mm-58 === The output tape will always start with a string of 1's> followed by a blank (or a 0).> by which you claim I can ?d that 0, and I can demonstrate> you cannot demonstrate that it is on the tape at any ?ite> location, then you cannot demonstrate that it is on the tape> _at all_, and this is the place where your intuition is failing,> and real mathematical logic must be used, instead.TM2 always makes sure there is a trailing blankThere must be at least one blank on the tape.If the blank is not at a ?ite position,then where is it?Russell- 2 many 2 count === > This is a variation of the same proof I give in> Cardinality of Computable Numbers.> I de?e two Turing machines.> The ?st machine (TM1) has these instructions:> 1) Write a 1> 2) Move right one position> Repeat> Assume we give this machine a tape that has> an in?ite string of 0's. It would seem that> TM1 will output an in?ite string of 1's.> Instructions for TM2:> 1) Scan right until a 0 is found> 2) Scan right until a second 0 is found> 3) Backup and write a 1 on the previous 0> Repeat> Assume we give TM2 a tape that contains> an in?ite string of 0's.> Even if we assume that TM2 performs an> in?ite number of operations, the tape> produced by TM2 will contain an initial> segment with a ?ite number of 1's> followed by a 0.> TM2 is incapable of writing an in?ite> number of 1's. If TM2 can not write> an in?ite number of 1's, we can not> assume that TM1 does.> No TM can write an in?ite string of symbols.>say X is the time unitsay TM2 is at position log Xand TM1 is at position X.even if X->oo, TM2 is always < TM1in other words, why cannot we assume TM1 has in?ite operations also.Herc === > Even if the input tape contains an in?ite number of 0's,> TM2 will think the tape contains a ?ite number.> TM2 is incapable of writing a number that represents> an in?ite number of 1's.Nonsense.First, two clari?ations to make your TM's meaningful.never comes back to that spot then that spot on the ?al output tapehas the symbol that was last written there.2. The ?al output tape resulting from a run (in?ite or not)of a TM is completely characterized by the symbols present at all?ite positions on the tape.Given an in?ite input tape ?led with all the integers encodedin unary as you have outlined (010110111011110111110...) and a TMthat erases the zeroes one at a time, always making sure that thereis still a zero to the right then the ?al output tape resultingfrom running this TM forever has 1's in every position.The zeroes to the right of the tape head disappear when you producethe ?al output tape by effectively taking the limit of theintermediate output tapes using a topology of pointwise convergence.Astute readers will note a similarity to the discussion aboutlabelled balls in buckets and convergence topologies and such.The two clari?ations above apply in that situation as well. Inthat environment, some people take the clari?ations as being soobvious and basic as to need no explicit mention. John Briggs === [...]> It is simple to show that the number of 1's> written by TM1 is some multiple of the number> of 1's written by TM2.> Inductively.>Turing doesn't say that computable numbers require induction.>It is easy to prove that the output of TM2 is ?ite.>TM2's tape will always have a string of 1's followed by a blank.>The blank must be at a ?ite position.>It is impossible for TM2 to write an in?ite string of 1's.As xanthian said, you need to be more speci? on the meaning o??ite. Mathematicians usually de?e the concept of in?ity interms of limits, and this is exactly how a TM can write an in?itetape: after any (?ite) number of steps, a TM has written a ?itepiece of tape, but it may be the case that as the number of steps grows,the sequence of (?ite) pieces of tape written tends to a speci?,in?ite piece of tape. This is the case, for example, if over timethe TM ceases to write on an ever increasing ?st part of the tape;it is undecidable whether an arbitrary TM does this, but particulartypes of TMs can be used that warrant this. E.g. you could write yourreal number generators in such a way that they systematically developthe numbers' decimals.>Russell>- 2 many 2 count-- Reinier === > [...]> It is simple to show that the number of 1's> written by TM1 is some multiple of the number> of 1's written by TM2.> Inductively.>Turing doesn't say that computable numbers require induction.>It is easy to prove that the output of TM2 is ?ite.>TM2's tape will always have a string of 1's followed by a blank.>The blank must be at a ?ite position.>It is impossible for TM2 to write an in?ite string of 1's.> As xanthian said, you need to be more speci? on the meaning of> in?ite. Mathematicians usually de?e the concept of in?ity in> terms of limits, and this is exactly how a TM can write an in?ite> tape: after any (?ite) number of steps, a TM has written a ?ite> piece of tape, but it may be the case that as the number of steps grows,> the sequence of (?ite) pieces of tape written tends to a speci?,> in?ite piece of tape. This is the case, for example, if over time> the TM ceases to write on an ever increasing ?st part of the tape;> it is undecidable whether an arbitrary TM does this, but particular> types of TMs can be used that warrant this. E.g. you could write your> real number generators in such a way that they systematically develop> the numbers' decimals.It makes sense to de?e in?ity in terms of limits.We can say that the limit of TM1 is a tape withan in?itely long string of 1's.TM2 makes sure there will be a blank left on thebe an in?ite string of 1's followed by a blank?The string produced by TM1 represents .111... (base 2).Does the tape produced by TM2 represent the same number?(I guess I should say the limit of TM1 represents a number.)Russell- 2 many 2 count === > [...]> It is simple to show that the number of 1's> written by TM1 is some multiple of the number> of 1's written by TM2.> Inductively.>Turing doesn't say that computable numbers require induction.>It is easy to prove that the output of TM2 is ?ite.>TM2's tape will always have a string of 1's followed by a blank.>The blank must be at a ?ite position.>It is impossible for TM2 to write an in?ite string of 1's.> As xanthian said, you need to be more speci? on the meaning of> in?ite. Mathematicians usually de?e the concept of in?ity in> terms of limits, and this is exactly how a TM can write an in?ite> tape: after any (?ite) number of steps, a TM has written a ?ite> piece of tape, but it may be the case that as the number of steps grows,> the sequence of (?ite) pieces of tape written tends to a speci?,> in?ite piece of tape. This is the case, for example, if over time> the TM ceases to write on an ever increasing ?st part of the tape;> it is undecidable whether an arbitrary TM does this, but particular> types of TMs can be used that warrant this. E.g. you could write your> real number generators in such a way that they systematically develop> the numbers' decimals. It makes sense to de?e in?ity in terms of limits.> We can say that the limit of TM1 is a tape with> an in?itely long string of 1's. TM2 makes sure there will be a blank left on the> be an in?ite string of 1's followed by a blank?No! In the limit, every position that once held a 0 has been overwritten by a 1, so no 0's are left.> The string produced by TM1 represents .111... (base 2).Unless the machine takes each step in a fraction the time of the previous step and the sum of these fractions converges to a ?ite limit, the machine's limit can never be achieved.> Does the tape produced by TM2 represent the same number?> (I guess I should say the limit of TM1 represents a number.)If the machines can take in?itely many steps in a ?ite interval of time, and so ?ish their in?ite sequences, then yes, they produce identical strings. === > TM2 makes sure there will be a blank left on theNo, TM2 merely makes sure that there is _currently_?st non-one. That in no way guarantees that thesecond non-one will be left on the tape, and inthat non-one as well, which it is guaranteed to beable to do, since you speci?d that the tape at thestart contained an in?ite string of zeros.> Would the limit of TM2> be an in?ite string of 1's followed by a blank?No, simply because that is a meaningless noise.xanthian.-- <8tmdnQVuXtGxs2qiRVn-sA@comcast.com> === > Inductively.>Turing doesn't say that computable numbers require induction.>It is easy to prove that the output of TM2 is ?ite.>TM2's tape will always have a string of 1's followed by a blank. Please de?e always. Then we'll see if your next two linesare a non sequitur or not. Here's a hint: If you de?e alwaysto mean after any given number of moves of the machine, thenyou haven't proved anything except that after any given number ofmoves, there is always a blank on the tape. Nobody disputesthat!>The blank must be at a ?ite position.>It is impossible for TM2 to write an in?ite string of 1's.> As xanthian said, you need to be more speci? on the meaning of> in?ite. Mathematicians usually de?e the concept of in?ity in> terms of limits, and this is exactly how a TM can write an in?ite> tape: after any (?ite) number of steps, a TM has written a ?ite> piece of tape, but it may be the case that as the number of steps grows,> the sequence of (?ite) pieces of tape written tends to a speci?,> in?ite piece of tape. [...]> It makes sense to de?e in?ity in terms of limits.> We can say that the limit of TM1 is a tape with> an in?itely long string of 1's. Indeed, Turing *does* say almost exactly this.> TM2 makes sure there will be a blank left on the> be an in?ite string of 1's followed by a blank? Of course not. Are you familiar with the phrase forever and a day?That's a phrase that uses paradox as a form of hyperbole -- there *is*no day after forever, because there's no after after forever!Likewise, there's no way to put a blank, or anything else, afteran in?ite number of anything, because an in?ite sequence by de?itionhas no end.> The string produced by TM1 represents .111... (base 2).> Does the tape produced by TM2 represent the same number? Of course.> (I guess I should say the limit of TM1 represents a number.) You should stick to Turing's terminology and say that thesequence computed by the machine corresponds to the real number0.11111... (base 2). Or you should de?e whatever terminology you *are* going to use,and then use it consistently. No more of this mucking around withalways and followed by. You need to de?e your terms and notstep outside the bounds of what can be deduced therefrom. True statements:http://www.abelard.org/turpap2/tp2-ie.asp Turing, in this paper, de?es the concept of computable number.Computable numbers are decimals that can be produced by a Turingmachine working on a one-direction-in?ite tape. They don't haveto be rational numbers. Being decimals, they do have to be between0 and 1 inclusive, but Turing lets X = X'+k be called computableif k is an integer and X' is computable. The set of computable numbers, like the set of Turing machines ingeneral, is countable. This can be observed by putting the set ofTuring machine speci?ations into a one-to-one correspondence witha subset of the natural numbers (e.g., using the ASCII representationsof the machines' state tables). The set of computable numbers is also an in?ite set, since thedecimals {.1, .11, .111, .1111, ...} are all members of the set. All rational numbers are trivially computable. Some irrational numbers are computable, as well. But as the setof real numbers is much larger than the set of computable numbers,there must exist some real number which is not computable. (In fact,the vast majority of irrational numbers are not computable.) Having written all this, I forget what Russell is still arguingabout. I don't see any problems with any of this trivial stuff,and I don't remember anything more complicated coming up.-Arthur === > TM2 always makes sure there is a trailing blankYes, for ?ite de?itions of always.> There must be at least one blank on the tape.After any ?ite step, yes, but you are consideringthe behavior after in?ite steps, which doesn'tfollow your intuition.> If the blank is not at a ?ite position,> then where is it?Exactly.That is the crux of the matter.You consider that an argument for its continued existence,but it is just the opposite.Your faulty intuition tells you it is there somewhere,because you just saw it, but a math proof by inductionshows that it has mysteriously vanished, alwaysoverwritten almost as soon as it is encountered.The _next_ zero merely enables the overwriting of the_previous_ zero, it _doesn't_ thereby protect its ownlocation, which remains squarely in the path of destruction.You are still trying to envision an in?ite _process_ as amere _product_, and that doesn't work because there is _nopoint in time_ at which that _product_ *exists* or is_complete_, so you never have a _product_ to consider.You can only treat or consider the process as _itself_, adynamic thing, and any argument trying to describe thatunattainable product must be an argument from dynamicbehavior, not one from behavior that has stopped. That zeroyou think you see can only exist if the process stops, butthe process is _in?ite_, so that never happens, and eachzero in turn, all in?itely many of them, gets overwritten.Since there is no last zero, there is no place the processstops and leaves one stray zero because there is not anenabling zero beyond it; the process eats them _all_.For any named value of any place, there is no any placethat zero can be, the _process_ will _always_ eat it at _any_any place, therefore it is no place, the only remainingchoice.xanthian.-- === The output tape will always start with a string of 1's> followed by a blank (or a 0).> by which you claim I can ?d that 0, and I can demonstrate> you cannot demonstrate that it is on the tape at any ?ite> location, then you cannot demonstrate that it is on the tape> _at all_, and this is the place where your intuition is failing,> and real mathematical logic must be used, instead. TM2 always makes sure there is a trailing blank> There must be at least one blank on the tape.> If the blank is not at a ?ite position,> then where is it?Always about to be overwritten? > Russell> - 2 many 2 count === why i can't use oe to open this group?is the url math.sci? === > why i can't use oe to open this group?> is the url math.sci?I use oe to open this group. It is sci.math === > why i can't use oe to open this group?> is the url math.sci? === > why i can't use oe to open this group?> is the url math.sci?-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/The question of copying music from the Internet is like a two-barreled sword. - Finnish rap artist Ezkimo === > why i can't use oe to open this group?> is the url math.sci?>url for newsgroup?This group is sci.math === On Sun, 4 Jan 2004 01:26:49 -0800, William Elliot why i can't use oe to open this group?> is the url math.sci?>url for newsgroup?like ftp sites start with ftp and web pagesstart with http.>This group is sci.mathDavid C. Ullrich === -------------------------------------------------- -------------------Wouldn't determining a sum for a in?ite series, be a paradox itself?OrA sum of in?ite series, would then classify an ?in?ite series', in fact a ?ite series, since a sum can be made.>Who proved that an in?ite number of numbers (i.e. in?ite series)>can have a ?ite sum? This goes back to the Greeks; they were the ?st> to prove anything. The Achilles and the tortoise> paradox was known to be a paradox of that type. -- > This address is for information only. I do not claim that these views> are those of the Statistics Department or of Purdue University.> Herman Rubin, Department of Statistics, Purdue University === > Wouldn't determining a sum for a in?ite series, be a paradox itself?> Or> A sum of in?ite series, would then classify an ?in?ite series', in fact a > ?ite series, since a sum can be made.> Not if the real numbers behave as advertized. The Cauchy construction of the real numbers guarantees that for every Cauchy sequence there is a real number. If the partial sums of an in?ite sum form a Cauchy sequence, q.v. , there MUST be a unique real number represented by that sequence. === In other word's, you can't work with real numbers, if your goal is to ?dsomething that is infact in?ite.By in?ite, there is no end.> Wouldn't determining a sum for a in?ite series, be a paradox itself?> Or> A sum of in?ite series, would then classify an ?in?ite series', infact a> ?ite series, since a sum can be made.> Not if the real numbers behave as advertized. The Cauchy construction of> the real numbers guarantees that for every Cauchy sequence there is a> real number. If the partial sums of an in?ite sum form a Cauchy> sequence, q.v. , there MUST be a unique real number represented by that> sequence. === > Wouldn't determining a sum for a in?ite series, be a paradox itself?> Or> A sum of in?ite series, would then classify an ?in?ite series', in> fact a> ?ite series, since a sum can be made.> Not if the real numbers behave as advertized. The Cauchy construction of> the real numbers guarantees that for every Cauchy sequence there is a> real number. If the partial sums of an in?ite sum form a Cauchy> sequence, q.v. , there MUST be a unique real number represented by that> sequence. > In other word's, you can't work with real numbers, if your goal is to ?d> something that is infact in?ite.> By in?ite, there is no end.Quite the reverse. If you want to work with the reals, you have to have a basic set that is already uncountably in?ite., and is the set of real numbers is based on an in?ite set of in?ite sets (Dedekind cuts or equivalence classes of Cauchy sequences). === > I get a result which can be used to ?d that your sum is also: sum{k|n} phi(n/k) x^k.> It looks much better. === sum[ n=0 , n=oo , (3^n) / ( (n+1)5^n ) ]Proving that the series converges is pretty simple but howdo i go about computing it's value? I suppose it starts withextending the fraction somehow but that's how far i get...-- KindlyKonrad------------------------------------------------- --May all spammers die an agonizing death; have no burial places; their souls be chased by demons in Gehenna from one room to another for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sense to be lazy--------------------------------------------------- === > sum[ n=0 , n=oo , (3^n) / ( (n+1)5^n ) ] Proving that the series converges is pretty simple but how> do i go about computing it's value? I suppose it starts with> extending the fraction somehow but that's how far i get...Write (3^n) / ((n+1) 5^n) as (3/5)^n/(n+1), and then you are asking forf(3/5), where f(x) = sum[n=0, n=oo, x^n/(n+1)].Multiply by x: x f(x) = sum[0,oo,x^(n+1)/(n+1)]Differentiate: x f'(x) + f(x) = sum[0, oo, x^n] = 1/(1-x)Notice that f(x) = -ln(1-x)/x satis?s this.-- --Tim Smith === >sum[ n=0 , n=oo , (3^n) / ( (n+1)5^n ) ]>Proving that the series converges is pretty simple but how>do i go about computing it's value? I suppose it starts with>extending the fraction somehow but that's how far i get...Consider the function oo --- 1 n f(x) = > --- x --- n+1 n=0Your sum is just f(3/5). Compute the derivative of xf(x): oo --- n (xf(x))' = > x --- n=0 1 = --- 1-xSolving for f(x), and noting that f(0) = 1, gives f(x) = -log(1-x)/xSo f(3/5) = 5/3 log(5/2) which is about 1.52715122.Rob Johnson take out the trash before replying === En el mensaje:bt8p7n$fbj$1@news.gu.se,Konrad Den Ende escribi.97:> sum[ n=0 , n=oo , (3^n) / ( (n+1)5^n ) ]> Proving that the series converges is pretty simple but how> do i go about computing it's value? I suppose it starts with> extending the fraction somehow but that's how far i get...Ln(1 + x) = Sum((-1)^nx^n/n, n, 1, inf) ==>-Ln(1 - x) = Sum(x^n/n, n, 1, inf) = x*Sum(x^(n-1)/n, n, 1, inf) = x*Sum(x^n/(n+1), n, 0, inf) ==>S(x) = Sum(x^n/(n+1), n, 0, inf) = -Ln(1 - x)/x (all if |x| < 1)Then your sum is S(3/5) = -Ln(1 - 3/5))/(3/5) = (5/3)Ln(5/2) ~= 1.527151219...-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === > What is the vector space structure> of ?ed rank (k), (n x n) matrices ?> The set of all nxn matrices of rank k with sum and multiplication by ascalar> is not a vector space. Indeed, the sum of 2 matrices of rank k is not, asa> general rule, a rank k matrix. (take a matrix A that has rank k. -A hasrank> k too, and A + (-A) = 0, so it has rank 0).>I think that it can be deduced frommy question that I have already knownthat fact. I was asking how can youstill consider it as a vector space,say by quotienting by the ?right' equivalencerelation (or is that impossible) ? === > What is the vector space structure> of ?ed rank (k), (n x n) matrices ? The set of all nxn matrices of rank k with sum and multiplication by a>scalar> is not a vector space. Indeed, the sum of 2 matrices of rank k is not, as>a> general rule, a rank k matrix. (take a matrix A that has rank k. -A has>rank> k too, and A + (-A) = 0, so it has rank 0).>I think that it can be deduced from>my question that I have already known>that fact. Certainly didn't look that way to me.>I was asking how can you>still consider it as a vector space,>say by quotienting by the ?right' equivalence>relation (or is that impossible) ?Well, for example you could say any two matricies of rankk are equivalent. Then they become the vector space {0}.That's probably not the answer you want, but it's hard tosee what other answer there is. Usually when whatever/~is a vector space it's because whatever is _already_ avector space, and then the addition in whatever/~ isde?ed using the addition in whatever. But the matriciesof rank k simply do not form a vector space - I have noidea why you think that there is a vector space structurehidden in there somewhere.(Of course the question is not precise enough to givea precise answer, until you specify what sort of additionyou had in mind for these equivalence classes.)>David C. Ullrich === John R Ramsden schrieb: > > I get B(6) = 8, > > No you'r wrong: > O > X X > X X X > X X X X > O X X X O > O O X X O O > > Darn. I thought my reply might be oversimplifying the problem. >I see it as an optimization problem of the following form:search minimal s_x ...for rows=2 s0 = 2*0 + 1for rows=3 s0 = 2*0 + 3 s1 = 2*1 + 0for rows=4 s0 = 2*0 + 6 s1 = 2*1 + 1for rows=5 s0 = 2*0 + 10 s1 = 2*1 + 3 s2 = 2*3 + 0for rows=6 s0 = 2*0 + 15 s1 = 2*1 + 6 s2 = 2*3 + 1for rows=7 s0 = 2*0 + 21 s1 = 2*1 + 10 s2 = 2*3 + 3 s3 = 2*6 + 0for rows=8, s0 = 2*0 + 28 s1 = 2*1 + 15 s2 = 2*3 + 6 s3 = 2*6 + 1for rows=9, s0 = 2*0 + 36 s1 = 2*1 + 21 s2 = 2*3 + 10 s3 = 2*6 + 3 s4 = 2*10+ 0Where in the formula sx = 2*a + ba is the sum of dots of the two lower triangles, which are movedand b is the sum of dots of the top triangle which is moved. b . . a . aSolutions for minimal overall-sum of dots: rows ! rows_a! rows_b ! sum_a ! sum_b ! overall_sum ----------------------------------------------------- 2 ! 0 ! 1 ! 0 ! 1 ! 1-------------------------------------------------- 3 ! 1 ! 0 ! 1 ! 0 ! 2 4 ! 1 ! 1 ! 1 ! 1 ! 3 5 ! 1 ! 2 ! 1 ! 3 ! 5-------------------------------------------------- 6 ! 2 ! 1 ! 3 ! 1 ! 7 7 ! 2 ! 2 ! 3 ! 3 ! 9 8 ! 2 ! 3 ! 3 ! 6 ! 12-------------------------------------------------- 9 ! 3 ! 2 ! 6 ! 3 ! 15I extend this just by eyeball-inspection.... 10 ! 3 ! 3 ! 6 ! 6 ! 18 11 ! 3 ! 4 ! 6 ! 10 ! 22(Hmm, too lazy now to put it into a formula... ;-) )Gottfried Helms === John R Ramsden schrieb:>I get B(6) = 8, No you'r wrong: O> X X> X X X> X X X X> O X X X O>O O X X O O > Darn. I thought my reply might be oversimplifying the problem.> I see it as an optimization problem of the following form:search minimal s_x ...for rows=2 s0 = 2*0 + 1for rows=3 s0 = 2*0 + 3 s1 = 2*1 + 0for rows=4 s0 = 2*0 + 6 s1 = 2*1 + 1for rows=5 s0 = 2*0 + 10 s1 = 2*1 + 3 s2 = 2*3 + 0for rows=6 s0 = 2*0 + 15 s1 = 2*1 + 6 s2 = 2*3 + 1for rows=7 s0 = 2*0 + 21 s1 = 2*1 + 10 s2 = 2*3 + 3 s3 = 2*6 + 0for rows=8, s0 = 2*0 + 28 s1 = 2*1 + 15 s2 = 2*3 + 6 s3 = 2*6 + 1for rows=9, s0 = 2*0 + 36 s1 = 2*1 + 21 s2 = 2*3 + 10 s3 = 2*6 + 3 s4 = 2*10+ 0Where in the formula sx = 2*a + ba is the sum of dots of the two lower triangles, which are movedand b is the sum of dots of the top triangle which is moved. b . . a . aSolutions for minimal overall-sum of dots: rows ! rows_a! rows_b ! sum_a ! sum_b ! overall_sum ----------------------------------------------------- 2 ! 0 ! 1 ! 0 ! 1 ! 1-------------------------------------------------- 3 ! 1 ! 0 ! 1 ! 0 ! 2 4 ! 1 ! 1 ! 1 ! 1 ! 3 5 ! 1 ! 2 ! 1 ! 3 ! 5-------------------------------------------------- 6 ! 2 ! 1 ! 3 ! 1 ! 7 7 ! 2 ! 2 ! 3 ! 3 ! 9 8 ! 2 ! 3 ! 3 ! 6 ! 12-------------------------------------------------- 9 ! 3 ! 2 ! 6 ! 3 ! 15I extend this just by eyeball-inspection.... 9 ! 3 ! 3 ! 6 ! 6 ! 18 9 ! 3 ! 4 ! 6 ! 10 ! 22(Hmm, too lazy now to put it into a formula... ;-) )Gottfried Helms === > The consistent way to say, There is no absolute truth> You said it already. Why do you need to say anything more?> (How about this one?) ;o)> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.> This statement is false if your ?st statement is true.> It seems you have been tortured by locking you up in a round room and> telling you that can only piss at a corner.> Clari?ation for You:> There is no absolute truth> How about There is no absolute truth ?> Is it absolute?> GeorgeOf course not.If we say ..For all propositions p (p is not absolutely true) then, (q isnot absolutely true) ..for any q.(including There is no absolute truth)Even if we were in possession of an absolute truth, how could we know thatit is absolutely true?Witt === > Of course not.It was attempt to ?d a non-explicit way to state consistently thereis no absolute truth.The post belowhttp://www.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8 &frame=right&th=3f55bc43b6983a5a&seekm=vvgbgvfrh8909a% 40corp.supernews.com#link21suggested the elegant way to destroy my hope ;o(-------------THE STATEMENT THE STATEMENT THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT IS WRONG IS WRONG-------------There is another way to look at this:If there is a way to state consistently there is no absolute truthOR there is no way to do so, the both scenarios cause the sameconclusion about impossibility of absolute truth. In case of the ?stscenario it is true because it is expressed explicitly, in case ofsecond scenario - because even statement about non-absoluteness is notabsolute that indicates the deep internal consistency ofnon-absolute assumption.I am not sure it sounds convincing.George === > The consistent way to say, There is no absolute truth (How about this one?) ;o) THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT. George Buyanovsky If there were no truth, neither would there be any lies. If reality> exists, then its existence is true, thus proving that truth exists. If> reality doesn't exist, how could we be deceived into experiencing it?> How can there be deception unless that deception masks a reality about> which we're being deceived? Either way, some kind of reality must> exist, thus proving something true. > One Day Tesshu, the famous swordsman and zen devotee, went > to Dokuon and told him triumphantly he believed all that exists > is empty, there is no you or me, and so on. The master, who had > listened in silence, suddenly snatched up his long tobacco pipe > and struck Tesshu's head. The infuriated swordsman would have > killed the master there and then, but Dokuon said calmly, > Emptiness is quick to show its anger, isn't it? > Forcing a smile, Tesshu left the room. (Soul Food -- Stories to Nourish the Spirit and the Heart> Ed. Jack Korn?ld & Christina Feldman) > Accept the terrible truth that all is illusion. All being One, > wherever you go, there you are. But that doesn't have to ruin > things. The Gnosis of illusion and what to do about it leads> the seeker to Moksha-- experiential knowledge of the liberation> from dualistic bondage. The choice after Moksha is your own. > Return to source? To nothingness? Or choose the road for the > sake of the undiscovered country and the experience of experience> for the sake of itself. Accept the terrible truth and return to yourself. Your path> after that crossroads is your own choice. =~) Students achieving oneness, will move ahead to twoness.> (Woody Allen) We shall not cease from exploration.> And the end of all our exploring> Will be to arrive where we started> Knowing the place for the ?st time.> (T.S. Eliot)So we agree then.- Tue === > <^> <()> <^> ---- <^> <()> <^> 2004 <^> <()> <^> --- <^> <()> <^> buyanovsky@attbi.com (George Buyanovsky)> The consistent way to say, There is no absolute truth (How about this one?) ;o)> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.> George Buyanovsky> If there were no truth, neither would there be any lies. If reality> exists, then its existence is true, thus proving that truth exists. If> reality doesn't exist, how could we be deceived into experiencing it?> How can there be deception unless that deception masks a reality about> which we're being deceived? Either way, some kind of reality must> exist, thus proving something true. Tue true!Absolutely! :-)- Tue === A ?deception' can be control itself. Whom is to say that the deception, isthat we have control. There is no truth or truth in a existence where weand everything else is controlled. To claim there is a truth, is to claimthere is an existence. Reality, an illusion of control.> <^> <()> <^> ---- <^> <()> <^> 2004 <^> <()> <^> --- <^><()> <^> buyanovsky@attbi.com (George Buyanovsky)> The consistent way to say, There is no absolute truth (How aboutthis one?) ;o)> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE STATEMENT.> George Buyanovsky> If there were no truth, neither would there be any lies. If reality> exists, then its existence is true, thus proving that truth exists. If> reality doesn't exist, how could we be deceived into experiencing it?> How can there be deception unless that deception masks a reality about> which we're being deceived? Either way, some kind of reality must> exist, thus proving something true.> Tue true!> Herc> === >[...]>|>If we were talking about formal proofs, it would be appropriate to>|>speak in hard-edged terms about the properties of the proof. But>|>this is an informal proof.>|>|>Take an analogous situation. Fermat's proof of his conjecture for>|>p=4 ?s the standard mold of proof by in?ite descent. He shows>|>that if it fails for some value of z, then it also fails for some>|>smaller value of z. How wrong is it to say he was doing a proof by>|>induction? Well, I tend to cover myself in situations like that by>|>saying essentially a proof by induction, but I don't think it's>|>just plain wrong to call it a proof by induction when it's>|>essentially a proof by induction.>|>|>Likewise, I don't think it's just plain wrong to call a proof>|>by proving the contrapositive a form of proof by contradiction,>|>since it's essentially a proof by contradiction.>|>|>For one thing, if we were to take this informal proof (or any>|>proof by proving the contrapositive) and convert it into a>|>formal proof, quite typically it would formalize as a proof by>|>contradiction.>|>|We may be getting somewhere:>|>|Depends on the formal system. Suppose we were talking>|about a formal system that had this rule of inference:>|>|[Contrapositive:]>|>|~B |- ~A>|_______>| |- A -> B.>|>|In that formal system a proof via the contrapositive would>|formalize precisely as a proof via Contrapositive.>Well, you could put together a system like that. People have been known>to write systems of logic that would do a lot more than just include a>proof-by-contrapositive. I gather there's a logic textbook by a guy named>Copi which has several more ways of disguising proof by contradiction>under several more obscure names, each distinct from the other one.>Proof by dilemma, etc. Not considered very elegant, though.Of course it's not very elegant.>One problem with adding proof by contrapositive as a special rule is that>it's redundant. Of _course_ it's redundant._If_ the aim is to give a system that allows direct faithful translations of the proofs that people actually write in Englishthen the system is going to be _hugely_ redundant, becausethe list of informal inference rules that people actually use_is_ redundant.Usually when people are setting up formal systems of logicthat's _not_ the aim. But _if_ we're going to attempt to sayprecise things about the logical struucture of a proof written in English then we need a system that directlyre?the informal system that people actually use.>If we were to drop another rule, probably we could make>it stop being redundant, but then it'd be essentially doing the job that>proof by contradiction had been doing.>[...]>If one were to append Zorn's lemma and the well-ordering principle to>ZFC, one could say of a lot of proofs that they are not using the axiom>of choice, but using the axiom of well-ordering. But that would be a>somewhat strange and redundant axiomatization.Yes, that would be strange. But when we're deciding what theaxioms for set theory should be the aim is _not_ to allow directtranslations of informal proofs into a formal system, the aimis to give a minimal set of axioms that actually suf?es to derivethe theorems we want.Really, it's a different context. Take the proof of Cauchy's Theoremin some complex analysis book. Suppose we want to say somethingabout the logical structure of the proof we read there. Would the?st step to be to translate the proof into ZFC? Of course not.The translation of the _statement_ of the theorem, written outin the language of set theory (with in as the only predicate, noabbreviations) would take I don't know how many pages, andwould not be recognizable as Cauchy's Theorem by humans;the proof would take an incredibly larger number of pages,and the logical structure of that proof would be much morecomplicated than the _apparent_ logical structure visible onthe pages of that complex book. Saying that the structureof the formal proof in ZFC has nothing to do with the structureof the informal proof in the book would of course be anexaggeration, but it's close enough to the truth that itseems to me that the translation into formal ZFC is simplynot appropriate _if_ we want to say something about thestructure of the proof in the book (eg uses proof bycontradiction).>Without turning Zorn's lemma and the well-ordering principle into axioms,>one could still reasonably categorize proofs in ZFC which are not proofs>in ZF by the form of the axiom of choice, or the consequence of the axiom>of choice, being used. Give me a pile of such proofs, and three bins>labelled axiom of choice, Zorn's lemma, and well-ordering principle,>and I can probably put nearly every such proof unequivocally into one of>the bins, by the principle *most directly* being used. (Well, some might>use more than one.) (On sci.math, Matthew Wiener once quoted Herstein and>Kaplansky as writing that the axiom of choice is intuitively obvious,>while as Zorn's lemma it's merely plausible, and as the well-ordering>principle it's obviously false.) This sort of categorization is probably>pretty sharp.If we actually wanted to imitate real proofs in a formal system ithas to be a system that allows for new de?itions, referencesto previously proved theorems, etc. In a system like that onecould make the distinction precise: The proof contains a referenceto Zorn's Lemma or it doesn't, and it contains a reference to ACor it doesn't.>Yet for someone to say of all three piles, that they are proofs using the>axiom of choice, would be appropriate.Yes. Right now we're writing English, which always contains implicitassumptions and implicit notions of what the context is. Although thetwo statements (i) It uses Zorn's lemma, not AC and (ii) It uses AC are logically contradictory, they can both be correct of a proof,in the sense that what the speaker _means_ by (i) is true andwhat the other speaker _means_ by (ii) is true. (In (i) we're trying to analyze the structure directly as above, while in(ii) we're thinking about what parts of ZFC are required tomake a formalization of the proof valid.)>[...]>I think I've seen proofs in textbooks that were essentially the same, but>with slightly different wording. This is ?e. And if it helps, if we>played this game of sorting proofs into bins, this time labelled proofs>using contradiction and proofs using contrapositive, I'm sure we'd>be reasonably consistent in how we sorted them out.Well then what's the problem? >But I still think it's a bit as if someone were to say, Here's a proof>by the axiom of choice that.... and we were to say, Wrong, it's a proof>by Zorn's lemma. It's at least not completely wrong to call it a proof>by the axiom of choice, if it's a proof by Zorn's lemma.Oh, that's the problem. Fine. I assert that (i) and (ii) above can both be correct; similarly, although it's not completely wrong tosay Amanda's proof is in fact a proof by contradiction it's alsonot completely wrong (in fact hardly wrong at all) to sayno, it's not a proof by contradiction, it's a proof of thecontrapositive.>[...]Sorry about the major snippage - I _think_ that most of whatI snipped was explanations and illustrations of points thatI think I got. As of your I'm sure we'd be reasonably consistent in how we sorted them out much of this seems moot to meanyway...>Keith RamsayDavid C. Ullrich === In a spherical triangle with dihedral angles A,B,C and side common toA,B angles subtending angle c at sphere center,we haveCos C= - Cos A Cos B + Sin A Sin B Cos c . This very same relationship would result if we take two taper wedgeswith taper angles [(A + 3Pi/2),B] and rotate one prismatic wedge blockwith respect to the other as follows: We place the second wedge bottomplane on the ?st wedge top plane,make a planar contact and rotatesecond wedge by rubbing/rotating it through an angle c in theinterface plane. Non contacting edges of these two wedges (consideronly edges containing angles [(A + 3Pi/2),B])now make a skewed angle(Pi/2- C) exactly .I am unable to imagine the identity of these two situations yieldingthe same formula. If the blocks visualization from the above is notclear enough,I could supply a link to drawing. TIA. === > of the solutions came across this little equation:> x^2 + x + 1 = 0 (i)> For fun, I started to push things around a bit, instead of just going> with x = (-b +- sqrt(b^2 - 4ac))/(2a):> x(x + 1) = -1> (x + 1) = -1/x (ii) <-- I though it would be OK to divide by x,> since x<>0> Substituting (ii) back into (i):> x^2 + (x + 1) = 0> x^2 - 1/x = 0> x^3 - 1 = 0 <-- again, multiplying by x, since x<>0> (x - 1)(x^2 + x + 1) = 0 (iii)> Somehow in my little manipulations I introduced a new factor onto the> original equation, but I can't ?ure out how. This seems pretty> ridiculous, but I'd appreciate another pair of eyes looking at it and> letting me know where the (x - 1) factor popped in.> What you have proved was that> x^2 + x + 1 = 0 => (x - 1)(x^2 + x + 1)> and this is obviously correct. It would still be correct if x - 1 was> replaced by anything else. But note that your computations do not prove> the reverse implication: you cannot devide by x - 1 since you are not> assuming that x <> 1.Oh, sure you are, since the original equation came up because you arelooking for the non-trivial solutions of x^3-1=0.Jon Miller === > I am looking for some web sites and references that deal with> algorithms of optimization that permit to arrange geometric shapes in> order to uniformly cover a surface (example of problem : How to> optimize the parameters of a sprayer in order to have an uniform> thickness coating?). Have you some informations about these sorts of> problems?> Jean> I'll give you a real world problem that I encountered a few days ago that> might be what you're looking for.> I work for a meteorology website. We are currently undergoing some changes> and want to promote our website even more. As a result, we considered> looking at making rain gauges. Being an environmental website, we want to> use the least amount of material, but we'd like it to hold 50 cc o?.> What dimensions do you make the gauge?You make it a sphere? Cool, how do I get one?Jon Miller === > I am looking for some web sites and references that deal with> algorithms of optimization that permit to arrange geometric shapes in> order to uniformly cover a surface (example of problem : How to> optimize the parameters of a sprayer in order to have an uniform> thickness coating?). Have you some informations about these sorts of> problems?> Jean> I'll give you a real world problem that I encountered a few days agothat> might be what you're looking for.> I work for a meteorology website. We are currently undergoing somechanges> and want to promote our website even more. As a result, we considered> looking at making rain gauges. Being an environmental website, we wantto> use the least amount of material, but we'd like it to hold 50 cc of> liquid.> What dimensions do you make the gauge?> You make it a sphere? Cool, how do I get one?> Jon Miller>We were working with cylinders. We probably won't do this promotion, but wewere messing with it a few days ago.-- David MoranChief MeteorologistOklahoma Storm Team === Number of positions...> After exactly two pawn moves:> ... after two moves by the same pawn: 16> ... after two pawns each moving 1 step: 28> ... after two pawns each moving 2 steps: 28> ... after two pawns, one moving 1 and one moving 2: 56> TOTAL: 128 > After a pawn move and a non-pawn move:> ... after a3: 4> ... after a4: 6> ... after b3: 6> ... after b4: 6> ... after c3: 6> ... after c4: 7> ... after d3: 12> ... after d4: 13> ... after e3: 15> ... after e4: 15> ... after f3: 4> ... after f4: 5> ... after g3: 6> ... after g4: 6> ... after h3: 4> ... after h4: 6> TOTAL: 121 After a knight move and a rook move: 4> After two moves by same knight: 11> After two moves by different knights: 4> TOTAL: 19 Grand total: 128+121+19=268 But i don`t really understand it!> the 16 is ok, but how do you get the 28?> I think after the two moves of the pawn you go back> to the initial boardposition, right? How do i get up> to 28 different positions by moving 2 pawns, each one step? complicated stuff...:-( > 28 is found by what mathematicians call 8 choose 2. Look up> combinations, or binomial coef?ients to understand. > But this is easy to count. We have the following combinations of> two pawns to move one square each (named after their ?e): ab> ac> ad> ae> af> ag> ah> bc> bd> be> bf> bg> bh> cd> ce> cf> cg> ch> de> df> dg> dh> ef> eg> eh> fg> fh> gh 28 possibilities. Note that this is 7+6+5+4+3+2+1.> ok, is see!so i have to setup the initial boardposition after each move, right? === > [snip] The analogy doesn't work. There are lots of ways to factor> an expression. If I said, What is the form of the> factorization of (125*x^3 + 75*x^2 + 45*x + 7)?, most people> might think I wanted a factorization as a polynomial in x.> That, however, is not what you have done with your expression.> You have factored it as a polynomial in 5. That is why it> is important to say that you *assumed* a factorization of> the speci?d form. It is a natural and even necessary> part of the discussion.>Let's say you assume that irrationals are not integers.>Yet someone else tells you, hey, why assume that, as they're just not?>Don't you think a *rational* response might be to go, oh, yeah, you're>right?> Again not a valid parallel. This argument is so goddamned silly> and petty. I have little patience with arguments about semantic trivia. > If you don't like my saying that you assumed a factorization of the > form given, you can replace assumed by speci?d. That conveys the > same meaning: that there were lots of other ways it could have been > speci?d (or assumed for that matter).So why are you arguing? Can you see my point?A *reasonable* person can see how the word assume has conotations,which don't apply if something is known to be true, like why assumethat irrationals are not rational, or why assume that 3 is a factor of6?What I'm exploring is how deep your problem in this area goes.Will you *ever* be able to acknowledge what most people take forgranted? >Now then, what if instead, you acted defensive and argued endlessly>about your use of words where there was an implication that something>might not be true by use of the word assume, which *reasonable*>people could see.>Is there any possibility that you might concede that saying assume>for a truth might be loaded or could look like you're trying to imply>that something might not be true?> And here, it's a distraction. You are trying desperately to focus> on a minor wording issue to take attention away from the fact> that you have not dealt with the mathematics.>I'm curious about how you manage to spend so much time rather>obviously ignoring the truth, twisting things a certain way, yet>acting as if you're aboveboard no matter how often you're caught.>Your defensive reaction here with the word assume is a case in>point.>You basically refuse to be rational, and refuse to concede even minor>points, arguing endlessly as if you're always right, no matter what,>or how much evidence is produced to challenge your world view.> No, you're simply wrong. You assumed a factorization of a> certain form. You could have assumed many other forms. It> is appropriate to note what you did and to convey the impression> that the choice involved a decision on your part. By no means> was I implying that your assumption was wrong. It's not.> There IS a factorization of that form. I was just describing that> that was the form you were specifying..Your original usage may have just been a minor mistake, but my pointis that you can't accept correction, or at a minimum, in the interestof moving forward, shift your wording on what you yourself claimed isa minor point.Your *actions* speak volumes as you've gone on now through severalreplies, ?hting over what you claim is a minor point, only to nowjust push the previous position! That's what I was checking on, as itlooks to be some deeply ingrained pattern of behavior.You have demonstrated unreasonableness to a high degree. It's uselessto present a rational position to you contrary to what you're saying,as you have demonstrated that you will simply reject it, hold on toyour own point of view, and eventually just repeat your originalposition, now matter what has come before.Therefore, it's pointless to discuss anything with you, unless someoneis willing to agree with you at the outset, as you will just continuewith your own position, no matter what.James Harris === > [snip] [inane drivel snipped] Your original usage may have just been a minor mistake, but my point> is that you can't accept correction, or at a minimum, in the interest> of moving forward, shift your wording on what you yourself claimed is> a minor point. Your *actions* speak volumes as you've gone on now through several> replies, ?hting over what you claim is a minor point, only to now> just push the previous position! That's what I was checking on, as it> looks to be some deeply ingrained pattern of behavior. You have demonstrated unreasonableness to a high degree. It's useless> to present a rational position to you contrary to what you're saying,> as you have demonstrated that you will simply reject it, hold on to> your own point of view, and eventually just repeat your original> position, now matter what has come before. Therefore, it's pointless to discuss anything with you, unless someone> is willing to agree with you at the outset, as you will just continue> with your own position, no matter what.> This is transparent pompous B.S. designed to provide you witha super?ial excuse to avoid talking about the mathematics. There was in fact one central problem with my previous post: I did not provide an adequate answer to your comment regarding the existence of the algebraic integers A, B, and C. I claimed they exist but I did not specify what they were or even give an existence proof. If I had been you, I would have pounced on that like a rooster on a June bug as evidence that I could not back up my claims. You were too focussed on your stupid semantic issue and trying to ?d an excuse to stop replying to the math (not that you ever did). Dik Winter has also claimed to have a way of de?ing A, B, and C but has recently backed off from his original de?ition (see his post in the thread ?Mathematical consistency, courage') and provided another one. Below I give an existence proof for A, B, and C. I refer to themas Dik has done before, i.e. A = w1 = w1(x), B = w2 = w2(x), and C = w3 = w3(x). All are dependent on x. In general when x <> 0,they are not, needless to say, equal to 7, 7, 1. === == === First, use the Magidin-Mckinnon result to factor P(x)/49 in the form [*] P(x)/49 = (c1*5 + d1)*(c2*5 + d2)*(c3*5 + d3), where c1, c2, c3, d1, d2, and d3 are all algebraic integers. Note that this is a factorization as a polynomial in 5,not of course as a polynomial in x. Note that d1*d2*d3 = 7. Note that c1 and d1 are coprime, as are c2, d2 and c3, d3. This is true because in general P(x)/49 has integer coef?ients and is primitive. Now de?e w1 = d2*d3, w2 = d1*d3, and w3 = d1*d2. Now go back to the Harris factorization of P(x) itself: [**] P(x) = (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7). Clearly 7 = d1*w1 = d2*w2 = d3*w3. Now you know that -7/a1 and -d1/c1 are roots of the same polynomial, a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). Therefore a1 = 7*c1/d1 = c1*(7/d1) = c1*d2*d3 = c1*w1. Similarly, a2 = c2*w2 and a3 = c3*w3. Finally, note that w1 = gcd(a1, 7) = gcd(c1*d2*d3, d1*d2*d3). Similarly, w2 = gcd(a2, 7), and w3 = gcd(a3, 7). All of these things - w1, w2, w3, c1, d1, ..., a1, b1, ... are algebraic integers that are dependent on x, and should be written as w1(x), ..., c1(x), ..., a1(x), ... etc. Thus w1, w2, and w3 give you the factors to go from the Harrisfactorization [**] to the Magidin-Mckinnon factorization [*], andof course w1*w2*w3 = (d2*d3)*(d1*d3)*(d1*d2) = (d1*d2*d3)*(d1*d2*d3) = 7*7 = 49. And all the coef?ients in the Magidin-Mckinnon factorization arealgebraic integers, which is what you wanted in the ?st place. QED. === === = The Magidin-Mckinnon result is constructive, i.e., with enoughwork, for any given integer x, you can compute the answers. Similarlythe gcd function in the algebraic integers. Actually carrying outthe calculations however is admittedly dif?ult. Nora B. James Harris === > Therefore, it's pointless to discuss anything with you, unless someone> is willing to agree with you at the outset, as you will just continue> with your own position, no matter what. > James HarrisJSH obviously avoids knowing himself.Doesn't this sound like something that everyone else has said about JSH? === > Again not a valid parallel. This argument is so goddamned silly> and petty. I have little patience with arguments about semantic trivia. > If you don't like my saying that you assumed a factorization of the > form given, you can replace assumed by speci?d. That conveys the > same meaning: that there were lots of other ways it could have been > speci?d (or assumed for that matter). So why are you arguing? Can you see my point? A *reasonable* person can see how the word assume has conotations,> which don't apply if something is known to be true, like why assume> that irrationals are not rational, or why assume that 3 is a factor of> 6? What I'm exploring is how deep your problem in this area goes. Will you *ever* be able to acknowledge what most people take for> granted?Most people can't do the math so what they take for granted is irrelevant here. Why does JSH keep on going about things which are of no mathematical import, unless he is unable or unwilling to face the mathematical issues directly? === > [snip] [snip] I have no interest in your theories of personality,psychology or social interaction or your inane hairsplittingabout semantics. If you cannot address the math, don't bother to post at all. Nora B. === > [snip] The analogy doesn't work. There are lots of ways to factor> an expression. If I said, What is the form of the> factorization of (125*x^3 + 75*x^2 + 45*x + 7)?, most people> might think I wanted a factorization as a polynomial in x.> That, however, is not what you have done with your expression.> You have factored it as a polynomial in 5. That is why it> is important to say that you *assumed* a factorization of> the speci?d form. It is a natural and even necessary> part of the discussion.>Let's say you assume that irrationals are not integers.>Yet someone else tells you, hey, why assume that, as they're just not?>Don't you think a *rational* response might be to go, oh, yeah, you're>right?> Again not a valid parallel. This argument is so goddamned silly> and petty. I have little patience with arguments about semantic trivia. > If you don't like my saying that you assumed a factorization of the > form given, you can replace assumed by speci?d. That conveys the > same meaning: that there were lots of other ways it could have been > speci?d (or assumed for that matter). So why are you arguing? Can you see my point?Nora sees your point.Do you see Nora's point?Are you reading her replies?You keep giving examples like Assume sqrt(2) is irrational andNora has repeatedly pointed out that such an example is notat all similar to the case being discussed. This indicates thatyou are not understanding what the discussion is about, whetheror not you are correct and Nora is wrong.> A *reasonable* person can see how the word assume has conotations,> which don't apply if something is known to be true, like why assume> that irrationals are not rational, or why assume that 3 is a factor of> 6?How many explanations and examples does Nora have to give you toindicate that 3 is a factor of 6 is not a good example of what isbeing discussed?The statement 3 is a factor of 6 is always true. However, thisis not the situation that we are in.Here are some more, better examples of what is being discussed:1) Let a deck of cards be shuf?ith the assumption thatthe top card is the three of clubs, then there are 26 redcards in the remaining portion of the deck.2) With the assumption that 30 is factored in the form 60 = 2*5*6,then there are an odd number of factors in the factorization.The point is that the top card of a shuf?k need not bethe three of clubs. Likewise, the factorization of 60 canbe in other forms, like 60 = 2*2*3*5 (where there are nowan even number of factors in the factorization).> What I'm exploring is how deep your problem in this area goes. Will you *ever* be able to acknowledge what most people take for> granted?It is not clear to me that you are right. As with Nora, I agreethat you might be right, but even if you are right, you are notexplaining it very well: you are giving cases where most peopleagree that the phrase assuming would be misleading, and avoidingthe cases where the problem area is (where the statement isnot a fact true in all situations). >Now then, what if instead, you acted defensive and argued endlessly>about your use of words where there was an implication that something>might not be true by use of the word assume, which *reasonable*>people could see.>Is there any possibility that you might concede that saying assume>for a truth might be loaded or could look like you're trying to imply>that something might not be true?> And here, it's a distraction. You are trying desperately to focus> on a minor wording issue to take attention away from the fact> that you have not dealt with the mathematics.>I'm curious about how you manage to spend so much time rather>obviously ignoring the truth, twisting things a certain way, yet>acting as if you're aboveboard no matter how often you're caught.>Your defensive reaction here with the word assume is a case in>point.>You basically refuse to be rational, and refuse to concede even minor>points, arguing endlessly as if you're always right, no matter what,>or how much evidence is produced to challenge your world view.> No, you're simply wrong. You assumed a factorization of a> certain form. You could have assumed many other forms. It> is appropriate to note what you did and to convey the impression> that the choice involved a decision on your part. By no means> was I implying that your assumption was wrong. It's not.> There IS a factorization of that form. I was just describing that> that was the form you were specifying.. > Your original usage may have just been a minor mistake, but my point> is that you can't accept correction, or at a minimum, in the interest> of moving forward, shift your wording on what you yourself claimed is> a minor point. Your *actions* speak volumes as you've gone on now through several> replies, ?hting over what you claim is a minor point, only to now> just push the previous position! That's what I was checking on, as it> looks to be some deeply ingrained pattern of behavior. You have demonstrated unreasonableness to a high degree. It's useless> to present a rational position to you contrary to what you're saying,> as you have demonstrated that you will simply reject it, hold on to> your own point of view, and eventually just repeat your original> position, now matter what has come before. Therefore, it's pointless to discuss anything with you, unless someone> is willing to agree with you at the outset, as you will just continue> with your own position, no matter what. > James Harris === > [snip] The analogy doesn't work. There are lots of ways to factor> an expression. If I said, What is the form of the> factorization of (125*x^3 + 75*x^2 + 45*x + 7)?, most people> might think I wanted a factorization as a polynomial in x.> That, however, is not what you have done with your expression.> You have factored it as a polynomial in 5. That is why it> is important to say that you *assumed* a factorization of> the speci?d form. It is a natural and even necessary> part of the discussion.>Let's say you assume that irrationals are not integers.>Yet someone else tells you, hey, why assume that, as they're just not?>Don't you think a *rational* response might be to go, oh, yeah, you're>right?> Again not a valid parallel. This argument is so goddamned silly> and petty. I have little patience with arguments about semantic trivia. > If you don't like my saying that you assumed a factorization of the > form given, you can replace assumed by speci?d. That conveys the > same meaning: that there were lots of other ways it could have been > speci?d (or assumed for that matter). So why are you arguing? Can you see my point? Nora sees your point.My point is that using assume for a fact is misleading. > Do you see Nora's point?tautologies.I hope you understand that is true.For instance, x^2 + 2x + 1 = (x+1)(x+1)is an identity or what I call a tautology.It's just true.Now then, what if I say, assume that x^2 + 2x + 1 has x+1 as a factor?I've replied to you here as *your* reply shows what I see is adeliberate attempt to deny basic facts, in a social need to promote aperson you see as a member of your group.The fact is that saying assume for a truth is just not the way mostpeople operate. > A *reasonable* person can see how the word assume has conotations,> which don't apply if something is known to be true, like why assume> that irrationals are not rational, or why assume that 3 is a factor of> 6? How many explanations and examples does Nora have to give you to> indicate that 3 is a factor of 6 is not a good example of what is> being discussed?How much time do I have to take to show that it is?First, do you accept that a factorization is a truth? > The statement 3 is a factor of 6 is always true. However, this> is not the situation that we are in.Yes it is, as I've repeatedly explained, as a factorization is atruth. > Here are some more, better examples of what is being discussed: 1) Let a deck of cards be shuf?ith the assumption that> the top card is the three of clubs, then there are 26 red> cards in the remaining portion of the deck.But what if the top card is NOT the three of clubs?Your example allows falsity. That is, implied with the wordassumption usually there is the possibility that something might notbe true.It's that implication that I'm faulting in the statement by NoraBaron.What if I assume that Nora Baron is a female? By saying that, am Inot also pointing out that possibly that poster is a man masqueradingas a woman? > 2) With the assumption that 30 is factored in the form 60 = 2*5*6,> then there are an odd number of factors in the factorization.That is a specious example given the actual statements by NoraBaron.More like that poster's statement would be something like, assume that30 has the factorization 2*3*5. > The point is that the top card of a shuf?ck need not be> the three of clubs. Likewise, the factorization of 60 can> be in other forms, like 60 = 2*2*3*5 (where there are now> an even number of factors in the factorization).And my point is that implicit with the word assume there's thepossibility that what is assumed is *false*, while withfactorizations, a factorization is a truth, so it can't be false.Now the poster Nora Baron further claimed that it was a minor point,but repeatedly argued that supposedly minor point through severalreplies.My major point is that Nora Baron is NOT reasonable, but simply hasa desire to push one point of view, without the willingness to concedeeven minor points.If you're dealing with such a person, there's no reason to debatethem, as the only thing they will accept is your acceptance of theirclaims, so what's the point of debating with them?James Harris === > For instance, > x^2 + 2x + 1 = (x+1)(x+1)> is an identity or what I call a tautology.It is not a tautology in any reasonable (and technical) sense of theword.It is certainly not a theorem of ?st order logic with equality,which is a fairly liberal sense of tautology. Instead, it depends on certain axioms regarding multiplication andaddition, including distributivity, commutativity of addition and themultiplicative identity laws. Any reason you insist on misusing every technical term you can ?d?-- Meaningless movies on the screen behind the band that's blowing Waterboys,throwing shapes My Love is My RockHalf of the music is on tape in the Weary Land === JSH's Bizarrre math position seems to be that the math is not as important as one's attitude, particularly one's attitude about JSH's claims.He is tha only one whose _math_ position is the least bit bizarre in any of his threads (though there are some bizarre math positions in threads to whichJSH has not coontributed).In sci.math postings, one's positions on non-mathematical issues should be entirely irrelevant, and omitted, however much that might constrain JSH's volubility. === > [...]>Therefore, it's pointless to discuss anything with you, unless someone>is willing to agree with you at the outset, as you will just continue>with your own position, no matter what.Sounds like you're ?ally starting to realize she's right about themath...>James HarrisDavid C. Ullrich === >[...]>As for your claim about ways to factor 49, that's just something>you're making up.>What I *say* is that the constant terms go from being 7, 7 and 22 to>being 1, 1, and 22, when 49 is divided off, which is an easily>veri?ble fact.Why don't you _prove_ it if it's so easy, instead of simply continuingto _assert_ it?Just for fun, I'm going to read Nora's mind: the reason she thinksyou're saying something about factoring 49 when you say whatyou say you say is that (i) _if_ there were only one way to factor49 then what you say you say would follow, (ii) she can't imaginewhy _else_ you think what you say you say is true.So _prove_ that the new constant terms must be 1, 1, and 22already. >[...]David C. Ullrich === Is there an explicit example of a function f: O->F (E, F are two Banachspaces, O is an open set of E) such that there is one point a in O, df(a) isa continuous isomorphism E->F, but f^(-1) is not differentiable at point a ?Note that such an example would require that both E and F are in?itedimensional, and that df(a) is not bicontinuous.--J.S === > Is there an explicit example of a function f: O->F (E, F are two Banach> spaces, O is an open set of E) such that there is one point a in O, df(a) is> a continuous isomorphism E->F, but f^(-1) is not differentiable at point a ?> Note that such an example would require that both E and F are in?ite> dimensional, and that df(a) is not bicontinuous.> I shouldn't have assumed E and F are Banach spaces; let's say they're only> normed vector spaces (the point of my question was to show explicitely that> df(a) can be a continuous isomorphism, and yet f^(-1) is not differentiable> at point a).Let E = l^2 and de?e f((x_n)) = (x_n/n). Set F = f(E). Then f is a linear continuous 1-1 map of E onto the normed linear space F. We have df(a) = f for every a in E, yet f^(-1) is continuous nowhere in F, hence differentiable nowhere in F. === On Sun, 4 Jan 2004 16:35:39 +0100, Julien Santini>Is there an explicit example of a function f: O->F (E, F are two Banach>spaces, O is an open set of E) such that there is one point a in O, df(a) is>a continuous isomorphism E->F, but f^(-1) is not differentiable at point a ?Presumably you meant to ask for the inverse to fail to be differentiable at f(a).>Note that such an example would require that both E and F are in?ite>dimensional, and that df(a) is not bicontinuous.Actually an example doesn't require that, nor does it require thatE and F be in?ite-dimensional. There is a map f:R^2 -> R^2 whichis differentiable at the origin, such that f(0) = 0, df(0) is theidentity, but such that the inverse is not differentiable at 0;it's easy to give such an example, for example arranging thatthe range of f is not even a neighborhood of 0!Probably that's not what you meant either, but if not maybeyou should re-state the question, including all the conditionsyou have in mind.David C. Ullrich === > There is a map f:R^2 -> R^2 which> is differentiable at the origin, such that f(0) = 0, df(0) is the> identity, but such that the inverse is not differentiable at 0;> it's easy to give such an example, for example arranging that> the range of f is not even a neighborhood of 0!No need to go to R^2 for that: On R^1, just let f jump back and forth between x and x - x^2 in an appropriate manner. === On Sun, 04 Jan 2004 13:21:22 -0800, The World Wide Wade> There is a map f:R^2 -> R^2 which> is differentiable at the origin, such that f(0) = 0, df(0) is the> identity, but such that the inverse is not differentiable at 0;> it's easy to give such an example, for example arranging that> the range of f is not even a neighborhood of 0!>No need to go to R^2 for that: So I realized a little later.>On R^1, just let f jump back and forth >between x and x - x^2 in an appropriate manner.David C. Ullrich === Is there an explicit example of a function f: O->F (E, F are two Banach> spaces, O is an open set of E) such that there is one point a in O, df(a) is> a continuous isomorphism E->F, but f^(-1) is not differentiable at point a ?> Note that such an example would require that both E and F are in?ite> dimensional, and that df(a) is not bicontinuous.> So df(a) is continuous and linear and bijective? Then it isbicontinuous by the closed graph theorem. === > So df(a) is continuous and linear and bijective? Then it is> bicontinuous by the closed graph theorem.I shouldn't have assumed E and F are Banach spaces; let's say they're onlynormed vector spaces (the point of my question was to show explicitely thatdf(a) can be a continuous isomorphism, and yet f^(-1) is not differentiableat point a). === The trivial answer is to compute the distance for each eadge separately andchoose the minimum.But, if you have a constant polygon and several points, you could probablydo faster at cost of space for storing the polygon's additional data or atcost of time needed to order the polygons data in some way, which is doneonly once. What I am thinking of is some arrangement of edges in pre-sortedarrays or tree-like structures so that for a ?ed given polygon you get adecision tree so that the number of quieries (hence, running time) is low.Maybe you could use the fact that in order to decide between edge A iscloser and edge B is closer, there exist one line (in rare cases 2 lines)which are at the same distance to both, so that the query (left/right ofthis line) could distinguish between them.Look into books about computational geometry.alex.> Say i have a point p, and an arbitrary n-sided polygon. How can i tell how> far p is from any point (either vertex or edge) of the polygon?> === > so I thought I'd talk about it in detail. Here>are some headers so you can ?d the post:>In his post Decker claimed to mirror my argument using a quadratic>instead of a cubic, where he has>(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >where his a's are roots of >a^2 - (x - 1)a + 7(x^2 + x).>Checking at x=0 reveals that the actual constant terms of the>factorization are 7 and 2, where Decker picked a_1(0) = 0 at x=0.>Now then, consider what happens if you divide both sides of>(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) >by 7, as then you end up with something like>(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2>where the b's are roots of some unknown quadratic, though the ?st>and last coef?ients ARE known:>b^2 + ? b + (x^2 + x).>Now then, it's just a quadratic people. SOME mathematician in all the>world should be able to give what the middle coef?ient is, right?>Right. The middle coef?ient in this case is (-3x + sqrt(-7x^2 - 8x))/4> Well I put that up kind of curious that someone might ?l it in, but> you failed badly here Decker, as my ?st check was at x=1. > a^2 - 7(1^2 + 1) = a^2 - 14, so a=+/-sqrt(-14). Now then, dividing off 7, should then give b=+/-sqrt(-2).> No. See below. > But with your claims, you get b^2 + (-3 + sqrt(-15))b/4 - 2 = 0.> Right, that's what I get. Here's how. Your speci?ation was that you wanted two functions, b_1(x) and b_2(x)> which satisfy P(x)/7 = 25x^2 + 30x + 2 = (5b_1(x) + 1)(5b_2(x) + 2) [1] and which are roots of b^2 + C(x)b + (x^2 + x) [2] Where C(x) is to be found. I remind you that those are your requirements, as stated above.> [Note in passing that there are in?itely many other b's> one could pick, but in each case you'd have a different> polynomial for which they would be roots.] At any rate, since [2] is de?ed by (b - b_1(x))(b - b_2(x)) = b^2 + C(x)b + (x^2 + x) we must have b_1(x)b_2(x) = x^2 + x and -(b_1(x) + b_2(x)) = C(x). Expand the RHS of [1] and we have 25x^2 + 30x + 2 = 25b_1(x)b_2(x) + (2b_1(x) + b_2(x))(5) + 2> = 25(x^2 + x) + (2b_1(x) + b_2(x))(5) + 2 so, rewriting the LHS and subtracting 2 from both sides we have 25(x^2 + x) + 5x = 25(x^2 + x) + (2b_1(x) + b_2(x))(5) so 2b_1(x) + b_2(x) = x from which we have b_2(x) = x - 2b_1(x) Since we have your requirement on the product of the b's> we must have x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x)) so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we> can solve for b_1(x): b_1(x) = (x + sqrt(-7x^2 - 8x))/4Nope. You dropped a 2, as you should have2b_1(x)^2 - x b_1(x) + x^2 + x = 0.I was curious to see if anyone on sci.math would correct your or evennotice your mistake.At this time, I'm curious to know if anyone read Decker's post, andthought he was right, or noticed his mistake.Or did no one else besides me bother to read his post?James Harris === > x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x))so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we>can solve for b_1(x): b_1(x) = (x + sqrt(-7x^2 - 8x))/4> Nope. You dropped a 2, as you should have 2b_1(x)^2 - x b_1(x) + x^2 + x = 0.>Quite right, I dropped a coef?ient of 2. Everything else iscorrect, however. Rick === > x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x))so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we>can solve for b_1(x): b_1(x) = (x + sqrt(-7x^2 - 8x))/4> Nope. You dropped a 2, as you should have 2b_1(x)^2 - x b_1(x) + x^2 + x = 0.> Quite right, I dropped a coef?ient of 2. Everything else is> correct, however.Yes, despite dropping the 2, you correctly solved for b_1(x), but thepoint here is that now it's clear that in general b_1(x) is not analgebraic integer function!!!That jumped out at me with that dropped 2 added back.Now then, do you understand why that's signi?ant?James Harris === > x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x))>so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we>can solve for b_1(x):> b_1(x) = (x + sqrt(-7x^2 - 8x))/4>Nope. You dropped a 2, as you should have>2b_1(x)^2 - x b_1(x) + x^2 + x = 0.>Quite right, I dropped a coef?ient of 2. Everything else is>correct, however.> Yes, despite dropping the 2, you correctly solved for b_1(x), but the> point here is that now it's clear that in general b_1(x) is not an> algebraic integer function!!! That jumped out at me with that dropped 2 added back. Now then, do you understand why that's signi?ant?> Yup. In fact, that underscores the point I've been makingall along--you can't in general split 7 as (in Nora's terms)7, 1 and expect to have a factorization in algebraic integers.Rick === > x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x))>so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we>can solve for b_1(x):> b_1(x) = (x + sqrt(-7x^2 - 8x))/4>Nope. You dropped a 2, as you should have>2b_1(x)^2 - x b_1(x) + x^2 + x = 0.>Quite right, I dropped a coef?ient of 2. Everything else is>correct, however.> Yes, despite dropping the 2, you correctly solved for b_1(x), but the> point here is that now it's clear that in general b_1(x) is not an> algebraic integer function!!! That jumped out at me with that dropped 2 added back. Now then, do you understand why that's signi?ant? Yup. In fact, that underscores the point I've been making all along--you can't in general split 7 as (in Nora's terms)> 7, 1 and expect to have a factorization in algebraic integers. RickRick, Harris is simply not going to understand this. He did not quiteget the point of your example in the ?st place. He is absolutelylocked into the idea that there is only one conceivable way tofactor such expressions. It is like talking to someone whohears some of what you say, but when certain words or phrasesare spoken he suddenly goes deaf. Some time ago both Dik and I created examples like yours except they were cubics, irreducible, but the roots had simple expressions. Harris understood for a while that they disproved his method. Thensomehow he rationalized them away. He may be so deeply into denialnow that he is unreachable. It is surprising that he replied to you at all. Nora B. === > x^2 + x = b_1(x)b_2(x) = b_1(x)(x - 2b_1(x))>so (b_1(x))^2 - xb_1(x) + (x^2 + x) = 0, which we>can solve for b_1(x):> b_1(x) = (x + sqrt(-7x^2 - 8x))/4>Nope. You dropped a 2, as you should have>2b_1(x)^2 - x b_1(x) + x^2 + x = 0.>Quite right, I dropped a coef?ient of 2. Everything else is>correct, however.> Yes, despite dropping the 2, you correctly solved for b_1(x), but the> point here is that now it's clear that in general b_1(x) is not an> algebraic integer function!!! That jumped out at me with that dropped 2 added back. Now then, do you understand why that's signi?ant? Yup. In fact, that underscores the point I've been making all along--you can't in general split 7 as (in Nora's terms)> 7, 1 and expect to have a factorization in algebraic integers.> Actually it makes my point, which I can outline quickly. Rememberyour example is(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x).The conventional thinking is that you can divide 7 from both sides andstill be in the ring of algebraic integers, but your own result Deckerdisproves that, as let's imagine algebraic integer functions w_1(x)and w_2(x), such thatw_1(x) w_2(x) = 7, and(5a_1(x)/w_1(x)+ 7/w_1(x))(5a_2(x)/w_2(x) +7/w_2(x)) = 25x^2 +30x +2.The assumption is that you're still in the ring of algebraic integerswith an algebraic integer x, so consider algebraic integer functionsf_1(x), and f_2(x), such thatf_1(x) f_2(x) = 25x^2 + 30x + 2,and f_1(x) = (5a_1(x)/w_1(x) + 7/w_1(x))andf_2(x) = (5a_2(x)/w_2(x) + 7/w_2(x)).Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2.Now then, replacing them withf_1(0) = b_1(x) + 1, and f_2(0) = b_2(x) + 2, I have(b_1(x) + 1)(b_2(x) + 2) = 25x^2 + 30x + 2which ties back in with your work Decker, as now we can take *your*result above which you acknowledged to get the result:2b_1(x)^2 - x b_1(x) + x^2 + x = 0,which is a non-monic, and not generally reducible over Q, with aninteger x, proving that b_1(x) is not in general an algebraic integer.That is, your own work proves *my* point.Do you understand?James Harris === ... > Yup. In fact, that underscores the point I've been making > all along--you can't in general split 7 as (in Nora's terms) > 7, 1 and expect to have a factorization in algebraic integers. > > Actually it makes my point, which I can outline quickly. Remember > your example isNo, James, it does not. > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where the a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > The conventional thinking is that you can divide 7 from both sides and > still be in the ring of algebraic integers, but your own result Decker > disproves that, as let's imagine algebraic integer functions w_1(x) > and w_2(x), such that > > w_1(x) w_2(x) = 7, and > (5a_1(x)/w_1(x)+ 7/w_1(x))(5a_2(x)/w_2(x) +7/w_2(x)) = 25x^2 +30x +2. > The assumption is that you're still in the ring of algebraic integers > with an algebraic integer x, so consider algebraic integer functions > f_1(x), and f_2(x), such that > f_1(x) f_2(x) = 25x^2 + 30x + 2, > and > f_1(x) = (5a_1(x)/w_1(x) + 7/w_1(x)) > and > f_2(x) = (5a_2(x)/w_2(x) + 7/w_2(x)). > Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2. > Now then, replacing them with > f_1(0) = b_1(x) + 1, and f_2(0) = b_2(x) + 2, I have > (b_1(x) + 1)(b_2(x) + 2) = 25x^2 + 30x + 2 > which ties back in with your work Decker, as now we can take *your*No, it does not tie back. Your b_1 and b_2 are other functions thanRick's b_1 and b_2. To be precise, your_b_1(x) = (5a_1(x)/w_1(x) + 7/w_1(x)) - 1 your_b_2(x) = (5a_2(x)/w_2(x) + 7/w_2(x)) - 2Rick's b_1(x) and b_2(x) are also solutions of the above functionalequation, but different, namely with the added requirement that theyare roots of b^2 + C(x).b.x + (x^2 + x), your b_1 and b_2 are *not*roots of this polynomial. Actually, your b_1 and b_2 are not rootsof *any* polynomial, unless the w's are very special, namelyw_1(x) = 7 and w_2(x) = 1, but that is something you have to prove. > result above which you acknowledged to get the result: > 2b_1(x)^2 - x b_1(x) + x^2 + x = 0,No. Your b_1 and b_2 just de?ed do *not* satisfy a polynomial. > which is a non-monic, and not generally reducible over Q, with an > integer x, proving that b_1(x) is not in general an algebraic integer. > That is, your own work proves *my* point.No. > Do you understand?Do you?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [snip]> Actually it makes my point, which I can outline quickly. Remember> your example is> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)> where the a's are roots of> a^2 - (x - 1)a + 7(x^2 + x).> The conventional thinking is that you can divide 7 from both sides and> still be in the ring of algebraic integers,No, it isn't. The product on the left may equal the algebraic integer on the right, but there's noreason to believe that division results in two algebraic integers on the left. On the contrary, thatis an *unexpected* result. If you stick with *integers* there will be a unique prime factorizationinvolved, but for algebraic integers or algebraic numbers it doesn't apply.Note that an algebraic integer may be factored as two non-algebraic integers or as one algebraicinteger and one non-algebraic integer.Consider: 2 * 3 = 6, 1/2 * 12 = 6, 17/13 * 78/17 = 6So why on earth would you *expect* the division of some product to result only in algebraic integerfactors?--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === >Dear all,>I am facing with this dif?ult problem. Please help me!>In this problem, I need to construct some matrices which satisfy the>following matrix equation:>(( A * A )V1 + ( B * B) V2) V = (D * D)>where * denotes Kronecker product, A, B, V1, V2, V are unknownmatrices>that needs solving; they are all square. Some structure needs to beimposed:>I have certain pattern for A and B; and V1, V2, and V are required to be>diagonal... Matrix D is given...>The task is to ?d the best approximation of the above-mentioned A, B,V1,>V2 and V... I have been thinking about this for long time... can anybody>give me some hints?>I guess it is dif?ult to ?d closed form analytical solution..> How dii?ult is it - it seems to me that the problem has an awful lot> of structure.>. How to>design iterative algorithm to let computer search for the answer? It is>really hard... please help me!> You are trying to solve the nonlinear system of equations> (( A * A )V1 + ( B * B) V2) V - (D * D) = 0> If the problem is not too large use Newton's method. If the problem> is too big for that look for a canned program that will solve> nonlinear systems without needing to generate and then solve the> jacobian.>Dear Joe,a) (( A * A )V1 + ( B * B) V2) V - (D * D) = 0These * are Kronecker products not the normal multiplier, how to designNewton's iteration algorithm for this equation?b) The worst thing is that this problem is not continuous; the elements in Vcan be arbitary; but the elements in A, B, V1, V2 need to be integer...hence I lost completely how to do iterative algorithm for these kind ofmix-integer-continuous problem...Could you please help me more?-Walalla <4de76b8.0312281437.5eadfee9@posting.google.com> <4de76b8.0312290119.45da2df9@posting.google.com> <20040101072404.G49311@agora.rdrop.com> <4de76b8.0401011321.73d2f58a@posting.google.com> <4de76b8.0401031036.7ffa639c@posting.google.com> === > I'm still puzzle showing the product topology {0,1}^S or actually the> power-set topology is the largest topology assuring the assigned limits.> You have shown that, in this topology, all the topological limits are> set-limits, no? Something to ponder: If you start throwing in more> open sets, do you lose some of these convergences?>Possibly, but when it's the largest assuring designated limits ofsequences, a ?er topologies will lose some of those convergences.> Another puzzle I'm musing upon is if a topology assuring the assigned> limits can be enlarged to a 1st countable topology while still assuring> the assigned limits. This I think likely not possible.> Do you really mean enlarge? The coinduced topology is already the> largest. Also, it seems to me that to make a space ?st countable,> you have to *remove* open sets, not add them, no? (Not sure about> that one.) For example, all sequences converge to every point in the> indiscrete topology, which is ?st countable. Thus, you could use as> a subbase the union of all ?st countable topologies where the> sequences converge to the designated limits. Trouble is, you get even> more convergent sequences than before.Is ?st countable a subbase property? No.A ?st countable subbase topology is stronger property than a?st countable base topology; the later being equivalent toa ?st countable topology.Thus, am I right there's no largest ?st countable topology?No, the discrete topology is the largest.The nascent feature being that both the indiscrete and the discretetopology are ?st countable and somewhere in between is the realm of thenot ?st countable. Thus to chose the smallest or the largest likelydepends where in the middle one starts.For example when I combined requirement for ?st countable with a secondproperty such as assuring designated limits, then it appears, tho thesecond (a subbase) property has a largest topology preserving itself, whenin combination with ?st countable, the sup of the topologies with bothproperties will preserve the second property but may not be ?stcountable. <4de76b8.0312281437.5eadfee9@posting.google.com> <4de76b8.0312290119.45da2df9@posting.google.com> <20040101072404.G49311@agora.rdrop.com> <4de76b8.0401011321.73d2f58a@posting.google.com> <4de76b8.0401031036.7ffa639c@posting.google.com> <20040104090005.Q77631@agora.rdrop.com> === > I'm still puzzle showing the product topology {0,1}^S or actually the> power-set topology is the largest topology assuring the assigned limits.> You have shown that, in this topology, all the topological limits are> set-limits, no? Something to ponder: If you start throwing in more> open sets, do you lose some of these convergences?> Possibly, but when it's the largest assuring designated limits of> sequences, ?er topologies will lose some of those convergences.> Another puzzle I'm musing upon is if a topology assuring the assigned> limits can be enlarged to a 1st countable topology while still assuring> the assigned limits. This I think likely not possible.> Do you really mean enlarge? The coinduced topology is already the> largest. Also, it seems to me that to make a space ?st countable,> you have to *remove* open sets, not add them, no? (Not sure about> that one.) For example, all sequences converge to every point in the> indiscrete topology, which is ?st countable. Thus, you could use as> a subbase the union of all ?st countable topologies where the> sequences converge to the designated limits. Trouble is, you get even> more convergent sequences than before.>Having personally met some of the trees in the forest below, I have foundthe forest isn't a ? forest but a pine forest.> Is ?st countable a subbase property? No.>Yes, if there's a countable bunch of open Uj nhoods of x, such thatfor every subbase U containing x, some Uj with Uj subset U thenx has a local base of ?ite intersections of Uj's> A ?st countable subbase topology is stronger property than aNo> ?st countable base topology; the later being equivalent to> a ?st countable topology.> Thus, am I right> there's no largest ?st countable topology?> No, the discrete topology is the largest.> The nascent feature being that both the indiscrete and the discrete> topology are ?st countable and somewhere in between is the realm of the> not ?st countable. Thus to chose the smallest or the largest likely> depends where in the middle one starts.> For example when I combined requirement for ?st countable with a second> property such as assuring designated limits, then it appears, tho the> second (a subbase) property has a largest topology preserving itself, when> in combination with ?st countable, the sup of the topologies with both> properties will preserve the second property but may not be ?st> countable.>The reason that ?st and second countable aren't preserved by arbitraryintersection or union of topologies is because of cardinality.Cardinality may not preserved upon intersections or unions.However the issue is even more, just the mention of a existence is enuf.For example there is no smallest topology with isolated points, thothere's many minimal topologies (infraspaces) with an isolated point.Conversely, there's no largest topology without isolated points, thothere are many maximal topologies without isolated points. Is not anyfree ultra?ter topology an example of such?So the preservation of a property by unions or intersections requires morethan being a subbase or base property. It involves some additionallogical structure perhaps akin the logic structure for varieties ofalgebraic structures. Alternatively a topological variety could bede?ed, like others do with algebraic varieties, as a property preservedby unions and intersections. === In the chees-board, we may considerer a coordinates system, such as{1,2,3,4,5,6,7,8}X{1,2,3,4,5,6,7,8}. So, a chess kniht in (1,1) -thewhite square of de left top of the ?st ?e- can go to (2,3) and(3,2). But we can consider an in?e board, that is, ZxZ. Then anatural cuestiopn is: Which is the minimal number of chees steps togo from (m,n) to (s,t) in this in?ite board? I have gotten anexpression for this, and my computer can use it, but is too large. Can === Jgissw2 scribbled the following:> Been a number of years since I took math - I was trying to remember how the> notation, O(p^n), and o(p^-n), if that is correct, ran. That is, suppose you> have a series of powers in, say, which are at least as great as n, or smaller> than -n. How does that go?> John GWAFAIK it is something like:O(something) means smaller than or equal to somethingo(something) means smaller than somethingTheta(something) means equal to somethingOmega(something) and omega(something) mean greater than or equal tosomething, and greater than something, but I forget which is which.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/ === > AFAIK it is something like:> O(something) means smaller than or equal to something> o(something) means smaller than something> Theta(something) means equal to something> Omega(something) and omega(something) mean greater than or equal to> something, and greater than something, but I forget which is which.I hope the original poster doesn't take this response too seriously,as it is fairly inaccurate, egf(n) = O(p^n) means f(n)/p^n is bounded, at least for n >= Nf(n) = o(p^n) means f(n)/p^n -> 0 as n -> infty-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === >Yes, it is continuous. Call it f. It is easy to see that, given x and y>in the Cantor set, |f(x) - f(y)| <= |x - y|. > Whoops, |f(1/3) - f(0)| = |1/2 - 0| = 1/2. (Also note that any f satisfying > |f(x) - f(y)| <= |x - y| takes sets of measure 0 to sets of measure 0.)Quite right. My mistake.Jose Carlos Santos === There exists a bijection from the Cantor set (middle thirds removed) to> [0,1]. (Ternary rep - 0's and 2's -> Binary rep 0's and 1's) Is this mapping continuous? It may be but I'm having trouble seeing why> this is so. Yes, it is continuous. Call it f. It is easy to see that, given x and y> in the Cantor set, |f(x) - f(y)| <= |x - y|. Whoops, |f(1/3) - f(0)| = |1/2 - 0| = 1/2. (Also note that any f satisfying > |f(x) - f(y)| <= |x - y| takes sets of measure 0 to sets of measure 0.) To see f is continuous, note that for x and y in the Cantor set, |x - y| < > 1/3^n => the ternary expansions of x and y agree in the ?st n digits => |f(x) - f(y)| <= 1/2^n.But it's not a bijection. f(1/3)=1/2=f(2/3) for example.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ or signal local activity? === >In one of my program, I need to compute>the variance of a 8x8 data matrix as>fast as possible... > What does it means ?> You just have 16 samples of a signal and you want the variance ? For me it is not clear why deviation from> the mean is a good indicator of activity. Obviously, there is no fast algorithm for computing a sum> of 16 terms ! BUT if you compute the variance on a moving windows, there> is a very fast algorithm, that you can ?ure out pretty easily. Your just have to perform :> 1. compute mean m and replace your image I by M=(I-m)^2> 2. compute moving average along X direction. This can be done> by a recursive algorithm, because if S(k) is the moving average> and F(k) is the initial function (e.g. a row of M), you have> S(k) = S(k-1)+F(k)-F(k-9) (here I assume window of size 8)> 3. compute moving average along Y direction.The computation I isolated above can have long-term drift away from the rue value when performed in ?g point. Think it through carefully.> All this stuff needs about O(n) operations where n is the number of pixels> in the original image and it's independant of the size of the window. GabrielJerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash] === [...]|Ax Ay( a e b <-> (|| Az( y c z -> x e z )| /| Ez( x e z / ~(y c z) ) ) )Do you mean x e y instead of a e b in the ?st line?[...]|Well, it is not a great theory. But, it does have some relation to|apartness.I don't have much to say about it, partly because I don't seewhat the motivation is. Is it supposed to be some kind ofset theory? Inventing your own home-brew set theory is ?e,but without anything more speci? you intend to accomplishwith it, it's not liable to go anywhere.Keith Ramsay X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In , on 01/03/2004 at 01:44 AM, Russell Easterly said:>I even give such a mapping for the set [0,1).No.In <9M-dnfIn3_DmB2uiRVn-gw@comcast.com>, on 01/03/2004 at 02:33 AM, Russell Easterly said:>I am de?ing a method to compute x.No. You've de?ed a method to compute a sequence of rational numbers,not a method to compute a speci? rational number.>Is this really an open set?Is what? Certainly {x: xS is supposed to contain ALL of the rational approximations of e-2.What is a rational approximation to e-2?>Isn't this the same thing as saying S contains e-2?No. Does {1/i:1=1 to oo} contain 0? No, but it contains numbersarbitrarily close.>I de?e a method to calculate the largest rational approximation of>e-2 in S:No. There is no such element.>For i=0 to ?:What is ?? If it's ?ite, then the method is wrong, and if it'sin?ite then the method is meaningless.>Clearly, if I can examine every member of S then I can compute x.FSVO Clearly. That may be clear to you, but it's wrong.>One of my three assumptions must be false.Number 3 is meaningless, but can be reworded as 3) S contains every rational number of the form .111...1 in base!, and e-2 is a limit point of that subset.The problem is not with your assumptions. The problem is that yourconclusion does not follow from your assumptions. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === >Well, it's not a proof since it contains the errors that others have>pointed out. But I do wish people would stop saying things like you>can't prove false things or you shouldn't attempt to. I'm not very sure>that mathematics is consistent. So, if there is a proof of statement>S and someone comes along with a proof of the negation of S and both>proofs are valid, then mathematics is doomed. Doomed? I would think ?ding two such proofs would be the begining of a most> enlightening period for mathematics. Unless, of course, *every* valid proof of> S could somehow be turned into a valid proof of ~S. A small dose of> inconsistency could be a very good thing. richIf we take some set theory as the foundation of mathematics (much more than 99% of current mathematics can be formulated in ZFC for example), thenthe existence of correct proofs of some statement S and its negation immediately implies that ALL statements formulated in the system areprovable. In that sense, there is no such thing as a small dose o?consistency. To reiterate:Read an introductory mathematical logic text, and you willsee that if a theory is inconsistent, then that theory contains ALLstatements in the language of that theory, i.e., if one contradictioncan be found then ALL statements are provable. So, if Russell's proofsare correct, then I can prove 1 = 2 and any other statement you coulddream of. Sounds like doom to me. But of course, we have already shown that Russell is wrong. === > If we take some set theory as the foundation of mathematics (much more than > 99% of current mathematics can be formulated in ZFC for example), then> the existence of correct proofs of some statement S and its negation > immediately implies that ALL statements formulated in the system are> provable. In that sense, there is no such thing as a small dose of> inconsistency. I don't see that an inconsistency in ZFC would necessarily yield aninconsistency in, say, Peano arithmetic. After all, the axioms of PAare independent of ZFC, aren't they? They can be interpreted in ZFC,but that doesn't mean that they inherit any of ZFC's inconsistencies,any more than PA somehow inherits ZFC's trans?ite hierarchy.> To reiterate:> Read an introductory mathematical logic text, and you will> see that if a theory is inconsistent, then that theory contains ALL> statements in the language of that theory, i.e., if one contradiction> can be found then ALL statements are provable. So that would mean big trouble for ZFC, no doubt. But it does notnecessarily spell big trouble for all of mathematics (although, ofcourse, it might). Theories which are interpretable in ZFC don'tnecessarily inherit ZFC's inconsistencies, near as I can ?ger.> So, if Russell's proofs are correct, then I can prove 1 = 2 and any> other statement you could dream of. Sounds like doom to me.Which of Russell's proofs?> But of course, we have already shown that Russell is wrong.Pardon?-- Jesse HughesBy de?ition m is a variable. By de?ition all then (sic) numbersrepresented by letters are variables--that's algebra[,] Magidin. -- James Harris shows deep understanding of algebra === > ... So, if there is a proof of statement S and someone> comes along with a proof of the negation of S and both> proofs are valid, then mathematics is doomed. > Doomed? I would think ?ding two such proofs would be the begining of a > most enlightening period for mathematics. Unless, of course, *every* valid> proof of S could somehow be turned into a valid proof of ~S. A small dose> of inconsistency could be a very good thing. > rich If we take some set theory as the foundation of mathematics (much more than > 99% of current mathematics can be formulated in ZFC for example), then> the existence of correct proofs of some statement S and its negation > immediately implies that ALL statements formulated in the system are> provable. In that sense, there is no such thing as a small dose of> inconsistency. Depends on your point of view. Suppose the shortest formal proofs of anytwo contradictory statements are very long. There might then still be alot of useful mathematics left -- provided we use a notion of boundedprovability. Whether such a theory can be worked so the close dependencyon details of the formalism can be abstracted (as the details of computingmachinery can be abstracted from the class of polynomial-time complexity ofrecursive functions), I don't know.Another thing that could happen is that we ?d out that one of the axiomsdoesn't really capture what we wanted to capture, and needs to be re?ed.This is what happened a century ago, when (unbounded) comprehension wasreplaced with separation, and other axioms were introduced to recapturemost of what would have been lost. I think I'm not alone in suspectingthat the Powerset axiom may be the most vulnerable in this regard. It isin this sense that discovery of an inconsistency would rejuvenate maths.Michel. === Well, it's not a proof since it contains the errors that others have>pointed out. But I do wish people would stop saying things like you>can't prove false things or you shouldn't attempt to. I'm not very sure>that mathematics is consistent. So, if there is a proof of statement>S and someone comes along with a proof of the negation of S and both>proofs are valid, then mathematics is doomed. > Doomed? I would think ?ding two such proofs would be the begining of a>most> enlightening period for mathematics. Unless, of course, *every* valid>proof of> S could somehow be turned into a valid proof of ~S. A small dose of> inconsistency could be a very good thing. > rich>If we take some set theory as the foundation of mathematics (much more than >99% of current mathematics can be formulated in ZFC for example), then>the existence of correct proofs of some statement S and its negation >immediately implies that ALL statements formulated in the system are>provable. In that sense, there is no such thing as a small dose of>inconsistency. >To reiterate:>Read an introductory mathematical logic text, and you will>see that if a theory is inconsistent, then that theory contains ALL>statements in the language of that theory, i.e., if one contradiction>can be found then ALL statements are provable. So, if Russell's proofs>are correct, then I can prove 1 = 2 and any other statement you could>dream of. Sounds like doom to me. Sure it is bad news for, say, ZFC. But is mathematics in its entirety doomed? There are *no* alternatives to ZFC within which 95% of current mathematicscouldn't be re-formulated (without the inconsistency)? I don't know. My guessis that mathematicians, being the very clever folk they are, would ?d somesystem that avoided the inconsistency. And rather quickly. Like I statedinitially, it would be a very enlightening period. Something to look forwardto even (unless your work is likely to be part of the 4% that gets lost!). A *little* inconsistency would be a good thing for mathematics, just as it isgood for all the other sciences. ;-) rich === > Rationals are Uncountable> Let S be the set of all rational numbers [0,1).> s is a member of S if 0.000... <= s < 1.000...> and s is rational.> Assume s is represented in base factorial (!).> Base ! is used because every rational number> has a ?ite representation in base !.> In base ! the allowable digits for> position k are (0,1,...,k).> (k starts at 1)> Every position, k, represents 1 / (k+1)!.> k> 1 1/2! = 1/2> 2 1/3! = 1/6> 3 1/4! = 1/24> .123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10).> Every rational number has an unique ?ite base !> representation. Any ?ite base ! number is rational.> We can create the set S de?ed above by taking the> set produced by counting in base !.> .0> .1> .01> .11> .02> .12> .001> .101> ...> There exists a rational number, x, not in S.> Asserted without basis.I give the following proof:> If S(i) is of the form .111...1 and> its length is equal to or greater than x> then set x to a string of 1's one longer than S(i).> i = ....So?I am assuming that every member of Scan be examined. Is there some reasonI shouldn't make this assumption?> x differs from every member of S.> x is a rational number because it has a ?ite number> of digits. The length of x is exactly one greater> than some member of S.> Lots of members of S have a length exactly one greater than some (other)> member of S. In fact, all but the ?st two members of your list have> such a length.> 0.0 <= x < 1> x = 1/2! + 1/3! + ... + 1/k!> Then x's position on your list is (k-1)! + (k-2)! + ... + 2! + 2 (and> obviously does not differ from that member of S).x is not in S so the position of S(i)=x is unde?ed.> and equals the largest rational number> less than the fractional part of e.> No such number exists. You might as well talk about the smallest> rational number greater than 0.You must mean the smallest rational number not in S.Easy enough to do.This proof can be easily converted to ?dthe smallest positive rational number not in S.I make three assumptions:1) Every member of S can be examined.2) I can determine if S(i) is less than, equal to, or greater than S(j)3) S contains every rational number in [0,1)If you think the proof is wrong please point outwhich of my assumptions is false.Simply stating the proof is wrong is not very convincing.Russell- 2 many 2 count === > No such number exists. You might as well talk about the smallest> rational number greater than 0. You must mean the smallest rational number not in S.> Easy enough to do.> This proof can be easily converted to ?d> the smallest positive rational number not in S.He means just what he says. There is no such thing as a smallest positive rational over all rationals since half of any positive rational is yet a smaller rational, but there is a smallest positive rational not in [0,1), namely 1.So the cases are quite different! === > If S(i) is of the form .111...1 and> its length is equal to or greater than x> then set x to a string of 1's one longer than S(i).> i = .... So?What, pray tell, is i supposed to be in the quoted sentence?> I am assuming that every member of S> can be examined. Is there some reason> I shouldn't make this assumption?Besides the fact that you have not de?ed examined?> 0.0 <= x < 1> x = 1/2! + 1/3! + ... + 1/k!> Then x's position on your list is (k-1)! + (k-2)! + ... + 2! + 2 (and> obviously does not differ from that member of S). x is not in S so the position of S(i)=x is unde?ed.True or false:S((k-1)! + (k-2)! + ... + 2! + 2) = 1/2! + 1/3! + ... + 1/k! > I make three assumptions: 1) Every member of S can be examined.> 2) I can determine if S(i) is less than, equal to, or greater than S(j)> 3) S contains every rational number in [0,1) If you think the proof is wrong please point out> which of my assumptions is false.x is not in S-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W === > I make three assumptions:> 1) Every member of S can be examined.> 2) I can determine if S(i) is less than, equal to, or greater than S(j)> 3) S contains every rational number in [0,1)> If you think the proof is wrong please point out> which of my assumptions is false.> x is not in SA truly stunning refutation.Russell- 2 many 2 count === >Let me make three assumptions:>1) I can examine every member, S(i), of S.> De?e examine. S is an in?ite set.>2) I can determine if S(i) is less than, equal to, or greater thanS(j).> Yes. The rationals in [0, 1) are well-ordered.> See Dave Seaman's reply.> I am not requiring S to be well-ordered.I mis-spoke.Assumption (1) requires S to be well ordered.I can not examine every member of S unless I give a well ordering.I DO give such an ordering.>3) S contains every rational approximation of e-2 (of form .111...1 in>base!)Russell- 2 many 2 count === >Let me make three assumptions:>1) I can examine every member, S(i), of S.> De?e examine. S is an in?ite set.>2) I can determine if S(i) is less than, equal to, or greater than> S(j).> Yes. The rationals in [0, 1) are well-ordered.> See Dave Seaman's reply.> I am not requiring S to be well-ordered. I mis-spoke.> Assumption (1) requires S to be well ordered.> I can not examine every member of S unless I give a well ordering.> I DO give such an ordering.Your well-ordering is an ill ordering, since it is incompatible with the standard ordering and with your own factorial-base representation. > Russell = 2 little 2 count === I collect pretty calendars and knew that one year I could revisit anold calendar and hang it up to use for that new year. I had to wait 6years because now I am re-using a 1998 calendar for year 2004. So thatI can basicly recycle my old calendars and have a stock of 6 calendarsto cover every future year.I collect calendars that have a botany interest. The 1998 calendar isfrom Sierra Club showing for January a Bloodroot from Illinois. It isa beautiful ?of luminescent white and in the middle a yellowthat is almost gold.What I need for my collection of calendars are SierraClub typepictures of pretty and outstanding trees.I have crop patterns calendar but I need calendars showing trees.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === > I collect pretty calendars and knew that one year I could revisit an> old calendar and hang it up to use for that new year. I had to wait 6> years because now I am re-using a 1998 calendar for year 2004. So that> I can basicly recycle my old calendars and have a stock of 6 calendars> to cover every future year.... You need 14, not 6. A year may begin on any day of the week (7 possibilities), and it may be either a leap year or not (2 possibilities), giving 14 possible year-patterns altogether. This ignores holidays such as Easter. The cycle of recurring year-patterns goes 6, 11, 11, 28.What I mean, for example, is thatyour 1993 calendar could be re-used in 1993 + 6 = 1999,your 1994 calendar could be re-used in 1994 + 11 = 2005,your 1995 calendar could be re-used in 1995 + 11 = 2006,your 1996 calendar could be re-used in 1996 + 28 = 2024. Like you, I keep old calendars. This 6, 11, 11, 28 rule shows when I can re-use the nicer ones. Since 1998 was two years after a leap year, your 1998 calendar can't actually be re-used this year, but rather in 1998 + 11 = 2009. Ken Pledger. === I collect pretty calendars and knew that one year I could revisit an> old calendar and hang it up to use for that new year. I had to wait 6> years because now I am re-using a 1998 calendar for year 2004. So that> I can basicly recycle my old calendars and have a stock of 6 calendars> to cover every future year.... You need 14, not 6. A year may begin on any day of the week (7 > possibilities), and it may be either a leap year or not (2 > possibilities), giving 14 possible year-patterns altogether. This > ignores holidays such as Easter.someone knows of calendars with tree pictures, not for themathematics.But since you layed the issue out mathematically, may as well partake.Granted it is 7 possible days for which January 1 can fall on. So Ineed 7 calendars in all for all possible future years, not 14.I do not accept the 2 possibilities of the 29Feb for leap years tomake 7 X 2 = 14. I consider Leap years as gimmick years for which the7 possible 1January can cover. I mean by this, that having bought 7calendars each starting with one of the 7 possible January 1st andwhich all 7 are nonleap years. Now, on all 7 calendars I write theleap years on the backside covering the total years which I expect tobe alive so that I do not forget, or just writing a few years of thesequence such as 2000.... 2004 ..... 2008 ....So, after each year I look at 31Dec and in my other 6 calendars I ?hout the one that harmonizes 31Dec with the correct 1January. If thatyear happens to be a leap year containing a 29Feb which none of my 7calendars possess, then I look at my backpage footnote to remember i? is a leap year and then in amongst the other 6 calendars in mypossession I write a post-it rememberance note to make another switchof calendars.So what I am saying is the total number of calendars needed to coverevery future year is 7, not 14, because on leap years two joints haveto be conformed in the Dec31 and 1Jan, but also Feb29 and 1Mar.But I am not certain of that method, Ken. I am not certain that if Iwere to collect 7 calendars all having nonleap years and all having adifferent day of the week for 1Jan, whether those 7 will cover allbases for a 29Feb and 1Marswitch.Would you know Ken, whether 7 calendars is the Minimum number ofcalendars to cover all future years, provided those 7 have a doubleswitch in leap years? You see, I am certain that 14 will cover allbases, but will 7 cover all bases if I make a double switch in leapyears. The cycle of recurring year-patterns goes 6, 11, 11, 28.> What I mean, for example, is that> your 1993 calendar could be re-used in 1993 + 6 = 1999,> your 1994 calendar could be re-used in 1994 + 11 = 2005,> your 1995 calendar could be re-used in 1995 + 11 = 2006,> your 1996 calendar could be re-used in 1996 + 28 = 2024. Like you, I keep old calendars. This 6, 11, 11, 28 rule shows > when I can re-use the nicer ones. Since 1998 was two years after a leap > year, your 1998 calendar can't actually be re-used this year, but rather > in 1998 + 11 = 2009. Ken Pledger.Ken, I was wondering if we can redesign the calendar such that thewinter months do not fall between two years. Where Jan 1 is autumnalequinox to relieve the confusion when saying winter of 2006 whetheryou mean december of 2005 or January of 2006.If we made the Autumnal equinox or if we made say 14Dec or 7Novemberas 1 January, by that way, we can have all the four seasons into oneyear without winter divided between two different years. And when wemake such a switch wehave all the months have either 30 or 31 days, including February. Anda leap year in such a revision would have December, the last monthhave either 30 or 31 days depending on whether it is leap year or not.The purpose for the change is that it is best to have all the seasonsrepresented in one year and not to have winter divided between twoyears.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Please do not use logic when responding to apost by Archimedes Plutonium. After all, hisname is an anagram of I pound his malerectum and Penis could mature him. === > You need 14, not 6. A year may begin on any day of the week (7 > possibilities), and it may be either a leap year or not (2 > possibilities), giving 14 possible year-patterns altogether. This > ignores holidays such as Easter.If you include Easter, you need about 70 calendars: about 35 possibleEaster dates times leap/non-leap years. That should also cover mostWestern holidays which have ?ed dates (New Year), or ?ed weekdayEaster (Mardi Gras).A real challenge would be to ?d out how many calendars you need when you want to cover the Jewish holidays as well: they are not directly related to the Western calendar but share some features: for instance, Easter is near or after a full moon, thus near or after the 15th of a Jewish month. And the weekdays are the same with not every weekday possible for a given Jewish date. So there are many restrictionson the possible offsets between the calendars.Helmut Richter === > I collect pretty calendars and knew that one year I could revisit an> old calendar and hang it up to use for that new year. I had to wait 6> years because now I am re-using a 1998 calendar for year 2004. So that> I can basicly recycle my old calendars and have a stock of 6 calendars> to cover every future year.> I collect calendars that have a botany interest. The 1998 calendar is> from Sierra Club showing for January a Bloodroot from Illinois. It is> a beautiful ?of luminescent white and in the middle a yellow> that is almost gold.> What I need for my collection of calendars are SierraClub type> pictures of pretty and outstanding trees.> I have crop patterns calendar but I need calendars showing trees.> Archimedes Plutonium> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxiesSince 1998 was not a leap year but 2004 was one, the two calendars cannot bein sync throughout the year. January and February match but after thatthey'll be one day out. You'll need a 1999 calendar for the rest of theyear.Of course, if you've got a 1976 calendar there....Naturally the dates of Easter and the other moveable feasts won't match--Paul V. S. TownsendInterchange the alphabetic elements to reply === >SNIP<1. Stop snipping the parts of the posts you don't agree with or don'twant others to see in your reply without indicating you have done so.This is dishonest.2. The dictionary is not the ?al say in de?ing everything there isto be de?ed. Several scholars put together a consensus on what theybelieve to be the best de?ition.3. The parts of my post you deleted show that I am most certainly amnot a bigot.4. Everyone has the right to live their life as they see ? as longas they do not initiate force, coercion, or fraud against any personor his property. This is known as a PRINCIPLE that I live by.5. I won't be responding to any further posts by you because you areobviously a schmuck who thinks that because he has been brainwashed ina statist educational institution he is somehow intelligent. PLONK! === It is not known if there exists a point in a unit square all of whosedistances from the corners are rational...It seems like such a simple problem, until one considers it.Has there been any progress on this?Barnaby === It is not known if there exists a point in a unit square all of whose> distances from the corners are rational... It seems like such a simple problem, until one considers it.An obvious start is to draw a pair of perpendicular lines through thesupposed point and parallel to the sides of the square.Then, if these lines divides the square sides into segments of lengthx, 1 - x, y, 1 - y, the problem amounts to four Pythagoras triangleconditions.A priori (?up front') there's no reason why x or y should be rational.But subtracting the x^2 + y^2 = a^2 condition from (1-x)^2 + y^2 = b^2and x^2 + (1-y)^2 = b^2, where a, b, c are rational, gives x and y asrational functions of a, b, c. So in fact x and y are rational.But at this point your problems are just beginning ... ;-)John Ramsden === Guy and Bremner did papers on this a while back[1,2,3], but you might like tocontact Guy directly on any addendums to D19 in UPINT since this problem is insection D of his book[4] and is considered the reference for these types ofproblems.[1] Guy, Richard K. Tiling the square with rational triangles. Number theory andapplications (Banff, AB, 1988), 45--101, NATO Adv. Sci. Inst. Ser. C Math.Phys. Sci., 265, Kluwer Acad. Publ., Dordrecht, 1989. MR1123070 (92f:11044)[2] See also Bremner, Guy - Supplement to Nu-con?urations in tiling the195-202[3] The Delta-Lambda Con?uration in Tiling the Square,Bremner, Guy, Journalof Number Theory, 1989 vol 32, #3 pp263-280 MR1006593 (90g:11031)[4] Unsolved Problems in Number Theory, ed 2, Springer-Verlag, NY 1994 Hardcover. 285 pages. ISBN 0-387-94289-0 (Amazon has a few copies left)Hopefully edition 3 is just around the corner. === Barnaby Finch scribbled the following:> It is not known if there exists a point in a unit square all of whose> distances from the corners are rational...> It seems like such a simple problem, until one considers it.> Has there been any progress on this?My ?st thought was of course there exists, pretty much any pointwhose one coordinate was rational and the other either 0 or 1 woulddo. But that would only count for three corners, not four. So it'ssneakier than I thought!-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/O pointy birds, O pointy-pointy. Anoint my head, anointy-nointy. - Dr. Michael Hfuhruhurr === > My ?st thought was of course there exists, pretty much any point> whose one coordinate was rational and the other either 0 or 1 would> do. But that would only count for three corners, not four. So it's> sneakier than I thought!Mathworld has:M.Guy found an in?ite numbers of solutions to the problem of three suchdistances being integers, which involves solving (s^2+b^2-a^2)^2+(s^2+b^2-c^2)^2=(2bs)^2where a, b, and c are the three distances and s is the side length of thesquare (Guy 1994, p.181).anonymous said:Guy and Bremner did papers on this a while back... Barnaby === >How about the Chinese Remainder Theorem. At least the name is>somewhat suggestive, but why can't we just call it the remainder>theorem. What makes it inherently Chinese.A special case of it was known from Chinese writings.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === >This is a problem that should be very easy to get rid off, yet it's tormenting me.>(warning: sorry for my broken english, I am from Italy)>suppose you've got a list of integers>N1, N2, N3, N4, N5... Nm>sorted from the smallest to the biggest one. The distribution is absolutely>random, so that the list could be>(1,2,2,3,3,4,5,5,5,6,7)>but also>(5,26,6246,20123400)>( 3,3,27235235,12347472479679234592646)>etc.>starting from such a list, I want to multiply its elements until I obtain another>list made up just of two integers, whose difference must be as little as possible.>Let me explain: for example>(2,2,2,3,3,5) may be grouped as (2*2*2*3,3*5) = (18,15) 24>are there known algorithms or this proceeding? Anything will work, heuristic>or deterministic... speed isn't important.One can compute all possible combinations. It can be speeded up by stopping the computation when the productexceeds the square root of the product of all. I believe the problem is NP, and may even be NP complete.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === >You may have trouble getting into a top ranked graduate program if>your background is de?ient. However, what's far more important is>what you do in the graduate school you do get into. Many will allow>you to take a few undergraduate courses to ?m up your background. > As this problem is rife, that is an understatement. They> often also have semi-remedial graduate courses, but there> is too much of a tendency to downplay the weakness of the> student's background.>Have I misunderstood? It sounds like you are saying that when a>program lets you take a few undergrad courses to ?m up, that's>downplaying the weakness of one's background? That is NOT what I said. They will often put you into aremedial course for graduate students which assumes toomuch, or a too advanced undergraduate course.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === >On 2 Jan 2004 14:00:00 -0500, hrubin@odds.stat.purdue.edu (Herman>Watch out! Many schools with MS programs have essentially>computational programs, and what you need for a PhD program>is the ability to do abstract mathematics and proofs.>I presume by computational you mean non-rigorous cookbook>mathematics, not actual scienti? computation and assorted numerical>methods.Most numerical methods course are of the non-rigorous type.Many years ago, I asked a strong numerical analyst if therewas a beginning numerical analysis text for those who hadthe requisite theoretical real analysis, complex analysis,and linear algebra. The answer then, and now, is no. Allthe beginning texts are at least to a large part cookbook.The beginning courses all do not require this, and one cannot give a sound introduction without at least realanalysis and linear algebra.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === On 4 Jan 2004 16:32:42 -0500, hrubin@odds.stat.purdue.edu (Herman>On 2 Jan 2004 14:00:00 -0500, hrubin@odds.stat.purdue.edu (Herman>Watch out! Many schools with MS programs have essentially>computational programs, and what you need for a PhD program>is the ability to do abstract mathematics and proofs.>I presume by computational you mean non-rigorous cookbook>mathematics, not actual scienti? computation and assorted numerical>methods.>Most numerical methods course are of the non-rigorous type.Blanket statement.>Many years ago, I asked a strong numerical analyst if there>was a beginning numerical analysis text for those who had>the requisite theoretical real analysis, complex analysis,>and linear algebra. The answer then, and now, is no. All>the beginning texts are at least to a large part cookbook.May be. The math department at my school (which has numerical analysisas one of its specialities) seems to prefer using texts written by theinstructors themselves, so I assume the numerical analysis courses arealso taught like this. No cookbooks or scribbled 20-page lea?buthundreds of pages of well-edited material including rigorous proofs ofalmost all results.>The beginning courses all do not require this, and one >cannot give a sound introduction without at least real>analysis and linear algebra.This sentence escapes me. They do not require theoretical realanalysis, complex analysis and linear algebra as prerequisites? Howcould such a course be anything but cookbook? Why aren't the numericalanalysis basics covered along with the more theoretic material duringthe rigorous introductory courses? === >On 4 Jan 2004 16:32:42 -0500, hrubin@odds.stat.purdue.edu (Herman>On 2 Jan 2004 14:00:00 -0500, hrubin@odds.stat.purdue.edu (Herman>Watch out! Many schools with MS programs have essentially>computational programs, and what you need for a PhD program>is the ability to do abstract mathematics and proofs.>I presume by computational you mean non-rigorous cookbook>mathematics, not actual scienti? computation and assorted numerical>methods.>Most numerical methods course are of the non-rigorous type.>Blanket statement.>Many years ago, I asked a strong numerical analyst if there>was a beginning numerical analysis text for those who had>the requisite theoretical real analysis, complex analysis,>and linear algebra. The answer then, and now, is no. All>the beginning texts are at least to a large part cookbook.>May be. The math department at my school (which has numerical analysis>as one of its specialities) seems to prefer using texts written by the>instructors themselves, so I assume the numerical analysis courses are>also taught like this. No cookbooks or scribbled 20-page lea?but>hundreds of pages of well-edited material including rigorous proofs of>almost all results.One can have rigorous proofs in cookbooks; it does nothelp that much. Also, using textbooks written by thoseteaching the course is generally not a good idea, unlessit is necessary. I was told that at a particular majoruniversity, to use a textbook written by one of the faculty requires special approval. I know that this isnot generally the case.An instructor should NOT follow a textbook; one of myformer colleagues told me that one of his instructors, whodid write a good textbook, told his classes that he would>The beginning courses all do not require this, and one >cannot give a sound introduction without at least real>analysis and linear algebra.>This sentence escapes me. They do not require theoretical real>analysis, complex analysis and linear algebra as prerequisites? How>could such a course be anything but cookbook? Why aren't the numerical>analysis basics covered along with the more theoretic material during>the rigorous introductory courses?All the beginning courses descriptions I have seen do nothave as prerequisites those theoretical courses. Mostuniversities have more than one cookbook linear algebracourse, with recipes and computational methods piled up inthem, and students coming out of them essentially need tostart all over. Computational courses do not provide abasis for theory; it has to essentially be unlearned andall except facts learned all over.We should teach abstract to concrete, instead of the wayit is being done now. We USED to do this in geometry, butfew places have decent Euclid courses available now.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === On 5 Jan 2004 11:06:42 -0500, hrubin@odds.stat.purdue.edu (Herman>One can have rigorous proofs in cookbooks; it does not>help that much.Well obviously; if the students don't understand the methods used orthe concepts behind the proofs, nor practice writing their own proofs.Any textbook is as good as a cookbook if one simply browses throughthe proofs and skips all the excercises.>Also, using textbooks written by those>teaching the course is generally not a good idea, unless>it is necessary. I was told that at a particular major>university, to use a textbook written by one of the >faculty requires special approval. I know that this is>not generally the case.I suspect this has something to do with the instructors getting paidfor selling textbooks, which is not a problem for us since thematerial is distributed at cost for each course. Of course often thereare additional textbook recommendations which can be used tosupplement the primary material.>An instructor should NOT follow a textbook; one of my>former colleagues told me that one of his instructors, who>did write a good textbook, told his classes that he wouldI agree in the sense that examples and results should also come fromoutside the ones in the text(s) used as material. === Leroy Quet> ....> And A(r) is, if I am right, ...> oo> --- (1+1/2+1/3+...+1/k)> -------------------> / k^(r+1)> ---> k=1> = sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^(1+r),> which is an Euler sum.> [ http://mathworld.wolfram.com/EulerSum.html ]> I wonder naively> if there is some kind of zeta-function-like re?n formula for the> Euler-sum analytical continuation which relates> E(r) and E(2-r),> where E(r) is> sum{k=1 to inf} (1+1/2+1/3+...+1/k)/ k^r.> (I know of no study regarding analytical continuating Euler sums at> all, nor anything beyond the consideration of such at r = integers >=> 2.)I looks to me as if these Euler sums just haven't been studied very much.You might try this question on the pros over at sci.math.research. Theremight be some vast literature to which most of us are quite oblivious :)LH === > Wheeler is correct, Einstein's GR is battle-tested both> philosophically and experimentally.> So is orthodox quantum theory.Newtonian mechanics was battle-tested in its day. Only when instrumentsbecame more accurate and tiny discrepancies showed up did it have to bere?ed, ?stly by SR then by GR.Who's to say that GR and QM won't need similar re?ement by whatever highertheory (string theory? superstring theory?) marries them together.Meanwhile, calculations made according to present-day GR and QM are goodenough. So there's no need to knock them off their separate pedestals,rather seek to build a pedestal big enough for two.--Paul V. S. TownsendInterchange the alphabetic elements to reply === In our ordinary real number system, we say that the number K withdecimal expansion .99999... is the samas 1. An informal argument forthis is sketched below: 10K = 9.999...- K = .9999...____________ 9K = 9 K = 1But maybe this argument is misleading. What if there is some number,call it 1 - 1/omega, that is greater than any ?ite string .9...9 ofnines, yet less than 1? If K were actually equal to 1 - 1/omega, theinformal argument used in the last paragraph would not work, for thisargument overlooks the fact that the difference between 10K and 10 isten times as great as the difference between K and 1. There is aresidual in?itesimal quantity below that does not get canceled out: 10K = 10 - 10/omega- K = 1 - 1/omega_________________ 9K = 9 - 9/omega K = 1 - 1/omegaIntuitively, nothing could be more natural than to go ahead and talkabout 1/omega, 1/Aleph-1, and so on. Just as we move from the naturalnumbers to the fractions and then on to the reals, should we not beable to move from the whole ordinal numbers to some richer number?ld?Curiously, Cantor himself was very much opposed to this step. When afellow mathematician attempted to use Cantors trans?ite numbers todevelop a theory of in?itely small quantities, Cantor accused him oftrying to ?infect mathematics with the Cholera-Bacillus o??itesimals'. Cantor even constructed a proof that no number can bein?itesimal. This proof, however, is just as circular and worthlessas ?itist attempts to prove that no number can be in?ite. In bothcases, the desired conclusion is smuggled in as part of the de?itionof ?number'.Why was Cantor so vehemently opposed to in?itesimals? In hisvaluable essay, ?The Metaphysics of the Calculus', Abraham Robinsonsuggests that Cantor already had enough problems trying to defendtrans?ite numbers. It seems likely that, consciously or otherwise,Cantor deemed it politically wise to go along with othodoxmathematicians on the question of in?itesimals. Cantors stancemight be compared to that of a pro-marijuana Congressional candidatewho advocates harsh penalties for the sale or use of heroin. Yet, aswe shall see, there is almost as much justi?ation for in?itesimalsas there is for Cantors trans?ite ordinals.Formally speaking, it is as consistent to say that there is a numberbetween all of .9, .99, .999, ... and 1 as it is to say that there isa number greater than all of 1, 2, 3, ... . And just as we go on to?d more and more ordinals piled atop one another, we can go on to?d more and more in?itesimals squeezed beneath each other.[...]But so great is the average persons fear of in?ity that to this daycalculus all over the world is being taught as a study of *limitprocesses* instead of what it really is: *in?itesimal analysis*.As someone who has spent a good portion of his adult life teachingcalculus courses for a living, I can tell you how weary one gets oftrying to explain the complex and ?dling theory of limits to waveafter wave of uncomprehending freshman.'How pleasant it would be to let pass away some of the verbiage Ilearnt at school--learnt because teachers must live, I suppose. Theapeing and prolonged caw called grammar, the cackling of the human henover the egg of language--I should like to unlearn grammar.'But there is hope for a brighter future. Robinsons investigations ofthe hyperreal numbers have put in?itesimals on a logicallyunimpeachable basis, and here and there calculus texts based onin?itesimals have appeared [*].[*: - Keisler, H. J. (1976). /Elementary Calculus/. Boston: Prindle,Weber & Schmidt.- Henle & Kleinberg (1978). /In?itesimal Calculus/. Cambridge,Mass.: MIT Press.][Rucker, Rudy (1995). /In?ity and the Mind/. Princeton, NJ:Princeton University Press. (pp. 79/80 + 87)]PH === > 1 - 1/omegaThere is another system, besides Robinson's hyperreals, namely Conway'ssurreals, in which such in?itesimal calculations are common.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > But there is hope for a brighter future. Robinsons investigations of the> hyperreal numbers have put in?itesimals on a logically unimpeachable> basis, and here and there calculus texts based on in?itesimals have> appeared [*].What is the current status of this? I read somewhere, but don't recallwhere that it was proven that Robinson's approach and the standard approachare exactly equivalent in what they can do, so that while Robinson'sapproach is more intuitive, it cannot lead to any more or less than thestandard approach. Thus, while the non-standard approach would be better ifeveryone was starting from scratch, there is so much already in standardterms, that learning the standard approach is necessary, and once you'vedone that, there is no need to learn the non-standard approach, since youcan do everything using the standard approach.Basically, a mathematical society gets to choose one way or the other, andthen they are stuck with it. We hit on the limit approach rather than therigorization of infenitesimals, and now are stuck.So, is that correct?-- --Tim Smith === > But there is hope for a brighter future. Robinsons investigations of the> hyperreal numbers have put in?itesimals on a logically unimpeachable> basis, and here and there calculus texts based on in?itesimals have> appeared [*]. What is the current status of this? I read somewhere, but don't recall> where that it was proven that Robinson's approach and the standard approach> are exactly equivalent in what they can do,correct> so that while Robinson's> approach is more intuitive,not clear> it cannot lead to any more or less than the> standard approach.There are no new theorems in standard mathematics that can be proved bynon-standard methods.[*] In some cases non-standard proofs may beshorter or easier in some other sense. In other cases longer orharder.> Thus, while the non-standard approach would be better if> everyone was starting from scratch,a wild assumption> there is so much already in standard> terms, that learning the standard approach is necessary, and once you've> done that, there is no need to learn the non-standard approach, since you> can do everything using the standard approach. Basically, a mathematical society gets to choose one way or the other, and> then they are stuck with it. We hit on the limit approach rather than the> rigorization of in?itesimals, and now are stuck.a rather extreme way to say it So, is that correct?Except for specialists, one need only learn one system. And that mightas well be the commonly-used system, so one can talk to other people. Specialists may learn, and use, two or more systems.. . .[*] This equivalence does depend on the Axiom of Choice. Proof of theexistence of non-standard models relies on the Compactness Theorem ofmodel theory. Non-standard analysis can prove the Boolean AlgebraPrime Ideal Theorem without any further use of AC.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > so that while> Robinson's approach is more intuitiveNo. it isn't.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Basically, a mathematical society gets to choose one way or the other, and> then they are stuck with it. We hit on the limit approach rather than the> rigorization of infenitesimals, and now are stuck. So, is that correct?No, because the original idea is the in?itesimal - Through P let therebe drawn to this [ spherical ] surface two lines HK, IL, interceptingvery small arcs HI, KL ; ... - Prop. LXX Theor. XXX of Newtons' Principia.This is his proof that there is no gravitational attraction withina hollow uniform sphere.I think nonstandard analysis is a more direct justi?ation of thistype of reasoning, as opposed to the limit justi?ation which castsit in somewhat different terms.In physics, one never really makes the distinction between delta xand dx, and I think if one understands delta x as ?st order inx the equation can be made rigorous, as witness the nonstandardconstruction of in?itesimals in terms of sequences.Lew Mammel, Jr. === > 881c8779.0401041419.1097372c@posting.google.com>...> [...]> But there is hope for a brighter future. Robinsons investigations of> the hyperreal numbers have put in?itesimals on a logically> unimpeachable basis, and here and there calculus texts based on> in?itesimals have appeared [*]. [*: - Keisler, H. J. (1976). /Elementary Calculus/. Boston: Prindle,> Weber & Schmidt.> - Henle & Kleinberg (1978). /In?itesimal Calculus/. Cambridge,> Mass.: MIT Press.] [Rucker, Rudy (1995). /In?ity and the Mind/. Princeton, NJ:> Princeton University Press. (pp. 79/80 + 87)]I have just discovered that the entire text of Keislers book, towhich Rucker refers, can be downloaded for free at the following link:http://www.math.wisc.edu/~keisler/calc.htmlPH === In sci.logic, Paul Holbachon 4 Jan 2004 14:19:31 -0800<881c8779.0401041419.1097372c@posting.google.com>:> In our ordinary real number system, we say that the number K with> decimal expansion .99999... is the samas 1. An informal argument for> this is sketched below: 10K = 9.999...> - K = .9999...> ____________> 9K = 9> K = 1 But maybe this argument is misleading.That argument suffers from a major problem, as one can readily seeby considering the ?ite case.Kn = .999...9910*Kn = 9.999...9010*Kn - Kn = 8.999...91> What if there is some number,> call it 1 - 1/omega, that is greater than any ?ite string .9...9 of> nines, yet less than 1? If K were actually equal to 1 - 1/omega, the> informal argument used in the last paragraph would not work, for this> argument overlooks the fact that the difference between 10K and 10 is> ten times as great as the difference between K and 1. There is a> residual in?itesimal quantity below that does not get canceled out: 10K = 10 - 10/omega> - K = 1 - 1/omega> _________________> 9K = 9 - 9/omega> K = 1 - 1/omega Intuitively, nothing could be more natural than to go ahead and talk> about 1/omega, 1/Aleph-1, and so on. Just as we move from the natural> numbers to the fractions and then on to the reals, should we not be> able to move from the whole ordinal numbers to some richer number> ?ld?K = .9999... = 1 - 1/inftyK/10 = .09999... = 0.1 - 1/(10*infty)0.9 + K/10 = 0.9999... = 1 - 1/(10*infty)Might work but then one would have to, as you state below,be very careful as to where the in?tesimals and trans?ites go.Some possible questions, for example, from college freshman (well,OK, one ex-college freshman :-) ).[1] What is 1/infty + 1/infty? [a] 2/infty [b] 1/infty [c] 0 [d] indeterminate [e] the expression cannot be simpli?d, but must be left in this form[2] What is 1/infty - 1/infty? [a] 0 [b] 1/infty [c] -1/infty [d] indeterminate [e] the expression cannot be simpli?d[3] What is (1/infty)/(1/infty)? [a] 1 [b] infty [c] any number you want [d] unde?ed [e] the expression cannot be simpli?d[4] What is 1/infty^2? [a] 1/infty [b] 0 [c] any number you want [d] unde?ed [e] the expression cannot be simpli?d[5] What is 1/sqrt(infty)? [a] 1/infty [b] 0 [c] any number you want [d] unde?ed [e] the expression cannot be simpli?d[6] What is 0.999... + 1/infty? [a] 1 [b] 0.999.... [c] any number you want [d] unde?ed [e] the expression cannot be simpli?d[7] What is the relationship between 1/infty and 0? [a] 1/infty > 0 [b] 1/infty = 0 [c] indeterminate[8] What does 1/3 equal? [a] 0.333.... [b] 0.333.... + 1/infty [c] 0.333.... + 1/(3*infty) [d] indeterminate [e] the expression cannot be expanded into an in?ite decimal[9] What does 3*(1/3) equal? [a] 1 [b] 0.999... [c] 1 - 1/infty [d] 1 - 3/infty [e] indeterminate [f] the expression cannot be simpli?d[10] What does sqrt(1 - 1/infty) equal? [a] 1 - 1/(2*infty) + 3/(8*infty^2) - 5/(16*infty^3) + ... [b] 1 - 1/(2*infty) [c] 1 - 1/infty [d] 1 [e] indeterminate [f] the expression cannot be simpli?d[11] What does lim(h->0+) (1+h)^(1/h) equal? [a] Euler's number, ?e' [b] (1+1/infty)^infty [c] the expression cannot be simpli?d[12] What does 1/(1 - 1/infty) equal? [a] 1 + 1/infty + 1/infty^2 + 1/infty^3 + ... [b] 1 + 1/infty [c] 1 [d] indeterminate [e] the expression cannot be simpli?d[13] What is lim(x->0-) (x^2/x)? [a] 0 [b] -1/infty [c] 1/infty [d] infty/infty^2 [e] indeterminate [f] the expression cannot be simpli?d[14] What is sum(i=1,+infty) (1/infty)? [a] 0 [b] 1 [c] e [d] indeterminate [e] the expression cannot be simpli?d[15] Does the trichotomy principle always hold? [a] yes [b] yes, but it may be tricky to determine when infty is in there [c] only when infty is not involved [d] no [e] unknownAll of these will have to be dealt with in some fashion.Standard mathematics uses the sequence1c2a3d4b5b6a7b8a9a10d11a12c13a14d15a, which implicitly sets 1/infty=0pretty much everywhere.> Curiously, Cantor himself was very much opposed to this step. When a> fellow mathematician attempted to use Cantors trans?ite numbers to> develop a theory of in?itely small quantities, Cantor accused him of> trying to ?infect mathematics with the Cholera-Bacillus of> in?itesimals'. Cantor even constructed a proof that no number can be> in?itesimal. This proof, however, is just as circular and worthless> as ?itist attempts to prove that no number can be in?ite. In both> cases, the desired conclusion is smuggled in as part of the de?ition> of ?number'.It depends on how one de?es number. If one uses Cauchysequences one has problems, for example, as Cauchy sequenceslead to limits.0.999... is such a sequence.f_1 = 0.9f_2 = 0.99f_3 = 0.999etc.The classical de?ition of limit for this case is:f = lim(s->+infty) f_s if, for any epsilon > 0 I pick, onecan show an N such that for all s > N, abs(f_s - f) < epsilon.If f = 1 - 1/infty, I pick epsilon = 1/2*infty, and then wonderwhat N satis?s this de?ition, bearing in mind 0 < 1/infty < 1/10^Nfor any integer N (although the nonpositive integers aren't all thatinteresting :-) )You may also recall Douglas Adams' number 2^(infty - 1).> Why was Cantor so vehemently opposed to in?itesimals? In his> valuable essay, ?The Metaphysics of the Calculus', Abraham Robinson> suggests that Cantor already had enough problems trying to defend> trans?ite numbers. It seems likely that, consciously or otherwise,> Cantor deemed it politically wise to go along with othodox> mathematicians on the question of in?itesimals. Cantors stance> might be compared to that of a pro-marijuana Congressional candidate> who advocates harsh penalties for the sale or use of heroin. Yet, as> we shall see, there is almost as much justi?ation for in?itesimals> as there is for Cantors trans?ite ordinals.> Formally speaking, it is as consistent to say that there is a number> between all of .9, .99, .999, ... and 1 as it is to say that there is> a number greater than all of 1, 2, 3, ... . And just as we go on to> ?d more and more ordinals piled atop one another, we can go on to> ?d more and more in?itesimals squeezed beneath each other.> [...]> But so great is the average persons fear of in?ity that to this day> calculus all over the world is being taught as a study of *limit> processes* instead of what it really is: *in?itesimal analysis*.> As someone who has spent a good portion of his adult life teaching> calculus courses for a living, I can tell you how weary one gets of> trying to explain the complex and ?dling theory of limits to wave> after wave of uncomprehending freshman. ?How pleasant it would be to let pass away some of the verbiage I> learnt at school--learnt because teachers must live, I suppose. The> apeing and prolonged caw called grammar, the cackling of the human hen> over the egg of language--I should like to unlearn grammar.' But there is hope for a brighter future. Robinsons investigations of> the hyperreal numbers have put in?itesimals on a logically> unimpeachable basis, and here and there calculus texts based on> in?itesimals have appeared [*].Let us hope Robinson has some answers for my questions.I've not read his works, though.The Leibnitz notation [dy/dx] is currently a limit,even if it does look like the division of two in?itesimals.If y(x) = x^2, then dy/dx = lim(h->0) (y(x+h) - y(x))/h = lim(h->0) (x^2 + 2hx + h^2 - x^2) / h = lim(h->0) (2hx + h^2)/h = lim(h->0) (2x + h) = 2x, for example.Integration is similar. This notation can be abused nastily, and is. [*: - Keisler, H. J. (1976). /Elementary Calculus/. Boston: Prindle,> Weber & Schmidt.> - Henle & Kleinberg (1978). /In?itesimal Calculus/. Cambridge,> Mass.: MIT Press.] [Rucker, Rudy (1995). /In?ity and the Mind/. Princeton, NJ:> Princeton University Press. (pp. 79/80 + 87)] > PH-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Or, 1/3 = .333333...3 * 1/3 = 1 = .99999...> In our ordinary real number system, we say that the number K with> decimal expansion .99999... is the samas 1. An informal argument for> this is sketched below: 10K = 9.999...> - K = .9999...> ____________> 9K = 9> K = 1 But maybe this argument is misleading. What if there is some number,> call it 1 - 1/omega, that is greater than any ?ite string .9...9 of> nines, yet less than 1? If K were actually equal to 1 - 1/omega, the> informal argument used in the last paragraph would not work, for this> argument overlooks the fact that the difference between 10K and 10 is> ten times as great as the difference between K and 1. There is a> residual in?itesimal quantity below that does not get canceled out: 10K = 10 - 10/omega> - K = 1 - 1/omega> _________________> 9K = 9 - 9/omega> K = 1 - 1/omega Intuitively, nothing could be more natural than to go ahead and talk> about 1/omega, 1/Aleph-1, and so on. Just as we move from the natural> numbers to the fractions and then on to the reals, should we not be> able to move from the whole ordinal numbers to some richer number> ?ld?> Curiously, Cantor himself was very much opposed to this step. When a> fellow mathematician attempted to use Cantors trans?ite numbers to> develop a theory of in?itely small quantities, Cantor accused him of> trying to ?infect mathematics with the Cholera-Bacillus of> in?itesimals'. Cantor even constructed a proof that no number can be> in?itesimal. This proof, however, is just as circular and worthless> as ?itist attempts to prove that no number can be in?ite. In both> cases, the desired conclusion is smuggled in as part of the de?ition> of ?number'.> Why was Cantor so vehemently opposed to in?itesimals? In his> valuable essay, ?The Metaphysics of the Calculus', Abraham Robinson> suggests that Cantor already had enough problems trying to defend> trans?ite numbers. It seems likely that, consciously or otherwise,> Cantor deemed it politically wise to go along with othodox> mathematicians on the question of in?itesimals. Cantors stance> might be compared to that of a pro-marijuana Congressional candidate> who advocates harsh penalties for the sale or use of heroin. Yet, as> we shall see, there is almost as much justi?ation for in?itesimals> as there is for Cantors trans?ite ordinals.> Formally speaking, it is as consistent to say that there is a number> between all of .9, .99, .999, ... and 1 as it is to say that there is> a number greater than all of 1, 2, 3, ... . And just as we go on to> ?d more and more ordinals piled atop one another, we can go on to> ?d more and more in?itesimals squeezed beneath each other.> [...]> But so great is the average persons fear of in?ity that to this day> calculus all over the world is being taught as a study of *limit> processes* instead of what it really is: *in?itesimal analysis*.> As someone who has spent a good portion of his adult life teaching> calculus courses for a living, I can tell you how weary one gets of> trying to explain the complex and ?dling theory of limits to wave> after wave of uncomprehending freshman. ?How pleasant it would be to let pass away some of the verbiage I> learnt at school--learnt because teachers must live, I suppose. The> apeing and prolonged caw called grammar, the cackling of the human hen> over the egg of language--I should like to unlearn grammar.' But there is hope for a brighter future. Robinsons investigations of> the hyperreal numbers have put in?itesimals on a logically> unimpeachable basis, and here and there calculus texts based on> in?itesimals have appeared [*]. [*: - Keisler, H. J. (1976). /Elementary Calculus/. Boston: Prindle,> Weber & Schmidt.> - Henle & Kleinberg (1978). /In?itesimal Calculus/. Cambridge,> Mass.: MIT Press.] [Rucker, Rudy (1995). /In?ity and the Mind/. Princeton, NJ:> Princeton University Press. (pp. 79/80 + 87)] > PHX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === In <8tEJb.2572$uF6.1092753@news1.news.adelphia.net>, on 01/03/2004 at 07:11 PM, Leonard M. Wapner said:>But does it make sense to call the function continuous?Yes.>(It's the gap in the domain that bothers me.)That might be relevant in a Calculus textbook. It's certainly notrelevant in Topology, and I'm not aware of any Analysis text that usesa de?ition of continuous inconsistent with that used in Topology.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === On Sat, 03 Jan 2004 15:37:53 -0500, A N Niel> Whether or not it makes sense to call it that, the given function> _is_ continuous.>You haven't been reading elementary calculus texts recently, have>you...???No.>Of course a mathematician will say it is continuous. But their>de?ition may not agree with the de?ition in all elementary texts.I was aware of that. If he'd asked whether calculus books wouldcall this function continuous I wouldn't have spoken up, cuz Iwouldn't have known the answer.A much worse example: Say as usual (i) when a function is de?edby a formula we take the domain to be the set of all reals for whichthe formula makes sense, (ii) we say a function is continuous if itis continuous at every point of its domain. Then the function f(x) = 1/xis continuous.David C. Ullrich === [snip]> A much worse example: Say as usual (i) when a function is de?ed> by a formula we take the domain to be the set of all reals for which> the formula makes sense,I'll also assume, in that case, that you're intending the range to bea subset of R. But even then there can be differences of opinion aboutthose reals for which the formula makes sense. As a simple example:If we were asked for the implied domain of f(x) = Sqrt(x^2 (x-1)),I would hope that everyone would say that it is {0} U [1, +oo). Butsuppose that we were asked for the implied domain of g(x) = |x| Sqrt(x-1).I suspect that some would say that 0 is in the implied domain of g (andalso that f and g are the same function), while others would say that 0is not in the implied domain of g (and thus that f and g are differentfunctions).BTW, is there a better term than implied domain for what I was talkingabout above?David Cantrell === >[snip]> A much worse example: Say as usual (i) when a function is de?ed> by a formula we take the domain to be the set of all reals for which> the formula makes sense,>I'll also assume, in that case, that you're intending the range to be>a subset of R. But even then there can be differences of opinion about>those reals for which the formula makes sense. As a simple example:>If we were asked for the implied domain of> f(x) = Sqrt(x^2 (x-1)),>I would hope that everyone would say that it is {0} U [1, +oo). But>suppose that we were asked for the implied domain of> g(x) = |x| Sqrt(x-1).>I suspect that some would say that 0 is in the implied domain of g (and>also that f and g are the same function), while others would say that 0>is not in the implied domain of g (and thus that f and g are different>functions).If we were talking about actual math we'd need to be a lot morecareful with all of this, or better yet simply throw out the notionof implied domain. I think we're talking about the way terminologyis used in a typical calculus class - in a context like that there areno complex numbers, so the domain does not include 0.I hope so, anyway.>BTW, is there a better term than implied domain for what I was talking>about above?>David CantrellDavid C. Ullrich === I have a question and would appreciate any help.> Let (f_n)_ be a sequence of real valued functions de?ed on [a, b]> that converges uniformly to f. For every x in [a,b], let F_n(x) => Integral (from a to x) f_n(t)dt. Then, (F_n) converges to F(x) => Integral (from a to x) f(t)dt. Is this convergence uniform on [a,b]?> If not, is there any condition that assures uniformity? Yes, because | F_n(x) - F(x) | can be majorized by the> product of (b-a) and the maximum difference between f_n(t) and f(t). If each f_n is de?ed on [a, inf) and its in?ite integral exists,> then does the sequence of such in?ite integrals converge to the> in?ite integral of f? No. Consider a sequence of functions f_n which converge to f(t) = 0> such that for each n, int (0,in?ity) f_n = 1, but they> grow shallower and wider ( e.g., lambda exp(-lambda t) , when> lambda = 1/n).> Artur> This had better not have been homework :-). Best wishes,> MikeArtur === >I have a question and would appreciate any help.>Let (f_n)_ be a sequence of real valued functions de?ed on [a, b]>that converges uniformly to f. For every x in [a,b], let F_n(x) =>Integral (from a to x) f_n(t)dt. You have to say a little more than you have or this integral> doesn't even exist. Possibly you meant to assume that f_n> was continuous - that would do it.Then, (F_n) converges to F(x) =>Integral (from a to x) f(t)dt. Is this convergence uniform on [a,b]? Yes. Presumably he meant to assume the f_n are Riemann integrable--the> minimum needed to make sense of Integral (from a to x) f_n(t)dt. > The uniform limit of Riemann integrable functions is also Riemann> integrable, so this seems OK.> Yes, exactly. I didn't asume continuity, but just Riemann integrability.Artur === Lode Vandevenne [CapitalEth][EDoubleDot][Micro] .b3 .b9[EDoubleDot].b9>[snip]> It looks like a quite analytical function in ]-2,oo[ to the eye.> Extending this to tetration of other values A than sqrt(2) looks quitehard> though, since if A > e^(1/e) there's no horizontal asymptote at all, and> else the horizontal asymptote is not at +2 so you can't use the mirroring.> Maybe this is possible after a suitable transformation, then mirror, then> transform back...The exact bounds where the (real) hyperpower function converges are wellknown: [(1/e)^e,e^(1/e)]As far as extending the hyperpower to reals, you might ?d Munafo's pagesquite interesting and a good place to start.http://home.earthlink.net/~mrob/pub/math/ln-notes1.html #real-hyper4--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------ ------------------------Eventually, _everything_ is understandable === > This concludes my attempt to de?e sqrt(2)^^x for real x, did it make any> sense?Absolutely.What happens with 3^(1/3)=1.442249570307408382321638311?Phil-- Unpatched IE vulnerability: XSS in Unparsable XML FilesDescription: Cross-Site Scripting on any site hosting ?es that can be misrendered in MSXMLReference: http://sec.greymagic.com/adv/gm013-ie/Exploit: http://sec.greymagic.com/adv/gm013-ie/ === > a^^n is convergent for n going to oo (in?ity) if a is positive and a <> e^(1/e). (I'm ignoring negative a here)You need to recheck a close to zero. For example, 0.1^^n does notconverge, but gets close to a cycle of period 2.> This concludes my attempt to de?e sqrt(2)^^x for real x, did it make> any> sense?Here is a mathematical problem it suggests.Describe all functions g:(-2,oo) -> (-oo,2) such that:g(x+1)=sqrt(2)^g(x)g(x)=y <==> g(-y)=-xg(0)=1In particular, is there a unique continuous solution? === And afterwards we collate the replies to measure my claim.> Herc > What, EXACTLY, is your claim?That seems to be where you fall apart. === ... > where b_3(x) = a_3(x) - 3 and the a's are roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and when x=0, a_1(0) = a_2(0) = b_3(0) = 0. > Yup, that has not been challenged of late. > That's important mathematical information that I can luckily include > for reference in my posts as it's short.Well, you could have included that information about a year ago. > Now dividing both sides by 49, you get differing functions like > (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where when x=0, b_1(0) = b_2(0) = c_3(0) = 0. > Possibly. Although you do not explain c_3 any further. > I replaced all the previous function names with new ones advancing to > the next letter in the alphabet. > So before there was a_1(x), a_2(x), and b_3(x), so I went to b_1(x), > b_2(x), and c_3(x).So you mean that c_3(x) is b_3(x) advanced one letter? I have no ideawhat that would mean. > The functions are to a large extent unknown, as there's no de?ing > cubic within the ring of algebraic integers, unlike with the a's, and > b_3(x). > Remember, I ended up with b_3(x) from b_3(x) = a_3(x) - 3, where the > a's are roots of the cubic: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)Still does not explain what c_3(x) is. > Now before the a's were de?ed by the cubic: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > > But what is the de?ing cubic now? > Well, the leading coef?ient is known, as is the last coef?ient but > what about the others? > b^3 + ? b^2 + ? b - 2401 x^3 - 147 x^2 + 3x. > What is c? > Explained above,No. How does c_3(x) relate to the b's as (possibly de?ed) by that cubic? > Explained above, now consider what I'm emphasizing by pointing out the > unknown coef?ients, which is that it's not as easy as before with > the a's and the *de?ing* cubic: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > Something has happened, which shows that the resultant functions--the > functions that result after 49 is divided from both sides--are NOT in > the ring of algebraic integers as they aren't de?able within that > ring.That is *only* because you want to divide off 49 in a speci? way.There are other ways to divide off 49 (depending on the value of x)that give perfect algebraic integers. The functions are indeed notsimple. But they exist. And they are de?able (and I have de?edthem) in the ring of algebraic integers. > So there's the challenge, which is to ?l in the other two > coef?ients. > > Now then, if posters disagreeing with me have been correct then there > must exist integers or integer functions that go there, and it should > be possible to show them! > That is *false*. There is *no* disagreement that a_1(x) and a_2(x) are > not divisible by 7 in the algebraic integers, which is what you are > claiming with the above statement. > Where? I am in fact not making that claim with that statement.You are claiming a disagreement about the divisibility by 7 with yourstatement. There is *no* disagreement about that. > The b's are just not algebraic integer functions, and I've made no > mention of factors of a_1(x) or a_2(x) in the ring of algebraic > integers as such are irrelevant to my point.You state that people disagreeing with you should show integer functions.But that means that you require them to show that a_1(x) and a_2(x) aredivisible by 7. But be careful. I have given de?itions of threefunctions w such that f1(x) = (5 a_1(x) + 7)/w1(x), f2(x) = (5 a_2(x) + 7)/w2(x), f3(x) = (5 a_3(x) + 7)/w3(x) = (5 b_3(x) + 22)/w3(x)are algebraic integer functions with w1(x)*w2(x)*w3(x) = 49. And so: f1(x)*f2(x)*f3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22for all algebraic integers and rational integers. >Because if there are integers or > integer functions, b_1 and b_2 would be algebraic integer functions. > You are again misrepresenting the claims of your opponents. > My point is that there just aren't algebraic integer functions that > exist to ful? the requirements, which is shown by the inability to > de?e them within the ring of algebraic integers.You now have a new requirement, namely that the factorisation shoould beof the form (5 b_1(x) + 1)(5 b_2(x) + 1)(5 c_3(x) + 22),but probably such functions indeed do not exist. But I do not see whysuch should be required. The requirement comes only by your blindlyassuming that two of the factors should be divisible by 7. And thatassumption has not yet been justi?d. > Now then, people like yourself with your own claims need algebraic > integer functions, so I'm not misrepresenting your position at all.But you are when you require us to show that b_1, b_2 and c_3 arealgebraic integer functions. That is *not* the disagreement. > It seems to me that you should have noticed your inability to ?d > algebraic integer functions to support your position as if you *could* > ?d them, then that would have ended arguments long ago in *your* > favor.What is this? I have given the de?itions, but you refuse to say whatpart of my de?itions is wrong. You only state that I *could* not ?dthem when I already have found them. There are however probably *no*algebraic integer functions that allow the distribution of 49 as 7, 7, 1. > Instead I think you relied on other people not paying attention to the > mathematical facts to promote your own views *against* mathematical > truth.What mathematical facts? That 49 should be distributed as 7, 7, 1? Inwhat way is that a mathematical fact?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Here is a program I quickly put together to try this out. For n=10 itran for 8 minutes and produced:Enter the size to test for:10Creating table...Done!Beginning search...Solution found: 22811131212931356164833861185611876294441366485156151289881636 19586569118418574426545164841464669441Solution found: 13368529693773153476372527122563541623698121434161557635562523 12455744942316732967266242569659154961Solution found: 53183473293339221796134828496189849545213229421584428527344471 44137529379554566429628426249611449444Solution found: 43325356843225217681324637852925621819245231484241317881716157 81473296665921281688224691844194116644Solution found: 35613443295713545744614122995613117435243527528449442424522545 93857284379542244924524284849464954944Solution found: 32339694242519437636315181188139198868819488318464631818316997 16833476468841478423886678764611496464Solution found: 12919711362367892921963126332117113114249823584996796181444112 31448464193494414432229464416114614416Solution found: 48554417618672265625575519476952517559614217463364469567562515 45354721767936742462666222441591451449Solution found: 14769417614975327296778435644165467517449337936689425534428917 616487847247627689694488889616149949699 solutions found!real 8m13.069suser 6m59.930ssys 0m5.160sInterestingly this continues the 13,12,11,10 pattern.Anyone have ideas on how I could speed it up? It inserts digits fromthe bottom-right, and it initialises a map that takes an end-sequenceof digits and returns what the digit before it can be.The code is as follows:#include #include #include #include #include #include #include using namespace std;#de?e i64 long long#de?e fu(i,n) for(int i=0; i<(n); i++)#de?e fd(i,n) for(int i=(n)-1; i>=0; i--)#de?e fr(i,b,e) for(typeof(e) i=(b); i!=(e); ++i)#de?e fa(i,x) fr(i,(x).begin(),(x).end())int N;i64 pows[30];i64 vals[30];vector< pair order;int cnt=0;template<> struct hash { size_t operator()(long long x) const { return x+(x>32); }};template<> struct hash { size_t operator()(const long long x) const { return x+(x>32); }};hash_map cache;void doit(int o) { int r = order[o].?st; int c = order[o].second; if(o==order.size()) { cout << Solution found:n; fd(i,N) cout << vals[i] << endl; cout << endl; ++cnt; return; } set::const_iterator A,B; typeof(cache.end()) X=cache.?d(vals[r]); if(X==cache.end()) return; hash_map::const_iterator Y=cache.?d(vals[c]); if(Y==cache.end()) return; A=X->second.begin(); B=Y->second.begin(); for(;;) { if(*A<*B) ++A; else if(*A>*B) ++B; else { if(*A==10) break; vals[c] += (*A)*pows[r]; if(r!=c) vals[r] += (*A)*pows[c]; doit(o+1); vals[c] -= (*A)*pows[r]; if(r!=c) vals[r] -= (*A)*pows[c]; ++A; ++B; } }}int main(void) { cout << Enter the size to test for:n; cin > N; cout << Creating table...n; pows[0]=1; for(int i=1; i<=N; i++) pows[i]=10*pows[i-1]; // Create the cache // Given a number representing the last digits, give a list of thepossible previous digits char digits[N+1]; for(i64 i=pows[(N-1)/2]; i*isecond.insert(10); cout << Done!n; cout << Beginning search...nn; doit(0); printf(%d solution%s found!n,cnt,cnt==1?:s); return 0;} === ... > However, that common sense view is wrong, as although algebraic > integers *are* in?itely decomposable into algebraic integers, it > turns out that it's rather easy to show that the peculiar construction > here doesn't allow for a decomposition *in general* in the ring of > algebraic integers. > So why do I emphasize in general? That's because you may indeed > ?d for some particular x that you can get an algebraic integer > decomposition, but you can't ?d ?d a de?ition like for a_1(x), > and a_2(x), where they're *guaranteed* to be algebraic integers for > algebraic integer x because they're the roots of > a^2 - (x - 1)a + 7(x^2 + x).This is your basic error James. I have de?ed algebraic integerfunctions w (with your original polynomial) that give algebraic integersfor *all* algebraic integer arguments. They do not satisfy a simplepolynomial indeed. *But they need not!*Consider the function: f(x) = 2 when x is integer and even, f(x) = 1 when x is integer and odd, f(x) = 0 when x is an algebraic integer, but not a rational integer.This is obviously an algebraic integer function, but there is *no*polynomial that it satis?s.Or do you think that the above function is not an algebraic integerfunction in general? It is guaranteed to give an algebraic integerresult for all algebraic integer arguments.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > I made a post about a quadratic example that Rick Decker, a professor> at Hamilton (I said University before but it might be College), gave> in a recent post. I haven't seen any replies to that yet in Google> Groups, so I'll leave it and make another thread to consider Decker's> example in more detail, again here are some headers to allow you toWell he did ?ally reply.> ?d his original post: > In his post Decker claimed to mirror my argument using a quadratic> instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).What I like here is how *simple* a quadratic should look to most ofyou.I want to emphasize that I'm making these points with quadratics!!! > So from a *common sense* perspective, that is, the common> understanding of the ability to decompose algebraic integers, you> might *reasonably* suppose that you can always ?d decompositions in> the ring of algebraic integers such that a_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x), where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,> so that you could have w_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x) in the ring of algebraic integers. However, that common sense view is wrong, as although algebraic> integers *are* in?itely decomposable into algebraic integers, it> turns out that it's rather easy to show that the peculiar construction> here doesn't allow for a decomposition *in general* in the ring of> algebraic integers. Now then, consider what happens if you divide both sides of (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2It's that expression which holds the key here, as the asymmetrybetween the factors (5b_1(x) + 1), and (5b_2(x) + 2) is what preventsb_1(x), and b_2(x) from being algebraic integer functions. > where the b's are roots of some unknown quadratic, though the ?st> and last coef?ients ARE known: b^2 + ? b + (x^2 + x). If you're a little confused on that point, remember that I had a_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x), where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,> so that you could have w_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x) and now dividing both sides by 7 gives b_1(x) b_2(x) = x^2 + x. If you assume that the b's like the a's before them are algebraic> integer functions (that common sense again) then there should be a> de?ing quadratic *in the ring of algebraic integers* like there IS> for the a's, as for the a's that de?ing quadratic is a^2 - (x - 1)a + 7(x^2 + x). Now then Decker probably picked his example not for me to outline some> of the ?er points of Advanced Polynomial Factorization Theory (just> made that up, being a discoverer is cool). As I mentioned in my other thread on his example, he probably picked> it because at x=1, you have *both* a's with sqrt(7) as a factor.Notice that if you *could* ?d the middle coef?ient of b^2 + ? b + (x^2 + x)as a function of x, it would necessarily equal 0, when x=1, where Imake use of what probably drew Decker to this example in the ?stplace!!!Some of you may realize that focusing on that quadratic is a key tounderstanding *why* the b's can't be algebraic integer functions,where the challenge to anyone wishing to deny that, or keep arguing,is to provide that middle coef?ient.James Harris === > Now then, consider what happens if you divide both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> It's that expression which holds the key here, as the asymmetry> between the factors (5b_1(x) + 1), and (5b_2(x) + 2) is what prevents> b_1(x), and b_2(x) from being algebraic integer functions.>where the b's are roots of some unknown quadratic, though the ?st>and last coef?ients ARE known:b^2 + ? b + (x^2 + x).If you're a little confused on that point, remember that I hada_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x),where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,>so that you could havew_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x)and now dividing both sides by 7 givesb_1(x) b_2(x) = x^2 + x.If you assume that the b's like the a's before them are algebraic>integer functions (that common sense again) then there should be a>de?ing quadratic *in the ring of algebraic integers* like there IS>for the a's, as for the a's that de?ing quadratic isa^2 - (x - 1)a + 7(x^2 + x).Now then Decker probably picked his example not for me to outline some>of the ?er points of Advanced Polynomial Factorization Theory (just>made that up, being a discoverer is cool).As I mentioned in my other thread on his example, he probably picked>it because at x=1, you have *both* a's with sqrt(7) as a factor.> Notice that if you *could* ?d the middle coef?ient of b^2 + ? b + (x^2 + x) as a function of x, it would necessarily equal 0, when x=1, where I> make use of what probably drew Decker to this example in the ?st> place!!!>Well, no. If you have a_1(x) and a_2(x) de?ed to be theroots of a^2 - (x - 1)a + (x^2 + x)then a_1(x) + a_2(x) = x - 1. Now if you write a_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x)as you did above you'll have the b's as roots of b^2 + C(x)b + x^2 + xwhere C(x) = -(a_1(x)/w_1(x) + a_2(x)/w_2(x)).It's certainly not obvious to me that we should have C(1) = 0. > Some of you may realize that focusing on that quadratic is a key to> understanding *why* the b's can't be algebraic integer functions,> where the challenge to anyone wishing to deny that, or keep arguing,> is to provide that middle coef?ient.>As I said in another thread, the middle coef?ient is (-3x + sqrt(-7x^2 - 8x))/4which, when x = 1, is (-3 + sqrt(-15))/4This doesn't quite look like 0 to me. James Harris> === > Now then, consider what happens if you divide both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) by 7, as then you end up with something like(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> It's that expression which holds the key here, as the asymmetry> between the factors (5b_1(x) + 1), and (5b_2(x) + 2) is what prevents> b_1(x), and b_2(x) from being algebraic integer functions.> >where the b's are roots of some unknown quadratic, though the ?st>and last coef?ients ARE known:b^2 + ? b + (x^2 + x).If you're a little confused on that point, remember that I hada_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x),where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,>so that you could havew_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x)and now dividing both sides by 7 givesb_1(x) b_2(x) = x^2 + x.If you assume that the b's like the a's before them are algebraic>integer functions (that common sense again) then there should be a>de?ing quadratic *in the ring of algebraic integers* like there IS>for the a's, as for the a's that de?ing quadratic isa^2 - (x - 1)a + 7(x^2 + x).Now then Decker probably picked his example not for me to outline some>of the ?er points of Advanced Polynomial Factorization Theory (just>made that up, being a discoverer is cool).As I mentioned in my other thread on his example, he probably picked>it because at x=1, you have *both* a's with sqrt(7) as a factor.> Notice that if you *could* ?d the middle coef?ient of b^2 + ? b + (x^2 + x) as a function of x, it would necessarily equal 0, when x=1, where I> make use of what probably drew Decker to this example in the ?st> place!!!> Well, no. If you have a_1(x) and a_2(x) de?ed to be the> roots of a^2 - (x - 1)a + (x^2 + x) then a_1(x) + a_2(x) = x - 1. Now if you write a_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x) as you did above you'll have the b's as roots of b^2 + C(x)b + x^2 + x where C(x) = -(a_1(x)/w_1(x) + a_2(x)/w_2(x)). It's certainly not obvious to me that we should have C(1) = 0.You're right. I forgot that the b's don't directly translate to thea's. > Some of you may realize that focusing on that quadratic is a key to> understanding *why* the b's can't be algebraic integer functions,> where the challenge to anyone wishing to deny that, or keep arguing,> is to provide that middle coef?ient.> As I said in another thread, the middle coef?ient is (-3x + sqrt(-7x^2 - 8x))/4 which, when x = 1, is (-3 + sqrt(-15))/4 This doesn't quite look like 0 to me.Yeah, my mistake before, but still you have a problem. Your problemis that you get *two* different quadratics when you try to solve forthe functions b_1(x) and b_2(x), for a total of 4 solutions.You may wish to choose the positive value of the sqrt() operator, butoperator ambiguity doesn't allow you to pick and choose!After all, what if you change your mind and instead decide to take thenegative result of taking the square root?The math doesn't bother caring what your mood is.Why don't you take the alternate path in your formulation, like takethe negative with your square roots, and see if you get the same C? James Harris === [snip]> You may wish to choose the positive value of the sqrt() operator, but> operator ambiguity doesn't allow you to pick and choose!This is *your* clearly stated interpretation of ?sqrt()'. Unfortunately, this position repudiates thecorrectness of your prime counting method.> After all, what if you change your mind and instead decide to take the> negative result of taking the square root?> The math doesn't bother caring what your mood is.Well, you clearly do NOT accept your own prime counting method, according to your stated position about'sqrt()'. Please do the honorable thing and remove your prime counting algorithm and all claims about its valuefrom public view.--A man of good character acknowledges his ?A man without it defends them.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === James Harris skrev i melding> Now then, consider what happens if you divide both sides of(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)by 7, as then you end up with something like(5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> It's that expression which holds the key here, as the asymmetry> between the factors (5b_1(x) + 1), and (5b_2(x) + 2) is what prevents> b_1(x), and b_2(x) from being algebraic integer functions.>where the b's are roots of some unknown quadratic, though the ?st>and last coef?ients ARE known:b^2 + ? b + (x^2 + x).If you're a little confused on that point, remember that I hada_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x),where w_1(x) w_2(x) = 7, and the w's are algebraic integer functions,>so that you could havew_1(x) b_1(x) w_2(x) b_2(x) = 7(x^2 + x)and now dividing both sides by 7 givesb_1(x) b_2(x) = x^2 + x.If you assume that the b's like the a's before them are algebraic>integer functions (that common sense again) then there should be a>de?ing quadratic *in the ring of algebraic integers* like there IS>for the a's, as for the a's that de?ing quadratic isa^2 - (x - 1)a + 7(x^2 + x).Now then Decker probably picked his example not for me to outline some>of the ?er points of Advanced Polynomial Factorization Theory (just>made that up, being a discoverer is cool).As I mentioned in my other thread on his example, he probably picked>it because at x=1, you have *both* a's with sqrt(7) as a factor.> Notice that if you *could* ?d the middle coef?ient of> b^2 + ? b + (x^2 + x)> as a function of x, it would necessarily equal 0, when x=1, where I> make use of what probably drew Decker to this example in the ?st> place!!!> Well, no. If you have a_1(x) and a_2(x) de?ed to be the> roots of> a^2 - (x - 1)a + (x^2 + x)> then a_1(x) + a_2(x) = x - 1. Now if you write> a_1(x) = w_1(x) b_1(x), and a_2(x) = w_2(x) b_2(x)> as you did above you'll have the b's as roots of> b^2 + C(x)b + x^2 + x> where> C(x) = -(a_1(x)/w_1(x) + a_2(x)/w_2(x)).> It's certainly not obvious to me that we should have C(1) = 0.> You're right. I forgot that the b's don't directly translate to the> a's.> Some of you may realize that focusing on that quadratic is a key to> understanding *why* the b's can't be algebraic integer functions,> where the challenge to anyone wishing to deny that, or keep arguing,> is to provide that middle coef?ient.> As I said in another thread, the middle coef?ient is> (-3x + sqrt(-7x^2 - 8x))/4> which, when x = 1, is> (-3 + sqrt(-15))/4> This doesn't quite look like 0 to me.> Yeah, my mistake before, but still you have a problem. Your problem> is that you get *two* different quadratics when you try to solve for> the functions b_1(x) and b_2(x), for a total of 4 solutions.> You may wish to choose the positive value of the sqrt() operator, but> operator ambiguity doesn't allow you to pick and choose!> After all, what if you change your mind and instead decide to take the> negative result of taking the square root?> The math doesn't bother caring what your mood is.> Why don't you take the alternate path in your formulation, like take> the negative with your square roots, and see if you get the same C?> James HarrisMr. HarrisHow do you de?e the squareroot of a real number?What is sqrt(a^2), where a is a real number?Karl-Olav Nyberg === Mr. Harris How do you de?e the squareroot of a real number?The square root operator returns that number which when multiplied byitself gives the operand.> What is sqrt(a^2), where a is a real number?The sqrt(a^2) is a or -a.James Harris === James Harris skrev i melding> Mr. Harris> How do you de?e the squareroot of a real number?> The square root operator returns that number which when multiplied by> itself gives the operand.> What is sqrt(a^2), where a is a real number?> The sqrt(a^2) is a or -a.> James HarrisOk, I shall re?e my question (as a controlquestion).What is sqrt(a^2), where a is a positive real number?Karl-Olav Nyberg <3FF86BDB.8090106@hamilton.edu> <3c65f87.0401041420.695fd51@posting.google.com> <3c65f87.0401050603.5746bc6c@posting.google.com> === > Mr. Harris> How do you de?e the squareroot of a real number?> The square root operator returns that number which when multiplied by> itself gives the operand.An operator is functional[1]. That means that, for each x, there is aunique y such that sqrt(x) = y. This is not the ?st time that thishas been mentioned. It's amazing that you still stupidly ignore thisfact.Even your own attempt to de?e the square root function aboveindicates that functions are deterministic. You refer to thatnumber, which implicitly suggests that there is one and only onenumber satisfying what follows (not true, of course, in your attemptedde?ition).As many here have repeatedly said, the square root of x is thatnon-negative real number y such that y^2 = x. (Of course, I onlyde?ed square root for non-negative reals here.)> What is sqrt(a^2), where a is a real number?> The sqrt(a^2) is a or -a.There is no technical term which you are capable of using properly.Footnotes: [1] In fact, I don't know why you use the term operator hereinstead of function. Well, yes I do. You don't know what anoperator is as opposed to a function (and you don't know what afunction is, either), so you arbitrarily select one term or theother. Or, more likely, you select the one that sounds cooler.-- Jesse F. HughesMy baby don't allow me in the kitchen and I've come to love her decision. -- Bad Livers === ... > My latest de?ition is a bit more > elaborate, but I think it is fool-proof (though probably not James-proof):My error, I omitted the 5's here, I insert them now: > v1(x) = GCD(5 a_1(x) + 7, 49) > v2(x) = GCD(5 a_2(x) + 7, 49) > v3(x) = GCD(5 a_3(x) + 7, 49) > k3(x) = v1(x)*v2(x)*v3(x) ; can be a multiple of 49 > g(x) = k3(x) / 49 ; the excess, must be distributed > k2(x) = GCD(v2(x), g(x)) ; the part in v2 of the excess > k1(x) = g(x) / k2(x) ; and the remainder in v1. > z3(x) = v3(x) ; this one is plain > w2(x) = v2(x) / k2(x) ; part of the excess removed > w1(x) = v1(x) / k1(x) ; the remaining excess removed > u(x) = w1(x)*w2(x)*z3(x)/49 ; must be a unit > w3(x) = z3(x) / u(x) ; force it off. > More elaborate than I would have guessed.Most of the eleboration comes because I wanted to disambiguate withthe occurrence of units. I see it still fails to do so... But atleast with this de?ition: w1(x)*w2(x)*w3(x) = 49without any unit involved.My steps in getting these de?itions. First I need GCD's with respectto 49, not to 7. The reason that for general integer ?x', 7 can bewritten as the product p*q*r with mutually co-prime p, q and r. Thethree factors in that case are (p*q)^2, (p*r)^2 and (q*r)^2. A GCDwith 7 can not show such factors.Second, the product can certainly be a multiple of 49. This is thecase when the polynomial value is a multiple of 49, and that occurseven for particular values of x.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > ...> Here is another way to view this. You say there is only> one way to factor 49 which makes sense. Dik and others> here say there is another. It is de?ed by:> w1(x) = GCD(a_1(x), 49)> w2(x) = GCD(a_2(x), 49)> w3(x) = GCD(a_3(x), 49), You must be careful. Strange enough, but with this de?ition it is not> certain that w1(x)*w2(x)*w3(x) = 49. My latest de?ition is a bit more> elaborate, but I think it is fool-proof (though probably not James-proof):> v1(x) = GCD(a_1(x) + 7, 49)> v2(x) = GCD(a_2(x) + 7, 49)> v3(x) = GCD(a_3(x) + 7, 49)> k3(x) = v1(x)*v2(x)*v3(x) ; can be a multiple of 49> g(x) = k3(x) / 49 ; the excess, must be distributed> k2(x) = GCD(v2(x), g(x)) ; the part in v2 of the excess> k1(x) = g(x) / k2(x) ; and the remainder in v1.> z3(x) = v3(x) ; this one is plain> w2(x) = v2(x) / k2(x) ; part of the excess removed> w1(x) = v1(x) / k1(x) ; the remaining excess removed> u(x) = z1(x)*z2(x)*z3(x)/49 ; must be a unit> w3(x) = z3(x) / u(x) ; force it off. Dik, Here is another way to specify w1(x), w2(x), and w3(x). Use the Magidin-Mckinnon theorem to factor P(x)/49 in the form P(x)/49 = (c1*5 + d1)*(c2*5 + d2)*(c3*5 + d3),where c1, c2, c3, d1, d2, and d3 are all algebraic integers. Note that this is a factorization as a polynomial in 5,not of course as a polynomial in x. Note that d1*d2*d3 = 7. Note that c1 and d1 are coprime, as are c2, d2 and c3, d3. This is true because P(x)/49 is primitive. Now de?e w1 = d2*d3, w2 = d1*d3, and w3 = d1*d2. Now go back to the Harris factorization of P(x) itself: P(x) = (a1*5 + 7)*(a2*5 + 7)*(a3*5 + 7). Clearly 7 = d1*w1 = d2*w2 = d3*w3. Now you know that -7/a1 and -d1/c1 are roots of the same polynomial; i.e. a1 = 7*c1/d1 = c1*(7/d1) = c1*d2*d3 = c1*w1. Similarly, a2 = c2*w2 and a3 = c3*w3. Finally, note that w1 = gcd(a1, 7) = gcd(c1*d2*d3, d1*d2*d3). Similarly, w2 = gcd(a2, 7), and w3 = gcd(a3, 7). Nora B. === > ... > Here is another way to view this. You say there is only > one way to factor 49 which makes sense. Dik and others > here say there is another. It is de?ed by: > > w1(x) = GCD(a_1(x), 49) > w2(x) = GCD(a_2(x), 49) > w3(x) = GCD(a_3(x), 49), > You must be careful. Strange enough, but with this de?ition it is not > certain that w1(x)*w2(x)*w3(x) = 49. My latest de?ition is a bit more > elaborate, but I think it is fool-proof (though probably not James-proof): > v1(x) = GCD(a_1(x) + 7, 49) > v2(x) = GCD(a_2(x) + 7, 49) > v3(x) = GCD(a_3(x) + 7, 49) > k3(x) = v1(x)*v2(x)*v3(x) ; can be a multiple of 49 > g(x) = k3(x) / 49 ; the excess, must be distributed > k2(x) = GCD(v2(x), g(x)) ; the part in v2 of the excess > k1(x) = g(x) / k2(x) ; and the remainder in v1. > z3(x) = v3(x) ; this one is plain > w2(x) = v2(x) / k2(x) ; part of the excess removed > w1(x) = v1(x) / k1(x) ; the remaining excess removed > u(x) = z1(x)*z2(x)*z3(x)/49 ; must be a unit > w3(x) = z3(x) / u(x) ; force it off. > Here is another way to specify w1(x), w2(x), and w3(x). > Use the Magidin-Mckinnon theorem to factor P(x)/49 in > the form > P(x)/49 = (c1*5 + d1)*(c2*5 + d2)*(c3*5 + d3), > where c1, c2, c3, d1, d2, and d3 are all algebraic integers.Of course that can also be done. But I thought that reasonablystraightforward de?itions (in terms of gcd) would work better.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > | Here is another way to view this. You say there is only> |one way to factor 49 which makes sense. Dik and others> |here say there is another. It is de?ed by:> |> | w1(x) = GCD(a_1(x), 49)> | w2(x) = GCD(a_2(x), 49)> | w3(x) = GCD(a_3(x), 49),> |> |where a_1(x), a_2(x), and a_3(x) are the roots of your> |auxiliary polynomial,> |> | a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x).> |> |> | Note that w1, w2, and w3 are perfectly well-de?ed: Keep in mind that the GCD is de?ed only up to multiplication> by units. |the > |roots a_1, a_2, and a_3 exist and can be computed, and the > |GCD function exists (by a deep theorem of Dedekind) and can > |be computed. For example, if x=1,> a_1(1) = -138.210434458... = 6375.47596375... * -0.0216784496160...> a_2(1) = -31.3299394631... = 998.524028548... * -0.0313762499122...> a_3(1) = 25.5403739215... = 0.00000769706133063... * 3318198.0530... where 6375.47596375..., 998.524028548..., and 0.00000769706133063...> are the three roots of t^3-7374t^2+6366066t-49=0, and they multiply> together to give 49,Interesting. Are you claiming to be supporting Dik Winter's claims?I assume that from how you've posted but I want you to be very clearabout your purpose in making your post.If you are stepping forward to try and support him at some speci?values, I hope you don't mind if I ask how you're making yourcalculations.Also, I now need you to do x=7, which I ask for because 7 is a factorof 49.That makes it a nice test of your methodology. Oh yeah, if you can dothat calculation can you then also say *which* of your a's does nothave algebraic integer factors in common with 7, at x=7?If you claim none do, then I'll elaborate on why I asked thatquestion. > and -0.0216784496160..., -0.0313762499122..., 33189198.0530... are the> three roots of t^3-3318198t^2-176046t-2257=0. Since the polynomials> t^3-7374t^2+6366066t-49 and t^3-3318198t^2-176046t-2257 are monic> polynomials with integer coef?ients, their roots are of course> algebraic integers. That was sort of fun; let me try to do another. If x=2, we get (ugh,> I'll mark places where a number is continued on the next line with > like in C code) > a_1(2) = -279.300354441...> = 1016597193845414091969293355102867957348> 409.048415...> *-2.74740434197... * 10^{-40} a_2(2) = -63.3122938187...> = 3574859564584023060416081070013097699574> 371117492591974367421061.951584...> *-1.77104282489... * 10^{-62} a_3(2) = 51.6126482600...> = 1.34830513302... * 10^{-104}> * 3827964975879016817625368936818067843946> 6426613668209951263900140296253120450185> 91493808304178435225834041.000... where the ?st factors in each of these is a root of x^3> - (3574859564584023060417097667206943113666> 340410847694842324769471) x^2> + (3634192201747556708576802697957676247053> 7456650480245362950679011731799555936769> 95172533270978579455876911) x> - 49 (and the three roots obviously are algebraic integers multiplying> together to give 49) and the other factor in each is a root of x^3 > - (3827964975879016817625368936818067843946> 6426613668209951263900140296253120450185> 91493808304178435225834041) x^2> - (1051696759566469898052937474234752487358> 853239615505188006778565836) x> - 18626. The last root a_3(2)/w_3(2)=20995126...4041.000... is a little> surprising to me, since it appears to be very close to being an> integer, although actually irrational. About 39 zeros after the> decimal point, apparently. Presumably there's a good reason for> that.Now there's a test to see. Do x=7. If it's too much for yourprogram, try x = sqrt(7). If that's too much for it, try x = 7^{1/3},which should be small enough.Oh yeah, now there's one other cubic you should be able to give aswell, which is the cubic whose roots are a_1(x)/w_1(x), a_2(x)/w_2(x),and a_3(x)/w_3(x). Note that so far I haven't run afoul of nonreal roots, or of> nonunique factorization in the algebraic integers of the form> r1*a^2+r2*a+r3, where r1,r2,r3 are rational numbers. That could> blow up the numbers one has to deal with even bigger! Please excuse any typos I may have made.Ok.James Harris === |Interesting. Are you claiming to be supporting Dik Winter's claims?I have only been skimming his postings, so I don't know exactly whathe's been saying. I don't want to say anything so open-ended assupporting his claims.I'm more interested in knowing what you make of this data. What doyou think the numbers are telling us?|I assume that from how you've posted but I want you to be very clear|about your purpose in making your post.I thought a numerical example might help to clear the air.|If you are stepping forward to try and support him at some speci?|values, I hope you don't mind if I ask how you're making your|calculations.I don't mind your asking... I hope you don't mind if I enjoy playingoracle for awhile, though. I'm sort of curious whether anybody elsewill bother to ?ure out how to do it on their own. Perhaps people willbe able to ?ure out how I might have done it. Think of it as anexercise for the reader. :-)|Also, I now need you to do x=7, which I ask for because 7 is a factor|of 49.Need? I think the word is want.|That makes it a nice test of your methodology. Oh yeah, if you can do|that calculation can you then also say *which* of your a's does not|have algebraic integer factors in common with 7, at x=7?||If you claim none do, then I'll elaborate on why I asked that|question.Each of the three roots has a common factor with 7 in the algebraicintegers (that is not a unit).[...]|Now there's a test to see. Do x=7. If it's too much for your|program, try x = sqrt(7). If that's too much for it, try x = 7^{1/3},|which should be small enough.Roots of 7 would be harder to compute with than just 7. The problem isnot the *size* of the numbers; it's the *complexity*.|Oh yeah, now there's one other cubic you should be able to give as|well, which is the cubic whose roots are a_1(x)/w_1(x), a_2(x)/w_2(x),|and a_3(x)/w_3(x).I think I had those in the last posting. Look again.Let's see, x=7:Unique factorization in the algebraic integers of Q(a_1(7)) fails...with class group of order 6. It will be easier to explain this if westart by dealing with the cubes of a_1(7), a_2(7) and a_3(7):a_1(7)=-984.74975149205592391...a_1^3(7)=- 954943417.96332121170...=w_1(7)^3 * (a_1(7)/w_1(7))^3=-1.7091596327006183680...*10^155 * 5.5872102271361801072...*10^(-147)w_1(7)^3 is rather close to being an integer, apparently. I must befailing to observe something here. In fact, it's really close to beingthe x^2 coef?ient of the ?st big polynomial below. (Why shouldtwo of them be so small?)a_2(7)=-223.22464971658329045...a_2(7)^3=- 11123115.591359682514...=w_2(7)^3 * (a_2(7)/w_2(7))^3= 7.0638530147534189164...*10^(-107) * -1.5746527522767207658...*10^113a_3(7)=181.974401208639214363 ...a_3(7)^3=6026024.5546808942165...=w_3(7)^3 * (a_3(7)/w_3(7))^3=-9.7445987101782209629...*10^(-45) * -6.1839637874330517387...*10^50where the w_1(7)^3, w_2(7)^3, and w_3(7)^3 are roots of: x^3 + x^2 * ( 1709159632700618 36801252617176736836 46811355105204799369 18092316426418976142 36566985178805793369 66540228556627452242 49189593582663638654 39584418358954402805) + x * ( 166550747523 03127640659037846440 40807150639772482245 86082618369089666087 03982611546880953070 33210044772061027411) - 117649(note 117649=7^6) and the (a_1(7)/w_1(7))^3 and so on are roots of: x^3+ x^2 * ( 15746527522767 20765848800802473662 60416442233792419056 74462198398676566043 46711000686190804491 48175997276599544325)+ x * ( 9737 59559786102913134816 03375452230999016288 72460051697150414298 55549052793507337905 69636130521308595124 61366728332214047662 01236195021069167474 28909340928079783347)- 544059937120853881It's unfortunate that the coef?ients are so big this time, but I don'tsee any way around it.Now we can use the fact that cube roots of algebraic integers are alsoalgebraic integers: if a number is a root of x^3+Ax^2+Bx+C, then itscube root is a root of x^9+Ax^6+Bx^3+C. We can let the w_1(7), w_2(7)and w_3(7) be the cube roots of the values given above. (The real cuberoots are okay.) The product of the values above is 7^6, so when we takecube roots, we get values multiplying together to w_1*w_2*w_3=7^2=49.The w_1, w_2, and w_3 are GCDs of a_1, a_2, and a_3 with 7,incidentally, even though they multiply together to give 49.Keith Ramsay === > |Interesting. Are you claiming to be supporting Dik Winter's claims?> I have only been skimming his postings, so I don't know exactly what> he's been saying. I don't want to say anything so open-ended as> supporting his claims.> I'm more interested in knowing what you make of this data. What do> you think the numbers are telling us?> |I assume that from how you've posted but I want you to be very clear> |about your purpose in making your post.> I thought a numerical example might help to clear the air.> |If you are stepping forward to try and support him at some speci?> |values, I hope you don't mind if I ask how you're making your> |calculations.> I don't mind your asking... I hope you don't mind if I enjoy playing> oracle for awhile, though. I'm sort of curious whether anybody else> will bother to ?ure out how to do it on their own. Perhaps people will> be able to ?ure out how I might have done it. Think of it as an> exercise for the reader. :-)> |Also, I now need you to do x=7, which I ask for because 7 is a factor> |of 49.> Need? I think the word is want.Yeah, when did James become a dictator? You would think a dictator wouldknow something about his ?ld.> |That makes it a nice test of your methodology. Oh yeah, if you can do> |that calculation can you then also say *which* of your a's does not> |have algebraic integer factors in common with 7, at x=7?> |> |If you claim none do, then I'll elaborate on why I asked that> |question.> Each of the three roots has a common factor with 7 in the algebraic> integers (that is not a unit).> [...]> |Now there's a test to see. Do x=7. If it's too much for your> |program, try x = sqrt(7). If that's too much for it, try x = 7^{1/3},> |which should be small enough.> Roots of 7 would be harder to compute with than just 7. The problem is> not the *size* of the numbers; it's the *complexity*.> |Oh yeah, now there's one other cubic you should be able to give as> |well, which is the cubic whose roots are a_1(x)/w_1(x), a_2(x)/w_2(x),> |and a_3(x)/w_3(x).> I think I had those in the last posting. Look again.> Let's see, x=7:> Unique factorization in the algebraic integers of Q(a_1(7)) fails...> with class group of order 6. It will be easier to explain this if we> start by dealing with the cubes of a_1(7), a_2(7) and a_3(7):> a_1(7)=-984.74975149205592391...> a_1^3(7)=-954943417.96332121170...> =w_1(7)^3 * (a_1(7)/w_1(7))^3> =-1.7091596327006183680...*10^155 * 5.5872102271361801072...*10^(-147)> w_1(7)^3 is rather close to being an integer, apparently. I must be> failing to observe something here. In fact, it's really close to being> the x^2 coef?ient of the ?st big polynomial below. (Why should> two of them be so small?)> a_2(7)=-223.22464971658329045...> a_2(7)^3=-11123115.591359682514...> =w_2(7)^3 * (a_2(7)/w_2(7))^3> = 7.0638530147534189164...*10^(-107) * -1.5746527522767207658...*10^113> a_3(7)=181.974401208639214363...> a_3(7)^3=6026024.5546808942165...> =w_3(7)^3 * (a_3(7)/w_3(7))^3> =-9.7445987101782209629...*10^(-45) * -6.1839637874330517387...*10^50> where the w_1(7)^3, w_2(7)^3, and w_3(7)^3 are roots of:> x^3> + x^2 * ( 1709159632700618> 36801252617176736836> 46811355105204799369> 18092316426418976142> 36566985178805793369> 66540228556627452242> 49189593582663638654> 39584418358954402805)> + x * ( 166550747523> 03127640659037846440> 40807150639772482245> 86082618369089666087> 03982611546880953070> 33210044772061027411)> - 117649> (note 117649=7^6) and the (a_1(7)/w_1(7))^3 and so on are roots of:> x^3> + x^2 * ( 15746527522767> 20765848800802473662> 60416442233792419056> 74462198398676566043> 46711000686190804491> 48175997276599544325)> + x * ( 9737> 59559786102913134816> 03375452230999016288> 72460051697150414298> 55549052793507337905> 69636130521308595124> 61366728332214047662> 01236195021069167474> 28909340928079783347)> - 544059937120853881> It's unfortunate that the coef?ients are so big this time, but I don't> see any way around it.> Now we can use the fact that cube roots of algebraic integers are also> algebraic integers: if a number is a root of x^3+Ax^2+Bx+C, then its> cube root is a root of x^9+Ax^6+Bx^3+C. We can let the w_1(7), w_2(7)> and w_3(7) be the cube roots of the values given above. (The real cube> roots are okay.) The product of the values above is 7^6, so when we take> cube roots, we get values multiplying together to w_1*w_2*w_3=7^2=49.> The w_1, w_2, and w_3 are GCDs of a_1, a_2, and a_3 with 7,> incidentally, even though they multiply together to give 49.> Keith Ramsay>David Moran === |w_1(7)^3 is rather close to being an integer, apparently. I must be|failing to observe something here. In fact, it's really close to being|the x^2 coef?ient of the ?st big polynomial below. (Why should|two of them be so small?)[...]|It's unfortunate that the coef?ients are so big this time, but I don't|see any way around it.D'oh! I plan to multiply them by appropriate units later....Keith Ramsay === > Interesting. Are you claiming to be supporting Dik Winter's claims? I assume that from how you've posted but I want you to be very clear> about your purpose in making your post. If you are stepping forward to try and support him at some speci?> values, I hope you don't mind if I ask how you're making your> calculations. Also, I now need you to do x=7, which I ask for because 7 is a factor> of 49. That makes it a nice test of your methodology. Oh yeah, if you can do> that calculation can you then also say *which* of your a's does not> have algebraic integer factors in common with 7, at x=7? If you claim none do, then I'll elaborate on why I asked that> question.JSH, as usual, misses the point. JSH has made a mathematical claim. The burden of proof for that mathematicalclaim is entirely on him. JSH keeps trying to shift that burden onto those who question his claim by refusing to deal with their mathematical objections with mathematical answers.The crux of the objections seem to center on which factorizations of 49, into 3 factors over the ring of algebraic integers, are relevant in a certain factorization of a cubic polynomial.JSH claims 49 = 7*7*1 is the only relevant factorization. Several others, dispute that claim and have provided detailed mathematical examples and analyses showing that 7*7*1 is generally not appropriate.JSH ignores the mathematical content ot these objections, relying only on agumentum ad hominem.Thus JSH demonstrates a remarkable consistency: this has been his ?st approach to every probing of his claims for well over 7 years now.Excuse me for putting this on other Math newsgroups also; only doing it forbetter points of view or reliability of information:Is it fair in studying Matrices and systems of linear equations from a collegealgebra textbook to resort to a graphing calculator to do the various problemswhich use 3 x 3 or higher matrices? When I follow the examples in the book, Iunderstand well about the matrix operations; WHEN I WORK THEM BY HAND, if theyare more than 2 x 2, they take a long time to do, and most of the work containserrors that are dif?ult to ?d. When I redo them, AGAIN BY HAND, I makeother errors which are just as dif?ult to ?d. Still, I understand theexample problems, and some of the exercises I handle successfully. When I usea graphing calculator on which I have been learning how to enter matrixelements and invoke inverse-matrix and reduced row echelon functions, thecorrect results are obtained. Is it common today in college courses of algebra involving the learning ofsystems of linear equations or matrices to let the students use a graphingcalculator during classtime on quizes and tests? If a matrix is 3 x 3 orlarger, it can become very time-consuming and as described, many errors canoccur doing manual problem solving. How is all this managed in instruction? Certainly, understanding the concepts and showing that one can perform themanual skill is important; just that for only the simplest problems time is notheavily taxed. For other problems, so much time is needed for an assessmentproblem.G C === > Suppose f is a representation of the group G in the vector space V over k.> Then f* is the representation of G in the dual vector space V* de?ed by> the action (gp)(v) = p((g^-1) v), where g is in G, p is in V*, and v is in> V. (Firstly, why is this action de?ed like this?) Now suppose f is given in matrix form : f : g |--> A(g) where g is in G and A(g) denotes the matrix g is sent to. Then f* (the dual representation) is the representation g |---> (A(g)^{-1})^T, the transpose of the inverse of the matrix. Can> someone tell me why the transpose of the inverse de?es the dual> representation?> Also, I wonder if this motivation might help you. Suppose one has a diffeomorphism between n-dimensional manifolds M and N:f:M->NThis induces a map on the tangent spaces:f:TM->TNf(v)(x) = Df v(f(x))where Df is the Jacobian of f, that is, the matrix of partial derivatives. One also has a map on the cotangent spaces:f*:T*N->T*Mf*(p)(f(x)) = (Df)^T p(x)which is the same as a map g:T*M->T*N given byg(p)(x) = (Df)^{T,-1} p(f(x)),that is, the same formula as for the dual representation. (If you think of the case M=N, then you could think of this as a representation of Diff(M) on TM.)Best, Stephen === >Suppose f is a representation of the group G in the vector space V over k.>Then f* is the representation of G in the dual vector space V* de?ed by>the action (gp)(v) = p((g^-1) v), where g is in G, p is in V*, and v is in>V. (Firstly, why is this action de?ed like this?)>We have f:G->L(V), where L(V) denotes the linear operators on V. We de?e>f*:G->L(V*) so that> = where = p(v) de?es the duality between p in V* and v in V. This gives> Why exactly do we want to de?e f*:G->L(V*) so that = ? Why> does this de?ition make sense? Is < > the standard inner product? What> do you mean by duality between p in V* and v in V?> You want to have some sensible de?ition. This is as good as any.< > will be the standard inner product if V and V^ are R^n in the usual way. It is quite common to write = p(v) - it is notation, nothing else. The word duality was just a word I through in there which really means nothing except === >Suppose f is a representation of the group G in the vector space V over> k.>Then f* is the representation of G in the dual vector space V* de?ed> by>the action (gp)(v) = p((g^-1) v), where g is in G, p is in V*, and v is> in>V. (Firstly, why is this action de?ed like this?)>We have f:G->L(V), where L(V) denotes the linear operators on V. We> de?e>f*:G->L(V*) so that> = where = p(v) de?es the duality between p in V* and v in V. This> gives> Why exactly do we want to de?e f*:G->L(V*) so that = ?Why> does this de?ition make sense? Is < > the standard inner product?What> do you mean by duality between p in V* and v in V?> You want to have some sensible de?ition. This is as good as any.> < > will be the standard inner product if V and V^ are R^n in the usualway. It> is quite common to write = p(v) - it is notation, nothing else. Theword> duality was just a word I through in there which really means nothingexcept>So then the de?ition where we must satisfy = means the sameas(gp)(gv) = p(v). i.e. gp is a map in the dual which is applied to gv, avector in V.I just don't really see why this de?ition is made in the ?st place, i.e.why it is the most sensible de?ition, but maybe I will understandit better when I work through some examples?According to Fulton/Harris, The dual V* of V is also a representation,though not in the most obvious way: we want the two representations of G torespect the natural paring denoted < , > between V* and V, so that if f :G --> GL(V) is a rep and f* : G --> GL(V*) is the dual rep, we should have = . This in turn FORCES us to de?e the dualrep by f*(g) = (f(g^-1))^T : V* --> V* for all g in G. forces? In that as= A(g)^{-1,T}?