mm-449 === Subject: Re: Co-ordinate representation !!! Could somebody tell me what the co-ordinate representation [x,y) means . I know it represents co-ordinates on a circle but what would be the range of values b/w x and y One interpretation of this notation is that it represents an interval, a square bracket implies that the endpoint of the interval is to be included and a round bracket implies that it is excluded. So the end point y in your example would not be in the interval, but values arbitrarily close to it, approached from the 'x-side', would be. === Subject: Re: Computer Algebra System CAS & Pocket PC a.. SymbMath.com: web-based symbolic calculator, math handbook and computer algebra system. It can run in any PC with Java, including Pocket PC, many years ago. www.symbMath.com Computer Algebra System CAS & Pocket PC ================================================================== Now, Pocket PC with a Linux OS (called Familiar) are able to run a real Computer Algebra System. This CAS is free and is called xcas This CAS runs on PC with Windows, Linux, ARM Pocket PC with Linux Familiar. It is very performant (as the calculator HP-49 is, because it has been developped by the same programmer). It can be compared with Maple, Mathematica, or Mupad. It is FREE (GNU). At now, no version of this software have been developped for the MS Pocket PC. The only thing missing is volunteers... People who knows eVC++ and are able of such a project... === Subject: Re: [MuPad]Declaring and using arrays > Please provide me with an example of a procedure creating a 1-dimensional > array, passing it to another procedure that sets its values and then in > the first procedure the array > values are printed. > The explanations and examples in the Help on the declaration and use of > arrays has not helped me. > Here's one of the many variations I have tried: > P1 := proc(Cn, n) > local > i; > begin > for i from 0 to n do > Cn[i] := i * i; > end_for; > end_proc; > P2 := proc(n) > local > Cn; > begin > Cn := array [0 .. n]; > P1(Cn, n); > print(Unquoted, Cn[4]); > end_proc; There is no call by reference in MuPAD (*). P1 has to return Cn; and P2 has to assign the return value (Cn:= P1(Cn, n)). (*)Exception: If Cn was a domain and P1 assigned to a slot of the domain, your solution would work because of the reference effect for domains. I don't believe that you are looking for this, and do not go into details. Referenced arrays is precisely what I need, especially for use in a function that returns boolean values and also changes an array. A technique for doing this was provided by Ralf Hillebrand in another reply. I'll be trying that code. -- === Subject: Re: Mathematica 4 symbolics Problem: TrigToExp[ArcTan[x]] gives a compex value 1/2 i Log(-(i+x)/(x-i)) but ExpToTrig[ ] does not return to the ArcTan[ ] (?) What kind of assignments one must use in symbolicprogramming: ComplexExpand, TagSetDelayed, Simplify[ , TransformationFunctions-> ], or something else ? Entering TrigToExp[ArcTan[x]] followed by ComplexExpand[%, TargetFunctions -> {Re, Im}] gets one back to ArcTan[x]. A better newsgroup for Mathematica-specific questions is comp.soft-sys.math.mathematica Cheers, Paul === Subject: Re: Mathematica 4 symbolics A little-bit unlogical for beginner. Tanel Paul Abbott skrev i meddelandet > Problem: > TrigToExp[ArcTan[x]] gives a compex value 1/2 i Log(-(i+x)/(x-i)) > but ExpToTrig[ ] does not return to the ArcTan[ ] (?) > What kind of assignments one must use in symbolicprogramming: > ComplexExpand, TagSetDelayed, > Simplify[ , TransformationFunctions-> ], or something else ? Entering TrigToExp[ArcTan[x]] followed by ComplexExpand[%, TargetFunctions -> {Re, Im}] gets one back to ArcTan[x]. A better newsgroup for Mathematica-specific questions is comp.soft-sys.math.mathematica Cheers, Paul -- A 6009 mailto:paul@physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul === Subject: What's six dollars? Nothing to loose by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id hAFNWnB21514 by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.10 $, legacy) id hAFNWmt03292 [ l followups ] [ l postfp ] [ http://boomersint.org/wwwboard/intboard.html ] [ http://boomersint.org/wwwboard/faq.html ] Subject: Re: Integral of [sin(x)/x]^2 The temptation's too great. I gotta do it. This is done within a distribution-theoretic context. integral (sin(x)/x)^2 dx = pi. Since the function is even, the integral of this from 0 to +infinity should be pi/2. Very beautiful, congratulations Sergio === Subject: Tough Integral : Please, Help I was wondering if any one knows a clever way to solve the following integral: K[x_,y_]= Integrate[ 1/(x^2+1+(x+z)^2) * 1/(y^2+1+(y+z)^2) * Sqrt[2 + z^2] / (2 + z^2 - lz^2), {z,-Infinity,Infinity}] I really only need K[x,y]+K[x,-y]. I think is is easier to find the latter. I can solve the problem with a lot of algebra and factoring the fraction, but if some one knows a more elgant way of doing this, I'd be greatful. === Subject: Re: Tough Integral : Please, Help I was wondering if any one knows a clever way to solve the following integral: K[x_,y_]= Integrate[ 1/(x^2+1+(x+z)^2) * 1/(y^2+1+(y+z)^2) * Sqrt[2 + z^2] / (2 + z^2 - lz^2), {z,-Infinity,Infinity}] -------------------------^^^^^^^^^^ This part seems really strange, so I'd like to ask if you made a typo before I work on the problem further. If that is really what you want, then why not just say a*z^2, where a is just as arbitrary as (1-l). === Subject: Re: Tough Integral : Please, Help > I was wondering if any one knows a clever way to solve the following > integral: > K[x_,y_]= > Integrate[ 1/(x^2+1+(x+z)^2) * 1/(y^2+1+(y+z)^2) > * Sqrt[2 + z^2] / (2 + z^2 - lz^2), {z,-Infinity,Infinity}] -------------------------^^^^^^^^^^ This part seems really strange, so I'd like to ask if you made a typo before I work on the problem further. If that is really what you want, then why not just say a*z^2, where a is just as arbitrary as (1-l). Sorry, several people have asked me about this. lz is another parameter like x and y. === Subject: Re: Tough Integral : Please, Help Here is what I can up with as a possible answer after factoring and using the substitution z= Sinh[t] !(* RowBox[{(-lz), *, RowBox[{(, FormBox[(((-2) @(-lz^2) @(lz^2 - 2) ((lz^2 + 2 x^2 - 1)) ((lz^2 + 2 y^2 - 1)) *ArcTan[lz/(( )(@(2 - lz^2)))]) /(((@2 - lz)) ((lz + @2)) ((lz^4 - 2 lz^2 + 4 x^4 + 4 x^2 + 1)) ((lz^4 - 2 lz^2 + 4 y^4 + 4 y^2 + 1))) + (@(x^4 + x^2) ((2 x^4 - lz^2 x^2 - 2 y^2 x^2 + 3 x^2 - lz^2 - lz^2 y^2 - y^2 + 1)) ((tanh^(-1))((2 @(x^4 + x^2))/(2 x^2 + 1))))/(2 ((x^2 + 1)) ((lz^4 - 2 lz^2 + 4 x^4 + 4 x^2 + 1) ) ((x^4 + x^2 - y^4 - y^2))) + (( )(@(y^4 + y^2) ((2 y^4 - lz^2 y^2 - 2 x^2 y^2 + 3 y^2 - lz^2 - lz^2 x^2 - x^2 + 1)) ((tanh^(-1))((2 @(y^4 + y^2))/(2 y^2 + 1)))))/(2 ((y^2 + 1)) (((-x^4) - x^2 + y^4 + y^2)) ((lz^4 - 2 lz^2 + 4 y^4 + 4 y^2 + 1)))), TraditionalForm], )}]}]) The trouble is that when I plug in values it does not exact agree with a numerical evaluation of the integral nor with an uglier expression I obtained via full factorization. === Subject: Substitution of cot x for a limit problem lim x->Pi for cot(x)/(Pi-x) What would be a general substitution method for this and the overall trigonometric functions for limit problems of this kind in this form. My book is lousy on this in specific. for instance: lim x-> for the above with cot replaced by cos, sin and tan. === Subject: Re: Substitution of cot x for a limit problem lim x->Pi for cot(x)/(Pi-x) What would be a general substitution method for this and the overall trigonometric functions for limit problems of this kind in this form. My book is lousy on this in specific. for instance: lim x-> for the above with cot replaced by cos, sin and tan. Convert all trig functions to sin/cos form, then use the identity sin(x) = sin(Pi-x), and then the fundamental limit sin(x)/x -> 1 as x -> 0. === Subject: Re: Substitution of cot x for a limit problem lim x->Pi for cot(x)/(Pi-x) What would be a general substitution method for this and the overall trigonometric functions for limit problems of this kind in this form. My book is lousy on this in specific. for instance: lim x-> for the above with cot replaced by cos, sin and tan. Nothing beats thinking. cot(x) blows up as x --> Pi (more precisely, |cot(x)| --> + infinity, because cos(x) --> -1 and sin(x) --> 0 through both negative and positive values). Dividing cot(x) by Pi-x makes it blow up even more. As for replacing cot by cos, sin, and tan, nothing will piss off your professor more than your asking this question. You're SUPPOSED to be able to figure this out. For example, sin(x) sin(x) - sin(Pi) ------ = - ---------------- Pi - x x - Pi and as x --> Pi this will, BY DEFINITION, converge to (the negative of) the derivative of sin at x = Pi. You can get it by L'Hospital's rule, too, but this is classified as L'H. abuse. What you're looking for is a purely formulaic approach. Sorry, no sympathy from me. There's a huge wet noodle inside your cranium. Consult it. --Ron Bruck === Subject: Re: Substitution of cot x for a limit problem I would solve it like this: lim x->Pi cot(x)/(Pi-x), 1. lim x->Pi cos(x)/sin(x)/(Pi-x), that is when Pi is counterclockwise: cos(x) reduces to -1. by algebraic calculation rules replacement by -1 is valid. sin(x) goes through extremely small positive values when x is approaching Pi so the overall numerator reduces to -infinity. x is close to Pi but the angle is smaller than Pi so it reduces to plus -> 0 We have -infinity/denom -> (+)0 leads to -infinity. 2. lim x->Pi cos(x)/sin(x)/(Pi-x), that is when Pi is clockwise: cos(x) reduces to -1 sin(x) goes through extremely small negative values when x is approaching Pi so the overall numrator reduces to +infinity. x is moving to -Pi so -Pi - -(~Pi) it reduces to negative -> 0 We have +infinity/denom -> (-)0 reduces to -infinity === Subject: Re: Substitution of cot x for a limit problem Since there's no answer I post it here: Point 2 illustrate wrong thinking. Apart from the obvious misspellings, it should say lim x->-Pi and approaching -Pi, it isn't the correct treatment of the original problem. It is only showed because the method displayed lend itself easily to wrong thinking, which it did. Since cot is an odd function cot(-x) <--> -cot(x) not cot(x). x->Pi isn't the same as x->-Pi however the cot function is periodic with k*Pi, so with respect to disregarding the entire expression and only look on the function cot(x) and -cot(x) the limits are the same. Looking at the wole expression: - cos(x)/sin(x)/(Pi-(-x)); then a right limit is +infinity and a corresponding left limit -infinity. the former task in the previous postage only use right limits of which the second is analytically in error by symbolic manipulation with respect to the original problem. Therefore I asked for a general approach. However I haven't got to Taylor series yet. === Subject: Re: Substitution of cot x for a limit problem Expand the function in a taylor series about the point x=pi. Its limiting behavior should then be clear. My guess is that you will not know how to do this, and your instructor would be unsympathetic to this algorithmic solution instead of the flailing techniques taught in a calculus course. so you should ask your instructor for assistance in the approved methods. For other homework problems you could visit the Dr. Math site. If you are not doing this for homework but want to write a program to do this task, see the first sentence and learn what it means. RJF lim x->Pi for cot(x)/(Pi-x) What would be a general substitution method for this and the overall trigonometric functions for limit problems of this kind in this form. My book is lousy on this in specific. for instance: lim x-> for the above with cot replaced by cos, sin and tan. === Subject: Help with Dixon's character table algorithm I'm working on implementing Dixon's character table algorithm, more-or-less as a matter of personal interest at this point. My intent is to get the basic (and, I gather, slower) algorithm down, and possibly implement the Dixon-Schneider algorithm down the road. The algorithm, which at first seemed entirely opaque and subsequently seemed pretty straightforward, now seems a little more subtle than my second impression had led me to believe... Gotta love that. I'm hoping somebody here can lead me in the direction of an In particular, a couple of things seem ambiguous to me right now... First, the algorithm requires the determination of the nullspace of an integer matrix mod p. Although I'm familiar with modular arithmetic at a basic level, it's not clear to me whether: (1) the matrix is operated on by the mod p, then the nullspace found, or (2) the whole operation is done mod p. If the latter, I am (apparently) entirely unclear on how it's done. Second, Dixon refers to considering the action of a matrix on a set in refining the eigenvectors. What is meant by considering is also unclear to me... Any help you guys could offer would be great; I'm at wits end === Subject: Re: Help with Dixon's character table algorithm now... First, the algorithm requires the determination of the nullspace of an integer matrix mod p. Although I'm familiar with modular arithmetic at a basic level, it's not clear to me whether: (1) the matrix is operated on by the mod p, then the nullspace found, or (2) the whole operation is done mod p. (2) is what is meant. If the latter, I am (apparently) entirely unclear on how it's done. Just do Gaussian elimination as normal, execpt replace all division operations with multiplication by the inverse mod p, and reduce results mod p when they exceed p-1. Actually, it is significantly faster than doing elimination over the rational numbers. === Subject: Eddington Monkey There was a physicist (a Nobel prize winner) on TV many years ago who said he spent all of his time after he retired studying the statistics of Shakespeares writing. I think Im a very determined deterministic Eddington monkey sometimes. Take a 26 by 26 keyboard; the height of each key is proportional to the letter pair frequencies in the first ten thousand characters of Hamlet. I hit the highest key which stays down and it prints the letter t. Next I hit the highest key in the h row because the first key was the letter pair th, and it prints an h. Then I print an e, etc.. It will take another kind of monkey to interpret what I type. Here is the function g to simulate the mon-key. g[n_, rowStart_, y_] := Module[{t, k, an, ma, ne}, an = {rowStart}; t = y; k = rowStart; Do[ne = First[Flatten[Position[t[[k]], Max[t[[k]]]]]]; t[[k, ne]] = -1; k = ne; an = Append[an, ne], {n}]; an] The 26 by 26 table c of the letter pair frequencies from the first ten thousand characters of Hamlet in the statement below is from the book Mr. Babbages Secret, The Tale of a Cypher and APL, by Ole Immanuel Franksen, 1985. alphabet = CharacterRange[a, z] alphabet[[g[48,First[Flatten[Position[c,Max[c]]]],c]]] {t,h,e,r,e,s,t,i,n,d,i,s,e,a,n,t,o,n,g,e,n,o,r,i,t,e,d,a,t,a,r ,t,t,s,a,l, e,t,r,a,s,o,u,r,o,f,t,u,t} I got a hit on an Internet search for the meaning of the word trasour and it is probably a tracery, something you might find in king tuts tomb (which was broken into twice in antiquity) and sold at an art sale. The 28 by 28 letter pair correlation matrix s from the book Scientific and engineering problem-solving with the computer, by William Ralph Bennett, Jr., 1976 based on the dialog from Act III of Hamlet with an alphabet that includes the space and the apostrophe: it is obvious the first letter must be a q because that row has a sum of one more than the sum of the corresponding column. alphabetTwo = Join[CharacterRange[a, z], { , }] alphabetTwo[[g[26,17,s]]] {q,u,r, ,t,h,e, ,a,n, ,m,e,r,e,n,d, ,w,h,a,t, ,s, ,i,n} They spell the Moslem religious book the Quran, sometimes. Does anybody know what this kind of characteristic run through a frequency table is called? Can anybody give a probability based estimate of what to expect from it from various text? Cliff Nelson === Subject: Phonetic research with Maple, Maple-Soundex X-ID: TnE17mZGgeaA5wp4mH+fAc0KrCfAqSDBZaTHozz5otGd8PBs3WO7od MUG, I have developed a Maple package that compares customer data phonetically and automatically classifies it. It takes use of an advanced Soundex algorithm incorporating my own analysis of real data, study of referring books on phonetics and linguistics, and even experiences stemming from the Phoenicians, who developed our ABC 3,000 years ago. Classifications: - persons being the same despite of typing errors, - family members being found, - persons not belonging to the same family, - abbreviations, - subsets, - phonetic analysis of strings representing human names, - etc. If you are interested, please contact me at: afw@algonetics.de. Alexander Walz with(algonetics): ana(Fris.9arparty, Vrisoerpartie); # Japser [true, EQ] ana(Birgitta Schmidt, B. Schmidt); [true, CUT] ana(Birgitta Schmidt, Schmidt, Beate); ana(Birgitta Schmidt, Anton Meier); [false] ana(Birgitta Schmidt, Meier, Birgitta); [true, WED] ana(Birgitta Schmidt-M.9fller, Birgitta M.9fller); [true, SUB] === 4577 === Subject: Re: Particular solution to differential equation I am a senior high-school. In this year's mathematics curriculum, we are introduced to differential equations in the form of y' = ay +b [...] I know perfectly well how to deal with such exercises, and usually don't do any mistakes. However I am wondering how it would be possible to find that particular solution, h(x). Do you have any information about it? I would just like to comment how refreshing it is to see that there are high school students who are not only studying differential equations but who are thinking past the exam questions and would like to know a little more of the material. Can I interest you in being a student where I teach? dave === Subject: Re: What is the number derivative ??? X-ID: rqjLyvZAwepTgRx46SSuPJPv7kER9G3A1rughp9sH3GCViIYS6jDYX Am 19.03.04 22:51 schrieb Steve Wildstrom: >Yep, may be I made my words unwanted sharp. I just wondered, that the >abstract/announcements of the two resources sounded very similar about It's not easy to find, but the paper is available on the Web at http://web.mit.edu/lwest/wwww/intmain.pdf Gottfried Helms === Subject: Conditions for being periodic For the two periodic signals with different period for each x1 with T1 and x2 with T2. If the sum of these 2 signals y(t)= x1+ x2 to be also periodic How could we determine the condition(s)? I think if there's existing the L.C.M. of each period T1, T2, that can be the period of the sum, but if the range if those periods are in irrational number like 3^0.5(square root 3) it's not possible to find L.C.M. -------------- .b3[CapitalYAc ute] .bd[P aragraph] KORNET ------------- === Subject: Re: Conditions for being periodic taejo For the two periodic signals with different period for each x1 with T1 and x2 with T2. If the sum of these 2 signals y(t)= x1+ x2 to be also periodic How could we determine the condition(s)? I think if there's existing the L.C.M. of each period T1, T2, that can be the period of the sum, ... True. If the sum is periodic, then each of the original periods divides the period of the sum, and conversely if W is a multiple of both periods, then the sum signal (or function) has W as a period. In more general cases I think the keyword is almost-periodic functions, but I'm no expert there. === Subject: easy...easy....problem.. lim n(x^n) = 0 n->00 --------------------- um.......how to prove it?? help.....me , please === Subject: Re: easy...easy....problem.. @news.hananet.net: lim n(x^n) = 0 n->00 --------------------- um.......how to prove it?? help.....me , please Off the top of my head, I'd try to L'hopital that bastard. === Subject: Re: easy...easy....problem.. > lim n(x^n) = 0 > n->00 Case x <= -1 Case x in (-1,1) Case 1 <= x Off the top of my head, I'd try to L'hopital that bastard. Do you know the dad? === Subject: Re: easy...easy....problem.. lim n(x^n) = 0 n->00 --------------------- um.......how to prove it?? help.....me , please If it's so easy, why are you asking? === Subject: Re: easy...easy....problem.. > lim n(x^n) = 0 > n->00 > --------------------- > um.......how to prove it?? > help.....me , please If it's so easy, why are you asking? Simple. Normal dude comes on here and asks for homework help, everyone says, do your own homework. Same dude comes on here with handle hot-girl and twenty guys fall all over themselves providing detailed solutions to simple homework problems. Look up all the hot-girl posts and you'll see what I mean. === Subject: Re: easy...easy....problem.. um.....sorry. but i am not student. my home-work is clean up my room and hang out the wash to dry. i don't have the math-homework. === Subject: Re: easy...easy....problem.. @news.hananet.net: um.....sorry. but i am not student. my home-work is clean up my room and hang out the wash to dry. i don't have the math-homework. If this is not homework, then why are you asking? If you are asking because you are learning on your own, then you should spend some time and effort in learning before you ask. Then, something better than just saying: A,B is compact but A intersection B is not compact find example is required. Try saying something like, The definition of compact that I am using is every open covering A of X contains a finite subcollection that also covers X. or The defintion of compact that I am using is a set K contained in R is compact is every sequence in K has a subsequence that converges to a limit that is also in K, or The definition of compact that I am using is a set K contained in R is compact if and only if it is closed and bounded. All three of these are equivalent definitions, but if we don't know which one you are using, it will be harder for us to give an example. Here's an example, I think that will work, though A = {1-1/(n+1):n in Z+} U {1} B = {0} U {1/n:n in Z+} Then A intersect B = {1/n:n in Z+} I'm not sure if this works, because it is late, I'm not a topologist or analyst, and I am drunk right now. === Subject: Re: easy...easy....problem.. A,B is compact but A intersection B is not compact find example Try saying something like, The definition of compact that I am using is every open covering A of X contains a finite subcollection that also covers X. Not needed, that's the definition. The definition of compact that I am using is a set K contained in R is compact is every sequence in K has a subsequence that converges to a limit that is also in K, That's sequentially compact. or The definition of compact that I am using is a set K contained in R is compact if and only if it is closed and bounded. All three of these are equivalent definitions Only for R^n like spaces. Here's an example, I think that will work, though A = {1-1/(n+1):n in Z+} U {1} B = {0} U {1/n:n in Z+} Then A intersect B = {1/n:n in Z+} I'm not sure if this works, because it is late, I'm not a topologist or analyst, and I am drunk right now. Compact sets of Hausdorff space are closed. Closed subsets of compact space are compact. Assume the space is Hausdorff, which R^n is. When A and B are compact, they are closed, so A / B is also closed. closed A / B subset compact B, thus A / B is compact. Hence examples of compact A,B and not compact A/B can only be found in non Hausdorff spaces Graduate Student Department of Mathematics Wake Forest University Sober up and quit drinking. === Subject: Re: oh.....sorry.... lim n(x^n) = 0 n->00 ==== 0 It crops up every time its like the 'witness' to uncountable. Is > this the only specified_irrational number that supports Cantors > argument? > Its like this statement has no proof forces us to discard > formal mathematics. There isn't only one witness number (which doesn't have to be irrational, by the way, only not on the list). Only one is mentioned (for a given list) because only one is needed to show that the given list is incomplete. If there were only one number missing from the list, then, yes, absolutely, there should be some other list that includes them all. However, there are actually as many Cantor-diagonals as there are reals. Rather than thinking of your job as almost completed (Just one more to cram in), you should think of it as barely begun (Still as many as we started with). There are just as many Cantor-diagonals as reals, that is, there is a bijection between the Cantor-diagonals and the reals: (In base ten) a real satisifies the Cantor-diagonal condition as long as its n-th place avoids three digits: 0, 9, and the n-th place digit of the n-th real in a specific list. That leaves a choice of seven digits for each place in the Cantor-diagonal. So it should be a simple matter to find a bijection between the Cantor-diagonals of any specific list and the reals in [0,1) when they're in base seven. Why do you say this statement has no proof forces us to discard formal mathematics? Goedel's results _surprised_ us, certainly, but, if anything has changed, mathematics has only grown _more_ formal. I think that's because formality is the only tool left to mathematicians to find their way out of the maze that their intuition led them into. Formal mathematics is just a cheap excuse to dodge the Axiom of Choice. Which is a just excuse for probabilty theory. Which is just a cheap excuse for physics, and Einstein previously stated. === Subject: Re: Cantor's Diagonal Argument [...] > It crops up every time its like the 'witness' to uncountable. Is > this the only specified_irrational number that supports Cantors > argument? > Its like this statement has no proof forces us to discard > formal mathematics. There isn't only one witness number (which doesn't have to be irrational, by the way, only not on the list). Only one is mentioned (for a given list) because only one is needed to show that the given list is incomplete. If there were only one number missing from the list, then, yes, absolutely, there should be some other list that includes them all. However, there are actually as many Cantor-diagonals as there are reals. Rather than thinking of your job as almost completed (Just one more to cram in), you should think of it as barely begun (Still as many as we started with). There are just as many Cantor-diagonals as reals, that is, there is a bijection between the Cantor-diagonals and the reals: (In base ten) a real satisifies the Cantor-diagonal condition as long as its n-th place avoids three digits: 0, 9, and the n-th place digit of the n-th real in a specific list. That leaves a choice of seven digits for each place in the Cantor-diagonal. So it should be a simple matter to find a bijection between the Cantor-diagonals of any specific list and the reals in [0,1) when they're in base seven. I'm talking about the halting number here : digit 1 = 1 if UTM(1) halts, digit 2 = 1 if UTM(2) halts... a non computable irrational. When the Cantor argument is questioned this number is used to support the argument. We've discarded the diagonal construct number(s) for now, considering they are 'unspecifiable'. I'm looking for more specified_irrationals that are not computable, like the halting number. Using the halting number to support uncountability is like using a design for an underwater helicopter, its proven not to work in Earth's oceans, so that proves we can land on Jupiter. Why do you say this statement has no proof forces us to discard formal mathematics? Goedel's results _surprised_ us, certainly, but, if anything has changed, mathematics has only grown _more_ formal. I think that's because formality is the only tool left to mathematicians to find their way out of the maze that their intuition led them into. Formal mathematical statements can be counted. A Univeral System of collected facts is no longer possible, because you believe the self paradoxical statement for no reason, other than it provides a cute theory of incompleteness that hasn't been countered yet. True in 1 system provable in another is not allowable in our universe, why do you accept it in mathematics? Herc === Subject: Re: Gaussian-Jordon Reduction C or C++ Library I am looking for a C or C++ library that supports Linear Systems of equations. I have a real-time application that needs to reduce 10's of thousands of simultaneous equations quickly. I am looking for speed optimized reduction algorithms like Gaussian-Jordon elimination or reduction techniques encapsulated in a library. Can you please assist me in finding this library or more appropriate news group. First off, I hope you are aware that there's other decompositions than Gauss-Jordan that are sometimes useful for solving systems of equations. Nevertheless, I use Gaussian elimination in Java in my LP solver. Anyways, I thought thre was something called the Basic Linear Algebra System (BLAS) in C. But this site seems to say it is a Fortran library with merely a C interface: Here's an interesting page: Are you aware of the book Numerical Recipes in C? For some editions, the code seemed buggy to me. You must read and understand the description of the math to get it work. But I have found the various language-based editions of use. Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Packing squares into a 1 x Zeta[2] rectangle I have seen your program failing on square 12,867 and I have figured out what is happening. It turns out that the programs diverge on handling the square 1/2838. And this is because the denominator n in your rectangle has reached about 308 (decimal) digits, and 10^308 is about the limit for the Java double type. So, java considers that the denominator to be infinite and your x gets set to zero, and two rectangles are lost. A few thousand squares later we run out of room! Here is the relevant part of my trace of your program... (A_B is the fraction A/B) Dear Clive, I should have thought of that. I'll go back to my homework ... Actually, I think the use of floating point arithmetic is unnecessary, as using BigIntegers throughout only increased the time marginally (45secs instead of 44secs with 10.000 squares). Or I could use the BigDecimal class to get around the problem. One other thing about your program that I noticed is that the method checkOrder does not update lastx each time around the loop so it will probably never find any mis-ordering. So my feeling of satisfaction that the order was always correct was misplaced ... as I think I read somewhere that there was a similar problem for 1/n x 1/n+1 rectangles fitting into a 1 x 1 container. I wonder if your work has led you to any ideas about how one might prove the result (for squares)? I thought when I'd got the Java right I'd see how critical it is to remove the square from the smallest possible rectangle at each point. I wondered too how the length of the list of rectangles varies as a function of the number of squares and the cut-off (epsilon) chosen? Can one predict from that that the algorithm is likely to work indefinitely? I had the impression from my mistaken figures that the descent was too steep to avoid a crash unless there was divine intervention. Tim Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: easy problem... 2^(x+1) + x = 4 x = ?? ---------------------- um.....i can't.... help...me...please....thank you. === Subject: Re: easy problem... X-ID: VI0JyvZpwezy6fZ9cT3YyxLOiL2DVf2BnMyOC0Jm6yau3C5GOt6fUH hot-girl schrieb: 2^(x+1) + x = 4 x = ?? You want to solve x = 4 - 2*2^x. LHS is strictly increasing, RHS strictly decreasing. Look what happens as x-> +/- oo and you'll see that the equation has a unique solution x0, 0 < x0 <1 (since 4 -2*2^1 =0). You can find x0 easily numerically (x0 ~ 0.72) You can also rewrite that equation, so that it can be solved using Lambert's W-function. But that seems to be a little too much for such a simple problem. === Subject: Re: easy problem... 2^(x+1) + x = 4 x = ?? Not easy. === Subject: easy...topology problem1 compact and open subspace find example ----------------------- help me...please === Subject: Re: easy...topology problem1 === Subject: easy...topology problem1 >not compact as closed subspace Find easy examples in the reals. -- >compact and open subspace >not closed set as compact subspace >A, B is compact but A intersection B is not compact let S = N/{x,y} have topology { U, {x} / K, {y} / K | U subset N, K cofinite subset N } A cofinite subset N ==> A / {x}, A / {y} open compact; A / {x,y} closed compact A/{x} / A/{y} = A not compact intersection compact sets === Subject: Re: easy...topology problem1 compact and open subspace find example Take any compact space K. Then K is a compact and open subspace of K. Best regards, Jose Carlos Santos === Subject: Re: easy...topology problem1 hot-girl compact and open subspace find example Consider a space which is compact but not connected. === Subject: easy...topology problem2 not compact as closed subspace find example ----------------------- help me...please === Subject: Re: easy...topology problem2 not compact as closed subspace find example It seems that you are asking for an example of a closed but not compact subspace of some topological space. It can't be that simple! Please explain your question. Best regards, Jose Carlos Santos === Subject: easy...topology problem3 not closed set as compact subspace find example ----------------------- help me...please === Subject: Re: easy...topology problem3 hot-girl not closed set as compact subspace find example In a Hausdorff space, any compact set is closed. But (informally speaking) consider a space like the interval [0,1] but containing two copies of the point 1. More formally, start with the union of the intervals [0,1] and [2,3], and identify x with x+2 for all x in [0,1), but not for x=1. ----------------------- help me...please === Subject: Re: Open set > what I gather from miscellinous discrete readings is this: > a topological space is defined as a collection of open sets, the term > open being sort of undefined, so that it follows the rules that any > intersection and (infinite) union of open sets is still open. > how is this definition going to make sense? along with the definition > of a continuous function as a function whose inverse image of an open > set is open? could you give me some clue as to how to get high at > this? The open sets of R^n have the properties: any finite intersection and any infinite union of open sets is open. So it is an example of a topology in the more abstract definition. I understand that. The thing is, for me, the idea of an open set is dependent on the idea of the neighborhood, points lying close to a point, and I don't seem to understand what plays that role in the abstract formulation. To be clear, I have read *some* graduate texts that define topological spaces, but none that has any worked out example, especially when it comes to the idea of closeness. They give the impression that open sets are more fundamental, and you don't *need* any idea analogous to a nbd in the basic topology. How is that? > to clarify my position here, I am confused about the ordering of the > real numbers that looks so unavoidable in the epsilon-delta > definition, which I understand is equivalent to the description in > terms of open sets as defined in my textbook, that is, along with > nbd's that make use of the ordering. the concept of choosing a very > small interval here to get a very small interval there does not seem > to make sense in the abstract definition I guess. So you will get to a theorem that says: a function f : R^n -> R^m is continuous if and only if the inverse image of an open set is open. So continuous functions defined in the epsilon-delta way are a special case of those defined in the abstract way. Yes, I see. But could you be more specific about what results are being successfully transferred to the abstract formulation and what are being left behind for, suppose, real analysis only, of the high school theorems? Just a curiosity, is the word theory the plural form of theorem? === Subject: Re: Open set > what I gather from miscellinous discrete readings is this: > a topological space is defined as a collection of open sets, the term > open being sort of undefined, so that it follows the rules that any > intersection and (infinite) union of open sets is still open. > how is this definition going to make sense? along with the definition > of a continuous function as a function whose inverse image of an open > set is open? could you give me some clue as to how to get high at > this? > The open sets of R^n have the properties: any finite intersection and > any infinite union of open sets is open. So it is an example of a > topology in the more abstract definition. I understand that. The thing is, for me, the idea of an open set is dependent on the idea of the neighborhood, points lying close to a point, and I don't seem to understand what plays that role in the abstract formulation. To be clear, I have read *some* graduate texts that define topological spaces, but none that has any worked out example, especially when it comes to the idea of closeness. They give the impression that open sets are more fundamental, and you don't *need* any idea analogous to a nbd in the basic topology. How is that? In general topology you can define everything in terms of neighborhoods. You can set up some axioms that neighborhoods satisfy. Then define: a set U is open iff for every x in U there is a neighborhood N of x contained in U. And in reverse: starting with open sets, define: N is a neighborhood of x iff there is an open set G such that x in G subset N. These two approaches are equivalent, but back in 1940 or so it was decided that the open set version was preferrable. For example, Hausdorff's definition was in terms of neighborhoods. When translated to the open set language you get a topology with the extra property that distinct points lie in disjoint neighborhoods, since that was one of Hausdorff's axioms. Drop that axiom in a suitable way and you get a set of axioms for neighborhoods that is equivalent to the open-set formulation of topology. There are other possible axiomitizations for topology as well. Specify which nets converge. Or which filters. Or use monads in non-standard analysis. Or a closure operator. I have an old, old paper on this... Three cryptoisomorphism theorems. In: Studies in Foundations and Combinatorics, G.-C. Rota (editor), Advances in Mathematics Supplementary Studies 1, Academic Press, 1978. pp. 49--60. MR 80c:54001 [...] Yes, I see. But could you be more specific about what results are being successfully transferred to the abstract formulation and what are being left behind for, suppose, real analysis only, of the high school theorems? Leave something for when you study point-set topology! Just a curiosity, is the word theory the plural form of theorem? No. G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Open set >what I gather from miscellinous discrete readings is this: >a topological space is defined as a collection of open sets, the term >open being sort of undefined, so that it follows the rules that any >intersection and (infinite) union of open sets is still open. >how is this definition going to make sense? along with the definition >of a continuous function as a function whose inverse image of an open >set is open? could you give me some clue as to how to get high at >this? > The open sets of R^n have the properties: any finite intersection and > any infinite union of open sets is open. So it is an example of a > topology in the more abstract definition. I understand that. The thing is, for me, the idea of an open set is dependent on the idea of the neighborhood, points lying close to a point, and I don't seem to understand what plays that role in the abstract formulation. To be clear, I have read *some* graduate texts that define topological spaces, but none that has any worked out example, especially when it comes to the idea of closeness. They give the impression that open sets are more fundamental, and you don't *need* any idea analogous to a nbd in the basic topology. How is that? In topology, a neighbourhood of a point P is a set A such that A has an open subset G which has P as an element. >to clarify my position here, I am confused about the ordering of the >real numbers that looks so unavoidable in the epsilon-delta >definition, which I understand is equivalent to the description in >terms of open sets as defined in my textbook, that is, along with >nbd's that make use of the ordering. the concept of choosing a very >small interval here to get a very small interval there does not seem >to make sense in the abstract definition I guess. > So you will get to a theorem that says: a function f : R^n -> R^m > is continuous if and only if the inverse image of an open set > is open. So continuous functions defined in the epsilon-delta > way are a special case of those defined in the abstract way. Yes, I see. But could you be more specific about what results are being successfully transferred to the abstract formulation and what are being left behind for, suppose, real analysis only, of the high school theorems? The intermediate value theorem becomes the theorem that the continuous image of a connected set is connected. The extreme value theorem becomes the theorem that the continuous image of a compact set is compact. Just a curiosity, is the word theory the plural form of theorem? No. David McAnally But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane === Subject: Re: Open set to clarify my position here, I am confused about the ordering of the > real numbers that looks so unavoidable in the epsilon-delta > definition, which I understand is equivalent to the description in > terms of open sets as defined in my textbook, that is, along with > nbd's that make use of the ordering. the concept of choosing a very > small interval here to get a very small interval there does not seem > to make sense in the abstract definition I guess. > So you will get to a theorem that says: a function f : R^n -> R^m > is continuous if and only if the inverse image of an open set > is open. So continuous functions defined in the epsilon-delta > way are a special case of those defined in the abstract way. >Yes, I see. But could you be more specific about what results are >being successfully transferred to the abstract formulation and what >are being left behind for, suppose, real analysis only, of the high >school theorems? The intermediate value theorem becomes the theorem that the continuous image of a connected set is connected. That is, if X and Y are topological spaces, f : X -> Y is continuous, and A is a connected subset of X, then {f(x) : x in A} is connected. The extreme value theorem becomes the theorem that the continuous image of a compact set is compact. That is, if X and Y are topological spaces, f : X -> Y is continuous, and A is a compact subset of X, then {f(x) : x in A} is compact. David McAnally But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane === Subject: easy...topology problem4 A, B is compact but A intersection B is not compact find example ----------------------- help me...please === Subject: Re: easy...topology problem4 A, B is compact but A intersection B is not compact find example A necessary condition is that A and B must be included in a non-Hausdorff space X. Take X = R (the reals), and consider the topology generated by the basis {R, empty_set, {x}; x >=0}. The two subspaces A = {-1} U {x; x>=0} and B = {-2} U {x; x>=0} are compact Hausdorff but A intersection B is not compact. -- Julien Santini === Subject: Question on sums of squares... Hello all, Back in January I asked about another sums of squares problem. This question is sort of a progression off of that one. I had so much help from that one I was hoping someone could help me with this one. I've done some searching on my own, but I was hoping someone could verify my results so far. I'm looking for a set (all sets?) of integers that satisfy the following sets of equations. a1^2 + a16^2 = x a2^2 + a15^2 = x a3^2 + a14^2 = x a4^2 + a13^2 = x a5^2 + a12^2 = x a6^2 + a11^2 = x a7^2 + a10^2 = x a8^2 + a9^2 = x (a16^2 - a1^2)^2 + (a9^2 - a8^2)^2 = y (a15^2 - a2^2)^2 + (a10^2 - a7^2)^2 = y (a14^2 - a3^2)^2 + (a11^2 - a6^2)^2 = y (a13^2 - a4^2)^2 + (a12^2 - a5^2)^2 = y ((a16^2 - a1^2)^2 - (a9^2 - a8^2)^2)^2 + ((a13^2 - a4^2)^2 - (a12^2 - a5^2)^2)^2 = z ((a15^2 - a2^2)^2 - (a10^2 - a7^2)^2)^2 + ((a14^2 - a3^2)^2 - (a11^2 - a6^2)^2)^2 = z Basically, the first 8 equations all equal each other, then the next 4 equations all equal each other, and the last 2 equations equal each other. I don't know if there's a way to generate all the solutions, but maybe there is some algorithm to create them. I've searched for x up to 55e6 but still haven't found anything that meets all of the criteria given above. I was hoping maybe someone here could independantly verify this, or maybe prove me wrong with a solution! ;) Anyway, thanks in advance for any help you can provide. -David C. === Subject: Re: Question on sums of squares... ... I'm looking for a set (all sets?) of integers that satisfy the following sets of equations. a1^2 + a16^2 = x ... (a16^2 - a1^2)^2 + (a9^2 - a8^2)^2 = y ... ((a16^2 - a1^2)^2 - (a9^2 - a8^2)^2)^2 + ((a13^2 - a4^2)^2 - (a12^2 - a5^2)^2)^2 = z ... Basically, the first 8 equations all equal each other, then the next 4 equations all equal each other, and the last 2 equations equal each other. [...] I've searched for x up to 55e6 but still haven't found anything that meets all of the criteria given above. I was hoping maybe someone here could independantly verify this, or maybe prove me wrong with a solution! ;) ^e There are infinitely many trivial solutions; do you have some criterion that rules them out? For example, choose any constant k and set all ai=k. -jiw === Subject: Re: Question on sums of squares... ... > I'm looking for a set (all sets?) of integers that satisfy the following > sets of equations. > a1^2 + a16^2 = x ... > (a16^2 - a1^2)^2 + (a9^2 - a8^2)^2 = y ... > ((a16^2 - a1^2)^2 - (a9^2 - a8^2)^2)^2 + ((a13^2 - a4^2)^2 - (a12^2 - > a5^2)^2)^2 = z ... > Basically, the first 8 equations all equal each other, then the next 4 > equations all equal each other, and the last 2 equations equal each > other. [...] I've searched for x > up to 55e6 but still haven't found anything that meets all of the > criteria given above. I was hoping maybe someone here could > independantly verify this, or maybe prove me wrong with a solution! ;) ^e There are infinitely many trivial solutions; do you have some criterion that rules them out? For example, choose any constant k and set all ai=k. -jiw Yeah, last night when I went to sleep it hit me that those trivial solutions existed. Unfortunately I didn't want to get up and re-post the extra criteria. But now that I'm awake the missing criteria is: a1 There are infinitely many trivial solutions; do you have > some criterion that rules them out? For example, choose > any constant k and set all ai=k. > -jiw Yeah, last night when I went to sleep it hit me that those trivial solutions existed. Unfortunately I didn't want to get up and re-post the extra criteria. But now that I'm awake the missing criteria is: I actually forgot one small detail (guess I'm not fully awake yet), the criteria should actually be: 0 a etc. No, the reciprocals of sin(A)/a = sin(B)/b = sin(C)/c ; To elaborate: a,b,c are geodesically drawn sides (in terms of sphere radius ) of a spherical triangle as angles subtended at center of sphere. When considering shrinking size of the spherical trigon,(a,b,c)-> 0 and tan[R]->R {or sphere radius -> Infinity}, the trigon gets flat towards limiting case Euclidean geometry.A,B,C are angles opposite of (a,b,c) on the sphere,(A+B+C-Pi) is the Spherical Excess in Gauss-Bonnet Theorem going to zero in flat case of plane trigonometry obtaining Sine Rule a/Sin[A]= ... = ... = 2R = D = circum-diameter. Also my typographical error above, without adverse bearing here, is correctly D = 2 tan[R] cos[a/2]Cos[b/2]Cos[c/2]. === Subject: Re: Randomness is a notion. [...] I won't entertain the notion notion again, and randomness as insufficient information is already being debated in another thread. However: > Because science is forever discovering more regularities > in nature, it's best to assume that > the future is perfectly fixed. > ( Even though the future is imperfectly known ) Hawking apparently shared this view for a time, and even held that in the big crunch everything would just stop and then reverse all events which had happenened in our universe since the Big Bang. Smashed dishes would come together whole again. Apples would fly from the ground and attach themselves to trees. The scattered molecules of Socrates would swoop from around the world to eventually converge in Greece. He would awaken from death, sit up on his bed, and hemlock would spew from his mouth into a cup. The tears of his friends would run up their faces and into their tear ducts. I'm not sure he still holds this view, since changing beliefs on the effects of expansion-contraction on entropy, and he seems uncertain of time-symmetry issues. It would certainly be an extreme-determinist view. I believe there are so many reasons to reject this view, and so few to recommend it, that it incurs a huge burden of proof. One of the more obvious questions: What would make the universe contract if gravity were to start working in reverse? The conclusive objection, to me, is that there would have to be a higher time framework in which this reversal could occur and be observed. In fact, the higher framework for the theory is just the imaginary god-like perspective of the person imagining it, so the conception is also imaginary. I mean, what would distinguish play from rewind? If there is no difference, then it's all one thing, no? -- john === Subject: Re: Going back home The task is having or assembling some marker from which to navigate. A random walk alone would get you home - you did specify indefinite time. Jerry This is basically of theoretical interest, but here goes: Assume one is dropped somewhere on the vicinity of the Milky way. Could a consistent and practical Mathematical model for navigating around be found assuming one has unlimited time to travel? Making the question more specific: Does there exist a sufficiently accurate (to be practical) Mathematical model that would allow one to calculate one's way back to Earth, once one was dropped, say, near a star which sits 6,000 light years away from Earth? It seems to me that an appropriate (linear(?) because the distances are big) transformation T: R^3->R^3 could be used to calculate the new star positions at any location, but how could one utilize such a device to find one's way back to Earth? -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Going back home charset=utf-8 jerry warner [CapitalEth][EDouble Dot][Micro] .b3 .b9[EDo ubleDot].b9 The task is having or assembling some marker from which to navigate. A random walk alone would get you home - you did specify indefinite time. Yes, but I changed that later to instantaneous travel. It was a wrong assumption based on infinite time to travel on the 'shortest' distance , whatever 'shortest' means in this case. Given the wrong assumption infinite time (plus relativity), a random walk would get me exactly where I started. Jerry -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Going back home Also, there are some considerations regarding dimensions. The Milky Way has an estimated radius of 60,000 light years, taking the black hole as the center. This is 5.67 * 10^20 m. It's worth mentioning that there's no more than one significant digit in the radius estimate, so the additional digits in these conversions are spurious. log2(5.67 * 10^20) = 68.94, which means standard IEEE-754 precision equipment won't quite cut it, as it can only handle 52 bits in the mantissa. An accuracy of 100 km is quite good, actually. It's about 0.00015 solar radii, small enough to represent over a trillion unique positions inside the sun, from a distance of 60,000 light-years. Even if it weren't sufficient, it'd be easy enough to store each number in two doubles rather than one. This is how time is represented in the code used by the U.S. Naval Observatory to calculate ephemerides, for example. One might also have said that 32 bits isn't enough, by a factor of 100, to assign a unique index (or memory address) to each star in the Milky Way, but again that's routinely handled by using more than one 32-bit integer per index. And the problem disappears entirely on 64-bit machines. - Ernie http://mywebpages.comcast.net/erniew === Subject: Re: Commutative algebra Hi all, does anyone know a good and entertaining math book on commutative algebra/ring theory ? I went through Neal McCoy 'Rings and ideals' and felt in love with it. It is both stimulating and progressive in the approach and i strangely (i actually don't know why) find it entertaining/amusing. But it lacks exercices (potentially with solutions/hints) to strengthen up what had might have been loosely understood throughout the text and a more advanced look. My aim would be to be as well prepared as possible to dig into books like Miles Reid's 'Undergraduate commutative algebra' (see thread called Please recommend some entertainment math books) or Matsurama's book on commutative algebra. The few first paragraphs of Reid's book are quite appealing to me (local algebra, ...). Hmmm. I'd regard Eisenbud's book (Comm Alg with a view towards Alg Geom) as quite entertaining. It gives lots of motivation and applications to other branches of maths as well as all the big theorems and proofs. In an other subject, do you have any comments on Humphreys's 'Introduction to lie algebra' ? Is it accessible to an undergraduate self-study folk ? That's quite readable. Now Jacobson's classic book has comparable scope, perhaps not as readable as Humphreys and a bit old-fashioned but it is much cheaper, being a Dover paperback. Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Hex Win Proof? > It is an old theorem that in Hex, once the board has been completely > filled in with two colours, there *must* be a winning path for one > or other of them. > Now, I can prove this easily enough mathematically, but I'm wondering if > there is a simple proof, or proof outline, that would be understandable > and reasonably convincing to the intelligent layman. > Can anyone help out please? Neither of the proofs (which are basically the same) posted so far is correct. Both would apparently conclude that a winning path would be formed on a squared board, whereas this is not the case - a squared board could end in a draw. An actual proof must use the hex nature of the board or, alternatively, that 3 cells meet at each vertex. A proof is given in Cameron Browne's book Hex Strategy, but whether it would convince an intelligent layman is not clear. Maybe a simpler proof could be achieved by induction? Im not sure what proof you are thinking of, but you make it sound like it's rather complicated. Here's a proof I think is pretty simple and straightforward (but as we've some evidence of, not simple to come up with!); you'll have to tell me if this is the same as Browne's proof. I think the following proof is understandable to the so-called intelligent layman: Suppose we've filled up the board with black and white colors. The boundaries of the hexagons form a graph with edges and vertices. We can look at the edges that separate hexagons of different colors. Because of the way the hexagons meet (three to a vertex), there is always a unique way to continue one of these special edges to another edge that separates different colors. In other words, we can't have more than two of these color separating edges meeting at a common vertex. So the boundary between the colors consists of either loops (that don't intersect each other or themselves) and/or paths (that don't intersect themselves) from a spot on the boundary of the board to another spot on the boundary of the board. Now we're not done since we haven't shown the boundary between colors has at least one of these paths, and furthermore, even if we knew that, we would have to show there's a path connecting one side of the board to the opposite side. Then we could conclude there's a chain of one color joining opposite sides, i.e. there is a winning path. Here's a simply way to accomplish both these goals. At each corner of the board, there is a hexagon that is basically shared by each side. We extend the graph (given by taking the boundaries of all the hexagons on the board) by adding an edge to each corner hexagon: there's only one way of doing this without favoring either side. We also pretend that each side of the board has an extra row of hexagons already colored by one color. So if one side belongs to white, then the extra row added adjacently is colored black, and vice versa. Note that the edge we added at each corner separates these newly added rows and so is one of the color separating edges of our graph. Now we've killed two birds with one stone. Since the boundary between colors now has edges that can't close up to be loops, we must have paths going from one side to one side (possibly the same side). By checking the possibilities of where a path starting at a corner can end up, we can easily see we have a winning path. I struggled a little in explaining the modification of the board; I can only wish I was adept in ASCII since with pictures it's pretty simple. Hopefully people can figure it out. Final observation, it still remains to show there is *only* one winner! This is harder (basically one needs to prove a piecewise-linear version of the Jordan separation theorem), so I'll stop here. In any case, it doesn't seem the OP was interested in more than showing there is at least one winner. === Subject: Re: Hex Win Proof? All related to the fact that the plane is 2-dimensional... You cannot tile 3-space with small tiles meeting no more than 3 at a time! But you can in 2-space. With hexagons (as here) or rectangles arranged like brick-work. Two opposite sides of a square (or paralellogram) can be separated from each other by a one-dimensional curve. But this cannot be done for two opposite faces of a cube... You have to use a 2-dimensional surface to separate it. G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Hex Win Proof? > Neither of the proofs (which are basically the same) posted so far is > correct. Both would apparently conclude that a winning path would be > formed on a squared board, whereas this is not the case - a squared > board could end in a draw. > An actual proof must use the hex nature of the board or, > alternatively, that 3 cells meet at each vertex. A proof is given in > Cameron Browne's book Hex Strategy, but whether it would convince an > intelligent layman is not clear. > Maybe a simpler proof could be achieved by induction? > Jonathan Welton I wasn't assuming a square board, I was imagining the board set up like a parallelogram. At least, that is how I orientate the board when I play. Then red goes top to bottom and blue goes left to right (red and blue because the board I made uses poker chips). What Jonathan is trying to point out is that you aren't using the fact that there are hexagons. If you took a checkerboard and squished it to form a parallelogram (with angles not 90 degrees), then you would have a board where every piece of the board looked like a little parallelogram (instead of a hexagon). Clearly we can color this checkerboard without a winning path by the usual checkerboard coloring. [I'm not considering two parallelograms that touch only in a corner to be part of a path] Your proof attempt makes no use of the specifics of the Hex board, and so would apply to any board like the one above. . === Subject: Re: Hex Win Proof? It is an old theorem that in Hex, once the board has been completely filled in with two colours, there *must* be a winning path for one or other of them. Now, I can prove this easily enough mathematically, but I'm wondering if there is a simple proof, or proof outline, that would be understandable and reasonably convincing to the intelligent layman. Can anyone help out please? Here's a pretty proof involving a closely related game called Y. (Don't hesitate about the length of this post; it consists mostly of descriptions of Y, the relation to Hex, and the definition of Y-reduction. The final proof is very short.) A Y board consists of a triangular configuration of hexagons. Play is the same as Hex except that the goal is to build a chain that connects all three sides (as in Hex, corner hexes belong to both sides). Y is a generalization of Hex as illustrated in the following ascii diagram. (view in a fixed font) _ _/* _/*_/ _/*_/* _/ _/*_/ _/ _/ _/* _/ _/ _/ _/ / _/ _/ _/ _/ _/ _/ _/ _/ _/ _/+ _/ _/+_/ _/+_/+ _/+_/ _/+ _/ * are pieces of one player and + are pieces of the other player. Playing Y from the above legal position is equivalent to playing Hex in the unoccupied hexes of this position because a chains wins for Y if and only if the chain restricted to originally unoccupied hexes wins for Hex (in the Hex version, player * is connecting the southwest border up to the northeast). Thus, to prove a completely filled-in Hex board has a winning chain for one of the two players, it suffices to prove that a filled-in Y board has a winning chain for one of the two players. The proof makes use of Y-reduction. a completely filled-in order n-1 Y-board as follows. For each triangle of three mutually adjacent hexes oriented the same way as the board, replace it with a single hexagon whose piece color is that of the majority of the colors in the triangle. Ex. _ _/4 _/31/ _/3 _/21/3 _/21/ /11/22/ /11/2 1/12/2 1/12/ 2/13/ 2/1 3/1 3/ 4/ The triangle of hexes with coordinates x x+1 x y y y+1 on the order-n board corresponds to the hexagon x y on the order n-1 board Lemma: If a player has a winning chain on the reduced board, then that player also has a winning chain on the original board. It's easy to verify (but ugly to write up rigorously) that if a player has a winning chain on the reduced board, then he also has a winning chain through the corresponding triangles on the original board. [note: the converse of the lemma is also true but we don't need it.] Thm. One of the player's has a winning chain on a completely filled-in Y board. Proof: Y-reductions to reduce the board to a filled-in order 1 board. The filled-in order 1 board obviously has a winning chain and hence by the lemma, each Y board in the sequence has a winning chain. I wish I could take credit for this proof but it is due to Ea Ea (his former name was Craige Schensted). === Subject: Re: Hex Win Proof? Now, I can prove this easily enough mathematically, but I'm wondering if there is a simple proof, or proof outline, that would be understandable and reasonably convincing to the intelligent layman. The following has a bunch of cases, but I think the intelligent layman can follow them. Let's assume Red has top/bottom, and Blue has the sides. We want to show that if Blue does not have a win when the board is full, the Red must have one. There must be at least one Red hex on the top, and at least one on the bottom, or else Blue would have a win along the top or bottom. Pick a Red hex on top, U, and a Red hex on the bottom, D, and consider any path between them. Go along that path from bottom to top. If Red doesn't have a win, that path must run into a Blue hex. Let S be the set of all Blue hexes connected to that Blue hex. Call the Red hex that ran into S A. As you continue along the path, you must eventually reach a Red hex that is not surrounded by S. Call that Red hex B. If S does not touch any edge, then there is an all Red path from A to B. Just go around the edge of S in either direction from A to B. If S touches one edge, there is still a Red path from A to B. Thus, we cannot reroute around S only if S connects to two or more edges. Let's consider the cases: 1. S connects to the top and bottom. Then the Red hexes adjacent to at least one of its sides form a win for Red, unless S connects is connected to opposite corners, in which case S would be a win for Blue. 2. S connects to left and right. S is a win for Blue. 3. S connects to the bottom and a side. WLOG we can assume S is connected to the left side. If the connection to the bottom is to the left of D, then there is a path from A to B around the boundary of S. If the connection to the bottom is to the right of D, then there is a connection from B to the bottom along the boundary of S (unless S connects to the bottom at the corner, in which case S is a win for Blue). 4. S connects to the top and a side. Similar to case #3. If the top connection is on the same side of U as the side connection, then there is a connection from A to B. Otherwise, there is a connection from A to the top. In all these cases, we are able to reroute the path such that it is still connected to the top and the bottom, but avoids S. Hence, if Blue does not have a win, there is a win for Red, and this procedure will even find such a win. === Subject: Re: Hex Win Proof? What would be more interesting is trying to explain to a lay person that whoever goes first should win, unless they screw it up. That is why whenever I play, I always go second. If I lose, it was destined. You mean you don't use the swap rule? It is near universal. Swap rule: After the first player's initial move, the second player has the option of either responding with his move or swapping sides and taking the initial play as his first turn. The idea is that the first player needs to make an initial play that leads to a roughly equal game. If the first player makes a very strong move opening move, then the second player will swap sides and start with an advantage. If the first player makes a weak opening move, then the second player won't swap and again will start with an advantage. Without the swap rule, the first player gets a strong advantage simply by opening in the center hex (or one of two central hexes for even sized boards). In practice, the swap rule does a good job of equalizing the game. It also leads to more varied opening play. === Subject: Re: Hex Win Proof? An actual proof must use the hex nature of the board or, alternatively, that 3 cells meet at each vertex. A proof is given in The key fact I need is that if a red path is blocked from going around blue hex A by blue hex B, then A and B are adjacent. That property does not hold on a checkerboard. Is this what you mean by using the hex nature of the board? === Subject: Re: Hex Win Proof? > An actual proof must use the hex nature of the board or, alternatively, > that 3 cells meet at each vertex. A proof is given in The key fact I need is that if a red path is blocked from going around blue hex A by blue hex B, then A and B are adjacent. That property does not hold on a checkerboard. Is this what you mean by using the hex nature of the board? I'm sure that's what he means. In simple terms, there's no corner where spaces touch diagonally. Spaces have an edge in common, or they don't touch at all. If you want to get a little more detailed, you can prove that there can't be any corner-touching if there's no place where more than three spaces come together. And with hexes, they don't. === Subject: Re: Hex Win Proof? I'm sure that's what he means. In simple terms, there's no corner where spaces touch diagonally. Spaces have an edge in common, or they don't touch at all. If you want to get a little more detailed, you can prove that there can't be any corner-touching if there's no place where more than three spaces come together. And with hexes, they don't. another way to think of it: a winning position allows you to build a highway as wide as the edge of a hexagon and drive all the way from one side to the other. if you cannot drive all the way, you dont win and the other guy can drive all the way on his hexagons. on a checkerboard, the diagonals pinch your car into a singularity. - nate === Subject: Re: Hex Win Proof? It is an old theorem that in Hex, once the board has been completely filled in with two colours, there *must* be a winning path for one or other of them. Now, I can prove this easily enough mathematically, but I'm wondering if there is a simple proof, or proof outline, that would be understandable and reasonably convincing to the intelligent layman. Can anyone help out please? Hello and everyone, Check http://web.cs.ualberta.ca/~javhar/hex/hex-yproof.html for a simple proof using the game of Y. The proof has been around for over thirty years but did not come to light until I discovered a closely related proof (which was published in Issue 12 of Abstract Games.) Steve === Subject: Re: Hex Win Proof? > It is an old theorem that in Hex, once the board has been completely > filled in with two colours, there *must* be a winning path for one > or other of them. > Now, I can prove this easily enough mathematically, but I'm wondering if > there is a simple proof, or proof outline, that would be understandable > and reasonably convincing to the intelligent layman. Check http://web.cs.ualberta.ca/~javhar/hex/hex-yproof.html for a simple proof using the game of Y. Nice proof and a nice page. Another way is to consider not the cells, but just the boundary between red and blue when the board is full. A component of the boundary must be either a closed curve (in which case we can contract it to a point, losing nothing) or a path from an edge to the same edge (contract again) or a path from one edge to another. In the last case, consider an enpoint of a boundary component which is closer than any other to an obtuse corner of board. According to where the other endpoint of that component is, we can determine who has won the game. Here's a related result in topology. Let T be the Euclidean triangle with vertices (1,0,0) (0,1,0) (0,0,1) in homogeneous coordinates. Let f be any continuous mapping T to T, and write f(x,y,z) = (A(x,y,z), B(x,y,z), C(x,y,z)) Let S be the (closed) subset of T consisting of the points (x,y,z) such that _at least two_ of these inequalities hold: A(x,y,z) >= x B(x,y,z) >= y C(x,y,z) >= z. Then S has a connected component which meets all three sides of T. === Subject: Re: Hex Win Proof? Hello and everyone, Check http://web.cs.ualberta.ca/~javhar/hex/hex-yproof.html for a simple proof using the game of Y. The proof has been around for over thirty years but did not come to light until I discovered a closely related proof (which was published in Issue 12 of Abstract Games.) - nate === Subject: Re: continuous maps === Subject: Re: continuous maps > The following statements are know to be equivalent: > f continuous > for all A, cl f^-1(A) subset f^-1(cl A) (a) > for all A, f^-1(int A) subset int f^-1(A) (b) -- reference > for all A, f(cl A) subset cl f(A) (1) > for all A, int f(A) subset f(int A) (2) -- f continuous iff (2) >Suppose both spaces are R in the usual topology and >f(x) = 0, x <= 0, f(x) = 1, x > 0. Then f is not continuous, >but it's range is two points so it satisfies (2) trivially. >In the other direction, let f : R^2 -> R be the continuous map given >by f(x,y) = x. Set A = R x {0}. Then f(A) = R, so int(f(A)) = R. But >A has empty interior, so (2) fails. Too real, the issue is fully settled. First example. Let A = (0,1); f(cl A) = f([0,1]) = {0,1} not subset {1} = cl f((0,1)) Adheres to: f continuous iff (1) Second example. p1(cl A) = R = cl p1(A) for the given A. But I see no quick way to show for any A p1(cl A) subset cl p1(A) so I'll defer to the proof in section below. Strange the asymmetry, that out of four statements the dual of just one is false. I suppose that comes of a bias open sets have for unions in preference to intersections. But then again a similar asymmetry appears with (where ' is ^-1) f(U / V) subset f(U) / f(V) f'(U / V) = f'(U) / f'(V) f(U / V) = f(U) / f(V) f'(U / V) = f'(U) / f'(V) so it shouldn't seem so out of place. -- proof: f continuous iff (1) > Starting with > for all A, cl f^-1(A) subset f^-1(cl A) > for all B, cl f^-1(f(B)) subset f^-1(cl f(B)) > f(cl B) subset f(cl f^-1(f(B))) subset ff^-1(cl f(B)) subset cl f(B) > Starting with > for all A, f(cl A) subset cl f(A) > for all B, f(cl f^-1(B)) subset cl ff^-1(B) > cl f^-1(B) subset f^-1f(cl f^-1(B)) subset f^-1(cl ff^-1(B)) > subset f^-1(cl f(B) === Subject: News in the site www.bymath.com News in www.bymath.com: Theoretical questions: ● reciprocal fractions ● repeating decimals ● coplanar vectors ● operations with complex numbers in trigonometric form ● common and natural logarithms ● mathematical induction ● de L'Hospital's rule ● integral with variable upper limit of integration Problems: ● more than 100 new problems with help, solutions and answers New pages: ● mathematical symbols ● jokes ● links to us ● about us ● site map Future news in the site: Theory: ● sets - basic notions - examples of sets - operations with sets ● probability - events - definition and basic properties of probability - conditional probability, independence of events - random variables - characteristics of random variables - normal ( Gaussian ) distribution ● analytic geometry - transformations of coordinates - straight line - circle - ellipse - hyperbola - parabola - plane - sphere Problems: ● 40-50 problems in all new sections === Subject: Re: Exercise on proper classes in ZF I have to solve following exercise, i.e. to prove that following classes are proper classes: 1 - The classe of all powersets Suppose the class of power sets is a set, which will be denoted by A. For a set x, denote the union of x by Ux. Since x uniquely determines Ux, then B = {Ux : x in A} is a set by the Axiom of Replacement. But every set is an element of B = {Ux : x in A}, since the union of the power set of x is x, which contradicts ZF, since the set C = {x in B : x not in x} leads to an immediate contradiction. So the class of power sets is a proper class. 2 - The classe of sets containing a given set a Suppose the class is a set, denoted by B. Investigate the set C = {x in B : x not in x}. 3 - The classe of equivalence relations Every set has an equivalence relation on it. Let A be the set of equivalence relations. Let B = {dom(x) : x in A}, i.e. the set of the domains of the elements of A. David McAnally But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane === Subject: Re: Exercise on proper classes in ZF It should be a hard day... regarding question 2), I tried to follow your hint considering C = {x in B : x not in x}. We have: A in B <--> A in A <--> B in B C in C <--> C in B and C not in C This is a contradiction, so C not in C C not in C <--> C not in B or C in C. So C not in B. C not in B implies A not in C, which implies A not in B or A in A A in C <--> A in B and A not in A. Contradiction. So A not in C, which means A not in B or A in A B in C <--> B in B and B not in B. Contradiction. So B not in C, which means B not in B or B in B Still I can't get a contradiction. What am I missing ? David McAnally a .8ecrit dans le message >I have to solve following exercise, i.e. to prove that following classes are >proper classes: >1 - The classe of all powersets > Suppose the class of power sets is a set, which will be denoted by A. > For a set x, denote the union of x by Ux. Since x uniquely determines > Ux, then B = {Ux : x in A} is a set by the Axiom of Replacement. But > every set is an element of B = {Ux : x in A}, since the union of the power > set of x is x, which contradicts ZF, since the set C = {x in B : x not in x} > leads to an immediate contradiction. So the class of power sets is a > proper class. >2 - The classe of sets containing a given set a > Suppose the class is a set, denoted by B. Investigate the set > C = {x in B : x not in x}. >3 - The classe of equivalence relations > Every set has an equivalence relation on it. Let A be the set of > equivalence relations. Let B = {dom(x) : x in A}, i.e. the set of > the domains of the elements of A. But I'm always true to you, darlin', in my fashion, > Yes, I'm always true to you, darlin', in my way. > -- Lois Lane > ---- === Subject: Re: Exercise on proper classes in ZF === Subject: Re: Exercise on proper classes in ZF >prove that following classes are proper classes: >1 - The classe of all powersets >2 - The classe of sets containing a given set a >3 - The classe of equivalence relations >If the power set of A is an element of A, since A is an element of >the power set, this would contradict Foundation, would it not? Yes and it proves, not that the class of all powersets is a proper class, but that there is no set of all powersets. So without foundation wouldn't those sets be possible? In some sort of nether land of undecidability? A = { P(X) | X = X } A in P(A) in A A = { X | A subset X } A in A A = { XxX | XxX equivalence relation } E = { (X,X) | X in A } equivalence relation E in A (E,E) in E using (a,b) = {{a}, {a,b}} E in {E} in (E,E) in E === Subject: Re: Exercise on proper classes in ZF You're write, I shall detail what is ZF for me, since it doesn't seem to be standardized. ZF is the set of following axioms: 1) Extensionality 2) Sum set 3) Power set 4) Replacement 5) Infinity Note: Foundation is NOT in. Jean-Pierre MERX a .8ecrit dans le > I have to solve following exercise, i.e. to prove that following classes are > proper classes: > 1 - The classe of all powersets > 2 - The classe of sets containing a given set a > 3 - The classe of equivalence relations > Please help giving some hints. > Regarding 1, I'm able to derive the fact (just using some definition...) > that if the given classe is a set A, then the powerset of A belongs to A, > which seems strange, but I wasn't able to derive a contradiction. > Regarding 2, I'm able to derive that if the classe is a set b then: a in b > <--> a in a <--> b in b. Then, so what ? === Subject: Re: Exercise on proper classes in ZF You're write, I shall detail what is ZF for me, since it doesn't seem to be standardized. ZF is the set of following axioms: 1) Extensionality 2) Sum set 3) Power set 4) Replacement 5) Infinity Note: Foundation is NOT in. This seems *exceedingly* restrictive for ZF. Assuming your Sum refers to Union, where are Pairing and Comprehension? Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Exercise on proper classes in ZF This seems *exceedingly* restrictive for ZF. Assuming your Sum refers to Union, where are Pairing and Comprehension? They can be derived from the other axioms. === Subject: News in the site www.bymath.com News in www.bymath.com: Theoretical questions: reciprocal fractions repeating decimals coplanar vectors operations with complex numbers in trigonometric form common and natural logarithms mathematical induction de L'Hospital's rule integral with variable upper limit of integration Problems: more than 100 new problems with help, solutions and answers New pages: mathematical symbols jokes links to us about us site map Future news in the site: Theory: sets - basic notions - examples of sets - operations with sets probability - events - definition and basic properties of probability - conditional probability, independence of events - random variables - characteristics of random variables - normal ( Gaussian ) distribution analytic geometry - transformations of coordinates - straight line - circle - ellipse - hyperbola - parabola - plane - sphere Problems: 40-50 problems in all new sections === Subject: Re: A rational X irrational game >Hello >I've been thinking about the following problem (or puzzle) but hasn't >come to a conclusion yet. Maybe someone can give a hint. A and B are playing the following game: On the real line, A chooses a >closed interval I1 of length 0interval I2, of length 0chooses a closed interval I3, of length 0and so on. We know there's one, and only one, real number x that >belongs to all of the intervals I1, I2, I3.... If x is rational, A >wins the game, and if x is irrational, then B wins. We are asked to >find an strategy that B should follow so that he will certainly win >the game no matter how B chooses his intervals. Since the rationals are countable and the irrationals are not, it >really seems that there is such a strategy that assures B will ever >win, but I couldn't find it so far. > Hint: B doesn't have to force a win on the first move. On > the first move B prevents one way he can lose, on the > next move he prevents another method of losing... Ummm... based on your hint, this may work: Once the interval I1 has been choosen by A, B enumerates the rationals in I1. Let {x_1, x_2....} be such enumeration. Then, B chooses I2 in such a way that x_1 is not in I2 (this is always possible, because the length of I2, though positive, is allowed to be arbitrarily small). In the next move, A chooses an I3 contained in I2. And, in his turn, B chooses an I4 that doesn't contain x_2 . Carrying on with this method of choices, B ensures the game will generate a sequence {Im} of closed intervals such that no x_n belongs to all of the Im's. Therefore, the element x, common to all Im's, is not covered by the enumeration {x_1, x_2....}, which implies x must be irrational. And B wins! Artur The important point here is that the rationals are countable and the irrationals are not. This same strategy works if we replace the words rational and irrational by algebraic and transcendental, because the algebraics are countable and the trascendental are not. Keeping the other rules, it seems we get really complicated puzzles if we replace the game's winning criterion by the following: a)Suppose the sets Da and Db, Da with measure zero but uncountable, both dense in R, form a partition of R. A wins if x is in Da and B wins if x is in Db. b) A wins if x is Liouville and B wins otherwise. Artur === Subject: Re: A rational X irrational game Your game doesn't terminate in a finite time. And at each time point the intersection contains intinitely many rational and irrational points. So the game is not well formulated. Best regards, Alex. PS. To email me, remove loeschedies from the email address given. === Subject: Re: A rational X irrational game Your game doesn't terminate in a finite time. No, it doesn't; hence it's probably not a game by some definition or other. But it's a perfectly standard sort of game in mathematics - yes it takes infinite time, so two people can't actually play it to the end, but so what? After infinitely many steps there's exactly one number in the intersection. And at each time point the intersection contains intinitely many rational and irrational points. So the game is not well formulated. Best regards, Alex. PS. To email me, remove loeschedies from the email address given. David C. Ullrich === Subject: Re: A rational X irrational game Your game doesn't terminate in a finite time. And at each time point the intersection contains intinitely many rational and irrational points. So the game is not well formulated. Best regards, Alex. PS. To email me, remove loeschedies from the email address given. No, actually, the game doesn't terminate in a finite time. But I think it's well defined, because there's one, and only one, real number common to all of the intervals. There's no ambiguity. Artur === Subject: Re: Complex Numbers, Equilateral Triangles Let z1, z2 and z3 be complex numbers and z1^2 + z2^2 + z3^3 = z1z2 + z2z3 + z3z1 Prove that z1, z2 and z3 form an equilateral triangle. Let me add another solution to the list. Let us show that the equation above is translation invariant. That is, if it holds, then it holds for z1+u, z2+u, and z3+u. (z1+u)^2 + (z2+u)^2 + (z3+u)^2 = z1^2 + z2^2 + z3^3 + 2u(z1 + z2 + z3) + 3u^2 = z1z2 + z2z3 + z3z1 + 2u(z1 + z2 + z3) + 3u^2 = (z1+u)(z2+u) + (z2+u)(z3+u) + (z3+u)(z1+u) Having shown that, translate the problem so that z3 = 0. All we need to show is that z1^2 + z2^2 = z1z2 iff z1 and z2 form an equilateral triangle with 0. Since z1^2 - z1z2 + z2^2 = (z1 + w z2)(z1 + w^2 z2), where w is a complex cube root of 1, we immediately get that either z1 + w z2 = 0 or z1 + w^2 z2 = 0. In either case, z1 and z2 form an equilateral triangle with 0. If z1 and z2 form an equilateral triangle with 0, one of those two equations holds. QED Rob Johnson take out the trash before replying === Subject: Re: pi, 2 pi and frivolity thusly: To me it seems rather curious that the human race has chosen pi rather than 2 pi as the important number that deserves its own special symbol. It is 2 pi and not pi that appears in the most fundamental equations. Primary school children are taught that C = pi D, but the diameter of the circle is not as fundamental as the radius and isn't used in any other commonly quoted equation. We use radians, not diametans. The equation which describes a circle uses the radius not the diameter. If you can remember back to when you first learnt about radians, didn't it seem rather odd and hard to remember that the angle you were used to thinking of as 60 degrees was pi/3 radians, that 90 degrees was pi/2 radians or that 30 degrees was pi/6 radians. I know I certainly did. Supposing that we refer to 2 pi as @, for want of a better symbol, then doesn't it seem more intuitive that 90 degrees which is also known as a quadrant - note the four there in that word - is @/4 radians, than pi/2 radians? Likewise that 30 degrees which is the angle at then centre between two hours on a clock face is @/12 radians? Surely a revolution is more important than a half revolution? What's the most salient thing about the periodic functions sin and cos? The fact that they are periodic, and the period of course is 2 pi. Consider the identity cos x = sin(x - pi/2). Isn't it more obvious and easier to remember that cos x = sin(x - @/4)? The identity shifts the origin a quarter of a period. In physics the ratio between frequency and angular frequency is 2 pi. We have centripetal acceleration equal to 4 pi^2 r / T^2. I realise that I'm saying the same thing in about 6 different ways, but surely the fact that it can be said in 6 different ways is some measure of its truth. Surely exp(i 2 pi) = 1 is just as mysterious and beautiful as exp(i pi) = -1 Of course the formula for the area of a circle uses pi and not 2 pi, but that's in two dimensions not one like C = 2 pi r, and its not fundamental. When I think back to the primary school demonstration of that equation though, having a divide by 2 in the formula actually seems more natural. The demonstration I mean is the one where the circle is divided up into a large number of sectors, and the sectors are rearranged in alternating fashion to end up with a rectangle that has dimensions r by C/2. To me the choice of pi and not 2 pi seems like a historical accident, and I wonder whether it was because people didn't immediately realise, centuries ago, that the radius and not the diameter of a circle was its more fundamental measure. The diameter gives a more primitive sense of the circle's size, like the side of a square or an equalateral triangle. So my question to the group is, is there general agreement with my points? If not why not and if so how many millions of times has this been pointed out before? Perhaps it's not an important point, but nevetheless one thing appalls me. All these people who have learnt pi to thousands of digits and all the college students who march up and down the halls chanting pi in unison - if I'm right they've been learning and chanting the wrong number! :-) In the series of formulae for the volume of N-dimensional spheres, the even more fundamental quantity gamma(1/2) appears, here represented by g: 2-D: V = g^2 r^2 3-D: V = 4/3 g^2 r^3 4-D: V = 1/2 g^4 r^4 5-D: V = 8/15 g^4 r^5 6-D: V = 1/6 g^6 r^6 7-D: V = 16/105 g^6 r^7 and so on[1]. Note that g always appears raised to an even power. Now only the 2-D and 3-D cases have immediate practical relevance in this three-dimensional world. The result was that gamma(1/2) itself was missed as a fundamental constant for a long time, but its more familiar square was given the conventional symbol pi. [1] To continue the series: in even dimensions the rational factors consist of the reciprocals of the factorials, in odd dimensions they consist of powers of 2 divided by oddorials (proper name anybody?) formed by multiplying together all the *odd* numbers from 1 to the number of dimensions. Paul Townsend I put it down there, and when I went back to it, there it was GONE! Interchange the alphabetic elements to reply === Subject: Re: pi, 2 pi and frivolity === Subject: Re: pi, 2 pi and frivolity Message-id: symbols. Using your arguments, why are there sixty minutes in an hour? The point is that maths would have been more convenient and more intuitive if 2 pi, rather than pi, had been chosen as the number with its own special symbol. I agree that having everyone using the same symbol is an important point. > Using your arguments, why are there sixty minutes in an hour? There are 60 minutes in an hour for historical reasons, which is exactly the same reason that I am claiming we use pi and not 2 pi. > a formula, sometimes just pi. How do you judge the fundamentality of a number? I judge how fundamental a number is by whether it appears in fundamental formulas and equations. I judge how fundamental an equation is by how often it is quoted in deriving other equations. A similar judgement is how early it appears in a course of study in maths. Geometrically speaking, to me C = 2 pi r is the fundamental equation involving pi, it appears with 2, so 2 pi is the fundamental number. I believe that C = 2 pi r is the fundamental equation The fundamental equation is pi * a * radius**b For a=2, b=1 you get the circumference of a circle. For a=1, b=2 you get the area of a circle. For a=4, b=2 you get the surface area of a sphere. For a=4/3, b=3 you get the volume of a sphere. The fundamental components are pi and the radius, not the variables that change depending on what is being measured. because I imagine that it is substituted in more proofs that any other formula involving pi, and because its surely the first one involving pi presented to a geometry student in any reasonable course of study, except possibly C = pi D. I regard r as a more fundamental measurement of a circle than D, because r and not D is used to define a circle, so I discard C = pi D as being relatively unimportant. Analytically speaking, to me the most fundamental use of pi is the period of the periodic functions cos and sin, and that period is 2 pi. Using 2 pi does make certain formulas easier to remember, in my opinion, but we can be more objective that that. Simply count the number of mathematical operations in the formula. You've got an extra 2 there, that's one more operation. -- Mensanator Ace of Clubs === Subject: Re: pi, 2 pi and frivolity >I realise that I'm saying the same thing in about 6 different ways, but >surely the fact that it can be said in 6 different ways is some measure of >its truth. Nope. Well the point is, we use C = 2 pi r over and over again, thats why its fundamental, and why its constant deserves the special symbol. === Subject: Re: pi, 2 pi and frivolity Well the point is, we use C = 2 pi r over and over again, thats why its fundamental, and why its constant deserves the special symbol. Physicists have a special symbol for h / 2Pi. But they have to work with a load of different constants in most of the equations, so it's understandable to shorten things out quite a bit. Why shouldn't we also have special symbols for e^i, Pi^2, Sqrt(Pi), log 2,...? I'm not interested in mathematics that might have anything to do with reality. -- Easterly, in sci.math === Subject: Re: pi, 2 pi and frivolity >Well the point is, we use C = 2 pi r over and over again, thats why its >fundamental, and why its constant deserves the special symbol. Physicists have a special symbol for h / 2Pi. But they have to work with a load of different constants in most of the equations, so it's understandable to shorten things out quite a bit. Why shouldn't we also have special symbols for e^i, Pi^2, Sqrt(Pi), log 2,...? Why? Because physicists do not have a special symbol for h/2pi. They have a retard's version for the letter E. E is spelled pi, and pi is spelled E, and everything is symmetric, and photons are spelled like most things in philosophy are spelled: Gibberish with a capitol G. === Subject: Phase of a Fourier Transform I've learned that the phase of a Fourier transformed signal can be given by either: 1) arctan(H_Imag/H_Real) or 2)H=|H|*e(j*phi), where phi is the wanted phase H is here an arbitrary Fourier transformed signal. But let H be real, due to a symmetric time signal, e.g H=1+2*cos(w). Then we have Imag(H)=0 => arctan(0) = 0 for all w. But according to 2) above we will get a non-zero phase when 2*cos(w) <-1. So what then happened with definition 1), does it not apply? Next, say one wants to find the phase of: exp(-jx)*sin(x)/x, how would that best be done? I was thinking of using 2), but I'm not sure of how to handle the modulo 2*pi part that is inherent in phase calculations, any tip? best regards HansO === Subject: Re: Ability In Mathematics >.. Even though I had understanding and competence in every other > subject I studied both at school and later, college.. > Fears/inferiority associated with early failures,long held belief > about self incapacity ...turning into a phobia of equations or symbols > and a refusal to recognize patterns in mathematics which faculty comes > so easily in other subjects or life situations. > It is never too late to reverse it. It needs some relaxed fortitude, > perseverance and perhaps an occasional word of encouragement from a > guru/mentor to help locate and remove old stumbling blocks hidden from > one's conscious experience. ' A joy of victory against self-doubt and pride in success gained later > are not impossible. There are many who were branded poor in the early > years but picked up miraculously later on, may be in different > environments, but not without a strong determination. Good luck ! You could have been describing my wife with this statement. She was told in high school by a teacher that she would NEVER be able to succeed in the sciences due to her ability in math. It turns out that she had teachers who simply didn't care enough to explain it to her. Later, she met a teacher who explained the math to her in clear terms. She now has an associate's degree in electronics and will begin her bachelor's degree in MATH this summer. Perception of ability combined with poor teachers have ruined many students. Well, that's given with mathematicians. They only thing they fear more than physics is electronics. They only thing they fear than more than electronics is software. The only thing they fear more than software is robots. The only thing they fear more than robots is that evolution was not only invented by morons, it's a stationary point in moron-ville. === Subject: Re: Ability In Mathematics >.. Even though I had understanding and competence in every other >subject I studied both at school and later, college.. >Fears/inferiority associated with early failures,long held belief >about self incapacity ...turning into a phobia of equations or symbols >and a refusal to recognize patterns in mathematics which faculty comes >so easily in other subjects or life situations. >It is never too late to reverse it. It needs some relaxed fortitude, >perseverance and perhaps an occasional word of encouragement from a >guru/mentor to help locate and remove old stumbling blocks hidden from >one's conscious experience. >A joy of victory against self-doubt and pride in success gained later >are not impossible. There are many who were branded poor in the early >years but picked up miraculously later on, may be in different >environments, but not without a strong determination. Good luck ! > You could have been describing my wife with this statement. She was > told in high school by a teacher that she would NEVER be able to succeed > in the sciences due to her ability in math. It turns out that she had > teachers who simply didn't care enough to explain it to her. Nope, it had everything to do with caring. If it was like my experience, it was because your wife was smarter than those teachers and they knew it; they cared a hell of a lot about that. ... Later, she > met a teacher who explained the math to her in clear terms. > She now has an associate's degree in electronics and will begin her > bachelor's degree in MATH this summer. Perception of ability combined > with poor teachers have ruined many students. It sounds like you are one of these teachers. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Ability In Mathematics charset=utf-8 > Suppose a person has been involved with endeavor y and is q% good at it. The > Theorem states that IF the person starts involving themselves with logical > endeavor z, after having devoted to it an equivalent amount of time as the > time for y, in the limit, he cannot be more than q% good at z. This is quite interesting. What kind of theorem is it? It doesn't sound like a mathematical one. I mean, can you prove this?!? As of now it's empirical knowledge. Only Statistics could conceivably prove it. It would assume that one has at one's disposal, similar perfomance tests for every logical endeavor. As of now, many logical endeavors have performance tests, but these tests are far from standardized (whatever this may mean) and sort of unrelated. For example, chess has one powerful and standard performance test, (the ELO rating), which gives one a very clear idea where one stands on the curve globally, but there are thousands of tests for Mathematics, so there is no clear choice which one should be picked. I think the GRE is a pretty good candidate. Programming has some good competitions which accurately grade on the curve, but my impression is that these competitions are more like show-off competitions. As different people are fluent in different programming languages, such a test would have to be made for every programming language. On the other hand, some endeavors which appear logical, like piano playing in my case, have very subjective performance tests, like piano competitions and these are biased, based on the judges' preferences. In my case the theorem checks for everything I've been involved with, including some non-logical endeavors. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Products and Factors [x^y means x raised to the power of y, ~= means approximately equal to] let t = the Target, some arbitrary integer, relatively large let c = the Components, a set of very small integers > 0 let p = the Product, a product of elements of c, such that p ~= t let f = the Factors, a set of all factors of p Is there a strategy which will provide the best chance that there will be members of f ~= p^(1/2)? By strategy, what should c consist of, and how should the elements be picked to get p? Never mind, I just realized that using c' = a set of squares of all elements in c, will do what I want for any and all non-trivial strategies. / === Subject: [HELP]measure theory Help me please. Let mu be measure, B be Borel set, {x} be singleton set such that mu(X) < infinity and {x} in B. Show that x which satisfies mu({x}) > 0 is at most countably infinite. === Subject: Re: [HELP]measure theory Help me please. Let mu be measure, B be Borel set, {x} be singleton set such that mu(X) < infinity and {x} in B. Show that x which satisfies mu({x}) > 0 is at most countably infinite. What have you done so far? === Subject: Re: [HELP]measure theory Let mu be measure, B be Borel set, {x} be singleton set such that mu(X) < infinity and {x} in B. Show that x which satisfies mu({x}) > 0 is at most countably infinite. Classic trick: consider the sets E_n = {x in X; m({x})>1/n}. Each of them is finite (because mu is finite), et their union (which is precisely the {x in X; mu({x})>0} is therefore (at most) countable). === Subject: seeking tutors in the following area: Functional Analysis; Wavelet; Information Thoery; Mathematical Statistics Tutor(s) needed in the following separate areas: 1) Functional Analysis; or 2) Wavelet Signal Processing; or 3) Information Theory; or 4) Mathematical Statistics. I am desperately interested in the above courses... I am willing to pay good compensation. But I have not got idea on how to conduct teaching/learning if you are not in my local area... maybe distance learning? === Subject: Weak derivative A question about weak derivative. If f is a bounded, continuous function on an open set of R^n, with continuous derivative, the strong derivative is the same of the weak derivative? Thank you. === Subject: Re: Proving associativity OP: The point is that showing that all groups of order less than 6 are abelian is an easy exercise once you've built up enough machinery, and I don't think it's very instructive to do this exercise after only just learning what a group is (though you could probably write a computer program to solve it for you). Leonard Blackburn Actually, this can be done fairly early in one's study of group theory: Let V = { e, a, b, c } be a group of order 4 (e the identity) that is _not_ cyclic. Since the order of any element of V must be one of 1, 2 or 4 and V is not cyclic, o(a) = o(b) = o(c) = 2. So we can begin building V's multiplication table: | e a b c --+------------ e | e a b c | a | a e | b | b e | c | c e To finish this off, note that each row of the completed table has to give a permutation of V and (except for the 'e'-row) none of these permutations can fix any points. So ab can _only_ be c (since it has to be either b or c) and similarly for the rest. So V's multiplication table is uniquely determined: | e a b c --+------------ e | e a b c | a | a e c b | b | b c e a | c | c b a e Since that array is symmetric about its main diagonal, V _is_ abelian. === Subject: Re: Proving associativity Back in 2002, I posted a question about how to prove that all groups of size <= 5 were Abelian. With the hints that I received then (right before losing my Usenet access), I made some progress. I had thought that I had proved the cases for groups of size 1, 2, and 3. However, I just realized that there was a biiig hole in my proofs. I hadn't shown that the objects with which I was working had the associative property, and therefore had no right to claim that they had anything to do with groups! FWIW, that's not really a hole, if you proved that all of your *candidates* were commutative, and that your set of candidates omitted none of the groups. I'm guessing you did something like list all the Cayley tables that have identity and cancellation. By so doing, you would certainly catch all groups in your net, plus possibly some other stuff. As it happens, up to order 4 there is *no* other stuff but you wouldn't have to prove that. OTOH, the problem asks for n<=5, and this simpleminded approach becomes unwieldy exactly at order 5. There are a lot more candidates, and worse, not all of them are commutative. Of course the only candidate of order 5 that is actually a group is the cyclic group; but if you're not allowed to use that result in your proof, you will have a lot of checking to do. I suppose you'd only have to take each noncommutative table and demonstrate one case for each where the associative property fails; but even so, I doubt this would be very instructive -- see Ted Hwa's comments elsewhere in the thread. === Subject: Want example of non-ergodicity Can anyone give an example which is not ergodic? === Subject: Re: Vector product [Q] (name this matrix) We are in R^3, with basis e_1, e_2, e_3. Let v be any vector in R^3. Is there a name for the matrix whose columns are the vectors e_i x v (i = 1, 2, 3)? (x here refers to the vector product in R^3, aka cross-product.) I'm looking for theorems and identities featuring this matrix, and it would be helpful to have a name to search by. Let's call this matrix M. For any vector u, we have uM = v x u, the cross product of v and u. For this reason, I have seen it called the cross product matrix. Since w . uM = w . v x u = u . w x v = - u . v x w = -u . wM we have that the adjoint of M is -M, that is it is skew-symmetric, or for complex matrices, skew-hermitian. For this reason, it is called SKEW(v) at http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/special.html#skew It is referred to as v-cross in http://www.cs.berkeley.edu/~wkahan/Math128/Cross.pdf While M^3 = -|v|^2 M, M is not invertible, so we cannot cancel M and deduce that M^2 = -|v|^2 I. Furthermore, e^M is a rotation matrix fixing the vector v and rotating by an angle of |v| radians. Rob Johnson take out the trash before replying === Subject: Re: Vector product [Q] (name this matrix) >We are in R^3, with basis e_1, e_2, e_3. Let v be any vector in >R^3. Is there a name for the matrix whose columns are the vectors >e_i x v (i = 1, 2, 3)? (x here refers to the vector product in >R^3, aka cross-product.) I'm looking for theorems and identities >featuring this matrix, and it would be helpful to have a name to >search by. Typically (if not always), statements about cross product in R^3 are more enlightening when translated into statements about wedge product (or exterior product). I think your question is an example. Here is a rephrasing, with extra padding, in exterior-algebraic terms. We are in R^3, with dual space (R^3)* and exterior algebra Lambda(R^3), where by construction (or definition) Lambda(R^3) is a graded algebra whose degree-0 part is identified with R, whose degree-1 part is (R^3)*, and which is generated as an exterior algebra by its degree-1 part; and where, therefore, by definition (or construction) Lambda(R^3) is the algebra of alternating multilinear forms on R^3. Let lambda be any covector in R^3, that is, any element of (R^3)*. Consider the transformation mu |--> mu ^ lambda from (R^3)* to (R^3)* ^ (R^3)* (that is, from the [necessarily alternating!] 1-forms to the alternating 2-forms; that is, from the degree-1 part of the exterior algebra to the degree-2 part). Then this transformation is linear. WHEN ONE USES (for instance) A METRIC (or a choice of basis, or ...) TO IDENTIFY R^3 with (R^3)*, and consequently to make many other identifications of various 3-dimensional vectorspaces, THEN THE MATRIX OF THIS wedge-with-lambda TRANSFORMATION becomes the matrix you're interested in. Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different. Johann Wolfgang von Goethe === Subject: Re: Vector product [Q] (name this matrix) >We are in R^3, with basis e_1, e_2, e_3. Let v be any vector in >R^3. Is there a name for the matrix whose columns are the vectors >e_i x v (i = 1, 2, 3)? (x here refers to the vector product in >R^3, aka cross-product.) I'm looking for theorems and identities >featuring this matrix, and it would be helpful to have a name to >search by. >Typically (if not always), statements about cross product in >R^3 are more enlightening when translated into statements about wedge product (or exterior product). I think your question >is an example. Here is a rephrasing, with extra padding, in >exterior-algebraic terms. We are in R^3, with dual space (R^3)* and exterior algebra >Lambda(R^3), where by construction (or definition) Lambda(R^3) >is a graded algebra whose degree-0 part is identified with R, >whose degree-1 part is (R^3)*, and which is generated as an >exterior algebra by its degree-1 part; and where, therefore, >by definition (or construction) Lambda(R^3) is the algebra >of alternating multilinear forms on R^3. Let lambda be any >covector in R^3, that is, any element of (R^3)*. Consider >the transformation mu |--> mu ^ lambda from (R^3)* >to (R^3)* ^ (R^3)* (that is, from the [necessarily alternating!] >1-forms to the alternating 2-forms; that is, from the degree-1 >part of the exterior algebra to the degree-2 part). Then >this transformation is linear. WHEN ONE USES (for instance) >A METRIC (or a choice of basis, or ...) TO IDENTIFY R^3 with >(R^3)*, and consequently to make many other identifications of >various 3-dimensional vectorspaces, THEN THE MATRIX OF THIS wedge-with-lambda TRANSFORMATION becomes the matrix you're >interested in. Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different. But, that's merely saying that mathematicians are like NAZIs, and that Germans are literate, since that's merely the principal of inflection. Invented by Romans rather than Frenchmen with one two many categories, not enough time, and only enough space to build monestaries in Venezula, rather than choosing choice timber land in Brazil. Johann Wolfgang von Goethe === Subject: Comapacticity in R^n I would like a hint to prove the following proposition: A subset A of R^n is compact if, and only if, whenever a subset of any metric space is homeomorphic to A, then this subset is closed. If A is compact, then it's immediate that every set homeomorphic to A is compact. Therefore, if this set is in a metric space it's closed. To prove the converse, I considered that, if A is not compact, then A has a sequence with no subsequence that converges in A. But based on this I couldn't find a homeomorphism between A and a non closed subset of a metric space. Thank you. Amanda === Subject: Re: Comapacticity in R^n I would like a hint to prove the following proposition: A subset A of R^n is compact if, and only if, whenever a subset of any metric space is homeomorphic to A, then this subset is closed. If A is compact, then it's immediate that every set homeomorphic to A is compact. Therefore, if this set is in a metric space it's closed. To prove the converse, I considered that, if A is not compact, then A has a sequence with no subsequence that converges in A. Forget sequences ..... Don't forget by Heine-Borel a subset of R^n is compact iff it is closed and bounded. So you only have to worry about closed unbounded subsets of R^n. For them consider how they embed in the one-point compactification of R^n. Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Geometry Problem http://www.e-pixel.net/pub/math/circles.pdf Answers please send to pkafara@hotmail.com === Subject: Re: Geometry Problem http://www.e-pixel.net/pub/math/circles.pdf which says (along with pictures of 2 disks A and B) x=100 the center of circle B is placed anywhere on circumference of circle A. What is the radius of circle B if area C=1/2 area A Apparently x is the radius of disk A, so its area is 10000*pi. Apparently C is 1/4 of disk B. If either of those apprehensions is wrong, you need to explain the problem better. Otherwise, area(C)=10000*pi/2 ==> area(B)=20000*pi ==> radius(B)=100*sqrt(2). === Subject: Re: Euclid Algorithm for GCD(a/b,c/d) > GCD(87/22,9/5) = ? Multiply by a common denominator (in this case 110, need not be LEAST common denominator): 87/22*110 = 435, 9/5*110 = 198 Compute gcd using integer method: GCD(435,198) = 3 Divide by that same common denominator: 3/110, answer. A more interesting theoretical approach, because it can be modified to deal with such number-sets as 'cube roots of positive rationals' is: GCD(87/22,9/5) = ? Express as powers of primes, then use pointwise minimum (because a divisor iff less than or equal to each prime-power; a common divisor iff less than or equal to each prime-power for both numbers) Value 87/22 9/5 Min ------------------------------- 2 -1 -1 3 1 2 1 5 -1 -1 7 11 -1 -1 . . 29 1 0 ------------------------------ Answer = 3 / 2 * 5 * 11 === Subject: Interleaving Sorted Sequences and Decks of Cards [Cross-posted as I'm not sure where best to post the problem. Please The total number of possible deals in contract bridge is 52!/13!^4. IINM (which may be the problem!), this is the same as the number of ways in which a standard deck of 52 cards can be arranged such that the cards of each suit are still in rank order. (That is, the number of ways to start with the cards of each suit in sorted order and then to merge them into a single sequence by selecting an in-order card one at a time from any of the four suits.) This equivalence bothers my intuition. Concatenating the four hands of a bridge deal generally does not produce a properly interleaved deck, and giving each player 13 cards at a time from an interleaved deck does not reproduce all possible deals. What am I missing about the correspondence? [FN42nn] === Subject: Re: What is inverse equator? I read some where that a meridian is the inverse equator. What does that mean, that there are an infinite number of inverse equators? Using the same logic, does that mean that inverse meridian is the regular equator (and therefore there is only one)? Nice try, but... P=) The transverse/inverse graticule is a globe's coordinate webbing shifted down (or up, depending on your approach/orientation) from the the pole to the equator--see left image: http://astronomy.swin.edu.au/~pbourke/surfaces/prolatespheroid /prolate1.gif Thus the latitude rings are now equatorially concentric and defined as rings of TRANSVERSE LATITUDE (which this writer abbrieviates as TvL and has coined the term TEAVAL--T-V-L-->T-VEL, get it? P=), with the top/bottom of the ring tangent to its corresponding, upright latitude, hence the teaval is positive above the equator and negative below it, with the teaval on the equator being 0 and the transverse equator--i.e., the meridian 90 away from the 0 teaval--is the +/-90 teaval. The longitude circles, in contrast, are now great-circles intersecting at the 0 teaval and are considered TRANSVERSE MERIDIANs/lines of longitude or ARC PATHs (APs--again, my coinage) and are measured/defined as azimuth at the equator--thus (like azimuth) can have multiple periodic possibilities (0-90, 0-180, 0-360, -180-+180), though this writer explicitly defines AP as being from 0-90 and is known as the true course. The distance between two teavals is known as the ANGULAR DISTANCE (AD), which, along with a specific azimuth (Az_q) for a specific (upright) latitude (Lat_q), comprise the basic elements for spherical calculations: Lat_q = asin{cos{AP_q}*sin{TvL_q}}; Long_q = atan{sin{AP_q}*tan{TvL_q}}; AP_q = asin{cos{Lat_q}}*sin{Az_q}}; TvL_q = atan{sec{Az_q}*tan{Lat_q}}; AD = TvL_2 - TvL_1, = acos{[cos{Lat_2-Lat_1}*cos{.5*[Long_2-Long_1]}^2] -[cos{Lat_2+Lat_1}*sin{.5*[Long_2-Long_1]}^2]}; (spherical cosine for sides) Elliptical/spheroidal calculations use the same concepts, except that parametric/auxiliary values are used for lengths, as well as involve elliptic integrals: Lat_q --> RLat_q = atan{sin{AP_q}*tan{TpL_q}}; (reduced latitude) Long_q --> .9fLong_q = atan{sin{AP_q}*tan{TpL_q}}; (auxiliary longitude, usually found and worked in difference form: Y = .9fLong_2 - .9fLong_1) Az_q, AP_q found using RLat and .9fLong; TvL_q --> TpL_q (parametric teaval, or TEAPAL); AD --> .9fAD = TpL_2 - TpL_1 So, bringing the discussion back to the original question: Yes, there are an infinite number of transverse equators (one for each upright meridian), but there are also an infinite number of trsnsverse meridians/APs, with the 90th being the upright equator. ~Kaimbridge M. GoldChild~ ----- WantedKaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ---------- DigitologyThe Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. ***** === Subject: Re: seeking a reference regarding matrix eigenvectors theory: the eigenvectors associated with distinct eigenvalues of a Hermitian matrix are orthogonal. I browsed the well-known book titled Matrix Analysis by Horn and Johnson and couldn't locate. But I'm not really familiar with this book. So maybe I just missed it. Can anyone help me find this theorem? Let H be Hermitian. Suppose Hv=rv, Hw=sw with r, s different. Any eigenvalue of a Hermitian operator is _real_: r = = = = = = conj(r) and is not 0, so r = conj(r) -- that is, r is real. Similarly, s is real. Then (by an analogous computation) r = = = conj(s) = s . Since r, s are different, it follows that = 0. === Subject: Rivals to Kline's Mathematics: The Loss of Certainty? I am looking for a recent semi-popular account of o Development of non-Euclidean geometry o Changes in understanding of what math is all about o Attempts to axiomize math and the discovery of antinomies o Hibert's and 's programs o Godel's proof o Bourbaki Any suggestions, other than the book in the subject header? Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html