>> This argument looks correct, but won't it work for any candidate p_0(x) >> which is bounded above by some constant c? What is it that makes p_0(x) >> so special? >> Call your function p_0(x) (I can't bring myself to use f for the >> independent variable!). For any other fixed admissible function p(x), >> define p_t(x) = p_0(x) + t*dp(x) where dp(x) = p(x) - p_0(x) and 0 <= t <= >> 1. Note that int_0^B dp(x) dx = 0. Compute the derivatives of G(t) = >> Q(p_t) with respect to t: G' = int_0^B dp(x) / (p_t(x) + u(x)) dx >> ... >> John Mitchell Oops. My mistake is ... My original argument went like this: G'(0) = int_0^B dp(x) / (p_0(x) + u(x)) dx = int_{u<=c} dp(x) / (p_0(x) + u(x)) dx + >> int_{u>c} dp(x) / (p_0(x) + u(x)) dx >> ... >> so the integrand >> in the last formula is negative, showing that G'(0) <= 0 as desired. The trouble is that the sets R1 = {x | u(x) <= c} and R2 = {x | u(x) > c} could be too wild for the integrals above to make sense, but maybe >> you can approximate them arbitrarily closely by suitably nice regions. John Mitchell Actually, you can just approximate u(x) by a piecewise-linear function > (the proof above applies in that case), and use the fact that your > functional Q will converge as the approximation is refined to get the > result for continuous functions. John Mitchell ==== Hallo all, I urgently need an answer to following: My task is to discretize (to sample) the surface of the unit sphere. In terms of simplicity I've chosen to do this by deviding the surrounding cubus (borderlenght=2) in N*N*N little cubi (borderlenght=2/N). Only those little cubi that intersect the surface of the unitsphere are of interest. My question now: How many little cubi will intersect the surface of the unitsphere in dependency of N ? Somehow my brain seems (temporarily) too small for sufficient thoughts. Thus I've investigated it experimentally. The result seems to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the unitsphere. If correct, why ?? Can anyone help me ? Greetings (mad) Tobias ==== > I urgently need an answer to following: My task is to discretize (to sample) the surface of the unit sphere. > In terms of simplicity I've chosen to do this by deviding the > surrounding cubus (borderlenght=2) in N*N*N little cubi > (borderlenght=2/N). Only those little cubi that intersect the surface > of the unitsphere are of interest. My question now: How many little cubi will intersect the surface of the unitsphere in > dependency of N ? Somehow my brain seems (temporarily) too small for sufficient > thoughts. Thus I've investigated it experimentally. The result seems > to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the > unitsphere. If correct, why ?? Can anyone help me ? Greetings > (mad) Tobias For N=2, your formula says that 16 (of 8) little cubi intersect the sphere. This is surely wrong. Seaman's reply in the thread starting at http://tinylink.com/?qAd8iCXKHK -- Clive Tooth http://www.clivetooth.dk ==== For N=2, your formula says that 16 (of 8) little cubi intersect the sphere. > This is surely wrong. Seaman's reply in the thread starting at > http://tinylink.com/?qAd8iCXKHK of course You're right. I forgot the constraint N>3. Tobias ==== > Hallo all, I urgently need an answer to following: My task is to discretize (to sample) the surface of the unit sphere. > In terms of simplicity I've chosen to do this by deviding the > surrounding cubus (borderlenght=2) in N*N*N little cubi > (borderlenght=2/N). Only those little cubi that intersect the surface > of the unitsphere are of interest. My question now: How many little cubi will intersect the surface of the unitsphere in > dependency of N ? Somehow my brain seems (temporarily) too small for sufficient > thoughts. Thus I've investigated it experimentally. The result seems > to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the > unitsphere. If correct, why ?? Can anyone help me ? Greetings > (mad) Tobias The first few values are n n_cut 1 1 2 8 3 26 (1 central cube of 27 not hit) 4 56 (8 central cubes of 64 not hit) Looking up [1,8,26,56] in the On-Line Encyclopedia of Integer Sequences gives http://www.research.att.com/projects/OEIS?Anum=A005897 : Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c. lattice). The continuation of the sequence is: 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,... Hugo Pfoertner http://www.pfoertner.org/ ==== For N=2, your formula says that 16 (of 8) little cubi intersect the sphere. > This is surely wrong. > Dave > Seaman's reply in the thread starting at > http://tinylink.com/?qAd8iCXKHK of course You're right. I forgot the constraint N>3. > Tobias Well... I take each cubus [there is really no such word in English, but the term seems useful here] to be a closed cube (that is, it includes all its faces, edges and vertices). So two adjacent cubi share a common square face. For N=4 your formula has 40 cubi intersecting the sphere. But it seems to me that only the central 8 escape intersection, leading to the conclusion that 56 (=4^3-2^3) intersect the sphere. For N=5, I find that 98 cubi intersect the sphere. Your formula gives a non-integer answer. For N=6, we must be very careful how we count the cubi. Several of them intersect our sphere in just a single point. I find that 176 of the cubi intersect the sphere (48 of them at only a single point). So if we took the cubi to be open cubes that number would go down to 128. Anyway, your formula gives 80. Of course, my numbers could be wrong. -- Clive Tooth http://www.clivetooth.dk ==== > Hallo all, I urgently need an answer to following: My task is to discretize (to sample) the surface of the unit sphere. > In terms of simplicity I've chosen to do this by deviding the > surrounding cubus (borderlenght=2) in N*N*N little cubi > (borderlenght=2/N). Only those little cubi that intersect the surface > of the unitsphere are of interest. My question now: How many little cubi will intersect the surface of the unitsphere in > dependency of N ? Somehow my brain seems (temporarily) too small for sufficient > thoughts. Thus I've investigated it experimentally. The result seems > to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the > unitsphere. If correct, why ?? Can anyone help me ? Greetings > (mad) Tobias The first few values are n n_cut > 1 1 > 2 8 > 3 26 (1 central cube of 27 not hit) > 4 56 (8 central cubes of 64 not hit) Looking up [1,8,26,56] in the On-Line Encyclopedia of Integer Sequences > gives > http://www.research.att.com/projects/OEIS?Anum=A005897 : > Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c. > lattice). The continuation of the sequence is: > 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,... Ah... for n=6 that sequence gives 152. This is the mean of two answers (176 and 128) that I gave in another post in this thread and corresponds to viewing the cubi as neither open nor closed. I have no idea how the expression 6n^2+2 arises, though. -- Clive Tooth http://www.clivetooth.dk ==== >> The first few values are >> n n_cut >> 1 1 >> 2 8 >> 3 26 (1 central cube of 27 not hit) >> 4 56 (8 central cubes of 64 not hit) >> Looking up [1,8,26,56] in the On-Line Encyclopedia of Integer Sequences >> gives >> http://www.research.att.com/projects/OEIS?Anum=A005897 : >> Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c. >> lattice). >> The continuation of the sequence is: >> 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,... > Ah... for n=6 that sequence gives 152. This is the mean of two answers (176 > and 128) that I gave in another post in this thread and corresponds to > viewing the cubi as neither open nor closed. I have no idea how the > expression 6n^2+2 arises, though. I suppose one could view each of the cubi as an n-dimensional interval that is a Cartesian product of half-open intervals, so that they exactly partition the space. This would lead to the half-way answer, I think. be in agreement with most of the results that have been posted, but there are some differences compared to the A005897 sequence for n > 5: n count - ----- 1 1 2 8 3 26 4 56 5 98 6 176 or 128 7 194 8 272 9 362 10 464 or 416 I also thought of looking up the sequence, but my sequence gave no matches. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > Hallo all, I urgently need an answer to following: My task is to discretize (to sample) the surface of the unit sphere. > In terms of simplicity I've chosen to do this by deviding the > surrounding cubus (borderlenght=2) in N*N*N little cubi > (borderlenght=2/N). Only those little cubi that intersect the surface > of the unitsphere are of interest. My question now: How many little cubi will intersect the surface of the unitsphere in > dependency of N ? Somehow my brain seems (temporarily) too small for sufficient > thoughts. Thus I've investigated it experimentally. The result seems > to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the > unitsphere. If correct, why ?? Can anyone help me ? Greetings > (mad) Tobias The first few values are n n_cut > 1 1 > 2 8 > 3 26 (1 central cube of 27 not hit) > 4 56 (8 central cubes of 64 not hit) Looking up [1,8,26,56] in the On-Line Encyclopedia of Integer Sequences > gives > http://www.research.att.com/projects/OEIS?Anum=A005897 : > Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c. > lattice). The continuation of the sequence is: > 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,... Ah... for n=6 that sequence gives 152. This is the mean of two answers (176 > and 128) that I gave in another post in this thread and corresponds to > viewing the cubi as neither open nor closed. I have no idea how the > expression 6n^2+2 arises, though. -- > Clive Tooth > http://www.clivetooth.dk 6*n^2+2 is exactly what the sequence title says: for clarifying this. A 1*1*1 cube has 6*1+2 (lattice) points 2*2*2 -> 26 points 3*3*3 -> 56 points n*n*n -> 8 + 12*(n-1) + 6*(n-1)^2 = 6*n^2 + 2 vertices points on edges pts on faces Number of cut sub-cubes (cut means not only touched) = the number of lattice points on the surface of the (n-1)-subdivided cube, but only for n<7. See Dave Seaman's message and my reply. Hugo Pfoertner ==== >> The first few values are >> n n_cut >> 1 1 >> 2 8 >> 3 26 (1 central cube of 27 not hit) >> 4 56 (8 central cubes of 64 not hit) >> Looking up [1,8,26,56] in the On-Line Encyclopedia of Integer Sequences >> gives >> http://www.research.att.com/projects/OEIS?Anum=A005897 : >> Points on surface of cube: 6n^2 + 2 (coordination sequence for b.c.c. >> lattice). >> The continuation of the sequence is: >> 1,8,26,56,98,152,218,296,386,488,602,728,866,1016,1178,1352,... Ah... for n=6 that sequence gives 152. This is the mean of two answers (176 > and 128) that I gave in another post in this thread and corresponds to > viewing the cubi as neither open nor closed. I have no idea how the > expression 6n^2+2 arises, though. I suppose one could view each of the cubi as an n-dimensional interval > that is a Cartesian product of half-open intervals, so that they exactly > partition the space. This would lead to the half-way answer, I think. be in agreement with most of the results that have been posted, but there > are some differences compared to the A005897 sequence for n > 5: n count > - ----- > 1 1 > 2 8 > 3 26 > 4 56 > 5 98 > 6 176 or 128 > 7 194 > 8 272 > 9 362 > 10 464 or 416 I also thought of looking up the sequence, but my sequence gave no matches. -- > Dave Seaman I have now written a little Fortran program available at http://www.randomwalk.de/sequences/cutcub.txt , including results for n<=100. Here are the first few terms: 3 26 4 56 5 98 6 152 7 194 8 272 9 362 10 440 11 530 12 656 13 746 14 872 15 1034 16 1160 17 1298 18 1496 19 1658 20 1856 I have counted cubes only touched by the cutting sphere, with all other vertices either inside or outside as 1/2, because they occur always as inside/outside pairs. If either > or >= is used in the radius comparisons then I can reproduce both variants of Dave's n=6 and n=10 results. Hugo Pfoertner http://www.pfoertner.org/ ==== > I have now written a little Fortran program available at > http://www.randomwalk.de/sequences/cutcub.txt , > including results for n<=100. > Here are the first few terms: > 3 26 > 4 56 > 5 98 > 6 152 > 7 194 > 8 272 > 9 362 > 10 440 > 11 530 > 12 656 > 13 746 > 14 872 > 15 1034 > 16 1160 > 17 1298 > 18 1496 > 19 1658 > 20 1856 > I have counted cubes only touched by the cutting sphere, with all > other vertices either inside or outside as 1/2, because they occur > always as inside/outside pairs. If either > or >= is used > in the radius comparisons then I can reproduce both variants of > Dave's n=6 and n=10 results. I have downloaded your Fortran program and checked it against my lisp program (modified to print the average of the upper and lower results). They agree. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== does anyone know the official abbreviation for the cycle decomposition number used in graph theory. I thought about something like cd-number but couldn't find anything with google... Jon ==== posted. Here is my second try. Bill James developed the Pythagorean win formula to estimate the number of wins a baseball team should have in one season given the formula, estimated win percent = RS^1.83/(RA^1.83 + RS^1.83) was empirically derived (http://www.baseball.reference.com/about/faq.shtml). This note describes a derivation for the formula and provides a formula for a good exponent in the Pythagorean win formula. DERIVATION OF WIN PERCENT One way to derive a win formula from runs is to assume a distribution for the runs scored by each team, assume the team run distributions are independent so that you can multiply them to get a joint distribution of runs, and then sum the probabilities of winning situations. Jim Ferry used this procedure to get a win formula by assuming a Poisson distribution for the runs scored (sci.math, it produces probabilities for all the possible runs: 0, 1, 2, 3, .... Ferry notes that the Poisson distribution does not fit the actual runs distribution in baseball, but he is able to work out a formula for win percentage based on this distribution. There are several other common distributions used in statistics: the binomial, normal, log normal, and Raleigh could all be used to model runs scored and thus each distribution could produce a estimated win formula. One of these distributions, the lognormal, produces a win formula which almost exactly matches James's Pythagorean formula. (The lognormal win formula differs from the Pythagorean by less than 0.0003 over the typical range of RS and RA). If you assume that the runs for each game are produced by independent continuous distributions with density functions Dx[x] and Dy[y] for each team, then the win percent is simply winper = Integrate[ Integrate[ Dx[x]*Dy[y], {y, 0, x}], {x, 0, Infinity}] (The notation Integrate[ f[x], {x,a,b}] means the integral of f[x] from x=a to x=b.) If we substitute log normal distributions, then resulting integral is winper = Integrate[ Integrate[ 1/(2 x y sigma^2 Pi) * Exp[ -(Log[x/mux]^2 + Log[y/muy]^2)/(2 sigma^2) ], {y, 0, x}], {x,0, Infinity}]. where 'mux' is the geometric average of the first team's runs and 'muy' is the geometric average of the other teams' runs. Changing variable to a = Log[x] and b=Log[y] gives winper = Integrate[ Integrate[ 1/(2 sigma^2 Pi) * Exp[ -((a- Log[mux])^2 + (b - Log[muy])^2)/(2 sigma^2) ], {b, -Infinity, a}], {a,-Infinity, Infinity}]. Another change of variable to delta = a-b and integrating gives winper = Integrate[ 1/(2 sigma Sqrt[Pi]) * Exp[-(delta-Log[mux/muy])^2 / (4 sigma^2)], {delta, 0, Infinity}] which simplifies to winper = 1/2 * (1 + Erf[ Log[mux/muy]/ (2 sigma)] ) which we will call the LogNormal Formula. COMPARISON OF JAMES PYTHAGOREAN FORMULA AND THE LOGNORMAL FORMULA If we approximate mux/muy by RS/RA and call that r, then the LogNormal Formula becomes winperlognorm = 1/2 * (1 + Erf[ Log[r]/ (2 sigma)] ). Rewriting the Pythagorean formula in terms of r and letting c be the James Pythagorean exponent gives winperpythag = r^c/(r^c + 1). Computing the first three terms of the Taylor series of each about r=1 : 2 1 r-1 (r - 1 ) winperlognorm = - + ---------------- - --------------- + ... 2 2 Sqrt[Pi] sigma 4 Sqrt[Pi] sigma 2 1 c (r - 1) c (r - 1) winperpythag = - + --------- - ---------- + ... 2 4 8 Notice that the first three terms of the Taylor Series will be equal if 2 James Pythagorean Exponent = c = -------------- Sqrt[Pi] sigma where sigma is the standard deviation of the Log of the runs distribution. sigma can be approximated by sigma ~= StandardDeviation(runs scored)/average(runs scored). For baseball, sigma = 0.617 and c = 2/Sqrt[Pi]/sigma ~= 1.83 give almost identical results. (They differ by less than 0.0003 between r=.8 and r=1.2). The formulas also match exactly at r = 0, r = 1, and r = Infinity. The Pythagorean Exponent formula also explains why a higher value for c is required to estimate win percent from basketball scores. In basketball the standard deviation of points scored divided by the average points scored is closer to sigma = 0.1 which give a much higher value for c. SUMMARY James pythagorean formula winper = RS^c/(RS^c + RA^c) is almost exactly the same as the win percent formula derived from the assumption of independent lognormal run distributions. These formulas are related by the exponent formula 2 James Pythagorean Exponent = c = -------------- Sqrt[Pi] sigma where sigma is the standard deviation the log of runs which is approximately equal to the standard deviation of runs scored divided by the average number of runs scored. ==== > While its true that all derivations are technically proofs, most people seem > to distinguish proofs that can be accomplished by applying a series of > mechanical operations in succession, from something that requires, say, > explanation in English. And yes I realize this is all very vague. At what > line does a proof go from being a derivation to being a 'proof'-proof? It > doesn't seem like there is one (that can defined formally at least). I hope > someone knows what I'm talking about. l8r, Mike N. Christoff I find that most of the times I'm asked to derive something, it turns out to be a formula or identity, like Derive the equation Sum (n = 1 to oo) 1/(n^2) = (pi^2)/6 The term 'proof' seems to be more general in meaning. This is in my limited experience, however. Alex Solla Junior Reed College ==== | While its true that all derivations are technically proofs, most people seem | to distinguish proofs that can be accomplished by applying a series of | mechanical operations in succession, from something that requires, say, | explanation in English. And yes I realize this is all very vague. At what | line does a proof go from being a derivation to being a 'proof'-proof? It | doesn't seem like there is one (that can defined formally at least). I hope | someone knows what I'm talking about. I read a Poincare quote today that doesn't make the line any sharper, but which you might find interesting anyway. He says that derivations are routine and also sterile. Keith Ramsay ==== >| While its true that all derivations are technically proofs, most people seem >| to distinguish proofs that can be accomplished by applying a series of >| mechanical operations in succession, from something that requires, say, >| explanation in English. And yes I realize this is all very vague. At what >| line does a proof go from being a derivation to being a 'proof'-proof? It >| doesn't seem like there is one (that can defined formally at least). I hope >| someone knows what I'm talking about. I read a Poincare quote today that doesn't make the line any sharper, but >which you might find interesting anyway. He says that derivations are >routine and also sterile. But I bet, I just bet, he was even less enamored of anything that required explanation in English. Lee Rudolph ==== Is it always possible to do differential equation modeling for each an every physical phenomenon under a given set of appropriate conditions. ==== > Is it always possible to do differential equation modeling for each an > every physical phenomenon under a given set of appropriate conditions. > I was asking the same question with very little response.. My feeling, I could be wrong, is that a differential equation can model *MOST* phenomena however an intrinsic limit to the number of systems that can be modeled does not exist. This poses an interesting question in itself, if there's a limit to the number of systems in which a differential equation is known valid, then mustn't the order of the equation be so limited? But there are always discrete models that forbid infitessimal values. Patrick Meuser ==== I want to learn the formula to calculate any digit in the decimal form of pi. ex: What is the 100th digit in Pi? ans: 0 thanks ==== > I want to learn the formula to calculate any digit in the decimal form of pi. > ex: What is the 100th digit in Pi? > ans: 0 To find the n'th digit in the decimal expansion of pi, I generally use the formula: floor(pi*10^n) mod 10 Where floor(x) is the largest integer which is less than, or equal to, x. I have tested this formula extensively (well, for n=4 and n=5) and it appears to work. By the way, my formula says that the 100th digit of pi is 9. This agrees with the 100th digit of pi computed by Shanks and Wrench in 1961. I have not checked it against any more recent computation. I think there may be a similar formula for the n'th digit of sqrt(2). -- Clive Tooth http://www.clivetooth.dk ==== > I want to learn the formula to calculate any digit in the decimal form of pi. > ex: What is the 100th digit in Pi? > ans: 0 There is a way to do this for the binary/octal/hexadecimal/etc expansions of Pi, but to my knowledge, no method exists (or at least, none is yet known) for the decimal expansion. -Gary ==== I want to learn the formula to calculate any digit in the decimal form of > pi. > ex: What is the 100th digit in Pi? > ans: 0 There is a way to do this for the binary/octal/hexadecimal/etc expansions of > Pi, but to my knowledge, no method exists (or at least, none is yet known) is it possible for a method that is not know to exist? > for the decimal expansion. -Gary ==== In sci.math, The Last Danish Pastry : > >> I want to learn the formula to calculate any digit in the decimal form of > pi. >> ex: What is the 100th digit in Pi? >> ans: 0 To find the n'th digit in the decimal expansion of pi, I generally use the > formula: floor(pi*10^n) mod 10 Where floor(x) is the largest integer which is less than, or equal to, x. I have tested this formula extensively (well, for n=4 and n=5) and it > appears to work. By the way, my formula says that the 100th digit of pi is 9. This agrees > with the 100th digit of pi computed by Shanks and Wrench in 1961. I have not > checked it against any more recent computation. My computational algorithm (a rather simple one based on multiprecision arithmetic and the formula 16 * atn(1/5) - 4 * atn(1/239)) suggests that digits 96 - 100 of pi are '70679'. I think there may be a similar formula for the n'th digit of sqrt(2). > There is. :-) Your formula is extremely general; the main problem of course is computing floor(X * 10^n), where X is the computational victim: pi, e, sqrt(2), log_e of 2, etc. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > There is a way to do this for the binary/octal/hexadecimal/etc expansions of > Pi, but to my knowledge, no method exists (or at least, none is yet known) > for the decimal expansion. > I believe Simon Plouffe has discovered a method around 1996 (according to a biography I've read). Sam -- People sometimes ask me if it is a sin in the Church of Emacs to use vi. Using a free version of vi is not a sin; it's a penance. - Richard Stallman ==== on There is a way to do this for the binary/octal/hexadecimal/etc expansions of > Pi, but to my knowledge, no method exists (or at least, none is yet known) > for the decimal expansion. I believe Simon Plouffe has discovered a method around 1996 (according to a > biography I've read). Yes: The URL has changed over the years, having previously been at: http://www.stud.enst.fr/~bellard/pi/pi_n2/pi_n2.html -- Clive Tooth http://www.clivetooth.dk ==== ? I want to learn the formula to calculate any digit in the decimal form of pi. ? ex: What is the 100th digit in Pi? ? ans: 0 ? ? thanks Look at http://mathworld.wolfram.com/Bailey-Borwein-PlouffeAlgorithm.html and the links given there. Don't forget to read ?The story behind a formula of Pi? http://mathforum.org/discuss/sci.math/t/517721 HP ==== >is it possible for a method that is not know to exist? Certainly. In mathematics in general, we are always looking for things that are not yet known, but that we believe exist. This is to some extent a question of what you mean by exist. Working mathematicians tend to be Platonists, who consider the objects of mathematics to have an existence independent of what may be known about them at a particular time. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Suppose we take the integer 60 in {0,1,2 ... }, and write: 60 = 4*15; I now take any factor of 60, say 10. I can factor 10 as 2*5, and note that 2 divides 4, and 5 divides 15. So I broke up 10 into 2 factors, one of which goes with or divides 4, and the other which goes with or divides 15. That this is generally true is a simple consequence, I believe, of the fundamental theorem of arithmetic, which really depends on Z being a UFD. What is the situation in algebraic integers? Please correct me if I'm wrong: the ring of algebraic has no irreducibles, hence it cannot be a UFD. Some were asking for a new DISH challenge following Uncle Al's ... How does James Harris know that the algebraic integers form a ring with unity? Especially, these need proof: (a) a algebraic integer implies -a is an algebaic integer (b) a, b, albebraic integers implies a+b and a*b are algebraic integers. Another challenge would be along the lines of the fundamental theorem of algebra, namely: ``Every polynomial f(z) with coefficients in C such that the degree of f is at least one has at least one root in C. [Gauss] (A consequence is that there are many algebraic integers, but still the set is a countable one.) David Bernier ==== >Suppose we take the integer 60 in {0,1,2 ... }, and write: > 60 = 4*15; I now take any factor of 60, say 10. I can factor 10 as 2*5, >and note that 2 divides 4, and 5 divides 15. So I broke up 10 into 2 factors, one of which goes with or divides >4, and the other which goes with or divides 15. That this is generally true is a simple consequence, I believe, >of the fundamental theorem of arithmetic, which really depends >on Z being a UFD. What is the situation in algebraic integers? They are a Bezout domain: every finitely generated ideal is principal. As a consequence of this, any two elements have a gcd (which is unique up to units). As a consequence of this, the ring of all algebraic integers is a pre-Schreier Domain, which is exactly the property you have isolated: If a,b,c are algebraic integers, and a divides b*c, then a factors as a=B*C, with both B and C algebraic integers such that B divides b and C divides c. >Please correct me if I'm wrong: the ring of algebraic has no >irreducibles, hence it cannot be a UFD. Correct. >How does James Harris know that the algebraic integers form a >ring with unity? He does not; he takes it on faith, sometimes; sometimes his arguments seem to imply he believes they are, in fact, not a ring at all. Of course, he doesn't know what a ring is either, so... ====================================================================== 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 ==== > What is the situation in algebraic integers? > The algebraic integers are solutions in C of p(x) = 0, where p is monic polynomial in Z[x] ? > Please correct me if I'm wrong: the ring of algebraic has no > irreducibles, hence it cannot be a UFD. (a) a algebraic integer implies -a is an algebaic integer > (b) a, b, albebraic integers implies a+b and a*b are algebraic integers. ``Every polynomial f(z) with coefficients in C such that the degree of > f is at least one has at least one root in C. [Gauss] (A consequence is that there are many algebraic integers, > but still the set is a countable one.) ==== >>What is the situation in algebraic integers? >> The algebraic integers are solutions in C of p(x) = 0, > where p is monic polynomial in Z[x] ? Yes, my number theory book gives the same definition. I copy here Arturo Magidin's reply earlier in this thread: ``They are a Bezout domain: every finitely generated ideal is principal. As a consequence of this, any two elements have a gcd (which is unique up to units). As a consequence of this, the ring of all algebraic integers is a pre-Schreier Domain, which is exactly the property you have isolated: If a,b,c are algebraic integers, and a divides b*c, then a factors as a=B*C, with both B and C algebraic integers such that B divides b and C divides c. The second paragraph above is the result I was interested in. >>Please correct me if I'm wrong: the ring of algebraic has no >>irreducibles, hence it cannot be a UFD. >>(a) a algebraic integer implies -a is an algebaic integer >>(b) a, b, albebraic integers implies a+b and a*b are algebraic integers. [cf. A. Magidin's reply ] Challenge 1 for James H.: prove that the algebraic integers form a ring. Does he take mathematcians' word on this? The people he calls liars and so forth? Or has he his own proof... >>``Every polynomial f(z) with coefficients in C such that the degree of >> f is at least one has at least one root in C. [Gauss] >>(A consequence is that there are many algebraic integers, >> but still the set is a countable one.) Challenge 2 for James Harris: Prove the Fundamental Theorem of Algebra. David Bernier ==== field is a finite algebraic extension of Q and the ring of integers of a number field is the intersection of the number field and the algebraic integers. Furthermore I'll speculate as the extension is algebraic Q(u1,..uk) = Q[u1,..uk] = Q/(p1,..pk) where = may be taken as isomorphic and (p1,..pk) is the ideal generated by the irreduciable polynomials p1..pk for which pj(uj) = 0, j = 1,..k >I copy here Arturo Magidin's reply earlier in this thread: All interesting stuff for which I'm unprepared. >>Please correct me if I'm wrong: the ring of algebraic has no >>irreducibles, hence it cannot be a UFD. >> > They are a Bezout domain: every finitely generated ideal is > principal. As a consequence of this, any two elements have a > gcd (which is unique up to units). As a consequence of this, the > ring of all algebraic integers is a pre-Schreier Domain, which > is exactly the property you have isolated: > If a,b,c are algebraic integers, and a divides b*c, then a > factors as a=B*C, with both B and C algebraic integers such > that B divides b and C divides c. >>(a) a algebraic integer implies -a is an algebaic integer >>(b) a, b, albebraic integers implies a+b and a*b are algebraic integers. >prove that the algebraic integers form a ring. Close to where I left off with my self imposed homework 6.2.3. Proposition. Let F be an extension field of K and u an element of F. The following conditions are equivalent: (1) u is algebraic over K (2) K(u) is a finite extension of K (3) u belongs to a finite extension of K. 1 ==> 2, not hard as K(u) isomorphic K/p, for some irreducible p in K[x] with p(u) = 0 base is u, u^2,.. u^n where n is degree of u or p. 2 ==> 3, obvious. 3 ==> 1 the significate part Let u1,..u_n be the finite base for the extension and u = sum kj uj for some kj in K, j = 1,..n Thus u,u^2,..u^n are all so expressible. To find a polynomial for which sum aj u^j = 0 gives a linear system with n equation and n unknowns So what's left is to show the matrix is invertible. I think I'm on the right track but sigh, now I need to brush up on linear algebra. Hm, with that now, additive and multiplicative closure of algebraic numbers isn't looking all that abstruse. >>``Every polynomial f(z) with coefficients in C such that the degree of >> f is at least one has at least one root in C. [Gauss] >>(A consequence is that there are many algebraic integers, >> but still the set is a countable one.) >Prove the Fundamental Theorem of Algebra. The proof I've seen uses Liouville's theorem which is a bit much. I've been assured Gauss' proofs are no easier than Liouville's theorem. ---- ==== > Furthermore I'll speculate as the extension is algebraic > Q(u1,..uk) = Q[u1,..uk] = Q/(p1,..pk) The last term should be Q[x]/(p1,...,pk). > 6.2.3. Proposition. Let F be an extension field of K and u an > element of F. The following conditions are equivalent: > (1) u is algebraic over K > (2) K(u) is a finite extension of K > (3) u belongs to a finite extension of K. > 1 ==> 2, not hard as K(u) isomorphic K/p, > for some irreducible p in K[x] with p(u) = 0 > base is u, u^2,.. u^n where n is degree of u or p. 2 ==> 3, obvious. 3 ==> 1 the significate part Can 1, u, u^2, u^3, ..., all be linearly independent over K? -- ==== >> (1) u is algebraic over K >> (2) K(u) is a finite extension of K >> (3) u belongs to a finite extension of K. >> 1 ==> 2, not hard as K(u) isomorphic K/p, >> for some irreducible p in K[x] with p(u) = 0 >> base is u, u^2,.. u^n where n is degree of u or p. > >> 2 ==> 3, obvious. 3 ==> 1 the significate part >Can 1, u, u^2, u^3, ..., all be linearly independent over K? Ah shucks, I'm feeling stupid. It wasn't a change of base problem at all. 6.2.7. Corollary. Let F be an extension field of K. The set of all elements of F that are algebraic over K forms a subfield of F. To prove this I'd note that if a,b in F are algebraic over K, then [K(a,b):K(b)] <= [K(a):K] and [K(a,b):K] = [K(a,b):K(b)] [K(b),K] Now as a+b, ab, -a, 1/a in K(a,b), they're algebraic over K and closure of field operations follow. More elementarily If p(x) = sum aj x^j = 0, then sum (-1)^j aj (-x)^j = 0 sum aj (1/x)^(n-j) = 0 where n = degree of p, x /= 0 last proposition. Would you give some insight, hint for 6.2.10. Proposition. Let F be an algebraic extension of E and let E be an algebraic extension of K. Then F is an algebraic extension of K. ---- ==== The World Wide Wade escreveu na > Proposition. If exists k such that |a_n / b_n| <= k for > all n in N and the series sum b_n converges, then the > series sum a_n converges. counterexample: b_n = (-1)^n/n and a_n = 1/n. Yes, you are right. I forgot the hypothesis a_n >= 0 and b_n > 0. And, given this error in the proposition (without the forgotten hypothesis), my argument that the series converge is wrong. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com ==== I'm trying to plot a histogram on some data that I receive at runtime. I'm trying to measure distances between two points, and these distances can be from the range 1 to 5000000. I want to use a logarithmic scale with a range -2^29 to +2^29. How can I determine at any point in time which strata (i.e which cohort or portion of scale) the value belongs to (without using the square or square-root functions)? I'm trying to implement this on a computer. I could have mask-bits of all possible scales, such as 2^2, 2^3, 2^4 upto 2^29 and AND each of these with the value to figure this out but is there a cleaner way of doing this? --Andre ==== I'm trying to plot a histogram on some data that I receive at runtime. > I'm trying to measure distances between two points, and these distances > can be from the range 1 to 5000000. I want to use a logarithmic scale > with a range -2^29 to +2^29. How can I determine at any point in time > which strata (i.e which cohort or portion of scale) the value belongs to > (without using the square or square-root functions)? I'm trying to > implement this on a computer. I could have mask-bits of all possible scales, such as 2^2, 2^3, 2^4 > upto 2^29 and AND each of these with the value to figure this out but is > there a cleaner way of doing this? A distance can't be less than zero... Maybe you meant to write: 1/(2^29) to 2^29? In some languages, like C, you can use bit-shift operations to get approximations of logs in base 2 of integers... Anyway, for integer inputs, maybe this can be of some help: David Bernier ----------------------------------------------- #include int main(void) { long input, inputcopy; int sign, power; printf(nplease enter your number:n); scanf(%ld, &input); inputcopy = input; if(input>=0) { sign = 1; } else { sign = -1; inputcopy = -input; } power = 0; while((inputcopy>>power)>0) /*** >> bitwise-shift ***/ { power++; } printf(ninput = %ldtsign = %dtpower = %dnn, input, sign, power); return 0; } ----------------------------------------------------------- ==== Actually in my implementation I get negative distances that denote the opposite direction. However, thanks a lot for the code excerpt :) --Andre > >> I'm trying to plot a histogram on some data that I receive at runtime. >> I'm trying to measure distances between two points, and these >> distances can be from the range 1 to 5000000. I want to use a >> logarithmic scale with a range -2^29 to +2^29. How can I determine at >> any point in time which strata (i.e which cohort or portion of scale) >> the value belongs to (without using the square or square-root >> functions)? I'm trying to implement this on a computer. >> I could have mask-bits of all possible scales, such as 2^2, 2^3, 2^4 >> upto 2^29 and AND each of these with the value to figure this out but >> is there a cleaner way of doing this? > A distance can't be less than zero... > Maybe you meant to write: 1/(2^29) to 2^29? In some languages, like C, you can use bit-shift operations to get > approximations of logs in base 2 of integers... Anyway, for integer inputs, maybe this can be of some help: David Bernier ----------------------------------------------- > #include { > long input, inputcopy; > int sign, power; printf(nplease enter your number:n); > scanf(%ld, &input); > inputcopy = input; > if(input>=0) > { > sign = 1; > } > else > { > sign = -1; > inputcopy = -input; > } > power = 0; > while((inputcopy>>power)>0) /*** >> bitwise-shift ***/ > { > power++; > } printf(ninput = %ldtsign = %dtpower = %dnn, input, sign, power); return 0; > } ----------------------------------------------------------- > ==== First I wanna say that I need an equation that I can use in a C++ program, so I might have problems doing some special calculations without sertain libraries. Anyway, to the math stuff. I have a linear number from 0.0 to 1.0. And from that number I need to produce a steep curve, like a sinus curve from 0 to 90 degrees ( ? ). So if the linear number is 0.0, then the output number is 0.0. And if it is 1.0, then the output is 1.0. It's the numbers in-between that is my problem. For example, the input number 0.2 might produce 0.5. So the curve is steep in the beginning and flattens out at the top. Again like a sinus curve. I know I have done this several times in the past. But I can't figure it out now. I tried with a normal sinus equation: sin( 90 * X ) sin( 90 ) * X ...etc. Isn't 90 degree the top of a sinus curve? So shouldn't sin( 90 ) give 1.0? Anyway, while I ponder this by myself, I thought I would ask here to see if someone could help me... TIA! , Espen ==== > First I wanna say that I need an equation that I can use in a C++ program, > so I might have problems doing some special calculations without sertain > libraries. Anyway, to the math stuff. I have a linear number from 0.0 to 1.0. And > from that number I need to produce a steep curve, like a sinus curve from 0 > to 90 degrees ( ? ). So if the linear number is 0.0, then the output number is 0.0. And if it is > 1.0, then the output is 1.0. It's the numbers in-between that is my > problem. For example, the input number 0.2 might produce 0.5. So the curve > is steep in the beginning and flattens out at the top. Again like a sinus > curve. I know I have done this several times in the past. But I can't figure it > out now. I tried with a normal sinus equation: sin( 90 * X ) > sin( 90 ) * X ...etc. Isn't 90 degree the top of a sinus curve? So shouldn't sin( 90 ) > give 1.0? Anyway, while I ponder this by myself, I thought I would ask here to see if > someone could help me... > Don't you think your maths library takes radians as an argument to sin, rather than degrees? So, sin(90degrees) is sin(0.5*pi), which is 1. -- Robert Bronsing Can't you see? It all makes perfect sense, expressed in dollars and cents, pounds, shillings and pence (R. Waters) ==== >> First I wanna say that I need an equation that I can use in a C++ >> program, so I might have problems doing some special calculations >> without sertain libraries. >> Anyway, to the math stuff. I have a linear number from 0.0 to 1.0. And >> from that number I need to produce a steep curve, like a sinus curve >> from 0 to 90 degrees ( ? ). >> So if the linear number is 0.0, then the output number is 0.0. And if >> it is >> 1.0, then the output is 1.0. It's the numbers in-between that is my >> problem. For example, the input number 0.2 might produce 0.5. So the >> curve is steep in the beginning and flattens out at the top. Again >> like a sinus curve. >> I know I have done this several times in the past. But I can't figure >> it out now. I tried with a normal sinus equation: >> sin( 90 * X ) >> sin( 90 ) * X >> ...etc. Isn't 90 degree the top of a sinus curve? So shouldn't sin( >> 90 ) give 1.0? >> Anyway, while I ponder this by myself, I thought I would ask here to >> see if someone could help me... > > Don't you think your maths library takes radians as an argument to sin, > rather than degrees? So, sin(90degrees) is sin(0.5*pi), which is 1. > Thanx, I knew it was something simple I had missed... , Espen ==== It is an elementary sets fact that all integer sequences can be enumerated (i.e. mapped to integers). A method enumerating all integer sequences can be easily adopted for enumerating all monomials. Assuming that there is a countable number of variables, just (arbitrarily) map variables into integers and then, use integer sequences enumeration. Is there enumeration E(m) that preserves multiplication? For example, given that x*y^2 multiplied by x^3*y is equal to x^4*y^3 we require E(x*y^2)+E(x^3*y)=E(x^4*y^3) If not, could the requirement be relaxed to some associative function agg, other than addition, that meets agg(E(x*y^2),E(x^3*y))=E(x^4*y^3) ? Is there an enumeration (relaxed, or strict power additive) that preserves multiplication and is invariant over variables enumeration? ==== >It is an elementary sets fact that all integer sequences can be >enumerated (i.e. mapped to integers). Ummm, no. There are uncountably many sequences of integers, in fact there are uncountably many sequences of 0 and 1. Or did you mean something different? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >It is an elementary sets fact that all integer sequences can be >enumerated (i.e. mapped to integers). Ummm, no. There are uncountably many sequences of integers, in fact > there are uncountably many sequences of 0 and 1. Or did you mean > something different? Perhaps he/she means finite sequences, as the application to monomials would suggest. -- ==== > It is an elementary sets fact that all integer sequences can be > enumerated (i.e. mapped to integers). A method enumerating all integer > sequences can be easily adopted for enumerating all monomials. > Assuming that there is a countable number of variables, just > (arbitrarily) map variables into integers and then, use integer > sequences enumeration. Is there enumeration E(m) that preserves multiplication? For example, given that x*y^2 multiplied by x^3*y is equal to x^4*y^3 we require E(x*y^2)+E(x^3*y)=E(x^4*y^3) Such an enumeration is not possible. Suppose one existed. Clearly E(x^n)=E(x)*n and E(y^n)=E(y)*n. But then E(x^E(y))=E(x)*E(y)=E(y)*E(x)=E(y^E(x)). So we would have two distinct polynomials, x^E(y) and y^E(x), with the same number. [The only way those two polynomials could not be distinct would be if E(x)=E(y)=0, but this is obviously not possible.] > If not, could the requirement be relaxed to some associative function > agg, other than addition, that meets agg(E(x*y^2),E(x^3*y))=E(x^4*y^3) > > Is there an enumeration (relaxed, or strict power additive) that > preserves multiplication and is invariant over variables enumeration? -- Clive Tooth http://www.clivetooth.dk ==== It is an elementary sets fact that all integer sequences can be > enumerated (i.e. mapped to integers). A method enumerating all integer > sequences can be easily adopted for enumerating all monomials. > Assuming that there is a countable number of variables, just > (arbitrarily) map variables into integers and then, use integer > sequences enumeration. Is there enumeration E(m) that preserves multiplication? For example, given that x*y^2 multiplied by x^3*y is equal to x^4*y^3 we require E(x*y^2)+E(x^3*y)=E(x^4*y^3) Such an enumeration is not possible. Suppose one existed. Clearly E(x^n)=E(x)*n and E(y^n)=E(y)*n. But then E(x^E(y))=E(x)*E(y)=E(y)*E(x)=E(y^E(x)). So we would have two distinct polynomials, x^E(y) and y^E(x), with the same > number. > [The only way those two polynomials could not be distinct would be if > E(x)=E(y)=0, but this is obviously not possible.] + is just some commutative and associative operation, then such E exists: x^k1*y^k2*z^k3*... ---E--> 2^k1*3^k2*5^k3*... The + is a multiplication! I was wrong when generalizing one-variable monomials case where linear E exists E(x^n)=n. ==== I cant even get started on this one. Anyone help me?? Evil Fuzzles may look cute, but they can really make a loud noise when they ROAR!!! A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound Units) on a sound-measuring device that is 5 feet away. Yellow Evil Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as their yellow counterparts! So anyway, one day Zorlon-IX the Grundo was out doing a bit of maintenance on the exterior of the Space Station, and before him he saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5 feet high, 10 feet away and red! It looked straight at Zorlon, and gave the loudest ROAR it could!!! How many NSU did Zorlon's sound-measuring device register? ==== > I cant even get started on this one. Anyone help me?? > Evil Fuzzles may look cute, but they can really make a loud noise when > they ROAR!!! > A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian Sound > Units) on a sound-measuring device that is 5 feet away. Yellow Evil > Fuzzles are twice as loud, and Red Fuzzles... THREE times as loud as > their yellow counterparts! > So anyway, one day Zorlon-IX the Grundo was out doing a bit of > maintenance on the exterior of the Space Station, and before him he > saw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5 > feet high, 10 feet away and red! It looked straight at Zorlon, and > gave the loudest ROAR it could!!! > How many NSU did Zorlon's sound-measuring device register? The information supplied is inadequate. What is the correlation of the height of an Evil Fuzzie and the amount of NSU it makes when roaring? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Shh! The maestro is decomposing! - Gary Larson ==== > Greetings, Could anyone give me an example of a stochastic process (or random > process) > that is ergodic but not stationary? No. As I understand it, a non-stationary stochastic process must be nonergodic. On the other hand, a Spherically Invariant Random Process (SIRP) is an example of a stationary stochastic process that is non-ergodic. -- 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 ==== I did some intense web searching, and I think I've found an example (albeit trivial) of an ergodic, non-stationary stochastic/random process: Consider two degenerate random variables X and Y: p(X=1)=1 and p(Y=-1)=1. Then form a random/stochastic process by alternating draws from each of these RVs: {... X1 Y1 X2 Y2 ...}. Note the process is clearly not stationary since, for example, the mean changes from time sample to time sample. However, the process is ergodic (in the mean, for example) since each sample path of the process (there is only one!) time-averages to the same value. There seem to be two competing definitions of ergodicity that I've seen: 1) Ergodic: time-averages for each sample path of the process converge (almost surely or with probability 1) to the same value. 2) Ergodic: time-averages for each sample path of the process converge (again, almost surely or with probability 1) to the same value, and this value is equal to the time-invariate ensemble averages of the process. Hence, stationarity is assumed in this definition. My above example uses definition 1. I think the second defintion is the definition of Birkhoff's Ergodic Theorem (see Tom Cover's text , Elements of Information Theory, p. 474, for example), which holds for stationary, ergodic sources. The formal definition given by Cover for ergodicity is this: To be precise, an ergodic source is defined on a probability space (Omega, script_B, P), where script_B is a sigma-algebra of subsets of Omega an P is a probability measure. A random variable X is defined as a function X(omega), omega is-an-element-of Omega, on the probability space. We also have a transformation T:Omega-->Omega, which plays the role of a time shift... The transformation is called ergodic if every set A (element-of script_B) such that TA = A, has measure either 0 or 1. If T is... ergodic, we say that the process defined by X_n(omega) = X(T^n omega) is ... ergodic. This is too mathematically intense for my small brain. But I hope it jives with the first definition I supplied above. I hope this hasn't been total nonsense. Feedback GREATLY appreciated. Sincerely, rjc. -- Ryan Cassidy Electrical Engineering Graduate Student Stanford University Greetings, Could anyone give me an example of a stochastic process (or random > process) > that is ergodic but not stationary? No. As I understand it, a non-stationary stochastic process must be > nonergodic. On the other hand, a Spherically Invariant Random Process (SIRP) is > an example of a stationary stochastic process that is non-ergodic. -- > 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 ==== If dy and dx are the errors in y and x respectively, given z = sqrt[y^(2) - x^(2)] what is the error dz associated with z? z^(2) = y^(2) - x^(2) 2dz/z = 2dy/y - 2dx/x dz = z(dy/y - dx/x) I have a feeling this is way out. James ==== > z^(2) = y^(2) - x^(2) 2dz/z = 2dy/y - 2dx/x You seem to be saying that the differential of z^2 is 2 dz/z. That is incorrect. It is 2 z dz. -- 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 ==== >If dy and dx are the errors in y and x respectively, given z = sqrt[y^(2) - x^(2)] what is the error dz associated with z? z^(2) = y^(2) - x^(2) 2dz/z = 2dy/y - 2dx/x I have no idea where that formula comes from. If the errors are small enough, and if one of x and y is substantially larger than the other you could use calculus to estimate the error in z. Or you could just do it: Let's say that x is the _true_ value and x' = x + dx is the approximate value; similarly for y and z. Then (z + dz)^2 = (y + dy)^2 - (x + dx)^2 and z^2 = y^2 - x^2, which gives 2z dz + dz^2 = 2y dy + dy^2 - (2x dx + dx^2). You could solve that for dz to get an exact and essentially useless formula for the error. Typically one would instead assume that the errors are small enough that the squares of the errors are negligible; this gives dz ~ (2y dy - 2x dx)/z. But there's a subtle point here because of the subtraction; in order for the square of the errors to be negligible it's not enough to have dx much smaller than x and dy much smaller than y, you need that the squares of the errors are much smaller than the _difference_ y dy - x dx. This will happen if y is much larger than x and dx and dy are about the same size; if x and y are about the same size then analyzing the error is trickier. >dz = z(dy/y - dx/x) I have a feeling this is way out. That's correct, as far as I can see (where did that formula above come from anyway?) James > ************************ David C. Ullrich ==== If dy and dx are the errors in y and x respectively, given z = sqrt[y^(2) - x^(2)] what is the error dz associated with z? z^(2) = y^(2) - x^(2) 2dz/z = 2dy/y - 2dx/x I have no idea where that formula comes from. If the errors are small enough, and if one of x and y > is substantially larger than the other you could use > calculus to estimate the error in z. Or you could just > do it: Let's say that x is the _true_ value and x' = x + dx is > the approximate value; similarly for y and z. Then (z + dz)^2 = (y + dy)^2 - (x + dx)^2 and z^2 = y^2 - x^2, which gives 2z dz + dz^2 = 2y dy + dy^2 - (2x dx + dx^2). You could solve that for dz to get an exact and essentially > useless formula for the error. Typically one would instead > assume that the errors are small enough that the > squares of the errors are negligible; this gives dz ~ (2y dy - 2x dx)/z. But there's a subtle point here because of the subtraction; > in order for the square of the errors to be negligible it's not > enough to have dx much smaller than x and dy much > smaller than y, you need that the squares of the errors > are much smaller than the _difference_ y dy - x dx. > This will happen if y is much larger than x and dx and > dy are about the same size; if x and y are about the > same size then analyzing the error is trickier. dz = z(dy/y - dx/x) I have a feeling this is way out. That's correct, as far as I can see (where did that > formula above come from anyway?) You mean this? dz = z(dy/y - dx/x) Check out. http://badger.physics.wisc.edu/lab/manual/node4.html According to this site and a number of others I dug up on google, if z = x^(n) then, dz/z = n dx/x That is, the relative error of z is the relative error x multiplied by the power n. Thus if, z^(2) = y^(2) - x^(2) then, 2 dz/z = 2dy/y + 2dx/x (I think the minus actually becomes a plus) Therefore, dz/z = dy/y + dx/x dz = z(dy/y + dx/x) Although this seems wrong, anyone have any ideas? James > ************************ David C. Ullrich ==== >>If dy and dx are the errors in y and x respectively, given >>z = sqrt[y^(2) - x^(2)] >>what is the error dz associated with z? >>z^(2) = y^(2) - x^(2) >>2dz/z = 2dy/y - 2dx/x >> I have no idea where that formula comes from. >> If the errors are small enough, and if one of x and y >> is substantially larger than the other you could use >> calculus to estimate the error in z. Or you could just >> do it: >> Let's say that x is the _true_ value and x' = x + dx is >> the approximate value; similarly for y and z. Then >> (z + dz)^2 = (y + dy)^2 - (x + dx)^2 >> and >> z^2 = y^2 - x^2, >> which gives >> 2z dz + dz^2 = 2y dy + dy^2 - (2x dx + dx^2). >> You could solve that for dz to get an exact and essentially >> useless formula for the error. Typically one would instead >> assume that the errors are small enough that the >> squares of the errors are negligible; this gives >> dz ~ (2y dy - 2x dx)/z. >> But there's a subtle point here because of the subtraction; >> in order for the square of the errors to be negligible it's not >> enough to have dx much smaller than x and dy much >> smaller than y, you need that the squares of the errors >> are much smaller than the _difference_ y dy - x dx. >> This will happen if y is much larger than x and dx and >> dy are about the same size; if x and y are about the >> same size then analyzing the error is trickier. >>dz = z(dy/y - dx/x) >>I have a feeling this is way out. >> That's correct, as far as I can see (where did that >> formula above come from anyway?) You mean this? dz = z(dy/y - dx/x) Yes. >Check out. http://badger.physics.wisc.edu/lab/manual/node4.html According to this site and a number of others I dug up on google, if z = x^(n) then, dz/z = n dx/x That is, the relative error of z is the relative error x multiplied by >the power n. Assuming that the relative error is small then yes this is approximately correct. >Thus if, z^(2) = y^(2) - x^(2) then, 2 dz/z = 2dy/y + 2dx/x Huh? How does this follow, exactly? sum of the relative errors...) >(I think the minus actually becomes a plus) Therefore, dz/z = dy/y + dx/x dz = z(dy/y + dx/x) Although this seems wrong, anyone have any ideas? >>James >> ************************ >> David C. Ullrich > ************************ David C. Ullrich ==== [snip] >2 dz/z = 2dy/y + 2dx/x Huh? How does this follow, exactly? sum of the relative errors...) That's what I thought. Any ideas on what does follow? ==== [snip] >2 dz/z = 2dy/y + 2dx/x >> Huh? How does this follow, exactly? >> sum of the relative errors...) That's what I thought. Huh? If that's what you thought then why did you say 2 dz/z = 2dy/y + 2dx/x in the first place? >Any ideas on what does follow? Yes. I could retype and resend my original post. Or you could just go back and read it. > ************************ David C. Ullrich ==== > If dy and dx are the errors in y and x respectively, given z = sqrt[y^(2) - x^(2)] what is the error dz associated with z? > It depends on what kind of errors are in y and x. According to Experiments in Physical Chemistry by Shoemaker, Garland, & Nibler: 1) If the errors dx and dy are SYSTEMATIC, where f is a function of x and y, then the error in f is: df = f_x * dx + f_y * dy assuming that dx and dy are small enough so that the partial derivatives of f (f_x and f_y) are nearly constant. For your f = z = sqrt[y^2 - x^2], we have dz = (y/z)dy - (x/z)dx 2) If the errors in x and y are RANDOM, where dx and dy are the confidence limits, then the propagated error in f is: (df)^2 = (f_x)^2 * (dx)^2 + (f_y)^2 * (dy)^2 For your f, (dz)^2 = (y/z)^2 * (dy)^2 + (x/z)^2 * (dx)^2 Various assumptions are made when deriving this formula. Consult the reference provided. Hope this helps. Brett ==== If dy and dx are the errors in y and x respectively, given z = sqrt[y^(2) - x^(2)] what is the error dz associated with z? z^(2) = y^(2) - x^(2) 2dz/z = 2dy/y - 2dx/x I have no idea where that formula comes from. If the errors are small enough, and if one of x and y > is substantially larger than the other you could use > calculus to estimate the error in z. Or you could just > do it: Let's say that x is the _true_ value and x' = x + dx is > the approximate value; similarly for y and z. Then (z + dz)^2 = (y + dy)^2 - (x + dx)^2 and z^2 = y^2 - x^2, which gives 2z dz + dz^2 = 2y dy + dy^2 - (2x dx + dx^2). You could solve that for dz to get an exact and essentially > useless formula for the error. Typically one would instead > assume that the errors are small enough that the > squares of the errors are negligible; this gives dz ~ (2y dy - 2x dx)/z. But there's a subtle point here because of the subtraction; > in order for the square of the errors to be negligible it's not > enough to have dx much smaller than x and dy much > smaller than y, you need that the squares of the errors > are much smaller than the _difference_ y dy - x dx. > This will happen if y is much larger than x and dx and > dy are about the same size; if x and y are about the > same size then analyzing the error is trickier. dz = z(dy/y - dx/x) I have a feeling this is way out. That's correct, as far as I can see (where did that > formula above come from anyway?) You mean this? dz = z(dy/y - dx/x) Check out. http://badger.physics.wisc.edu/lab/manual/node4.html According to this site and a number of others I dug up on google, if z = x^(n) then, dz/z = n dx/x According to that site, the general prescription for estimating the uncertainty in z, given measurement error in x and y, is this: z = f(x,y) dz = (@f/@x)dx + (@f/@y)dy then work it out ignoring signs of each term. What you're really trying to do is figure out how far z could vary over the entire possible range of x and y values within x+-dx and y+-dy. That's why you ignore signs: you're estimating the worst case. For your function z = sqrt(y^2 - x^2) @f/@x = -(1/2)*[y^2-x^2]^(-1/2)*2x = -x/z @f/@y = (1/2)*[y^2-x^2]^(-1/2)*2y = y/z So you have, ignoring the signs, dz = (x/z)dx + (y/z)dy This doesn't work out into a nice relative error formula. It does give you z dz = x dx + y dy which differs from David's result by a factor of 2. I'm not sure why. Note that this is not really very rigorous mathematics. - Randy ==== >[snip] >2 dz/z = 2dy/y + 2dx/x >> Huh? How does this follow, exactly? >> sum of the relative errors...) That's what I thought. Huh? If that's what you thought then why did you say > 2 dz/z = 2dy/y + 2dx/x in the first place? >Any ideas on what does follow? Yes. I could retype and resend my original post. Or you could > just go back and read it. I'm questionable about the validity of your equation dz ~ (2y dy - 2x dx)/z . I found that an alternative way to evaluate the error in z is to find the change in h when x + dx is substituted in place of x z + dz_x = sqrt[(x + dx)^(2) + y^(2)] and likewise, the change in z due to y can be found from z + dz_y = sqrt[(x)^(2) + (y + dy)^(2)] Thus dz = dz_x + dz_y Interestingly, this method produces an error almost exactly the same as Randy Poe's equation dz = (x/z)dx + (y/z)dy, but differs from yours by a factor of two. > ************************ David C. Ullrich ==== [snip] > According to that site, the general prescription for estimating > the uncertainty in z, given measurement error in x and y, is > this: z = f(x,y) dz = (@f/@x)dx + (@f/@y)dy then work it out ignoring signs of each term. What you're > really trying to do is figure out how far z could vary over > the entire possible range of x and y values within x+-dx and > y+-dy. That's why you ignore signs: you're estimating the > worst case. For your function > z = sqrt(y^2 - x^2) @f/@x = -(1/2)*[y^2-x^2]^(-1/2)*2x > = -x/z > @f/@y = (1/2)*[y^2-x^2]^(-1/2)*2y > = y/z Interesting. Just out of curiousity, why is it that dz = (@f/@x)dx + (@f/@y)dy Furthermore, how did you get from -(1/2)*[y^2-x^2]^(-1/2)*2x to -x/z ? Sorry if these seem like dumb questions, but I'm still in high school and we have yet to study partial derivatives. > So you have, ignoring the signs, dz = (x/z)dx + (y/z)dy This doesn't work out into a nice relative error > formula. It does give you z dz = x dx + y dy which differs from David's result by a factor of 2. I'm > not sure why. Note that this is not really very rigorous mathematics. - Randy ==== > [snip] > According to that site, the general prescription for estimating > the uncertainty in z, given measurement error in x and y, is > this: z = f(x,y) dz = (@f/@x)dx + (@f/@y)dy then work it out ignoring signs of each term. What you're > really trying to do is figure out how far z could vary over > the entire possible range of x and y values within x+-dx and > y+-dy. That's why you ignore signs: you're estimating the > worst case. For your function > z = sqrt(y^2 - x^2) @f/@x = -(1/2)*[y^2-x^2]^(-1/2)*2x > = -x/z > @f/@y = (1/2)*[y^2-x^2]^(-1/2)*2y > = y/z Interesting. Just out of curiousity, why is it that dz = (@f/@x)dx + (@f/@y)dy That's just the expression for the total differential dz in terms of the partials with respect to x and y. The @ sign is standing in for the curly d used for partial derivatives. > Furthermore, how did you get from -(1/2)*[y^2-x^2]^(-1/2)*2x to -x/z ? There are three terms multiplied together here. A factor of (1/2) out front, a middle term raised to a this out on paper, you'd probably write it as a fraction with sqrt(y^2-x^2) in the denominator. That's what raising to the negative 1/2 power means. Multiplying the (1/2) by the (2x) gives me x. The sqrt(y^2-x^2) in the denominator is the same as z. > Sorry if these seem like dumb questions, but I'm still in high school and we > have yet to study partial derivatives. Ah. I misunderstood your math level. As you can see, the theory of error propagation is based on partial derivatives. It's really not that much harder than ordinary derivatives, you just have to learn how to treat one variable as a constant temporarily. - Randy ==== >>[snip] >2 dz/z = 2dy/y + 2dx/x >> Huh? How does this follow, exactly? >> sum of the relative errors...) >>That's what I thought. >> Huh? If that's what you thought then why did you say >> 2 dz/z = 2dy/y + 2dx/x in the first place? > Yes you did. That doesn't answer the question of why you also thought it seemed right... never mind. >>Any ideas on what does follow? >> Yes. I could retype and resend my original post. Or you could >> just go back and read it. I'm questionable about the validity of your equation dz ~ (2y dy - 2x dx)/z >. I found that an alternative way to evaluate the error in z is to find >the change in h when x + dx is substituted in place of x z + dz_x = sqrt[(x + dx)^(2) + y^(2)] and likewise, the change in z due to y can be found from z + dz_y = sqrt[(x)^(2) + (y + dy)^(2)] Thus dz = dz_x + dz_y Interestingly, this method produces an error almost exactly the same as >Randy Poe's equation dz = (x/z)dx + (y/z)dy, but differs from yours by a >factor of two. Well duh - if you look at how I derived that equation you see it was just a typo - I divided one side of an equation by 2 without dividing the other side. Giving an estimate exactly the same as what you say he did. That _is_ interesting. (Make certain to note the conditions under which that estimate is valid.) >> ************************ >> David C. Ullrich > ************************ David C. Ullrich ==== >> [...] This doesn't work out into a nice relative error >formula. It does give you z dz = x dx + y dy which differs from David's result by a factor of 2. I'm >not sure why. If you look at how I derived that equation you see I just dropped a 2. (Was actually a typo, or rather a cut&pasto.) >Note that this is not really very rigorous mathematics. - Randy ************************ David C. Ullrich ==== I have a set of x,y coordinates with bi-directional error bars, the data is as follows: X Y 0.62996825316836 1.267 0.62233431530006 1.255 0.60893349390553 1.2315 0.58855755878249 1.19 0.55879334283794 1.125 X error Y error 8.1893e-4 1.5342e-3 1.1619e-3 1.0801e-3 1.5507e-3 1.5000e-3 2.0207e-3 9.6225e-4 2.6421e-3 0.0104 How could one reliably compute the uncertainty in the slope m = 2.00704 given this data? James ==== >If we choose randomly and uniformly two points p,q in the n-dimensional >unit cube in R^n what is the expected Euclidean distance d(p,q) ? I can give you a method of computing it numerically with > one integration and a reasonable amount of computing. The moment generating function of the square of the distance > is m(t) = (6(exp(t) - 1 - t - t^2/2)/t^3)^n, and the Laplace > transform is just m(-t). Now compute int -m'(-t)/sqrt(t) dt/sqrt(pi), the integral going from 0 to infinity, and this is the answer. Is it possible to express this integral using the Gamma function ? ==== Here's an interesting little puzzle: Prove that if x^2 + y^2 = z^2 for integers x, y, z with (x, y) = 1 then every factor of x^2 + xy + y^2 is of the form 4.Z + 1. (and no, I haven't forgotten about negative solutions either ;) This isn't trivially true even without the condition x^2 + y^2 = z^2, since for example 2^2 + 2.1 + 1^2 = 7. Also of course I'm not claiming the converse holds. I'll post my solution in a few days in the unlikely event no one solves it. --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen ==== > Prove that if x^2 + y^2 = z^2 for integers x, y, z > with (x, y) = 1 then every factor of x^2 + xy + y^2 > is of the form 4.Z + 1. As x^2 + y^2 = z^2 and gcd(x, y) =1, there are m, n relatively prime, of opposite parity, so that x = m^2 - n^2 and y = 2*m*n (relabeling if needed). Then x^2 + xy + y^2 = (m(m+n))^2 + (n(m-n))^2. The gcd of m(m+n) and n(m-n) is 1, so x^2 + xy + y^2 is a sum of two relatively prime squares. As such, every prime factor is of the form 4Z+1. What more restrictive congruential criterion do the primes dividing x^2 + xy + y^2 satisfy? ==== Prove that if x^2 + y^2 = z^2 for integers x, y, z > with (x, y) = 1 then every factor of x^2 + xy + y^2 > is of the form 4.Z + 1. As x^2 + y^2 = z^2 and gcd(x, y) =1, there are m, n relatively prime, of > opposite parity, so that x = m^2 - n^2 and y = 2*m*n (relabeling if needed). Then x^2 + xy + y^2 = (m(m+n))^2 + (n(m-n))^2. The gcd of m(m+n) and n(m-n) > is 1, so x^2 + xy + y^2 is a sum of two relatively prime squares. As such, every > prime factor is of the form 4Z+1. Congrats. My method was slightly different, as I plugged the m, n relations into (x^2 + y^2)^2 - (x.y)^2 to get (m^4 - n^4)^2 + 16.m^4.n^4, in which m^4 - n^4 is odd, so that (m^4 - n^4, 4.m^4.n^4) = 1. > What more restrictive congruential criterion do the primes dividing x^2 + xy + > y^2 satisfy? SPOILER p = 3.Z + 1 in general, if (x, y) = 1. (hence 12.Z + 1 subject to x^2 + y^2 = z^2) John Ramsden ==== > SPOILER >hence 12.Z + 1 subject to x^2 + y^2 = z^2 Exactly what I had in mind. John Robertson ==== > Prove that if x^2 + y^2 = z^2 for integers x, y, z >> with (x, y) = 1 then every factor of x^2 + xy + y^2 >> is of the form 4.Z + 1. As x^2 + y^2 = z^2 and gcd(x, y) =1, there are m, n relatively prime, of >opposite parity, so that x = m^2 - n^2 and y = 2*m*n (relabeling if needed). Then x^2 + xy + y^2 = (m(m+n))^2 + (n(m-n))^2. The gcd of m(m+n) and n(m-n) is >1, so x^2 + xy + y^2 is a sum of two relatively prime squares. As such, every >prime factor is of the form 4Z+1. What more restrictive congruential criterion do the primes dividing x^2 + xy + >y^2 satisfy? It's easier to note x^2 + xy + y^2 = ((x + y)^2 + z^2)/2. Verify that GCD(x + y, z) = GCD(x, y) when x^2 + y^2 = z^2. -- Spammers: I don't want a small digital camera to post photos of a large, low weight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI ==== I'm looking for algorithm to determine the set of edges that form all circuits in a directed graph. Or equivalently the set of edges that are not present in any circuit in a digraph. I am aware of papers on enumerating all the circuits in digraph, but is there faster algorithms for finding just the set of edges? Jack Middleton ==== >I'm looking for algorithm to determine the set of edges that form all >circuits in a directed graph. Or equivalently the set of edges that >are not present in any circuit in a digraph. I am aware of papers on >enumerating all the circuits in digraph, but is there faster >algorithms for finding just the set of edges? An edge x->y is in a circuit iff there is a path from y to x in the digraph. So my suggestion for an algorithm is this: For each vertex y find the set of vertices x such that there is a path from y to x and intersect this set with the set of vertices x such that x -> y is an edge. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > I'm looking for algorithm to determine the set of edges that form all > circuits in a directed graph. Or equivalently the set of edges that > are not present in any circuit in a digraph. I am aware of papers on > enumerating all the circuits in digraph, but is there faster > algorithms for finding just the set of edges? Look up algorithms for finding strongly connected components in any algorithms book. That should be what you want... -- Steve Tate - srt[At]cs.unt.edu | A computer lets you make more mistakes faster Dept. of Computer Sciences | than any invention in human history with the University of North Texas | possible exceptions of handguns and tequila. Denton, TX 76201 | -- Mitch Ratliffe, April 1992 ==== Many thanks for those who provided an approach or pointed out the connection with SCC. I have found the following references: R. Tarjan. Depth first search and linear graph algorithms. SIAM Journal on Computing, 1(2):146--160, 1972. M. Sharir. A strong-connectivity algorithm and its application in data flow analysis. Computers and Mathematics with Applications, 7(1):67--72, 1981. Jack Middleton ==== These are variations of a result I posted a while back, but I mention them here because the equal finite-sums are free of floor-functions. Also, there MIGHT be some interesting number-theory results which could be related to this. (see below.) Let {a(k)} = any sequence defined for all positive integer indexes k. Let m = any positive integer. Then: m!m --- 1 |(m+1)!| ----- / a(|------|) (m+1)! --- |_j+m!_| j=1 = m --- a(j) / ------ --- j(j+1) j=1 In linear-mode: (1/(m+1)!) sum{j=1 to m!m} a(floor((m+1)!/(j+m!))) = sum{j=1 to m} a(j)/(j(j+1)) And related: m j --- --- m! / / a(k) = --- --- j=1 k=1 |(m+1)!| |------| (m+1)! |_ j _| --- --- / / k a(k) --- --- j=1+m! k=1 In linear-mode: m! sum{j=1 to m} sum{k=1 to j} a(k) = sum{j=1+m! to (m+1)!} sum{k=1 to floor((m+1)!/j)} k a(k) A specific example: (m+1)! --- |(m+1)!| |------| = / |_ j _| --- j=1+m! (m+1)!H(m) - m! m, where H(m) = 1+1/2+1/3+...+1/m, the m_th harmonic number. The sum in linear-mode: sum{j=1+m! to (m+1)!} floor((m+1)!/j) Now this last floor-involving sum is congruent to: m! H(r,m) (-1)^(m+1) (mod{m+r+1}), where H(r,m) = sum{k=1 to m} H(r-1,k), and H(0,m) = 1/m. (H(1,m) = H(m).) (H(r,m) also = binomial(m+r-1,r-1)(H(m+r-1)-H(r-1)).) The floor-sum itself is m!*H(2,m). Does the congruency imply any immediately interesting number theory-results?? Leroy Quet ==== > I have some 3D data points and I would like to find the way to calculate > the smallest sphere which would include all of them inside the sphere > (not necessarily on its surface). > It is easy enough to find the smallest cube that includes all your >> points. The smallest sphere that encloses this cube also includes all >> your points. But that doesn't answer the question. The question was to find the >*smallest* sphere that contains the given points. The analogous problem in 2D (minimum enclosing circle) is a famous result of Megiddo: it can be done in linear time. I believe this generalises to a linear-time algorithm in any fixed dimension. This might be a useful reference: http://www.cgal.org/Manual/doc_html/basic_lib/Optimisation_ref/Class_Min_sph ere_d.html -- Erick ==== Does this apply to mathematical journals??? Leroy Quet ==== > Does this apply to mathematical journals??? There's a number of expensive journals (eyeballing it, the rates for publishing appear to be on the order of $0.25 to $2 per page for most journals). The problem with this complaint is that the US library and books. Ie, the need just isn't there. For example, I live next to a library with a total staff of ten or less (well at least the equivalent of ten full time people). I've acquired half a dozen papers on theoretically general relativity (I was looking up some Kaluza Klein theory) from them through interlibrary loan. The price was free experience the problems that are being claimed in the story above. in question. Perhaps, it is a local thing and doesn't hold true for most of the US. Karl Hallowell ==== > Does this apply to mathematical journals??? Leroy Quet Absolutely. http://www.ams.org/notices/200007/forum-birman.ps http://www.ams.org/notices/199804/branin.pdf http://www.firstmonday.dk/issues/issue2_8/odlyzko/ For Dale Alspach's ruminations on the future of mathematical publishing, see section 2 of http://www.math.okstate.edu/~alspach/banach/banacharchist.ps And here is one group that's actually doing something about it: http://www.maths.warwick.ac.uk/gt/gtp.html ==== Nemo escribi.97 en el mensaje > Does this apply to mathematical journals??? Leroy Quet Absolutely. http://www.ams.org/notices/200007/forum-birman.ps > http://www.ams.org/notices/199804/branin.pdf > http://www.firstmonday.dk/issues/issue2_8/odlyzko/ For Dale Alspach's ruminations on the future of mathematical publishing, > see section 2 of > http://www.math.okstate.edu/~alspach/banach/banacharchist.ps And here is one group that's actually doing something about it: > http://www.maths.warwick.ac.uk/gt/gtp.html > See also http://www.emis.de/journals/ Jos.8e H. Nieto http://mipagina.cantv.net/jhnieto1 ==== Consider the following data: For n=24; 3n-6 = 66 Minimum number of labeled 4-color graphs with 66 edges = 3.22*10^56 Maximum number of labeled planar graphs with 66 edges = 4.18*10^21 Intuitively, one might expect the discrepancy between the number of planar graphs and the total number of 4C graphs to be within a few orders of magnitude. But the data shows that the discrepancy is many, many orders. (appx. 35) As n increases, the discrepancy becomes much larger! If anyone is interested, I can provide the formaulas I used to calculate these numbers? ==== Saluton, suppose f(r) is a rotationally symmetric real valued function, and F(k) is its Fourier transform [in my application, f(r) is the perturbational part either in the sense of vanishing identically for r > R, or in the sense of sufficiently fast decay for f -> infinity, with leading terms going like, e.g., 1/r^6 or exp(-z r)/r. Suppose now that F(k) is known for all k > Q: to what extent is F(k), k < Q, determined by the above properties? In my intuition the requirement of short-rangedness should be sufficient for fixing all of F(k) as the difference f1(r)-f2(r) between two candidate solutions, F1(k) = F2(k) for k > Q, F1(k) /= F2(k) for k < Q, must be short ranged, too, so that I would not expect F1(k)-F2(k) to vanish identically for k > Q unless the f1=f2. Could anyone shed some more light on this question, the validity of the above line of thought in particular, or point out some relevant results on Fourier transformations? Albert. please use . ==== > Saluton, suppose f(r) is a rotationally symmetric real valued function, and F(k) > is > its Fourier transform [in my application, f(r) is the perturbational > part > either > in the sense of vanishing identically for r > R, or in the sense of > sufficiently > fast decay for f -> infinity, with leading terms going like, e.g., 1/r^6 > or > exp(-z r)/r. Suppose now that F(k) is known for all k > Q: to what extent is F(k), k > < Q, > determined by the above properties? In my intuition the requirement of short-rangedness should be sufficient > for fixing > all of F(k) as the difference f1(r)-f2(r) between two candidate > solutions, F1(k) = > F2(k) for k > Q, F1(k) /= F2(k) for k < Q, must be short ranged, too, so > that I > would not expect F1(k)-F2(k) to vanish identically for k > Q unless the > f1=f2. Could anyone shed some more light on this question, the validity of the > above > line of thought in particular, or point out some relevant results on > Fourier > transformations? > Albert. > Here is my two cents. Let me discuss the 1 dimensional case, but I don't think the higher dimensions will make a great difference. If f(r) decays very rapidly (like exp(-a r) as r->infinity, a>0 is fixed), then this would be enough to make its Fourier transform F(k) analytic in a strip Im(k) < a (or a/pi or whatever depending on your definition of the Fourier transform). An analytic function must have isolated zeroes, so in that case I think that you would get the uniquness you want. If the decay is like 1/r^6, I don't think my argument will work so well. Well, I know that there are better experts in this forum, so hopefully you will get better answers. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen ==== > suppose f(r) is a rotationally symmetric real valued function, and F(k) > is > its Fourier transform [in my application, f(r) is the perturbational > part > either > in the sense of vanishing identically for r > R, or in the sense of > sufficiently > fast decay for f -> infinity, with leading terms going like, e.g., 1/r^6 > or > exp(-z r)/r. Suppose now that F(k) is known for all k > Q: to what extent is F(k), k > < Q, > determined by the above properties? > In my intuition the requirement of short-rangedness should be sufficient > for fixing > all of F(k) as the difference f1(r)-f2(r) between two candidate > solutions, F1(k) = > F2(k) for k > Q, F1(k) /= F2(k) for k < Q, must be short ranged, too, so > that I > would not expect F1(k)-F2(k) to vanish identically for k > Q unless the > f1=f2. Certainly F1(k) can equal F2(k) for k > Q without F1 and F2 being identical, even for f1 and f2 vanishing rapidly at oo. Take F1 to be identically 0 and let F2 be any C^oo function with compact support, not identically 0, that is rotationally symmetric. Let f1, f2 be the corresponding inverse Fourier transforms. Then we're done: each fj is rotationally symmetric, and each is in the Schwarz space, ie, each is a C^oo function that vanishes faster at oo than any 1/r^n (as do their derivatives). ==== Verify n Integral Log^n x dx = Sum (-1)^(j+n)*n!/j!*x*Log^j x j=0 by math induction and integration by parts. You will first need to do the subproblem: Verify n!=[(n-1)!+(n-2)!]*(n-1) ==== Can anyone prove: m m! (-1)^(n+m) x^(k+1) Log^n x Integral x^kLog^m x dx = Sum -------------------------------------------------------------- n=0 n! (k+1)^(m+1-n) I have proved k=0 and verified for k=1 to 7 m=1 to 25 with mathematica. It looks like this gives us an antiderivative of x^x = sum over a (xlnx)^a/a! ==== Can anyone prove: Integral x^kLog^m x dx =sum n=0 to m m! (-1)^(n+m)x^(k+1) Log^n x ------------------------------------------------------- n! (k+1)^(m+1-n) I have proved k=0 and verified for k=1 to 7 m=1 to 25 with mathematica. It looks like this gives us an antiderivative of x^x = sum over a (xlnx)^a/a! ==== > Can anyone prove: Integral x^kLog^m x dx =sum n=0 to m m! (-1)^(n+m)x^(k+1) Log^n x > ------------------------------------------------------- > n! (k+1)^(m+1-n) I presume that Log^m x is meant to denote (log x)^m and not log log ... log x (m terms) as is standard in analytic number theory. The way to prove such things is simply to differentiate the putative integral and hope all but one term cancels. Does it? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== What I think you might mean is that you wish to reconcile your > intuition with the result you've gotten. Well, given that the Riemann > integral is the limit of approximations of rectangles under the curve, > and that as x -> oo, f' -> 0, we can come up with an approximating > rectangle of arbitrary length. But that doesn't matter since f = 0 > where this approximation is acurate. So the area under the curve in > that interval is zero. That makes no sense at all to me. :( That's what I get for walking away for a few minutes in the middle of a write up. (Lost my train of thought) Well, basically (I'm sure you know this), since f -> 0, we can approximate f at infinity, that is, far from 0 by 0. So let's form an approximating rectangle, in the spirit of the Riemann integral: At oo, the height is 0. Since the formula for a rectangle's area is a = base*height, we have that the area is zero, no matter how large the base is. So there's a point on the curve such that to the right of this point, the area is zero. Hence nothing to the right of the point contributes any area, so there's no reason to expect the area to be infinite, even if the arclength is. Better? This is painfully naive, but it's intuitive as hell. This is why I prefer just coming to grips with a result by proving it to myself. Alex Solla Junior Reed College ==== Could someone please recommend a basic but rigorous book on Euclidean geometry. I have tried searching on amazon, but too many of the titles have no reviews and are rather expensive to be buying on spec. TIA -- Edmond Walsh ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements ==== Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements Golly. Somebody did a lot of work. Dover has Eudlid's Elements at a reasonable price. Three volume paperback. Our local Barnes and Noble has a generous return policy. They even let you return books that you bought from their web site, so you avoid paying return postage. ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. I suggest Robin Hartshorne's Geometry: Euclid and Beyond. Jose Carlos Santos ==== >>Could someone please recommend a basic but rigorous book on Euclidean >>geometry. I have tried searching on amazon, but too many of the titles >>have no reviews and are rather expensive to be buying on spec. > Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements Here is something more recent --- less than 100 years old: http://historical.library.cornell.edu/Dienst/UI/1.0/Display/cul.math/0179000 1 ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. > TIA Euclid's Elements is the classic. A more modern treatment is classic by reworking the subject, beginning with the axioms. I've recently been reading both. As a teacher of beginning & intermediate algebra, I find the usual (introductory textbook) treatment of such topics as angles and plane figures seriously wanting: One book on introductory math -- Basic Mathematics for College Students by Tussy and Gustafson -- says that A rectangle is a four-sided figure (like a dollar bill) whose opposite sides are of equal length, from which one could perhaps conclude that all rectangles are green or something. Abominable. These poor students have to unlearn so much crap they've been taught. But I digress. I suggest you visit the local library and check these out (if on the shelves) or get them on Interlibrary Loan (if not). Hope this helps. ==== Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements but rigorous, but unfortunately seems to be out of print (at least in English translation). -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. > TIA Euclid's Elements is the classic. A more modern treatment is > classic by reworking the subject, beginning with the axioms. I've recently been reading both. As a teacher of beginning & > intermediate algebra, I find the usual (introductory textbook) > treatment of such topics as angles and plane figures seriously > wanting: One book on introductory math -- Basic Mathematics for > College Students by Tussy and Gustafson -- says that A rectangle is > a four-sided figure (like a dollar bill) whose opposite sides are of > equal length, from which one could perhaps conclude that all > rectangles are green or something. Abominable. These poor students > have to unlearn so much crap they've been taught. But I digress. And have you considered what happens if that dollar bill happens to be crumpled? Seriously, I agree with you that dollar bills are to be kept out of mathematical definitions. Felix. ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements but rigorous, but unfortunately seems to be out of print (at least in > English translation). Wow. I wasn't aware it was out of print. I snagged a 2001 reprint (from Amazon, I believe) a couple years back. But, indeed, the book is no longer available there. I can't get into abebooks.com right now, but alibris.com has a hardbound copy currently available. Incidentally, the translated text has LOTS of errors. I checked at amazon.de, and if my German serves me correctly, Grundlagen der Geometrie is also currently out of print. That's really too bad. :( ==== Wow. I wasn't aware it was out of print. I snagged a 2001 reprint > (from Amazon, I believe) a couple years back. But, indeed, the book is > no longer available there. I can't get into abebooks.com right now, > but alibris.com has a hardbound copy currently available. There is a copy available for $25.00 (usd) from Aracana Books, Huntsville UT, USA Bob Kolker ==== > Wow. I wasn't aware it was out of print. I snagged a 2001 reprint > (from Amazon, I believe) a couple years back. But, indeed, the book is > no longer available there. I can't get into abebooks.com right now, > but alibris.com has a hardbound copy currently available. > Incidentally, the translated text has LOTS of errors. I checked at > amazon.de, and if my German serves me correctly, Grundlagen der > Geometrie is also currently out of print. That's really too bad. :( You can get a copy from K. Rosenthal, Berlin, k.A., Germany For about$57.00 (usd) Bob Kolker ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. > TIA Coxeter, Geometry Revisited is very good and not too expensive, and elementary but not entirely basic. The target audience I would say is undergrad math majors, or capable HS students. $21 at Amazon. Moise, Elementary Geometry from an Advanced Standpoint covers basic material in considerable depth; it's excellent IMO -- Moise is a skilled expositor, and the exercises are on point and frequently challenging. The nominal target audience is potential HS math teachers; I wish I believed even 10% of current HS math teachers had worked through it. It /is/ rather expensive though -- $95 (new) at Amazon. ==== the online version of Euclid's doesn't have Book 14, which is actually by Hypsicles, and full of stuff about the basic (spatial shapes) -- it's in the Dover books, though. I recommend trying to go at it from this end. or, try finding Altshiller-Court's _Modern Pure Solid Geometry_, which is the only thing that is like it. > Euclid's Elements http://aleph0.clarku.edu/~djoyce/java/elements > http://historical.library.cornell.edu/Dienst/UI/1.0/Display/cul.math/01790001 --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 ==== > Could someone please recommend a basic but rigorous book on Euclidean > geometry. I have tried searching on amazon, but too many of the titles > have no reviews and are rather expensive to be buying on spec. > TIA > -- > Edmond Walsh EW ==== Coxeter's book is co-authored with Greitzer. he has a few, other elementary texts, two. Bucky Fuller's dedication to him in _Synergetics_ is the only clue, that you have to actually know this stuff, firstly. I mean, it helps, even though there's nothing beyond Pyhtagoras' theorem in it! > Coxeter, Geometry Revisited is very good and not too expensive, and > elementary but not entirely basic. The target audience I would > say is undergrad math majors, or capable HS students. $21 at Amazon. Moise, Elementary Geometry from an Advanced Standpoint covers basic --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81