mm-1055 === Subject: Re: solvable group days. My association with the Department is that of an alumnus. [.snip quoted text.] For your next trick, remember that even when you are very excited, you may want to trim the quoted text from your response. (-: -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Heeelp pleeease F(X) = x*sqr(x) F(X)= 1*sqr(x)+x*(1/2sqr(X)) = sqr(x) + x/(2sqr(x)) (according to (VU) = Vu+Uv Now according to: (UV) = V * U F = x*(sqr(x)) = 1*(1/2sqr(x))= 1/sqr(x) Whats my prob.? thank you very very much *-----------------------* www.GroupSrv.com *-----------------------* === Subject: re:Heeelp pleeease I was thinking about the kind of functions that you can divide them to 2 different functions and multiply thair derivatives...I dont know the term (I am not native english speaker) but in my language its called intricate functions, so I suposed it may be something like this intricatecomplexentangled function I hope you catch on to what I mean ;) *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Heeelp pleeease >F(X) = x*sqr(x) >F(X)= 1*sqr(x)+x*(1/2sqr(X)) = sqr(x) + x/(2sqr(x)) >(according to (VU) = Vu+Uv >Now according to: (UV) = V * U >F = x*(sqr(x)) = 1*(1/2sqr(x))= 1/sqr(x) >Whats my prob.? That (UV) = V*U is false. Where did you get that? -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Heeelp pleeease [snipped the posting by OP] > That (UV) = V*U is false. Where did you get that? The OP probably gets this because he/she is unaware of the context when this formula is given. If the symbol Ô means matrix transpose, the above is actually correct. I always find the multiple meanings of the symbol Ô confusing. It heavily depends on the context. It can mean derivative, matrix transpose, or auxiliary variable as in the equation p_{i} = a_{i} / sum_{i=1}^n a_{i} . Is there any other mathematical use for the symbol Ô, btw? === Subject: Re: Heeelp pleeease > F(X) = x*sqr(x) > F(X)= 1*sqr(x)+x*(1/2sqr(X)) = sqr(x) + x/(2sqr(x)) This looks fairly nice so far, but we usually refer to x/sqr(x) as sqr(x), so things can be simplified. > (according to (VU) = Vu+Uv > Now according to: (UV) = V * U Huh! > F = x*(sqr(x)) = 1*(1/2sqr(x))= 1/sqr(x) Well Im not entirely sure where you arrived at (UV) = V * U, it being somewhat less than correct in almost all cases, including this one, but a 1/2 seems to have slunk away without permission in the subsequent algebra. > Whats my prob.? > thank you very very much > *-----------------------* > www.GroupSrv.com > *-----------------------* === Subject: Re: Heeelp pleeease > F(X) = x*sqr(x) > F(X)= 1*sqr(x)+x*(1/2sqr(X)) = sqr(x) + x/(2sqr(x)) > (according to (VU) = Vu+Uv > Now according to: (UV) = V * U this statement is simply not true. consider (x^2) = (x*x), the above says that (x^2) = 1*1 = 1. which is clearly false. > F = x*(sqr(x)) = 1*(1/2sqr(x))= 1/sqr(x) > Whats my prob.? > thank you very very much -- Justin Young http://web.syr.edu/~jryoun04/ === Subject: eXtreme Proving? (math research processes) Im a professional programmer, who hopes to be a math grad student. Ive been wondering lately about the processes mathematicians use to generate theorems. It seems to me that software development and mathematics may share some common difficulties, such as limiting complexity to ease understanding, and the possibility of bugs (or gaps) arising as one attempts to simplify to reduce complexity. A common approach in software development is to set up automated tests to ensure that the code is always working as expected. One of the great advantages of this is that the coder is freed to make considerable changes in code, and then very quickly see that the resulting still works. This allows faster modification of programs to add features, but also allows continual simplification of the existing code. Since a programmer can only track so much complexity, tests allow them to quickly eliminate unneeded complexity, allowing the program to grow. It seems it has become somewhat trendy for software developers to try to export their development processes to other domains - think of John Carmacks Armadillo Aerospace. I was specifically wondering if anyone had tried applying features of agile software development processes like extreme programming to mathematical research. Would pair proving at a computer terminal running automated proof-checking software make any sense? I was involved in a summer undergraduate research program which used Mathematica extensively as a fancy calculator, but Im not sure it changed the fundamental process, in that students tried out various ideas and presented them to each other, formally or informally, at intervals. In general, has much work been done looking into research processes other than the traditional paper, pen, wastebasket model? Or can anything else even work? -- Kevin === Subject: Re: eXtreme Proving? (math research processes) Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) >generate theorems. It seems to me that software development and >mathematics may share some common difficulties, such as limiting >complexity to ease understanding, and the possibility of bugs (or >gaps) arising as one attempts to simplify to reduce complexity. yes, there are many conceptual similarities. >It seems it has become somewhat trendy for software developers to try to >export their development processes to other domains - think of John >Carmacks Armadillo Aerospace. I was specifically wondering if anyone >had tried applying features of agile software development processes >like extreme programming to mathematical research. I thought that the trendy extreme programming had already been discredited, especially for high safety areas (or maybe this is just wishful thinking on my part). >Would pair >proving at a computer terminal running automated proof-checking >software make any sense? Sort of. Lots of mathematical collaboration works face-to-face with a blackboard, a fancy calculator might be nice to have around. But as to automated proof-checking, no one uses that for actually helping to produce new mathematics. Maybe for checking your work -after- a lot of intuitive work has already been done. Automated proof checking is probably much more useful for checking -programs- (something that is already formally specified and electronically recorded). >In general, has much work been done looking into research >processes other than the traditional paper, pen, wastebasket model? If you mean sociological research, into how successful alternatives are, I As to working mathematicians trying out alternatives, sure theres lots of alternatives: long walks, chemical enhancement, conferences and workshops, email, napkins at coffeehouses, computer experimentation. Most mathematicians are highly computer literate, but currently proof checkers are just not appropriate for mathematicians to bother with to get their work done. -- Mitch === Subject: Re: Science, Philosophy, Mysticism, Art, Mathematics, and Physics <41bd1833.17476667@netnews.att.net> <41bd9f23.21123889@netnews.att.net> <41be9010.22849242@netnews.att.net> <41bfb1af.31103987@netnews.att.net> posting-account=RkUO3A0AAAAo6FJoWWMb916-xRvUxBKb >>[...]Im mostly interested in definitions that hold true all the time. >>They do, Lester, they do. By definition. Whether a definition refers to >>anything real is another question. > On the other hand, Wolf, it might be interesting to explore this > definition of definition a little more objectively to decide how they > can refer to anything unreal. > A unicorn is.... A unicorn is a mythical creature. It does exist, like God, in the human mind and has an observable effect on behaviour. Apparently Queen Elizabeth in the 16th century paid 10,000 pounds for the horn of a unicorn. It was of course the spiraled tusk of the narwhal, an artic whale. -john === Subject: Re: Groups as galois groups <327qqoF3jfnrgU1@individual.net> posting-account=xslVFw0AAACcw30rwkkdKYhs8AhLZVeJ yes I was aware of that, and I was even able to fight my way through the first couple of sections of Serres book (although I had never heard of the other). The parts I read were mostly foundational (and highly algebraic), so I was hoping to learn some of the algebraic topology ßavors in the subject. Perhaps something a little deeper than simply seeing all of the analogies between galois theory and covering space theory. Michael Williams === Subject: Re: Groups as galois groups days. My association with the Department is that of an alumnus. >I am interested in learning more about the problem of determning which >groups arise as galois groups (over Q of course). Its called the Inverse Galois Problem. A quick search reveals: MR1352262 (96c:00033) Recent developments in the inverse Galois problem. Papers from the Joint Summer Research Conference held at the University of Washington, Seattle, Washington, July 17--23, 1993. Edited by Michael D. Fried, Shreeram S. Abhyankar, Walter Feit, Yasutaka Ihara and Helmut Voelklein The (positive) solution for solvable groups is due to Shafarevich; the original is in Russian (and the Math Review text in french), but its a safe bet that theres a translation or survey somewhere: MR0071469 (17,131d) Safarevich, I. R. Construction of fields of algebraic numbers with given solvable Galois group. (Russian) Izv. Akad. Nauk SSSR. Ser. Mat. 18 (1954), 525--578. I dont think I can help much with the geometric side, though. The details are very dim in my head. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Prime numbers problem days. My association with the Department is that of an alumnus. >Phil, >> Primes>30 are coprime to 30. >This is true, as primes are coprime to any number. No. If p is a prime and n is any integer, then either p divides n or p is coprime to n. But primes are not coprime to any number. The only numbers which are coprime to every number are 1 and -1. >> So are of the form 30n+{1,5,7,11,13,17,91,23,29} >I think 35 from above formula is not a prime even if the remainder is prime. >This shows also that composite number divided by 30 can give prime >remainder. So? The original question was: if p is a prime, and we divide p by 30, then the remainder is prime. This is certainly not the same thing as saying: if n is a number, and when we divide n by 30 the remainder is prime, then n is prime. What Phil is saying is NOT that every number of the form 30n + {1, 5, 7, 11, 13, 17, 19, 23, 29} [note the typo in the original] is a prime, but rather that every prime larger than 30 is of that form. The implication was only asserted in one direction. You are correct that 5 is superßuous, as 3 would be, for p>30; but it ->doesnt matter<-, because you were only interested in the implication if p is a prime, then the remainder of dividing p by 30 is a prime; even though the remainder will never be 5 for p>30, it is still true that every prime will have a remainder in the set {1, 5, 7, 11, 13, 17, 19, 23, 29}. 35 is not a counterexample to the assertion (which is correct). >30=2*3*5 >I think there should be a list of all numbers between 0 and 29 that are >coprime to each prime factor of 30: (2 and 3 and 5). Here is the list: >{1,7,11,13,17,19,23,29} >Lets assume you make small mistake forgetting to remove 5 from your list >and mistyping 19 as 91. >The formula: >p=30n+m where m={1,7,11,13,17,19,23,29} >generates all coprime to 30. Nobody was saying that you would get all number coprime to 30 through the formula. They were addressing your original question on remainders of ->primes<- (not of numbers coprime to 30). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Santa Claus is the god of math !!!! === >Subject: Re: Santa Claus is the god of math !!!! >Message-id: <3283krF3h3876U1@individual.net> There is NOTHING that Santa cannot prove !!!! >> Wish for a proof for Christmas and youll get it! >> We already had a thread discussing acceptance of Gods proofs in this >> group recently. I vainly refer to >> after which it petered out. Go back from there. >Im God and I have numerous proofs. Everyone thinks >Im joking year after year with proof of god posted in >front of them..... oh haha. > Looks like youve found the one thing god cant prove. > The only stupdity I see here is ignorance. There is NOT ONE SCIENTIST among this whole group. NOT ONE of you displays any credibility of using scientific method. YOU DONT ACCEPT HYPOTHESIS YOU IGNORE VALID MATHEMATICAL PROOFS This is proof that Hollywood bases movies on real people without saying that it is based on a true story. http://www.supernerd.com.au/~gray77/enemy-of-the-state.gif Youre just a bunch of wining idiots who only make complaints and dont operate in any other paradigm. Herc === Subject: Re: Santa Claus is the god of math !!!! >Youre just a bunch of wining idiots who only make complaints and ^^^^^^ I think what you mean is whining - very few people that are wining start whining. That usually happens if youre beering. Thomas [Skol] >dont operate in any other paradigm. >Herc === Subject: Re: Santa Claus is the god of math !!!! >Youre just a bunch of wining idiots who only make complaints and > ^^^^^^ > I think what you mean is whining - very few people that are wining > start whining. That usually happens if youre beering. SEE? Herc === Subject: Re: Santa Claus is the god of math !!!! <3297phF3jelotU1@individual.net> posting-account=_-j7cgwAAADnQK9-r68QgRsgfV-jhA3A === >Subject: Re: Santa Claus is the god of math !!!! >Message-id: <3283krF3h3876U1@individual.net >> There is NOTHING that Santa cannot prove !!!! >> Wish for a proof for Christmas and youll get it! >> We already had a thread discussing acceptance of Gods proofs in this >> group recently. I vainly refer to >> after which it petered out. Go back from there. > Im God and I have numerous proofs. Everyone thinks >Im joking year after year with proof of god posted in >front of them..... oh haha. Looks like youve found the one thing god cant prove. > There is NOT ONE SCIENTIST among this whole group. > NOT ONE of you displays any credibility of using scientific method. > YOU DONT ACCEPT HYPOTHESIS > YOU IGNORE VALID MATHEMATICAL PROOFS What do we need the scientific method for? Were math (and logic) guys (and gals). > This is proof that Hollywood bases movies on real people without > saying that it is based on a true story. > http://www.supernerd.com.au/~gray77/enemy-of-the-state.gif What does this have to do with your claim of being god? Ôcid Ôooh === Subject: Re: Santa Claus is the god of math !!!! === >Subject: Re: Santa Claus is the god of math !!!! >Message-id: <3283krF3h3876U1@individual.net >> There is NOTHING that Santa cannot prove !!!! >> Wish for a proof for Christmas and youll get it! >> We already had a thread discussing acceptance of Gods proofs in > this >> group recently. I vainly refer to >> after which it petered out. Go back from there. > Im God and I have numerous proofs. Everyone thinks >Im joking year after year with proof of god posted in >front of them..... oh haha. > Looks like youve found the one thing god cant prove. > There is NOT ONE SCIENTIST among this whole group. > NOT ONE of you displays any credibility of using scientific method. > YOU DONT ACCEPT HYPOTHESIS > YOU IGNORE VALID MATHEMATICAL PROOFS > What do we need the scientific method for? Were math (and logic) guys > (and gals). youre math-parrots guys and gals. you cant substantiate truth unless you take opposing points of view to what you are taught. its like taking the conscience out from the Daleks, sure it makes them reliable, strong, single purpose, but they are destined to a void existence as the rest of the Universe develops further. Herc === Subject: Re: Please help quickly! > please help me with the following problems ...by showing me the work and how > to do them..i am tryin to check my answers and work If youre really in a hurry, wouldnt it be better to show us your answers & your work and let us tell you how youve done? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Moments over a Simplex > The midpoint of a line segment (x1,x2) is (x1+x2)/2 > The midpoint of a triangle (r1,r2,r3) is (r1+r2+r3)/3 > The midpoint of a tetrahedron (r0,r1,r2,r3) is (r0+r1+r2+r3)/4 > The midpoint of a simplex in N dimensions is sum(k<=N) r_k/N ? > The variance in x over a line segment in 1-D is: (x2-x1)^2/12 > The variance in x over a triangle in 2-D is, if I made no errors: > ((x2-x1)^2 + (x3-x1)^2 + (x3-x2)^2)/36 > Question: What is the variance in x over a simplex in N dimensions? > Is there a sensible way to make these calculations (multiple integrals) > more amenable to treatment? Letting m be the vector average, sum(i=1..n, r_i). You are apparently trying to calculate the average over the simplex of the squared distance from m; it is incorrect to call this scalar the variance. The desired average is the trace of the variance-covariance matrix. The uniform distribution over a simplex is the Dirichlet distribution with parameters 1, 1, 1,..., 1 (n times for the (n-1)-dimensional simplex). If random vector X has D(a1, a2,..., a_n) distribution, X has n nonnegative components which sum to unity. Letting s = sum(i=1..n, a_i), EX = a / s Var(Xi) = a_i (s - a_i) / [s^2 (s+1)] Cov(Xi, Xj) = -a_i a_j / [s^2 (s+1)] for i != j, If random vector Y is uniformly distributed over the (n-1)-dimensional simplex with vertices r1, r2,..., r_n, then Y = [r1 r2 r3 ... r_n] X, where the vectors are written in columns and X has D(1, 1,..., 1) distribution. Thus, EY = [r1 r2 ... r_n] EX = sum(i=1..n, r_i) / n Yj = sum(i=1..n, (r_i)_j Xi) Var(Yj) = sum(i=1..n, [(r_i)_i]^2 Var(Xj)) + 2 sum( (i,k): i < k, (r_i)_j (r_k)_j Cov(X_i, X_k)) = [(n-1) sum(i=1..n. [(r_i)_i]^2) - 2 sum( (i,k): i < k, (r_i)_j (r_k)_j )] / [n^2 (n+1)] The average you want is sum(j=1..n, Var(Yj)). -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: Moments over a Simplex Correction: > Var(Yj) = sum(i=1..n, [(r_i)_i]^2 Var(Xj)) + 2 sum( (i,k): i < k, > (r_i)_j (r_k)_j Cov(X_i, X_k)) > = [(n-1) sum(i=1..n. [(r_i)_i]^2) - 2 sum( (i,k): i < k, (r_i)_j > (r_k)_j )] / [n^2 (n+1)] In the first sum in each formula, the coefficients should be [(r_i)_j)]^2. -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: SVD Algorithm > Im searching for an SVD algorithm (in C, C++ or Java). > I have tried SVD algorithm of Numerical Recipes, but Singular Values > are not in descending order. Use the Numerical Recipes code. It works well. If you need the singular values in decending order, you can either re-arrange the output, or use an indexing array. If you swap two singular values, you must also swap two rows of U and V. The following code will swap w[i] and w[k] and the corresponding rows of u and v: swp = w[k]; w[k] = w[i]; w[i] = swp; for (j=1; j ha scritto nel messaggio > If you swap two singular values, you must also swap two rows of U and V. > The following code will swap w[i] and w[k] and the corresponding rows of u > and v: Exactly... > An alternative would be to create an indexing array and sort that. Fill > the indexing array like this: > The values w[idx[1 ...n]] are then in decending order. The original w[1 > ... n] values are unchanged, and there is no need to re-arrange the U and > V matricies. - W with elements in descending order; - U re-arrange; - V re-arrange. So: U (re-arrange) * W (descending) * V transpose (re arrange) = A. (I will take the first k columns of U, the first k rows of V transpose, the first k elements of W.) Any idea? === Subject: Re: Poll: Are PCs Turing Machines? >> All functions possible in the real world can be represented / simulated >by >> TMs >> Herc >> What about functions on reals, like f(x)=sqrt(x)? Or did you mean to >assert >> a tautology? > The relationship between Turing Machines and real numbers is rather > interesting. > I think a Turing machine corresponds to a real number if, given any rational > number as input (in the form of a pair of integers) the TM will always > stopeither in state L (assigning the rational to set L of the Dedekind section > for the real) or in state U. > Given any obviously computable real number (such as an algebraic) it is > obviously possible to build a TM that corresponds to it. There will be in > fact an infinite number of such TMs (Find one and add a lot of irrelevant > states to it.) > Given an arbitrary TM, it is undecidable whether it corresponds to a real or > not. > Can one develop a meaningful theory of computability along these lines? That IS the theory of mathematics, take an objective stance, theres nothing you_can_do with reals that computers cant. Just ignore the uncountable theory its not a HIGHER infinity at all. If space was logarithmic it wouldnt even be a result because the length of combinations would grow as fast as the list, and the SQUARE segment that diag relies on would be contradictory to its completeness. TM1 = 0 TM2 = 0.5 TM3 = ? TM4 = pi TM5 = e TM6 = cos(e) TM7 = 0.333333.. All possible sequences of digits TO INFINITE LENGTH appear. Diagonalisation as it turns out does NOT produce a new sequence, it takes a naive view of a COMPLETE list. Why throw away computers and rely on your mind just because |_0_|_._|_1_|_2_|_3_|_1_|_2_| V |_0_|_._|_2_|_3_|_4_|_2_|_3_| if you are naive enough, it looks like a whole new number. also mathematicians would have to accept null values, welcome to the real world. {Reals} = UTM(x, y) mod 10 AI = UTM(x, y) mod 27 Herc === Subject: Re: Poll: Are PCs Turing Machines? > infinite people each ßip coins infinite times >= all sequenes of heads and tails have been tossed to infinite length > [rest snipped] > Produce the list. Read the assumptions. Herc === Subject: Re: (-1)^n(n^(1/n) - 1) conditionally convergent? posting-account=2s1rcw0AAABZtBqaUPbD5ZJN1V2LKUsY > n^(1/n) = exp((log(n)/n) > 1 + ... > I tried that already. > Well, whats the first term in ...? I think I am starting to undestand... (exp(log(n)/n) - 1) > (((1 + 1/n)^n)^(log(n)/n) - 1) = = ((1 + 1/n)^log(n) - 1) > (1 + 1/n - 1) = 1/n (which diverge) Is something like this is feasible? Boris. === Subject: Re: (-1)^n(n^(1/n) - 1) conditionally convergent? >> n^(1/n) = exp((log(n)/n) > 1 + ... >> I tried that already. >> Well, whats the first term in ...? >I think I am starting to undestand... >(exp(log(n)/n) - 1) > (((1 + 1/n)^n)^(log(n)/n) - 1) = >= ((1 + 1/n)^log(n) - 1) > (1 + 1/n - 1) = 1/n (which diverge) >Is something like this is feasible? Why so complicated? Let u be a positive number. Then exp(u) = 1 + u + u^2/2 + ... > 1 + u, since youre discarding positive numbers. Clear so far? Now let u = log(n)/n which clearly is positive for n>1. Therefore exp(log(n)/n) > 1 + log(n)/n. Subtract 1 on both sides: exp(log(n)/n - 1 > log(n)/n. Next: log(n) > 1 if n>3 (>e=2.718...) Therefore: exp(log(n)/n - 1 > 1/n if n > 3. Thomas >Boris. === Subject: Re: (-1)^n(n^(1/n) - 1) conditionally convergent? posting-account=2s1rcw0AAABZtBqaUPbD5ZJN1V2LKUsY > n^(1/n) = exp((log(n)/n) > 1 + ... > I tried that already. > Well, whats the first term in ...? I think I am starting to undestand... (exp(log(n)/n) - 1) > (((1 + 1/n)^n)^(log(n)/n) - 1) = = ((1 + 1/n)^log(n) - 1) > (1 + 1/n - 1) = 1/n (which diverge) Is something like this is feasible? Boris. === Subject: Re: Where am I going wrong? >>The nodes represent only a countable subset of the reals, namely, the >>rationals with denominators that are powers of 2. The real number 1/3, >>for example, does not lie in any node at any depth. Each real number in >>[0,1] can be represented as the limit of the values lying along a >>particular path, and there are uncountably many paths. >1) This tree: > a > / >b c >has 9 paths: >> This is a finite tree, meaning that each path terminates in a node. In >> the infinite tree, a path need not terminate. There are uncountably many >> paths, but only countably many that terminate. > That the paths do not terminate is the point. Are you saying that the > reals do not exist within the tree as defined or that a breadth first > traversal would never touch the reals within it, or would not do so in > countably many steps? My answer can be found in the first quoted paragraph above. The only reals that exist within the tree are the rationals with denominators that are powers of 2. Do you disagree? If so, then explain on which level we can find the real number 1/3. You have been asked this question many times by several different people, but you keep ignoring it. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Where am I going wrong? days. My association with the Department is that of an alumnus. >> The nodes represent only a countable subset of the reals, namely, the >> rationals with denominators that are powers of 2. The real number 1/3, >> for example, does not lie in any node at any depth. Each real number in >> [0,1] can be represented as the limit of the values lying along a >> particular path, and there are uncountably many paths. >1) This tree: > a > / >b c >has 9 paths: >(a) >(b) >(c) >(a b) >(a c) >(b a) >(c a) >(b a c) >(c a b) >Does it contain 9 values? Actually, in this instance, path is only meant to refer to downward paths, meaning that you start at the root and at each node select one of its descending branches. So you only have three paths here: (a), (a,b), and (a,c). If we assume this is the top of your original tree, they correspond to the three values 1/2 (the path (a)); 1/4 (the path (a,b)); and 3/4 (the path (a,c)). The tree, of course, does not contain any values at all. The tree can be made to correspond to values if we interpret each path in a suitable way. The way you chose makes each FINITE path (which naturally corresponds to nodes on the tree) into a finite sum, which represents exactly the rationals between 0 and 1 which can be expressed with a denominator which is a power of 2. Your representation also associates to every INFINITE path an infinite sum; if the path contains an infinite number of take the right branch-s, then this is truly an infinite series (i.e., it has an infinite number of nonzero terms). In order to get ALL real numbers, you must consider both nodes and infinite paths; this is not the case in a finite tree, since every path is finite and can be identified with its terminal node. >2) The point is that the tree is traversed in breadth first fashion. >This means that paths are irrelevant. No. The point is that in order to identify your tree with ALL real numbers between 0 and 1, you must consider infinite paths as well as finite ones (nodes). >3) How do you show that there are uncountably many paths? There are as many as there are sequences of 0s and 1s (read a 0 as take the left branch and a 1 as take the right branch); this is the same as the number of subsets of N (via the characteristic function), which of course is uncountable. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Where am I going wrong? > I am not claiming the tree is finite, or does my definition of the > tree fail for a depth of Aleph_0. >> Note that I did not say you were claiming the tree to be finite. What >> I said was that you do not get 1/3 as a finite sum of the form that >> you are using, namely as a sum of n/2^m for integer numerators n. >> Surely you realize that each node corresponds to a *finite* sum? > Not totally sure that I do. >> Note that the number of binary digits required to express the number >> tells us both a lower and upper bound for the natural number you intend >> to assign to the node: >> k+1 binary digits <--> number assigned is between >> 2^k and 2^(k+1). >> Now, consider 1/3 = 0.0101010101... >> What number do you suggest will be assigned to 1/3? > One in the set {x:2^k<=x<2^(k+1) as k->inf}? > Is this a bad definition, or is |{x:0<=x<2^(k+1), k->inf}| > Aleph_0? > Or do we actually have to treat k as Aleph_0 giving us 2^Aleph_0? Is there a number in that set? After all, the sets X_k = {x:2^k<=x<2^(k+1)} are distinct, and for every integer N, there is an X_k for which all elements are greater than N, namely by taking any k > log_2(N). If you were to name an integer label N for the node containing 1/3, that would place an upper bound (again, roughly log_2(N)) on the number of binary digits in its binary expansion. As for the concept of taking k = aleph_0, that is not a number, and your tree doesnt have any nodes at that depth. There may be a way to understand trees with transfinite depth (with nodes at such locations), but Im not competent to discuss such things. All I know is that the procedure youve described achieves the infinite sums required only by passing to monotonic paths from the root, including infinite paths. Dale === Subject: Re: Where am I going wrong? days. My association with the Department is that of an alumnus. >Ok, its another one of the cardinality of the reals things, but Im not >going to claim Im right I just hope some one can point out my mistake. >So here goes: >If we form a binary tree with 1/2 as the root node, 1/4 and 3/4 as its >children, 1/8, 3/8, 5/8 and 7/8 as their children, etc. Extended to >Aleph_0 levels we have every real >0 and <1 in the tree (i.e. every >number representable as sum_i(f(i).2^-i) where f is any function mapping >the natural numbers to {0,1}). If we traverse this tree in breadth first >fashion it would seem that we can assign a natural number to each node >in the tree as its order in breadth first traversal. Given that this >would mean a contradiction in ZF I assume Im wrong. Can anyone tell me why? >> The number of nodes is countable. The number of possible paths isnt. >The number of possible paths is irrelevant. The nodes represent every >real number >0 and <1, No. You are taking a sum; the nodes represent FINITE sums, and those only give you the rationals whose denominators are powers of 2. For infinite sums, what you are taking is a path, represented by taking the left branch at the i-th node if f(i)=0, and the right branch if f(i)=1. You can identify a node with the path that is always go to the left branch from that node, but you cannot identify a node with other kinds of sums. So the nodes only give you the countable number of rationals between 0 and 1 which are expressible with denominator a power of 2. The paths give you all real numbers between 0 and 1. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Where am I going wrong? > I am not claiming the tree is finite, or does my definition of the tree > fail for a depth of Aleph_0. >> Note that I did not say you were claiming the tree to be finite. What >> I said was that you do not get 1/3 as a finite sum of the form that >> you are using, namely as a sum of n/2^m for integer numerators n. >> Surely you realize that each node corresponds to a *finite* sum? > Not totally sure that I do. >> Note that the number of binary digits required to express the number >> tells us both a lower and upper bound for the natural number you intend >> to assign to the node: >> k+1 binary digits <--> number assigned is between >> 2^k and 2^(k+1). >> Now, consider 1/3 = 0.0101010101... >> What number do you suggest will be assigned to 1/3? > One in the set {x:2^k<=x<2^(k+1) as k->inf}? > Is this a bad definition, or is |{x:0<=x<2^(k+1), k->inf}| > Aleph_0? > Or do we actually have to treat k as Aleph_0 giving us 2^Aleph_0? You need to get straight what objects you are trying to enumerate. The nodes or the paths? Since one is countable and the other is uncountable, this is quite important. -- Justin Young http://web.syr.edu/~jryoun04/ === Subject: Re: Where am I going wrong? > You need to get straight what objects you are trying to enumerate. > The nodes or the paths? > Since one is countable and the other is uncountable, this is quite > important. I thought I was trying to enumerate the nodes, but I have been told these will not include real labeled nodes. Im afraid I dont understand why this is so. If sum(f(i)1/2^i) summed over the natural numbers can be an irrational, why will that node never appear on the tree? Im not attempting to construct the tree. === Subject: Re: Where am I going wrong? days. My association with the Department is that of an alumnus. >> You need to get straight what objects you are trying to enumerate. >> The nodes or the paths? >> Since one is countable and the other is uncountable, this is quite >> important. >I thought I was trying to enumerate the nodes, but I have been told >these will not include real labeled nodes. Im afraid I dont understand >why this is so. If sum(f(i)1/2^i) summed over the natural numbers can be >an irrational, It is only an irrational if f(i)=1 for infinitely many 1s. > why will that node never appear on the tree? Because each node has only finitely many nodes above it. It represents a sum which has f(i)=1 if and only if you took the right branch at the i-th level to get from the root to this node. Since the node is at a finite depth, say at the k-th level below the root, f(i)=0 for all i>k; so the sum has only finitely many values of f(i) equal to 1, so it cannot represent an irrational, only a rational (and only one with a denominator which is a power of 2, at that). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Where am I going wrong? >> You need to get straight what objects you are trying to enumerate. >> The nodes or the paths? >> Since one is countable and the other is uncountable, this is quite >> important. > I thought I was trying to enumerate the nodes, but I have been told > these will not include real labeled nodes. Im afraid I dont understand > why this is so. If sum(f(i)1/2^i) summed over the natural numbers can be > an irrational, why will that node never appear on the tree? Im not > attempting to construct the tree. Every node in the tree is a finite distance from the root. Consequently, each node represents a finite sum of the form youre describing, and thus a rational number with denominator some (finite) power of 2. It seems youre convinced that you can find a node that is an infinite distance from the root. Perhaps you could address that point to find the error in your reasoning. Dale === Subject: Re: JSH: But what if it works? > Well theres a serious problem on this newsgroup with people who dont > understand the idea process. > Brainstorming. You dont know a brainstorm from a brainfart. > Its been around for a while but you people act clueless about it. Nope. They make distinctions between viable ideas and worthless crap. > So I talk about having a lot of ideas where usually they dont pan out, > and get what? Criticized. You get criticized, not for having lots of ideas, but for defending bad ones and insulting your critics when they offer valid criticisms. > Errors are part of the discovery process. Errors are part of > brainstorming. No, the arent. Errors are anathema to the process. Identifying errors and correcting themis the remedy. > If you act like every single little idea is the grand great idea for > all time that will shine like a bright light throughout the world, then > you will waste a lot of time. As have you. Consider your own appraisal of your Ôresearch. > If you worry about every little mistake, then you will waste time. Wrong! If you *ignore* mistakes, you will waste time. > You people betray your own inconsistency by caring more about my > mistakes than I do, and then claiming that my work is not important. The essence of truth is freedom from error. How dare you criticize others by having higher standards than your own? It betrays your true character. [snip standard JSH diatribe -- no math content, only hysterical nonsense] -- A good scientist or technologist must have a high tolerance for frustration, somewhat less for confusion, very little for errors and none at all for dishonesty. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: A vector calc related question posting-account=2OYlAwwAAAAzuGZHzY8fB1XqLzeo4Yd5 > Back in the day, a visiting professor challenged my class to find an > exception to the following general statement: > F sub xy = F sub yx (Partial to x then y is equal to the partial of y > then x). > This is used, as yall may know, to find relative extrema where d = > Fxx*Fyy - (Fxy)^2. The reason it came up was the lecturer said that > they are one in the same. The visitor challenged us to find an > exception. Later, when my lecturer came back, he admited there were > cases where it isnt equal. Anyone have an idea? > A standard example is > F(x,y) = xy(x^2-y^2)/(x^2+y^2) if x =/=0 or y=/=0, > F(x,y) = 0 if x = y = 0. > The two mixed partial derivatives at (0,0) are different. > -- > 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_ I submitted the result. He was so impressed I pursued it that he asked my professor to grant me a load of extra credit. :) Yeehaaw D === Subject: Re: A vector calc related question posting-account=2OYlAwwAAAAzuGZHzY8fB1XqLzeo4Yd5 I found this function a minute ago at mathworld. I still manage to get matching mixed partial derivatives when I work them on the board. I am at a loss why. But the point is, one must drop the requirement that the function be continuous in order to use the example. So sorry, you were correct. === Subject: Re: A vector calc related question posting-account=2OYlAwwAAAAzuGZHzY8fB1XqLzeo4Yd5 === Subject: Re: A vector calc related question Ill betcha a nickel you didnt try the function. What did you get for the two mixed partial derivatives _at (0,0)_ (as per R.C.s suggestion?) Hint: if you just manipulated formulas and didnt use the definition of a derivative, you did it wrong. dave (I mean wrongly, but that just sounds wrongly.) === Subject: Re: A vector calc related question posting-account=2OYlAwwAAAAzuGZHzY8fB1XqLzeo4Yd5 I didnt manipulate the formula. I did it all by hand on a few boards. But I posted a msg a bit ago to appologize. I found the example at mathworld and it said one must drop the requirement of continuity in order to make it work. === Subject: Re: Hilbert Spaces and L^p > I am sorry, I was very unclear. I meant to say that the > only L^p, l^p spaces that are also Hilbert Spaces are, resp. > L^2 and l^2 > I was hoping you could give me a hint of the role that the > paralellogram rule plays in being able to define a norm in these > spaces. If youve got an inner product on a real vector space you can define a norm, and you discover that the parallelogram law holds. Conversely, if youve got a norm and the parallelogram law holds, you can define an inner product. In fact a little more is true: you can define the inner product from a norm-like function which satisfies the parallelogram law --- you dont have to _assume_ the triangle inequality holds for your norm; you can prove it. We looked at these a couple of years ago: http://www.math-atlas.org/02/norm-euclidean dave === Subject: ext. field Hi I was wondering if someone could help me out. Im trying to figure out how to construct an extention field for lets say g(x) = x^3 + 2x + 1 and I have to create the extention field GF(27) = Z3[x]