mm-3029 === Subject: Splitting [0,1] in half n many times. Consider the closed interval [0,1]. Now split it in half, and throw away the rightmost interval. We're left with [0,1/2]. Split this interval in half, but this time keep the rightmost interval. Now we're left with [1/4, 1/2]. Continue this alternating sequence as follows... C0=[0,1] C1=[0,1/2] C2=[1/4,1/2] C3=[1/4,3/8] ect. The question is; is there any number that will exist in every interval Cn as n goes to infinity? In other words, are there any numbers in the intersection of all sets Cn? I went all the way to C13. Every left endpoint is less than 1/3, and every right endpoint is greater than 1/3. I believe that 1/3 is the only real number that will belong to Cn for all n, but I can't prove it. As n goes to infinity, the length of the interval Cn will shrink down to zero. This means that as n goes to infinity, the left and right endpoints of Cn will become the same value. In this way the endpoints are bounded by each other. It is also easy to show that Sup {Cn} is always decreasing, and Inf {Cn} is always increasing. Because the endpoints are monotonic and bounded by each other, they must share a limit. I think this limit is 1/3. How do I prove it? === Subject: Re: Splitting [0,1] in half n many times. On Wed, 04 Jan 2006 11:45:11 EST, David Angeli >Consider the closed interval [0,1]. Now split it in half, and throw away the rightmost interval. We're left with [0,1/2]. Split this interval in half, but this time keep the rightmost interval. Now we're left with [1/4, 1/2]. Continue this alternating sequence as follows... >C0=[0,1] >C1=[0,1/2] >C2=[1/4,1/2] >C3=[1/4,3/8] >ect. >The question is; is there any number that will exist in every interval Cn as n goes to infinity? In other words, are there any numbers in the intersection of all sets Cn? Yes. In fact if you do the same thing except at each stage choose to keep the left half or the right half at random (or by any rule whatever, not necessarily alternating left and right) then there will be a number in the intersection of all the Cn. Proof (or sketch of proof, depending on what you know): Say Cn is the interval [a_n, b_n]. Then the a_n are a non-decreasing sequence, and a_n <= 1 for every n, so the a_n converge to something, say a_n -> x. Now fix k. It's clear that a_n < b_k for every n and k, and hence if you take the limit as n -> infinity, with k fixed, you see that x <= b_k for every k. So a_k <= x <= b_k for every k, showing that x is in every Ck. QED. (This is a special case of an important fact about compact sets: If each C_n is a compact set and the intersection of any finite number of C_n's is nonempty then the intersection of all the C_n's is nonempty.) >I went all the way to C13. Every left endpoint is less than 1/3, and every right endpoint is greater than 1/3. I believe that 1/3 is the only real number that will belong to Cn for all n, but I can't prove it. >As n goes to infinity, the length of the interval Cn will shrink down to zero. This means that as n goes to infinity, the left and right endpoints of Cn will become the same value. In this way the endpoints are bounded by each other. It is also easy to show that Sup {Cn} is always decreasing, and Inf {Cn} is always increasing. Because the endpoints are monotonic and bounded by each other, they must share a limit. I think this limit is 1/3. How do I prove it? ************************ David C. Ullrich === Subject: Re: Splitting [0,1] in half n many times. > Consider the closed interval [0,1]. Now split it in half, and throw away the > rightmost interval. We're left with [0,1/2]. Split this interval in half, > but this time keep the rightmost interval. Now we're left with [1/4, 1/2]. > Continue this alternating sequence as follows... > C0=[0,1] > C1=[0,1/2] > C2=[1/4,1/2] > C3=[1/4,3/8] > ect. > The question is; is there any number that will exist in every interval Cn as > n goes to infinity? In other words, are there any numbers in the > intersection of all sets Cn? Let Nth interval be [L(N), R(N)]. Since interval is halved every time, R(N) - L(N) = 1/2^N L(0) = 0 L(2N+1) = L(2N) L(2N) = L(2N-1) + ((R(2N-1) - L(2N-1))/2) = L(2N-2) + 1/2^2N = L(2(N-1)) + 1/4^N series that L(2N) = L(2N+1) = 1/3 * (1 - 1/4^N) R(0) = 1 R(2N-1) = R(2N) R(2N+1) = R(2N) - ((R(2N) - L(2N)) / 2) = R(2N-1) - 1/2^(2N+1) = R(2(N-1)+1) - 1/(2*4^N) series that R(2N) = R(2N-1) = 1 - 2/3 * (1 - 1/4^N) = 1/3 * (1 + 2/4^N) It's pretty clear from here that L(N) is monotonically non-decreasing and converges to 1/3 and that R(N) is monotonically non-increasing and also converges to 1/3. hth Ben -- If this message helped you, consider buying an item from my wish list: I changed my name: === Subject: Re: Splitting [0,1] in half n many times. > Consider the closed interval [0,1]. Now split it in half, and throw away > the rightmost interval. We're left with [0,1/2]. Split this interval in > half, but this time keep the rightmost interval. Now we're left with > [1/4, 1/2]. Continue this alternating sequence as follows... > C0=[0,1] > C1=[0,1/2] > C2=[1/4,1/2] > C3=[1/4,3/8] > ect. > The question is; is there any number that will exist in every interval Cn > as n goes to infinity? In other words, are there any numbers in the > intersection of all sets Cn? > I went all the way to C13. Every left endpoint is less than 1/3, and > every right endpoint is greater than 1/3. I believe that 1/3 is the only > real number that will belong to Cn for all n, but I can't prove it. > As n goes to infinity, the length of the interval Cn will shrink down to > zero. This means that as n goes to infinity, the left and right endpoints > of Cn will become the same value. In this way the endpoints are bounded > by each other. It is also easy to show that Sup {Cn} is always > decreasing, and Inf {Cn} is always increasing. Because the endpoints are > monotonic and bounded by each other, they must share a limit. I think > this limit is 1/3. How do I prove it? I'd do something like this. Lets first get rid of this horribly complicated number system we use today and switch to the much simpler binary numbers. :) Every binary number in [0,1] can be written as 0. b_1 b_2 b_3 .... which is equal to b_1 * (1/2) + b_2 * (1/4) + b_3 * (1/8) + ...., an ininite series. When we throw away all the numbers above 0.1 we're left with only the numbers that are smaller than 0.1, note that all of those have b_1 = 0. Now we throw away the right half and are left with the numbers that start with 0.01. If we keep this up we will end up with the binary number 0.0101010101010101.... = 0.01 * 1.01010101....) if we switch back to decimal number this is easily calculated to be (1/4) * (sum{i=0..inf} 1/(2^i)) = (1/4) * (1 / (1 - 1/4)) = 1/3 /T === Subject: Vector Spaces Reading the definition of a vetor space it makes me wonder why can't you define vector spaces over rings. For example, why can't you define a vector space over a ring I (integers)? Can anyone tell my why you can't have it defined over a ring as long as it has multiplicative associativity and a multiplicative identity? Usually deffinitions are as general as possilble. What am I missing? A vector space over a field F (such as the field of real or of complex numbers) is a set V together with two operations: * vector addition: defined on the Cartesian product V .81~ V with values in V and denoted v + w, where v, w .81ü V, and * scalar multiplication: defined on the Cartesian product F .81~ V with values in V and denoted a v, where a .81ü F and v .81ü V. which satisfy the following axioms (for all a, b .81ü F and u, v, and w .81ü V): 1. Vector addition is associative: u + (v + w) = (u + v) + w. 2. Vector addition is commutative: v + w = w + v. 3. There exists an additive identity element 0 in V, such that for all elements v in V, v + 0 = v. 4. For all v in V, there exists an element w in V, such that v + w = 0. 5. Scalar multiplication is associative: a(bv) = (ab)v. 6. 1 v = v, where 1 denotes the multiplicative identity in F. 7. Scalar multiplication distributes over vector addition: a(v + w) = a v + a w. 8. Scalar multiplication distributes over scalar addition: (a + b)v = a v + b v. === Subject: Re: Vector Spaces >Reading the definition of a vetor space it makes me wonder why can't >you define vector spaces over rings. For example, why can't you define >a vector space over a ring I (integers)? As other posters have pointed out already, you can define a 'vector space' over a ring, but it's called a module. The specific example you ask about (modules over the integers) is particularly interesting, as finitely generated modules over Z turn out to be the same thing as finitely generated abelian groups. >Usually deffinitions are as general as possilble. What am I missing? You're not really missing anything, but maybe coming at it from the wrong angle. Mathematical definitions tend to be quite abstract, so that they are not tied to any particular system, and so we can get general results. But some results that are true with a restricted definition will not be true for a more general one. For example, the definition of a ring is more general than the definition of a field (every field is a ring, but not vice-versa), but both definitions are useful - the ring one because there are plenty of interesting results that we can prove for all rings, and the field one because there are also interesting results that are only true for fields. In fact, mathematics is full of examples of hierarchies of progressively stricter definitions just like module -> vector space and ring -> field, e.g. semi-group -> monoid -> group -> abelian group. Joe. === Subject: Re: Vector Spaces > Reading the definition of a vetor space it makes me wonder why can't > you define vector spaces over rings. You can have something very much like vector spaces over arbitrary rings, but they are called /modules/ instead, and can behave quite differently from vector spaces. === Subject: Re: Vector Spaces On 4 Jan 2006 16:34:00 -0800, solarmist in alt.math.undergrad: > Reading the definition of a vetor space it makes me wonder why can't > you define vector spaces over rings. You can, but such objects are called modules. Vector spaces are special cases of modules; see . Historically, of course vector spaces came first. [...] Brian === Subject: Fermat's Method of Infinite Descent This is regarding my last lesson in the number theory correspondence course. The exercises are too involved to place on USENET so I made a web page: http://www.geocities.com/coalition_for_national_day_care/m328k/ I'm still on Exercise 1 though it seems to me that the descent given is for the equations in (13) and while the proof does create another version of (5), it fails to set up the needed requirements to descend again using (13). Plus, there are solutions to (5) so it is really a moot point. As for Exercise 2, I'm stumped. The idea, I guess, is to generate two primitive Pythagorean triplets, say: t^2 = r^2 + s^2 w^2 = u^2 + v^2 but... I'm not sure about that or how to come up with that. -e === Subject: Re: Fermat's Method of Infinite Descent On 05 Jan 2006 00:50:29 +0000, lizzy2 alt.math.undergrad: > This is regarding my last lesson in the number theory > correspondence course. The exercises are too involved to place on > USENET so I made a web page: > http://www.geocities.com/coalition_for_national_day_care/m328k/ > I'm still on Exercise 1 though it seems to me that the descent > given is for the equations in (13) and while the proof does create > another version of (5), it fails to set up the needed requirements to > descend again using (13). Yes. Can you be more precise about exactly what fails here? A specific example would be good; you can produce one by starting with 3^2 + 4^2 + 12^2 = 13^2, which *can* be rewritten as in (13), and carrying out the descent procedure. What solution of (5) do you get? Can it be written as in (13)? > Plus, there are solutions to (5) so it is really a moot > point. No, because the question deals with the logic, not the result itself. > As for Exercise 2, I'm stumped. The idea, I guess, is to generate > two primitive Pythagorean triplets, say: > t^2 = r^2 + s^2 > w^2 = u^2 + v^2 > but... I'm not sure about that or how to come up with that. I can't tell from this whether you've correctly understood what the question asks. It deals with the passage from the text that starts 'Further, it should be clear how to go in the other direction' and ends 'Thus (13) has infinitely many solutions'. If you examine the procedure illustrated in that paragraph, you'll see that it starts with a solution to (5), say a^2 + b^2 + c^2 = d^2. You then set u = d and pick one of a, b, and c to be v; how? The other two of a, b, and c have to be r and s, but which is which? The method then sets x = r^2 - s^2, y = 2rs, t = r^2 + s^2 = u^2 - v^2, z = 2uv, and w = u^2 + v^2 to get the solution x^2 + y^2 + z^2 = w^2, with x^2 + y^2 = t^2. The problem is that unless you assign v, r, and s correctly to a, b, and c, you might not end up with (x,y,t) and (t,z,w) both equal to 1. You may need to look at the examples in that passage (and perhaps a few more) to see how the assignment should be made and why it's always possible to make it properly. [...] Brian === Subject: Re: relation b/n poem and calculus On Tue, 3 Jan 2006 02:39:15 -0500, Stan Brown >Mon, 02 Jan 2006 19:38:40 -0800 from Bob : >> But, writing in the journal Nature, Leonardo Ricci, of the >> University of Trento, northern Italy, says Dante spotted the same >> thing early in the 14th century. He did not pursue the logic but >> did describe it in canto 17 of his epic work Inferno. >> http://tinyurl.com/4d59g >It's not unusual for a non-scientist or an early scientist to >propound some doctrine that later science shows to be correct. We >tend to focus on that and overlook all the other doctrines that later >science shows to be incorrect. >For instance, Dante taught that the earth was hollow (or at least >that a huge spherical sector of it was hollow, right down to the >center). Within the realm of poetry and metaphor, it was a brilliant >figure to make the circles of Hell that way. But of course as physics >it's nonsense. You make a good point. I tend to take such findings as fun. But it something to it. He might respond to your point that describing the sensation of flying correctly is more significant than the description of the earth.89s innards. There are a couple of other points of possible interest. One is whether Galileo was -- consciously or otherwise -- influenced by Dante's earlier report. (The author suggests that he was aware of it.) The other is whether the point is a natural one to make -- though it took until Galileo for someone to formalize it as a scientific point. email me privately, preferably at bbruner@berkeley.edu) bob === Subject: Re: discrete mathematic On Sun, 25 Dec 2005 12:08:28 EST, Petr Kalafatic in alt.math.undergrad: > Could some one please help me ? > You are given a 2^n * 2^n checkerboard with > one of the squares missing. > Is possible to solve how many times can be this > checkerboard completely tiled by a L-trominoes > (different ways) ? With a specific square missing, or with any square missing? That is, for n = 1 do you consider the correct number to be 1, the number of ways to tile the 2 x 2 checkerboard leaving a specific square uncovered, or 4, the number of ways to tile it leaving any single square uncovered? If you mean with a specific square missing, I'm not even sure that the answer is the same for all deleted squares. This appears to be a hard problem. I've thought about it a bit in odd moments and don't see any good approach to a general solution. I can tell you that the number grows very rapidly, however, for n > 2. Let t(n) be the number of possible tilings of the 2^n x 2^n checkerboard with the upper lefthand corner cell deleted. The usual inductive proof that the deficient 2^n x 2^n checkerboard can be tiled with L-trominoes shows that t(n+1) >= t(n)^4, since there are t(n) ways to tile each quadrant of the deficient 2^(n+1) x 2^(n+1) board. Clearly t(1) = 1, and it's not hard to verify that t(2) = 1 as well. At n = 3, however, matter change drastically: t(3) is at least 6145. This is obtained as follows. First, note that a 2 x 3 or 3 x 2 block can be covered with two L-trominoes in exactly two ways. Surround the deleted corner square with an L-tromino, and note that there are 60 squares yet to be covered. They can be covered by 2 x 3 and 3 x 2 blocks in at least six different ways. Each uses two 3 x 2 blocks to finish covering the first two columns of the board. Then: (1) Cover the upper right 6 x 6 region with two rows of three 3 x 2 blocks, and cover the remaining 2 x 6 region in the lower right with two 2 x 3 blocks. (2) Cover the upper right and lower right 3 x 6 regions with three 3 x 2 blocks each, and cover the 2 x 6 strip between them with two 2 x 3 blocks. (3) Cover the righthand 8 x 6 region with eight 2 x 3 blocks arranged in four rows of two blocks each. The six ways are (1), (2), (3), and their reflections in the main diagonal (upper left to lower right). In each of these six ways, each of the ten 2 x 3 or 3 x 2 blocks has two distinct tilings by L-trominoes, so each of the six ways gives rise to 2^10 distinct tilings of the deficient 8 x 8 checkerboard with L-trominoes. This gives rise to 6 * 2^10 = 6144 tilings. In addition there is the tiling used in the inductive proof that a tiling exists: in the upper lefthand quadrant it uses the unique tiling of the 4 x 4 board with the upper lefthand corner square missing, and in the other three quadrants it uses rotations of that tiling. Of course it now follows that t(4) >= 4^6145, t(5) >= 4^4^6145, etc. Brian === Subject: Re: Tutte The factorization of linear graphs Do you not have access to any kind of document delivery service? If you are a student at university, undergrad, grad, or other, then you probably have access to one. Ask around. === Subject: Re: How bad is this? > My lazy professor chose this to teach us a point-set > topology course in The Moore Method... if you work hard. > I am already thinking of dropping this course, even > though I really wanted to take a course in topology.. Why? > but thats the deal with cheap universities.. Which university? It sounds as if you're getting a pretty good deal. -- Mike Hardy === Subject: Re: Craps modeled as a Poisson Process >Can the casino game of craps be properly modeled as a Poisson Process? Does >it have all the essential elements to qualify it (or betting within the >game) as a Poisson Process? Perhaps in a certain elementary model. If you count the epochs where either a point is made or the shooter craps out, it might be apporximated by a Poisson process. While the interarrival times are surely independent, it is not clear whether an exponential distribution is an accurate appoximation. Rather, the interarrival time is a geometric sum of independent random variables. (In practice, the time period before the first roll after the establishment of the point is longer than the other inter-roll periods.) In any case, the counting process of (crap out or point made) can be modeled by a renewal process. And you can model the cash-flow process at a table as a regenerative process Just counting the crap-outs/points-made loses a lot of the details about other things going on in the game. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Craps modeled as a Poisson Process <43C47EDA.4000807@netscape.net> It would be easier to approach the problem using a simpler bet like a place 6 or 8 bet. The passline bet is a complex bet as illustrated here: http://img40.imageshack.us/img40/6320/markovcrapspl11ur.gif You would first note that the transitions between states follow an exponential [the cdf for a decision is = 1-e^[-N x Ln(1/(1-p))], where N=# of rolls=t, and lambda=Ln(1/(1-p)), and you could treat each transition (the arrows, or decision) as a seperate poisson arrival process. The superposition, decomposition properties could then be used as desired. If you were to pick a longer interval of hours, then you could say hours/interval>rolls/interval>decisions/interval. For a place 6 or 8 bet, you could consider the winning and losing decisions as independent poisson arrival processes (decomposition). Or, by superposition, as a single decision. I set up the problem I was working out in this manner because I wanted to use the nonstationary and decomposition/superposition properties to adjust tau or time. (It's a long story). === Subject: Re: Craps modeled as a Poisson Process I'm having this argument with a bunch of Crapsters over on another ng. I think that I can explain the basics if you are interested. Pubkey got the basics ok. Catnawe forget to mention that decisions in craps have an exponential cdf. so the answer is yes. They have the exponential, memoryless, independence...etc. properties. You can think of a decision (winning or losing) as a poisson arrival process. === Subject: Re: Yet Another 1= 0 proof > I was under the impression that algebraic simplifications were allowed > within the parenthesis and not subject to the infinity - infinity > problem. For example lim (2*N / N) = lim(2) because the N's cancel, but > that is not infinity / infinity. Yes, but the original post did something like breaking up pieces to create an indeterminate form. (He actually did the reverse of creating it.) ] All limits (lim) are N -> infinity ] ] lim (N+1)/N = 1 = lim(1) ] ] lim (N+1) = lim(N) To get to this point, the OP did the following: lim(N+1)/N = lim(N+1) / lim(N) and then inserted it in the equation: lim(N+1) / lim(N) = lim(1) lim(N+1) = lim(N) * lim(1) = lim(N) Technically, this is wrong, but it's not the source of the error. Then the OP subtracted lim(N) from both sides, which yields an indeterminate form (infinity - infinity) on both sides. lim(N+1) - lim(N) = lim(N) - lim(N). Now comes the manipulation that gives the actual error: saying that lim(f) - lim(g) = lim(f-g) when lim(f) and lim(g) are both +infinity. Then the OP continues. ] lim (N+1 - N) = lim (N - N) ] ] lim (1) = lim(0) ] ] 1 = 0. --- Christopher Heckman === Subject: Re: local extrema - continuity required? >Let h(x) = x, if 0 < x >= 1 + x, if x <= 0 >h has a local maximum at 0, no local minimum at 0. >>h has a local minimum at 0. > No, it doesn't. I mistook g for h. -- Darrell === Subject: RGB to HSL colorspace conversion, wikipedia entry Wikipedia's entry at (http://en.wikipedia.org/wiki/HSL_color_space). However my saturation values do not appear correct. For instance if I input R=250,G=10,B=10, the saturation result of PaintShop Pro (www.jasc.com) is 245 but in my conversion I'm getting 238. Here is a literal equivalent of the code being used: // calculate Saturation (N32) depending on Luminance (N33) being >=128 // N29 = int, minColor (0-255) // N30 = int, maxColor (0-255) // N32 = int, Saturation (0-255) // N33 = int, Luminance (0-255) // D34 = flo, temporary // (maxColor - minColor) %N32% = %N30% - %N29% // (maxColor + minColor) %D34% = %N30% + %N29% If %N33% >= 128 // S = (maxColor - minColor) / (2 - (maxColor + minColor)) // ((maxColor + minColor) - 2) produces a positive, equal-magnitude result.) %D34% = %D34% - 2.00000000 Else // S = (maxColor - minColor) / (maxColor + minColor), does not need -2 End If // divide divisor by dividend, result as float %D34% = %N32% / %D34% // scale result to 0-255 range %D34% = %D34% * 256 Round %D34% to 0 decimal places Truncate %D34% to integer %N32% Text Box Display: Saturation %N32% As far as I can tell, there is no computational error. The issue is likely in my implementation of the formula and not the formula itself. Lower saturation conversion values DO match those created by PaintShop Pro. R=25,G=50,B=25 returns H=85, S=85, L=38 which match my returned values of H=120Á, S=85, L=38 (PaintShop's hue is 0-255 based where mine is 0-360 degrees.) This leads me to believe that only the >=128 luminance portion of the saturation Mark Jones === Subject: McNemar's hypothesis test HI all, i am having some problems with finding literature on the Internet but they only have a few words about it. DO you know of any McNemar's test? Would any of you be interested in posting a description of it here, I think that would be of interest to this group. thx === Subject: Teaching notes on Integers posted... I just posted some teaching notes for Integers here... http://k12math.com/math-concepts/numbers/integer.htm - Russ Lewis === Subject: Efficient gcd-tuple construction? given a prime p, a natural number N, and two integers x,y in {0,...,N-1}, please give an efficient way to describe all integers a,b such that x=a mod N y=b mod N y =< b =< x, (here =< means lower equal) x =< a =< sqrt(p), gcd(a,b)=1 Especially: is there a way to reduce the number of gcd-computations? Best, Oswald === Subject: Math Problem, Sort of. Hi. I hope you don't mind me just coming in here and asking a question. I'm lookign for a little help, it's not important just something that been puzzling me. BTW. I'm useless at math. I'm trying to figure out how a certain newsreader generates it's message ID. I already know that the first part of the ID consists of mmddyyyyhhmm. Then it's followed by a sequence of 6 numbers. As a lay man I can see a sequence is developing but that's as far as I can get. This is a sample of those 6 numbers taken from a bulk post. 145573 256228 366898 447319 557987 068644 179302 289961 350384 461047 571723 082388 193053 263490 384164 494841 005513 116179 186611 297278 417944 528601 039287 109705 210369 321035 431698 542367 012799 123453 234108 344776 455428 525848 046521 157187 267842 Any ideas or help would be appreciated. Dave === Subject: Re: Math Problem, Sort of. A sequence? A sequence shown with just 5 terms is mind-boggling enough and you thought those hundreds of terms are part of a sequence? Could be. By the way, the 666666 is missing in your sequence. === Subject: Re: Math Problem, Sort of. On Sat, 14 Jan 2006 20:03:18 -0000, Dave Gordon I hope you don't mind me just coming in here and asking a question. >I'm lookign for a little help, it's not important just something that been >puzzling me. BTW. I'm useless at math. I'm trying to figure out how a >certain newsreader generates it's message ID. >I already know that the first part of the ID consists of mmddyyyyhhmm. >Then it's followed by a sequence of 6 numbers. As a lay man I can >see a sequence is developing but that's as far as I can get. >This is a sample of those 6 numbers taken from a bulk post. >145573 >256228 >366898 >447319 >557987 >068644 >179302 >289961 >350384 >461047 >571723 >082388 >193053 >263490 >384164 >494841 >005513 >116179 >186611 >297278 >417944 >528601 >039287 >109705 >210369 >321035 >431698 >542367 >012799 >123453 >234108 >344776 >455428 >525848 >046521 >157187 >267842 >Any ideas or help would be appreciated. >Dave Sounds like a problem for Marilyn Vos Savant. Try sending it in to Parade Magazine. :-) --Lynn