mm-73 === > While sur?g the Web, I stumbled upon the> following site:> http://www.mathpages.com/Oooh, thanks for that!> The *.com made me wonder who might> be af?iated to this site.Unless it is hidden in the HTML source, you areapparently doomed to keep wondering; I couldn't ?da trace of ownership claim.> and so on, I thought it might be worth mentioning> it in this NG.> Would anyone care to share their views/reviews> of this site?I am most struck by how _sane_ and _readable_ the siteis. The parts I found weren't overwhelmingly technical,either, a relief for someone with my modest math skills.The reader will gain a lot of historical perspectivefairly painlessly.xanthian.-- === David Bernier> While sur?g the Web, I stumbled upon the> following site:> http://www.mathpages.com/> The *.com made me wonder who might> be af?iated to this site.Indeed. He has a lot to say but is not trying to fatten his CV or sell usany book on which he makes a commission. Most unusual :)> Would anyone care to share their views/reviews> of this site?I've seen some but not all of these interesting bits and pieces before. IMOit's a good site for any teacher who is looking for enrichment material.Larry === >While sur?g the Web, I stumbled upon the>following site:> http://www.mathpages.com/>The *.com made me wonder who might>be af?iated to this site.http://www.coolwhois.com/?d=mathpages.comis a way to answer that question. (Who oh why do so many authors put no contact information on their sites?)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You ?d yourself amusing, Blackadder.I try not to ?the face of public opinion. === > While sur?g the Web, I stumbled upon the> following site:> http://www.mathpages.com/> The *.com made me wonder who might> be af?iated to this site.This was set up by Kevin S. Brown. It was started asa conributor in the early to mid nineties. He I believeholds the dubious honor of being the ?st person on sci.mathto attempt to explain math to JSH.> and so on, I thought it might be worth mentioning> it in this NG.> Would anyone care to share their views/reviews> of this site?> David Bernier> ___________________________________________________________ Then assuredly the world was made, not in time, but> simultaneously with time.> --St. Augustine> === (Frequently Exhorted Exhortations :-)I recently posted a message to newbies exhorting them to be more thoughtfulin their choice of subject headings. What with its ephemeral nature, Ididn't expect the message to do any good, and it didn't, as witnessrecently: math question.Now I wish to exhort oldies not to work homework problems for people, butmerely to provide hints. In general, sci.math has a remarkablegentlepersons'-agreement on this. In fact, it's one of the most impressiveaspects of sci.math. But often it seems that 5 repliers give hints, and the6th provides the complete solution.I know this is all WAY OLD HAT, but perhaps it bears repeating. If not, youcan send some heat my way. I will fry eggs on my monitor.Lest I be hoisted by my own petard, let me hasten to throw in a few meaculpas. === >I recently posted a message to newbies>Now I wish to exhort oldies not to work homework problems for peopleExCUSE me, but I prefer the term knowbies for the yang to the newbies' yin.dave === In sci.math, Dave Rusin:>I recently posted a message to newbies>Now I wish to exhort oldies not to work homework problems for people> ExCUSE me, but I prefer the term knowbies > for the yang to the newbies' yin.> dave> So if one's just graduated from college is he a newbie knowbie,or a knowbie noveau?*dodges tomatoes* :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > I recently posted a message to newbies> Now I wish to exhort oldies not to work homework problems for people> ExCUSE me, but I prefer the term knowbies> for the yang to the newbies' yin.Certainly it has to be at least oldbies, not oldies.I agree with the exhortation, BTW. Not because I particularlycare if somebody somewhere gets an unearned A; that isn't reallymy problem. I'm concerned, rather, that the more people go alongwith it, the more requests of this sort we'll get. === > Certainly it has to be at least oldbies, not oldies.Thereby creating a grammatical rule for changing adjectives intonouns. Maybe the rule would apply only to people. Or people anddogs, say. Hungrybies, Smartbies, Dumbbies, Stinkybies, Seedybies... hum, I could go on a long time, yukking like a Nerd all the while.Max, Mr. Maximally Maxed === > I wish to exhort oldies not to work homework problems for people> I agree with the exhortation, BTW. Not because I particularly> care if somebody somewhere gets an unearned A; that isn't really> my problem. I'm concerned, rather, that the more people go along> with it, the more requests of this sort we'll get.This looks like a wonderful opportunity to start a new sub-group - alt.math.homework-help. If it's getting to be that big a hassle, there's obviously a pent-up demand for it. Then let those people work out the question of whether to give hints or complete answers. === > I agree with the exhortation, BTW. Not because I particularly> care if somebody somewhere gets an unearned A; that isn't really> my problem. I'm concerned, rather, that the more people go along> with it, the more requests of this sort we'll get.We is not a ?ed entity.I, for one, hope the homework police ?dit dif?ult to regulate what people post. === possibly of interest,http://www.geocities.com/jongiff2000/aa_index_ math.htmlI haven't quite ?ured out yet the meaning of theangle between two vectors in n-space. If projectionsstill hold, then the cosine must still hold, but exactlywhat does that angle measure? === > possibly of interest,> http://www.geocities.com/jongiff2000/aa_index_math.html> I haven't quite ?ured out yet the meaning of the> angle between two vectors in n-space. If projections> still hold, then the cosine must still hold, but exactly> what does that angle measure?> Two non-parallel vectors in n-space determine a two dimensional subspace. The angle is measured and meaningful in that two dimensional subspace, i. e., in a plane. === > possibly of interest,> http://www.geocities.com/jongiff2000/aa_index_math.html> I haven't quite ?ured out yet the meaning of the> angle between two vectors in n-space. If projections> still hold, then the cosine must still hold, but exactly> what does that angle measure?Do you suspect that Equivalence of Zero and In?ity casts somedoubt on the balance of the content?-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === In sci.physics, Jon: > possibly of interest,> http://www.geocities.com/jongiff2000/aa_index_math.html> I haven't quite ?ured out yet the meaning of the> angle between two vectors in n-space. If projections> still hold, then the cosine must still hold, but exactly> what does that angle measure?> What does any angle measure?In N-dimensional Euclidean space (N > 2), two linesintersecting will describe a plane. That plane is2-dimensional regardless of N.Rotate that plane onto your dissertation paper. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Meissel's formula is given in the CRC Concise Encyclopedia of Mathematics and on the webpage http://mathworld.wolfram.com/MeisselsFormula.htmlI am not at all convinced that formula (4) is correct. I think in the second sum, the indices should be 1 <= i < j <= c and not 1 <= i <= j <= c, and in the third sum the indices should be c < i <= b and not c <= i <= b. I would appreciate it if anyone who has a different source or who is willing to derive the formula could check this. The formula given there for pi (x) = number of primes <= x: pi (x) = ?x) - sum (?x / p(i)), 1 <= i <= c) + sum (?x / p(i) / p(j)), 1 <= i <= j <= c) - ... + ... + (1/2) (b + c - 2) (b - c + 1) - sum (pi (x / p (i)), c <= i <= b)where b = pi (x ^ (1/2)) and c = pi (x ^ (1/3)), and p (i) is the i-th prime number, that is p(1) = 2, p(2) = 3, p(3) = 5 etc. === > Meissel's formula is given in the CRC Concise Encyclopedia of > Mathematics and on the webpage> http://mathworld.wolfram.com/MeisselsFormula.html I am not at all convinced that formula (4) is correct. I think in the > second sum, the indices should be 1 <= i < j <= c and not 1 <= i <= j <= > c, and in the third sum the indices should be c < i <= b and not c <= i > <= b. > I would appreciate it if anyone who has a different source or who is > willing to derive the formula could check this. For reference, Riesel gives a permitaion of the top formula identically, butit deviates from Mathworld later on, exactly as you indicate. If you didn'tcrib the above from Riesel, then I'd put my money on you and Hans being right.Phil === Suppose f maps the reals to the reals, f( 0 ) = 0, and for all reala and b, f( a + b ) = f( a ) + f( b ).Is f necessarily linear?Can you recommend a text that discusses this problem?Note: 1) For any rational q, f( qa ) = qf( a ).2) If f is continuous, it is linear.3) If, in the above problem, the real ?ld is replaced by the rationals, then f is linear.-- === >Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real>a and b, f( a + b ) = f( a ) + f( b ).>Is f necessarily linear?No. For example, let B be a Hamel basis of the reals over the rationals,take some b_1 in B and de?e f(sum_{b in B} r_b b) = r_{b_1}.On the other hand, if f is Lebesgue measurable, or if it is bounded on some set of positive Lebesgue measure, then it is linear.See e.g. the thread Dif?ult Problem from February 1996.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real> a and b, f( a + b ) = f( a ) + f( b ).> Is f necessarily linear?> Can you recommend a text that discusses this problem?> Note:> 1) For any rational q, f( qa ) = qf( a ).> 2) If f is continuous, it is linear.> 3) If, in the above problem, the real ?ld is replaced by the > rationals, then f is linear.First write down a proof for (1), (2) or (3). If you ?d this dif?ult then ask for help, but don't expect that you can solve the original problem if you cannot prove these three. Then use the axiom of choice. === > Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real> a and b, f( a + b ) = f( a ) + f( b ).>Sidenote: f(0) = 0 is redundant.> Is f necessarily linear?> Note:> 1) For any rational q, f( qa ) = qf( a ).> 2) If f is continuous, it is linear.> 3) If, in the above problem, the real ?ld is replaced by the> rationals, then f is linear.> === >Suppose f maps the reals to the reals, f( 0 ) = 0, and for all real>a and b, f( a + b ) = f( a ) + f( b ).>Is f necessarily linear?> No.> For example, let B be a Hamel basis of the reals over the rationals,> take some b_1 in B and de?e f(sum_{b in B} r_b b) = r_{b_1}.> On the other hand, if f is Lebesgue measurable, or if it is bounded on> some set of positive Lebesgue measure, then it is linear.> See e.g. the thread Dif?ult Problem from February 1996.>I suppose those results have dual or parallel form for g:R -> R with g(ab) = g(a)g(b), g(1) = 1Similarly, if g is continuous, g(x) = |x|^n for some n in R. === |I suppose those results have dual or parallel form for g:R -> R with| g(ab) = g(a)g(b), g(1) = 1|Similarly, if g is continuous, g(x) = |x|^n for some n in R.Incidentally, the only possibility ruled out by g(1)=1 is gidentically equal to 0.Your question is closely related to the one of the O.P., since ifg is such a function, then f(x) = log g(e^x) satis?s f(x+y)= log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y))= log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind offunction originally considered.Thus if g satis?s your conditions, then there exists a function fsatisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x)= e^f(log x) for x>0.That just leaves the question of what values g has when x<=0. We knowg(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. Soif g is not identically 1, then g(0)=0. The constant function 1 doessatisfy your conditions.satisfy the original conditions if f satis?s the functionalequation f(x+y)=f(x)+f(y).If g is continuous (or even measurable) then f is also continuous(measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, org(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0(since for c<=0, sgn(x)*|x|^c is discontinuous at x=0).If f is discontinuous, then it's one of these peculiar functionswhose graph is dense in the plane, and the graph of g is either densein the upper half plane y>=0, or in both the ?st quadrant x>=0, y>=0and the third quadrant x<=0, y<=0.Keith Ramsay === > |I suppose those results have dual or parallel form for g:R -> R with> | g(ab) = g(a)g(b), g(1) = 1> |Similarly, if g is continuous, g(x) = |x|^n for some n in R.> Incidentally, the only possibility ruled out by g(1)=1 is g> identically equal to 0.>Indeed, it was stated that way in parallel contrast to OP's redundant f(0) = 0> Your question is closely related to the one of the O.P., since if> g is such a function, then f(x) = log g(e^x) satis?s f(x+y)> = log g(e^{x+y}) = log g(e^x e^y) = log (g(e^x) g(e^y))> = log g(e^x) + log g(e^y) = f(x)+f(y), which means f is the kind of> function originally considered.> Thus if g satis?s your conditions, then there exists a function f> satisfying f(x+y)=f(x)+f(y) for every x, y, for which g(x)> = e^f(log x) for x>0.>That certainly demonstrates constructively the duality.It's taking a somewhat familiar form of other dualities. log g(x) = f(log x); e^f(x) = g(e^x) cl SA = Sint A; Scl A = int SA -lub a,b = glb -a,-b; lub -a,-b = -glb a,b -sup A = inf -A; sup -A = -inf A /{ S-X | X in C } = S - /{ X | X in C } S - /{ X | X in C } = /{ S-X | X in C }> That just leaves the question of what values g has when x<=0. We know> g(0)=g(0)g(x) for every x. So if g(0)<>0, then g(x)=1 for every x. So> if g is not identically 1, then g(0)=0. The constant function 1 does> satisfy your conditions.>Yup, noticed that. I'll take your word for the rest which pointsout some grit in an otherwise smooth duality. A duality expressedwith continuous functions, yet applicable to discontinuous ones also.> g(x) = -g(-x) or g(x)=g(-x) for every x. It's easy to check that> both possibilities,> g(x) = e^f(log x) if x > 0> = 0 if x = 0> = e^f(log -x) if x < 0> and> g(x) = e^f(log x) if x > 0> = 0 if x = 0> = e^f(log -x) if x < 0> satisfy the original conditions if f satis?s the functional> equation f(x+y)=f(x)+f(y).> If g is continuous (or even measurable) then f is also continuous> (measurable) and f(x)=cx for some c. Then either g(x)=1=|x|^0, or> g(x)=|x|^c, or the possibility you missed, g(x)=sgn(x)*|x|^c for c>0> (since for c<=0, sgn(x)*|x|^c is discontinuous at x=0).> If f is discontinuous, then it's one of these peculiar functions> whose graph is dense in the plane, and the graph of g is either dense> in the upper half plane y>=0, or in both the ?st quadrant x>=0, y>=0> and the third quadrant x<=0, y<=0.> === Robert,I've been puzzling over what you mean by the reals over the rationals. I guess you mean to somehow decompose the reals into algebraically closed subsets. Do you mean some kind of quotient space? I know what is meant by the quotient of a vector space with a vector subspace, but therationals aren't a vector subspace of the reals (in any sense I know), Can you expand on this please?> For example, let B be a Hamel basis of the reals over the rationals,> take some b_1 in B and de?e f(sum_{b in B} r_b b) = r_{b_1}.> -- === > I guess you mean to somehow decompose the reals into algebraically closed> subsets. Do you mean some kind of quotient space? I know what is meant> by the quotient of a vector space with a vector subspace, but the> rationals aren't a vector subspace of the reals (in any sense I know),Sure they are. Any ?ld K with a sub?ld k is a vector space over k. All thevector space axioms are satis?d if the elements in K are vectors, the elementsin k are scalars, and addition and multiplication are as de?ed in K. Many ofthe basic results in ?ld theory rely on this fact. === >Robert,>I've been puzzling over what you mean by the reals over the rationals. >I guess you mean to somehow decompose the reals into algebraically closed >subsets. Do you mean some kind of quotient space? I know what is meant >by the quotient of a vector space with a vector subspace, but the>rationals aren't a vector subspace of the reals (in any sense I know), >Can you expand on this please?> For example, let B be a Hamel basis of the reals over the rationals,> take some b_1 in B and de?e f(sum_{b in B} r_b b) = r_{b_1}.> I think you have missed something essential. The reals are a vectorspace over the rationals. Check the de?ition of a vectors space over a ?ld.Larry(this space unintentially left blank ..... === Let S be the set of natural numbers n such that mu(n) = 1 .What is the density of S ? === Let S be the set of natural numbers n such that mu(n) = 1 .What is the density of S ? === > Let S be the set of natural numbers n such that mu(n) = 1 .> What is the density of S ?The set of natural numbers n such that |mu(n)| = 1 is 6/(pi^2). The sum of mu(n), n from 1 to N, is little-oh of N. Does it not follow that the number you're after is 3/(pi^2)?-- === Dear reader,for my research I have programmed a combinatorial optimization problem inaimms 2 format. Unfortunately running this program takes a lot of time, so Iwant to try a more ef?ient solver like C-plex. However, in order to useC-plex I need to convert my aimms program to an mps format program. Cananyone tell me whether this is possible in aimms?Dion Bongaerts === > to this question, which has been driving me a bit nuts. Some quick> background: as a role-player, I use a lot of dice. At times, the> rules call for one to roll multiple dice (say, ?e six-sided dice, or> 5d6), then to drop the lowest two and total the other three. The> basic question is: is there a formula for determining the probability> of rolling a certain result, given these conditions?> Determining the probability of a particular outcome when just rolling> multiple dice is relatively straightforward (there's a brief> discussion here: http://mathforum.org/library/drmath/view/52207.html).> I can ?d a pattern to the summation needed when dropping a single> die from a set; but once I try to remove two dice from the set, the> pattern disappears and I ?d myself lost again. (The numbers can be> determined by brute force, of course, but that's neither practical nor> interesting.)> So, I guess the base question is: Is there a formula for calculating> the probability of achieving a result R on n dice with d sides,> dropping the k lowest dice?show us your pattern for dropping one dice and we might be able towork it out.Herc === De?e S:= {(x,y) in R : 0 De?e S:= {(x,y) in R : 0 De?e T:= {x in R : 0 Problem: Use the fact that every real number has a decimal expansion> to produce an injective function that maps S into T.Think about how you can take two in?ite decimal expansions and turn them that might give you an idea.)-- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman === > De?e S:= {(x,y) in R : 0 De?e T:= {x in R : 0 Problem: Use the fact that every real number has a decimal expansion> to produce an injective function that maps S into T.> thanks.For the reals which can have terminating decimal representations, require them to be so represented, then for x with decimal representation .abcdefg... and y with decimal representation .ABCDEFG...map(x,y) onto z with decimal representation .aAbBcCdDeEfFgG... === >De?e S:= {(x,y) in R : 0De?e T:= {x in R : 0Problem: Use the fact that every real number has a decimal expansion>to produce an injective function that maps S into T.The usual solution is a bit of a trick, hard to hint at withoutsimply giving it away. Given (x,y), take the decimal expansionsof x and y and do a rif?f?an === Can anyone please help me with this dif?ult question on probability?The probability that there is ?x' amount of items in a box (where x is in1000's):P(x) = -0.03x^2 + 0.12x + 0.15What is the probability that the box contains at least 4000 items? === > Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is in> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?I don't think that anybody can help. For large x, your probability will become negative (???), and the sum of P over all possible x should equal one, but isn't even ?ite in this case. Weird question, who asked you? === Since there are 4 thousands in 4000 simply calculate P(4)> Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is in> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?> === >Can anyone please help me with this dif?ult question on probability?>The probability that there is ?x' amount of items in a box (where x is in>1000's):>P(x) = -0.03x^2 + 0.12x + 0.15>What is the probability that the box contains at least 4000 items?>P(x) = -0.03 * (x^2 -4x -5) = -0.03*(x+1)(x-5) because probabilitieshave to be non-negative and numbers of items in a box also, P(x) isde?ed for 0<=x<=5.Because for x in [0,5] 0<=P(x)<=1 and int[0,5]P(x)dx=1 is P(x) indeed a density function and Prob(x>=4) = int [4,5]P(x)dx.I leave that to you. === Sorry, I misread the problem.P(at least 4000) = P(x > 4) = 1- P( x <= 4).Now you can solve either of the probabilities listed in the line above byintegration.But wait a minute: integral (-.03x^2 + .12x + .15) from 0 to oo isNEGATIVE. This problem makes no sense> Since there are 4 thousands in 4000 simply calculate P(4)> Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x isin> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?> === > Can anyone please help me with this dif?ult question on probability?> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?Sure. Show us the work you did on the problem ?st.Also, should we call you Marc Rice, Mary Watts, David Jones,or Frank Dewhurst?Paul Guertinpg@sff.net === >Can anyone please help me with this dif?ult question on probability?>The probability that there is ?x' amount of items in a box (where x is in>1000's):>P(x) = -0.03x^2 + 0.12x + 0.15>What is the probability that the box contains at least 4000 items?>Well, here's one interpretation. Find the zeros of P(x) (-1 and 5). Figure that x>=0, so integrate P from 0 to 5 to see if the total probability is 1. Then integrate P from 4 to 5 to get the probability that x>=4.This is not too far removed from what would happen with real data -- you'd have a model to ? it into (not a polynomial, simply because we don't do that in probabilistic models, but something) and some data, and you'd see how well the data ? the model, and adjust it (either the model or the data) to get something useful (the probability that total losses exceed a billion dollars, for example, or that the lifetime of an appliance will exceed 5 years).Jon Miller === > Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is in> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?> Probabilities must be values between 0 and 1, and the sum of all probabilities must add up to 1.You alleged probability function, as de?ed, does not have either of these properties. === > Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is in> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?Who knows if this is the real intention or not, it's just a guess (the OPshould clarify). Probability, in general, must be between 0 and 1inclusive. Thus, the largest possible range of P is [0,1]. This doesn'tnecessarily mean that's the actual range. It just means the range of P willbe a subset of [0,1] ie the actual range is taken from a codomain of[0,1]. (who knows where this model came from, apparently more is known tothem about P than simply it must be between 0 and 1 in this particularcase).Also, x is clearly >=0, possibly even just greater (noninclusive). Underthis very reasonable assumption, we can reasonably assume the domain of P,with just the given information, to be implied [0,5] (or possibly (0,5]).The range of P is therefore [0,.27] accordingly.We have:P(x) = -0.03x^2 + 0.12x + 0.15 domain: [0,5] range: [0,.27]IOW, the graph of P is just that part of the parabola in Q1.The question asks, I think, for:integral(x=4 to x=5) P(x) dx. Hope that helps.-- Darrell === Note: Apparently I'm too far off base, since integral(0 to 5) P(x)dx = 1.Makes perfect sense, right?-- Darrell> Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x isin> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?> Who knows if this is the real intention or not, it's just a guess (the OP> should clarify). Probability, in general, must be between 0 and 1> inclusive. Thus, the largest possible range of P is [0,1]. This doesn't> necessarily mean that's the actual range. It just means the range of Pwill> be a subset of [0,1] ie the actual range is taken from a codomain of> [0,1]. (who knows where this model came from, apparently more is known to> them about P than simply it must be between 0 and 1 in this particular> case).> Also, x is clearly >=0, possibly even just greater (noninclusive). Under> this very reasonable assumption, we can reasonably assume the domain of P,> with just the given information, to be implied [0,5] (or possibly (0,5]).> The range of P is therefore [0,.27] accordingly.> We have:> P(x) = -0.03x^2 + 0.12x + 0.15> domain: [0,5]> range: [0,.27]> IOW, the graph of P is just that part of the parabola in Q1.> The question asks, I think, for:> integral(x=4 to x=5) P(x) dx. Hope that helps.> -- > Darrell> === Oops.... should be _not_ too far off base.> Note: Apparently I'm too far off base, since integral(0 to 5) P(x)dx = 1.> Makes perfect sense, right?> -- > Darrell> Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is> in 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15 What is the probability that the box contains at least 4000 items?> Who knows if this is the real intention or not, it's just a guess (theOP> should clarify). Probability, in general, must be between 0 and 1> inclusive. Thus, the largest possible range of P is [0,1]. Thisdoesn't> necessarily mean that's the actual range. It just means the range of P> will> be a subset of [0,1] ie the actual range is taken from a codomain of> [0,1]. (who knows where this model came from, apparently more is knownto> them about P than simply it must be between 0 and 1 in this particular> case).> Also, x is clearly >=0, possibly even just greater (noninclusive).Under> this very reasonable assumption, we can reasonably assume the domain ofP,> with just the given information, to be implied [0,5] (or possibly(0,5]).> The range of P is therefore [0,.27] accordingly.> We have:> P(x) = -0.03x^2 + 0.12x + 0.15> domain: [0,5]> range: [0,.27]> IOW, the graph of P is just that part of the parabola in Q1.> The question asks, I think, for:> integral(x=4 to x=5) P(x) dx. Hope that helps.> -- > Darrell === > Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x is in> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?Two answers have been given (integrate this function, or add up some valuesof P(x) for some values of x), but neither is correct, because you haven'tsaid what the possible values of x are. In fact, if all the possible valuesof x are less than 4, the correct answer is zero. If the set of possible values is ?ite (like {1, 2, 3}), you will needto add up P(x) over all _possible_ values of x which are at least 4. If the set of possible values of x is an interval (like [0, 5]), thenyou have to have a probability _density_ function, and you integrate it from4 to whatever the upper limit is. Without further information, people can only guess at the answer. -- Christopher Heckman === > Can anyone please help me with this dif?ult question on probability?> The probability that there is ?x' amount of items in a box (where x isin> 1000's):> P(x) = -0.03x^2 + 0.12x + 0.15> What is the probability that the box contains at least 4000 items?> Two answers have been given (integrate this function, or add up somevalues> of P(x) for some values of x), but neither is correct, because you haven't> said what the possible values of x are. In fact, if all the possiblevalues> of x are less than 4, the correct answer is zero.I count no less than 5 answers (maybe your news feed is slow).> If the set of possible values is ?ite (like {1, 2, 3}), you willneed> to add up P(x) over all _possible_ values of x which are at least 4.> If the set of possible values of x is an interval (like [0, 5]), then> you have to have a probability _density_ function, and you integrate itfrom> 4 to whatever the upper limit is.I don't pretend to know much about probability, much less much about aprobability density function, but if it's what I think it is (?), thenthat's what we have with P.> Without further information, people can only guess at the answer....But we can make an educated guess. I, and at least one other, guessedthat integral(0,5)P(x)dx=1 is no coincidence. Yeah, practically speakingsince these are items then x can in reality take on only increments of1/1000 (x was in thousands of items) but I see the continuous paraboliccurve as, well, just a model. It ?s the data well, so we pretend xtakes on all values in a certain interval. Close enough.Under this interpretation (which is the only one that seems to make anysense out of the problem), the question asks for integral(4,5)P(x)dx. Thequestion could be better worded.-- Darrell === [...]> I don't pretend to know much about probability, much less much about a> probability density function, but if it's what I think it is (?), then> that's what we have with P.>If the distributed variable X is discrete, it is meaningful to have anequation forP(X=x). The summation of all the probabilities will be one. If thedistributed variable X is continuous a probability distribution function(pdf) is useful. In this case P(X=x) is always zero. The integral over thedomain of the pdf is one. For a uniform distribution we could have a pdfsuch that:f(x) = (1/2 0 If the distributed variable X is discrete, it is meaningful to have an> equation for> P(X=x). The summation of all the probabilities will be one.I stated earlier. The variable X is indeed discrete (is given inthousands of units; makes no sense to speak of a fractional units, so it canvary _at least_ in increments of 1/1000). At least that's a reasonableassumption. The notation P(X=x), which I may very well be misinterpreting,suggests to me that we have ? the data to some continuous function ofx. and are using P(x) as a model for the actual P(X) where X isdiscrete. The summation of all (discrete) probabilities is 1, as inP(1)+P(2)+P(3)+...+P(n) for some n. Am I interpreting correctly? If so,then I think I can see where the OP's problem, as stated, doesn't make muchsense, since the discrete functional values of the given equation don'tseem to add up to 1 under different reasonable assumptions on how much X canreally vary by (discretely).> If the> distributed variable X is continuous a probability distribution function> (pdf) is useful. In this case P(X=x) is always zero.> The integral over the> domain of the pdf is one. For a uniform distribution we could have a pdf> such that:> f(x) = (1/2 0 ( 0 elsewhereAre you saying that the given equation, with assumed domain [0,5], is or isnot a probability distribution function? It's integral is 1, but there isonly one point where P(x)=0. If I had to guess, it appears to be saying byway of your example that for all reals except [0,5] the function shouldindeed be de?ed and integrable, but with integral=0. IOW, I interpret youto imply that the (nonuniform) probability distribution function for thisproblem may look something like:f(x) = -.03x^2 + .12x + .15 on [0,5] = 0 on (-oo,0)U(5,oo)..since it's 0 elsewhere yet the integral over the entire domain is 1.Please straighten this dummy out ;-).-- Darrell === I am a hardware logic designer and I would greatly appreciate some help on de?ing a boolean function from the given conditionsbelow. I am trying to formulate a boolean function gwhich maps a 8-bit variable a[7:0] to another 8 bit variable b[7:0], i.e. g(a[7:0]) = b[7:0] ... (i)and, (b[0] + b[1].x + b[2].(x^2) + b[3].(x^3)) + (b[4] + b[5].x +b[6].(x^2) + b[7].(x^3)).y= 1/((a[0] + a[1].x + a[2].(x^2) + a[3].(x^3)) + (a[4] + a[5].x +a[6].(x^2) + a[7].(x^3)).y) ... (ii)y is an element in GF(2^8) that satis?sy^8 + y^6 + y^5 + y^3 + 1 = 0 ... (iii)andx = y^238 = y^6 + y^5 + y^3 + y^2 ... (iv)is an element in GF(2^4) that satis?s x^4 + x + 1 = 0 ... (v) Apparently, you can also assume 1/0 = 0, though not sure why (or how) that is true. - Swapnajit. boundary=------------5F60FDBAB57B35E6585F0C83 === ---------- ----------------------------------------------------------- x-no-archive: yes Please excuse my ignorance ! I'm the messenger ..my brother is having abuilding constructed and called me on his cell.. he's caught without histrigonometric function table book ! he needs to know the TANGENT of a 2o angle ! Please respond ASAP ..so I can call him back. remove (NOSPAM) of course oh thank you ..thank you === > x-no-archive: yes> Please excuse my ignorance ! I'm the messenger ..my brother is having a> building constructed and called me on his cell.. he's caught without his> trigonometric function table book !> he needs to know the TANGENT of a 2o angle !> Please respond ASAP ..so I can call him back.> remove (NOSPAM) of course> oh thank you ..thank you>This is pretty funny, actually, but anyway, here it is:0.034920769491747730500402625773725315879174297784615 approximately boundary=------------56E0B04682C893DE51A74A90 === ---------- ----------------------------------------------------------- x-no-archive: yes oh ..so funny and oh .. so appreciated ! THANK YOU SO MUCH === I'm sure the proof is just under my nose and I just can't seem tosee it. The problem is to prove that for p prime, no nonzero integersa,b exist such that a^2 = pb^2. This is supposed to be done withoutassuming that the square roots of the odd primes are irrational (youcan still assume sqrt(2) is irrational). So for p=2 it's trivial. For the other p, we know p is odd, which means that pb^2 is oddthat if any such a and b exist, they have equal parity. I've managed to prove that no even a and b exist which satisfythe equation: for, assume they do (for a given odd prime p), andconstruct the set M de?ed as {a+b | a^2 = pb^2 and a,b even andpositive} ie, M is the set of the sums of the positive even a,b whichsolve the equation. (Note that there's no loss of generality inrestricting it to the positive solutions due to the squaring). Clearly M is a set of positive integers, thus by the well orderingprinciple it has a lowest member m, which by the de?ition of M canbe written m=i+j for some i,j such that i^2=pj^2 (note that i and jare not necessarily unique, but it turns out any choice thereof whichsatis?s the equation will suf?e for the proof). But since i and jare even, we have i=2x, j=2y, for some positive integers x and y, andthus 4x^2=4py^2, dividing by four we get x^2=py^2, but this is theoriginal equation, and thus since x and y solve it, we must have x+yan element of M... but x+y is clearly smaller than i+j, whichcontradicts the fact that m=i+j was the smallest member of M. However, try as I might, I have been unable to prove it for thesolve it on my own. P.S. this is not homework. === > I'm sure the proof is just under my nose and I just can't seem to> see it. The problem is to prove that for p prime, no nonzero integers> a,b exist such that a^2 = pb^2. This is supposed to be done without> assuming that the square roots of the odd primes are irrational (you> can still assume sqrt(2) is irrational). So for p=2 it's trivial.Suppose to the contrary that such a and b exist. Then we can assume that they have no common prime factor, for if they did, we could just divide both sides with the squares of those common prime factors.So p divides the right-hand side.But then p divides a^2.So p divides a.So a=pc for some integer c.So (pc)^2=pb^2.So pc^2=b^2.So p divides b.So p divides both a and b.But we said a and b had no common prime factor. DANG! === Trishia Rose a .8ecrit dans le message de> I'm sure the proof is just under my nose and I just can't seem to> see it. The problem is to prove that for p prime, no nonzero integers> a,b exist such that a^2 = pb^2. This is supposed to be done without> assuming that the square roots of the odd primes areif a such a and b exist, then each prime number in the decomposition of a^2is raised to an even power, whereas p is raised to an odd power in thedecomposition of pb^2.-- Julien Santini,CMI Technop.99le de Ch.89teau-Gombert, FranceHome page: http://www.analgebra.com === > I'm sure the proof is just under my nose and I just can't seem to> see it. The problem is to prove that for p prime, no nonzero integers> a,b exist such that a^2 = pb^2. This is supposed to be done without> assuming that the square roots of the odd primes are irrational (you> can still assume sqrt(2) is irrational). So for p=2 it's trivial.> For the other p, we know p is odd, which means that pb^2 is odd> that if any such a and b exist, they have equal parity.> I've managed to prove that no even a and b exist which satisfy> the equation: for, assume they do (for a given odd prime p), and> construct the set M de?ed as {a+b | a^2 = pb^2 and a,b even and> positive} ie, M is the set of the sums of the positive even a,b which> solve the equation. (Note that there's no loss of generality in> restricting it to the positive solutions due to the squaring). > Clearly M is a set of positive integers, thus by the well ordering> principle it has a lowest member m, which by the de?ition of M can> be written m=i+j for some i,j such that i^2=pj^2 (note that i and j> are not necessarily unique, but it turns out any choice thereof which> satis?s the equation will suf?e for the proof). But since i and j> are even, we have i=2x, j=2y, for some positive integers x and y, and> thus 4x^2=4py^2, dividing by four we get x^2=py^2, but this is the> original equation, and thus since x and y solve it, we must have x+y> an element of M... but x+y is clearly smaller than i+j, which> contradicts the fact that m=i+j was the smallest member of M.> However, try as I might, I have been unable to prove it for the> solve it on my own. P.S. this is not homework.Trishia,The probable intention of the problem is to encourage you to examineclosely the proof of the irrationality of sqrt(2), and then to adaptthis proof to the situation where 2 is replaced by a general prime. Your problem is a rewritten form of the problem of proving thatsqrt(p) is irrational.Paul Epstein === > Dears> Can you give me the formula to get the sum of below series with only> avalable is ?n'.> ??(n) = (1/1 + 1/2 + 1/3 + 1/4.....+ 1/n) This is a harmonic number ... it is approxaimately ln(n) ...but for an exact value you cannot do it with an elementary function.Maple uses the digamma function Psi:> Hn := sum(1/k,k=1..n); Hn := Psi(n + 1) + gammaAsymptotically, we have:> asympt(Hn,n,15); 1/2 1 1/120 1 1/240 ln(n) + gamma + --- - 1/12 ---- + ----- - 1/252 ---- + ----- n 2 4 6 8 n n n n 691 ----- 1 32760 1 1 - 1/132 --- + ----- - 1/12 --- + O(---) 10 12 14 15 n n n n> gamma is Euler's constant-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Still trying to teach myself some math. :)I have no problem, really, with the procedure(s) for ?ding determinants. My actual problem is that I'm sort of number-dyslexic and not very good with arithmetic. So, if I work through any procedure that has lots of arithmetic operations, I will probably make one or more errors and have to go root it (them) out. It's a pain, but I'm used to it.My present problem with determinants is that I have no clue how to lookat the matrix and ?ure out, ballpark, what kind of answer to expect,so I know when I might have a correct one. I can check and recheck anduse different methods and see if I duplicate the same answer, but Ihope there might be a better way.Is there any way to look at a matrix and ?ure out *approximately* what its determinant is? (I was recently off by over 100,000 so really any clue would help...)-Laurel(Yes, I know I could use a calculator, but that takes all the fun out of it.) === |Still trying to teach myself some math. :)||I have no problem, really, with the procedure(s) for ?ding |determinants. My actual problem is that I'm sort of number-dyslexic |and not very good with arithmetic. So, if I work through any procedure |that has lots of arithmetic operations, I will probably make one or |more errors and have to go root it (them) out. It's a pain, but I'm |used to it.||My present problem with determinants is that I have no clue how to look|at the matrix and ?ure out, ballpark, what kind of answer to expect,|so I know when I might have a correct one. I can check and recheck and|use different methods and see if I duplicate the same answer, but I|hope there might be a better way.||Is there any way to look at a matrix and ?ure out *approximately* |what its determinant is? (I was recently off by over 100,000 so |really any clue would help...)the ability to estimate the determinant of a matrix by quicklyglancing at the matrix is probably a pretty silly thing to want tohave. on the other hand, the ability to estimate the determinant of amatrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) byquickly glancing at a geometric depiction of the transformationrepresented by the matrix is very important to have if you really wantto understand determinants. is this the sort of thing that youactually wanted to learn? it's not very dif?ult to learn how to doit.-- === |Is there any way to look at a matrix and ?ure out *approximately* >|what its determinant is? (I was recently off by over 100,000 so >|really any clue would help...)> the ability to estimate the determinant of a matrix by quickly> glancing at the matrix is probably a pretty silly thing to want to> have. Whoops.> on the other hand, the ability to estimate the determinant of a> matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by> quickly glancing at a geometric depiction of the transformation> represented by the matrix is very important to have if you really want> to understand determinants. is this the sort of thing that you> actually wanted to learn? it's not very dif?ult to learn how to do> it.What you describe sounds like an excellent thing to learn; I didn't even know the concept. In the (VERY introductory) linear algebra book I am following matrices don't represent anything in particular, so the determinant isn't anything useful, just a bunch of steps to go through to churn out some practically unrelated-seeming number. So it's hard to tell if I'm doing it right without going off and looking at the answer. (Which is, of course, lame, and I don't want to do that until I think I'm right.)Any help would be nice, but I am very uneducated, so there's a good chance I won't understand any of it. :(-Laurel === > Still trying to teach myself some math. :)> I have no problem, really, with the procedure(s) for ?ding > determinants. My actual problem is that I'm sort of number-dyslexic > and not very good with arithmetic. So, if I work through any procedure > that has lots of arithmetic operations, I will probably make one or > more errors and have to go root it (them) out. We all do. Especially with something as boring as calculating adeterminant.> It's a pain, but I'm used to it.> My present problem with determinants is that I have no clue how to look> at the matrix and ?ure out, ballpark, what kind of answer to expect,> so I know when I might have a correct one. I can check and recheck and> use different methods and see if I duplicate the same answer, but I> hope there might be a better way.> Is there any way to look at a matrix and ?ure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...)Except in very special cases, eg. triangular matrices for which youknow the determinant immediately, there isn't any general way toapproximate the determinant by eyeballing it. Just changing onenumber, slightly, can make major changes in the determinant.If you don't want to or can't use a computer to check, you cancalculate the determinant two ways; Gaussian elimination and expansionby cofactors for example.-- Paul SperryColumbia, SC (USA) === |> on the other hand, the ability to estimate the determinant of a|> matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by|> quickly glancing at a geometric depiction of the transformation|> represented by the matrix is very important to have if you really want|> to understand determinants. is this the sort of thing that you|> actually wanted to learn? it's not very dif?ult to learn how to do|> it.||What you describe sounds like an excellent thing to learn; I didn't |even know the concept. In the (VERY introductory) linear algebra book |I am following matrices don't represent anything in particular, so the |determinant isn't anything useful, just a bunch of steps to go through |to churn out some practically unrelated-seeming number. So it's hard |to tell if I'm doing it right without going off and looking at the |answer. (Which is, of course, lame, and I don't want to do that until |I think I'm right.)||Any help would be nice, but I am very uneducated, so there's a good |chance I won't understand any of it. :(the most important thing to understand about matrixes is that theyrepresent geometric transformations of a certain kind, namelyso-called linear transformations or linear operators. let me tryto demonstrate an example of a linear transformation: u t g t r t o n ? p i y w i n o w s e g t n o i m n f a k l o i o h l t s oW n o n d , e f a t s v n u m i e m i o ? ) a m i h . r n c t I e r f i h e t u i h g c e h w i iroughly, distorted by a geometric transformation represented by acertain matrix. (this is an ascii graphic which probably only worksif you're viewing this with a ?ed font.) now, eyeballing thedistorted text, i would estimate that the determinant of the matrixis, oh, say, somewhere around +5. that's an extremely rough estimate.what i'm really estimating is that text printed out in that distortedway will use up about 5 times as much paper as text printed out in thenormal way.ok, now let's check how bad my estimate was, by actually calculatingthe matrix that represents this geometric transformation. that matrixis:3 -11 3which if i calculate correctly has a determinant of 3*3 - 1*(-1) = 10.ok, so my estimate was way off, off by a factor of 2 in fact. thatis, according to this calculation, text printed out in that distortedway uses up 10 times as much paper as normal text.is the connection between the above matrix and the above geometrictransformation clear? notice that moving one step to the right in theoriginal text corresponds to 3 right, 1 up in the distorted text,while moving one step up in the original text corresponds to 1 left,3 up in the distorted text.-- === Laurel Amberdine> Still trying to teach myself some math. :)> I have no problem, really, with the procedure(s) for ?ding> determinants. My actual problem is that I'm sort of number-dyslexic> and not very good with arithmetic. So, if I work through any procedure> that has lots of arithmetic operations, I will probably make one or> more errors and have to go root it (them) out. It's a pain, but I'm> used to it.> My present problem with determinants is that I have no clue how to look> at the matrix and ?ure out, ballpark, what kind of answer to expect,> so I know when I might have a correct one. I can check and recheck and> use different methods and see if I duplicate the same answer, but I> hope there might be a better way.> Is there any way to look at a matrix and ?ure out *approximately*> what its determinant is?If each coef?ient is, say, between -10 and 10, then there are n! summandsall in the range(-10)^n to 10^n, which I guess is better than nothing. Better is a resultcalled the [drum roll] Hadamard Determinant Theorem: If the column vectorsin the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1 k_2... k_n.Larry === > Still trying to teach myself some math. :)> I have no problem, really, with the procedure(s) for ?ding > determinants. My actual problem is that I'm sort of number-dyslexic > and not very good with arithmetic. So, if I work through any procedure > that has lots of arithmetic operations, I will probably make one or > more errors and have to go root it (them) out. It's a pain, but I'm > used to it.> My present problem with determinants is that I have no clue how to look> at the matrix and ?ure out, ballpark, what kind of answer to expect,> so I know when I might have a correct one. I can check and recheck and> use different methods and see if I duplicate the same answer, but I> hope there might be a better way.> Is there any way to look at a matrix and ?ure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...)> -Laurel> (Yes, I know I could use a calculator, but that takes all the fun out > of it.)If you just want a ballpark ?ure, the following works.Let det A equal d.choose a random vector v.Work out A.v.The entries in A.v should be about d^(1/n) times the entries of v.(n is the length of the vector, nxn is the size of the matrix).note to ? :-He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!!Stop ? already, ok? ball - ouch! - park, ok??Anyway, this method should detect errors of order 100000. === >Still trying to teach myself some math. :)>I have no problem, really, with the procedure(s) for ?ding >determinants. My actual problem is that I'm sort of number-dyslexic >and not very good with arithmetic. So, if I work through any procedure >that has lots of arithmetic operations, I will probably make one or >more errors and have to go root it (them) out. It's a pain, but I'm >used to it.>My present problem with determinants is that I have no clue how to look>at the matrix and ?ure out, ballpark, what kind of answer to expect,>so I know when I might have a correct one. I can check and recheck and>use different methods and see if I duplicate the same answer, but I>hope there might be a better way.>Is there any way to look at a matrix and ?ure out *approximately* >what its determinant is? (I was recently off by over 100,000 so >really any clue would help...)In general, no. I am supposed to be rather good at arithmetic,and can do more mentally than most, but unless the matrix was of a special form, no luck about the derivative.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === > |Is there any way to look at a matrix and ?ure out *approximately* >|what its determinant is? (I was recently off by over 100,000 so |really any clue would help...)> ...> ... In the (VERY introductory) linear algebra book > I am following matrices don't represent anything in particular, so the > determinant isn't anything useful, just a bunch of steps to go through > to churn out some practically unrelated-seeming number. ...How horrifying. Find a new textbook, fast.John Mitchell === > Laurel Amberdine Is there any way to look at a matrix and ?ure out *approximately*> what its determinant is?> If each coef?ient is, say, between -10 and 10, then there are n! summands> all in the range> (-10)^n to 10^n, which I guess is better than nothing.Doesn't narrow it down a whole lot. :)> Better is a result> called the [drum roll] Hadamard Determinant Theorem: If the column vectors> in the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1 k_2> ... k_n.I had to look up norm: the square root of the sum of the squares of the absolute values of the elements of a matrix or the components of a vector. Is that right?-Laurel === Is there any way to look at a matrix and ?ure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...)> Except in very special cases, eg. triangular matrices for which you> know the determinant immediately, there isn't any general way to> approximate the determinant by eyeballing it. Just changing one> number, slightly, can make major changes in the determinant.> If you don't want to or can't use a computer to check, you can> calculate the determinant two ways; Gaussian elimination and expansion> by cofactors for example.At least there *are* multiple ways, so I can just keep trying until I get consistently the same answer. Bleagh. :)-Laurel === > |Is there any way to look at a matrix and ?ure out *approximately* >|what its determinant is? (I was recently off by over 100,000 so |really any clue would help...)> ...> ... In the (VERY introductory) linear algebra book > I am following matrices don't represent anything in particular, so the > determinant isn't anything useful, just a bunch of steps to go through > to churn out some practically unrelated-seeming number. ...> How horrifying. Find a new textbook, fast.I have two textbooks that are better, but they're too sophisticated for meto start with. Since I'm teaching myself without any formal help, I haveterrible trouble ?uring out terminology and symbols. I get completelystuck on even the simplest things.Once I already know what a book is (generally) supposed to say, it's a loteasier to read. :)I think I need a math dictionary.-Laurel === Is there any way to look at a matrix and ?ure out *approximately* > what its determinant is? (I was recently off by over 100,000 so > really any clue would help...)> If you just want a ballpark ?ure, the following works.> Let det A equal d.> choose a random vector v.> Work out A.v.> The entries in A.v should be about d^(1/n) times the entries of v.> (n is the length of the vector, nxn is the size of the matrix).Okay, I will give that a try. Should be interesting. > note to ? :-> He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!!> Stop ? already, ok? ball - ouch! - park, ok??She said, actually, but that's okay. :)> Anyway, this method should detect errors of order 100000.-Laurel === >|> on the other hand, the ability to estimate the determinant of a>|> matrix (particularly a 2-by-2 one, or maybe also a 3-by-3 one) by>|> quickly glancing at a geometric depiction of the transformation>|> represented by the matrix is very important to have if you really want>|> to understand determinants. is this the sort of thing that you>|> actually wanted to learn? it's not very dif?ult to learn how to do>|> it.>|>|What you describe sounds like an excellent thing to learn; I didn't >|even know the concept. In the (VERY introductory) linear algebra book >|I am following matrices don't represent anything in particular, so the >|determinant isn't anything useful, just a bunch of steps to go through >|to churn out some practically unrelated-seeming number. So it's hard >|to tell if I'm doing it right without going off and looking at the >|answer. (Which is, of course, lame, and I don't want to do that until >|I think I'm right.)>|>|Any help would be nice, but I am very uneducated, so there's a good >|chance I won't understand any of it. :(> the most important thing to understand about matrixes is that they> represent geometric transformations of a certain kind, namely> so-called linear transformations or linear operators. let me try> to demonstrate an example of a linear transformation:> u t g t r t> o n ? p i> y w i n> o w s e g> t n o i m n f> a k l o i o> h l t s o> W n o n d ,> e f a t s> v n u m i> e m i o ? )> a m i h .> r n c t> I e r f i h e> t u i h g c> e h w i i roughly, distorted by a geometric transformation represented by a> certain matrix. (this is an ascii graphic which probably only works> if you're viewing this with a ?ed font.)No problem there.> now, eyeballing the> distorted text, i would estimate that the determinant of the matrix> is, oh, say, somewhere around +5. that's an extremely rough estimate.> what i'm really estimating is that text printed out in that distorted> way will use up about 5 times as much paper as text printed out in the> normal way.> ok, now let's check how bad my estimate was, by actually calculating> the matrix that represents this geometric transformation. that matrix> is:> 3 -1> 1 3> which if i calculate correctly has a determinant of 3*3 - 1*(-1) = 10.> ok, so my estimate was way off, off by a factor of 2 in fact.A lot closer than 100,000 though. :)> that> is, according to this calculation, text printed out in that distorted> way uses up 10 times as much paper as normal text.Hm. Okay.> is the connection between the above matrix and the above geometric> transformation clear? notice that moving one step to the right in the> original text corresponds to 3 right, 1 up in the distorted text,> while moving one step up in the original text corresponds to 1 left,> 3 up in the distorted text.Aha. I see...practicing on. Hope I can get to something more practical soon.-Laurel === Laurel Amberdine> Laurel Amberdine...> Better is a result> called the [drum roll] Hadamard Determinant Theorem: If the columnvectors> in the square matrix have norms k_1, ..., k_n, then |determinant| <= k_1k_2> ... k_n.> I had to look up norm: the square root of the sum of the squares of the> absolute values of the elements of a matrix or the components of a vector.> Is that right?Yes. Norm just means length in this setting. The determinant can bethought of as the volume of a certain n-dimensional parallelogram, and HDTjust says that the volume is <= the product of the lengths of the edges thatde?e it. HDT is easy to prove along those lines, too.Larry === > He said.. ouch! he said ballpark... ouch! BALLPARK ouch!! ouch!!> Stop ? already, ok? ball - ouch! - park, ok??> She said, actually, but that's okay. :)> Oops! Sorry abt that... Blame centuries of culturally-ingrained genderbias still as manifest in western civilisation as elsewhere.... :-) === Some of the readers of this list might be interested in the following bookPierre Baldi, Paolo Frasconi, and Padhraic Smyth, Modeling theISBN: 0-470-84906-1.It covers various models and algorithms for the Web includinggenerative models of networks, IR and machine learning algorithms fortext analysis, link analysis, focused crawling, methods for modelinguser behavior, and for mining Web e-commerce data.1. Mathematical Background - 2. Basic WWW Technologies - 3. Web Graphs -4. Text Analysis - 5. Link analysis - 6. Advanced Crawling Techniques -7. Modeling and Understanding Human Behavior on the Web - 8. Commerceon the Web: Models and Applications - Appendix A MathematicalComplements - Appendix B List of Main Symbols and AbbreviationsThe webpage http://ibook.ics.uci.edu/ contains more details, ahyperlinked bibliography, and a sample chapter in pdf.Paolo Frasconihttp://www.dsi.uni?it/~paolo/ === Jean-Paul Allouche and I are pleased to announce the publicationof our book (AS)^2, also known as Automatic Sequences: Theory, Applications, Generalizationscurrently selling for US $50 on amazon.com.This book is about the class of sequences generated by ?ite automata,their generalizations, and applications to number theory andtheoretical physics. The book has 571+xvi pages, 1600 citations to theliterature, 460 exercises, 85 open problems, 1 musical score, and 2jokes in the index. It will be of interest to number theorists andtheoretical computer scientists.The web page for the book, http://www.math.uwaterloo.ca/~shallit/asas.htmlhas a table of contents and other information.Jeffrey Shallit, Computer Science, University of Waterloo,Waterloo, Ontario N2L 3G1 Canada shallit@graceland.uwaterloo.caURL = http://www.math.uwaterloo.ca/~shallit/ === > The book has 571+xvi pages,?teen imaginary pages? Cool!-- J.97n Fairbairn Jon.Fairbairn@cl.cam.ac.uk === > The book has 571+xvi pages,> ?teen imaginary pages? Cool!But only two jokes, a de?ite downside.xanthian, would have loved to see what a joke looks like in the Complexplane.[Probably a little out of phase.]-- === In sci.math, Kent Paul Dolan The book has 571+xvi pages, ?teen imaginary pages? Cool!> But only two jokes, a de?ite downside.> xanthian, would have loved to see what a joke looks like in the Complex> plane.> [Probably a little out of phase.]> Either that, or highly twisted. :-)-- #191, ewill3@earthlink.net -- we could spin this a number of waysIt's still legal to go .sigless. === [trim self-serving bollocks]Wrong newsgroup sonny. I trimmed out the one ng this ?should' have gone to.-- Patrick Hamlyn posting from Perth, Western AustraliaWindsur?g capital of the Southern HemisphereModerator: polyforms group (polyforms-subscribe@egroups.com) === Could anyone point me to a reference for the noncentralt-distribution, that is its probability density as well as itscumulative distribution function?I did ?d the probability density p(t) -- no original author given --as follows:... (p + t^2) ^ (-(p+1)/2) series ... (2 t^2 / (p + t^2)) ^ (i/2),where p is the degree of freedom and i the index of the series.However, this function is even, although the probability density isnot. A missing sign?Also, I did ?d a cumulative distribution function P(-t0 <= t <= t0)-- likewise no original author given --, which basically is a seriesin the incomplete beta function. That's ?e, but is there a way tocompute the left tail P(-oo < t < t0) and right tail P (t0 < t < +oo)other than by numerical integration? After all the distributionfunction is not even around its maximum.Bjoern === > The question is: if we let U, V be skewsymmetric and let> A = exp(U), B = exp(V) and C = exp(U + V) must (Cx, ABx) = (Cx, BAx)> and must Cx, ABx and BAx be linearly dependent? I don't think so:> We don't in general have C = AB. As long as U and V are small> C will be given by the Campbell-Baker-Hausdorff formula which> starts C = (AB + BA)/2 + lots of increasingly nasty terms> involving iterated commutators. If we ignore the later terms.> we see that for small U and V, C is approximately (AB + BA)/2, so> approximately dependent on AB but BA, but exactly? .... a bit unlikely> I think.Well, let me give you a hand-waving argument (which is what motivated myquestion):In the limit t->0, Z1 = (A^tB^t)^(1/t) should be equal to Z2 =(B^tA^t)^(1/t). That means that as t gets small, I expect that xtransformed by Z1 must approach x transformed by Z2. Where might Ireasonably expect to ?d Zx? Well, half-way between ABx and BAx.I've tested a few numbers, and what's interesting is that Zx _doesn't_ seemto converge to a linear combination of ABx and BAx, in general, although(Zx, ABx) does seem to equal (Zx, BAx).David Turner === If, and only if (for positive integers n), n = a composite >= 6, then:n divides (n-1)!.So, what can be said about the reals or complex numbers x, whereGamma(x)/x is an integer?I guess we could de?e the set of x's which satisfy this ascomposites >= 6, a set which contains noninteger values.And related to Wilsons theorem, we could de?e the set ofcomplex/real x's where:(Gamma(x)+1)/x = integeras a set of Wilson-derived generalized primes.This must be a well-known topic. Anything interesting about this kindof continuation, or is this all useless?(It is useless to say useless, to paraphrase a cliche...)Leroy Quet === ...> So, what can be said about the reals or complex numbers x, where> Gamma(x)/x is an integer?> I guess we could de?e the set of x's which satisfy this as> composites >= 6, a set which contains noninteger values.> And related to Wilsons theorem, we could de?e the set of> complex/real x's where:> (Gamma(x)+1)/x = integer> as a set of Wilson-derived generalized primes....Since for positive reals 2,3,4... we have Gamma(x)=(x-1)!and Gamma is strictly increasing for reals >= 2 , apparentlythere are no other x>=2 (other than integer values) where (Gamma(x)+1)/x is an integer.For complex values of x, in my ignorance it isn't obvious to me whether (Gamma(x)+1)/x is ever an integer. Do youhave some simple examples?-jiw === >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)!>and Gamma is strictly increasing for reals >= 2 , apparently>there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer.??? What do you mean? (Gamma(x)+1)/x = 1, 5, 103, 329891, ..., for x = 3, 5, 7, 11, .... By the Intermediate Value Theorem, theremust be non-integer reals where (Gamma(x)+1)/x takes the other positiveinteger values 2, 3, 4, 6, 7, .... Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >...> So, what can be said about the reals or complex numbers x, where> Gamma(x)/x is an integer?> I guess we could de?e the set of x's which satisfy this as> composites >= 6, a set which contains noninteger values.> And related to Wilsons theorem, we could de?e the set of> complex/real x's where:> (Gamma(x)+1)/x = integer> as a set of Wilson-derived generalized primes.>...>Since for positive reals 2,3,4... we have Gamma(x)=(x-1)!>and Gamma is strictly increasing for reals >= 2 , apparently>there are no other x>=2 (other than integer values) where >(Gamma(x)+1)/x is an integer.On the contrary, there are lots of them. (Gamma(x)+1)/x is 1.75 for x = 4, and is 5 for x=5. There are solutionsfor 2, 3, and 4 which are between 4 and 5.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === >Since for positive reals 2,3,4... we have Gamma(x)=(x-1)!>and Gamma is strictly increasing for reals >= 2 , apparently>there are no other x>=2 (other than integer values) where>(Gamma(x)+1)/x is an integer.> ??? What do you mean? (Gamma(x)+1)/x = 1, 5, 103, 329891, ...,> for x = 3, 5, 7, 11, .... By the Intermediate Value Theorem, there> must be non-integer reals where (Gamma(x)+1)/x takes the other positive> integer values 2, 3, 4, 6, 7, ....Yes, sorry, missed the obvious.-jiw === >...> So, what can be said about the reals or complex numbers x, where> Gamma(x)/x is an integer?> I guess we could de?e the set of x's which satisfy this as> composites >= 6, a set which contains noninteger values.> And related to Wilsons theorem, we could de?e the set of> complex/real x's where:> (Gamma(x)+1)/x = integer> as a set of Wilson-derived generalized primes.>...>Since for positive reals 2,3,4... we have Gamma(x)=(x-1)!>and Gamma is strictly increasing for reals >= 2 , apparently>there are no other x>=2 (other than integer values) where (Gamma(x)+1)/x is an integer.> On the contrary, there are lots of them. (Gamma(x)+1)/x > is 1.75 for x = 4, and is 5 for x=5. There are solutions> for 2, 3, and 4 which are between 4 and 5.For reference (and because I have nothing better to do withmy afternoon):In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0];4.154444.562154.81616I've checked the ?st one in Mathematica, and I presume(read ?hope') that the others are right as well.-- Dave TaylorOh yeah? Well I'll build my own theme park! With Blackjack, andhookers ... in fact, forget the theme park![Futurama] === > If, and only if (for positive integers n), n = a composite >= 6, then:> n divides (n-1)!.> So, what can be said about the reals or complex numbers x, where Gamma(x)/x is an integer?> I guess we could de?e the set of x's which satisfy this as> composites >= 6, a set which contains noninteger values.> And related to Wilsons theorem, we could de?e the set of> complex/real x's where: (Gamma(x)+1)/x = integer> as a set of Wilson-derived generalized primes.> This must be a well-known topic. Anything interesting about this kind> of continuation, or is this all useless?> (It is useless to say useless, to paraphrase a cliche...)> A few things:1) Even considering only x = positive integers, x = 1 satis?s both equations above. So, in a way, 1 is both a composite >=6 and a Wilson-derivedgeneralized prime.(snicker.)2) What if we, for x = composite, allow the integers (that theequations above are equal to) to be Gaussian?2.5) Is (Gamma(x)+1)/x, if we let x = a Gaussian (integer) prime, an (real orGaussian)integer?3) If we take the maximum real root, x(n), of(Gamma(x)+1)/x = n,then, what can be said about the Wilson-zeta function:WZ(y) = product{n=1 to oo} 1/(1 -1/x(n)^y) ?Does the WZ() product converge for any y?Leroy Quet === >For complex values of x, in my ignorance it isn't obvious >to me whether (Gamma(x)+1)/x is ever an integer. Do you>have some simple examples?For example, (Gamma(x)+1)/x = 3 for x = 5.7584660619435256965 + 3.9675461457510952995 iapproximately.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > (Gamma(x)+1)/x = integer> For reference (and because I have nothing better to do with> my afternoon):> In[27]:= WilsonRoot[2.0]; WilsonRoot[3.0]; WilsonRoot[4.0];> 4.15444> 4.56215> 4.81616> I've checked the ?st one in Mathematica, and I presume> (read ?hope') that the others are right as well.All pucker:(21:50) gp > p100 realprecision = 105 signi?ant digits (100 digits displayed)(21:51) gp > solve(x=4.1,4.2,(gamma(x)+1)/x-2)%8 = 4.154439060271062694827261583664629151822185286325427363404753 300368340020397814797012023001544277754(21:51) gp > solve(x=4.6,4.2,(gamma(x)+1)/x-3)%9 = 4.562151433950973261499668111682367676594713372717728076006767 266218451425330181354003523067887979890(21:51) gp > solve(x=4.6,4.9,(gamma(x)+1)/x-4)%10 = 4.816164962269824928459405177621703579279411919117658752833716 798354235922037604118825483457603002219(21:52) gp > solve(x=5,5.2,(gamma(x)+1)/x-6)%13 = 5.143587135664545182467227708447357783039049810591881586494699 613779238822779092252706929290717662694Phil === >For complex values of x, in my ignorance it isn't obvious >to me whether (Gamma(x)+1)/x is ever an integer. Do you>have some simple examples?>For example, (Gamma(x)+1)/x = 3 for > x = 5.7584660619435256965 + 3.9675461457510952995 iBut... is anything known about Leroy's question?I'm asking becausesome years ago I formulated the very same question and I'm verycurious about it.To be fair I'm only thinking to its second part i.e. that involvinggeneralized primes as those numbers satisfying the obviouscontinuous version of Wilson's (necessary and suf?ient) conditionfor primality.Well, maybe something interesting can be said about a suitablefunction having those primes as zeroes or poles...Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc === For complex values of x, in my ignorance it isn't obvious to me whether (Gamma(x)+1)/x is ever an integer. Do youhave some simple examples?>For example, (Gamma(x)+1)/x = 3 for > x = 5.7584660619435256965 + 3.9675461457510952995 i> But... is anything known about Leroy's question?I'm asking because> some years ago I formulated the very same question and I'm very> curious about it.> To be fair I'm only thinking to its second part i.e. that involving> generalized primes as those numbers satisfying the obvious> continuous version of Wilson's (necessary and suf?ient) condition> for primality.> Well, maybe something interesting can be said about a suitable> function having those primes as zeroes or poles... You mean like sin(pi (Gamma(x)+1)/x)?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 === > Well, maybe something interesting can be said about a suitable> function having those primes as zeroes or poles...>You mean like sin(pi (Gamma(x)+1)/x)?Well, this is exactly the function I had considered in my youth,along with the *real* functionsin^2(pi x)+sin^2(pi(Gamma(x)+1)/x).But now I'm more sort-of-thinking of something likeprod_p 1/(1-p^{-x})with p ranging over the zeroes of the function you supplied...provided that the product does make sense.Michele-- >It's because the universe was programmed in C++.No, no, it was programmed in Forth. See Genesis 1:12:And the earth brought Forth ...- Robert Israel on sci.math, thread Why numbers? === I have to solve the following problem:Assume that A and B are square matrixes with||A||=mand||B||=dand that(I-A)and(I+B)are non-singular, show that:||((I-A)^-1)B((I+B)^-1)<=d/(1-d-m(1+d))wherem<(1-d)/(1+d )X=((I-A)^-1)B((I+B)^-1)Premultiply then both sides ?st by (I-A) and then postmultiply both sidesby (I+B).Whatever I try in my result I have a wrong sign in the denominator and Ihave a wrong relation symbol.I would be very grateful for some hints how can I solve this task.Michaelp.s. I put the task in the WWW in a more convenient manner:http://www.rdg.ac.uk/~sip02mm2/matrix.html === > I have to solve the following problem:> Assume that A and B are square matrixes with> ||A||=m> and> ||B||=d> and that> (I-A)> and> (I+B)> are non-singular, show that:> ||((I-A)^-1)B((I+B)^-1)<=d/(1-d-m(1+d))> where> m<(1-d)/(1+d)> X=((I-A)^-1)B((I+B)^-1)> Premultiply then both sides ?st by (I-A) and then postmultiply bothsides> by (I+B).> Whatever I try in my result I have a wrong sign in the denominator and I> have a wrong relation symbol.> I would be very grateful for some hints how can I solve this task.> Michael> p.s. I put the task in the WWW in a more convenient manner:> http://www.rdg.ac.uk/~sip02mm2/matrix.html> === Would some kind soul possibly post a complete solution to thisproblem, as none of us on the course can fathom it and our exam istomorrow... *gulp* :)Darren. === > Would some kind soul possibly post a complete solution to this> problem, as none of us on the course can fathom it and our exam is> tomorrow... *gulp* :)> Darren.My reply was essentially complete. Just follow through with a detail ortwo.Here it is again:------------------------This question occurred to be the other day; it's been a while since Itook calculus, and I can't come up with an answer.Does there exist a function from R to R that is nowhere continuous,but that is de?ed everywhere and limited everywhere (i.e. that has a?ite limit at each real number)?Foghorn Leghornmoc.enecswen @ nrohgof === A function f(x) is said to be continuous at x=a if 1a)limx-->a- exist,1b)limx-->a+ exists, 2) these two limits are = to one another ,say k. and3)f(a)=k.Now you say that your function,f(x), is limited everywhere.Let's consider xat a. So limx-->a+=lim x-->a-(=k). If f(a)=k then of course we can't saythat f(x) is continuous nowhere. What type of function violates justcondition 3 of the de?ition of a continuous function? Functions that haveremovable discontinuity come to mind.Now the real question is whether or nota function can have an 00 number of removable discontinuities so thefunction is nowhere continuous. I think that you can come up with one.> This question occurred to be the other day; it's been a while since I> took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere continuous,> but that is de?ed everywhere and limited everywhere (i.e. that has a> ?ite limit at each real number)?> Foghorn Leghorn> moc.enecswen @ nrohgof === >This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a>?ite limit at each real number)?>Foghorn Leghorn>moc.enecswen @ nrohgof>Hmm... I don't know, but what about The Dirichlet Function: f(x) = 1 [if x is rational] = 0 [if x is irrational]Plot some points... It looks like two straight, horizontal lines...Obviously, it's de?ied everywhere, but I don't know if it'scontinuous and limited (sic). Maybe someone else does.cheerio, === > This question occurred to be the other day; it's been a while since I> took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere continuous,> but that is de?ed everywhere and limited everywhere (i.e. that has a> ?ite limit at each real number)?Consider the functionf:R->Rwhere :(a) f(x) = 0 if x is irrational, and(b) if x is rational and we de?e: denominator(x):= the least integer n>=1 such that n*x is an integer, and then let f(x):= 1/denominator(x) .I think Prof. Herman Rubin recently mentioned this functionin another thread.David Bernier === > Does there exist a function from R to R that is nowhere continuous,> but that is de?ed everywhere and limited everywhere (i.e. that has a> ?ite limit at each real number)?> Consider the function> f:R->R> where :> (a) f(x) = 0 if x is irrational, and> (b) if x is rational and we de?e:> denominator(x):= the least integer n>=1 such that n*x is an integer,> and then let f(x):= 1/denominator(x) .The f above _is_ continuous at all irrational points in R, right?David Bernier === > Does there exist a function from R to R that is nowhere continuous,> but that is de?ed everywhere and limited everywhere (i.e. that has a> ?ite limit at each real number)?> Consider the function> f:R->R> where :> (a) f(x) = 0 if x is irrational, and> (b) if x is rational and we de?e:> denominator(x):= the least integer n>=1 such that n*x is an integer,> and then let f(x):= 1/denominator(x) .> The f above _is_ continuous at all irrational points in R, right?> David BernierRight. As for the original question, I have a feeling that no such functionexists. I wonder if you could prove this using a Baire category argument,maybe something like the proof that there is no function continuous only onthe rationals. === >This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a>?ite limit at each real number)?>Foghorn Leghorn>moc.enecswen @ nrohgof>Hmm... I don't know, but what about The Dirichlet Function:>f(x) = 1 [if x is rational]> = 0 [if x is irrational]>Plot some points... It looks like two straight, horizontal lines...>Obviously, it's de?ied everywhere, but I don't know if it's>continuous and limited (sic). Maybe someone else does.That function is obviously nowhere continuous. But italso obviously has a limit at no point, so it doesn't help.>cheerio,David C. Ullrich === >This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a>?ite limit at each real number)?No. If f has a limit at every point then there is a dense set ofpoints at which it is continuous:Say the variation of f on S is V(f, S) = sup {|f(x) - f(y)| : x, y in S}.The fact that f has a limit at 0 shows that there is a closedinterval I_1 such that V(f, I_1) < 1. Now the fact that f hasa limit at the midpoint of I_1 shows that there is a closedinterval I_2, contained in the _interior_ of I_1, such thatV(f, I_2) < 1/2. Repeat. Now if x is in the intersection ofthe I_n it foilows that f is continuous at x.(The fact that I_{n+1} is in the interior of I_n shows thatx is in the interior of I_n, and |f(x) - f(y)| < 1/n for ally in I_n.)>Foghorn Leghorn>moc.enecswen @ nrohgofDavid C. Ullrich === Foghorn Leghorn http://mathforum.org/discuss/sci.math/m /523425/523425> This question occurred to be the other day; it's been a while> since I took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere> continuous, but that is de?ed everywhere and limited> everywhere (i.e. that has a ?ite limit at each real number)?Given a function f: R --> R and a real number x in R, letL(f,x) be the limit of f(x') as x' --> x, when this limitexists. Then the set (I'm using /= for not equal) P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}is countable. Thus, what you're looking for doesn't exist by along shot (a function f such that L(f,x) exists for each x in Rand P(f) = R). On the other hand, I'm pretty sure that for eachcountable set Z there exists a function f such that L(f,x) existsfor each x in R and P(f) = Z (i.e. no restriction besidescountable can be proved, even when L(f,x) exists for each x in R).One way to prove this is to ?st prove for each n = 1, 2, 3, ...that P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n}is an isolated set, and hence each P(f,n) must be countable. Thisimmediately implies that P(f) is countable, since P(f) = UNION(n=1 to oo) P(f,n).A much stronger version of this result holds, by the way. Given afunction f: R --> R and a real number x in R, let C(f,x) be the setof all extended real numbers y (i.e. y can be -oo or +oo) suchthat there exists a sequence {x_k} with x_k --> x and f(x_k) --> y.In other words, C(f,x) is the set of all numbers (including -oo and+oo) that can be obtained as the limit of some sequence convergingto x. Then for each f and x, C(f,x) is a nonempty closed set inthe extended real line, the minimum number in C(f,x) islim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) islim-sup(x'--> x) of f(x'). Note that L(f,x) exists means thatC(f,x) = {y} for some y in R.Then the set Q(f) = {x in R: f(x) does not belong to C(f,x)}is countable. This can be proved in the same way as the result above.First, a bit of notation. If y belongs to R and E is a subset of R,let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. Now de?e Q(f,n) for each n = 1, 2, 3, ... by Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}.Then each Q(f,n) winds up being an isolated set, and hence Q(f)is countable since Q(f) = UNION(n=1 to oo) Q(f,n).Dave L. Renfro === >This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a?ite limit at each real number)?> No. If f has a limit at every point then there is a dense set of> points at which it is continuous:> Say the variation of f on S is> V(f, S) = sup {|f(x) - f(y)| : x, y in S}.> The fact that f has a limit at 0 shows that there is a closed> interval I_1 such that V(f, I_1) < 1. Now the fact that f has> a limit at the midpoint of I_1 shows that there is a closed> interval I_2, contained in the _interior_ of I_1, such that> V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of> the I_n it foilows that f is continuous at x.> (The fact that I_{n+1} is in the interior of I_n shows that> x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all> y in I_n.)Very nice argument. Compact and to the point :-) === > Foghorn Leghorn http://mathforum.org/discuss/sci.math/m/523425/523425> This question occurred to be the other day; it's been a while> since I took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere> continuous, but that is de?ed everywhere and limited> everywhere (i.e. that has a ?ite limit at each real number)?> Given a function f: R --> R and a real number x in R, let> L(f,x) be the limit of f(x') as x' --> x, when this limit> exists. Then the set (I'm using /= for not equal)> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}> is countable. Thus, what you're looking for doesn't exist by a> long shot (a function f such that L(f,x) exists for each x in R> and P(f) = R). On the other hand, I'm pretty sure that for each> countable set Z there exists a function f such that L(f,x) exists> for each x in R and P(f) = Z (i.e. no restriction besides> countable can be proved, even when L(f,x) exists for each x in R).> One way to prove this is to ?st prove for each n = 1, 2, 3, ...> that> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n}> is an isolated setDave, can you ?l this in? I don't see why it's an isolated set.Sorry to be so dense :-) === An attempt at clarifying for calculus students.>This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a>?ite limit at each real number)?>No. If f has a limit at every point then there is a dense set of>points at which it is continuous:>Say the variation of f on S is> V(f, S) = sup {|f(x) - f(y)| : x, y in S}.>The fact that f has a limit at 0 shows that there is a closed>interval I_1 such that V(f, I_1) < 1. Now the fact that f has>a limit at the midpoint of I_1 shows that there is a closed>interval I_2, contained in the _interior_ of I_1, such that>V(f, I_2) < 1/2. Repeat.>Which you can do, because there's a point in I_2 (possibly different from the point 0 above) where f has a limit.> Now if x is in the intersection of the I_n it foilows that f is continuous at x.>And there is at least one x in the intersection because the sets are closed (and bounded). I'm at a loss as to how to explain this in a short note to someone whose background doesn't extend beyond calculus. So I'll beg off for today (but I hope you'll be interested enough to make me [or someone else] do it Monday.>(The fact that I_{n+1} is in the interior of I_n shows that x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all y in I_n.)So now you've got a point x such that f is continuous at x. Now, pick any point z, and any tolerance level epsilon. By simply picking the ?st I_1 to be within the interval (z-epsilon, z+epsilon), we can ?d a point of continuity of f that is close enough to z. Since we can do this for any z, the set of points of continuity of f are arbitrarily close to any point of R. That's the de?ition of a dense set.Jon Miller === >This question occurred to be the other day; it's been a while since I>took calculus, and I can't come up with an answer.>Does there exist a function from R to R that is nowhere continuous,>but that is de?ed everywhere and limited everywhere (i.e. that has a>?ite limit at each real number)?> No. If f has a limit at every point then there is a dense set of> points at which it is continuous:> Say the variation of f on S is> V(f, S) = sup {|f(x) - f(y)| : x, y in S}.> The fact that f has a limit at 0 shows that there is a closed> interval I_1 such that V(f, I_1) < 1. Now the fact that f has> a limit at the midpoint of I_1 shows that there is a closed> interval I_2, contained in the _interior_ of I_1, such that> V(f, I_2) < 1/2. Repeat. Now if x is in the intersection of> the I_n it foilows that f is continuous at x.> (The fact that I_{n+1} is in the interior of I_n shows that> x is in the interior of I_n, and |f(x) - f(y)| < 1/n for all> y in I_n.)>Very nice argument. Compact and to the point :-)heh-heh.David C. Ullrich === >Foghorn Leghorn http://mathforum.org/discuss/sci.math/ m/523425/523425> This question occurred to be the other day; it's been a while> since I took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere> continuous, but that is de?ed everywhere and limited> everywhere (i.e. that has a ?ite limit at each real number)?>Given a function f: R --> R and a real number x in R, let>L(f,x) be the limit of f(x') as x' --> x, when this limit>exists. Then the set (I'm using /= for not equal)> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}>is countable. Well duh. I considered conjecturing that the worddense in the result I gave could be strengthened,but I didn't want to ?ure out by how much.Has countable complement is much strongerthan what I would have guessed.>Thus, what you're looking for doesn't exist by a>long shot (a function f such that L(f,x) exists for each x in R>and P(f) = R). On the other hand, I'm pretty sure that for each>countable set Z there exists a function f such that L(f,x) exists>for each x in R and P(f) = Z (i.e. no restriction besides>countable can be proved, even when L(f,x) exists for each x in R).This is clear - if Z = {x_1, ...} let f(x_n) = 1/n and f(x) = 0 for other x.>One way to prove this is to ?st prove for each n = 1, 2, 3, ...>that> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n}>is an isolated set, and hence each P(f,n) must be countable. This>immediately implies that P(f) is countable, since> P(f) = UNION(n=1 to oo) P(f,n).>A much stronger version of this result holds, by the way. Given a>function f: R --> R and a real number x in R, let C(f,x) be the set>of all extended real numbers y (i.e. y can be -oo or +oo) such>that there exists a sequence {x_k} with x_k --> x and f(x_k) --> y.>In other words, C(f,x) is the set of all numbers (including -oo and>+oo) that can be obtained as the limit of some sequence converging>to x. Then for each f and x, C(f,x) is a nonempty closed set in>the extended real line, the minimum number in C(f,x) is>lim-inf(x'--> x) of f(x'), and the maximum number in C(f,x) is>lim-sup(x'--> x) of f(x'). Note that L(f,x) exists means that>C(f,x) = {y} for some y in R.>Then the set> Q(f) = {x in R: f(x) does not belong to C(f,x)}>is countable. This can be proved in the same way as the result above.>First, a bit of notation. If y belongs to R and E is a subset of R,>let DIST(y,E) be the extended real number inf{|y-e|: e is in E}. >Now de?e Q(f,n) for each n = 1, 2, 3, ... by> Q(f,n) = {x in R: DIST[f(x), C(f,x)] > 1/n}.>Then each Q(f,n) winds up being an isolated set, and hence Q(f)>is countable since> Q(f) = UNION(n=1 to oo) Q(f,n).>Dave L. RenfroDavid C. Ullrich === > Foghorn Leghorn http://mathforum.org/discuss/sci.math/m/523425/523425> This question occurred to be the other day; it's been a while> since I took calculus, and I can't come up with an answer.> Does there exist a function from R to R that is nowhere> continuous, but that is de?ed everywhere and limited> everywhere (i.e. that has a ?ite limit at each real number)?> Given a function f: R --> R and a real number x in R, let> L(f,x) be the limit of f(x') as x' --> x, when this limit> exists. Then the set (I'm using /= for not equal)> P(f) = {x in R: L(f,x) exists and L(f,x) /= f(x)}> is countable. Thus, what you're looking for doesn't exist by a> long shot (a function f such that L(f,x) exists for each x in R> and P(f) = R). On the other hand, I'm pretty sure that for each> countable set Z there exists a function f such that L(f,x) exists> for each x in R and P(f) = Z (i.e. no restriction besides> countable can be proved, even when L(f,x) exists for each x in R).> One way to prove this is to ?st prove for each n = 1, 2, 3, ...> that> P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n}> is an isolated set>Dave, can you ?l this in? I don't see why it's an isolated set.>Sorry to be so dense :-)Let epsilon = 1/(2n) in the de?ition of L(f,x) exists...David C. Ullrich === [snip] P(f,n) = {x in R: L(f,x) exists and | L(f,x) - f(x) | > 1/n}> is an isolated set>Dave, can you ?l this in? I don't see why it's an isolated set.>Sorry to be so dense :-)> Let epsilon = 1/(2n) in the de?ition of L(f,x) exists...> David C. Ullrich === Simple question.Let A' denote the complement of a set A, thenOxtoby in Measure and Category says A is nowhere dense iff A' contains adense open set,..I understand the condition that A' contains a dense set, since A' is denseand A' contains itself. I don't understand the condition of the dense subsethaving to be _open_. Why is this condition of having to be open necessary?Matthew Breneman === >Simple question.>Let A' denote the complement of a set A, then>Oxtoby in Measure and Category says A is nowhere dense iff A' contains a>dense open set,..>I understand the condition that A' contains a dense set, since A' is dense>and A' contains itself. I don't understand the condition of the dense subset>having to be _open_. Why is this condition of having to be open necessary?Clearly A' is dense is not nearly strong enough: the set of irrationalsis really really dense in R, even though its complement is dense :).An equivalent de?ition that may be helpful is this: A is nowhere denseiff it is not dense on any open set, that is cl(A) (the closure of A)contains no open neighbourhoods. The motivation for this de?itionshould be pretty clear.It's quite easy to see the equivalence: if cl(A) has empty interior, thenits complement is dense. But the complement of cl(A) is int(A'), so A'contains a dense open set, namely its own interior. -- Erick === >Let A' denote the complement of a set A, then>Oxtoby in Measure and Category says A is nowhere dense iff A' contains a>dense open set,..>I understand the condition that A' contains a dense set, since A' is dense>and A' contains itself. I don't understand the condition of the dense subset>having to be _open_. Why is this condition of having to be open necessary?Saying that A is nowhere dense means that there is no nonempty open set inwhich A is dense. So every nonempty open set U must have a nonempty opensubset V(U) disjoint from A. The union of these V(U) for all nonempty open U is a dense open set contained in A'.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > Let A' denote the complement of a set A, then> Oxtoby in Measure and Category says A is nowhere dense iff A' contains a> dense open set,..>Theorem: For a topological space S SA nowhere dense iff some open dense U with U subset AProof:If SA nowhere dense: S = S - int cl SA = cl int A; int A subset A; let U = int AConversely if some open dense U with U subset A: cl U = S int cl SA subset int cl SU = int SU = int(cl U - U) = int cl U - cl U = nulset-- Another theorem in same vein A nowhere dense iff cl A = bd cl A-- Previous thoughts upon nowhere denseDef: A somewhere dense when A not nowhere denseTheorem: A somewhere dense iff some open nonnul U with cl U = cl U/A iff some open nonnul U with U subset cl A(The second equivalence gleaned from this thread.)---- === Just wondering if anybody knows an approach to the following:An n-prime set is a set of n integers such that each element can becombined in some way with 1,2,3...(n-1)other elements such that their sum is a prime. Can you form an n-primeset for any n ?It seems like the answer should be no, because the primes are such anintractable set. But for any n I've tried, I've been able to produce aset.Any thoughts would be appreciated.David === >Just wondering if anybody knows an approach to the following:>An n-prime set is a set of n integers such that each element can be>combined in some way with 1,2,3...(n-1)>other elements such that their sum is a prime. Can you form an n-prime>set for any n ?I'm not sure what you mean by 1,2,3...(n-1) other elements. Please state it more clearly. And perhaps it would help if you gave an examplefor, say, n=5.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === An example of such a set for n=5 would be 22111, for we have1+2=3 (combines 1 (or 2) with some other element to make a prime)1+2+2=5 (combines 1 (or 2) with two other elements to make a prime)1+1+1+2=5 1+1+1+2+2=7Of course, this is a very simple example because we don't need todisplay any more than one sum for each length. === Need help to complete the following number sequence:1 4 4 20 23 15 16 15 9 14 20 23 5 14 20 25 6 9 22 5 20 15 20 8 5 14 1518 20 8 9 14 7 9 14 1 16 16 12 5 29 1 14 4 25 15 21 1 18 5 7 15 15 420 15 7 15Any help? Thanx === A=1===Consider the following ODE system:y_1^(n_1) = f_1 [x, y_1, y_2, ..., y_k, ..., y_1^(n_1 - 1), ..., y_k^(n_1 -1)]y_2^(n_2) = f_2 [x, y_1, y_2, ..., y_k, ..., y_1^(n_2 - 1), ..., y_k^(n_2 -1)]... ... ... ...... ... ... ...... ... ... ...y_k^(n_k) = f_2 [x, y_1, y_2, ..., y_k, ..., y_1^(n_k - 1), ..., y_k^(n_k -1)](y^(w) means the w-th derivative of y).Could you shom me (in detail) how to reduce it to the following of ?storderY_1' = F_1 [x, Y_1, Y_2, ..., Y_n]Y_2' = F_2 [x, Y_1, Y_2, ..., Y_n]... ... ... ...... ... ... ...... ... ... ...Y_n' = F_n [x, Y_1, Y_2, ..., Y_n](where n= max{n_1, n_2, ..., n_k) ????? === The problem is, if we have the number 1^.5, is it 1 or -1. Going evenfarther, 1^.25 could be 1, -1, i, or -i. How do you ? this? Simple,convert all numbers to polar coordinates. Now, if we say (1<0)^.5, theanswer is 1^0 (0*2 = 0). If we try (1<360), the answer 1<180 (180*2 = 360)which is -1. A number like (1<453)^.5 would have an answer of 1<226.5or -.68835 + -.7253 i as opposed to getting the other possible without polarcoordinates as 1<406.5 or .68835 + .7253 i.With this logic, (1<0)^.25 becomes 1 (1<0)^.5 is 1, (1<360)^.25 is i,(1<360)^.5 is -1, (1<720)^.25 is -1, (1<720)^.5 is 1, (1<1080)^.25 is -i,and (1<1080)^.5 is 1Their are two small problems with this.The ?st problem is when you add two polar vectors together head to tailthat are on or over 360 degree apart. Say we add the numbers 1<15 and 1<375together. Under currently accepted mathematics, we would mod 375 by 360then add them together to get 2<15 degrees. This works perfectly foraddition and subtraction, but what if we had the problem (1<15 +1<375)^.25. I'll tell you right now, the solution is not (2^.5)<7.5,although that is what nearly every single calculator in the world would giveyou.The second problem is the conversion between normal number system andpolar coordinates. The number 5<390 is 5*cos(390) + 5*sin(390) i. Thisconverts it to nice angles, which are 0, 90, 180, and 270 so it can beeasily added to something like 3<0 without getting a geometry book out.Unfortunately, when the number is converted back to the normal numbersystem, 5<30 will result, losing all additional information about the numberof revolutions that have occurred. === > The problem is, if we have the number 1^.5, is it 1 or -1. Going even> farther, 1^.25 could be 1, -1, i, or -i. How do you ? this?CASs like Maple and Mathematica de?e a principal value and alwayscompute with it. 1^(1/2) = 1, 1^(1/4) = 1, etc.But also (-1)^(1/3) is complex, not real. So some beginners areconfused by it. === > The problem is, if we have the number 1^.5, is it 1 or -1. Going even> farther, 1^.25 could be 1, -1, i, or -i. How do you ? this?> CASs like Maple and Mathematica de?e a principal value and always> compute with it. 1^(1/2) = 1, 1^(1/4) = 1, etc.> But also (-1)^(1/3) is complex, not real. So some beginners are> confused by it.No, you are wrong. (1<0)^1/4 = 1 always. Since we are dealing withpolar coordinates, this theorem no longer applies. Via. -1^.5 = 1<180= i. You made a very common mistake and do not understand theproblem.Basically, with that theorem (1 1<01<45 -> 1<22.51<90 -> 1<451<135 -> 1<67.51<180 -> 1<901<225 -> 1<112.51<270 -> 1<1351<315 -> 1<157.51<360 -> ? 1<0 ? != 1<180. Uh oh, your theorem just failed. Dare you to graph it with your theorem.A = 1<360 is like saying A went around the circle one time and takingthe square root would be asking where would I be one half revolutionago. B = 1<0 is like saying I have not had any revolutions yet. Theymay be at the same position but in theory, they are two numbers,unless you mod them by 360 like you did and said they are exactly thesame.They are not. PS: Don't treat people like they are idiots and don't have aneducation because you think they are wrong. You just haven'tthought about it long enough to realize the de?ition is wrong injust a few cases. It is bad judgement. === > How do you do induction on the reals?> Given a real, what is the next real?> Back around 1900 or so, the Heine-Borel Theorem was viewed as acontinuous form of induction. So instead of the situation withnatural numbers (if P holds for n, then it holds for the next integer)we have something like: if P holds up to a value x, then it holds forsome small neighborhood of x, so it holds up to x+epsilon, for someepsilon>0, possibly depending on x.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > I have just ?ished a course in Representation Theory of ?ite> groups, ending with the classi?ation of all representations of the> symmetric group (using the young diagrams and all that). (I am a> (pure)math honours undergraduate having completed 2 years.)> My questioin is -> What are the applications of representation theory of groups (in> general)? What are the reasons for studying it apart from its beauty?> It can't be denied that this theory is beautiful enough to merit a> study solely for enjoying the same, but whichever book I go to tells> me that it ?ds applications as varied and far as physics and quantum> chemistry. Prefaces after prefaces say that most mathematicians would> at some point or the other come accross it. I can't readily see that.> Studying representations of a ?ite group does seem to magically> deduce a few properties abt the group itself (for instance, the> Burnside's p^a.q^b theorem).> But, all said and done, I really dont know how or where these> things could get applied. And, for instance, how do they show up> their head in physics?The groups whose representations are important in modern physics aremainly, but not solely, Lie groups. A Lie group is an in?ite groupthat is also a topological space and a differentiable manifold, withthe operations of group multiplication and inverse taking bothdifferentiable.As Greg said, representations of various ?ite groups are useful forclassifying the symmtries of crystals. Interestingly, representationsof the symmetric group, including Young diagrams, are useful forconstructing representations of Lie groups on tensor spaces. The repsof the symmetric group give the symmetries of the (?ite number of)tensor indices involved.As a simple example, consider the function T : R^3 -> R that gives thetemperature at all points in (normal, 3-dim) space, where one point inspace is chosen as the origin. Then x in R^3 represents a posiition inspace with respect to the origin and 3 orthogonal axes, and T(x) isthetemperature at position x (with respect to the chosen origin andaxes).If we chose the same origin for space, but chose a set of orthogonalaxes that are rotated with respect to the original axes, then we geta new temperature function T' : R^3 -> R.Now, the set of rotations in 3-space forms a group that is isomorphictoSO(3), the group of real 3x3 (this what the 3 means) that havedeterminant +1 (this is what the S, means, i.e., they are Special),andare Orthogonal (this is what the O means).It turns out that T'(x) = T( g^(-1) x). This gives a representationon the in?ite-dimensional vector space of functions that map R^3 toR.Other examples of important matrix Lie groups are SU(2), SU(3), U(1),etc. SU(2) is the group of 2x2 complex Unitary matrices thatdeterminant+1. I think you can see the pattern.A fairy introductory test, written in a style amenable to puremathies,that gives myriad applications of reps of both ?ite and Lie groupsisGroup Theory and Physics by S. Sternberg. A pure math book thatgivesa readable pure math introduction to the representation theory of Liegroups and Lie algebras is Representation Theory A First Course W.Fulton and J. Harris. These are both excellent books. The second bookcontains no applications to physics, even though much material, withnotational changes, is directly applicable to physics.I am very interested in the applications of group representationtheoryin physics. Here are some links to some old posts of mine onapplications. These posts require moderate background in physics, ormath, or both, and all have signi?ant typos.Georgehttp://groups.google.ca/groups?hl=en&lr=&ie=UTF-8 &selm=3C990919.D50FED01%40yahoo.com&rnum=5http:// groups.google.ca/groups?q=rotation+author:George+author:Jones &hl=en&lr=&ie=UTF-8&scoring=d&selm= 7c74e370.0305120951.21c52059%40posting.google.com&rnum=1http ://groups.google.ca/groups?q=quark+author:George+author:Jones &hl=en&lr=&ie=UTF-8&scoring=d&selm=3ADA0F4F.97ACB477% 40uvi.edu&rnum= 51234567890123456789012345678901234567890123456789012345678901 23456789012 === > I have just ?ished a course in Representation Theory of ?ite> groups, ending with the classi?ation of all representations of the> symmetric group (using the young diagrams and all that). (I am a> (pure)math honours undergraduate having completed 2 years.) My questioin is -> What are the applications of representation theory of groups (in> general)? What are the reasons for studying it apart from its beauty?> It can't be denied that this theory is beautiful enough to merit a> study solely for enjoying the same, but whichever book I go to tells> me that it ?ds applications as varied and far as physics and quantum> chemistry. Prefaces after prefaces say that most mathematicians would> at some point or the other come accross it. I can't readily see that.> Studying representations of a ?ite group does seem to magically> deduce a few properties abt the group itself (for instance, the> Burnside's p^a.q^b theorem).> But, all said and done, I really dont know how or where these> things could get applied. And, for instance, how do they show up> their head in physics?> Anshul> A S K,> your questions got me thinking:> You have ?ished a course on group representations, and you are asking > very natural questions. In fact, I would hope that most students taking > the same course would want to know about the origins of representation > theory and its uses.> So let me ask you: Did you put your questions to the professor who > taught the course? If so, what did (s)he say?> You don't have to reveal the name of the university and the professor. I > am just curious to know what answer you get when you ask those who teach > the subject.Why should a pure math course give physical applications?Why should a pure math prof know physical applications?George === > You have ?ished a course on group representations, and you are asking > very natural questions. In fact, I would hope that most students taking > the same course would want to know about the origins of representation > theory and its uses.> So let me ask you: Did you put your questions to the professor who > taught the course? If so, what did (s)he say?> You don't have to reveal the name of the university and the professor. I > am just curious to know what answer you get when you ask those who teach > the subject.>Why should a pure math course give physical applications?To engage and edify the students. If you have homework problems that work with the material anyway, some effort should be made, when possible, to relate it to something useful.>Why should a pure math prof know physical applications?Because he's a math prof teaching stuff that people use in physical applications.-- Is that plutonium on your gums?Shut up and kiss me! -- Marge and Homer Simpson === > A S K, your questions got me thinking: > You have ?ished a course on group representations, and you are > asking very natural questions. In fact, I would hope that most > students taking the same course would want to know about the > origins of representation theory and its uses. > So let me ask you: Did you put your questions to the professor who > taught the course? If so, what did (s)he say? > You don't have to reveal the name of the university and the > professor. I am just curious to know what answer you get when you > ask those who teach the subject. > Why should a pure math course give physical applications? >A pure math course doesn't have to give physical applications (and Ididn't state it should), but pointing out connections and relationshipsto anything familiar helps in learning. Students learn new material better if they can relate it to something they already know. It has to do with the structure of the brain and learning mechanisms.That something familiar could be in other areas of mathematics, inphysics, or in their daily life. Seeing multiple connections helpsstudents develop intuition, which is just as important for theirmathematical ability as formal derivation of results. > Why should a pure math prof know physical applications?I don't know, maybe (s)he is curious? Maybe (s)he wants to know wherethe subject came from, and why the researchers who started the areathought it was important?I know there are pure math profs who don't care. That's ?e. It justmeans that students like A S K have to take their questions elsewhere.Nemo === > [...]>their head in physics?>You must know by now that transformations form a group. Something like a >Dirac spinor isn't a vector, but you can ?ure out how it transforms >under boosts or rotations by starting with the properties of the Lorentz >group.> Ya, invertible vector space transformations do form a group. I don't> know what a Dirac Spinor is. I don't know what the Lorentz group is> either. (I am a pure math undergrad:|)> It's not a bad idea to acquire the habit of Googling for terms you don'tunderstand. As for spinors, there are two trails of references to lookfor: function, a solution to the Schr.9adinger equation. Its standard interpretation is as a probability density function; The question of the behavior of this function to the observation that two types of symmetry may take place: even [swapping preserves the wave function] and odd [swapping reverses the sign of the wave function]. Solutions of the latter sort lead to the consideration of a phenomenon requiring a 4 pi rotation to return to its original state; that idea turns out to coincide with: 2. The so-called spinor group Spin(n), a particular double cover of the special orthogonal group (SO(n) is the Lie of proper rotations of R^n, i.e., the set of orthogonal matrices of determinant +1). See E. Cartan, The Theory of Spinors, originally published in 1966 by Hermann; currently (1981), Dover Press.The relevance of Dirac's name here is due largely to the DiracEquation, a relativistic spin-1/2 wave equation; the solutions tothe Dirac Equation are spinors in that their transformationproperties are given by representations of the spinor group Spin(3),often referred to as the so-called spin representations of SO(3).The second question, as to the Lorentz group, there are generalizationsof the orthogonal groups O(n) as those matrices that preserve thequadratic form x1^2 + x2^2 + ... + xn^2; if A is a non-degeneratenxn symmetric matrix, one can de?e O(n;A) as the set of all nxnmatrices K that preserve the inner product de?ed by the matrix A: x'Ax = (Kx)'A(Kx) for all n-vectors x,or K'AK = A.Special relativity determines that the quantity x^2 + y^2 + z^2 - (ct)^2is conserved under transformations between inertial coordinate systems.By suitable normalization, one can reduce this to the requirement thatthe matrix [ 1 0 0 0 ] [ 0 1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 -1 ]The group of 4x4 matrices that preserve this form is called O(3,1). Thesubgroup of elements of determinant 1 is called SO(3,1), or the Lorentzgroup, and the subgroup that preserves the orientation of the 4thcoordinate is the orthochronous Lorentz group.The standard model appears by applying certain symmetries to a generic >Lagrangian. Decide you want it invariant under U(1), ?ure out what >extra terms you need to make it so, and there's electromagnetism. >Theorists ?d that more satisfying than just putting a bunch of terms in >way before they were discovered.> I vaguely remember from my mechanics course that a lagrangian used to> be the kinitec energy less the potential energy. What is U(1)? I> would be grateful if you could explain that keeping my background in> mind.> U(n) is the group of unitary nxn matrices: those (complex) nxn matricesA for which A^* A = A A^* = Id(nxn), the nxn identity matrix,where (...)^* is the complex-conjugate transpose, also called theHermitian adjoint. SU(n) is the subgroup of U(n) consisting ofmatrices of determinant 1.If n = 1, then you've got a 1x1 complex matrix, or a complex number z.Unitarity requires z^* z = 1, so the square magnitude of z is 1. Thus,U(1) is the set of complex numbers of modulus 1, or the circle group.> AnshulDale. === >It's not a bad idea to acquire the habit of Googling for terms you don't>understand. However (as Dale surely knows; this is addressed to the same newbiehe was addressing, whose name I seem already to have deleted) using this habit wisely involves also developing a good nose for cranks, frauds, and those who are simply deluded (or ignorant), and thus post misinformation which--on the face of it--looks like it might be right, but isn't.You also have to watch out for physicists (who aren't *necessarily*subsumed under any or all of the previous categories), whose useof mathematical terms doesn't always agree with the mathematicians'use of the same terms. (The asymmetry in this statement isn't purely generated by my prejudices; the newbie was a self-describedmaths undergraduate.)Lee Rudolph === >It's not a bad idea to acquire the habit of Googling for terms you don't>understand. > However (as Dale surely knows; this is addressed to the same newbie> he was addressing, whose name I seem already to have deleted) using > this habit wisely involves also developing a good nose for cranks, > frauds, and those who are simply deluded (or ignorant), and thus post > misinformation which--on the face of it--looks like it might be right, > but isn't.> You also have to watch out for physicists (who aren't *necessarily*> subsumed under any or all of the previous categories), whose use> of mathematical terms doesn't always agree with the mathematicians'> use of the same terms. (The asymmetry in this statement isn't > purely generated by my prejudices; the newbie was a self-described> maths undergraduate.)> Lee Rudolph> blind search process.Dale. === Hey sci/alt.math, I'm a physics grad student, but I have been doing some summer readingin order to get a more rigorous background in mathematics... long storyshort, I'm started reading Calculus on Manifold by Spivak, and I'mwondering if people can critique my answer to one of the problemsI'm working on. The problem is very similar to one found in the Spivak book, thougha bit less general. Here it is:(Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the?st component of the n-tuple of numbers, x) Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < x2} is Open.So, basically, this is the part of a circle of radius ?r' located in the?st quadrent, and our job is to prove that it is open.Now, I understand that all I have to do is ?d some open rectangle, A,that is a subset of S such that ( y within S ) --> ( y within A )So, what I came up with is this. Given any old y within S choosethe open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a )where ?a' satisifys0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) }where min{ } means, choose the smallest number in the set.and 0 < b = (r^2) - (y1^2) - (y2^2) Obviously, the y1 and y2 are in there to keep the rectangles lowerleft corner inside S,and the third term is in there to keep the rectangles upperright corner inside S.It is true that y1 and y2 are greater than zero, and so isthe third term because b is greater than zero.Okay, So I have found that S is open. Because for any y within Sthere is an open rectangle, A, for which y is within A and A is a subset of S.So, am I done? Is this a good and rigorous proof? Comments andCritisism (constructive, hopefully) would be very muchappreciated.adrock === I didn't go into the computations, but you have the right. You can also useopen circles as well. For example,if (x_0,y_0) is any point in the region, then let d_1 be the distance from(x_0,y_0) to the edge of the quarter-cirlce, let d_2 be the distance from(x_0,y_0) to the x-axis, and let d_3 be the distance from (x_0,y_0) to they-axis (you can come up with formulas for each of these quite easily for youI think).In any event, let d = min (d_1, d_2, d_3). Then,( x - x_0)^2 + (y - y_0)^2 = (d/2)^2gives the equation of a circle, centered at (x_0,y_0) that is completelyinside of the region at hand.As I stated before, I didn't go into your computations, but your idea isright on (get some open neighborhood around any point and you've shown thatthe set is open). I just thought I would throw in the open circle to giveyou another example to use either now or in the future.I hope this helps,Brian> Hey sci/alt.math,> I'm a physics grad student, but I have been doing some summer reading> in order to get a more rigorous background in mathematics... long story> short, I'm started reading Calculus on Manifold by Spivak, and I'm> wondering if people can critique my answer to one of the problems> I'm working on.> The problem is very similar to one found in the Spivak book, though> a bit less general. Here it is:> (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and0 < x2}> is Open.> So, basically, this is the part of a circle of radius ?r' located in the> ?st quadrent, and our job is to prove that it is open.> Now, I understand that all I have to do is ?d some open rectangle, A,> that is a subset of S such that ( y within S ) --> ( y within A )> So, what I came up with is this. Given any old y within S choose> the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a )> where ?a' satisifys> 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) }> where min{ } means, choose the smallest number in the set.> and 0 < b = (r^2) - (y1^2) - (y2^2)> Obviously, the y1 and y2 are in there to keep the rectangles lower> left corner inside S,> and the third term is in there to keep the rectangles upper> right corner inside S.> It is true that y1 and y2 are greater than zero, and so is> the third term because b is greater than zero.> Okay, So I have found that S is open. Because for any y within S> there is an open rectangle, A, for which y is within A and A is a> subset of S.> So, am I done? Is this a good and rigorous proof? Comments and> Critisism (constructive, hopefully) would be very much> appreciated.> adrock> === > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and> 0 < x2} is Open.>Here's an approach you might like.B = { x : ||x|| < r } is an open ball.A = { x : 0 < x } is open line segement in R.Thus AxA is open in RxRS = B / AxA is the intersection of two open sets, hence open. === > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and> 0 < x2} is Open.>Here's an approach you might like.>B = { x : ||x|| < r } is an open ball.>A = { x : 0 < x } is open line segement in R.>Thus AxA is open in RxR>S = B / AxA is the intersection of two open sets, hence open.>Ah, that is just great... to bad I have not proven that the open ballis open. Could you enclude a proof that the open ball is open? === > Hey sci/alt.math,> I'm a physics grad student, but I have been doing some summer reading> in order to get a more rigorous background in mathematics... long story> short, I'm started reading Calculus on Manifold by Spivak,An excellent choice for summer reading. and I'm> wondering if people can critique my answer to one of the problems> I'm working on.> The problem is very similar to one found in the Spivak book, though> a bit less general. Here it is:> (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Note that Spivak's use of superscripts in this context is a little nonstandard. Most books prefer subscripts to denote the components of an element of n-space. The standard ASCII convention is x_1 for x-subscript-1, so that an n-vector is denoted (x_1, x_2, . . ., x_n).> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < > x2}> is Open.> So, basically, this is the part of a circle of radius ?r' located in the> ?st quadrent, and our job is to prove that it is open.> Note that most books de?e an open set in terms of balls, not rectangles. In other words if x is a point (in n-space) then we de?e the *open ball* centered at x of radius d, B(x,d) = {b element of R^n such that |x-b| < d}. Then we say that a set U is open if each point of U is contained in an open ball that is a subset of U.With this de?ition it's very easy to show that the interior of a circle is an open set. Here Spivak is having a little fun with us. By de?ing open sets in terms of open rectangles and then asking us to show that the interior of a (quarter) circle is open, he's forcing us to mess around with square roots and ?ure out the distance from a point inside a circle, to the circle.You might ?d it instructive to convince yourself that a set is open using the rectangle de?ition if and only if it is open using the ball de?ition. Actually if you do that ?st, this problem gets a lot easier, since (using the ball de?ition) if x is an element of the set S = {b such that |b| < r}, then B(x, r/2} is an open ball that contains x and is a subset of S.Ok, back to the original problem.> Now, I understand that all I have to do is ?d some open rectangle, A,> that is a subset of S such that ( y within S ) --> ( y within A )> statement ( y within S ) --> ( y within A )is the DEFINITION of ?S is a subset of A', which is not what you mean to say.Spivak's de?ition of an open set U is that if x is any arbitrary element of U, there exists an open rectangle A with (x element of A) and (A subset of U).choose some open rectangle A subset of S that DOES NOT contain y. Then ( y within S ) --> ( y within A ) is VACUOUSLY TRUE since (y within S) is false. Therefore you have just proven that ANY SET that contains an open rectangle is an open set, which is false.So again, just to be clear, here is Spivak's de?ition of an open set.A set U is open if for any x in U, there exists an open rectangle A with x in A and A subset of U. And as a reminder, an open rectangle is the cartesian product of open intervals.> So, what I came up with is this. Given any old y within S choose> the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a )> where ?a' satisifys> 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) }> where min{ } means, choose the smallest number in the set.> and 0 < b = (r^2) - (y1^2) - (y2^2) > Obviously, the y1 and y2 are in there to keep the rectangles lower> left corner inside S,> and the third term is in there to keep the rectangles upper> right corner inside S.> It is true that y1 and y2 are greater than zero, and so is> the third term because b is greater than zero.> I have to admit I didn't work all the way through your solution. It seems too complicated. That doesn't mean it's wrong, just that I found it easier to work this out for myself than work through your reasoning. The idea is that for a point inside the circle {|x| < r} we want to ?d the horizontal and vertical distance to the circle, and use those distances to construct the open intervals.The other thing I did was to use the standard notational convention for R^2, which is to label points (x,y) instead of (x_1, x_2). This lets us use the familiar concepts of the x-axis, the y-axis, and functions denoted as y = f(x). Why make life complicated?As you're following along with this it's incredibly helpful to have a diagram in front of you, a big circle centered at the origin, with a random point (a,b) in the ?st quadrant inside the circle. We need to ?ure the horizontal and vertical distances from (a,b) to the circle.If (a,b) is a point inside the circle, what is the horizontal distance to the circle? The horizontal line through (a,b) has equation y=b, and the circle has equation x^2 + y^2 = r^2 => x^2 + b^2 = r^2 => x = sqrt(r^2 - b^2).In that last equation we know r^2 - b^2 is positive since a^2 + b^2 < r^2 is given, therefore the sqrt exists in the reals.Then the distance between (a,b) and (sqrt(r^2 - b^2), b) is sqrt(r^2 - b^2) - a. We know the sign is positive from the geometry: sqrt(r^2 - b^2) is the x-coordinate of a point on the circle horizontally to the right of (a,b).On the other hand, the distance between (a,b) and the y-axis is just a. So we can choose our x-interval around a to be (a/2, a + (sqrt(r^2 - b^2) - a) / 2).Likewise, going through the same reasoning in the vertical direction, a workable y-interval is (b/2, b + (sqrt(r^2 - a^2) - b)/2).The open rectangle we need is the cartesian product of the x- and y-intervals.If this is the same answer you had, that's good. On the other hand yours doesn't look quite right to me, but I haven't taken the time to ?ure out why.> Okay, So I have found that S is open. Because for any y within S> there is an open rectangle, A, for which y is within A and A is a > subset of S.> Yes, that's the correct statement that you mis-stated above.> So, am I done? Is this a good and rigorous proof? Comments and> Critisism (constructive, hopefully) would be very much> appreciated.> Another way to go would be to prove that the interior of a circle is open, the upper open half plane {y>0} is open, and the right open half plane {x>0} is open, and a ?ite intersection of open sets is open (a Spivak exercise, at least in my ancient edition). Therefore the original set is open. I hope my long-winded post helped. === > (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and> 0 < x2} is Open.>Here's an approach you might like.>B = { x : ||x|| < r } is an open ball.>A = { x : 0 < x } is open line segement in R.>Thus AxA is open in RxRS = B / AxA is the intersection of two open sets, hence open.> Ah, that is just great... to bad I have not proven that the open ball> is open. Could you enclude a proof that the open ball is open?> Ah, but that's the whole point of the exercise, as I pointed out in my detailed response in the original thread. If you de?e open sets in terms of open balls, it's trivial to show that the interior of a circle is open. If you de?e open sets in terms of open rectangles, you have to do a little work, and that's what Spivak wants the reader to do. === > Hey sci/alt.math,> I'm a physics grad student, but I have been doing some summer reading> in order to get a more rigorous background in mathematics... long story> short, I'm started reading Calculus on Manifold by Spivak, and I'm> wondering if people can critique my answer to one of the problems> I'm working on.> The problem is very similar to one found in the Spivak book, though> a bit less general. Here it is:> (Note: x1 would be written as (x superscript 1) in Spivak, so x1 means the> ?st component of the n-tuple of numbers, x)> Prove that S = { x within R^2, such that: ||x|| < r and 0 < x1 and 0 < x2}> is Open.> So, basically, this is the part of a circle of radius ?r' located in the> ?st quadrent, and our job is to prove that it is open.> Now, I understand that all I have to do is ?d some open rectangle, A,> that is a subset of S such that ( y within S ) --> ( y within A )I don't think that last phrase such that ... is what you want to say.I suspect that this is your own phraseolgy (which is not standard).In brief, you need to get the quanti?rs correct. That is,for each y within S, there is some open rectangle A that isa subset of S, such that y is in A. In particular, A depends on the y.> So, what I came up with is this. Given any old y within S choose> the open rectangle A = ( y1-a , y1+a )X( y2-a , y2+a )Right. The y is given ?st and then you choose the A.> where ?a' satisifys> 0< a < min{ y1, y2, (1/2) ( Sqrt[ (y1^2) + (y2^2) + (2b) ] - y1 - y2 ) }> where min{ } means, choose the smallest number in the set.> and 0 < b = (r^2) - (y1^2) - (y2^2) > Obviously, the y1 and y2 are in there to keep the rectangles lower> left corner inside S,> and the third term is in there to keep the rectangles upper> right corner inside S.> It is true that y1 and y2 are greater than zero, and so is> the third term because b is greater than zero.The above paragraph gives a geometric reasoning for the proof.I didn't check it out. But, you may want to give a moreanalytic reasoning for the proof. That is, you want toshow that A is a subset of S. To do this, proceed thestandard way. That is, something like the following: Now let (u1, u2) be in A. I claim that (u1, u2) is in S. For ?st, 0 < u1 because ... Seconding, 0 < u2 because ... Finally, u1^2 + u2^2 < r because ... Hence, A is a subset of S.As an aside, the last sentence of the above paragraph canbe improved by stating explicitly the motivation for it.That is, it justi?s that a can indeed be found. Thus,you can say something like: I can ?d an a since themim is greater than zero because y1 and y2 are greater than zero,and so is the third term because b is greater than zero.> Okay, So I have found that S is open. Because for any y within S> there is an open rectangle, A, for which y is within A and A is a > subset of S.This is the correct statement of what you are trying to prove.> So, am I done? Is this a good and rigorous proof? Comments and> Critisism (constructive, hopefully) would be very much> appreciated.> adrock === > You might ?d it instructive to convince yourself that a set is open > using the rectangle de?ition if and only if it is open using the ball > de?ition. Actually if you do that ?st, this problem gets a lot > easier, since (using the ball de?ition) if x is an element of the set > S = {b such that |b| < r}, then B(x, r/2} is an open ball that contains > x and is a subset of S.> Correction. Of course what I meant to write was that B(x, 1/2 * min(|x|, r - |x|) is the required open ball. === I think that the neatest suggestion was to show that the open ball is open under Spivak's de?ition (which isbasically Spivak problem 1-15), and show that the 1st quadrentis open, and then that the interesction of two (or a ?ite number) of open sets is open...Then my problemis solved. get stuck again...ciao,adrock === > I think that the neatest suggestion was to show that> the open ball is open under Spivak's de?ition (which is> basically Spivak problem 1-15), and show that the 1st quadrent> is open, and then that the interesction of two> (or a ?ite number) of open sets is open...Then my problem> is solved.Indeed you really learned something from that problem.Fishfry pointed out what was important.Another uniformally equivalent metric to circles and rectangles isdiamonds or rhomboids D((x,y) - (a,b)) = |x-a| + |y-b|Also there's the box metric which is about the same except instead ofopen rectangles it uses open squares.> get stuck again...ciao,>You're welcome. You don't have to wait to get stuck before returning. Theapproach I suggested that was to your liking was a topological approachthe topology of metric spaces which is easier handled just topologically.Of course, there is some basic transitions to understand, the rectangle,circle topologies being equivalent is one, another being the equivalenceof the rectangle topology to the product topology which seemed instantlyclear to you. === I have to minimize something like tr(X'AX) where A in R^{n*n} is a de?itepositive matrix andX is a n*k binary matrix, such that x_ij={0,1} and the sum of each row is 1(i.e. sum_i x_ij=1).The problem is well known to be NP-complete. Do you know any reference to acontinuous approximation? Or an ef?ient way to solve the problem when X isa big matrix? === > William Elliot ha scritto nel messaggio> Consider the subposet ({a,b}, <=) and the function F such that> * * ------------> * F(a)=F(b)> a b F> (a and b are not comparable). Is F order preserving? I think so.> a,b not comparable, written a || b,> when not a <= b & not b <= a> What's order preserving, an order isomorphism?> F is order preserving iff x<=y ==> F(x)<=F(y);> F is an order embedding iff x<=y <==> F(x)<=F(y);> F is an order isomorphism iff x<=y <==> F(x)<=F(y) and F is surjective.> 1 for a,b, (a || b ==> f(a) = f(b))> imply> 2 f is order preserving.> Now I know that if F:(S,<=) --> (S',<=') is a function such that F(x)=K for> all x in S, F is an order preserving function. Consider x < y, x < z, y || z; f(x) = a, f(y) = f(z) = b; b < a> f has 1 but lacks 2, does it not?> Also consider same f with a < b, b || a or a = b instead.> You mean (the ?st case)> *y *z *f(x)> /> / ----------> / f> /> *x *f(y)=f(z)> so f is not order preserving.>Exactly, nor in the case a || b> Now consider the following:> a* |> | -----------> * G(a)=G(b)> | G> b*> Is G order embedding? It is true that we have G(b)<=G(a) === > b<=a, but> also> G(a)<=G(b) & a<=b.> Does the diagram mean> for all a,b, (a <= b ==> g(a) = g(b)) ?> Now g(a) = g(b) ==> g(a) <= g(b). Thus> for all a,b, (a <= b ==> g(a) <= g(b))> Is that last property order embedding?> I think this last is not an order embedding because:> 1) b<=a ==> G(b)<=G(a) is true> 2) G(a)<=G(b) ==> a<=b is false;> so the equivalence 1)<==>2) does not hold.>Sometimes it would, but in general it wouldn't.What's needed to cinch that is a counter example.As you're understanding this stuff, have you acounter example to present?Easy theorem to prove, if you haven't already order embedding maps are injective.Exercise: when is an order preserving injection, order embedding? Include proofs and/or counterexamples of claims. === As I understand it, probably the most important open question in ?ite geometry is: which numbers occur as the order of a ?ite (af?e or projective) plane? In particular, a proof is sought for what might be called the Order Conjecture (OC).OC: If n is the order of a ?ite af?e/projective plane, then n is a prime power.In the case that an af?e plane A^2 does have an order q that is a prime power, then A^2 can easily be coordinatized by a ?ite ?ld of the same order (i.e., A^2 = GF(q)^2, where GF(q) is the Galois ?ld of order q). Conversely, if it could be proved that _every_ ?ite af?e plane can be coordinatized by a ?ite ?ld, then OC would follow, since the only orders for GF(q) are prime powers.Has this approach been taken? I know Bruck and Ryser's result on numbers of the form 4k+1 and 4k+2, and I've heard of the work at Concordia University (Montreal) to eliminate 10 as a possibility; neither of these seems to use this line of reasoning. Can someone point me in the right direction?-- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman === > As I understand it, probably the most important open question in ?ite> geometry is: which numbers occur as the order of a ?ite (af?e or> projective) plane? In particular, a proof is sought for what might be> called the Order Conjecture (OC).> OC: If n is the order of a ?ite af?e/projective plane, then n is a> prime power.The n = 10 case was dif?ult enough ....> In the case that an af?e plane A^2 does have an order q that is a prime> power, then A^2 can easily be coordinatized by a ?ite ?ld of the same> order (i.e., A^2 = GF(q)^2, where GF(q) is the Galois ?ld of order q).> Conversely, if it could be proved that _every_ ?ite af?e plane can be> coordinatized by a ?ite ?ld, then OC would follow, since the only> orders for GF(q) are prime powers.But there are ?ite af?e/projective planes which are *not* isomorphicto the ones constructed in the classical way from ?ite ?lds.The smallest order for which such exist is 9. Seehttp://www.math.uni-kiel.de/geometrie/klein/math/geometry/ smallproj.html-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === > But there are ?ite af?e/projective planes which are *not* isomorphic> to the ones constructed in the classical way from ?ite ?lds.> The smallest order for which such exist is 9. See> http://www.math.uni-kiel.de/geometrie/klein/math/geometry/ smallproj.htmlAha. I was unaware of the different isomorphism types -- hadn't seen them number of new (to me) ideas to explore. So I have a couple more questions. Is there a reference that describes _how_ the order-10 case was eliminated (one that says more than The proof took thousands of hours of computer veri?ation.?) Secondly, in the above linked site's description of quasi?lds and semi?lds, it uses two distinct existence symbols: the usual backwards E, and backwards E^1. Do these have different meanings?-- The above address is intended to prevent spam. Please change the capital Joshua P. Bowman === > But there are ?ite af?e/projective planes which are *not* isomorphic> to the ones constructed in the classical way from ?ite ?lds.> The smallest order for which such exist is 9. See> http://www.math.uni-kiel.de/geometrie/klein/math/geometry/ smallproj.html> Aha. I was unaware of the different isomorphism types -- hadn't seen them> number of new (to me) ideas to explore. So I have a couple more questions.> Is there a reference that describes _how_ the order-10 case was eliminated> (one that says more than The proof took thousands of hours of computer> veri?ation.?)Pehaps you should consult this account by one of the main protagonists.92b:51013Lam, C. W. H.(3-CONC-C)The search for a ?ite projective plane of order $10$.Amer. Math. Monthly 98 (1991), no. 4, 305--318.> Secondly, in the above linked site's description of> quasi?lds and semi?lds, it uses two distinct existence symbols: the> usual backwards E, and backwards E^1. Do these have different> meanings?Dunno, but I would guess that the second means there exists exactly one.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === I'm trying to calculate the expected value of the magnitude of a Gaussian random vector with given mean (let's assume zero for simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma).Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear function. (I just wanted to address a common answer I've received from people I've asked already.)Note also that if sigma is the identity matrix, the problem is easily solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is relatively simple to deal with because x'*x appears as a group everywhere. Contrast this with what I'd like, which is of the form: integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that helps anyone when making substitutions, but it's something I noticed.Afsheen === >I'm trying to calculate the expected value of the magnitude of a >Gaussian random vector with given mean (let's assume zero for >simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma).>Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >function. (I just wanted to address a common answer I've received from >people I've asked already.)>Note also that if sigma is the identity matrix, the problem is easily >solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >relatively simple to deal with because x'*x appears as a group >everywhere. Contrast this with what I'd like, which is of the form: >integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >helps anyone when making substitutions, but it's something I noticed.About the only way to get accurate numerical answersis to numerically integrate the derivative of the Laplace transform of the distribution of x'x. If L is the Laplace transform, the -int sqrt(1/t) L'(t) dt/sqrt(pi)is the expected value of sqrt(x'x).-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === Is there no way to solve this analytically? Since we're given that x is a Gaussian random vector of known mean and covariance, I would think that there's a nice expression for E(sqrt(x'x)).Afsheen>I'm trying to calculate the expected value of the magnitude of a >Gaussian random vector with given mean (let's assume zero for >simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma).>Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >function. (I just wanted to address a common answer I've received from >people I've asked already.)Note also that if sigma is the identity matrix, the problem is easily >solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >relatively simple to deal with because x'*x appears as a group >everywhere. Contrast this with what I'd like, which is of the form: >integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >helps anyone when making substitutions, but it's something I noticed.> About the only way to get accurate numerical answers> is to numerically integrate the derivative of the > Laplace transform of the distribution of x'x. If > L is the Laplace transform, the> -int sqrt(1/t) L'(t) dt/sqrt(pi)> is the expected value of sqrt(x'x).> === >Is there no way to solve this analytically? Since we're given that x is >a Gaussian random vector of known mean and covariance, I would think >that there's a nice expression for E(sqrt(x'x)).>AfsheenIf the characteristic roots of the covariance matrix,assuming the mean vector is 0, all occur with evenmultiplicities, the distribution is a linear combination,some coef?ients being negative, of Gamma distributionswith integer exponents. In this case, the integrationcan be done in closed form. I am not sure if the integral representation given can be done in elementaryterms if there is one root of odd multiplicity.If there are two such roots, we are already at leastat the stage of elliptic integrals.>I'm trying to calculate the expected value of the magnitude of a >Gaussian random vector with given mean (let's assume zero for >simplicity) and covariance. So, in essence, I want E(sqrt(x'*x)) where >x' is transpose of x, E is expectation, and x has a pdf of normal(0,sigma).>Note that although E(x'*x) = trace(E(x*x')) = trace(sigma), this doesn't >imply that E(sqrt(x'*x)) = sqrt(trace(sigma)). This is because >E(sqrt(a)) does not equal sqrt(E(a)) because sqrt() is not a linear >function. (I just wanted to address a common answer I've received from >people I've asked already.)>Note also that if sigma is the identity matrix, the problem is easily >solvable, since the integral of (sqrt(x'*x) * exp(-.5*x'*x)) is >relatively simple to deal with because x'*x appears as a group >everywhere. Contrast this with what I'd like, which is of the form: >integral of (sqrt(x'*x) * exp(-.5*x'*sigma^-1*x). I don't know if that >helps anyone when making substitutions, but it's something I noticed.> About the only way to get accurate numerical answers> is to numerically integrate the derivative of the > Laplace transform of the distribution of x'x. If > L is the Laplace transform, the> -int sqrt(1/t) L'(t) dt/sqrt(pi)> is the expected value of sqrt(x'x).-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === >Wow! 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That might even be funny.Euclidean space : Non-Euclidean space :: Conventional math : Non-conventional math [ 2^(7/12) = 3/2 ](cross-posted to sci.math)ObPuzzle: Why did I choose this particular equation? === > consider the following statement, where * is some arbitrary operation: if a and b are in the empty set, then a*b is in the empty set> Why not>Because * isn't de?ed at a,b, that's why.> if a and b are in the empty set, then c is in the empty set> ?>As you've bestowed existence upon c, yes, c in nulset.> Immediately a philosophical/logical/set theoretical nightmare arises> as one attempts to decide whether the statement is true or false or> whether it depends on one's choice of *.> logic's equation FALSE->FALSE = TRUE unequivically says the> statement is true, but all common sense screams otherwise.> Mine doesn't? So why does yours?>Philosophical conundrum: are non existence things in the nulset? nulset = { x | x /= x } if x doesn't exist, does x /= x ? === > if a and b are in the empty set, then a*b is in the empty set> logic's equation FALSE->FALSE = TRUE unequivically says the> statement is true, but all common sense screams otherwise.> Mine doesn't? So why does yours?So you would hold, then, that if a and b are both prime numbersdivisible by 4, then a+b is also a prime number divisible by 4?if a ?ite set M of integers has two distinct largest elements, a andb, then a^b is also the largest element of M?if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? (Ironically if we replace 5 here with aleph_0, it *almost* becomestrue)OK, but more seriously, if indeed the empty set possesses the closureproperty, can it also include an identity with regard to *? thisseems to depend on how one phrases the de?ition of a group... if onesays, the set in question must possess some ?identity' i... thenclearly the empty set does not possess an identity. But if one says,for any a in the set in question, there exists some ?identity' i,which is the same for all other elements in the set... then rightaway it seems the statement is true, since the false statement theempty set contains an identity i is cancelled out by the false forany a in the empty set...See, what I'm really getting at here is, is the empty set a group? === > Philosophical conundrum: are non existence things in the nulset?Yes! No element of the emptyset exists. So... all nonexistent thingsare in the emptyset. [Phrased more technically: The emptyset is asubset of the emptyset.] === > consider the following statement, where * is some arbitrary operation:> if a and b are in the empty set, then a*b is in the empty set> Immediately a philosophical/logical/set theoretical nightmare arises> as one attempts to decide whether the statement is true or false or> whether it depends on one's choice of *.> logic's equation FALSE->FALSE = TRUE unequivically says the> statement is true, but all common sense screams otherwise.I prefer to say that the sentence a*b is in the empty setis not meaningful, rather than that it is true (or false).I believe the logicians have a name for what I am thinking of(rathen than just meaningful).In more detail, ?st I require that you tell me the type ofa and b. For example, you could say that a and b are in theset Q of rational numbers. Then, your statement is: If a in Q and b in Q are in the empty set, then a*b is in the empty setSecond, I require that you tell me the type of operation * is.I need to know the domain and range of the operation *. I thinkthat you are willing to say that the operation is a binary operationfrom the same set to the same set: that is, * : AxA -> A where Ais some set. For example, you could say that A is a cyclic group oforder 2 and the operation * is the group product. Then, yourstatement is: Let A be the cyclic group of order 2 with multiplication *. If a in Q and b in Q are in the empty set, then a*b is in the empty setI prefer to say that the statement a*b is in the empty setis not meaningful, since the rational numbers a and b are notin the domain of * (that is, (a,b) is not in AxA). Since theconclusion is not a meaningful statement, I take the wholeimplication to not be a meaningful statement.But, I suspect that some (many?) may wish to say that the statementa*b is in the empty set is just false.As an aside, I would take the following to be a meaningful statement: Let A be the cyclic group of order 2 with multiplication *. If a in Q and b in Q are in the empty set, then if a is in A and b is in A, then a*b is in the empty setI take it to be meaningful since the hypothesis a is in A and b in is Aallows be to do the group multiplication a*b to get an element of A.-- Bill Hale === > if a and b are in the empty set, then a*b is in the empty set> logic's equation FALSE->FALSE = TRUE unequivically says the> statement is true, but all common sense screams otherwise.> Mine doesn't? So why does yours?> So you would hold, then, that if a and b are both prime numbers> divisible by 4, then a+b is also a prime number divisible by 4?Yes.> if a ?ite set M of integers has two distinct largest elements, a and> b, then a^b is also the largest element of M?Yes.> if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well?Yes.The above three statements are not that strange. In fact, they occurall the times. Consider the standard epsilon-delta de?ition ofcontinuity. The condition statement |x - x0| < delta is falsefor a great many x's, but the implication statement if |x-x0| < delta,then |f(x) - f(x0)| < epsilon is still true for those x's. Indeed,we need to prove the implication even for those x's because thatwhat the de?ition says.> (Ironically if we replace 5 here with aleph_0, it *almost* becomes> true)> OK, but more seriously, if indeed the empty set possesses the closure> property, can it also include an identity with regard to *? this> seems to depend on how one phrases the de?ition of a group... if one> says, the set in question must possess some ?identity' i... then> clearly the empty set does not possess an identity.And the de?ition of group *does* require such existence of the unit element.> But if one says,> for any a in the set in question, there exists some ?identity' i,> which is the same for all other elements in the set... then right> away it seems the statement is true, since the false statement the> empty set contains an identity i is cancelled out by the false for> any a in the empty set...> See, what I'm really getting at here is, is the empty set a group?But, that is why the de?ition of group is not phrase in your secondway.-- Bill Hale === > OK, but more seriously, if indeed the empty set possesses the closure> property, can it also include an identity with regard to *? this> seems to depend on how one phrases the de?ition of a group... if one> says, the set in question must possess some ?identity' i... then> clearly the empty set does not possess an identity. But if one says,> for any a in the set in question, there exists some ?identity' i,> which is the same for all other elements in the set... then right> away it seems the statement is true, since the false statement the> empty set contains an identity i is cancelled out by the false for> any a in the empty set...But who would be perverse enough to use the second formulation? Thisis simply not a real problem.> See, what I'm really getting at here is, is the empty set a group?Of course it is not a group, because no one would ever (even byaccident) de?e a group using the clause for any a in the set in question, there exists some ?identity' i, which is the same for all other elements in the set...-- Jesse Hughes How come there's still apes running around loose and there arehumans? Why did some of them decide to evolve and some did not? Didthey choose to stay as a monkey or what? -Kans. Board of Ed member === > consider the following statement, where * is some arbitrary operation:> if a and b are in the empty set, then a*b is in the empty set> Why not> if a and b are in the empty set, then c is in the empty set> ?Indeed, why notif a and b are members of the empty set, then a*b is the King ofFrance?--------------------------------------------------- C Brown Systems DesignsMultimedia Environments for Museums and Theme Parks--------------------------------------------------- === if a and b are in the empty set, then a*b is in the empty set> logic's equation FALSE->FALSE = TRUE unequivically says the> statement is true, but all common sense screams otherwise.> Mine doesn't? So why does yours?> So you would hold, then, that if a and b are both prime numbers> divisible by 4, then a+b is also a prime number divisible by 4?Yes. > if a ?ite set M of integers has two distinct largest elements, a and> b, then a^b is also the largest element of M?Yes. > if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well?> (Ironically if we replace 5 here with aleph_0, it *almost* becomes> true)Yes. > OK, but more seriously, if indeed the empty set possesses the closure> property, can it also include an identity with regard to *? this> seems to depend on how one phrases the de?ition of a group... if one> says, the set in question must possess some ?identity' i... then> clearly the empty set does not possess an identity. But if one says,> for any a in the set in question, there exists some ?identity' i,> which is the same for all other elements in the set... then right> away it seems the statement is true, since the false statement the> empty set contains an identity i is cancelled out by the false for> any a in the empty set...> See, what I'm really getting at here is, is the empty set a group?No. A group has an identity element.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === > consider the following statement, where * is some arbitrary operation:> if a and b are in the empty set, then a*b is in the empty set> Why not> if a and b are in the empty set, then c is in the empty set> ?> Indeed, why not> if a and b are members of the empty set, then a*b is the King of> France> ?That empty set better not accidentally get any members, because if itdoes then /all/ kinds of weird is going to happen!Phil === > That empty set better not accidentally get any members, because> if it does then /all/ kinds of weird is going to happen!lolThis seems to be related to not playing with a full deck,but I don't know which is the cause and which the effec'. === > So you would hold, then, that if a and b are both prime numbers> divisible by 4, then a+b is also a prime number divisible by 4?> Right.> if a ?ite set M of integers has two distinct largest elements, a and> b, then a^b is also the largest element of M?> Right.> if some a exists such that a+1 = 5 and a+2 = 5, then a+3 = 5 as well? > (Ironically if we replace 5 here with aleph_0, it *almost* becomes> true)> Right.If 0 = 1 then I'm gonna eat my heat! (No, I'm really not eager to eat myheat!)> OK, but more seriously, if indeed the empty set possesses the closure> property, can it also include an identity with regard to *?>No.Note: its certainly true that (if our domain is R and * is the usualmultiplication de?ed for real numbers) (x)(y)(x e 0 & y e 0 -> x*y e 0)BUT this does in no way imply the EXISTENCE of some element in 0.> this seems to depend on how one phrases the de?ition of a group...> if one says, the set in question must possess some ?identity' i...> then clearly the empty set does not possess an identity.>Exactly.> But if one says, for any a in the set in question, there exists some > ?identity' i [in the set], which is the same for all other elements > in the set... then right away it seems the statement is true [...]>Right.BUT a group just ISN'T de?ed this way. (Know why? ;-)> See, what I'm really getting at here is, is the empty set a group?>The answer is simply NO, just because... (see explanation above).F. === > consider the following statement, where * is some arbitrary operation:> if a and b are in the empty set, then a*b is in the empty set> Why not> Because * isn't de?ed at a,b, that's why.> Well... of course it should (better) read:For any to real numbers a,b: if a and b are in the empty set, then a*b is in the empty set ,where * is the usual multiplication de?ed for real numbers. But Iguess this was meant. (For example, a,b could be quali?d variablesalways meant to denote real numbers, etc.)> Philosophical conundrum: are non existence things in the nulset?>No.In Free Logic we actually can deal with such things. If there existssuch a thing as t, we write E!tAnd we might adopt the following Rule of Atomic Denotation: A(t) where A(t) is atomic ---- E!tNow we might actually do set theory in a framework of free logic. Thenwe would have t e u |- E!t.With other words, if something is an element of a set, it exists.The other way round: ~E!t |- t !e ufor any set u. Hence, if u is the empty set, 0, we have: ~E!t |- t !e 0.or |- ~E!t -> t !e 0. If t doesn't exist, it's not an element of the empty set.> 0 = {x | x =/= x}>Right.> if x doesn't exist, does x =/= x ?>Yes. But it's still not an element of 0. (You surely know that nativecomprehension usually won't work in a properly de?ed set theory. ;-)F. === Let d be a positive integer. What are the suf?ient and necessarycondition that the Pell's equation x^2-dy^2=-1 has integer solutions?It's easy to see that there are no integer solutions when d is perfectsquare or when d contain a prime divisor of the form 4k+3. But whatpositive integer value of d give rises to integers solutions? === sinisa escribi.97 en elmensaje|n22401dcf.0307130403.6994fe3d@posting.google.com:> Let d be a positive integer. What are the suf?ient and necessary> condition that the Pell's equation x^2-dy^2=-1 has integer solutions?> It's easy to see that there are no integer solutions when d is perfect> square or when d contain a prime divisor of the form 4k+3. But what> positive integer value of d give rises to integers solutions?It must be a complex question. At least it seems yet from the title of theLagarias, J.C. On the Computational Complexity of Determining theSolvability or Unsolvability of the Equation X^2-DY^2=-1. Trans. Amer. MathSoc. 260, 485-508, 1980.Althought some things can be easily saw. By example,If d is prime and equal to 1 (mod 4) the equation has solutions.If d has some factor equal to 3 (mod 4) there aren't solutions.If d = 0 (mod 4) there aren't solutions.Then, in order to exist solutions, d must be equal to 1, 2 or 5 (mod 8) andhasn't factors equals to 3 (mod 4).Furthermore, studuing the length of the period of the develop in continuedfraction of Sqrt(d), is easy to see that:If d=m^2+1, there are solutions.If d=m^2 + k (2m/k integer and k > 1), there aren't solutions.I?d=(m+1)^2 - k, m>1 and 2(m+1)/k integer, there aren't solutions.Also it can got conditions progresivaly more complex for d equal to(m^2 +a*2m) and ((m+1)^2-a*(m+1)), where a is a fraction as 2/3, 3/4, 2/5,3/5, 4/5, ....In other way, if d can be expressed asd = ((4h^2 + 1)k + h)^2 + (4hk + 1))the period length is 3 and there are solutions.But for a any large d, the more easy way to decide if there are solutions ornot must be develop Sqrt(d) in continued fraction and see if the length ofthe period is odd or even.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === > Let d be a positive integer. What are the suf?ient and> necessary conditions that the Pell's equation x^2-dy^2=-1> has integer solutions?Many suf?ient conditions are known. See:A Numerical Investigation of the Diophantine Equation x^2 - dy^2 = -1R. D. Beach and H. C. WilliamsDepartment of Computer Science, University of Manitobawhich is inCongressus Numerantium VIProceedings of the Third Southeastern Conference of Combinatorics, GraphTheory, and ComputingFlorida Atlantic UniversityBoca RatonFebruary 28 - March 2, 1972Edited by F. Hoffman, R. B. Levow, and R. S. D. ThomasUtilitas Mathematica Publishing Inc.P. O. Box 7, University CentreUniversity of ManitobaWinnipeg, Manitoba, CanadaR3T 2N2ISBN 0 919628 06 0Gives tables of values of d for which x^2 - dy^2 = -1 has a solution, otherthan those that can be predicted by a set of criteria that are listed.One criterion listed is when d = p and p == 1 (mod 4). In fact, if d is aprime, then x^2 - dy^2 = -1 has a solution if and only if p is 2 or p==1 (mod4).Another criterion is d=pq, where p==1 mod 4, q==1 mod 4, and (p/q)=-1. Here,x^2 - dy^2 = -1 has a solution when the criterion is met, but there aresolutions to (1) for some pq that do not meet this criterion.Yet another criterion is d=pq, where p==1 mod 4, q==1 mod 4, and (p/q)=1,(p/q)_4=(q/p)_4=-1.(No other criteria are listed for pq.)Some d of the form pq for which the equation has a solution, but not covered bythe above criteria are 2305, 2605, 2705, 5305, 5713, 6757, 6953, 8005, 8653,9305, 9953, 10001, 10405, 10817, 10961, 11029, 11629, 12181, 12961, 13105,14305, 15529, 16601, 17305, 17693, 20005.A criterion given for d=2p is p==5 (mod 8).Another criterion for d=2p is p==9 (mod 16) and (2/p)_4 = -1.A few d's of the form 2p for which the equation has a solution, but not coveredby the above criteria are 226, 2402, 3202, 3554, 4226, 6178, 6242, 6274, 6626,7522.[(*/*) is the Legendre symbol, and (*/*)_4 is the biquadr symbol)Other pq:5305 - 5, 10615713 - 29, 1976757 - 29, 2336953 - 17, 4098005 - 5, 16012p:1. p==5 (mod 8)2. p==9 (mod 16), (2/p)_4=-1John Robertson === Of course the main criterion is whether or not the length of the period of thecontinued fraction expansion of d is odd or even.John Robertson === Dear All,Sorry I can not solve this, as what I have learnt in school seems thatI have given them back to my teacher :..)It is about my project:3 possible workloads, running on 4 processors,how many combinations (not permutation, as it is 3^4) are there?What is the formula for this?e.g.111112122123111222223331332333Many thanks!!Jialin === > Dear All,> Sorry I can not solve this, as what I have learnt in school seems that> I have given them back to my teacher :..)> It is about my project:> 3 possible workloads, running on 4 processors,> how many combinations (not permutation, as it is 3^4) are there?> What is the formula for this?If you mean you have 4 processors, each of which can be running one of 3 tasks, then you will use the multiplication principle. 3*3*3*3 = 3^4.If you wish to use combinations in this problem, then there is some information missing, but it isn't clearly stated (in my mind).Note: your example below does not have an obvious connection with the problem stated above.> e.g.> 111> 112> 122> 123> 111> 222> 223> 331> 332> 333> Many thanks!!> Jialin> -- Will Twentyman === www.YeOldeCoffeeShoppe.com has a great many visitors and sectionsfor talk, tv, reviews and romance.Now add free, open to anyone to post but signing up ?st is free and you can use anavatar. Dont let my avatar scare you off, I trained in the gym 3 days running beforethat photo!Herc--www.winternet.com/~mikelr/?.html __ / / / / / / / / / / / /__/__ /________ ___________/ === > www.YeOldeCoffeeShoppe.com has a great many visitors and sections> for talk, tv, reviews and romance.but I've already got 1000s dropping in, all with nothing to read,just jump the buck, I'll have to pay a dozen people just to postsomething there for a few months otherwise, give the site a KICK,its a GREAT traf? getter a few minutes of telling people drawsreal crowds, but without the ?st couple of people standing round thebusker nooone will gather. And if you like usenet you'll love forums.Herc === >just jump the buck, Cool band name.>without the ?st couple of people standing round the>busker nooone will gather. Cool album name.--Seanhttp://www.livejournal.com/users/spclsd223/ === just jump the buck,> Cool band name.>without the ?st couple of people standing round the>busker nooone will gather.> Cool album name.> --Sean> http://www.livejournal.com/users/spclsd223/Dude.I love it when two worlds within my ?ld of interest collide and celebrate with each other.Do you like Standing Outside Imaginations Door as a book title?ps need some writing talent at www.Yeoldecoffeeshoppe.comHerc === >Dude.>I love it when two worlds within my ?ld of interest collide and celebrate with each other.Quoting my journal?!My plan is working perfectly!>Do you like Standing Outside Imaginations Door as a book title?No, but ?Standing Outside Imagination's Door' would be cool.Even better would be ?Ringing the Bell of Life and Running'.>ps need some writing talent at www.Yeoldecoffeeshoppe.comI'll let you know as soon as I have the motivation and time toactually look at the site and see what the hell it is.--Seanhttp://www.livejournal.com/users/spclsd223/ === Task: Put n different points so that there areas many lines with 3 points on it as possible.(example for 7 points - the familiar con?uration * * * * * * *with 6 lines)Does it help allowing higher dimensions (Leechlattice and all that funny stuff)? Can oneeven overcome the Sylvester theorem (therewill always be a line with only 2 points) -the proof in the 2D case was a metric oneafter all.-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === For a given point (x,y) lying on a cartesian plane what would the vertecies of the corners of a regular ploygon be?(In the simple case I'm assuming that the ?base' of the ploygonis parralel to the x-axis.)If the baseline is then rotated what adjustment would need to bemade to the ?ed points identifed above?(I concede that if then drawing the lines on a computer screenBresnham or simmilar would be used for actual drawing.)ii)There are 4 towns a,b,c,d located on a ?ane.A highway has been proposed between the two major towns a and b.and with only these towns it would be a straight line.Towns c and d are perpendicular to AB and it is proposedthat the highway routing should also serve thse communities(which are comparitivly smaller than A or B)c and and d are not of equal size, but are the same distancefrom the highway.re?the relative'size/traf?/distance from highway #' of c and d.Ideally the engineers want an ideal route.(I'm assuimng they would have access to likley traf? usage.)What model could be used to determine the routing of the new highway?iii)What would be a good model to use when attempting to plan atimetiablkefor an out and back metro line? (Assuming an unequal traf?distribution over time.)FMNT80 === >For a given point (x,y) lying on a cartesian plane what >would the vertecies of the corners of a regular ploygon be?Um, that depends on which polygon you want. How many sides,is (x,y) the center or one of the vertices, how long are thesides (or how far is it from the center to the vertices...)I mean really, if you're going to post homework problemsyou could at least include the statement of the problem...>(In the simple case I'm assuming that the ?base' of the ploygon>is parralel to the x-axis.)>If the baseline is then rotated what adjustment would need to be>made to the ?ed points identifed above?>(I concede that if then drawing the lines on a computer screen>Bresnham or simmilar would be used for actual drawing.)>ii)>There are 4 towns a,b,c,d located on a ?ane.>A highway has been proposed between the two major towns a and b.>and with only these towns it would be a straight line.>Towns c and d are perpendicular to AB and it is proposed>that the highway routing should also serve thse communities>(which are comparitivly smaller than A or B)>c and and d are not of equal size, but are the same distance>from the highway.>re?the relative'size/traf?/distance from highway >#' of c and d.>Ideally the engineers want an ideal route.>(I'm assuimng they would have access to likley traf? usage.)>What model could be used to determine the routing of the new highway?>iii)>What would be a good model to use when attempting to plan a>timetiablke>for an out and back metro line? (Assuming an unequal traf?>distribution over time.)>FMNT80David C. Ullrich === Why are certain polyhedra that are self intersecting considered REALpolyhedra? Clearly polyhedrons which are self intersecting shouldn'tbe considered REAL polyhedrons. Then we could do all sorts of weirdstuff to them. Take a look athttp://mathworld.wolfram.com/GreatDodecahedron.html andhttp://mathworld.wolfram.com/ SmallStellatedDodecahedron.html forexample.Why do we consider ?ures which intersect themselves to bepolyhedrons or polyhedra in the ?st place? Doesn't that kindaviolate the rules, about being able to be constructed as if fromboundaries from algebraic expressions, and such?This is what I don't get about this weird geometry mumbo jumbo aboutthings which are self intersecting being considered actual shapes.What is wrong with these theoretical geometricians?(...Starblade Riven Darksquall...) === > Why are certain polyhedra that are self intersecting considered REAL> polyhedra? Clearly polyhedrons which are self intersecting shouldn't> be considered REAL polyhedrons. Then we could do all sorts of weird> stuff to them. Take a look at> http://mathworld.wolfram.com/GreatDodecahedron.html and> http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for> example.> Why do we consider ?ures which intersect themselves to be> polyhedrons or polyhedra in the ?st place? Doesn't that kinda> violate the rules, about being able to be constructed as if from> boundaries from algebraic expressions, and such?> This is what I don't get about this weird geometry mumbo jumbo about> things which are self intersecting being considered actual shapes.> What is wrong with these theoretical geometricians?> (...Starblade Riven Darksquall...)Some work I've done with polygons started out with a restriction that they be convex polygons. The results I ended up with were easier to work with if I dropped that restriction and allowed them to be convex or self intersecting. All I was doing was working with the vertices, and dropping the restrictions allowed me to do less work to prove my result. I suspect there are similar reasons for working with self-intersecting polyhedra.-- Will Twentyman === > Why are certain polyhedra that are self intersecting considered REAL> polyhedra? See Imre Lakatos' _Proofs and Refutations_ for a variety of detailedviews about this exact question.Question to all - am I much mistaken in thinking that nowadays, mostmathematicians' views on such issues are of the whatever genre?cdjClearly polyhedrons which are self intersecting shouldn't> be considered REAL polyhedrons. Then we could do all sorts of weird> stuff to them. Take a look at> http://mathworld.wolfram.com/GreatDodecahedron.html and> http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for> example.> Why do we consider ?ures which intersect themselves to be> polyhedrons or polyhedra in the ?st place? Doesn't that kinda> violate the rules, about being able to be constructed as if from> boundaries from algebraic expressions, and such?> This is what I don't get about this weird geometry mumbo jumbo about> things which are self intersecting being considered actual shapes.> What is wrong with these theoretical geometricians?> (...Starblade Riven Darksquall...) === there's a much-simpler example, buti can't recall the particulars. last that I saw itwas in Stillman's book on 3-manifolds,which is introductory. was it a tetrahedron?> Why are certain polyhedra that are self intersecting considered REAL> polyhedra? Clearly polyhedrons which are self intersecting shouldn't> be considered REAL polyhedrons. Then we could do all sorts of weird> stuff to them. Take a look at> http://mathworld.wolfram.com/GreatDodecahedron.html and> http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for> example.--Dec.2000 ?WAND' Chairman Paul O'Neill, reelectedto Board. Newsish?http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanac === > Why are certain polyhedra that are self intersecting considered REAL> polyhedra? > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed> views about this exact question.And where will I ?d that?> Question to all - am I much mistaken in thinking that nowadays, most> mathematicians' views on such issues are of the whatever genre?> How can they not care? Isn't it their job to, you know, care?> cdj> Alright.> Clearly polyhedrons which are self intersecting shouldn't> be considered REAL polyhedrons. Then we could do all sorts of weird> stuff to them. Take a look at> http://mathworld.wolfram.com/GreatDodecahedron.html and> http://mathworld.wolfram.com/SmallStellatedDodecahedron.html for> example.> Why do we consider ?ures which intersect themselves to be> polyhedrons or polyhedra in the ?st place? Doesn't that kinda> violate the rules, about being able to be constructed as if from> boundaries from algebraic expressions, and such?> This is what I don't get about this weird geometry mumbo jumbo about> things which are self intersecting being considered actual shapes. What is wrong with these theoretical geometricians?> (...Starblade Riven Darksquall...)(...Starblade Riven Darksquall...) === > Why are certain polyhedra that are self intersecting considered REAL> polyhedra? > See Imre Lakatos' _Proofs and Refutations_ for a variety of detailed> views about this exact question.> And where will I ?d that?Any decent library.-- === I am trying to derive a very simple equation but for some reason I amhaving trouble with it.I want to derive the equationS = So + vo*(t-to) + 1/2*a*(t-to)^2 where o = notI can derive the equation if I dont use ?t' and ?to' but just use ?t'to represent delta t.Any help? === Differentiate S with respect to t,dS/dt = vo + a*(t-to)Michael Leung???????:d93c7061.0307252129.29ebe8b0@posting.google.com. ..> I am trying to derive a very simple equation but for some reason I am> having trouble with it.> I want to derive the equation> S = So + vo*(t-to) + 1/2*a*(t-to)^2> where o = not> I can derive the equation if I dont use ?t' and ?to' but just use ?t'> to represent delta t.> Any help? === This argument works.We know that v = ds/dt = vo + at = (vo + ato) + (at - ato) = (vo + ato) +a(t - to)-- just rename vo + ato by v'oIE ds/dt = v'o + a(t - to). Integrating gives S= So +v'ot + (1/2)a(t - to)^2which is your desired result.> I am trying to derive a very simple equation but for some reason I am> having trouble with it.> I want to derive the equation> S = So + vo*(t-to) + 1/2*a*(t-to)^2> where o = not> I can derive the equation if I dont use ?t' and ?to' but just use ?t'> to represent delta t.> Any help? === Oops, it is not your desired result.> This argument works.> We know that v = ds/dt = vo + at = (vo + ato) + (at - ato) = (vo + ato) +> a(t - to)-- just rename vo + ato by v'o> IE ds/dt = v'o + a(t - to). Integrating gives S= So +v'ot + (1/2)a(t -to)^2> which is your desired result.> I am trying to derive a very simple equation but for some reason I am> having trouble with it.> I want to derive the equation> S = So + vo*(t-to) + 1/2*a*(t-to)^2> where o = not> I can derive the equation if I dont use ?t' and ?to' but just use ?t'> to represent delta t.> Any help?> === If a is constant: a = (v- vo)/( t - to)=>a( t- to) = v - vo => v = vo +a( t - to) => S = So + vot + (1/2)a( t - to )^2 = (So + voto) + vo( t - to)+ (1/2)a( t - to)^2. Now let S'o = So + voto.> I am trying to derive a very simple equation but for some reason I am> having trouble with it.> I want to derive the equation> S = So + vo*(t-to) + 1/2*a*(t-to)^2> where o = not> I can derive the equation if I dont use ?t' and ?to' but just use ?t'> to represent delta t.> Any help? === Does anybody know if the following problem has been solved, or if anyprogress has been made towards a solution?Let k be a positive integer.Let G(k) be the least n such that all suf?iently large positiveintegers are the sum of <= n kth powers.Let H(k) be the least n such that (the set of positive integers whichcan not be expressed as the sum of <= n kth powers) has Schnirelmanndensity 0.Does G(k) = H(k) ?Paul Epstein === Suppose you have a ?ite abelian cyclic group . Is there acanonical way (or any way, really) to express this alternatively interms of two generators and two relations as:Generators = {g,h}Relations = {ag=0, bh=cg} where a, b, and c are integers.?If so, this would be very useful to me, and I'd appreciate it ifanyone could post with a way to do this (if it indeed can be done). Isthere any kind of _Mathematica_ or Maple command that does thisautomatically, perhaps? (say once you enter the generator of thecyclic group and its order)Or perhaps if it can be done in certain special cases, maybe someonecould point to those? === > Suppose you have a ?ite abelian cyclic group . Is there a> canonical way (or any way, really) to express this alternatively in> terms of two generators and two relations as:> Generators = {g,h}> Relations = {ag=0, bh=cg} where a, b, and c are integers.>I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, withgenerator (1,1) as prototype of theorem. That should work unless order ofgroup is power of prime. === > Suppose you have a ?ite abelian cyclic group . Is there a> canonical way (or any way, really) to express this alternatively in> terms of two generators and two relations as:> Generators = {g,h}> Relations = {ag=0, bh=cg} where a, b, and c are integers.> ?> It's not clear to me exactly what you're given in this problem.If by this you mean we are given n, and told only that the group G =, where g and h are unspeci?d elements of G, is isomorphic toZ_n; then it's not clear that a non-trivial presentation of the givenform exists.By non-trivial here, I mean a presentation where neither = G nor = G; clearly, isalways going to give a representation of Z_a, but that's probably notwhat you want.Consider the cyclic group Z_6. We must select g and h from {2,3,4},since otherwise one of g or h will generate Z_6 on its own; so either{g,h} = {2,3} or {g,h} = {3,4}.If we take g = 2 or 4, then the only possible presentation of yourgiven form is ; if we take g = 3, then .But these are both equivalent to . There is ahomomorphism from this group onto D_3 (dihedral group) withpresentation , and onto Z_6 withpresentation ; so it would seemthere is no non-trivial representation of Z_6 of your form.> If so, this would be very useful to me, and I'd appreciate it if> anyone could post with a way to do this (if it indeed can be done). Is> there any kind of _Mathematica_ or Maple command that does this> automatically, perhaps? (say once you enter the generator of the> cyclic group and its order)> Or perhaps if it can be done in certain special cases, maybe someone> could point to those? === > Suppose you have a ?ite abelian cyclic group . Is there a> canonical way (or any way, really) to express this alternatively in> terms of two generators and two relations as:> Generators = {g,h}> Relations = {ag=0, bh=cg} where a, b, and c are integers.> It's not clear to me exactly what you're given in this problem.See my other post where I try to work this out in general, startingfrom the hints I gave in my ?st post. In working thru this, thereason I disallowed the order of the group to be a power of a primeis elucidated. More comments below.> If by this you mean we are given n, and told only that the group G => , where g and h are unspeci?d elements of G, is isomorphic to> Z_n; then it's not clear that a non-trivial presentation of the given> form exists.> By non-trivial here, I mean a presentation where neither = G nor = G; clearly, is> always going to give a representation of Z_a, but that's probably not> what you want.> Consider the cyclic group Z_6. We must select g and h from {2,3,4},> since otherwise one of g or h will generate Z_6 on its own; so either> {g,h} = {2,3} or {g,h} = {3,4}.> If we take g = 2 or 4, then the only possible presentation of your> given form is ; if we take g = 3, then 2g = 0, 2g = 3h>.>And h = 2 or h = 4. Don't take h = 2, take h = 4,as in accordance to my work in the other post.Thus Z_6 = = <-1>, does it not?> But these are both equivalent to . There is a> homomorphism from this group onto D_3 (dihedral group) with> presentation , and onto Z_6 with> presentation ; so it would seem> there is no non-trivial representation of Z_6 of your form.> === > Suppose you have a ?ite abelian cyclic group . Is there a> canonical way (or any way, really) to express this alternatively in> terms of two generators and two relations as:> Generators = {g,h}> Relations = {ag=0, bh=cg} where a, b, and c are integers.> I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with> generator (1,1) as prototype of theorem. That should work unless order of> group is power of prime.>Let G be a ?ite cyclic group with order n.Assume n > 1 isn't a power of a prime.Thus some co?ite j,k > 1 with n = rsLet (1,0) and (0,1) be generators of Z/r & Z/s.r(1,0) = (0,0) = s(0,1)If I've done this right, G = <(1,1)> = <(1,0) + (0,1)>is a cyclic group of order n with generator (1,1)Oh, if you want it in the - form then use (1,1) = (1,0) - (0,s-1)Whence (0,s-1) is a generator of Z/s and r(1,0) = (0,0) = s(0,s-1)When n the order of G is a power of a prime,then be contented with trivial solution,g = 0, h = 1, G = <0-1> = <-1>, 1g = 0 = nh. === I am looking for software to carry out primality testing of positiveintegers. I have the following requirements: 1) It must be an exact test. Probabilistic tests commonly used incryptographic applications are not an option. 2) It must be able to tackle integers up to about one thousand decimaldigits long. 3) It should return an answer within a reasonable time when running onamodern PC - say, at worst no more than a few hours for any arbitrary integer asabove. Any pointers would be much appreciated. === >I am looking for software to carry out primality testing of positive>integers. I have the following requirements:> 1) It must be an exact test. Probabilistic tests commonly used in>cryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand decimal>digits long.> 3) It should return an answer within a reasonable time when running ona>modern PC - say, at worst no more than a few hours for any arbitrary integer as>above.If you perform an exact primality test on a computer there is always apositive probability that the answer will be wrong because of a random hardware error, cosmic rays, etc. A probabilistic testthat says n is prime with probability p can hence be just asreliable as an exact test, depending on the value of p...> Any pointers would be much appreciated.David C. Ullrich === >I am looking for software to carry out primality testing of positive>integers. I have the following requirements:> 1) It must be an exact test. Probabilistic tests commonly used in>cryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand decimal>digits long.> 3) It should return an answer within a reasonable time when running ona>modern PC - say, at worst no more than a few hours for any arbitraryinteger as>above.> If you perform an exact primality test on a computer there is always a> positive probability that the answer will be wrong because of a> random hardware error, cosmic rays, etc. A probabilistic test> that says n is prime with probability p can hence be just as> reliable as an exact test, depending on the value of p...>David's point is a good one. Apart from exhaustively testingan implementation with all possible inputs of length less than1000 digits, I see no way to prove an implementation attainsyour goal of establishing primality in the given time limit of nomore than a few hours.Bear in mind that assuming a generalized Riemann hypothesisin 1976, Miller showed that testing strong pseudoprimality forO((log N)^2) bases is deterministic.For this reason the tests commonly used in cryptographicapplications might be said to be probabilistic in name only.A good summary of primality testing algorithms is provided bythe Primes in P FAQ found here:http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_ FAQ.htmlHere's a page with links to various primality testing programs:http://directory.google.com/Top/Science/Math/Number_ Theory/Prime_Numbers/Primality_Tests/Primality_Proving/ Software/I've used the PRIMO program by Marcel Martin, based on theECPP algorithm. This approach is attractive for many purposesbecause (for prime inputs) the algorithm outputs a primalitycerti?ate whose validity is much easier to check in practicethan running the ECPP implementation itself. In my experiencethe software handles inputs of a couple of hundred digits veryquickly, but I have no experience in the range of inputs withseveral hundred digits contemplated by you.(replace bellsouth by mpinet and .net by .com to reply) === > I am looking for software to carry out primality testing of positive> integers. I have the following requirements:> 1) It must be an exact test. Probabilistic tests commonly used in> cryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand decimal> digits long.> 3) It should return an answer within a reasonable time when running ona> modern PC - say, at worst no more than a few hours for any arbitrary integer as> above.> Any pointers would be much appreciated.Look at PRIMO - Primality proving using elliptic curves.http://www.ellipsa.net/Hugo Pfoertner === >I am looking for software to carry out primality testing of positive>integers. I have the following requirements: 1) It must be an exact test. Probabilistic tests commonly used in>cryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand decimal>digits long.> 3) It should return an answer within a reasonable time when running ona>modern PC - say, at worst no more than a few hours for any arbitrary> integer as>above.> If you perform an exact primality test on a computer there is always a> positive probability that the answer will be wrong because of a> random hardware error, cosmic rays, etc. A probabilistic test> that says n is prime with probability p can hence be just as> reliable as an exact test, depending on the value of p...> David's point is a good one. Apart from exhaustively testing> an implementation with all possible inputs of length less than> 1000 digits, I see no way to prove an implementation attains> your goal of establishing primality in the given time limit of no> more than a few hours.> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> For this reason the tests commonly used in cryptographic> applications might be said to be probabilistic in name only.> A good summary of primality testing algorithms is provided by> the Primes in P FAQ found here: http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html> Here's a page with links to various primality testing programs:> http://directory.google.com/Top/Science/Math/Number_Theory/ Prime_Numbers/Primality_Tests/Primality_Proving/Software/> I've used the PRIMO program by Marcel Martin, based on the> ECPP algorithm. This approach is attractive for many purposes> because (for prime inputs) the algorithm outputs a primality> certi?ate whose validity is much easier to check in practice> than running the ECPP implementation itself. In my experience> the software handles inputs of a couple of hundred digits very> quickly, but I have no experience in the range of inputs with> several hundred digits contemplated by you.> (replace bellsouth by mpinet and .net by .com to reply)To give an example, PRIMO takes 53min CPU time on an 800 MHz PC toprove (5^1301-2^1301)/3 (908 decimal digits) prime.A strong probable prime test as used in PFGWhttp://www.primeform.net/openpfgw/takes 0.49s on the same PC:Primality testing (5^1301-2^1301)/3 [N-1, Brillhart-Lehmer-Selfridge](5^1301-2^1301)/3 is PRP! (0.490000 seconds)The probability that a random 1000 digit PRP is composite is1.2*10^(-123)(from http://www.utm.edu/research/primes/notes/prp_prob.html )So David Ullrich's comment on hardware reliablility is really justi?d.Hugo Pfoertner === >I am looking for software to carry out primality testing of positive>integers. I have the following requirements:> 1) It must be an exact test. Probabilistic tests commonly used incryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand decimal>digits long.> 3) It should return an answer within a reasonable time when running ona>modern PC - say, at worst no more than a few hours for any arbitrary>integer as>above.> If you perform an exact primality test on a computer there is always a> positive probability that the answer will be wrong because of a> random hardware error, cosmic rays, etc. A probabilistic test> that says n is prime with probability p can hence be just as> reliable as an exact test, depending on the value of p...>David's point is a good one. Apart from exhaustively testing>an implementation with all possible inputs of length less than>1000 digits, I see no way to prove an implementation attains>your goal of establishing primality in the given time limit of no>more than a few hours.>Bear in mind that assuming a generalized Riemann hypothesis>in 1976, Miller showed that testing strong pseudoprimality for>O((log N)^2) bases is deterministic.Couldn't ?ure out what that sentence meant at ?st - thenext sentence clari?d it. Is the result that checking justany old set of that many bases is conclusive, or that thereexists such a set of bases?>For this reason the tests commonly used in cryptographic>applications might be said to be probabilistic in name only.>A good summary of primality testing algorithms is provided by>the Primes in P FAQ found here:>http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.htmlHere's a page with links to various primality testing programs:>http://directory.google.com/Top/Science/Math/ Number_Theory/Prime_Numbers/Primality_Tests/Primality_ Proving/Software/>I've used the PRIMO program by Marcel Martin, based on the>ECPP algorithm. This approach is attractive for many purposes>because (for prime inputs) the algorithm outputs a primality>certi?ate whose validity is much easier to check in practice>than running the ECPP implementation itself. In my experience>the software handles inputs of a couple of hundred digits very>quickly, but I have no experience in the range of inputs with>several hundred digits contemplated by you.>(replace bellsouth by mpinet and .net by .com to reply)>David C. Ullrich === David C. Ullrich escribi.97 en elmensaje|n7?bdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: I am looking for software to carry out primality testing of> positive> integers. I have the following requirements:> 1) It must be an exact test. Probabilistic tests commonly used in> cryptographic applications are not an option.> 2) It must be able to tackle integers up to about one thousand> decimal> digits long.> 3) It should return an answer within a reasonable time when> running ona> modern PC - say, at worst no more than a few hours for any> arbitrary> integer as> above.> If you perform an exact primality test on a computer there is> always a> positive probability that the answer will be wrong because of a> random hardware error, cosmic rays, etc. A probabilistic test> that says n is prime with probability p can hence be just as> reliable as an exact test, depending on the value of p...> David's point is a good one. Apart from exhaustively testing> an implementation with all possible inputs of length less than> 1000 digits, I see no way to prove an implementation attains> your goal of establishing primality in the given time limit of no> more than a few hours.> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> Couldn't ?ure out what that sentence meant at ?st - the> next sentence clari?d it. Is the result that checking just> any old set of that many bases is conclusive, or that there> exists such a set of bases?> === >David C. Ullrich escribi.97 en el>mensaje|n7?bdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: [...]> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> Couldn't ?ure out what that sentence meant at ?st - the> next sentence clari?d it. Is the result that checking just> any old set of that many bases is conclusive, or that there> exists such a set of bases?>Millers Test: If the extended Riemann hypothesis is true, then if n is an>a-SPRP for all integers a with 1 < a < 2(log n)^2, then n is prime.Keen.David C. Ullrich === David C. Ullrich escribi.97 en el>mensaje|n7?bdl0c0pem1fdg8jvmsm0ap4clo@4ax.com: [...]> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> Couldn't ?ure out what that sentence meant at ?st - the> next sentence clari?d it. Is the result that checking just> any old set of that many bases is conclusive, or that there> exists such a set of bases?>Millers Test: If the extended Riemann hypothesis is true, then if n is an>a-SPRP for all integers a with 1 < a < 2(log n)^2, then n is prime.> Keen.>Yes. === > David's point is a good one. Apart from exhaustively testing> an implementation with all possible inputs of length less than> 1000 digits, I see no way to prove an implementation attains> your goal of establishing primality in the given time limit of no> more than a few hours.>Are you testing an implementation of an algorithm or are you testing anumber for primality ? Algorithms are not proven with all possible inputsbut proven primes are tested will all necessary imputs...with the algorithmitself not usually in question.> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> For this reason the tests commonly used in cryptographic> applications might be said to be probabilistic in name only. === > David's point is a good one. Apart from exhaustively testing> an implementation with all possible inputs of length less than> 1000 digits, I see no way to prove an implementation attains> your goal of establishing primality in the given time limit of no> more than a few hours.> Are you testing an implementation of an algorithm or are you testing a> number for primality ? Algorithms are not proven with all possible inputs> but proven primes are tested will all necessary imputs...with thealgorithm> itself not usually in question.> Bear in mind that assuming a generalized Riemann hypothesis> in 1976, Miller showed that testing strong pseudoprimality for> O((log N)^2) bases is deterministic.> For this reason the tests commonly used in cryptographic> applications might be said to be probabilistic in name only.>If you're asking me, I was responding to the original poster's questfor primality testing software. See the OP's requirements, whichwere left in my response, specifying that inputs up to a thousanddigits should be handled in no more than a few hours.In terse fashion I was making a case for Miller's test, involving atmost 94 strong pseudoprime computations to prove primalityin the given range, as being more deterministic with respect torunning time than ECPP. However the suf?iency of Miller's testrests on the extended Riemann hypothesis, and ECPP has nosuch theoretical dependence.My brevity was no doubt confusing. It seemed likely to me thatthe OP knew about strong pseudoprime tests and was explicitlyrejecting them, so in that case one of the ECPP implementationswould be my best suggestion.-- chip === > David's point is a good one. Apart from exhaustively testing> an implementation with all possible inputs of length less than> 1000 digits, I see no way to prove an implementation attains> your goal of establishing primality in the given time limit of no> more than a few hours.> Are you testing an implementation of an algorithm or are you testing a> number for primality ? Algorithms are not proven with all possibleinputs> but proven primes are tested will all necessary imputs...with the> algorithm> itself not usually in question.> If you're asking me, I was responding to the original poster's quest> for primality testing software. See the OP's requirements, which> were left in my response, specifying that inputs up to a thousand> digits should be handled in no more than a few hours.>Yeah, I was responding to your wording of a situation. I guess it can be ===