mm-15 This argument looks correct, but wonfit 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 canfit 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: Gfi = int_0^B dp(x) / (p_t(x) + u(x)) dx>> ...>> John Mitchell Oops. My mistake is ... My original argument went like this: Gfi(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 Gfi(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 Ifive chosen to do this by deviding thesurrounding cubus (borderlenght=2) in N*N*N little cubi(borderlenght=2/N). Only those little cubi that intersect the surfaceof the unitsphere are of interest.My question now:How many little cubi will intersect the surface of the unitsphere independency of N ?Somehow my brain seems (temporarily) too small for sufficientthoughts. Thus Ifive investigated it experimentally. The result seemsto be:((N/2)*(N/2)+1)*8 little cubi intersect the surface of theunitsphere.If correct, why ?? Can anyone I urgently need an answer to following:>> My task is to discretize (to sample) the surface of the unit sphere.> In terms of simplicity Ifive 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 Ifive investigated it experimentally. The result seems> to be:>> ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the> unitsphere.>> If TobiasFor N=2, your formula says that 16 (of 8) little cubi intersect the sphere.This is surely wrong.Seamanfis reply in the thread starting athttp://tinylink.com/?qAd8iCXKHK-- Clive Toothhttp://www.clivetooth.dk = For N=2, your formula says that 16 (of 8) little cubi intersect the sphere.> This is surely wrong. Seamanfis reply in the thread starting at> http://tinylink.com/?qAd8iCXKHKof course Youfire 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 Ifive 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 Ifive investigated it experimentally. The result seems> to be: ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the> (mad) TobiasThe first few values aren n_cut1 12 83 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 Sequencesgiveshttp://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 Pfoertnerhttp://www.pfoertner.org/ > For N=2, your formula says that 16 (of 8) little cubi intersect thesphere.> This is surely wrong.>Dave> Seamanfis reply in the thread starting at> http://tinylink.com/?qAd8iCXKHK>> of course Youfire right. I forgot the constraint N>3.> TobiasWell...I take each cubus [there is really no such word in English, but the termseems 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 tothe conclusion that 56 (=4^3-2^3) intersect the sphere.For N=5, I find that 98 cubi intersect the sphere. Your formula gives anon-integer answer.For N=6, we must be very careful how we count the cubi. Several of themintersect our sphere in just a single point. I find that 176 of the cubiintersect the sphere (48 of them at only a single point). So if we took thecubi to be open cubes that number would go down to 128. Anyway, your formulagives 80.Of course, my numbers could be wrong.-- Clive Toothhttp://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 Ifive 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 Ifive 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 ?>> 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 (176and 128) that I gave in another post in this thread and corresponds toviewing the cubi as neither open nor closed. I have no idea how theexpression 6n^2+2 arises, though.-- Clive Toothhttp://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 intervalthat is a Cartesian product of half-open intervals, so that they exactlypartition 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 thereare 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 416I also thought of looking up the sequence, but my sequence gave no matches.-- Dave SeamanJudge Yohnfis 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 Ifive 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 Ifive investigated it experimentally. The result seems> to be:>> ((N/2)*(N/2)+1)*8 little cubi intersect the surface of the> unitsphere.>> If 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.dk6*n^2+2 is exactly what the sequence title says:for clarifying this.A 1*1*1 cube has 6*1+2 (lattice) points2*2*2 -> 26 points3*3*3 -> 56 pointsn*n*n -> 8 + 12*(n-1) + 6*(n-1)^2 = 6*n^2 + 2 vertices points on edges pts on facesNumber 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 Seamanfis 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 SeamanI have now written a little Fortran program available athttp://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 1856I have counted cubes only touched by the cutting sphere, with allother vertices either inside or outside as 1/2, because they occuralways as inside/outside pairs. If either > or >= is usedin the radius comparisons then I can reproduce both variants ofDavefis n=6 and n=10 results.Hugo Pfoertnerhttp://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> Davefis n=6 and n=10 results.I have downloaded your Fortran program and checked it against my lispprogram (modified to print the average of the upper and lower results).They agree.-- Dave SeamanJudge Yohnfis mistakes revealed in Mumia Abu-Jamal ruling. =does anyone know the official abbreviation for the cycle decompositionnumber used in graph theory. I thought about something like cd-number butcouldnfit find =posted. Here is my second try.Bill James developed the Pythagorean win formula to estimate thenumber of wins a baseball team should have in one season given theformula,estimated win percent = RS^1.83/(RA^1.83 + RS^1.83)was empirically derived(http://www.baseball.reference.com/about/faq.shtml). This notedescribes a derivation for the formula and provides a formula for agood exponent in the Pythagorean win formula.DERIVATION OF WIN PERCENTOne way to derive a win formula from runs is to assume a distributionfor the runs scored by each team, assume the team run distributionsare independent so that you can multiply them to get a jointdistribution of runs, and then sum the probabilities of winningsituations. Jim Ferry used this procedure to get a win formula byassuming 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 runsdistribution in baseball, but he is able to work out a formula for winpercentage based on this distribution.There are several other common distributions used in statistics: thebinomial, normal, log normal, and Raleigh could all be used to modelruns scored and thus each distribution could produce a estimated winformula. One of these distributions, the lognormal, produces a winformula which almost exactly matches Jamesfis Pythagorean formula.(The lognormal win formula differs from the Pythagorean by less than0.0003 over the typical range of RS and RA).If you assume that the runs for each game are produced by independentcontinuous distributions with density functions Dx[x] and Dy[y] foreach team, then the win percent is simplywinper = 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 iswinper = 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 Omuxfi is the geometric average of the first teamfis runs andfimuyfi is the geometric average of the other teamsfi runs. Changingvariable to a = Log[x] and b=Log[y] giveswinper = 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 giveswinper = Integrate[ 1/(2 sigma Sqrt[Pi]) * Exp[-(delta-Log[mux/muy])^2 / (4 sigma^2)], {delta, 0, Infinity}]which simplifies towinper = 1/2 * (1 + Erf[ Log[mux/muy]/ (2 sigma)] )which we will call AND THE LOGNORMAL FORMULAIf we approximate mux/muy by RS/RA and call that r, then the LogNormalFormula becomeswinperlognorm = 1/2 * (1 + Erf[ Log[r]/ (2 sigma)] ).Rewriting the Pythagorean formula in terms of r and letting c be theJames Pythagorean exponent giveswinperpythag = 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 8Notice that the first three terms of the Taylor Series will be equalif 2James Pythagorean Exponent = c = -------------- Sqrt[Pi] sigmawhere sigma is the standard deviation of the Log of the runsdistribution. sigma can be approximated bysigma ~= StandardDeviation(runs scored)/average(runs scored).For baseball, sigma = 0.617 and c = 2/Sqrt[Pi]/sigma ~= 1.83 givealmost identical results. (They differ by less than 0.0003 betweenr=.8 and r=1.2). The formulas also match exactly at r = 0, r = 1, andr = Infinity.The Pythagorean Exponent formula also explains why a higher value forc is required to estimate win percent from basketball scores. Inbasketball the standard deviation of points scored divided by theaverage points scored is closer to sigma = 0.1 which give a muchhigher value for c.SUMMARYJames pythagorean formulawinper = RS^c/(RS^c + RA^c)is almost exactly the same as the win percent formula derived from theassumption of independent lognormal run distributions. These formulasare related by the exponent formula 2James Pythagorean Exponent = c = -------------- Sqrt[Pi] sigmawhere sigma is the standard deviation the log of runs which isapproximately equal to the standard deviation of runs scored dividedby 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 Oprooffi-proof? It> doesnfit seem like there is one (that can defined formally at least). I hope> someone knows what Ifim talking about. l8r, Mike N. ChristoffI find that most of the times Ifim asked to derive something, it turnsout to be a formula or identity, likeDerive the equation Sum (n = 1 to oo) 1/(n^2) = (pi^2)/6 The term Oprooffi seems to be more general in meaning. This is in mylimited experience, however.Alex SollaJuniorReed 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 Oprooffi-proof? It| doesnfit seem like there is one (that can defined formally at least). I hope| someone knows what Ifim talking about.I read a Poincare quote today that doesnfit make the line any sharper, butwhich you might find interesting anyway. He says that derivations areroutine 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 Oprooffi-proof? It>| doesnfit seem like there is one (that can defined formally at least). I hope>| someone knows what Ifim talking about.>>I read a Poincare quote today that doesnfit 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 thatrequired explanation in English.Lee Rudolph =Is it always possible to do differential equation modeling for each anevery 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, Icould be wrong, is that a differential equation can model *MOST* phenomenahowever an intrinsic limit to the number of systems that can be modeled doesnot exist. This poses an interesting question in itself, if therefis a limitto the number of systems in which a differential equation is known valid,then mustnfit the order of the equation be so limited? But there are alwaysdiscrete models that forbid infitessimal values.Patrick Meuser =I want to learn the formula to calculate any digit in the decimal form I want to learn the formula to calculate any digit in the decimal form ofpi.> ex: What is the 100th digit in Pi?> ans: 0To find the nfith digit in the decimal expansion of pi, I generally use theformula: ?or(pi*10^n) mod 10Where ?or(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 itappears to work.By the way, my formula says that the 100th digit of pi is 9. This agreeswith the 100th digit of pi computed by Shanks and Wrench in 1961. I have notchecked it against any more recent computation.I think there may be a similar formula for the nfith digit of sqrt(2).-- Clive Toothhttp://www.clivetooth.dk I want to learn the formula to calculate any digit in the decimal form ofpi.> ex: What is the 100th digit in Pi?> ans: 0There is a way to do this for the binary/octal/hexadecimal/etc expansions ofPi, 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 formof> pi.> ex: What is the 100th digit in Pi?> ans: 0>> There is a way to do this for the binary/octal/hexadecimal/etc expansionsof> 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 nfith digit in the decimal expansion of pi, I generally use the> formula: ?or(pi*10^n) mod 10 Where ?or(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 multiprecisionarithmetic and the formula 16 * atn(1/5) - 4 * atn(1/239)) suggeststhat digits 96 - 100 of pi are O70679fi. I think there may be a similar formula for the nfith digit of sqrt(2).> There is. :-) Your formula is extremely general; the mainproblem of course is computing ?or(X * 10^n), where X isthe computational victim: pi, e, sqrt(2), log_e of 2, etc.-- #191, ewill3@earthlink.netItfis 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 abiography Ifive 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; itfis a penance. - Richard Stallman =on>> There is a way to do this for the binary/octal/hexadecimal/etcexpansions of> Pi, but to my knowledge, no method exists (or at least, none is yetknown)> for the decimal expansion.>> I believe Simon Plouffe has discovered a method around 1996 (according toa> biography Ifive 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 Toothhttp://www.clivetooth.dk =? I want to learn the formula to calculate any digit in the decimal form of pi.? ex: athttp://mathworld.wolfram.com/ Bailey-Borwein-PlouffeAlgorithm.htmland the links given there. Donfit forget to read ?The story behind a formula of Pi?http://mathforum.org/discuss/sci.math/t/517721HP is it possible for a method that is not know to exist?Certainly. In mathematics in general, we are always looking for thingsthat are not yet known, but that we believe exist. This is to someextent a question of what you mean by exist. Working mathematicianstend to be Platonists, who consider the objects of mathematics to havean existence independent of what may be known about them at a particulartime.Robert Israel israel@math.ubc.caDepartment 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 divides4, 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 dependson Z being a UFD.What is the situation in algebraic integers?Please correct me if Ifim wrong: the ring of algebraic has noirreducibles, hence it cannot be a UFD.Some were asking for a new DISH challenge following Uncle Alfis ...How does James Harris know that the algebraic integers form aring 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 theoremof 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 isprincipal. As a consequence of this, any two elements have a gcd(which is unique up to units). As a consequence of this, the ring ofall algebraic integers is a pre-Schreier Domain, which is exactly theproperty you have isolated:If a,b,c are algebraic integers, and a divides b*c, then a factors asa=B*C, with both B and C algebraic integers such that B divides b andC divides c.>Please correct me if Ifim 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 argumentsseem to imply he believes they are, in fact, not a ring at all. Ofcourse, he doesnfit know what a ring is either, so... selective about what I accept as reality. --- Calvin (Calvin and Hobbes) 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 Ifim 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 Magidinfis 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 Ifim 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. Magidinfis reply ]Challenge 1 for James H.: prove that the algebraic integers form a ring. Does he take mathematciansfi 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 Theoremof Algebra.David Bernier =field is a finite algebraic extension of Q and the ring of integers of anumber field is the intersection of the number field and the algebraicintegers.Furthermore Ifill 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 bythe irreduciable polynomials p1..pk for which pj(uj) = 0, j = 1,..k >I copy here Arturo Magidinfis reply earlier in this thread:All interesting stuff for which Ifim unprepared.>>Please correct me if Ifim 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) = 0base is u, u^2,.. u^n where n is degree of u or p.2 = 3, obvious. 3 = 1 the significate partLet u1,..u_n be the finite base for the extensionand u = sum kj uj for some kj in K, j = 1,..nThus 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 unknownsSo whatfis left is to show the matrix is invertible.I think Ifim on the right track but sigh, now I need to brush up on linear algebra.Hm, with that now, additive and multiplicative closure of algebraicnumbers isnfit 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 Ifive seen uses Liouvillefis theorem which is a bit much.Ifive been assured Gaussfi proofs are no easier than Liouvillefis theorem.---- Furthermore Ifill 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 partCan 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, Ifim feeling stupid.It wasnfit 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 Ifid 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), theyfire algebraic over K and closure offield operations follow. More elementarilyIf 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 /= 0last 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, giventhis error in the proposition (without the forgotten hypothesis), myargument that the series converge is wrong. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com =Ifim trying to plot a histogram on some data that I receive at runtime. Ifim 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)? Ifim 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 = Ifim trying to plot a histogram on some data that I receive at runtime. > Ifim 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)? Ifim 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 canfit 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 getapproximations 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 the code excerpt :)--Andre >> Ifim trying to plot a histogram on some data that I receive at runtime. >> Ifim 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)? Ifim 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 canfit 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;> } ----------------------------------------------------------- > =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 sertainlibraries.Anyway, to the math stuff. I have a linear number from 0.0 to 1.0. Andfrom that number I need to produce a steep curve, like a sinus curve from 0to 90 degrees ( ? ).So if the linear number is 0.0, then the output number is 0.0. And if it is1.0, then the output is 1.0. Itfis the numbers in-between that is myproblem. For example, the input number 0.2 might produce 0.5. So the curveis steep in the beginning and ?ttens out at the top. Again like a sinuscurve.I know I have done this several times in the past. But I canfit figure itout now. I tried with a normal sinus equation: sin( 90 * X ) sin( 90 ) * X...etc. Isnfit 90 degree the top of a sinus curve? So shouldnfit sin( 90 )give 1.0?Anyway, while I ponder this by myself, I thought I would ask here to see ifsomeone 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 from0> to 90 degrees ( ? ).>> So if the linear number is 0.0, then the output number is 0.0. And if itis> 1.0, then the output is 1.0. Itfis the numbers in-between that is my> problem. For example, the input number 0.2 might produce 0.5. So thecurve> is steep in the beginning and ?ttens out at the top. Again like a sinus> curve.>> I know I have done this several times in the past. But I canfit figure it> out now. I tried with a normal sinus equation:>> sin( 90 * X )> sin( 90 ) * X>> ...etc. Isnfit 90 degree the top of a sinus curve? So shouldnfit sin( 90 )> give 1.0?>> Anyway, while I ponder this by myself, I thought I would ask here to seeif> someone could help me...>Donfit 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 BronsingCanfit 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. Itfis 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 ?ttens out at the top. Again>> like a sinus curve.>> I know I have done this several times in the past. But I canfit figure>> it out now. I tried with a normal sinus equation:>> sin( 90 * X )>> sin( 90 ) * X>> ...etc. Isnfit 90 degree the top of a sinus curve? So shouldnfit 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...>> Donfit 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 beenumerated (i.e. mapped to integers). A method enumerating all integersequences 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 integersequences 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^3we requireE(x*y^2)+E(x^3*y)=E(x^4*y^3) If not, could the requirement be relaxed to some associative functionagg, other than addition, that meetsagg(E(x*y^2),E(x^3*y))=E(x^4*y^3) ?Is there an enumeration (relaxed, or strict power additive) thatpreserves 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 factthere are uncountably many sequences of 0 and 1. Or did you meansomething different?Robert Israel israel@math.ubc.caDepartment 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 samenumber.[The only way those two polynomials could not be distinct would be ifE(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 Toothhttp://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 thesame> 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-variablemonomials 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 whenthey ROAR!!!A 1-foot tall Green Evil Fuzzle can register 10 NSU (Neopian SoundUnits) on a sound-measuring device that is 5 feet away. Yellow EvilFuzzles are twice as loud, and Red Fuzzles... THREE times as loud astheir yellow counterparts! So anyway, one day Zorlon-IX the Grundo was out doing a bit ofmaintenance on the exterior of the Space Station, and before him hesaw, to his fright, the biggest Evil Fuzzle he had ever seen. It was 5feet high, 10 feet away and red! It looked straight at Zorlon, andgave the loudest ROAR it could!!!How many NSU did Zorlonfis 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 Zorlonfis sound-measuring device register?The information supplied is inadequate. What is the correlation ofthe height of an Evil Fuzzie and the amount of NSU it makes whenroaring?-- /-- 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 stochastic process (or random > process)> that is ergodic but not stationary?No. As I understand it, a non-stationary stochastic process must benonergodic.On the other hand, a Spherically Invariant Random Process (SIRP) isan 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.htmlr 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 Ifive found an example (albeittrivial) 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 ofthese RVs: {... X1 Y1 X2 Y2 ...}. Note the process is clearly notstationary since, for example, the mean changes from time sample to timesample. However, the process is ergodic (in the mean, for example) sinceeach sample path of the process (there is only one!) time-averages to thesame value.There seem to be two competing definitions of ergodicity that Ifive 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 thisvalue 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 thedefinition of Birkhofffis Ergodic Theorem (see Tom Coverfis text , Elementsof Information Theory, p. 474, for example), which holds for stationary,ergodic sources. The formal definition given by Cover for ergodicity isthis: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 isa probability measure. A random variable X is defined as a functionX(omega), omega is-an-element-of Omega, on the probability space. We alsohave a transformation T:Omega-->Omega, which plays the role of a timeshift... The transformation is called ergodic if every set A (element-ofscript_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 itjives with the first definition I supplied above.I hope this hasnfit rjc.--Ryan CassidyElectrical Engineering Graduate 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 orwealth> e a e of a man is a question fit perhaps to bediscussed by> n e . slaves in the hearing of their masters, buthighly> @ r c m unbecoming to reasonable and free men in searchof> d o the truth. -- Rousseau =If dy and dx are the errors in y and x respectively, givenz = sqrt[y^(2) - x^(2)]what is the error dz associated with z?z^(2) = y^(2) - x^(2)2dz/z = 2dy/y - 2dx/xdz = 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/xYou seem to be saying that the differential of z^2 is 2 dz/z. Thatis incorrect. It is 2 z dz.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr 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/xI have no idea where that formula comes from.If the errors are small enough, and if one of x and yis substantially larger than the other you could usecalculus to estimate the error in z. Or you could justdo it:Letfis say that x is the _true_ value and xfi = x + dx isthe approximate value; similarly for y and z. Then(z + dz)^2 = (y + dy)^2 - (x + dx)^2andz^2 = y^2 - x^2, which gives2z dz + dz^2 = 2y dy + dy^2 - (2x dx + dx^2).You could solve that for dz to get an exact and essentiallyuseless formula for the error. Typically one would insteadassume that the errors are small enough that thesquares of the errors are negligible; this gives dz ~ (2y dy - 2x dx)/z.But therefis a subtle point here because of the subtraction;in order for the square of the errors to be negligible itfis notenough to have dx much smaller than x and dy muchsmaller than y, you need that the squares of the errorsare much smaller than the _difference_ y dy - x dx.This will happen if y is much larger than x and dx anddy are about the same size; if x and y are about thesame size then analyzing the error is trickier.>dz = z(dy/y - dx/x)>>I have a feeling this is way out.Thatfis correct, as far as I can see (where did thatformula 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:>> Letfis say that x is the _true_ value and xfi = 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 therefis a subtle point here because of the subtraction;> in order for the square of the errors to be negligible itfis 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.>> Thatfis 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. htmlAccording to this site and a number of others I dug up on error of z is the relative error x multiplied bythe 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/xdz = 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:>> Letfis say that x is the _true_ value and xfi = 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 therefis a subtle point here because of the subtraction;>> in order for the square of the errors to be negligible itfis 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.>> Thatfis 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 relative error of z is the relative error x multiplied by>the power n. Assuming that the relative error is small then yes this isapproximately correct.>Thus if,>>z^(2) = y^(2) - x^(2)>>then,>>2 dz/z = 2dy/y + 2dx/xHuh? 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 dz/z = 2dy/y + 2dx/x>> Huh? How does this follow, exactly?>> sum of the relative errors...)Thatfis what I thought. Any ideas on what does follow? Huh? How does this follow, exactly?>> sum of the relative errors...)>>Thatfis what I thought. Huh? If thatfis what you thought then why did you say2 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 couldjust 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 xand y, then the error in f is:df = f_x * dx + f_y * dyassuming that dx and dy are small enough so that the partialderivatives 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)dx2) If the errors in x and y are RANDOM, where dx and dy are theconfidence limits, then the propagated error in f is:(df)^2 = (f_x)^2 * (dx)^2 + (f_y)^2 * (dy)^2For your f, (dz)^2 = (y/z)^2 * (dy)^2 + (x/z)^2 * (dx)^2Various assumptions are made when deriving this formula. Consult thereference 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:>> Letfis say that x is the _true_ value and xfi = 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 therefis a subtle point here because of the subtraction;> in order for the square of the errors to be negligible itfis 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.>> Thatfis 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 x^(n) then, dz/z = n dx/xAccording to that site, the general prescription for estimatingthe uncertainty in z, given measurement error in x and y, isthis:z = f(x,y)dz = (@f/@x)dx + (@f/@y)dythen work it out ignoring signs of each term. What youfirereally trying to do is figure out how far z could vary overthe entire possible range of x and y values within x+-dx andy+-dy. Thatfis why you ignore signs: youfire estimating theworst case.For your functionz = 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/zSo you have, ignoring the signs,dz = (x/z)dx + (y/z)dyThis doesnfit work out into a nice relative errorformula. It does give you z dz = x dx + y dywhich differs from Davidfis result by a factor of 2. Ifimnot sure why.Note that this is not really very rigorous mathematics. - Randy Huh? How does this follow, exactly?>> sum of the relative errors...)>>Thatfis what I thought.>> Huh? If thatfis 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.Ifim 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 findthe change in h when x + dx is substituted in place of xz + dz_x = sqrt[(x + dx)^(2) + y^(2)]and likewise, the change in z due to y can be found fromz + dz_y = sqrt[(x)^(2) + (y + dy)^(2)]Thusdz = dz_x + dz_yInterestingly, this method produces an error almost exactly the same asRandy Poefis equation dz = (x/z)dx + (y/z)dy, but differs from yours by afactor of two.> ************************>> David C. Ullrich 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 youfire> 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. Thatfis why you ignore signs: youfire 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/zInteresting. Just out of curiousity, why is it thatdz = (@f/@x)dx + (@f/@y)dyFurthermore, 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 Ifim still in high school and wehave yet to study partial derivatives.> So you have, ignoring the signs,>> dz = (x/z)dx + (y/z)dy>> This doesnfit work out into a nice relative error> formula. It does give you>> z dz = x dx + y dy>> which differs from Davidfis result by a factor of 2. Ifim> not sure why.>> Note that this is not really very rigorous mathematics.>> - Randy 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 youfire> 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. Thatfis why you ignore signs: youfire 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)dyThatfis just the expression for the total differential dzin terms of the partials with respect to x and y. The @sign is standing in for the curly d used for partialderivatives.> 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, youfid probably write it as a fractionwith sqrt(y^2-x^2) in the denominator. Thatfis what raisingto 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 sameas z.> Sorry if these seem like dumb questions, but Ifim still in high school and we> have yet to study partial derivatives.Ah. I misunderstood your math level. As you can see, thetheory of error propagation is based on partial derivatives.Itfis really not that much harder than ordinary derivatives,you just have to learn how to treat one variable as aconstant temporarily. - Randy does this follow, exactly?>> sum of the relative errors...)>>Thatfis what I thought.>> Huh? If thatfis what you thought then why did you say>> 2 dz/z = 2dy/y + 2dx/x in the first place?>Yes you did. That doesnfit answer the question of why you alsothought 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.>>Ifim 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 Poefis 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 wasjust a typo - I divided one side of an equation by 2 without dividingthe other side.Giving an estimate exactly the same as what you say he did.That _is_ interesting. (Make certain to note the conditionsunder which that estimate is valid.)>> ************************>> David C. Ullrich>************************David C. Ullrich > [...]>>This doesnfit work out into a nice relative error>formula. It does give you>> z dz = x dx + y dy>>which differs from Davidfis result by a factor of 2. Ifim>not sure why.If you look at how I derived that equation you see I justdropped 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 isas follows:X Y0.62996825316836 1.2670.62233431530006 1.2550.60893349390553 1.23150.58855755878249 1.190.55879334283794 1.125X error Y error8.1893e-4 1.5342e-31.1619e-3 1.0801e-31.5507e-3 1.5000e-32.0207e-3 9.6225e-42.6421e-3 0.0104How could one reliably compute the uncertainty in the slope m = 2.00704given 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 -mfi(-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 ? =Herefis an interesting little puzzle:Prove that if x^2 + y^2 = z^2 for integers x, y, zwith (x, y) = 1 then every factor of x^2 + xy + y^2is of the form 4.Z + 1.(and no, I havenfit forgotten about negative solutionseither ;)This isnfit trivially true even without the conditionx^2 + y^2 = z^2, since for example 2^2 + 2.1 + 1^2 = 7.Also of course Ifim not claiming the converse holds.Ifill post my solution in a few days in the unlikelyevent 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, ofopposite 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) is1, so x^2 + xy + y^2 is a sum of two relatively prime squares. As such, everyprime 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 ifneeded).>> Then x^2 + xy + y^2 = (m(m+n))^2 + (n(m-n))^2. The gcd of m(m+n) andn(m-n)> is 1, so x^2 + xy + y^2 is a sum of two relatively prime squares. Assuch, every> prime factor is of the form 4Z+1.Congrats. My method was slightly different, as I plugged the m, n relationsinto (x^2 + y^2)^2 - (x.y)^2 to get (m^4 - n^4)^2 + 16.m^4.n^4, in whichm^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?SPOILERp = 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^2Exactly 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? Itfis 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 donfit want a small digital camera to post photos of a large, lowweight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI =Ifim looking for algorithm to determine the set of edges that form allcircuits in a directed graph. Or equivalently the set of edges thatare not present in any circuit in a digraph. I am aware of papers onenumerating all the circuits in digraph, but is there fasteralgorithms for finding just the set of edges?Jack Middleton Ifim 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.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 Ifim 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 anyalgorithms book. That should be what you want...-- Steve Tate - srt[At]cs.unt.edu | A computer lets you make more mistakes fasterDept. of Computer Sciences | than any invention in human history with theUniversity of North Texas | possible exceptions of handguns and tequila.Denton, TX 76201 | -- Mitch Ratliffe, April 1992 or pointed out theconnection with SCC.I have found the following references:R. Tarjan. Depth first search and linear 1972.M. Sharir. A strong-connectivity algorithm and its application in data?w 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 mentionthem here because the equal finite-sums are free of ?or-functions.Also, there MIGHT be some interesting number-theory results whichcould 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=1In linear-mode:(1/(m+1)!) sum{j=1 to m!m} a(?or((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=1In 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 ?or((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)!} ?or((m+1)!/j) Now this last ?or-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 ?or-sum itself is m!*H(2,m).Does the congruency imply any immediately interesting numbertheory-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 doesnfit 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 resultof Megiddo: it can be done in linear time. I believe this generalises toa linear-time algorithm in any fixed dimension. This might be a usefulreference:http://www.cgal.org/Manual/doc_html/basic_lib/ Optimisation_ref/Class_Min_sphere_d.html -- Erick = Does this apply to mathematical journals???Leroy Quet Does this apply to mathematical journals???Therefis a number of expensive journals (eyeballing it, the rates forpublishing appear to be on the order of $0.25 to $2 per page for mostjournals). The problem with this complaint is that the US libraryand books. Ie, the need just isnfit there. For example, I live next toa library with a total staff of ten or less (well at least theequivalent of ten full time people). Ifive acquired half a dozen paperson theoretically general relativity (I was looking up some KaluzaKlein theory) from them through interlibrary loan. The price was freeexperience the problems that are being claimed in the story above.in question. Perhaps, it is a local thing and doesnfit hold true formost of the US.Karl Hallowell Does this apply to mathematical journals??? Leroy QuetAbsolutely.http://www.ams.org/notices/200007/ forum-birman.pshttp://www.ams.org/notices/199804/ branin.pdfhttp://www.firstmonday.dk/issues/issue2_8/odlyzko/For Dale Alspachfis ruminations on the future of mathematical publishing, see section 2 ofhttp://www.math.okstate.edu/~alspach/banach/ banacharchist.psAnd here is one group thatfis actually doing something about it:http://www.maths.warwick.ac.uk/gt/gtp.html =Nemo 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 Alspachfis 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 thatfis actually doing something about it:> http://www.maths.warwick.ac.uk/gt/gtp.html>See alsohttp://www.emis.de/journals/Jos.8e H. Nietohttp://mipagina.cantv.net/jhnieto1 =Consider the following data:For n=24; 3n-6 = 66Minimum number of labeled 4-color graphs with 66 edges = 3.22*10^56Maximum number of labeled planar graphs with 66 edges = 4.18*10^21Intuitively, one might expect the discrepancy between the number ofplanar graphs and the total number of 4C graphs to be within a feworders of magnitude. But the data shows that the discrepancy is many,many orders. (appx. 35) As n increases, the discrepancy becomes muchlarger!If anyone is interested, I can provide the formaulas I used tocalculate these numbers? =Saluton,suppose f(r) is a rotationally symmetric real valued function, and F(k)isits Fourier transform [in my application, f(r) is the perturbationalparteitherin the sense of vanishing identically for r > R, or in the sense ofsufficientlyfast decay for f -> infinity, with leading terms going like, e.g., 1/r^6orexp(-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 sufficientfor fixingall of F(k) as the difference f1(r)-f2(r) between two candidatesolutions, F1(k) =F2(k) for k > Q, F1(k) /= F2(k) for k < Q, must be short ranged, too, sothat Iwould not expect F1(k)-F2(k) to vanish identically for k > Q unless thef1=f2.Could anyone shed some more light on this question, the validity of theaboveline of thought in particular, or point out some relevant results onFouriertransformations?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 donfit 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 donfit 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-Smithstephen@math.missouri.eduhttp:// 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 wefire 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 nIntegral Log^n x dx = Sum (-1)^(j+n)*n!/j!*x*Log^j x j=0by math induction and integration by parts. You will first need to do thesubproblem:Verify n!=[(n-1)!+(n-2)!]*(n-1) =Can anyone prove: m m! (-1)^(n+m)x^(k+1) Log^n xIntegral 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. Itlooks 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 (k+1)^(m+1-n)I have proved k=0 and verified for k=1 to 7 m=1 to 25 with mathematica. Itlooks 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 notlog log ... log x (m terms) as is standard in analytic numbertheory.The way to prove such things is simply to differentiatethe putative integral and hope all but one 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 youfive gotten. Well, given that the Riemann> integral is the limit of approximations of rectangles under the curve,> and that as x -> oo, ffi -> 0, we can come up with an approximating> rectangle of arbitrary length. But that doesnfit 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.:(Thatfis what I get for walking away for a few minutes in the middle ofa write up. (Lost my train of thought)Well, basically (Ifim sure you know this), since f -> 0, we canapproximate f at infinity, that is, far from 0 by 0. So letfis forman approximating rectangle, in the spirit of the Riemann integral: Atoo, the height is 0. Since the formula for a rectanglefis area is a =base*height, we have that the area is zero, no matter how large thebase is. So therefis a point on the curve such that to the right ofthis point, the area is zero. Hence nothing to the right of the pointcontributes any area, so therefis no reason to expect the area to beinfinite, even if the arclength is.Better? This is painfully naive, but itfis intuitive as hell. This iswhy I prefer just coming to grips with a result by proving it tomyself.Alex SollaJuniorReed College =Could someone please recommend a basic but rigorous book on Euclideangeometry. I have tried searching on amazon, but too many of the titleshave 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.Euclidfis 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.>> Euclidfis Elements http://aleph0.clarku.edu/~djoyce/java/elementsGolly. Somebody did a lot of work.Dover has Eudlidfis Elements at a reasonable price. Three volume paperback.Our local Barnes and Noble has a generous return policy. They even let youreturn books that you bought from their web site, so you avoid paying returnpostage. 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 Hartshornefis 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. > Euclidfis Elements http://aleph0.clarku.edu/~djoyce/java/elementsHere is something more recent --- less than 100 years old:http://historical.library.cornell.edu/Dienst/UI/1.0/ Display/cul.math/01790001 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.> TIAEuclidfis Elements is the classic. A more modern treatment isclassic by reworking the subject, beginning with the axioms.Ifive 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 seriouslywanting: One book on introductory math -- Basic Mathematics forCollege Students by Tussy and Gustafson -- says that A rectangle isa four-sided figure (like a dollar bill) whose opposite sides are ofequal length, from which one could perhaps conclude that allrectangles are green or something. Abominable. These poor studentshave to unlearn so much crap theyfive been taught. But I digress.I suggest you visit the local library and check these out (if on theshelves) 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. Euclidfis Elements http://aleph0.clarku.edu/~djoyce/java/elementsbut 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 Euclidfis Elements is the classic. A more modern treatment is> classic by reworking the subject, beginning with the axioms. Ifive 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 theyfive been taught. But I digress.And have you considered what happens if that dollar bill happens to becrumpled?Seriously, I agree with you that dollar bills are to be kept out ofmathematical 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. Euclidfis 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 wasnfit aware it was out of print. I snagged a 2001 reprint(from Amazon, I believe) a couple years back. But, indeed, the book isno longer available there. I canfit 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 atamazon.de, and if my German serves me correctly, Grundlagen derGeometrie is also currently out of print. Thatfis really too bad. :( = Wow. I wasnfit 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 canfit 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, USABob Kolker = > Wow. I wasnfit 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 canfit 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. Thatfis really too bad. :(You can get a copy from K. Rosenthal, Berlin, k.A., GermanyFor 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.> TIACoxeter, Geometry Revisited is very good and not too expensive, and elementary but not entirely basic. The target audience I wouldsay is undergrad math majors, or capable HS students. $21 at Amazon.Moise, Elementary Geometry from an Advanced Standpoint covers basicmaterial in considerable depth; itfis excellent IMO -- Moise is a skilledexpositor, and the exercises are on point and frequently challenging.The nominal target audience is potential HS math teachers; I wish Ibelieved even 10% of current HS math teachers had worked through it.It /is/ rather expensive though -- $95 (new) at Amazon. =the online version of Euclidfis doesnfit have Book 14,which is actually by Hypsicles, and full of stuffabout the basic (spatial shapes) -- itfis in the Dover books,though. I recommend trying to go at itfrom this end. or,try finding Altshiller-Courtfis _Modern Pure Solid Geometry_,which is the only thing that is like it. > Euclidfis 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 dfiExploitation 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 -- LfiATTEMPTER de COUP DfiETAT, 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 WalshEW =Coxeterfis book is co-authored with Greitzer. he has a few, other elementary texts, two.Bucky Fullerfis dedication to him in _Synergetics_ is the only clue,that you have to actually know this stuff, firstly. I mean,it helps, even though therefis nothing beyond Pyhtagorasfi theoremin 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 dfiExploitation 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 -- LfiATTEMPTER de COUP DfiETAT, 3/30/81