL : http://home.earthlink.net/~tftn > URL : http://victorian.fortunecity.com/carmelita/435/ Is there a PhD shortage? Why would an organization pay to evaluate applicants when they could simply ask for PhD which acts to evaluate them instead? I agree that a non-PhD can do high quality math, but I'm not going to hire them when I can hire someone else without having to spend more money up front just to see if they're worth keeping, so I suspect it's an economic decision not a philosophical one. === Subject: Re: PROOF that emulators are impossible > The Pretender with video memory, immortality, instant stud appeal, guess > anything correct, infinite knowledge, channeling for knowledge, > ambidextrous, > freckles on accupressure points, strength of 3 men, regeneration, .......... >halve my heart rate consciously, automatic pelvis gyration, perfect >coordiation, >telepathy, accelerated learning, author of G.U.T., Cyberchrist, Adam, >thousands >of powers, not one to make you listen. You must find that annoying. You know you have all these near-miraculous powers, and yet you lack the power to convince most of us mortals of that fact. Maybe condescending to tidy up your G.U.T. enough to get it published in Physical Review Letters, and then probably getting a Nobel Prize, would convince a lot of us. Or maybe getting an Olympic gold for weightlifting would be a start. I would however strongly suggest that you NOT try any direct demonstration of your alleged immortality. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- === Subject: David Ullrich and Oklahoma State University by James Harris === Subject: David Ullrich and Oklahoma State University Author: James Harris Now David Ullrich is a professor at Oklahoma State University, but I keep seeing people acting like I'm wrong for believing that if he posts using his university affiliation that he should be held to some standard. My thinking is that if he just posts as a netizen that's one thing, but if he posts as a professor at Oklahoma State University, then some standards should apply, but here's a posting from Ullrich to the newsgroup. alt.math.undergrad among others, where he uses expletives and still doesn't give a math argument. Notice his organization listing, which is Oklahoma State University. Now I talked once with officials from that school but they told me it was a private matter. Am I silly for thinking this professor of their's is representing that school badly? Maybe out in Oklahoma they don't care about such things, eh? Well that would actually be kind of ok in my opinion as it'd be nice to know they're more liberal in Oklahoma than I'd guessed. James Harris -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: David Ullrich and Oklahoma State University by James Harris > [...] What's your aim in rehashing these old arguments? -- G.C. === Subject: Re: David Ullrich and Oklahoma State University by James Harris--> troll alert I'm sorry that my former experiences with Dr. Ullrich taint my opinion. I am concerned that people associated with Mathematics may come to believe he is representative of Mathematicians in general. But it should be known that he is one of the worst sci.math trolls in it's history. One of the worst trolls in newsgroup history. He is deadly and very destructive. Mostly because he is a genius level mathematician with a large amount of knowledge and as far as I can tell no moral code at all. It is a sorry state of affairs when he can beat out numerologist Daryl S. Kabatoff in this category. but it is his evil attacks that single him out as dangerous. I once quoted a theorem from a recognized text word for word ( typed it in) and he denied it's truth. I get scared when he posts in the same thread as he has attacked me and personally hurt me in the past. I have nightmares about this guy! Praise from him is like being approved by the devil??? I just felt that I should warn others and make it clear he is known for his ßame wars that hurt people. I couldn't in conscience go without saying this. He shouldn't be allowed near people's children to teach them in my opinion. === > Subject: David Ullrich and Oklahoma State University > Author: James Harris > Now David Ullrich is a professor at Oklahoma State University, but I > keep seeing people acting like I'm wrong for believing that if he > posts using his university affiliation that he should be held to some > standard. > My thinking is that if he just posts as a netizen that's one thing, > but if he posts as a professor at Oklahoma State University, then some > standards should apply, but here's a posting from Ullrich to the > newsgroup. alt.math.undergrad among others, where he uses expletives > and still doesn't give a math argument. > Notice his organization listing, which is Oklahoma State University. > Now I talked once with officials from that school but they told me it > was a private matter. Am I silly for thinking this professor of > their's is representing that school badly? > Maybe out in Oklahoma they don't care about such things, eh? > Well that would actually be kind of ok in my opinion as it'd be nice > to know they're more liberal in Oklahoma than I'd guessed. > James Harris === Subject: Re: David Ullrich and Oklahoma State University by James Harris--> troll alert >I'm sorry that my former experiences > with Dr. Ullrich taint my opinion. >I am concerned that people >associated with Mathematics may come > to believe he is representative of >Mathematicians in general. >But it should be known that he is one of the worst sci.math trolls in >it's history. >One of the worst trolls in newsgroup history. >He is deadly and very destructive. >Mostly because he is a genius level mathematician with a large amount of >knowledge and as far as I can tell >no moral code at all. It's time that Usenet replenished its supply of highly moral morons. Lee Rudolph === Subject: TI-89 Programs for Calculus III Anyone recommend/have some good TI-89 programs for help with verifying Calculus III problems? Or multivariable problems like tangent plane to a surface, green's theorem, line integrals, etc? Netsniper === Subject: Re: I need help with proof ETAtAhRi19VDAna15qReFKfCYS1f3CfdxAIVAIQ9ojOhLqXV6/S9jW/ CsYlTnWNX It depends on how you count. x^2-2x+1 = 0 has one root, not two, unless you count the double root twice. The theorem assumes that repeated roots are counted as many times as the exponent on the corresponding factor of the polynomial whose zeroes are being sought. Start with the Fundamental Theorem of Algebra, which says that there is at least ONE root. Call this root a. The Factor Theorem, applicable to complex numbers, guarantees that the given polynomial P(x) equals Q(x)*(x-a). Thus any remaining zeroes are zeroes of Q(x) which is of degree n-1. So the theorem allpies to polynoials of degree n if it applies to polynomials of degree n-1, and you have a mathematical induction. I leave the grounding of the induction up to you. --OL === Subject: Re: I need help with proof >Are there any easy way to proove that a polynomial of order n has >got n and only n complex roots? >That the polynomial has n complex roots follows from the Fundamental >Theorem of Algebra. >Yes, this is what confuses me a little. The Fundamantal Theorem says >that any polynomial of order n has at least one complex root. I dont >understand how this implies the existence of exactly Ôn' roots. Of course you don't understand it, because it isn't true. The polynomial x^2 - 2x + 1 has only one complex root, 1. Mathematicians are supposed to be precise when making mathematical statements, but it never ceases to amaze me how imprecise they are in practice. They defend their imprecision by saying that everybody knows what they really mean i.e. there are n roots up to multiplicity. The fundamental theorem implies that a polynomial of degree n over the complex numbers has factorizes into n not necessarily distinct linear factors. Derek Holt. > That it has only n comple roots follows from the >fact that the complex numbers form a field. Specifically, suppose f >is a polynomial of degree n, and its roots include x_0,...,x_n (counting >multiplicities), then f(x) should be divisible by a polynomial of degree >n+1, i.e. prod_{j=0}^n (x-x_j). Since f has degree n, then it is not >divisible by a polynomial of degree n+1, or higher. >Ok thank you. === Subject: Re: I need help with proof <3jtha01f7cbt1hmur0m0gse8j4cqvkl8dn@4ax.com> > Never mind I think I understand why > a polynomial f(x) of order n has a root z_o from the fundamental > theorem, which means that f(x) can be factored : > f(x) = (z - z_0)g(x) where g(x) has order n-1. g(x) has at least one > root z_1 and f(x) can be factored : > f(x) = (z - z_0)(z - z_1)h(x) where h(x) har order n - 2. Induction > shows that f(x) has n and only n complex roots. Some of which may be duplicates. For example x^3 (x - 1)^2 (x - 2) has six roots and three distinct roots. > This should in other words also imply that f(x) cannot have more than > n complex roots, or am I totally wrong? Yes. === Subject: Re: Primes and Diophantine Equations > If I have a simple Diophantine equation, is there any way to check to see if > the primes are a subset of it? > For example: > 2x+2y (2 is the only prime in the solution set) > xy-1 (All of the primes are in the solution set) > x^2 + y^2 (All of the primes = 4k+1 are in the solution set; none of the > primes = 4k+3 are) x^2+xy+y^2 : All the primes = 3k+1 in solution set; none of primes 3k-1 are; 3 is in solution set. === Subject: Re: Primes and Diophantine Equations > If I have a simple Diophantine equation, is there any way to check to see if > the primes are a subset of it? > For example: > 2x+2y (2 is the only prime in the solution set) > xy-1 (All of the primes are in the solution set) > x^2 + y^2 (All of the primes = 4k+1 are in the solution set; none of the > primes = 4k+3 are) None of 2x+2y, xy-1, x^2 + y^2 is an equation. Do you mean: Can the primes be represented as 2x+2y, xy-1, x^2 + y^2 (or whatever)? Note xy - 1 cannot represent the prime 2. -- G.C. === Subject: Re: Primes and Diophantine Equations > If I have a simple Diophantine equation, is there any way to check to see if > the primes are a subset of it? > For example: > 2x+2y (2 is the only prime in the solution set) > xy-1 (All of the primes are in the solution set) > x^2 + y^2 (All of the primes = 4k+1 are in the solution set; none of the > primes = 4k+3 are) > None of 2x+2y, xy-1, x^2 + y^2 is an equation. Do you mean: Can the > primes be represented as 2x+2y, xy-1, x^2 + y^2 (or whatever)? Note xy > - 1 cannot represent the prime 2. Yes Given set S=f(x,y) x,y e Z, which primes are elements of S By the way xy-1 = 2 when x=1 and y=3 === Subject: Re: Primes and Diophantine Equations > ... Note xy > - 1 cannot represent the prime 2. .. > By the way xy-1 = 2 when x=1 and y=3 Silly me! -- G.C. === Subject: Re: Primes and Diophantine Equations > > If I have a simple Diophantine equation, is there any way to check to > see if > the primes are a subset of it? > > For example: > 2x+2y (2 is the only prime in the solution set) > xy-1 (All of the primes are in the solution set) > x^2 + y^2 (All of the primes = 4k+1 are in the solution set; none of the > primes = 4k+3 are) > None of 2x+2y, xy-1, x^2 + y^2 is an equation. Do you mean: Can the > primes be represented as 2x+2y, xy-1, x^2 + y^2 (or whatever)? Note xy > - 1 cannot represent the prime 2. > Yes > Given set S=f(x,y) x,y e Z, which primes are elements of S > By the way xy-1 = 2 when x=1 and y=3 I think there are more questions here than answers. E.g., no one knows whether there are infinitely many primes of the form n^2 + 1. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Primes and Diophantine Equations > > If I have a simple Diophantine equation, is there any way to check to > see if > the primes are a subset of it? > > For example: > 2x+2y (2 is the only prime in the solution set) > xy-1 (All of the primes are in the solution set) > x^2 + y^2 (All of the primes = 4k+1 are in the solution set; none of the > primes = 4k+3 are) > > None of 2x+2y, xy-1, x^2 + y^2 is an equation. Do you mean: Can the > primes be represented as 2x+2y, xy-1, x^2 + y^2 (or whatever)? Note xy > - 1 cannot represent the prime 2. > > Yes > Given set S=f(x,y) x,y e Z, which primes are elements of S > By the way xy-1 = 2 when x=1 and y=3 > I think there are more questions here than answers. E.g., > no one knows whether there are infinitely many primes > of the form n^2 + 1. As Gerry says, there are very few proven facts. There's Dirichlet's theorem (if gcd(a,b)=1, then ax+b is prime infinitely often) and an analogous theorem for binary quadratic forms ax^2+bxy+cy^2. If you're willing to settle for conjectures, there are some very general ones. More or less, f(x_1,...,x_n) should be prime infinitely often unless there is some obvious reason that it can't be. [An obvious reason would be that there is an integer d > 1 that divides every value, as in your example 2x+2y, or in the less obvious example x^2+x, which is also always divisible by 2.] You might take a look at Schinzel's Hypothesis, see http://mathworld.wolfram.com/SchinzelsHypothesis.html The version on MathWorld just deals with polynomials of one variable, but it asks for them all to be prime simultaneously. I believe that Schinzel's most general conjecture also deals with the multivariable Joe Silverman === Subject: Re: Primes and Diophantine Equations > If you're willing to settle for conjectures, there are some very > general ones. More or less, f(x_1,...,x_n) should be prime infinitely > often unless there is some obvious reason that it can't be. [An > obvious reason would be that there is an integer d > 1 that divides > every value, as in your example 2x+2y, or in the less obvious example > x^2+x, which is also always divisible by 2.] Perhaps better to cite x^2 + x + 2, as x^2 + x = x(x + 1) can be seen to be generally composite without invoking divisibility by 2. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Need Euler Angle help Suppose I have a body-axis in a 3-axis system in a coordinate system: X = [0.5774 0.5774 0.5774] Y = [-0.5774 0.5774 0.5774] Z = [0 -0.7071 0.7071] How do I calculate the euler angles YPR to the original system? I have the yaw, pitch and roll matrices but I dont know how to solve for the yaw pitch and roll values. P = [cos(pitch) 0 sin(pitch); 0 1 0; -sin(pitch) 0 cos(pitch)]; Y = [cos(yaw) -sin(yaw) 0; sin(yaw)*pi/180) cos(yaw) 0; 0 0 1]; === === Subject: definition of Gamma function? I am trying to learn about the Gamma function. I found a website that gives three definitions (presumably all consistent!) of the Gamma function, given that the real component is positive. What is the definition of the Gamma function for the real component nonpositive? Ted Shoemaker === Subject: Re: definition of Gamma function? > I am trying to learn about the Gamma function. I found a website that > gives three definitions (presumably all consistent!) of the Gamma > function, given that the real component is positive. > What is the definition of the Gamma function for the real component > nonpositive? > Ted Shoemaker Does this help? http://mathworld.wolfram.com/GammaFunction.html === Subject: Re: RRRRIIINNNNGGG -> Re: And the rantings go on -> Re: 911 Aftermath: Scalar Weapons Used in World Trade Tower Attacks X-Meow: Approved X-Tra: time X-Files: off Kirk >posted: > >http://scienceworld.wolfram.com/physics/ LorentzInvariant.html > > > Equation number 4 can't equal zero. >It's the canonical form of the famous: >E=mc^2 >equation >http://www.aip.org/history/einstein/emc1.htm >BTW: you can learn something about interferometers too, that word you >love to interject sometimes, in the hope to sound scientifical. > Heh, I'm into keeping it simple. Einstein was correct, but the equation > CAN NOT equal zero. Nothing can equal zero, if it does then something's > missing. There's more to energy than that. the map exp: x|->exp(x) maps every number in to a non-zero positive number, so where is the problem? Mathematics could survive with positive and non zero numbers, but introducing a zero, and the notion of negative numbers seemed to be way more easy. In't only a question of setting a framework. -- mhm 27x12 smeeter #28 Usenet Valhalla Circle #19 & #21 Bartlo's hate lits #1: <40376AD8.C83FBF5A@enter.net> CEO Alcatroll Labs Inc. Alexa knows everything about PSI, electromagnetic radiation and quantum mechanics: They are beaming earth with EMR which is interfering with natural human PSI abilities, which is likely quantum. in alt.alien.visitors === Subject: Re: RRRRIIINNNNGGG -> Re: And the rantings go on -> Re: 911 Aftermath: Scalar Weapons Used in World Trade Tower Attacks >posted: > >http://scienceworld.wolfram.com/physics/ LorentzInvariant.html > > > Equation number 4 can't equal zero. >It's the canonical form of the famous: >E=mc^2 >equation >http://www.aip.org/history/einstein/emc1.htm >BTW: you can learn something about interferometers too, that word you >love to interject sometimes, in the hope to sound scientifical. > Heh, I'm into keeping it simple. Einstein was correct, but the equation > CAN NOT equal zero. Nothing can equal zero, if it does then something's > missing. There's more to energy than that. >the map exp: x|->exp(x) maps every number in to a non-zero positive >number, so where is the problem? >Mathematics could survive with positive and non zero numbers, but >introducing a zero, and the notion of negative numbers seemed to be way >more easy. >In't only a question of setting a framework. Reducing all the numbers to 1 and 2 using computers can solve the equation without introducing zero. === Subject: Re: RRRRIIINNNNGGG -> Re: And the rantings go on -> Re: 911 Aftermath: Scalar Weapons Used in World Trade Tower Attacks X-Meow: Approved X-Tra: time X-Files: off Kirk >posted: > >posted: > > >http://scienceworld.wolfram.com/physics/ LorentzInvariant.html > > > Equation number 4 can't equal zero. > >It's the canonical form of the famous: > >E=mc^2 >equation > >http://www.aip.org/history/einstein/emc1.htm > >BTW: you can learn something about interferometers too, that word you >love to interject sometimes, in the hope to sound scientifical. > > Heh, I'm into keeping it simple. Einstein was correct, but the equation > CAN NOT equal zero. Nothing can equal zero, if it does then > something's missing. There's more to energy than that. >the map exp: x|->exp(x) maps every number in to a non-zero positive >number, so where is the problem? >Mathematics could survive with positive and non zero numbers, but >introducing a zero, and the notion of negative numbers seemed to be way >more easy. >In't only a question of setting a framework. > Reducing all the numbers to 1 and 2 using computers can solve the equation > without introducing zero. You'll only exclude all irrational numbers there, not honest! -- mhm 27x12 smeeter #28 Usenet Valhalla Circle #19 & #21 Bartlo's hate lits #1: <40376AD8.C83FBF5A@enter.net> CEO Alcatroll Labs Inc. Alexa knows everything about PSI, electromagnetic radiation and quantum mechanics: They are beaming earth with EMR which is interfering with natural human PSI abilities, which is likely quantum. in alt.alien.visitors === Subject: Re: RRRRIIINNNNGGG -> Re: And the rantings go on -> Re: 911 Aftermath: Scalar Weapons Used in World Trade Tower Attacks >posted: > >posted: > > >http://scienceworld.wolfram.com/physics/ LorentzInvariant.html > > > Equation number 4 can't equal zero. > >It's the canonical form of the famous: > >E=mc^2 >equation > >http://www.aip.org/history/einstein/emc1.htm > >BTW: you can learn something about interferometers too, that word you >love to interject sometimes, in the hope to sound scientifical. > > Heh, I'm into keeping it simple. Einstein was correct, but the equation > CAN NOT equal zero. Nothing can equal zero, if it does then > something's missing. There's more to energy than that. >the map exp: x|->exp(x) maps every number in to a non-zero positive >number, so where is the problem? >Mathematics could survive with positive and non zero numbers, but >introducing a zero, and the notion of negative numbers seemed to be way >more easy. >In't only a question of setting a framework. > Reducing all the numbers to 1 and 2 using computers can solve the equation > without introducing zero. >You'll only exclude all irrational numbers there, not honest! Computers are non-emotional, but the people operating them many not like the result. === Subject: Re: RRRRIIINNNNGGG -> Re: And the rantings go on -> Re: 911 Aftermath: Scalar Weapons Used in World Trade Tower Attacks X-Meow: Approved X-Tra: time X-Files: off Kirk >posted: > >posted: > > >posted: > > >http://scienceworld.wolfram.com/physics/ LorentzInvariant.html > > > Equation number 4 can't equal zero. > >It's the canonical form of the famous: > >E=mc^2 >equation > >http://www.aip.org/history/einstein/emc1.htm > >BTW: you can learn something about interferometers too, that word you >love to interject sometimes, in the hope to sound scientifical. > > Heh, I'm into keeping it simple. Einstein was correct, but the > equation CAN NOT equal zero. Nothing can equal zero, if it does then > something's missing. There's more to energy than that. > >the map exp: x|->exp(x) maps every number in to a non-zero positive >number, so where is the problem? > >Mathematics could survive with positive and non zero numbers, but >introducing a zero, and the notion of negative numbers seemed to be way >more easy. > >In't only a question of setting a framework. > > Reducing all the numbers to 1 and 2 using computers can solve the > equation without introducing zero. >You'll only exclude all irrational numbers there, not honest! > Computers are non-emotional, but the people operating them many not like > the result. They might feel debunked for talking nonsense and using scientific terms they have not the slightest idea about, like your retarded uncle Tom, who pretends to be a nuclear scientist, though a military man, only learned how he could impress the guillible with fancy combined pseudo scientifical terms. Cold explosions. Free energy from the quantum ßucuations of the vacuum Scalar Electromagnetic Radiation weapons EMR used for mind control Black Hole implosion at 9/11 conspired by ....... and so and so on...... -- mhm 27x12 smeeter #28 Usenet Valhalla Circle #19 & #21 Bartlo's hate lits #1: <40376AD8.C83FBF5A@enter.net> CEO Alcatroll Labs Inc. Alexa knows everything about PSI, electromagnetic radiation and quantum mechanics: They are beaming earth with EMR which is interfering with natural human PSI abilities, which is likely quantum. in alt.alien.visitors === Subject: Another circular definition: Like f = ma MASS = DENSITY*VOLUME or VOLUME=MASS/DENSITY) Newton said at the beginning of the Principia: that MASS = DENSITY x VOLUME: Now HEAR THIS< !! Since DENSITY = MASS/VOLUME, and VOLUME= MASS/DENSITY; He must have meant the product: DENSITY x VOLUME to mean [MASS/VOLUME] x [MASS/DENSITY] = [MASS/VOLUME] x VOLUME = MASS. A circular definition if there ever was one! It would have been better, and true if he'd said: MASS = DENSITY/VOLUME: Just as it's true that m = f/a = w/g. As it is the metric system has erroneously taken Mass to be fundamental; which it's not. That's what I think. === Subject: Re: Another circular definition: Like f = ma > MASS = DENSITY*VOLUME or VOLUME=MASS/DENSITY) > That's what I think. Dumb Donny Head, you don't think. You don't even think you think. You are too ing stoooopid to suspect that the whole world around you, built by physics by the book, even exists. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Another circular definition: Like f = ma > Now HEAR THIS No you listen--Newtons Second Law http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html A force F acting on a body gives it an acceleration a which is in the direction of the force and has magnitude inversely proportional to the mass m of the body: F = ma F = ma is a differential equation and therein lies its power. Some examples of application: Case 1. The force is constant: Assuming that the mass remains constant, we have constant acceleration: F/m = dv/dt = a = constant direct integration (with respect to t) gives formulas such as: v - v_o = at v = at + v_o A second integration results in: x - xo = 1/2 at^2 + v_ot x = 1/2 at^2 + v_ot + x_o eliminating t, we get 2a(x-x_o) = v^2 - v_o^2 Case 2. The force is a function of position only: F(x) = m d^2x/dt^2 In a great many instances the force that an object experiences depends only on the object's position with respect to other bodies, for example, electrostatic and gravitational forces. The equation can be rewritten as: F(x) = mv dv/dx = m/2 d(v^2)/dx = dT/dx where the quantity T = 1/2 mv^2 The integral [ F(x)dx ] + constant = T oh it gets good from here.... the upshot for a freely falling body under the inßuence of gravity is: v = dx/dt = sqrt (v_o^2 - 2gx), where g is the acceleration due to gravity. Gosh I didn't even cover the equations for the force of a spring! Case 3. The force as a function of velocity only: F(v) = m dv/dt It often happens that the force on a object depends only on the velocity of the object. This is, for example, true of viscous A single integration t = integral [ m dv/F(v)] = t(v) results in: v = v(t), and a second integration gives: x = integral [ v(t) dt] = x(t) You see, Shead, if you know a bit of calculus and the circumstances of some rectilinear motion problem, you can figure out a hell of a lot of physics just from the relationship between Force and motion that Isaac Newton figured out: F = m d^2x/dt^2 = ma === Subject: Re: Another circular definition: Like f = ma Shead is, of course, full of . But he does have one *sort of* point. f=ma by itself is pretty much useless, because it really just acts as a definition of force. Your examples below hold f to be constant, but what does that mean? What is force? Without something other then f=ma, the definition of f is circular. Of course, Newton created another equation that used f .... f=Gm/r^2, and indeed Hooke had already (?) created an equation that used f (f = kL for springs); either of these provides another definition of f which means it isn't circular. > Now HEAR THIS > No you listen--Newtons Second Law > http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html > A force F acting on a body gives it an acceleration a which is > in the direction of the force and has magnitude inversely > proportional to the mass m of the body: F = ma > F = ma is a differential equation and therein lies its power. > Some examples of application: > Case 1. The force is constant: > Assuming that the mass remains constant, we have constant > acceleration: F/m = dv/dt = a = constant > direct integration (with respect to t) gives formulas such as: > v - v_o = at > v = at + v_o > A second integration results in: > x - xo = 1/2 at^2 + v_ot > x = 1/2 at^2 + v_ot + x_o > eliminating t, we get > 2a(x-x_o) = v^2 - v_o^2 > Case 2. The force is a function of position only: > F(x) = m d^2x/dt^2 > In a great many instances the force that an object experiences > depends only on the object's position with respect to other bodies, > for example, electrostatic and gravitational forces. > The equation can be rewritten as: > F(x) = mv dv/dx = m/2 d(v^2)/dx = dT/dx where the quantity T = 1/2 mv^2 > The integral [ F(x)dx ] + constant = T > oh it gets good from here.... the upshot for a freely falling body > under the inßuence of gravity is: > v = dx/dt = sqrt (v_o^2 - 2gx), where g is the acceleration due to > gravity. Gosh I didn't even cover the equations for the force of a > spring! > Case 3. The force as a function of velocity only: > F(v) = m dv/dt > It often happens that the force on a object depends only on the > velocity of the object. This is, for example, true of viscous > A single integration t = integral [ m dv/F(v)] = t(v) results in: > v = v(t), and a second integration gives: > x = integral [ v(t) dt] = x(t) > You see, Shead, if you know a bit of calculus and the circumstances of some > rectilinear motion problem, you can figure out a hell of a lot of physics > just from the relationship between Force and motion that Isaac Newton > figured out: > F = m d^2x/dt^2 = ma === Subject: Re: Another circular definition: Like f = ma > Shead is, of course, full of . > But he does have one *sort of* point. f=ma by itself is pretty much useless, > because it really just acts as a definition of force. Actually Newton clearly defines all parts of F = ma = dp/dt providing a definition for mass, position, displacement, deivatives, velocity, momemtum, acceleration, integration, etc... a complete package. I see nothing circular in that. Fundamentals concepts included mass, length, time.... the calculus. Newton incorporated the work of Galileo, Kepler, etc. Peter--In one of the Annenberg CPB Mechanical Universe series episodes, Goodstein point out that among the gems of western culture including Beethoven Symphonies, Michelangelo's painting on the Sistine Chapel Ceiling, Shakespeare's drama, is Isaac Newton's Mechanics. === Subject: Re: Another circular definition: Like f = ma > Shead is, of course, full of . > But he does have one *sort of* point. f=ma by itself is pretty much useless, > because it really just acts as a definition of force. > Actually Newton clearly defines all parts of F = ma = dp/dt providing > a definition for mass, position, displacement, deivatives, velocity, > momemtum, acceleration, integration, etc... a complete package. I see > nothing circular in that. Fundamentals concepts included mass, length, > time.... the calculus. Newton incorporated the work of Galileo, Kepler, > etc. > Peter--In one of the Annenberg CPB Mechanical Universe series > episodes, Goodstein point out that among the gems of western culture > including Beethoven Symphonies, Michelangelo's painting on the Sistine > Chapel Ceiling, Shakespeare's drama, is Isaac Newton's Mechanics. OK. Allow me to define the concept of borce, for which I will use the symbol b. b = md where m=mass and d=distance (length) Its easy to see that if we keep b constant, then mass must vary inversely with distance. A borce of 0 implies either a mass of zero or a disctnce of zero (or both). And plenty of other results, analogous to what you gave for force. Point is, on its own f=ma is useless, just as b=md is useless. Its a definition of force (or borce), and any results that you can get from either of these equations alone are circular. The difference between force and borse is that there are other equations which use f, which prevent it being circular. I have no other equations for borce that aren't just the same thing re-packaged, or introduce some other new variable. Which is why Newton was agreat scientist and f=ma is the basis of dynamics, and I'm just a newsgroup pedant and borce will be forgotten tomorrow ... BTW, if you don't agree, tell me how force is conceptually different to borce WITHOUT using some other definition of force. === Subject: Re: Another circular definition: Like f = ma > MASS = DENSITY*VOLUME or VOLUME=MASS/DENSITY) > Newton said at the beginning of the Principia: that MASS = DENSITY x > VOLUME: > Now HEAR THIS< !! > Since DENSITY = MASS/VOLUME, and VOLUME= MASS/DENSITY; He must have > meant the product: DENSITY x VOLUME to mean [MASS/VOLUME] x > [MASS/DENSITY] = [MASS/VOLUME] x VOLUME = MASS. > A circular definition if there ever was one! > It would have been better, and true if he'd said: MASS = > DENSITY/VOLUME: Just as it's true that m = f/a = w/g. > As it is the metric system has erroneously taken Mass to be > fundamental; which it's not. > That's what I think. What you think as no bearing on reality, crackpot. === Subject: Job Interview Question I have a job interview next week for a community college teaching position where I need to give a 15 minute teaching demonstration. I know that I should treat the search committee as students and should not say things like if I had more time I would do this or that. My question is do you just end at the 15 minute mark where ever that may be (of course I will have an idea of where the 15 minutes will bring me to) or do you have to have reached some high point? Mkajuma === Subject: Re: Job Interview Question Here is my own opinion. I had to do the exact same thing myself, in material and covered it until they stopped me. There did not appear to be any negative aspect at all. In fact, I got hired. This also happens to be for a community college. Also, I have sat on a hiring committee and watched 15 minute demonstrations/lectures. Most of the candidates went over and neither I nor anyone else on the committee judged the lecture negatively because they were too long. If there was a main point to be made, some didn't get to it because we stopped them after about 20 minutes (although most will make you stop after the 15 minutes if there are a lot of candidates to interview). Additionally, we didn't have any that were too short (but we wouldn't have minded if they had been done early - after several in a row you do get tired). I wouldn't imagine that there will be any problems if you are a little short, but I would plan on having more than a 15 minute lecture, and let them stop you when they think the time is up. Let me also mention that, being on a hiring committee is very enlightening. Although there are usually less jobs than candidates, you find yourself rooting for each candidate. I personally did not want to see any of the candidates not get the job. Of course, none of our recommendations had anything to do with how long there lecture was. We wanted to see if they were competent teachers who could relay the material to their students. Brian >I have a job interview next week for a community college teaching position >where I need to give a 15 minute teaching demonstration. I know that I >should treat the search committee as students and should not say things like >if I had more time I would do this or that. My question is do you just end >at the 15 minute mark where ever that may be (of course I will have an idea >of where the 15 minutes will bring me to) or do you have to have reached >some high point? >Mkajuma === Subject: Re: Job Interview Question >I have a job interview next week for a community college teaching position >where I need to give a 15 minute teaching demonstration. I know that I >should treat the search committee as students and should not say things like >if I had more time I would do this or that. My question is do you just end >at the 15 minute mark where ever that may be (of course I will have an idea >of where the 15 minutes will bring me to) or do you have to have reached >some high point? If you are asked to give a 15 minute lecture, then give a 15 minute lecture. If you have a choice in picking your own topic, pick something that can be presented in 15 minutes in a coherent and self-contained way. For instance, you may want to demonstrate why the derivative of x^n is n x^(n-1). Or why derivative of sin x is cos x. Each of these may be packaged to fit in a 15 minute lesson. -- Rouben Rostamian === Subject: Re: Job Interview Question Give it a practice run and time it, have extra material if you need to extend it some. Be Confident, they look for that. >I have a job interview next week for a community college teaching position >where I need to give a 15 minute teaching demonstration. I know that I >should treat the search committee as students and should not say things like >if I had more time I would do this or that. My question is do you just end >at the 15 minute mark where ever that may be (of course I will have an idea >of where the 15 minutes will bring me to) or do you have to have reached >some high point? > If you are asked to give a 15 minute lecture, then give a 15 > minute lecture. If you have a choice in picking your own topic, > pick something that can be presented in 15 minutes in a coherent and > self-contained way. For instance, you may want to demonstrate why > the derivative of x^n is n x^(n-1). Or why derivative of sin x is > cos x. Each of these may be packaged to fit in a 15 minute lesson. > -- > Rouben Rostamian === Subject: left regular representation of a group if |G|=m.n, and x in G has order |x|=n, then why is it that in the left regular representation of G->Symmetric group on m.n, x is a product of m n-cycles? I can see that x acts on powers of x as an n-cycle, i.e., the stabilizer of is , and that the action on cosets of is transitive (as it should be), but then why is the action by x a product of m n-cycles? === Subject: Re: left regular representation of a group : if |G|=m.n, and x in G has order |x|=n, then why is it that in the : left regular representation of G->Symmetric group on m.n, x is a : product of m n-cycles? I can see that x acts on powers of x as an : n-cycle, i.e., the stabilizer of is , and that the action on : cosets of is transitive (as it should be), but then why is the : action by x a product of m n-cycles? Pick any element y in G. What is the orbit of y with respect to the action of left multiplication by x? Ted