mm-364 === Subject: : Re: Stupid question on notation (norm of a matrix) >> But then the good news are that they are all equivalent, thanks to the >> finite-dimensionality of the space. >Under the operator norm, the identity n x n matrix has norm equal to 1, >under the other norm, it equals square root of n. >How do you mean by equivalent? > I mean they are equivalent in the sense of the definition of > *equivalent* norms that is not to say that they take the same values > on the same matrices, in which case they would be *identical*... > It should suffice to you to know that they generate the same > topologies.That may be true, when you need only the topology. But there are partsof mathematics where they are used for different purposes (e.g. numericalalgebra). In that case you also have to do with vector norms, and youbetter chose a matrix norm that is at least consistent with the vectornorm you use (i.e. ||Ax|| <= ||A||.|x|). Or better still, a matrixnorm that is subordinate to the vector number (i.e., there is at leastone A != 0 and x != 0 such that ||Ax|| = ||A||.|x|).The matrix p-norms are subordinate to the vector p-norms. The (whatis called here) operator norm is consistent with the vector 2-norm,but not subordinate. The matrix norm defined by ||A|| = max |a_ij|is not consistent with any vector norm, etc.This is important when you use them in error analysis.-- === === Subject: : Re: GCD proof...>> 2. Show that (a, b, c) = ax + by + cz for some integers x, y, z.> It's essentially the same proof that works for (a,b). So:Interesting. Another approach is to just reuse the result for (a,b),after showing ((a,b),c) = (a,b,c): (a,b) = au + bv for some integers u, v ((a,b),c) = (a,b)w + cz for some integers w, ztherefore (a,b,c) = auw + bvw + czI think your approach is more mathematician-like and mine is moreprogrammer-like.-- === === Subject: : Re: GCD proof...That's a good way to do it, too, assuming that one already knows theresult for two elements. And since we know how to actually compute theanswer for two elements using the Euclidean algorithm, your suggestiongives an algorithm for computing the answer in general. The proof Ipos is just an existance proof, so undoubly not very satisfyingto someone interes in algorithms.JS>> 2. Show that (a, b, c) = ax + by + cz for some integers x, y, z.> It's essentially the same proof that works for (a,b). So:> Interesting. Another approach is to just reuse the result for (a,b),> after showing ((a,b),c) = (a,b,c):> (a,b) = au + bv for some integers u, v> ((a,b),c) = (a,b)w + cz for some integers w, z> therefore> (a,b,c) = auw + bvw + cz> I think your approach is more mathematician-like and mine is more> programmer-like.=== === Subject: : Development of LogicI am trying to better understand how logic can be developed either by Boolean algebra's axioms or by LF propositions, i.e. by(p / p) -> pq -> (p / q)(p / q) -> (q / p)(q -> r) -> [(p / q) -> (p / r)]and the rule of detachment.I thought I would try to show how to derive the distributive laws of Boolean algebra using the LF propositions, but got nowhere.I tried something simpler without success:(a / ~b) -> (a / (~a / ~b))I checked some books and looked on the Internet, but am still stumped. Can somebody give me a jumpstart?Tom Adams=== === Subject: : Re: Development of Logic> I am trying to better understand how logic can be developed either by> Boolean algebra's axioms or by LF propositions, i.e. by> (p / p) -> p> q -> (p / q)> (p / q) -> (q / p)> (q -> r) -> [(p / q) -> (p / r)]> and the rule of detachment.With substitution and -> defined from / and ~, these are Whitehead andRussell's as simplified by Lukasiewicz and Bernays.> I thought I would try to show how to derive the distributive laws of> Boolean algebra using the LF propositions, but got nowhere.> I tried something simpler without success:> (a / ~b) -> (a / (~a / ~b))> I checked some books and looked on the Internet, but am still stumped.> Can somebody give me a jumpstart?> Tom Adams-- G.C.=== === Subject: : Re: Development of Logic>>I am trying to better understand how logic can be developed either by>>Boolean algebra's axioms or by LF propositions, i.e. by>>(p / p) -> p>>q -> (p / q)>>(p / q) -> (q / p)>>(q -> r) -> [(p / q) -> (p / r)]>>and the rule of detachment.> With substitution and -> defined from / and ~, these are Whitehead and> Russell's as simplified by Lukasiewicz and Bernays.>>I thought I would try to show how to derive the distributive laws of>>Boolean algebra using the LF propositions, but got nowhere.>>I tried something simpler without success:>>(a / ~b) -> (a / (~a / ~b))>>I checked some books and looked on the Internet, but am still stumped.>>Can somebody give me a jumpstart?>>Tom AdamsAny hints on how to prove:(a / ~b) -> (a / (~a / ~b))Tom Adams=== === Subject: : Re: Development of Logic>I am trying to better understand how logic can be developed either by> Boolean algebra's axioms or by LF propositions, i.e. by> (p / p) -> p> q -> (p / q)> (p / q) -> (q / p)> (q -> r) -> [(p / q) -> (p / r)]> and the rule of detachment.>>With substitution and -> defined from / and ~, these are Whitehead and>> Russell's as simplified by Lukasiewicz and Bernays.>I thought I would try to show how to derive the distributive laws of> Boolean algebra using the LF propositions, but got nowhere.I tried something simpler without success:> (a / ~b) -> (a / (~a / ~b))I checked some books and looked on the Internet, but am still stumped.> Can somebody give me a jumpstart?Tom Adams>Any hints on how to prove:> (a / ~b) -> (a / (~a / ~b))> Tom AdamsActually, I found the proof by searching for Lukasiewicz and distributive:http://osophy.wisc.edu/fitelson/beavers.pdfIt's not as simple as I thought.Thanks George.Tom Adams=== === Subject: : Re: Development of Logic> I am trying to better understand how logic can be developed either by> Boolean algebra's axioms or by LF propositions, i.e. by> (p / p) -> p> q -> (p / q)> (p / q) -> (q / p)> (q -> r) -> [(p / q) -> (p / r)]> and the rule of detachment.I guess that -> is defined in terms of / and ~, or that / is definedin terms of -> and ~. Which?Also, either these are axiom schemata or you need a rule of substitutionas well.> I thought I would try to show how to derive the distributive laws of> Boolean algebra using the LF propositions, but got nowhere.> I tried something simpler without success:> (a / ~b) -> (a / (~a / ~b))> I checked some books and looked on the Internet, but am still stumped.> Can somebody give me a jumpstart?> Tom Adams-- G.C.=== === Subject: : Re: Continued-Fraction-Genera Sequence Also Equals...> Consider the continued fraction:> [1; 1,1, 1/2,1/2, 1/3,1/3, 1/4,1/4,...,ceiling((m-2)/2)]> = c(m), for m >= 3, > where the total number of (rational) CF-terms is (m-1).> Let c(1) = 1, and c(2) = 2, so we have:> {c(j)} -> 1, 2, 2, 3/2, 7/4, 19/12, 61/36,... > Now, let us take 'another' sequence {c'(j)},> where:> c'(1) = 1, c'(2) = 2.> And c'(1+m) => (sum{j=0 to floor((m-1)/2)} c'(m- 2j)) /ceiling(m/2).> In other words, > c'(1+m) is the average of every-other previous c'(),> the average of {c'(m),c'(m-2),c'(m-4),...,c'(1 or 2)}.> So, you guessed it,> each c(k) = c'(k).> Also,> c(1+m) = c(m)/ceiling(m/2) + c(m-1)(1 -1/ceiling(m/2)),I forgot to mention that the above is true ONLY for m >= 4.I also will rewrite the above recursion as to be in-terms of only integers.Let n(m) = c(m)*(floor((m-1)/2))!*(ceiling((m-1)/2))!.So, then, for m >= 3,n(m+2) = n(1+m) + n(m) *floor(m/2) *ceiling(m/2),with n(3) = 2, n(4) = 3.n(m) (first term is n(3))..2, 3, 7, 19, 61, ...(Not in the EIS, actually.)Anyone out there making progress on a closed-form for {c(m)} or {n(m)},or on the limit of the c's?> which should (easily?) lead to a closed-form for c(m).> But I have not the time now to explore further...> By the way, what is the closed-form for> limit{m-> oo} c(m),> anyway?Leroy Quet=== === Subject: : Matrices identified as vectors?Let ln(2) > epsilon > 0. Let U_epsilon = {X in M_n(C) | ||X|| 1, 1, 1, 1, 0, 1,...)Let f(y) =sum{m=0 to oo} a(m) y^m /m!.I believe that f(y) =1 + integral{0 to y} f(x) f((y-x)/(1+x))/(1+x) dx,which is in ascii-art mode:f(y) = /y y-x dx1 + | f(x) f( --- ) ----- /0 1+x (1+x) (unless I erred) But what is a closed-form for {a(m)} and/or f(y) ??Leroy Quet=== === Subject: : Re: An Unusual Generating Function'Vladeta' has written me that {a(m)} is the EIS-sequence A086672.There it gives f(y) as:>E.g.f.: hypergeom([1/2],[2],4*ln(1+x)) => (1+x)^2*(BesselI(0,2*ln(1+x))-BesselI(1,2*ln(1+x)))Leroy Quet> Let C(m) be the mth Catalan number,> C(m) = binomial(2m,m)/(1+m).)> Let S(m,n) = an unsigned Stirling number of the first kind.> (S(0,0)=1, S(0,n)=0 for n not 0,> S(m+1,n) = m*S(m,n) + s(m,n-1).)> OK, let> a(m) => sum{k=0 to m} S(m,k) C(k) (-1)^(k+m).> (If I did not error figuring the first few terms by hand,> a(m) -> 1, 1, 1, 1, 0, 1,...)> Let f(y) => sum{m=0 to oo} a(m) y^m /m!.> I believe that f(y) => 1 + integral{0 to y} f(x) f((y-x)/(1+x))/(1+x) dx,> which is in ascii-art mode:> f(y) => /y y-x dx> 1 + | f(x) f( --- ) -----> /0 1+x (1+x)> (unless I erred)> But what is a closed-form for {a(m)} and/or f(y) ??> Leroy> Quet=== === Subject: : Re: Chicken Nugget Problem (Number Theory)>* mareg@mimosa.csv.warwick.ac.uk>> A better upper bound is 43.Here is a straightforward solution:Using 6's and 9's, you can get 3n for any for n>=2Hence, using one 20 + 6's and 9's, you can get 3n+2 for n>=8Using two 20's + 6's and 9's, you can get 3n+1 for any n>=15You cannot get 43, because 43 = 1 (mod 3), and so you would need at>> least two 20's to get 43.>> True, but you do not need that fact.> To get any 3n+1 you actuall _need_ two 20s (or 5, 8, ... 20s).> Therefore you cannot make 43, which is 3*14+1 so you need to make a> 3 which is clearly impossible.> I said that! So, it remains to show how you can make any number _higher_ than 43> which you indicate, but just indicate.> No, I gave a complete proof! Worthy of full marks, I would say!I know my Analysis prof would not have given full marks. Fill in thegaps.On the other hand, when I took p-adic number theory, this proof would havegotten full marks. Number theorists (and a fair number of logicians) seemto be willing to believe that students understand the obvious gaps.Jon Miller=== === Subject: : Re: Fibonacci PHI, PI, e RelationHere is a bigger list, this one goes to 200,000 digits.{12, 99, 169, 395, 499, 595, 606, 693, 824, 827, 840, 940, 1282, 1291,1384, 1594, 1705, 1742, 1905, 2020, 2060, 2153, 2257, 2302, 2359,2367, 2507, 2546, 2557, 2710, 2724, 2791, 2832, 2857, 3036, 3051,3280, 3309, 3429, 3497, 3518, 3591, 3651, 3709, 3867, 4210, 4292,4390, 4493, 4719, 4826, 4859, 4862, 4892, 4934, 4940, 5087, 5315,5427, 5480, 5488, 5653, 5699, 6155, 6426, 6617, 6838, 6854, 7113,7155, 7202, 7358, 7390, 7659, 7685, 7721, 7761, 7816, 7833, 7867,7923, 8417, 8570, 8611, 8653, 8731, 8914, 9051, 9077, 9133, 9283,9286, 9310, 9704, 9717, 9724, 9805, 9897, 10086, 10127, 10349, 10696,10727, 10791, 10818, 10860, 10896, 11070, 11112, 11223, 11280, 11285,11326, 11345, 11392, 11393, 11428, 11449, 11451, 11566, 11579, 11643,11904, 12035, 12047, 12135, 12243, 12621, 12635, 12651, 13093, 13159,13195, 13354, 13370, 13430, 13501, 13593, 13745, 13777, 13810, 13839,13915, 13922, 13942, 14061, 14116, 14165, 14349, 14419, 14428, 14462,14481, 14482, 14611, 14641, 14717, 14801, 14812, 14873, 14973, 15240,15285, 15413, 15428, 15513, 15683, 15883, 15890, 16117, 16133, 16135,16152, 16223, 16452, 16472, 16528, 16539, 16562, 16639, 16641, 16911,16922, 16925, 16998, 17136, 17178, 17277, 17285, 17320, 17360, 17449,17553, 17723, 17759, 17791, 18187, 18229, 18262, 18283, 18292, 18395,18410, 18444, 18446, 18512, 18685, 18718, 18746, 18871, 19080, 19108,19237, 19265, 19280, 19324, 19385, 19513, 19663, 19690, 19783, 20090,20188, 20275, 20290, 20311, 20416, 20508, 20785, 20855, 20894, 20929,20938, 20943, 21028, 21125, 21285, 21289, 21300, 21610, 21631, 21677,21680, 21805, 22146, 22193, 22487, 22509, 22569, 22656, 22762, 22926,22955, 23116, 23167, 23203, 23261, 23377, 23423, 23611, 23673, 23788,23834, 23844, 23915, 23937, 23977, 24132, 24148, 24196, 24412, 24681,24698, 24712, 24726, 24775, 24991, 25067, 25123, 25182, 25271, 25424,25439, 25515, 25566, 25581, 25593, 26163, 26283, 26340, 26416, 26506,26514, 26630, 26652, 26742, 26984, 27029, 27393, 27409, 27476, 27552,27586, 27655, 27704, 27796, 27893, 27908, 27980, 28196, 28247, 28250,28383, 28723, 28830, 28908, 28952, 29012, 29110, 29200, 29331, 29363,29372, 29415, 29420, 29453, 29507, 29581, 29699, 29836, 29876, 30049,30209, 30278, 30449, 30571, 30646, 30712, 30819, 30878, 30900, 31070,31094, 31150, 31189, 31321, 31373, 31536, 31538, 31547, 31625, 31858,31918, 31934, 32043, 32086, 32218, 32234, 32242, 32351, 32446, 32479,32658, 32744, 32909, 32923, 33052, 33173, 33886, 34029, 34033, 34070,34096, 34097, 34106, 34151, 34295, 34379, 34406, 34553, 34650, 34732,34815, 34832, 35062, 35191, 35192, 35326, 35333, 35357, 35686, 35717,35829, 35964, 35985, 36445, 36470, 36744, 36777, 36838, 36971, 37102,37266, 37271, 37323, 37537, 37577, 37647, 37692, 37731, 37892, 38030,38127, 38235, 38256, 38337, 38398, 38474, 38550, 38746, 38851, 38861,38927, 38930, 39090, 39150, 39206, 39252, 39294, 39327, 39416, 39593,39720, 40339, 40346, 40352, 40387, 40497, 40502, 40669, 40682, 40699,40767, 40768, 40965, 41196, 41223, 41374, 41455, 41528, 41535, 41694,41700, 41707, 41819, 41896, 41945, 42067, 42218, 42245, 42263, 42306,42386, 42444, 42462, 42502, 42619, 42815, 42823, 42949, 43040, 43304,43444, 43603, 43641, 43681, 43730, 43809, 44069, 44252, 44288, 44325,44346, 44357, 44384, 44496, 44510, 44592, 44630, 44746, 44929, 44983,44992, 45062, 45095, 45131, 45309, 45316, 45380, 45528, 45680, 45706,45756, 45934, 46192, 46335, 46573, 46585, 46663, 47084, 47261, 47401,47473, 47988, 48063, 48148, 48900, 48965, 49130, 49154, 49427, 49590,49821, 49876, 50077, 50324, 50572, 50918, 51593, 51659, 51808, 52062,52120, 52151, 52154, 52331, 52547, 52638, 52705, 52826, 52906, 53098,53200, 53510, 53665, 53695, 53738, 53755, 53919, 53944, 53963, 54055,54412, 54456, 54541, 54597, 54705, 54805, 54907, 54922, 54971, 55047,55188, 55251, 55442, 55502, 55594, 55607, 55635, 55674, 55824, 55833,55905, 55950, 56155, 56257, 56280, 56719, 56824, 56837, 56844, 56868,57065, 57086, 57120, 57126, 57184, 57446, 57468, 57483, 57493, 57592,57603, 57646, 57773, 57973, 57983, 58071, 58152, 58242, 58356, 58357,58609, 58630, 59283, 59294, 59389, 59457, 59539, 59593, 59678, 59756,59772, 59818, 59841, 59869, 59987, 60097, 60134, 60156, 60352, 60361,60738, 60819, 61012, 61457, 61692, 61694, 61795, 61828, 61832, 61967,62037, 62058, 62195, 62266, 62286, 62309, 62374, 62420, 62749, 62825,62861, 62934, 62954, 62976, 63073, 63122, 63316, 63463, 63640, 63859,64096, 64185, 64234, 64246, 64265, 64419, 64435, 64509, 64558, 64565,64763, 64878, 64927, 65115, 65252, 65325, 65443, 65494, 65495, 65496,65612, 65891, 65939, 66058, 66102, 66104, 66236, 66246, 66309, 66445,66555, 66597, 66780, 66842, 66879, 67096, 67154, 67198, 67225, 67405,67508, 67625, 67723, 67793, 67796, 67800, 67898, 67948, 67976, 68028,68105, 68334, 68362, 68649, 68849, 68854, 68875, 68938, 69021, 69074,69124, 69181, 69499, 69532, 69704, 69965, 70102, 70328, 70451, 70503,70910, 70955, 71004, 71010, 71155, 71356, 71397, 71448, 71771, 71782,71887, 71955, 72187, 72283, 72288, 72300, 72321, 72334, 72513, 72933,72944, 72968, 73033, 73042, 73125, 73223, 73246, 73301, 73385, 73393,73491, 73531, 73644, 73859, 73952, 74308, 74826, 74838, 74935, 75456,75530, 75549, 75763, 75782, 75819, 75939, 75995, 76049, 76128, 76235,76313, 76460, 76484, 76567, 76598, 76666, 76687, 76713, 76817, 77039,77088, 77203, 77265, 77386, 77504, 77617, 77631, 77705, 77859, 77888,78423, 78493, 78523, 78639, 78651, 78655, 78765, 79053, 79099, 79108,79141, 79254, 79285, 79373, 79387, 79532, 79541, 79588, 79638, 79671,79713, 79810, 79843, 79910, 80049, 80080, 80087, 80141, 80163, 80172,80348, 80394, 80597, 80641, 80653, 80794, 80878, 80896, 80958, 80976,80981, 81185, 81422, 81424, 81427, 81428, 82067, 82118, 82130, 82358,82413, 82449, 82541, 82707, 82741, 82898, 82994, 83028, 83097, 83141,83183, 83462, 83682, 83711, 83821, 83948, 84001, 84041, 84055, 84259,84436, 84534, 84547, 84581, 84636, 84733, 84801, 84938, 84972, 85084,85181, 85317, 85373, 85592, 85755, 85781, 85848, 85892, 86072, 86077,86089, 86187, 86256, 86321, 86327, 86330, 86350, 86490, 86585, 86750,86789, 86820, 86920, 86959, 87037, 87233, 87337, 87360, 87370, 87378,87434, 87449, 87644, 87683, 87686, 87692, 87798, 87877, 88133, 88266,88848, 89080, 89165, 89183, 89520, 89563, 89981, 89988, 90319, 90423,90633, 90645, 90735, 90764, 90832, 90843, 90874, 91040, 91089, 91491,91561, 91609, 91626, 91672, 91734, 91759, 91797, 91861, 91931, 91996,92050, 92080, 92581, 92639, 92672, 92676, 92779, 92802, 92854, 92942,92972, 92977, 93113, 93164, 93438, 93447, 93459, 93509, 93546, 93560,93646, 93677, 93804, 93909, 93992, 94007, 94053, 94126, 94130, 94170,94278, 94389, 94896, 95015, 95018, 95063, 95113, 95293, 95582, 95595,95711, 95770, 95811, 96018, 96097, 96230, 96270, 96439, 96532, 96555,96581, 96638, 96662, 96686, 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111138,111222, 111253, 111279, 111313, 111334, 111534, 111569, 111639,111780, 111861, 111961, 111971, 112094, 112296, 112335, 112352,112646, 112661, 112736, 112817, 113061, 113083, 113191, 113204,113455, 113576, 113703, 113848, 113909, 113970, 114116, 114197,114349, 114394, 114405, 114451, 114490, 114548, 114653, 114804,114871, 114872, 114941, 115018, 115096, 115115, 115213, 115224,115310, 115356, 115435, 115465, 115846, 116049, 116056, 116284,116368, 116619, 117189, 117219, 117675, 117729, 117792, 117838,118142, 118278, 118307, 118353, 118422, 118462, 118650, 118784,118891, 119140, 119365, 119397, 119418, 119459, 119469, 119476,119487, 119538, 119541, 119601, 119879, 119897, 119955, 119980,120269, 120311, 120914, 121021, 121027, 121110, 121262, 121329,121397, 121492, 121506, 121619, 121639, 121819, 121864, 122207,122232, 122377, 122380, 122473, 122552, 122569, 122659, 122714,122718, 122954, 122955, 123130, 123189, 123201, 123215, 123383,123441, 123583, 123660, 123673, 123682, 123708, 123729, 123769,123894, 123897, 124048, 124065, 124249, 124304, 124307, 124402,124426, 124540, 124580, 124776, 124987, 125113, 125209, 125386,125422, 125711, 125835, 125868, 125944, 126008, 126057, 126128,126271, 126278, 126344, 126535, 126713, 127000, 127030, 127034,127190, 127228, 127265, 127390, 127816, 127828, 128209, 128250,128449, 128759, 128983, 129067, 129103, 129138, 129184, 129257,129442, 129461, 129543, 129571, 129606, 129658, 129742, 129908,129979, 130034, 130149, 130171, 130388, 130632, 130732, 130751,130937, 130961, 131121, 131359, 131453, 131454, 131649, 131669,131966, 132011, 132110, 132160, 132229, 132254, 132298, 132489,132530, 132620, 132738, 133074, 133409, 133433, 133767, 134075,134122, 134132, 134257, 134448, 134568, 134655, 134687, 134693,134783, 134840, 134843, 134889, 135040, 135091, 135151, 135235,135449, 135537, 135623, 135678, 135681, 136174, 136275, 136415,136468, 136611, 136642, 136722, 136739, 136747, 136889, 137009,137107, 137270, 137308, 137337, 137539, 137658, 137801, 137826,137876, 137893, 137991, 138193, 138423, 138605, 138671, 138740,138934, 138942, 139066, 139087, 139282, 139310, 139367, 139461,139559, 139672, 139888, 139948, 140041, 140261, 140302, 140343,140495, 140530, 140587, 140602, 140933, 141001, 141079, 141094,141132, 141141, 141182, 141200, 141210, 141630, 141758, 141785,141971, 142126, 142635, 142653, 142722, 142808, 143108, 143144,143205, 143231, 143264, 143315, 143316, 143397, 143430, 143621,143678, 143929, 143989, 144246, 144289, 144295, 144338, 144374,144454, 144583, 144588, 144690, 144698, 144939, 145215, 145424,145616, 145669, 145727, 145779, 145864, 145879, 145931, 145970,146090, 146228, 146474, 146684, 146742, 146926, 147041, 147053,147274, 147811, 148146, 148242, 148657, 148693, 148705, 148850,148891, 148898, 148920, 148969, 149080, 149147, 149311, 149353,149376, 149449, 149479, 149631, 149678, 149716, 149894, 149919,150032, 150161, 150177, 150288, 150752, 150933, 151014, 151047,151244, 151316, 151356, 151381, 151817, 151873, 151893, 152111,152207, 152210, 152287, 152429, 152598, 152603, 152778, 152790,152799, 152845, 152879, 152881, 153175, 153176, 153261, 153341,153362, 153405, 153407, 153444, 153468, 153672, 153761, 153799,153807, 153946, 154071, 154483, 154490, 154603, 154628, 154667,154893, 154973, 155154, 155252, 155343, 155349, 155496, 155699,155751, 155789, 155865, 155877, 155933, 156005, 156100, 156203,156231, 156382, 156515, 156614, 156656, 156681, 156714, 156726,156734, 156909, 156974, 157044, 157099, 157202, 157207, 157227,157232, 157464, 157519, 157783, 157809, 158014, 158103, 158173,158360, 158478, 158499, 159169, 159400, 159628, 159637, 159685,159850, 160126, 160160, 160201, 160345, 160426, 160491, 160532,160555, 160594, 160753, 160965, 161012, 161072, 161260, 161620,161754, 161943, 161998, 162072, 162295, 162299, 162402, 162438,162452, 162467, 162814, 163830, 163873, 163959, 163961, 164010,164014, 164475, 164489, 164492, 164536, 164664, 164774, 164845,164859, 165090, 165125, 165427, 166044, 166156, 166441, 166532,166590, 166762, 166888, 166975, 167203, 167317, 167432, 167489,167524, 167584, 167630, 167642, 167643, 167752, 167794, 167991,168186, 168192, 168526, 168700, 168789, 168849, 168879, 168982,169043, 169117, 169123, 169276, 169347, 169348, 169486, 169495,169568, 169616, 169699, 169727, 169744, 169746, 169982, 170022,170083, 170265, 170281, 170395, 170580, 170671, 170698, 170816,170996, 171054, 171200, 171222, 171233, 171332, 171385, 171526,171617, 171717, 172058, 172246, 172328, 172433, 172453, 172642,172786, 172861, 172961, 173292, 173345, 173348, 173416, 173652,173673, 173802, 173885, 173905, 174060, 174088, 174105, 174231,174399, 174524, 174526, 174559, 174562, 174679, 174769, 174956,175144, 175175, 175259, 175287, 175536, 175599, 175663, 175854,175908, 175915, 176021, 176045, 176465, 176493, 176515, 176718,177523, 177557, 177614, 177702, 177726, 177898, 177942, 178007,178068, 178235, 178282, 178356, 178578, 178668, 178784, 178850,178952, 178979, 179043, 179091, 179105, 179120, 179207, 179229,179254, 179405, 179510, 179530, 179583, 179632, 179653, 179744,179801, 179929, 180337, 180379, 180430, 180442, 180520, 180635,180864, 180912, 181032, 181243, 181347, 181431, 181458, 181562,181578, 181637, 181677, 181840, 182085, 182183, 182686, 182792,182869, 182891, 182985, 183012, 183076, 183266, 183454, 183476,183599, 183873, 184142, 184237, 184297, 184404, 184529, 184553,184556, 184606, 184705, 184711, 184800, 184996, 185006, 185044,185136, 185138, 185338, 185403, 185462, 185507, 185511, 185751,185809, 185840, 185873, 185927, 185953, 185983, 186093, 186407,186452, 186478, 186488, 186559, 186664, 186670, 186983, 187070,187093, 187097, 187129, 187240, 187371, 187671, 187726, 187782,187910, 187939, 188061, 188108, 188344, 188394, 188540, 188553,188558, 188764, 188771, 189085, 189189, 189275, 189327, 189556,189718, 189737, 189910, 190071, 190118, 190325, 190412, 190456,190557, 190580, 190764, 190841, 191004, 191018, 191036, 191248,191322, 191360, 191457, 191470, 191505, 191610, 191793, 192135,192572, 192631, 192662, 192838, 192858, 192886, 193033, 193302,193400, 193533, 193559, 193572, 193574, 193693, 193710, 193819,193857, 193863, 193936, 193978, 194062, 194066, 194114, 194141,194150, 194212, 194384, 194598, 194873, 194933, 195023, 195325,195457, 195542, 195669, 195713, 195748, 195752, 195863, 195914,196030, 196148, 196186, 196251, 196382, 196536, 196636, 196715,196779, 196928, 197042, 197228, 197762, 197882, 197975, 198100,198109, 198221, 198227, 198290, 198342, 198427, 198445, 198928,199064, 199139, 199200, 199369, 199376, 199389, 199584, 199649,199727, 199817, 199878, 199905, 199907, 199918}=== === Subject: : Re: Fibonacci PHI, PI, e Relation> Here is a bigger list, this one goes to 200,000 digits.1973 of them, when we expect 2000. Short by 27.Let's see, sqrt(2000)=44, so weare still within what we could likely see with random digits.-- http://www.math.ohio-state.edu/~edgar/=== === Subject: : Re: Fibonacci PHI, PI, e Relation> Here is a bigger list, this one goes to 200,000 digits.>> 1973 of them, when we expect 2000. Short by 27.> Let's see, sqrt(2000)=44, so we> are still within what we could likely see with random digits.Break it up into 200 lists of 1000 digits. How many of those lists appearnon-random at the 5% confidence level?Jon Miller=== === Subject: : Re: Fibonacci PHI, PI, e RelationI used Mathematica to do the calculations.Here are all the results out to 30,000 places.{12, 99, 169, 395, 499, 595, 606, 693, 824, 827, 840, 940, 1282, 1291,1384, 1594, 1705, 1742, 1905, 2020, 2060, 2153, 2257, 2302, 2359,2367, 2507, 2546, 2557, 2710, 2724, 2791, 2832, 2857, 3036, 3051,3280, 3309, 3429, 3497, 3518, 3591, 3651, 3709, 3867, 4210, 4292,4390, 4493, 4719, 4826, 4859, 4862, 4892, 4934, 4940, 5087, 5315,5427, 5480, 5488, 5653, 5699, 6155, 6426, 6617, 6838, 6854, 7113,7155, 7202, 7358, 7390, 7659, 7685, 7721, 7761, 7816, 7833, 7867,7923, 8417, 8570, 8611, 8653, 8731, 8914, 9051, 9077, 9133, 9283,9286, 9310, 9704, 9717, 9724, 9805, 9897, 10086, 10127, 10349, 10696,10727, 10791, 10818, 10860, 10896, 11070, 11112, 11223, 11280, 11285,11326, 11345, 11392, 11393, 11428, 11449, 11451, 11566, 11579, 11643,11904, 12035, 12047, 12135, 12243, 12621, 12635, 12651, 13093, 13159,13195, 13354, 13370, 13430, 13501, 13593, 13745, 13777, 13810, 13839,13915, 13922, 13942, 14061, 14116, 14165, 14349, 14419, 14428, 14462,14481, 14482, 14611, 14641, 14717, 14801, 14812, 14873, 14973, 15240,15285, 15413, 15428, 15513, 15683, 15883, 15890, 16117, 16133, 16135,16152, 16223, 16452, 16472, 16528, 16539, 16562, 16639, 16641, 16911,16922, 16925, 16998, 17136, 17178, 17277, 17285, 17320, 17360, 17449,17553, 17723, 17759, 17791, 18187, 18229, 18262, 18283, 18292, 18395,18410, 18444, 18446, 18512, 18685, 18718, 18746, 18871, 19080, 19108,19237, 19265, 19280, 19324, 19385, 19513, 19663, 19690, 19783, 20090,20188, 20275, 20290, 20311, 20416, 20508, 20785, 20855, 20894, 20929,20938, 20943, 21028, 21125, 21285, 21289, 21300, 21610, 21631, 21677,21680, 21805, 22146, 22193, 22487, 22509, 22569, 22656, 22762, 22926,22955, 23116, 23167, 23203, 23261, 23377, 23423, 23611, 23673, 23788,23834, 23844, 23915, 23937, 23977, 24132, 24148, 24196, 24412, 24681,24698, 24712, 24726, 24775, 24991, 25067, 25123, 25182, 25271, 25424,25439, 25515, 25566, 25581, 25593, 26163, 26283, 26340, 26416, 26506,26514, 26630, 26652, 26742, 26984, 27029, 27393, 27409, 27476, 27552,27586, 27655, 27704, 27796, 27893, 27908, 27980, 28196, 28247, 28250,28383, 28723, 28830, 28908, 28952, 29012, 29110, 29200, 29331, 29363,29372, 29415, 29420, 29453, 29507, 29581, 29699, 29836, 29876}=== === Subject: : Re: Automobile Speeds.> Exercise 2. Describe a physical meaning for the unit of area associa> to your automobile's efficiency. The cross-section of a stream of petrol your car is guzzling up as it drivesalong to get just the right amount to keep going at the speed it's guzzling.Michel.=== === Subject: : Re: More Accessible Algorithm-MazeBoolbar got the right answer, as I no in my other reply.I have pos the solution graphically below, after this message fromour sponsor...----I also pos in my reply to Boolbar:Besides being a source of mindless fun (FUN!), I made this maze (aswellas the other algorithm mazes I have pos) as an inspiration forothersto make such puzzles themselves, perhaps making them a lot more difficult and creating them using a lot more creativity (inspiredby,say, Game's Calculatrivia, for example, or inspired by actual computeralgorithms, or by mathematics or by word-puzzles, or by image puzzles,orby...).;)----Answer:||V||V||V||V||V!---------------- ---------!| # | # | P | U | O - W |!--|-+---+---+---+-|-+-|--!| R - 3 | 7 - * - D | N |!----+-|-+-|-+---+---+-|--!| 2 | A | 4 - 3 - % | * |!----+-|-+---+---+-|-+-|--!| @ | 5 - 8 | + | @ | L |!----+---+-|-+---+-|-+-|--!| 1 | R - * | T - % | E |!----+-|-+---+-|-+---+-|--!| 3 | I - G - H | T | F |!-------------------------! which ends on the F in lower-right square.Leroy Quet=== === Subject: : Re: theorem vs. proposition>What is the difference between a theorem and a proposition?The difference is explained in most elementary texts of logic.> A proposition is a sentence which makes an assertion which might be trueor> false but not both.> Oh. Yes. That particular usage slipped by me. That is a distinct technical> usage of the term in logic. Instead, I am specifically referring to the> other usage, that of general mathematicians in presenting their results.> --> Mitch Harris> (remove q to reply)I do not understand what you mean by 'the other usage'I thought it was a universal usage, assuming that mathematical statementssay anything at all.In had thought that mathematical proofs comprised a list of truepropositions successively rela to each other by equivalences orimplications and that the final step was a statement of the proposition tobe proved derived from the initial axioms or theorems used in the proof.I guess that what mathematicians do in practice is adumbrate a proofiteratively until they are intuitively satisfied that it is correct. Onlythen do they crawl over it for days, weeks, months, years even to make surethey haven't included an erroneous step.Logic, therefore, is a technique for validating or accurately expressingthoughts which have been arrived at by more general means..=== === Subject: : Re: An Interesting Recursively-Genera Integer Sequence> This sequence has multiple definitions.> I began wondering about the sequence today using this definition:> n(1) = 1;> n(m+1) is the sum of the numerator and denominator in the (reduced)> rational:> sum{k=1 to m} 1/n(k).> {n(j)} -> 1, 2, 5, 27, 739, ...> This is sequence A057438, without the initial term, of the> Encyclopedia of Integer Sequences, which gives the definition:>a(1) = 1; a(n+1) = product_{k = 1 to n} [a(k)] *sum_{j = 1 to n}> [1/a(j)]> ({a(j}) -> 1, 1, 2, 5, 27, 739,...)> a(j+1) does in-fact = n(j), since both are defined by the same> recursion:> n(m+2) = (product{k=1 to m} n(k)) + n(1+m)^2> = n(m) *(n(1+m) - n(m)^2) + n(1+m)^2.> But what I am wondering is, what is the sum (which does converge)> x = sum {k=1 to oo} 1/n(k)> equal to??> (One more thing: 1 + x > => limit{m-> oo} n(1+m)/(n(2+m)-n(m)^2).)> Anything anyone can add to this discussion??>I mention this sequence in the sci.math thread:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe =off&threadm=glfacbn6rlg7%40forum.mathforum.com&rnum=2&prev=I forgot to mention here thateach n(m) is coprime with each n(k), for every m not = k.Leroy Quet=== === Subject: : Re: functions needed> I need to construct such a family of functions which> satisfy following constrains:> f(a,b) = f(c,d) if and only if a=b, c=d or a=c, b=d.> for all positive integers a,b,c,d. > ThanksStart with a function for which (f(a,b)=f(c,d) iff (a,b)=(c,d).Restrict to a subset that has just one of (a,b) and (b,a) foreach pair, then extend ... .=== === Subject: : Re: sqrt(-1)=0/0How else am I to demonstrate your intellect?=== === Subject: : Re: sqrt(-1)=0/0Thus for Socrates, nothing = everything, for Denke, nothing/nothing = everything. As for myself, I know nothing, 0.=== === Subject: : Re: sqrt(-1)=0/0> Concerning the sqrt(-1)=0/0(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.> -Doctor Garry Whilhelm Denke, 1655> solving every equation; it was of 0 use, making 0 sense.> -/-> Are you some sort of idiot? > sqrt(-1) = 0/0> Square both sides.> -1 = 0/0> How can -1 be 0? The real numbers form a field and its trivial that> there can only be one number such that a * 0 = 0, for any a. But you> are saying -1 * 0 = 0. Contradiction.Aha! But that depends on the convention that 0/0 = 0. What ifsomeone adopts the convention that 0/0 = 1?> So are you using some different definition of real numbers, one that> isn't a field? I think the Completeness Axiom has something to say> about that.Yeah, it's pretty obvious that he's talking about something other thanthe real numbers. (I bet that terminology gets to him, too... realnumbers.)'cid 'ooh=== === Subject: : Re: sqrt(-1)=0/0> Concerning the sqrt(-1)=0/0(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.> -Doctor Garry Whilhelm Denke, 1655> solving every equation; it was of 0 use, making 0 sense.> -/-> Are you some sort of idiot? > sqrt(-1) = 0/0> Square both sides.> -1 = 0/0> How can -1 be 0? The real numbers form a field and its trivial that> there can only be one number such that a * 0 = 0, for any a. But you> are saying -1 * 0 = 0. Contradiction.I agree that the original poster lacks mathematical understanding in this particular area. But calling someone an idiot, and then claiming that -1 * 0 = 0 is a contradiction, does not reflect well on you today.=== === Subject: : Re: sqrt(-1)=0/0> Concerning the sqrt(-1)=0/0(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.> -Doctor Garry Whilhelm Denke, 1655solving every equation; it was of 0 use, making 0 sense.-/-> Are you some sort of idiot? > sqrt(-1) = 0/0i agree=== === Subject: : Re: sqrt(-1)=0/0> Concerning the sqrt(-1)=0/0(x^2+1)/0=(0)/0, Sokrates ist' Nichts Alles.> -Doctor Garry Whilhelm Denke, 1655solving every equation; it was of 0 use, making 0 sense.-/-Are you some sort of idiot?sqrt(-1) = 0/0Square both sides.-1 = 0/0How can -1 be 0? The real numbers form a field and its trivial that> there can only be one number such that a * 0 = 0, for any a. But you> are saying -1 * 0 = 0. Contradiction.>> I agree that the original poster lacks mathematical understanding in> this particular area. But calling someone an idiot, and then claiming> that -1 * 0 = 0 is a contradiction, does not reflect well on you today.I imagine that was to read -1 * -1 == 0, he was trying to say there is onlyone k for ak = 0, a = -1 and k = -1 is not it.Tom=== === Subject: : Re: What kind of math?> I would like to do math problems with a prison inmate I visit. How> should I go about determining what kind of math he will enjoy, and> evaluating his level of ability?> You might start with Prisoners Dilemma. Seriously, I would first askwhether> HE would enjoy the idea at all and then focus on simple real worldproblems> and working up.If he likes math he may not enjoy real world problems. It depends onhis and your mathematical background. Find this out and ask the questionagain. Maybe he already knows the answer to your question.=== === Subject: : Re: Factoring Conjecture: Triplets> === Subject: : Re: Factoring Conjecture: Triplets>Message-id: In short, I trust Nora, and I don't trust you. If you took one of>>your idiotic newsgroup polls on this issue I'm certain you'd find an>>overwhelming majority who feel the same way. (Not that it means her math>>is infallible, of course; it just means I trust her not to lie about it,>>whereas I *expect* you to lie at every opportunity.)>>If someone doesn't know the truth, is it possible for him to lie?>> There are differences between not knowing the truth, refusing to face the>> truth, and knowing the truth but lying about it. James has provided us>> with examples of all three on many occasions -- often all three within>I think in the case of JSH his need to believe in himself is so powerful >that it clouds the picture. His state of delusion is such that he >doesn't know the truth, about himself, about others, or about >mathematics. That's not the impression I came away with when we had a live conversation in his chat room. I got to see the that stands behind the curtain. The blithering ignorance and pathetic plea for help made it obvious that there is no delusion. What you see here is all a calcula act.> We are dealing with a mentally disturbed person here, I agree that he's disturbed, but not the way you think.>and I think the word lie does not accurately describe what he does.Every bit of it is a lie. I know it and most importantly, JSH knows it.>Gib--=== === Subject: : Re: Factoring Conjecture: Triplets> I agree that he's disturbed, but not the way you think.I'm intrigued. What kind of disturbed do you think he is?Gib=== === Subject: : Re: JSH: Understanding the algebra> Note also that for MOST integer values of x, w1 and w2 are not equal> to either 7 or 1. This happens only when x = 0. Note that> x = 0 is really a special case, because when x = 0,> a^2 - (x - 1)*a + 7*(x^2 + x)> is reducible. In general it is not.One may check that, for a^2 - (x - 1)*a + 7*(x^2 + x), the discriminant in the quadratic formula expression for the values of a, will be negative for all integer values of x except 0 and -1.So that, a^2 - (x - 1)*a + 7*(x^2 + x) is reducible, as a quadratic polynomial in a over the integers, for x = 0 and x = -1 and for no other values of x.Thus the in general statement above means whenever x^2+x is not zero.And, of course, in those 2 special cases, the presence of 7 in a^2 - (x - 1)*a + 7*(x^2 + x) is irrelevant.=== === Subject: : Re: Rational Limits> For intersections of any finite family of nes sets, meaning each is> a proper subset of all its predecessors, that intersection equals the> last, and therefore smallest, set in the family.For intersections infinite families of nes sets, this cannot be the> case because there is no last or smallest.> There must have been a smallest if you are claiming it was the empty set.> It doesn't matter if it was the last set.Then what is the smallest member ( or last member) of {(0,1/n):n in N}? Notice that I said n in N, so n must be finite and not one of your unicorn numbers.=== === Subject: : Re: Rational Limits> For intersections of any finite family of nes sets, meaning eachis> a proper subset of all its predecessors, that intersection equals the> last, and therefore smallest, set in the family.For intersections infinite families of nes sets, this cannot be the> case because there is no last or smallest.There must have been a smallest if you are claiming it was the emptyset.> It doesn't matter if it was the last set.> Then what is the smallest member ( or last member) of {(0,1/n):n in N}?I'm claiming there is no smallest set.I think the intersection of {(0,1/n):n in N} is always an open infiniteset.You are the one claiming the intersection is empty.> Notice that I said n in N, so n must be finite and not one of your> unicorn numbers.You mean like the n such that (0,1/n) is an empty set?You tell me. I don't think there is such an n.Russell- 2 many 2 count=== === Subject: : Re: Rational Limits <4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <877jzkxzdf.fsf@phiwumbda.org> Discussion, linux)For intersections of any finite family of nes sets, meaning each> is>> a proper subset of all its predecessors, that intersection equals the>> last, and therefore smallest, set in the family.For intersections infinite families of nes sets, this cannot be the>> case because there is no last or smallest.There must have been a smallest if you are claiming it was the empty> set.>> It doesn't matter if it was the last set.Then what is the smallest member ( or last member) of {(0,1/n):n in N}?> I'm claiming there is no smallest set.> I think the intersection of {(0,1/n):n in N} is always an open infinite> set.> You are the one claiming the intersection is empty.What you're doing is equivocating.Virgil says (correctly) that the intersection of {(0,1/n) | n in N} isempty. You claim that *therefore* there is a smallest interval (0,1/n). You give no reason for believing so, aside from yourprevious pronouncement that the intersection of nes intervals {A_n}is always some A_n. Now you are denying that claim, evidently.Which is just as well, since the claim is false.>> Notice that I said n in N, so n must be finite and not one of your>> unicorn numbers.> You mean like the n such that (0,1/n) is an empty set?> You tell me. I don't think there is such an n.Don't be disingenuous. I understand you're stupid. I forgive you forstupidity. Hell, I read you for the entertaining bits of yourstupidity. But now you're putting words into others' mouths knowingfull well that they deny (1) that the specific claim is true and (2) that your proposed derivation of the specific claim is a valid proof,given their claims.No one in this group, aside from you, is commit to the claim thatthe intersection of a family {A_n} of nes intervals is exactly A_nfor some n. Consequently, no one believes that, if the intersectionof {(0,1/n)} is empty, therefore some interval (0,1/n) is also empty.Shocked, shocked I tell you, to discover that your proofs ofinconsistency of ZFC always depend on some other claim which itself isinconsistent with ZFC. Who'da thunk it?-- This page contains information of a type (text/html) that can only be viewed with the appropriate Plug-in. Click OK to download Plugin. --- Netscape 4.7 error message=== === Subject: : Re: Rational LimitsFor intersections of any finite family of nes sets, meaning each> is> a proper subset of all its predecessors, that intersection equals the> last, and therefore smallest, set in the family.For intersections infinite families of nes sets, this cannot be the> case because there is no last or smallest.There must have been a smallest if you are claiming it was the empty> set.> It doesn't matter if it was the last set.Then what is the smallest member ( or last member) of {(0,1/n):n in N}?> I'm claiming there is no smallest set.There cdertainly is no smallest set in teh sequence (0,1/n), so you are right about that.> I think the intersection of {(0,1/n):n in N} is always an open infinite> set.Any open infinite set should have members. Name one.> You are the one claiming the intersection is empty.A proof for my claim:For any positive real number x, there is an integer n such that x > 1/n, so that EVERY positive x is outside at least of one of the (0,1/n) and therefore outside of the intersection of all of them. Since 0 and all negative reals are outside all of them and thus outside the intersection, that means ALL reals are outside the intersection.What numbers do you allege are inside that intersection?> Notice that I said n in N, so n must be finite and not one of your> unicorn numbers.> You mean like the n such that (0,1/n) is an empty set?Why do you think I would say anything so stupid. It is possible for the intersection of infinitely many non-empty sets to be empty.For eacn natural number n, let A_n be the set of all those natural numbers greater than or equal to n. What is the intersection of all of the A_n? If that intersection is not empty it must have a smallest member. If you claim the intersection is not empty, provide us with that number. If you cannot then you have been wrong all along and are just too stupid to reallize it.> You tell me. I don't think there is such an n.> Russell> - 2 many 2 count=== === Subject: : Re: Rational Limits permission for an emailed response.X-Zippy-Says: NEWARK has been REZONED!! DES MOINES has been REZONED!!> I think the intersection of {(0,1/n):n in N} is always an open infinite> set.Can you give me an element in this set? If there are infinitely many, it shouldn't be hard to name just one.=== === Subject: : Re: Rational Limits <4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <877jzkxzdf.fsf@phiwumbda.org> <87wu7hiygd.fsf@becket.becket.net> Discussion, linux)> Can you give me an element in this set? > If there are infinitely many, it shouldn't be hard to name just one.That's not entirely fair.The proof that there are infinitely many uncomputable functions isconsiderably simpler than exhibiting a particular uncomputablefunction (not that the proof that halting problem is uncomputable isparticularly hard, but it's harder than proving there are uncountablymany uncomputable functions in my estimation).On the other hand, we *do* have a proof that the intersection of {(0, 1/n) | n in N} is empty, so he should either (1) find an explicitflaw in that proof or (2) find a valid proof of the negation of thatclaim.Or he can go on saying, Well, I think .... and you expect me tobelieve ...? Very compelling, that line is.-- Jesse F. Hughes[I]t's the damndest thing. There's something wrong with every lastone of you, and I *never* thought that was a possibility. But now Ifeel it's the only reasonable conclusion. --JSH sees some sorta light=== === Subject: : Re: Rational Limits> Can you give me an element in this set?If there are infinitely many, it shouldn't be hard to name just one.> That's not entirely fair.> The proof that there are infinitely many uncomputable functions is> considerably simpler than exhibiting a particular uncomputable> function (not that the proof that halting problem is uncomputable is> particularly hard, but it's harder than proving there are uncountably> many uncomputable functions in my estimation).> On the other hand, we *do* have a proof that the intersection of> {(0, 1/n) | n in N} is empty, so he should either (1) find an explicit> flaw in that proof or (2) find a valid proof of the negation of that> claim.> Or he can go on saying, Well, I think .... and you expect me to> believe ...? Very compelling, that line is.I want to give a variation of Cantor's proof:Let X be the set of all real numbers in [0,1].Assume there exists a sequence, x_n, thatincludes every member of X.Construct two new sequences, a_n and b_n.a_n = 0b_n = x_m if x_m is of the form 1/k (k integer) and x_m < b_(n-1)Is b_n a finite sequence?Assume b_n is finite.This means there exists an n such that (a_n, b_n) is the empty set.We assumed x_n contains every member of X.a_n and b_n are non-equal rational numbers.There must be an infinite number of rationals not in x_n.Assume b_n is infinite.This is the basis of Cantor's proof.If b_n is infinite, there are an infinite number of realsnot in the sequence x_n.How does this relate to the intersection of (0,1/n)?Proofs have been given showing the intersection of(0,1/n) must be empty. Here is an example.For any real x there is an n such that x > 1/n.Therefore, there can be no x such that 1/n > x for all n.Compare that to this:For any natural, n, there is a real x such that 1/n > x.Therefore, there can be no n such that x > 1/n for all x.Let X be the set of rational number in [0,1].Let x_n be a sequence that includes every member of X.Many examples of x_n have been given in other threads.Applying Cantor's algorithm, will b_n be a finite sequence?Assume b_n is finite. This proves x_n does not contain any rationalsbetween (a_n, b_n) for some n.Assume b_n is infinite. There must be an infinite number ofrationals not in the range of x_n.Saying the intersection of (0,1/n) is empty is equivalent to sayingthe reals are countable.Russell- 2 many 2 count=== === Subject: : Re: Rational Limits permission for an emailed response.X-Tom-Swiftie: I just don't understand the number seventeen, Tom said randomly> The proof that there are infinitely many uncomputable functions is> considerably simpler than exhibiting a particular uncomputable> function (not that the proof that halting problem is uncomputable is> particularly hard, but it's harder than proving there are uncountably> many uncomputable functions in my estimation).Hrm, proving the halting problem is uncomputable is easy, it seems tome, once you know how to quote a program within itself. Somelanguages make that easy, but you can do it in any.(define (foo) (if (halts? foo) (foo)))This proves that halts? cannot accurately tell whether its argumenthalts.But different people find different things hard, so it doesn't seemwacko that someone might take it your way.Thomas=== === Subject: : Re: Rational Limits <4eCdnaUn0MJfppLdRVn-sQ@comcast.com> <877jzkxzdf.fsf@phiwumbda.org> <87wu7hiygd.fsf@becket.becket.net> <8765f15z9o.fsf@phiwumbda.org> <87ptd91695.fsf@becket.becket.net> Discussion, linux)>> The proof that there are infinitely many uncomputable functions is>> considerably simpler than exhibiting a particular uncomputable>> function (not that the proof that halting problem is uncomputable is>> particularly hard, but it's harder than proving there are uncountably>> many uncomputable functions in my estimation).> Hrm, proving the halting problem is uncomputable is easy, it seems to> me, once you know how to quote a program within itself. Some> languages make that easy, but you can do it in any.> (define (foo)> (if (halts? foo) > (foo)))> This proves that halts? cannot accurately tell whether its argument> halts.Proving the halting problem is easy, if someone says: Prove that thereis no function such that...Now imagine that Turing has just outlined his notion of computabilityfor you. One of the most obvious things in the whole theory is thatthere are countably many Turing machines and hence uncountably manyfunctions are uncomputable.Finding a particular uncomputable function is not, I think, easy givenonly this background knowledge. Once you know *which* function oneought to look at, it's pretty simple to prove it's uncomputable.Discovering that function in the first place is not too simple.But even if I'm wrong and the halting problem function was the firstfunction anyone ever thought of showing was uncomputable -- that partof the proof is *still* harder than proving that there are uncountablymany uncomputable functions. {f:N -> N} = {f:N -> N | f is computable} u {f:N -> N | f is uncomputable}The left hand side is uncountable. The first set of the right handside is countable. Therefore, the second set of the right hand sideis uncountable.You can't tell me that discovering and proving that the haltingproblem is uncomputable is comparably easy to *that*. -- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. --E.A. Poe=== === Subject: : The Right Stuff?Think of this as a combination of Jazz and Chamber Music. Paola Zizzi playing the cello with me on the Jazz saxophone improvising in Caffe Beat.Excerpts, with my Commentaries, from:HOLOGRAPHY, QUANTUM GEOMETRY, AND QUANTUMINFORMATION THEORYP. A. ZizziDipartimento di Astronomia dell Universit.88 di Padova, Vicolo dell Osservatorio, 535122 Padova, Italye-mail: zizzi@pd.astro.itABSTRACTWe interpret the Holographic Conjecture in terms of quantum bits (qubits). N-qubit states areassocia with surfaces that are punctured in N points by spin networks edges labelled by the spin-1/2 representations of SU(2) , which are in a superposed quantum state of spin up and spindown. The formalism is applied in particular to de Sitter horizons, and leads to a picture of theearly inflationary universe in terms of quantum computation. A discrete micro-causality emerges,where the time parameter is being defined by the discrete increase of entropy.Then, the model is analysed in the framework of the theory of presheaves (varying sets on a causalset) and we get a quantum history. A (bosonic) Fock space of the whole history is considered.The Fock space wavefunction, which resembles a Bose-Einstein condensate, undergoes decoherenceat the end of inflation. This fact seems to be responsible for the rather low entropy of our universe.Contribution to the 8th UK Foundations of Physics Meeting, 13-17 September 1999, ImperialCollege, London, U K..1 INTRODUCTIONToday, the main challenge of theoretical physics is to settle a theory of quantum gravity that willreconcile General Relativity with Quantum Mechanics.There are three main approaches to quantum gravity: the canonical approach, the histories approach,and string theory. In what follows, we will focus mainly on the canonical approach; however, wewill consider also quantum histories in the context of topos theory [1-2].A novel interest in the canonical approach emerged, about ten years ago, when Ashtekar introducedhis formalism [3] which lead to the theory of loop quantum gravity [4].In loop quantum gravity, non-perturbative techniques have led to a picture of quantum geometry. Inthe non-perturbative approach, there is no background metric, but only a bare manifold. It followsthat at the Planck scale geometry is rather of a polymer type, and geometrical observables like areaand volume have discrete spectra.Spin networks are relevant for quantum geometry. They were inven by Penrose [5] in order toapproach a drastic change in the concept of space-time, going from that of a smooth manifold to thatof a discrete, purely combinatorial structure. Then, spin networks were re-discovered by Rovelli andSmolin [6] in the context of loop quantum gravity. Basically, spin networks are graphs embedded in3-space, with edges labelled by spins j = 0, 1/2, 1, 3/2...and vertices labeled by intertwiningoperators. In loop quantum gravity, spin networks are eigenstates of the area and volume-operatorsHowever, this theory of quantum geometry, does not reproduce classical General Relativity in thecontinuum limit.JS: Note that.PZ: For this reason, Reisenberg and Rovelli [8] proposed that the dynamics of spinnetworks could be described in terms of spacetime histories, to overcome the difficulties of thecanonical approach. In this context, the spacetime histories are represen by discretecombinatorial structures that can be visualised as triangulations. The merging of dynamicaltriangulations with topological quantum field theory (TQFT) has given rise to models of quantumgravity called spin foam models [9] which seem to have continuum limits.JS: What is the relation of Hagen Kleinerts world crystal lattice to spin foam dynamical triangulations?PZ: Anyway, as spin foam models are Euclidean, they are not suitable to recover causality at the Planck scale.For this purpose, the theory should be intrinsically Lorentzian. However, the very concept of causalitybecomes uncertain at the Planck scale, when the metric undergoes quantum fluctuations as Penrose [10] argued.So, one should consider a discrete alternative to the Lorentzian metric, which is the causal set(a partially ordered set-or poset- whose elements are events of a discrete spacetime).Such theories of quantum gravity based on the casual set were formula by Sorkin et al. [11].Rather recently, a further effort in trying to recover causality at the Planck scale, has been undertaken by Markopoulou and Smolin [12]. They considered the evolution of spin networks in discrete time steps, and they claimed that the evolution is causal because the history of evolving spin networks is a causal set. However, it is not clear yet whether causality can really be achieved at the Planck scale, at least in the context of causal sets.In this paper, we also consider the issue of causality at the Planck scale in the framework of causal sets, although our results do not promote this aspect as decisively as in [12]. In our opinion, the very concept of causality makes sense in the quasi-classical limit, in relation with the fact that an observer is needed, and this observer cannot stand on the event at the Planck scale. Moreover, our picture is rela to some other issues, mainly the holographic conjecture [13-14] and quantum information theory.The issue of quantum information star with Neumann [15], who gave a mathematical expression of quantum entropy in photons and other However, Neumann s entropy lacked a clear interpretation in terms of information theory. It wasSchumacher [16], who showed that Neumanns entropy has indeed a rela meaning. Moreover,Holevo [17], and Levitin [18], found that the value of quantum entropy is the upper limit of theamount of quantum information that can be recovered from a quantum [19], in several aspects.The elementary unit of classical information is the bit, which is in one of the two states true = 1 andfalse = 0, and obeys Boolean logic. A classical bit is contextual-free. i.e. it does not depend on whatother information is present. Finally, a classical bit can be perfectly copied.In quantum information, the elementary unit is the quantum bit (or qubit), a coherent superpositionof the two basis states 0 and 1. The underlying logic is non-Boolean. Different from classical bits, aqubit is contextual [20], and cannot be copied, or cloned [21].JS: Clearly our inner thoughts are contextual more like quantum bits than classical bits. But that is not enough. Cloning or copying the thought is needed for intelligent non-random creative evolution. The inability to clone or copy a qubit comes from the linearity of superposition in micro-quantum theory and is associa with signal locality, i.e. the inability in micro-quantum theory to use controlled entanglement as a stand-alone conscious-command-control-communication channel C^4 that is both its own lock and key. Antony Valentini showed that signal locality is an artifact of sub-quantal heat death or sub-quantal thermodynamic equilibrium manifes in the Born configuration space. Bohm and Hiley showed that signal locality comes from the fragility of the quantum potential, i.e. from lack of generalized phase rigidity (P.W. Anderson). The rules of the game change in the More is different ground state phase transition from micro-quantum to MACRO-QUANTUM theory where there is a tradeoff:Nonlocality -> localityQuantum entropy as log of phase space volume of ground state decreases.Linearity of nonlocal Schrodinger eq. -> nonlinearity of local Landau-Ginsburg eq.Breakdown of Born probability game.Fragile micro-quantum potential -> robust MACRO-QUANTUM POTENTIALwith generalized phase rigidity and the possibility of signal nonlocality.PZ: The emerging fields of quantum computation [22], quantum communication and quantumcryptography [23], quantum dense coding [24], and quantum teleportation [25], are all based onquantum information theory. Moreover, quantum information theory is expec to illuminate someconceptual aspects of the foundations of Quantum Mechanics.In this paper, we shall illustrate, in particular, the interconnections between quantum informationtheory and quantum gravity.As spin networks are purely mathematical entities, they do not carry any information of their ownbecause information is physical [26]. However, they do carry information when they puncture asurface transversely. Each puncture contributes to a pixel of area (a pixel is one unit of Planck area)and, because of the holographic principle, this pixel will encode one unit of classical information (abit).JS: The idea of puncture is really 3D duality. A string theory line is dual to a loop quantum gravity area, and a point is dual to a 3D volume. One can generalize this to N-Dim bosonic hyperspace. The SU(2) qubit transformations are 2x2 Pauli spintronic matrices that introduce the hypercomplex fermion dimensions of supersymmetry (fermions <-> bosons).The idea then is that each component of 4-vector Xu is really a quaternion rather than a real number xu. This gives a non-commutative space-time geometry. This is not the same as representing an ordinary first rank tensor as a quaternion. That is we do not have a 4D real vector space, but rather a 16 D real vector space or an 8D complex vector space or a 4D quaternion vector space.PZ: The starting idea of our paper is that a surface can be punctured by a spin networks edge (labeled by the j = 1/2 representation of SU (2)), which is in a superposed quantum state of spin up and spin down. This induces the pixel to be in the superposed quantum state of on and off, which encodes one unit of quantum information, or qubit. By the use of the Bekenstein bound [27], we show that the entropy of an N-punctured surface is in fact the entropy of an N-qubit.Moreover, because of the Holevo-Levitin theorem [17-18], our picture provides a time parameter interms of a discrete increase of entropy. Thus the time parameter is discrete, and quantized inPlancks units. JS: The problem here is how to make this Diff(4) symmetric independent on how the space-time in classical limit is sliced. On the other hand R. Kiehn says that Diff(4) is reversible. You need something else to get arrow of time. Pfaff dimension 4?PZ: This is in agreement with Penrose s argument [28], that a theory of quantum gravityshould be time-asymmetric. However, our result is not strong enough to claim that causality isactually being recovered at the Planck scale. In fact, the thermodynamic arrow of time is rela tothe psychological arrow of time, and at the Planck scale, the latter is missing because there are noobservers. This is not just a matter of osophy, but it is a real handicap.The events of our causal set are not just points of a discrete spacetime, but they are the elementsof a poset whose basic set is the set of qubits and whose order relation follows directly from thequantum entropy of the qubits.Events that are not causally rela (qubits with the same entropy) form space-like surfaces (orantichains).The use of topos theory (in particular the theory of presheaves [29]) makes it possible to attach aHilbert space to each event, and to build up discrete evolution operators [30], between differentHilbert spaces. In this way, we get a quantum history, which turns out to be a quantum informationinterpretation of the theory of inflation.However, as we already poin out, we find an internal contradiction, in considering causality inthe framework of causal sets: an observer seems to be required for consistency.In fact presheaves (or varying sets on a causal set) obey the Heyting algebra which implies theintuitionistic logic [31-32], which in turn is rela to the concept of time flow. One can thenimagine a Boolean-minded observer who has to move in time [33], in order to grasp the underlyingquantum logic of the universe.JS: Why a Boolean-minded observer? This is an error. While my motor activity my body is Boolean, my inner consciousness is non-Boolean even without psycho-active drugs.PZ: Obviously, this picture has no meaning at the Planck time, when there was no observer at all. The above picture strictly depends on the fact that in the theory of presheaves we consider Boolean sub-lattices of the quantum lattice. To drop the Boolean observer, one should discard Boolean sub-lattices and consider the entire quantum lattice as a whole, endowed with a non-Boolean logic. However, we cannot get the Hilbert space of the entire history, because of the existence of unitary evolution operators among different Hilbert spaces. Thus, in order to escape the problem of the observer we perform the tensor sum of all the Hilbert spaces attached to the events of the causal set, and we get a (bosonic) Fock space. Although we find that the logic associa with the Fock space is Boolean, we are able to eliminate the observer at the Planck scale by depriving him of time evolution. In fact, the Fock space wave function Y , which is the coherent quantum superposition of all the events (qubits), leads to an atemporal picture of the early universe.The wavefunction Y is the coherent quantum state of multiple bosonic qubits, and resembles aBose-Einstein condensate [34].JS: Only resembles? How about is? Like Feynman reading Dirac on the Lagrangian in quantum theory.PZ: It can maintain coherence as far as thermal noise is absent, that is, as far as inflation is running: a cosmological era when the universe is cool and vacuum-domina. We find that the coherent quantum state decoheres at the end of inflation, giving rise to the (rather) low entropy of our universe. In fact, the present entropy seems to be rela to the quantum information stored in Y , which was lost to the emerging environment when thermal decoherence took place [35]. Only after Y has decohered, one can in principle retrodict the quantum past [36], as it was recorded by some ancient observer. It seems to us that only in this sense causality can be restored at the Planck scale.JS: John Wheeler in his Delayed Choice Law without law Universe as a self-exci circuit seems to invoke the Once and Future Observer-Participator. So did Sir Fred Hoyle in The Intelligent Universe.PZ: This paper is organised as follows.In Sect.2, we review the holographic conjecture by t Hooft and Susskind, and we interpret theinformation stored on a boundary surface in terms of quantum bits rather than classical bits.In Sect.3, we illustrate the relation between spin networks that puncture a boundary surface, and thequbits stored on the surface.In Sect.4, we put forward the mathematical formalism. We select the subspace of un-entangledsymmetric qubits of the full 2 ^N-dimensional Hilbert space of N qubits. We find that the symmetric1-qubit acts as a creation operator in this subspace. Then the formalism is applied to de Sitterhorizons, which are interpre as qubits. This leads to an interpretation of inflation in terms ofquantum information.In Sect.5, we show how the vacuum (the classical bit) at the unphysical time t=0, evolved to the 1-qubit at the Planck time by passing through a quantum logic gate.In Sect.6, we show that in this model a discrete micro-causality emerges, where the time parameteris given in terms of the discrete increase of entropy.In Sect.7, we formalise our model within the framework of the theory of presheaves. Qubits areinterpre as events of a causal set, where the order relation is given in terms of the entropy. Weget thus a quantum history which is the ensemble of all the finite-dimensional Hilbert spaces of N qubits.We find that the wavefunction of the entire history resembles a Bose-Einstein condensate.In Sect.8, we consider decoherence of the Bose-Einstein condensate. We find that the rather lowvalue of the entropy of our present universe can be achieved only if the wavefunction collapsed atthe end of inflation.In Sect.9, we interpret the N-qubits as N quantum harmonic oscillators. This allows us to find thediscrete energy spectrum, and in particular, the energy at the time when inflation ended. This energyis interpre as the reheating temperature.JS: In my theory, a local macro-coherent vacuum order parameter survives.The reheating is still tiny because matter is only 4% of all the stuff in our at leastTegmark Level 1 Hubble bubble universe confined to our past light cone.Gravity is from the phase ripples in this coherent order and dark energy/matter that is 96% of The Right Stuff of The World is from the amplitude ripples inthis coherent order. The oil calming the waters.This is Sarfattis decoding of The Cipher of Genesis.=== === Subject: : Re: The Right Stuff?> This is Sarfattis decoding of The Cipher of Genesis.Pure crap actually. Try stringing together a single sentence that makes sense.=== === Subject: : Re: JSH: Algebra is supreme> The algebra is simple. And algebra IS supreme, whether this or that> poster wishes to challenge it.> Rick Decker, a professor at Hamilton College, gave this example, which> I like because it's a quadratic, which is a lot easier to play with> than the cubics I've used before. If you want to see his original> post here are some headers which also show that he is indeed at> Hamilton College:> === Subject: : Re: Mathematical consistency, courage> Decker put forward the quadratic> (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).> I think that is what leaves some of you guessing and hesitant because> it's not a factorization like you're used to seeing.> Normally you see something like> (x+2)(x+1) = x^2 + 3x + 2> and there's nothing hidden, it's easy. There's no need to trust the> algebra, as you can trace everything out *easily*, while the Decker> example forces you to accept algebra not because you can see every> detail, but because it's true.> For instance, algebraically, factorizations multiply out in a certain> way demonstra by> (a+b)(c+d) = ac + ad + bc + bd> and that's just a FACT which is not open to debate.> Looking at> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > it IS true that> 25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).> Now subtracting 14 from both sides gives> 25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x).> Now notice that you have 35 on the left side, but 0 on the right in> terms of *constants* i.e. terms that don't vary as x varies.> That is just algebra. It's simple, and direct.> What I do is to use a_2(x) = b_2(x) - 1, so that I have> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > which again is JUST ALGEBRA, so there isn't anything weird.> Now multiplying out I get> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)> and subtracting 14 from both sides gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> which again is JUST ALGEBRA, so there's NO ROOM FOR DEBATE. OK up to this point ...> Now then, clearly with> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> the 7 and 2 visible to the naked eye on the left, are factors of 7(2)> on the right, which is verified by multiplying out!!!> It's just algebra. There's no room for debate. There's no poll to> take, and no reason to go look for votes either way!> The simple fact is that 7 and 2 on the left are factors of 7(2) on the> right, and you can multiply out, and subtract 14 from both sides to> get> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> and it's not magic that the *constants* are now gone from both> sides!!! Still OK.> It's ALGEBRA. Yes.> Now then, if 7 and 2 on the left are the factors of 7(2) on the right,> then what happens if you divide by 7?> Well you get 1 and 2 on the left as the factors of 2 on the right. There are infinitely many ways to factor 7. You are assuming there is only one which works in this case. True in a sense,but it is not the one you have chosen.> That looks like> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 Yes. That is the wrong factorization (unless x = 0), and as you well know, it doesn't accomplish what you wantbecause a_1(x)/7 is not an algebraic integer. That factin itself is trying very hard to tell you something. Here's what it's trying to tell you: THAT IS THE WRONG FACTORIZATION.> and here's where certain posters decide to attack algebra, as if it's> a debate, as if it's a democracy, as if mathematics gives a damn that> they don't like a conclusion! No, no, no. It's much more impersonal than that. What youare doing is not the right way to factor 7 out of the leftside. Yesterday you almost had it. You were talking about factoring7 in the form w1*w2, where w1 and w2 are algebraic integers, and (5 a_1(x) + 7)/w1 and (5 b_2(x) + 2)/w2 are both algebraic integers. Such w1 and w2 exist. You called it the mushing position. You thought it lookedkind of funny. You said it isn't mathematics. Wrong. If you choose w1 and w2 in the right way (they dependon x, incidentally), it is EXACTLY the right way to factor7 out of both sides. The expression you get on the left sideis the product of two algebraic integers. That is your goal in this case. You discussion yesterday was weird. Clearly you understood about 97% of what we have been trying to tell you. You werevery close to getting the whole thing. Then you abruptly turnedyour back on it, almost as though you were afraid. You turnedand ran away, back into your denial-cave and your blather aboutconstant terms, etc. You're not too dumb to understand this. Your not understanding it is an act of will, not a deficiency inlogic.> That's because the two factors (5a_1(x)/7 + 1) and (5b_2(x) + 2) are> not necessarily algebraic integers, and for some values, like with> x=1, or x=1 mod 7, you can *see* that they aren't algebraic integers! Correct. True for almost all integers: true whenever a^2 - (x - 1)*a + 7*(x^2 + x)is irreducible. One value for which it IS reducible is x = 0:a very special, singular case. You have made the mistake of tryingto generalize from that case to all others. It is exactly thewrong thing to do. Again, this fact is trying to send you a message. The messageis: YOU ARE USING THE WRONG FACTORIZATION OF 7. THERE IS ANOTHER FACTORIZATION THAT PRODUCES ALGEBRAIC INTEGER COEFFICIENTS ON THE LEFT SIDE. 7 * 1 CANNOT WORK UNLESS X = 0.> The problem is that these posters believe that there's SOME WAY you> can divide 7 from both sides *in general* and get algebraic integer> factors on the left side! Yes, you have got it exactly. Not only do we believe it, butalso we have *specified* it: w1(x) = gcd(a_1(x), 7) and w2(x) = gcd(a_2(x), 7). Check it out. First of all, by the properties of a gcd, w1(x) | a_1(x) and w1(x) | 7 w2(x) | a_2(x) and w2(x) | 7 This means that 5 a_1(x)/w1(x) + 7/w1(x) and 5 a_2(x)/w2(x) + 7/w2(x)are both sums of algebraic integers, and are algebraic integersthemselves. Moreover, because in general 25*x^2 + 30*x + 2is not divisible by 7, you can show that w1(x) * w2(x) = 7.> That's it!!! Yep. You got it.> That's what all the arguing is about because the algebra say NO, and> they say YES, and I keep saying what the algebra says, and they keep> saying what they say, and for some reason most of you seem to> disbelieve the algebra!!! We both use algebra. Our algebra is no less valid and correctthan your algebra. The two sides of the equation balance, whether you divide by 7*1 or by w1(x)*w2(x). The difference iswith your algebra, you end up with a_1(x)/7 not being an algebraicinteger. Whereas with my (or Rich Decker's) algebra, a_1(x)/w1(x)is GUARANTEED to be an algebraic integer. You run into a contradiction;we do not. Again, this is trying very hard to tell you something: YOU HAVE CHOSEN THE WRONG FACTORIZATION OF 7 !!!> The problem is that if you divide both sides by 7, and decide that one> factor has sqrt(7) as well as the other then you get> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2 sqrt(7) is appropriate ONLY when x = 1. In other cases, 7 = sqrt(7)*sqrt(7) is the wrong factorization also. A keyfact: in the correct factorization, 7 = w1(x)*w2(x), bothw1(x) and w2(x) are ***dependent on x***. There is noCONSTANT solution. In general the expressions for w1(x) and w2(x) are very complica. showed thatwhen x = 2, w1(x) is the root of an 11th degree monicpolynomial.> and that's JUST ALGEBRA, there's NO ROOM FOR DEBATE!!!> No point in calling your congressman to promote your right to have a> mathematical conclusion that fits your needs!!! You are the one around here who makes appeals to congressmen,employers, FBI, Army, the President, etc.. The rest of us justargue the math.> The FACT is that if both factors have sqrt(7) as a factor for ANY> VALUE of x, No one has said that. sqrt(7) is the right choice for w1(x) ONLY WHEN x = 1. The choice of factors is NOT CONSTANT. Maybe youshould write that down and think about it. NO ONE SAYS THE FACTORSARE ALWAYS sqrt(7)! > then what you have is> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2> and that gives the factors of 2 on the right as> sqrt(7) and 2/sqrt(7) The value of 2/sqrt(7), or more generally, 2/w2(x), is NOT IMPORTANT. Agreed, in general it is NOT an algebraic integer. Neither Dik nor Rick Decker nor I disagree with you about that. The important thing here is that (5 b_1(x) + 2)/w2(x)is an algebraic integer. See above for the correct definition of w2(x) which guarantees that this will happen. And note thatit is DEPENDENT ON x ! Note also that (5 b_1(x)/w2(x) + 2/w2(x)) = (5 a_2(x)/w2(x) + 7/w2(x)),and that both of the terms on the RIGHT side of this equationare algebraic integers. Therefore the whole expression on theright side is an algebraic integer. Therefore the whole expression on the LEFT side is also an algebraic integer, *even though 2/w2(x) is not*. There is no paradox here. Neither 5 b_2(x)/w2(x) nor 2/w2(x)are algebraic integers. But their sum *is* one. That is not strange. Neither 5 nor 9 is divisible by 7, but their sum is.The sum of two non-algebraic-integers can easily be an algebraicinteger. Another example: let A = sqrt(2) - 1/3 andB = sqrt(2) + 1/3. Neither A nor B is an algebraic integer.But A + B = 2*sqrt(2) is one. > and if you argue with that ALGEBRAIC FACT then you are debasing> yourself from the realm of sentient debate into an animal realm where> you're showing a weak need to push whatever the hell makes you happy> over LOGIC!!!> IF BOTH FACTORS HAVE sqrt(7) AS A FACTOR THEN ALGEBRA TELLS YOU THAT> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2> which tells you that the factors of 2 on the right are > sqrt(7) and 2/sqrt(7)> which works to give you 2, > BUT NOT IN THE RING OF ALGEBRAIC INTEGERS!!! Absolutely correct (for x = 1). And absolutely irrelevant to what you actually want. The value of 2/sqrt(7) by itself is useless.It is the value of the whole expression, (5 b_2(x) + 2)/w2(x)which has to be an algebraic integer.> The logical conclusion which follows mathematically is that given> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> dividng both sides by 7 is not possible within the ring of algebraic> integers in general. Totally false. Choose w1(x) and w2(x) as gcds, as specified above. Note that w1(x)*w2(x) = 7. Everything works out as I described above.> That is, there is no decomposition of 7 into algebraic integer factors> of> (5a_1(x) + 7) and (5b_2(x) + 2)> in general in the ring of algebraic integers. No, that's just wrong. The correct factorization is givenabove. It always works.> That's it! Sorry if it hurts your feelings. Sorry if it makes you> want to go cry to God or your mother because it JUST IS THAT WAY!> I didn't make it that way!!! It's NOT that way. Your algebra is fine. It's your factorization that is wrong.> I'm not forcing algebra to be something else just to ruin your life,> hurt your feelings, make you mad, or do anything at all because IT'S> NOT POSSIBLE TO CHANGE THE ALGEBRA!> ALGEBRA IS SUPREME!> It's the supreme authority here. Algebra *is* the key here, but more than just high-school algebraand factorization-by-inspection, which is what you mean when yousay algebra. You need to go beyond that into some nontrivial algebraic number theory here to understand the right way to factorthe Decker polynomial (and your own cubic polynomial). You have badlyhandicapped yourself here by refusing to learn any theory at all. You are like a one-legged man in a foot race. > === === Subject: : Re: JSH: Algebra is supreme > The FACT is that if both factors have sqrt(7) as a factor for ANY > VALUE of x, then what you have is > (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = No James, you do *not* have that, you have: [(5a_1(x) + 7)/sqrt(7)][(5b_2(x) + 2)/sqrt(7)] = > 25x^2 + 30x + 2Whenever the first is completely valid in the algebraic integers, thesecond is also completely valid. The reverse does *not* hold. > and if you argue with that ALGEBRAIC FACT then you are debasingWhat algebraic fact? That the second formulation being valid in thealgebraic integers implies that the first formulation is also validin the algebraic integers? That is not an algebraic fact, it isactually false in general. Even when you stay in the integers. > (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2 which tells you that the factors of 2 on the right are > sqrt(7) and 2/sqrt(7)No, it does not tell you that. There are no factors of 2 anymore whenyou chose a particular factor of x. > The logical conclusion which follows mathematically is that given > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > dividng both sides by 7 is not possible within the ring of algebraic > integers in general.It is, but not with a fixed decomposition of 7. > That is, there is no decomposition of 7 into algebraic integer factors > of > (5a_1(x) + 7) and (5b_2(x) + 2) > in general in the ring of algebraic integers.Not a fixed decomposition, but there is one dependent on x. > That's it! Sorry if it hurts your feelings. Sorry if it makes you > want to go cry to God or your mother because it JUST IS THAT WAY!Yes, sorry.Do you know what a gcd is? Do you understand it? I will define it.First some notation: a | b means a divides b in the ring we are working in.Now the definition: gcd(a, b) is defined as a number such that all of the following hold: (1) gcd(a, b) | a (2) gcd(a, b) | b (3) for any x such that x | a and x | b we have x | gcd(a, b).(Note that the definition says a number, there are multiple number, butthey are rela in that you can get from one to the other by multiplicationwith a unit.)Dedekind has shown that the algebraic integers form a gcd domain (actuallya bit more: it is a Bezout domain), so the gcd exists. It is quitesimilar to the standard gcd in the integers, except that this definitionis not concerned with the sign, so in the integers, the gcd of 6 and 14is +2 or -2.Now a little theorem: gcd(a * b, c) | gcd(a, c) * gcd(b, c).You might try to prove it. The prove is easy in a unique factorisationdomain, less easy in other domains.Now, lessee. For once give me some answers. Which of the followingsteps is false in the algebraic integers?1. define g1(x) = gcd(5a_1(x) + 7, 7)2. define g2(x) = gcd(5b_2(x) + 2, 7)3. define v(x) = g1(x) * g2(x) / 74. define h1(x) = gcd(v(x), g1(x))5. define h2(x) = v(x) / h1(x)6. define w1(x) = g1(x) / h1(x)7. define w2(x) = g2(x) / h2(x)8. (5a_1(x) + 7)/w1(x) is an algebraic integer function9. (5b_2(x) + 2)/w2(x) is an algebraic integer function10. w1(x) * w2(x) = 711. [(5a_1(x) + 7)/w1(x)]*[(5b_2(x) + 2)/w2(x)] = 25x^2 + 30x + 2. > I'm not forcing algebra to be something else just to ruin your life, > hurt your feelings, make you mad, or do anything at all because IT'S > NOT POSSIBLE TO CHANGE THE ALGEBRA! > ALGEBRA IS SUPREME! > It's the supreme authority here.I could just as well reciprocate this.-- === === Subject: : Re: Convergent subsequence> Can there be a (non-random) non-convergent sequence {a[n], n in N} for which> NO convergent subsequence exists?> Thanks in advance,> --a[n] = n springs to mind.=== === Subject: : Re: Convergent subsequence> Can there be a (non-random) non-convergent sequence {a[n], n in N} for> which> NO convergent subsequence exists?> Duh,> take an eventually constant sequence.What do you mean by non-random? Why is a sequence that is eventually constant an example of a non-convergent sequence?> Sorry, let me rephrase that:> Can there be a (non-random) non-convergent sequence {a[n], n in N} for which> NO non-eventually constant convergent subsequence exists?I'm still not sure what you're asking. Every bounded sequence of real numbers contains a subsequence converging to a real number. If the sequence is not bounded above, it contains a subsequence -> oo. If the sequence is not bounded below, it contains a subsequence -> -oo.=== === Subject: : Re: Convergent subsequence charset=iso-8859-7=== The World Wide Wade [snip]> Sorry, let me rephrase that:Can there be a (non-random) non-convergent sequence {a[n], n in N} forwhich> NO non-eventually constant convergent subsequence exists?> I'm still not sure what you're asking. Every bounded sequence of real> numbers contains a subsequence converging to a real number. If thesequence> is not bounded above, it contains a subsequence -> oo. If the sequence is> not bounded below, it contains a subsequence -> -oo.Yes, sorry, I wasn't entirely clear. Here's my situation:I have a sequence {a[n], n in N} of Complex numbers. Upon plotting thesequence with Maple it appears to be chaotic (it looks like randomsprinklings on the Complex plane), but I wasn't able to find on yahoo anyofficial definitions of what chaotic might mean.I know that it doesn't converge. What would be the best way to prove that?Perhaps by finding two separate subsequences which converge to differentlimits? What if the sequence is unbounded on C?PS: Thanks for the two other answers, re: a[n] = n from Zdislav. I thoughtof that too, as soon as I went to bed. :*(--------------------------------------------Eventually, _everything_ is understandable=== === Subject: : Re: Convergent subsequence> The World Wide Wade [NonBreakingSpace].8b.96.87.8c .97.99.95>[snip]>> Sorry, let me rephrase that:Can there be a (non-random) non-convergent sequence {a[n], n in N} for>which>> NO non-eventually constant convergent subsequence exists?I'm still not sure what you're asking. Every bounded sequence of real>> numbers contains a subsequence converging to a real number. If the>sequence>> is not bounded above, it contains a subsequence -> oo. If the sequence is>> not bounded below, it contains a subsequence -> -oo.>Yes, sorry, I wasn't entirely clear. Here's my situation:>I have a sequence {a[n], n in N} of Complex numbers. Upon plotting the>sequence with Maple it appears to be chaotic (it looks like random>sprinklings on the Complex plane), but I wasn't able to find on yahoo any>official definitions of what chaotic might mean.>I know that it doesn't converge. What would be the best way to prove that?Hard to see what this has to do with your original question. Anyway,it's impossible to say what the best way to prove it's non-convergentis without knowing what the sequence is - looks like randomsprinklings is a little too vague to lead to a proof.>Perhaps by finding two separate subsequences which converge to different>limits? What if the sequence is unbounded on C?>PS: Thanks for the two other answers, re: a[n] = n from Zdislav. I thought>of that too, as soon as I went to bed. :*(************************=== === Subject: : Chaos Question re Strange AttractorsI suppose, the answer is no: your condition implies that the attractorconsists of only one orbit, so this is not a strange attractor. Onthe other hand, if the system is ergodic on the attractor with respectto some good measure, then the trace of any (nonempty) open subset isdense and has full measure. Simeon> Will any given subset (other than the null set) of a strange attractor> ultimately flow to equal and fill out the entire strange attractor?> Thanks -> L=== === Subject: : Re: goldbach's and fermat's proofs>You may view my proofs at the Florida State web page:>www.math.fsu/Science/Specialized under Goldbach Proof and In>Defense of Mr. Fermat. Please mail me any questions. Kerry>Evans.> I have tried to read both of your papers twice. But I could not> follow your notations. Could you rewrite your paper in MS Word, usingObject; Math Equation? (Hopefully converting to PDF)> I would be glad to read your paper. I have been a little depressed> with my Differential Equation class whose textbook is written for> machines that need to be programmed.> erdos fan> You might like to know that I am rewriting some segments of my> Goldbach proof. Meanwhile there is a pdf listing on> www.mathprints.com. KerryYou repealy say things like 'let n and m be as defined' which isreally confusing. Either explain what holds in every instance, or makea definition to define things.In the golbach one for Lemma I-b you seem to claim that if n-m=1, andn+m is an odd prime then [(n+m-1)!]^4=(n+m+1)(mod 2*(n+m))Let n=2, m=3[(2+3-1)!]^4 (mod 2(2+3))=4^4 (mod 10)=4 (mod 10)(2+3+1) (mod 2(2+3))=6 (mod 10)Unfortunately 6 does not equal 4 in mod 10.For the Fermat one, the definition of F(x,y), and the definition of {}are really confusing. It could also would probably help make thingsclearer if you explain upfront why they are defined that way.I also don't understand the following paragraph>Suppose now that (1) has been rewritten for some n, where n is equalto or>greater than n' so that [r,a,b and n] is transformed to a reducedform,>redefining [r,a, b and n] to be [u,v,w and n']. Thus n' is greaterthan or>equal to 2 implies [v>u>w>0]. Consider the transformation (undern'), T,>such that T:v to v, u to w, and w to u. Respecting (1). T(v,u,w) =Tnot>(v,u,w) where Tnot signifies not T. It follows that [u>w] can be >transformed to [w>u] such that [u>w] is unaltered.! This can happenonly for>the case,u=w, which is moot. Conclusively, order must be assignablewhen it>exists. ( ! denotes contradiction.).I don't understand what exactly you mean by t, and I don't understandwhat you mean by tnot. It almost looks like you are trying to say thatu^n+v^n=w^n implies u^n+w^n=v^n under some conditions, but I don't seethat justified. You also go on about ordering, although I'm not surewhat you mean. Obviously u^n+v^n=w^n is equivalent to v^n+u^n=w^n,although u>v implies u^n>v^n for real n>0.Gershon Bialer=== === Subject: : Re: Google's rankings>> We perfectly agree on this: it doesn't matter the intrinsic value of the>> website, which makes Google's ranking totally useless and their>> proprietary ranking algorithm a joke.>Experiment proves you wrong: millions of people have made Google the first>place they go to search for information, usually find what they want on the>first page of results, and rarely find a need to go to other search engines.>People have done this because Google *works*, and works much *better* than>any of the previous search engines.I would have agreed with you before the Florida update. Google usedto be really good for searches. Of course, in an effort to thwartSEOs, google made a drastic algorithm change. While it sucks for those of us who make our money on the net, its alsonegative for the users of google. Have you tried finding somethinglately? It seems most searches are quite skewed now. I find myselfhaving a harder time finding things on google now. And the funniestpart is, those sites that used to be top for terms like hardcore sexare still at the top, so apparently the algorithm isn't affectingspammers as much as they liked. This is probably the reason some have concoc conspiracy theoriesthat google did it to sell more adwords. After all, those most likelyto buy adwords (legitimate small businesses) are the ones mostaffec by florida. Just my two cents. :: Jeremy Morgan :: Self Proclaimed ExpertWeb Developerhttp://www.webfootcentral.com=== === Subject: : Re: Google's rankings>> We perfectly agree on this: it doesn't matter the intrinsic value of the>> website, which makes Google's ranking totally useless and their>> proprietary ranking algorithm a joke.>>Experiment proves you wrong: millions of people have made Google the first>place they go to search for information, usually find what they want on the>first page of results, and rarely find a need to go to other search engines.>>People have done this because Google *works*, and works much *better* than>any of the previous search engines.> I would have agreed with you before the Florida update. Google used> to be really good for searches. Of course, in an effort to thwart> SEOs, google made a drastic algorithm change.> While it sucks for those of us who make our money on the net, its also> negative for the users of google. Have you tried finding something> lately? It seems most searches are quite skewed now. I find myself> having a harder time finding things on google now. And the funniest> part is, those sites that used to be top for terms like hardcore sex> are still at the top, so apparently the algorithm isn't affecting> spammers as much as they liked.> This is probably the reason some have concoc conspiracy theories> that google did it to sell more adwords. After all, those most likely> to buy adwords (legitimate small businesses) are the ones most> affec by florida.Good point. They do not ban my paid advertising after all. I think there's apotential for an interesting lawsuit based on unfair business practice.=== === Subject: : Re: Google's rankings>> http://www.scientology.com ==> 4/10>> http://www.datashaping.com ==> 0/10This gives you an idea on how biased Google's ranking is, favoring>> criminals over legitimate businesses.Hey stooopid - Google ranking is decided by software based on>> empirically valida statistical elements. Google uses their model>> because it works. Google favors nobody, and better is orthognal to>> their analysis. If you were looking for a whole lot of strange and>> ended up at the math site, you'd be holding your little wee-wee in>> your palsied hand and screaming that Google had chea you.Tell ya what, stooopid, Google Amanda Peet. One is an actress who>> takes off her clothes a lot, the other is a respec string>> theorist. Why don't you get up on your hind legs and tell us which>> one is the ripoff?Actually, Google is ripping you off.http://www.google-watch.org/gifs/gscrew.gif> http://www.google-watch.org> http://www.scroogle.org> Google-watch is run by one Mr. Daniel Brandt, who, while he has some > interesting things to say, was likewise bitten by Google when his massive > site wasn't indexed all the way and his content pages got a low PageRank. > He thinks he *deserves* higher.> http://www.salon.com/tech/feature/2002/08/29/google_watch/> The dispute is personal which, IMHO, diminishes Mr. Brandt's credibility.> GreyJust got this error from google: Your browser's cookie functionalityis turned off. Please turn it on., spook.JS=== === Subject: : Re: when NaturalNumbers = p-adics what alters in the Riemann Hypothesis Re: proof of the Riemann Hypothesis(snips)> I would have had to consider two extreme cases: (1) that the RH is> completely false just as FLT is completely false when NaturalNumbers => p-adics. If taking that extreme route would have had me find out what> was flawed in my claimed two proofs below.> (2) the second extreme case is to say that something lies on the 1/2> Real line but not the NaturalNumbers of the illdefined notion of> finite-integers but rather instead the p-adics. Only the line is not a> straight line. And my second *alleged proof below* using a spiral sort> of touches or hints of a curved line.(big snip)> TWO PROOFS OF THE RIEMANN HYPOTHESIS> PROOFS: Two proofs of the Riemann Hypothesis follows as A> and B.> Proof (A) is a geometrical proof. It was proved that the Riemann> Hypothesis is equivalent to the following-- the Moebius function mu of> x, m(x), and adding-up the values of m(x) for all n less than or equal> to N giving M(N). Then M(N) grows no faster than a constant multiple k> of (N^1/2)(N^E) as N goes to infinity (E is arbitrary but greater than> 0). Figure1, by setting-up a logarithmic spiral in a rectangle of> whirling squares where the squares are the sequences:> 1,1,2,3,5,8,13,21,34,55,89, . . . 2,2,4,6,10,16,26, . . .> 3,3,6,9,15,24,39, . . . then every number appears in at least one of> these sequences because every number will start a sequence. Since all> numbers are represen uniquely by prime factors (the unique prime> factorization theorem or called the fundamental theorem of arithmetic)> and The Prime Number Theorem: the distribution of prime numbers is> governed by a logarithmic function, where (An/n)/(1/Ln of n) tends to> 1> as n increases, where An denotes the number of primes below the> positive integer n, and where An/n is called the density of the primes> in the first n positive integers. The density of the primes, An/n, is> approxima by 1/(Ln of n), and as n increases, the approximation> gets> better. The distribution of prime numbers is governed by a> logarithmic function where these two math concepts-- one of prime> numbers, and the other, logarithms seem unconnec at first> appearance, but in reality they are totally connec. Geometrically,> the logarithmic spiral exhausts every positive integer, see figure 1.> The area of the rectangles containing the logarithmic spiral is always> greater, since the spiral is always inside the rectangles. Thus the> Moebius function k (N^1/2)(N^E) is satisfied since the area of the> logarithmic spiral is less than the rectangle whose area represents> the> number N, and whose sides represent its factors. The area of a> logarithmic spiral is represen by A=(r)(e^(Hj)) , and so depending> on where the point of origin for the spiral is taken rsubO determines> k, and depending on the value of H, H determines the E value for N,> when H=0 then the curve is a circle. The logarithmic spiral inside> rectangles of whirling squares implies that for any number N then> N^1/2> is the limit of the factors for N, for example, given the number 28,> then 28^1/2 = 5.2915. . and so looking for the factors of 28, it is> useless to try beyond 5 because the factors repeat, 4x7 then repeats> as> 7x4. But if the Moebius function was false then there must exist a> number M such that M^1/2 is not the limit of the factors for M and the> spiral is outside of the square, which is impossible, hence the> Moebius> function is true. Therefore the Riemann Hypothesis is proved. Q.E.D.> My second proof (B) of the Riemann Hypothesis uses a reductio> ad absurdum argument. Euler proved that a formula encoding the> multiplication of primes was equal to the zeta function. Euler's> formula in complex variable form is as follows:> (1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1 /(7^c))))x> (1/(1-(1/(11^c))))x . . . , where c is a complex variable, c=u+iv. The> Riemann zeta function is as follows. Re(c) => 1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex variable,> c=u+iv. Euler's formula involves multiplication of terms and the> Riemann zeta function involves addition of terms of a sequence. Taking> Re(c) > 0, suppose the Riemann Hypothesis is false then there is a 0> such that Re(c)=0 and c not equal 1/2 +iy, which implies there is> another 0 which is not on the 1/2 real line. Which means another real> number other than 1/2 works as an exponent resulting in a zero for the> Riemann zeta function, and a zero in the Euler formula. Thus, Riemann> zeta function subtract Euler formula must equal zero. This implies> for> any other real number exponent, either rational or irrational numbers,> such as for example the rational exponents: 1/3,1/4,1/5, . . . (Note:> any other exponent y/x , where y and x are Real numbers and where the> Real number of A^(y/x) such that y not equal 1, immediately transforms> to a number A^y(1/x), so that exponents with a 1 in the numerator> entail all of the Real exponents). To make clear of the above, for> example, 2^2/3 is 4^1/3. So then back to the proof. Then for exponent> 1/3 there has to exist a number M not equal 0 where (M+M+M)^1/M => (MXMXM)^1/M = M. Then for exponent 1/4 there has to exist a number M> not equal 0 where (M+M+M+M)^1/M = (MXMXMXM)^1/M = M, and so on.> Including the infinite number of cases where the x denominator is> irrational are impossible. Only the real number 1/2 works since 2 does> not equal 0, and (2+2)^1/2 = (2X2)^1/2 = 2, and so (2+2)^1/2> - (2X2)^1/2 = 0. In all of Reals and the Complex numbers, 2 is the> only number N which has the encoding ((N+N)^1/N) = ((NxN)^1/N) = N.> Unlike 0, the number 2, its sum equals its product and where the sum> and product is a new number 4. If RH were false, then another number> other than 2 would satisfy a generalized encoding formula ((N+N)^1/N)> = ((NxN)^1/N) = N. False, hence the proof. QED (Quantum> Electrodynamics)I have always rather felt that in doing science whether physics, orbiology or chemistry, the best way is to jump around and not stay puton one topic especially when in a jam or difficulty, then jump tosomething else and then the mind, like a photographic silver solutionwill make the image appear (the answer appear). This behaviour on mypart has served me extremely well for the past 15 years and my poststo the internet prove my position on this in that I rapidly jump fromone topic to another. I jump not only because it becomes dull stickingto one topic, but also for this reason that the mind when givenrespite and restfrom a difficulty and coming back fresh leads to better success.I have not thought of RH for the past 7 years, and that truly is anice rest.In the above algebraic proof of RH the linkage is addition tomultiplication and that is linked to the encoding formula using 2. Andsince the P-adic of ....00002is unique solution for that encoding formula, allows the possibilitythat this is truly a proof of RH and that RH is indeed a truestatment.In the above geometric proof of RH we have two geometric objects oflogarithmic spiral versus rectangles (or squares). We have all noticedhow the RH seems to be different for negative Reals as compared topositive Reals. We all have a sense that P-adics are not the geometryof straight lines, that is, p-adics are not Euclidean. We know thatthe positive Reals form Riemannian geometry and the negative Realsform Lobacheskian geometry.We know that if all the NaturalNumbers according to RH, lie on the 1/2Real line then the NaturalNumbers have Euclidean geometry. But theP-adics are not Euclidean geometry.Can we also run this type of argument over FLT? Are not the objects inFLT those of Euclidean geometry and indeed they are with thehypotenuse. But FLT is false according to P-adics.What a difference a day can make. I suspect my two proofs are correctand that the Riemann Hypothesis is a true statement. A true statementnot of NaturalNumbers = FiniteIntegers but a true statement ofNaturalNumbers = P-adics.We must be able to tie together a true statement of FLT onto a truestatement of RH.Yesterday I thought that since FLT was false to p-adics that RH wasalso a false statement. But today, I believe that these twomathematical statements connect to each other given that FLT is falseand RH is true.If RH were false, then there would have to exist another p-adic otherthan ...0002 that can solve that encoding formula. But the p-adic....00002 is unique.The dilemna I thought I was facing was that p-adics can never conformto a Euclidean Straight line geometry and that the p-adics could neverbe so well behaved as to fall all on the 1/2 Real line. But then,there are no p-adics falling on the negative1/2 Real line is there.There is no symmetry of RH to the negative Reals. The positive Realsare Riemannian geometry and a Logarithmic spiral is Riemanniangeometry.So, my difficulty, my trap as to reconcile how p-adics could ever beso very *regular* as to all fall on a straight line that is 1/2 Realis solved by understanding that the Logarithmic Spiral straightens outthe curvature of Riemann geometry of the positive Reals.So, if a Hypotenuse instead of Logarithmic spiral in my abovegeometrical proof of RH could ever serve as a proof thenmathematicians ever since Bernard Riemann had ever offered his RiemannHypothesis would have easily proven RH, and no doubt in my mind thatthe genius of Bernard Riemann would have proved his own RH because ifhypotenuses in whirling rectangles were constrained and restric asthe Logarithmic spiral, then NaturalNumbers = FiniteIntegers makessense and are a well defined set. But FiniteIntegers are not. They area ill-defined set.You see, the Logarithmic Spiral is the one and only curve thatsatisfies the Riemann Hypothesis and because it is aRiemann-geometry-curve embedded in the positive Reals where 1/2 Realline that the curve straightens out the Riem geometry leaving theEuclidean 1/2 Real line and thus the P-adics all fall on this 1/2 Realline.So, yesterday I was feeling assured that FLT was false due to p-adicsand that RH was false due also to p-adics. But today, my mind**maybe** clearer in that RH is true while FLT is false due both top-adics. In the above, we get a sense that we can use FLT to helpprove RH and vice versa and that is how mathematics has to work wheretruth in one room of mathematics has to agree with other rooms inmathematics.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies=== === Subject: : digital logic === Subject: : Digital LogicLevel: BeginnerI'd be very much obliged if someone would clue me in about sequential circuits.I had to drop a course because I couldn't grok this subject, and I've invesa superhuman amount of effort since then in trying to get it to make sense, butconsistently come up empty.I'm posting here because the answer I'm seeking is mathematical, even if thequestion isn't (I don't want to bungle this by imposing a wrong interpretationof my own onto things).Essentially what I want to know is this: How can you tell how a circuit willbehave without knowing the lengths of the gate delays? For example, how dosimulation programs (eg: LogicWorks, Multimedia Logic) determine how circuitsimulations should behave?My dream answer would be an analogue of boolean algebra [ footnote: in thecomputer-science sense of that term; ie: the set {0,1} with operations ~, & and/ defined by (~x) = (1-x), (x & y) = xy, and (x / y) = (x+y-xy) ], only withthe element of time added in somehow; a respectable mathematical theory. Infact, I don't see how I could be satisfied with anything else: the knowledge I'mlooking for is sharp enough to be a computer program, so it must be sharp enoughto be an axiomatic theory.Apparently it's very strange that this doesn't make sense to me. In the course Itook the question of time was never explicitly addressed. Maybe I'm overlookingsomething really obvious?=== === Subject: : Re: digital logic> === Subject: : Digital Logic>Level: Beginner>Essentially what I want to know is this: How can you tell how a circuit will>behave without knowing the lengths of the gate delays? For example, how do>simulation programs (eg: LogicWorks, Multimedia Logic) determine how circuit>simulations should behave?Others have given good answers but I will throw in my 2 cents worthanyway. Simple state machines generally consist of flip-flops, whichmay be either J-K or D input types, logic gates, and a clock. We mightdenote the current state of a flip-flop by Q and its next state by Q+.In the case of a JK flip-flop, for example, the outputs are immune tothe inputs when the clock is low. When the clock goes high the inputsare locked in and read. On the transition from high to low the stateof the flip-flop changes from its current state Q to its next stateQ+. During this low clock period, the flip-flop is immune to thechanges in inputs. This gives time for all the logic gates to receivetheir new inputs, including any fed-back values of the Q+ outputswhich now represent the new current state. All the gate delays havetime to settle to their new values and set the new J-K inputs, so theflip flop is settled and ready to go when the clock goes back high andthe cycle repeats.This happens for all flip-flops in the circuit at the same time, sowhat you get is a machine that proceeds from one state to the nexteach time the clock goes from high to low for a JK flip flop.The boolean logic is used to specify the next state equations, that isQ+ is a function of J and K, which are functions of the inputs and thecurrent state of the machine. Figuring out the logic equations for theJ and K inputs is where you usually get to dealing with Karnaugh maps.Machines of this type are called synchronous machines. They may appearto respond immediately to input changes, but that is only because theclocks are generally running very fast.Machines with no clock are called asynchronous machines and may indeedhave problems with gate delays and races between outputs and inputswhich must be carefully guarded against.Hope that helps.--Lynn=== === Subject: : Re: digital logic>Subject: digital logic>Message-id: > === Subject: : Digital Logic>Level: Beginner>I'd be very much obliged if someone would clue me in about sequential>circuits.>I had to drop a course because I couldn't grok this subject, and I've>inves>a superhuman amount of effort since then in trying to get it to make sense,>but>consistently come up empty.>I'm posting here because the answer I'm seeking is mathematical, even if the>question isn't (I don't want to bungle this by imposing a wrong>interpretation>of my own onto things).>Essentially what I want to know is this: How can you tell how a circuit will>behave without knowing the lengths of the gate delays? For example, how do>simulation programs (eg: LogicWorks, Multimedia Logic) determine how circuit>simulations should behave?>My dream answer would be an analogue of boolean algebra [ footnote: in the>computer-science sense of that term; ie: the set {0,1} with operations ~, &>and>/ defined by (~x) = (1-x), (x & y) = xy, and (x / y) = (x+y-xy) ], only>with>the element of time added in somehow; a respectable mathematical theory. In>fact, I don't see how I could be satisfied with anything else: the knowledge>I'm>looking for is sharp enough to be a computer program, so it must be sharp>enough>to be an axiomatic theory.>Apparently it's very strange that this doesn't make sense to me. In the>course I>took the question of time was never explicitly addressed. Maybe I'm>overlooking>something really obvious?Possibly what you're overlooking is the clock. Time isn't continuous,it's a series of discrete states, each of which can be consideredas a seperate set of input conditions to the combinatorial logic.The same Boolean algebra is applicable.But sequential logic introduces the multivibrators: astable, monostable and bistable (commonly known as flip-flops). Theseelements have truth tables just like gates (they are, in fact,construc from gates), but the outputs are clock driven. For example, the truth table for a D-flip flop is simply (where D isthe input and Q is the output)D Q-- --0 01 1but Q only changes in response to a clock transition (which canbe levels or edges, usually the latter). A D flip-flop remembersthe state of it's input from one clock transition to the next, sochanges in the state of D (in the absence of the clock) do not effect Q.So what you (or your simulation program) do is evaluate the initial state of your circuit using standard boolean algebra,apply the clock transition, re-evaluate the boolean algebrabased on which elements were affec by the clock, applythe next clock transition, re-evaluate...and so on.Sequential logic is not about time, it's about transitioning fromone logic state to another. In the case of my clock radio, thereis a direct correlation between the state of the digital displayand time because the state transitions are driven by thesynchronous 60 Hz power line. In the case of my garage dooropener, there is no correlation with time because its logicstate is driven by the random asynchronous signals fromthe remote control.That's my view in a nutshell. Lots of detail was omit, but Idon't intend for this post to be a course in sequential logic.--=== === Subject: : Re: digital logicThis is hardly my speciality, but I will give it a go.Maybe you are missing something. Digital circuits are almost always designedto operate in the same way, completely irrespective of gate delay. On almostall large systems (like computers) , the system changes with a clock signal,and this is designed to eliminate problems of actual gate delays.I don't know what the two programs you mention actually do. Do they justanalyse the operations of NAND gates, clocks, counters, memory units? Or dothey look at circuit lengths and chip/circuit layout? If so, this is moreelectronic engineering than computer science.> === Subject: : Digital Logic> Level: Beginner> I'd be very much obliged if someone would clue me in about sequentialcircuits.> I had to drop a course because I couldn't grok this subject, and I'veinves> a superhuman amount of effort since then in trying to get it to makesense, but> consistently come up empty.> I'm posting here because the answer I'm seeking is mathematical, even ifthe> question isn't (I don't want to bungle this by imposing a wronginterpretation> of my own onto things).> Essentially what I want to know is this: How can you tell how a circuitwill> behave without knowing the lengths of the gate delays? For example, how do> simulation programs (eg: LogicWorks, Multimedia Logic) determine howcircuit> simulations should behave?> My dream answer would be an analogue of boolean algebra [ footnote: in the> computer-science sense of that term; ie: the set {0,1} with operations ~,& and> / defined by (~x) = (1-x), (x & y) = xy, and (x / y) = (x+y-xy) ], onlywith> the element of time added in somehow; a respectable mathematical theory.In> fact, I don't see how I could be satisfied with anything else: theknowledge I'm> looking for is sharp enough to be a computer program, so it must be sharpenough> to be an axiomatic theory.> Apparently it's very strange that this doesn't make sense to me. In thecourse I> took the question of time was never explicitly addressed. Maybe I'moverlooking> something really obvious?=== === Subject: : hello...anaysis problem...let continuous function f:[0,1] -> Rshow that lim (int |f(x)|^n dx)^(1/n) = max {|f(x)| | x in [0,1]} n->00 0~1 --------------------- hello...geniusi think .....by mean thm of integral,int |f(x)|^n dx = (1-0)|f(c)|^n , c in (0,1)0~1 um....thusint |f(x)|^n dx <= max {|f(x)| | x in [0,1]}0~1 um...i can't progress.....to the result...i will seemed to be find other method.help me ...please...my genius...=== === Subject: : Re: hello...anaysis problem...> let continuous function f:[0,1] -> R> show that > lim (int |f(x)|^n dx)^(1/n) = max {|f(x)| | x in [0,1]} > n->00 0~1 Let M be equal to max {|f(x)| | x in [0,1]}. Since it is obviousthat, for each n, (int |f(x)|^n dx)^(1/n) <= M, all that has to beproved is that the limit that you are interes in cannot besmaller than M.Le N be an element of ]0,M[. Since |f| is continuous, there is someinterval [a,b] in [0,1] such that |f(x)| > N for every x in [a,b].Therefore, int(|f(x)|^n,x in [0,1]) >= (b - a).N^n;so (int(|f(x)|^n,x in [0,1]))^(1/n) >= (b - a)^(1/n).Nand this implies that lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= N.Since this is true for each N < M, lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= M.=== === Subject: : Re: hello...anaysis problem...> so> (int(|f(x)|^n,x in [0,1]))^(1/n) >= (b - a)^(1/n).N> and this implies that> lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= N.You haven't shown the limit exists yet. > Since this is true for each N < M,> lim (int(|f(x)|^n,x in [0,1]))^(1/n) >= M.Same problem.By the way, this is probably homework, so don't you think it would be better to give a nice hint rather than the entire solution?=== === Subject: : Re: hello...anaysis problem...Hint:Let M < max |f|. Then there exist A = (a,b) non empty subset [0,1]s.t. |f(x)|>M for all x in A.nojb.> let continuous function f:[0,1] -> R> show that > lim (int |f(x)|^n dx)^(1/n) = max {|f(x)| | x in [0,1]} > n->00 0~1 > --------------------- > hello...genius> i think .....> by mean thm of integral,> int |f(x)|^n dx = (1-0)|f(c)|^n , c in (0,1)> 0~1 > um....thus> int |f(x)|^n dx <= max {|f(x)| | x in [0,1]}> 0~1 > um...i can't progress.....to the result...> i will seemed to be find other method.> help me ...please...my genius...=== === Subject: : Re: hello...anaysis problem...>let continuous function f:[0,1] -> R>show that >lim (int |f(x)|^n dx)^(1/n) = max {|f(x)| | x in [0,1]} >n->00 0~1 Think about what an integral actually *is*. When you sum together lotsof terms that are raised to increasingly higher powers, which termdominates the sum? What happens after you take the n:th root of thesum of powers of n?=== === Subject: : Re: Isoperimetric ZeppJim Ferry a .8ecrit dans le message deJim Ferry a .8ecrit dans le message de> Hello sci.math,Consider the problem of maximizing the integral of some functionF(x,y)> over a simply connec region whose boundary C is allowed to vary> subject to these two constraints:1) C contains the origin (0,0), and> 2) C has length 1.> In general, the extremal curve C must satisfy F(x(s),y(s)) = alpha kappa(s), for some constant alpha.I derived this rather laboriously using the calculus of variationsCan you give hints for the proof of result.> I am very interes by this proof. I have very searched but withoutresult> Many thanks.> Excuse my bad English> Well, I don't want to attempt a rigorous proof where I have be carefulabout> just how smooth everything is. The elegant, simple route to a proof would> rely on this physical intuition (which I sta in my original post):> Finally, we rearrange each equation to get> cos(phi(s)) (d/ds) (F(x(s),y(s)) / phi'(s)) = 0, and> sin(phi(s)) (d/ds) (F(x(s),y(s)) / phi'(s)) = 0,> and we conclude that F(x(s),y(s)) / phi'(s) must be constant.> -Jim FerryMany thanks for your proof.=== === Subject: : Characterization of irrational numberslet x be a positive number. Then x is rational if and only if min { n^a* (nx - INT(nx)) } = 0, where the minimum is taken over all integers n>0. It is assumed that a=1.QUESTION: is this result correct? Is it still correct if a=0.5 or a=0.2?=== === Subject: : Re: Characterization of irrational numbersIn-reply-to: Vincent Granville >let x be a positive number. Then x is rational if and only if>min { n^a * (nx - INT(nx)) } = 0, where the minimum is taken>over all integers n>0. It is assumed that a=1.>QUESTION: is this result correct? Is it still correct if a=0.5 or a=0.2?Certainly, if x is a rational, p/q, then q * (qx - INT(qx)) = 0.However, for irrational x, n * (nx - INT(nx)) can never equal 0; but,for certain x, it can come as close to 0 as you want. For those x,inf { n * (nx - INT(nx)) } = 0. Note that I use inf instead of minsince there may be no smallest element of an infinite set. Suppose thethe partial quotients of the continued fraction of x are { c_k } and thecorresponding convergents are { p_k/q_k }. If k is even, then 0 < q ( q x - p ) < 1/c [1] k k k k+1That is, not only is p_k/q_k very close to x, but it is slightly smallerthan x so that INT(q_k x) = p_k. Therefore, when k is even, q ( q x - INT(q x) ) < 1/c [2] k k k k+1Thus, if the odd indexed partial quotients of x are arbitrarily large,the infimum of the quantities in [2] is 0.Consider the continued fraction for e: {2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...}where c_{3k+2} = 2k+2. Whenever k is odd, 3k+1 is even and therefore, q ( q x - INT(q x) ) < 1/(2k+2) 3k+1 3k+1 3k+1For example, let's look at k = 1; q_4 = 7 and 7 ( 7e - 19 ) = .1958... < 1/4.Look at k = 3; q_10 = 1001 and 1001 ( 1001e - 2721 ) = .1104... < 1/8e is just one of many numbers whose continued fractions behave this way,and for all such numbers, which are irrational simply because they haveinfinite continued fractions, inf { n * (nx - INT(nx)) } = 0.Rob Johnson take out the trash before replying=== === Subject: : Re: Characterization of irrational numbers>let x be a positive number. Then x is rational if and only if min { n^a>* (nx - INT(nx)) } = 0, where the minimum is taken over all integers n>>0. It is assumed that a=1.>>QUESTION: is this result correct? Is it still correct if a=0.5 or a=0.2?Assuming the min is taken over all *positive* integers, yes. Allow zero in the set, no. Allow negative integers, it doesn's even make sense for a = 1/2 or 1/10. With positive integers and irrational x, I am not sure the minimum exists. The interesting question is whether the infimum can be, or is necessarily, 0 in this case. I think this relates to another topic discussed here earlier under the subject, Approximating Pi by Rationals. [Was the comma preceding the quote appropriate?]-- === === Subject: : Re: Pi 3.14In sci.math, Jason P<51566c78.0401231238.69fd6871@ posting.google.com>:>> pi pie ( P ) Pronunciation Key (p)>> n. >> 1.A baked food composed of a pastry shell filled with fruit, meat,>> cheese, or other ingredients, and usually covered with a pastry crust.>> 2.A layer cake having cream, custard, or jelly filling. >> 3.A whole that can be shared: That would... enlarge the economic pie>> by making the most productive use of every investment dollar (New>> York Times). Idiom:>> pie in the sky>> An empty wish or promise: To outlaw deficits... is pie in the sky>> (Howard H. Baker, Jr.).>> Good thing a pie is usually circular in shape otherwise we never could>> have figured out how pie ties to 3.1415926... Also, mathmeticians are>> basically lazy, always looking for the simple solutions; they even>> leave the e off pie!> So they can have their pi and e it too! Ha ha ha ha!> Jasoni yi yi!Mix them properly and one gets e^(i * pi) = -1, a niceedible recipe for mathematicians. Yum. Euler was a very prolific cook.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Pi 3.14> Also, mathmeticians are> basically lazy, always looking for the simple solutions; they even> leave the e off pie!Parsimony, not sloth; that e got used.=== === Subject: : Re: Squaring A Circle...(almost)In sci.math, Rick Decker<40117927.5080803@hamilton.edu>:>> We all (should) know it is impossible to square a circle with only a>>straight-edge and compass and pencil.>I'm sorry, but you never make it clear what you mean by square a>circle. Please explain.> It is a well known expression (as others have explained to you). You>> did no real harm asking, but you would have better wai a while to>> see if he really was missing some fundamental detail or if others>> could understand his question unambiguously, and if so, had you still>> problems, then you could ask: what is this 'squaring of the circle'>> you all seem to be so familiar with?>> And since we're on sci.math, I'll add my informal answer too: it's one>> of the top-three crank attracting topics in mathematics >> > Although that and duplicating the cube seem to have fallen out> of fashion with the crank fringe lately. BTW, what did you> have in mind for the third entry in your top-three list?The traditional third one IIRC is trisecting an arbitrary anglewith naught but a compass and an unmarkable straighge,a la Euclid.(One can trisect it if one is allowed to mark the straighge.Unfortunately I forget the specifics.)> Rick-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Squaring A Circle...(almost)) The traditional third one IIRC is trisecting an arbitrary angle) with naught but a compass and an unmarkable straighge,) a la Euclid.)) (One can trisect it if one is allowed to mark the straighge.) Unfortunately I forget the specifics.)I thought you didn't need to mark the straige, but it involved slidingthe compass along the straighge or something like that.SaSW, Willem-- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOT=== === Subject: : Re: Squaring A Circle...(almost)In sci.math, Willem: > ) The traditional third one IIRC is trisecting an arbitrary angle> ) with naught but a compass and an unmarkable straighge,> ) a la Euclid.> )> ) (One can trisect it if one is allowed to mark the straighge.> ) Unfortunately I forget the specifics.)> I thought you didn't need to mark the straige, but it involved sliding> the compass along the straighge or something like that.That might work too. I'd have to look.> SaSW, Willem-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Squaring A Circle...(almost)> I thought you didn't need to mark the straige, but it involved sliding> the compass along the straighge or something like that.The purpose of the marks is to have one specified distance along thestraighge. Holding a modern compass to the desired distance(Euclid's compasses couldn't do that) would be equivalent.And the construction appears early on this page:http://mathworld.wolfram.com/AngleTrisection.html-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W=== === Subject: : Re: Squaring A Circle...(almost)Willem a .8ecrit:> ) The traditional third one IIRC is trisecting an arbitrary angle> ) with naught but a compass and an unmarkable straighge,> ) a la Euclid.> )> ) (One can trisect it if one is allowed to mark the straighge.> ) Unfortunately I forget the specifics.)> I thought you didn't need to mark the straige, but it involved sliding> the compass along the straighge or something like that.I think you have to slide simultaneously the straighge and the compass, which is not Euclid compliant at all.BTW there is a nice anima gadget of that at :http://www.cut-the-knot.org/pythagoras/archi.shtmlregards-- === === Subject: : Re: Squaring A Circle...(almost)>I'm sorry, but you never make it clear what you mean by square a>circle. Please explain.> It is a well known expression (as others have explained to you). You> did no real harm asking, but you would have better wai a while> Sorry, I'm actually reading and posting on rec.puzzles, where we're> less math savvy :)> MosheMy fault, perhaps, for cross-posting. But I do find that some postsare *both* mathematical and puzzling, and deserve to be cross-posto these 2 groups.(And I want sometimes to very much encourage cross-dialog betweenrepliers to both groups.)As for squaring the circle, I believe this phrase is odd, actually.(I believe I have also heard rectifying the circle, as well.)So, what are you doing when you take a compass, put its point in themiddle of a square, and use it to draw a circle around the square?...Circling the Square, of course!...;)Leroy Quet=== === Subject: : Re: Squaring A Circle...(almost)>>I'm sorry, but you never make it clear what you mean by square a>>circle. Please explain.>>It is a well known expression (as others have explained to you). You>did no real harm asking, but you would have better wai a while>>Sorry, I'm actually reading and posting on rec.puzzles, where we're>>less math savvy :)>>Moshe> My fault, perhaps, for cross-posting. But I do find that some posts> are *both* mathematical and puzzling, and deserve to be cross-pos> to these 2 groups.> (And I want sometimes to very much encourage cross-dialog between> repliers to both groups.)> As for squaring the circle, I believe this phrase is odd, actually.> (I believe I have also heard rectifying the circle, as well.)Slightly different, although the problem is equivalent...squaring the circle is construct a square with the same area as a given circle.rectifying the circle is construct a segment whose length equals the perimeter of a given circle.Both problems are equivalent to construction of number pi.> So, what are you doing when you take a compass, put its point in the> middle of a square, and use it to draw a circle around the square?...> Circling the Square, of course!...> ;)Yes... and No...You do not construct a circle of same area...Squaring the circle does not mean construct a square around it...A bit strange these math idiomatic expressions...-- === === Subject: : Re: Newbie Questions: SeriesEn el mensaje:8447d48a.0401231624.93650f9@posting.google.com, omniscient idiot escribi.97:> Dear all,> I have two newbie questions. First, if one sums the following:> 1*k + 2*k^2 + 3*k^3 + ..... + (n-1)*k^(n-1) + n*k^n> with k = some constant number, what will the result be? Anyone has any> idea? Does this have a special name?Without calculus, letS = 1*k + 2*k^2 + 3*k^3 + ..... + (n-1)*k^(n-1) + n*k^nkS = 1*k^2 + 2*k^3 + 3*k^4 + ..... + (n-1)*k^n + n*k^(n+1)(1 - k)S = k + k^2 + k^3 + ... + k^n - n*k^(n+1) = (k - k^(n+1))/(1 - k) - n*k^(n+1)S = (k - k^(n+1))/(1 - k)^2 - n*k^(n+1)/(1 - k)-- Saludos,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Why is this a proof?Hello.I am considering a class of finite modules indexed by a set A. To an a in A let M(a) denote the corresponding module. Let a and b be in A and letsigma(a,b) = number of submodules of M(b) isomorphic to M(a).eta(a,b) = number of submodules N of M(b) with M(b)/N isomorphic to M(a)alpha(a) = number of automorphisms of M(a).I have shown that given a submodule U of M(a) with M(a)/U isomorphic to M(c) and a submodule V of M(b) isomorphic to M(c) there is a bijection between Aut(M(c)) and the set of homomorphisms M(a)-->M(b) with kernel U and image V.Why does this show that eta(c,a)*alpha(c)*sigma(c,b) equals the number of homomorphisms M(a)-->M(b) with image isomorphic to M(c)?-- Michael Knudsen=== === Subject: : Re: Why is this a proof?> Hello.> I am considering a class of finite modules indexed by a set A. To an a> in A let M(a) denote the corresponding module. Let a and b be in A and let> sigma(a,b) = number of submodules of M(b) isomorphic to M(a).> eta(a,b) = number of submodules N of M(b) with M(b)/N isomorphic to M(a)> alpha(a) = number of automorphisms of M(a).> I have shown that given a submodule U of M(a) with M(a)/U isomorphic to> M(c) and a submodule V of M(b) isomorphic to M(c) there is a bijection> between Aut(M(c)) and the set of homomorphisms M(a)-->M(b) with kernel U> and image V.> Why does this show that eta(c,a)*alpha(c)*sigma(c,b) equals the number> of homomorphisms M(a)-->M(b) with image isomorphic to M(c)?You have complica notation. Consider a map f:M -> N of modules.We ask, how many such maps have image isomorphic to another module P.We count them according to their kernel and image. Let these be K and I.Then M/K and I are isomorphic to P. Fix such a K and I. Then f isin effect an isomorphism from M/K to I. As these are both isomorphic toP there are |Aut(P)| such maps. Allowing K and I to vary, we get|Aut P| x number of submodules of N isomorphic to P xnumber of submodules K of M with M/K isomorphic to Pas the number of f:M -> N with image isomorphic to P.-- === === Subject: : Re: Why is this a proof?> You have complica notation. Consider a map f:M -> N of modules.> We ask, how many such maps have image isomorphic to another module P.> We count them according to their kernel and image.Hmmm...do you say that the map f is completely determined by its kernel and its image?!-- Michael Knudsen=== === Subject: : Re: The Lost Proof of FermatUsing good ol' Fermat's equation:A^n + B^n = C^nonly has integer solutions for n = 2ButA^aleph_0 + B^aleph_0 = C^aleph_0Then againForC > B > A >= 2Limit as n--->infinity[A^n + B^n ]^[1/n] = B, not C A^aleph_0 + B^aleph_0 = C^aleph_0is false?Therefore 2^aleph_0 is not aleph_1=== === Subject: : Re: The Lost Proof of Fermat> But> A^aleph_0 + B^aleph_0 = C^aleph_0Ah, this explains it! You don't even know what FLT says, so why should weexpect you to be able to prove it? Very illuminating!=== === Subject: : Re: The Lost Proof of Fermat> Using good ol' Fermat's equation:> A^n + B^n = C^n> only has integer solutions for n = 2> But> A^aleph_0 + B^aleph_0 = C^aleph_0> Then again> For> C > B > A >= 2> Limit as n--->infinity> [A^n + B^n ]^[1/n] = B, not C > A^aleph_0 + B^aleph_0 = C^aleph_0> is false?> Therefore 2^aleph_0 is not aleph_1Damn, you're nuts.Nathan=== === Subject: : Re: The Lost Proof of Fermatx-no-archive: yes windows-nt)> A^aleph_0 + B^aleph_0 = C^aleph_0It makes no sense to connect this statement with Fermat's conjecture;the domain of Fermat's conjecture is the finite cardinals. Among otherthings that don't work as you expect, A^n does not converge toA^aleph_0 as n-->infinity; there is no metric w.r.t which suchconversion can be said to be taking place.> C > B > A >= 2> Limit as n--->infinity> [A^n + B^n ]^[1/n] = B, not C Of course! (A^n+B^n) = C^n for only a particular value of n--there isno expectation that this should converge to C. You are observing thatthe length of the vector (A,B) is equal to C in some l^n norm, but notin the l^infinity norm. So what?> A^aleph_0 + B^aleph_0 = C^aleph_0 is false?No such conclusion can be drawn.> Therefore 2^aleph_0 is not aleph_12^aleph_0 is set-isomorphic to the power set of a set of cardinalityaleph_0, which equals the cardinality of the reals. The continuumhypothesis is required to decide whether it's equal to aleph_1 or not.Len.=== === Subject: : Re: The Lost Proof of Fermat[(x!)/(x-1)!]^p! + [(y!)/(y-1)!]^p! = [(z!)/(z-1)!]^p!x^p + y^p = z^px^[p+n] + y^[p+n] = z^[np]x^[p/n+1] + y^[p/n+1] = z^p= x^p + y^pp/n + 1 = pp = n = 2 = p!=== === Subject: : Re: The Lost Proof of Fermatx-no-archive: yes windows-nt)Russ,Two questions. You do realize that you don't know anything about math,right? And, you do realize that REAL proofs contain explanatory prose,right? For example, here's an important annotation you left out:> x^p + y^p = z^p> [Then a miracle happens and the following false statement becomes> true:]> x^[p+n] + y^[p+n] = z^[np]--Len.=== === Subject: : Re: The Lost Proof of Fermat>> x^p + y^p = z^p[Then a miracle happens and the following false statement becomes>> true:]x^[p+n] + y^[p+n] = z^[np]=== === Subject: : Re: The Lost Proof of FermatKeep working at it, Slappy. You'll prove it someday.=== === Subject: : Re: The Lost Proof of Fermat (3^2 + 4^2)^3 = (5^2)^3> 'cos then w = n = 3 and p = 2.> Brilliant!http://www.guinness1759society.com/offer/Brann1759/ 33351392/EN/welcome.asp:) Called the worst Guinness ad ever on another site, but I couldn'tdownload the appropriate version of Windows Media Player to verify that it'sreally the same ad.Jon Miller=== === Subject: : Re: The Lost Proof of FermatNow I know why it was lost.=== === Subject: : Re: Dilemma with undergraduate real analysis>Right now I'm taking the undergraduate year-long 'real analysis' class. To be>honest the professor is terrible and the way he goes about things is to make>the class so easy that even the worst students do great. The main thing that>has happened is that I'm not learning anything. The problem is that I'm going>to graduate school next year and I bet they will expect me to be taking their>graduate level analysis class with Lebesgue Integration and the rest.>I have two questions. Do I need to know this before taking a class at the>graduate level (I assume so but it could be radically different..right?). What>are books that are good for self-study that teach all the basics about metric>spaces, riemann/riemann-stieltjes integrals, and the rest?What's wrong with the text for the class? (What book is it?)>Randy************************=== === Subject: : Re: Dilemma with undergraduate real analysis>>Right now I'm taking the undergraduate year-long 'real analysis' class. To>be>>honest the professor is terrible and the way he goes about things is to make>>the class so easy that even the worst students do great. The main thing>that>>has happened is that I'm not learning anything. The problem is that I'm>going>>to graduate school next year and I bet they will expect me to be taking>their>>graduate level analysis class with Lebesgue Integration and the rest.>>I have two questions. Do I need to know this before taking a class at the>>graduate level (I assume so but it could be radically different..right?).>What>>are books that are good for self-study that teach all the basics about>metric>>spaces, riemann/riemann-stieltjes integrals, and the rest?>What's wrong with the text for the class? (What book is it?)>>Randy>************************>The 'text' is a short workbook-style thing that the Professor made. We are notusing any standard textbook so he covers what he thinks is important.=== === Subject: : Re: Dilemma with undergraduate real analysis>>Right now I'm taking the undergraduate year-long 'real analysis' class. To> be>>honest the professor is terrible and the way he goes about things is to make>>the class so easy that even the worst students do great. The main thing> that>>has happened is that I'm not learning anything. The problem is that I'm> going>>to graduate school next year and I bet they will expect me to be taking> their>>graduate level analysis class with Lebesgue Integration and the rest.>>I have two questions. Do I need to know this before taking a class at the>>graduate level (I assume so but it could be radically different..right?).> What>>are books that are good for self-study that teach all the basics about> metric>>spaces, riemann/riemann-stieltjes integrals, and the rest?>Some comments on previous posts.1. For myself, I never learned anything by reading books; all the mathI know I learned either from lectures or from personal conversation. I had wonderful teachers (and some clunkers too). But if you canlearn from a book (Rudin is the best I know) by all means do. Thefirst six chapters ought to suffice. Leave Lebesgue measure for gradschool.2. This is the inevitable result of student evaluation, which resulin grade inflation. One thing was clear was that there was a strongcorrelation between the ease of the midterm and the evaluations.3. Although I never taught the graduate analysis course, in algebra wedo sort of start from the beginning and cover a large part of Lang ina year. Generally the requirements for graduate school in math are ayear of analysis and a year of algebra; everything else is windowdressing. But they must be honest courses.4. I just cannot relate to the claim that mathematicians suffer moremath anxiety than anyone else. While there are things I find hard tolearn, I am not anxious about it. The person who said it is veryinsecure.=== === Subject: : Re: Dilemma with undergraduate real analysis > I have two questions. Do I need to know this before taking a class at the> graduate level (I assume so but it could be radically different..right?).What> are books that are good for self-study that teach all the basics aboutmetric> spaces, riemann/riemann-stieltjes integrals, and the rest?You don't need to know anything. You start all over in grad school. Thesethings are often called maturity prerequisites.With that in mind, what you remember from other sources may give you hintsabout which way to go when exploring and when proving theorems. But thequestion really is, how confident are you?Which raises the question, so what if the prof is lousy? Why aren't youlearning this stuff anyway? Or are you just having a crisis of confidence?(One of my profs said fairly frequently that no one suffers more from mathanxiety than professional mathematicians.)Jon Miller=== === Subject: : Re: Dilemma with undergraduate real analysis> You don't need to know anything.That could not be further from the truth.=== === Subject: : Re: Dilemma with undergraduate real analysis >> I have two questions. Do I need to know this before taking a class at the>> graduate level (I assume so but it could be radically different..right?).>What>> are books that are good for self-study that teach all the basics about>metric>> spaces, riemann/riemann-stieltjes integrals, and the rest?>You don't need to know anything. You start all over in grad school. These>things are often called maturity prerequisites.>With that in mind, what you remember from other sources may give you hints>about which way to go when exploring and when proving theorems. But the>question really is, how confident are you?>Which raises the question, so what if the prof is lousy? Why aren't you>learning this stuff anyway? Or are you just having a crisis of confidence?>(One of my profs said fairly frequently that no one suffers more from math>anxiety than professional mathematicians.)>Jon MillerI guess part of it is a crisis of confidence but I know that our material isincomplete and a lot easier than a regular book. I went to the library andcompared it to other books and there are big chunks left out. His style iscompletely different than what seems standard. That is why I asked fortextbook recommendations.=== === Subject: : Re: Dilemma with undergraduate real analysis> I guess part of it is a crisis of confidence but I know that our material is> incomplete and a lot easier than a regular book. I went to the library and> compared it to other books and there are big chunks left out. His style is> completely different than what seems standard. That is why I asked for> textbook recommendations.I wonder if the situation is really as bad as you think it is. And Ireally wonder whether your prof's put-together workbook is reallyinferior. The simple fact of the matter is that you aren't in a position to judgewhether his approach is worse than the standard one. If you can'ttrust your prof to teach you well, there's very little you can do,except study on your own. The catch-22 is that if he really is doing adecent job of teaching, you're screwing yourself by studying othermaterial; you're losing out on the chance to ask your professorquestions and making inefficient use of your time. Actually, if you want more challenging material, why can't you ask theprof? It's also a good idea to ask the prof for books that willsupplement his workbook.=== === Subject: : Re: Dilemma with undergraduate real analysis> I have two questions. Do I need to know this before taking a class at the> graduate level (I assume so but it could be radically different..right?).> What> are books that are good for self-study that teach all the basics about> metric> spaces, riemann/riemann-stieltjes integrals, and the rest?> You don't need to know anything. You start all over in grad school. That wasn't true in my experience. In my Real Analysis and Topology andManifolds courses in grad school, there certainly were prerequisites. Ineeded to know a lot of stuff that is usually taught in an undergrad analysiscourse and a point-set topology course. However, in my Algebra courses,things did start from scratch, but the pace was so quick (covered most ofLang's Algebra in addition to other material in one year) that I would neverhave been able to keep up without having remembered the algebra I learnedas an undergrad.Anyway, to the original poster, I say forget about your professor. It is more natural to learn mathematics from books than from spoken lectures anyway. Things are writtenmore precisely and you can learn at exactly your own pace. While all booksare not equal, you can probably learn the analysis you need from just aboutany half-way decent text. I recommend Elementary Classical Analysis byJerrold E. Marsden.Good luck,Leonard (email defunct)> These> things are often called maturity prerequisites.> With that in mind, what you remember from other sources may give you hints> about which way to go when exploring and when proving theorems. But the> question really is, how confident are you?> Which raises the question, so what if the prof is lousy? Why aren't you> learning this stuff anyway? Or are you just having a crisis of confidence?> (One of my profs said fairly frequently that no one suffers more from math> anxiety than professional mathematicians.)> Jon Miller=== === Subject: : Re: Dilemma with undergraduate real analysis> Anyway, to the original poster, > I say forget about your professor. It is more natural to learn > mathematics from books than from spoken lectures anyway. Things are written> more precisely and you can learn at exactly your own pace. While all books> are not equal, you can probably learn the analysis you need from just about> any half-way decent text. I recommend Elementary Classical Analysis by> Jerrold E. Marsden.Yes he should forget about this particular dismal professor (who should probably be fired), but I disagree about written versus a live learning experience. A good teacher will supplement the text in important ways, giving intuition and motivation, especially in working problems. The book is there for all the details. One profits greatly from both.=== === Subject: : Re: Dilemma with undergraduate real analysis> === Subject: : Re: Dilemma with undergraduate real analysis>That wasn't true in my experience. In my Real Analysis and Topology and>Manifolds courses in grad school, there certainly were prerequisites. I>needed to know a lot of stuff that is usually taught in an undergrad analysis>course and a point-set topology course. However, in my Algebra courses,>things did start from scratch, but the pace was so quick (covered most of>Lang's Algebra in addition to other material in one year) that I would>never>have been able to keep up without having remembered the algebra I learned>as an undergrad.>Anyway, to the original poster, >I say forget about your professor. It is more natural to learn >mathematics from books than from spoken lectures anyway. Things are written>more precisely and you can learn at exactly your own pace. While all books>are not equal, you can probably learn the analysis you need from just about>any half-way decent text. I recommend Elementary Classical Analysis by>Jerrold E. Marsden.>Good luck,>Leonard (email defunct)Thanks for the recommendation.=== === Subject: : Re: Dilemma with undergraduate real analysis >>I have two questions. Do I need to know this before taking a class at the>>graduate level (I assume so but it could be radically different..right?).>>What>>are books that are good for self-study that teach all the basics about>>metric>spaces, riemann/riemann-stieltjes integrals, and the rest?>You don't need to know anything. You start all over in grad school. These>things are often called maturity prerequisites.>With that in mind, what you remember from other sources may give you hints>about which way to go when exploring and when proving theorems. But the>question really is, how confident are you?>Which raises the question, so what if the prof is lousy? Why aren't you>learning this stuff anyway? Or are you just having a crisis of confidence?>(One of my profs said fairly frequently that no one suffers more from math>anxiety than professional mathematicians.)Or, if you are really that concerned, you probably learned enough to be able to go through Rudin, Principles of Mathematical Analysis on your own. I think the first seven chapters (of the second edition - through Sequences and Series of Functions) should suffice. Don't worry about Lebesgue integration - you'll learn that in graduate school.-- === === Subject: : Re: Dilemma with undergraduate real analysis > Right now I'm taking the undergraduate year-long 'real analysis' class. To be> honest the professor is terrible and the way he goes about things is to make> the class so easy that even the worst students do great. The main thing that> has happened is that I'm not learning anything. The problem is that I'm going> to graduate school next year and I bet they will expect me to be taking their> graduate level analysis class with Lebesgue Integration and the rest.> I have two questions. Do I need to know this before taking a class at the> graduate level (I assume so but it could be radically different..right?). What> are books that are good for self-study that teach all the basics about metric> spaces, riemann/riemann-stieltjes integrals, and the rest?> RandyIt depends which graduate school you will be going to, but many graduate schools will allow you to retake these classes.=== === Subject: : Re: Dilemma with undergraduate real analysis> It depends which graduate school you will be going to, but many graduate> schools > will allow you to retake these classes.Well, sure, if you don't mind looking like an idiot. Some willdefinitely see retaking classes as a sign of weakness and maybe evenstupidity. Even if they aren't going to be interacting with you verymuch, they may be one of the people that will determine your fundingsituation. Looking like an idiot is something to definitely be avoidedif possible. I'm not trying to alarm the OP (but I think if I did, that's betterthan instilling a sense of complacency); however, being prepared isdefinitely a huge advantage, and an important part of that is beingfamiliar with the first-year material like Lebesgue integration.=== === Subject: : Re: Trigonometric IntegralIn-reply-to: N. Karjanto >How to solve the following integral>Int (cos nx /(a - b cos x), x = 0.. Pi ), n = 0,1,2,3,4,........,>a, b are real numbersAn integral we will need later is |pi dx | ------------ ( z = tan(x/2) ) | 0 1 - a cos(x) |oo 2 dz 2 1+a = | ------------------ ( b = --- ) | 0 (1+z^2) - a(1-z^2) 1-a 2 |oo b dz = ----------- | ---------- ( bz = tan(w) ) sqrt(1-a^2) | 0 1 + (bz)^2 pi = ----------- [1] sqrt(1-a^2)An identity that we will use is oo --- n > r (cos(nx) + i sin(nx)) --- n=0 oo --- ix n = > ( r e ) --- n=0 1 = ------------ 1 - r e^{ix} 1 - r e^{-ix} = ------------------- [2] 1 - 2r cos(x) + r^2Taking the real part of [2], we get oo --- n > r cos(nx) --- n=0 1 - r cos(x) = ------------------- ( r = tan(s/2) ) 1 - 2r cos(x) + r^2 1 cos(s) = - ( 1 + -------------- ) [3] 2 1-sin(s)cos(x)Let |pi dx a = | cos(nx) ------------ [4] n | 0 a - b cos(x)Then, if we let b/a = sin(t) and apply [3], we get oo --- n > a r --- n n=0 |pi 1 cos(s) dx = | - ( 1 + -------------- ) ------------ | 0 2 1-sin(s)cos(x) a - b cos(x) 1 |pi cos(s) dx = -- | ( 1 + -------------- ) -------------- 2a | 0 1-sin(s)cos(x) 1-sin(t)cos(x) 1 pi = -- ------ 2a cos(t) cos(s) |pi 1 dx + ------ | -------------- -------------- [5] 2a | 0 1-sin(s)cos(x) 1-sin(t)cos(x)Using partial fractions, we get 1 1 -------------- -------------- 1-sin(s)cos(x) 1-sin(t)cos(x) 1 sin(s) sin(t) = ------------- ( -------------- - -------------- ) [6] sin(s)-sin(t) 1-sin(s)cos(x) 1-sin(t)cos(x)Plugging [6] into [5] and applying [1], we get [5] pi 1 cos(s) tan(s)-tan(t) = -- ------ + ------ pi ------------- 2a cos(t) 2a sin(s)-sin(t) pi 1 tan(s)-tan(t) = -- ( ------ + cos(s) ------------- ) ( r = tan(s/2) ) 2a cos(t) sin(s)-sin(t) pi 1 2r cos(t) - (1-r^2) sin(t) = -- ------ ( 1 + -------------------------- ) 2a cos(t) 2r - (1+r^2)sin(t) pi 1 2r + 2r cos(t) - 2 sin(t) = -- ------ ------------------------- 2a cos(t) 2r - (1+r^2)sin(t) pi 1 r(1+cos(t)) - sin(t) = -- ------ -------------------- a cos(t) 2r - (1+r^2)sin(t) pi 1 1 + cos(t) = -- ------ --------------------- [7] a cos(t) 1 + cos(t) - r sin(t)The coefficient of r^n in [7] is pi 1 sin(t) n a = -- ------ ( -------- ) ( b/a = sin(t) ) n a cos(t) 1+cos(t) pi b n = ------------- ( --------------- ) [8] sqrt(a^2-b^2) a+sqrt(a^2-b^2)Therefore, combining [4] and [8], we get |pi dx | cos(nx) ------------ | 0 a - b cos(x) pi b n = ------------- ( --------------- ) [9] sqrt(a^2-b^2) a+sqrt(a^2-b^2)Rob Johnson take out the trash before replying=== === Subject: : Re: Rationals are Uncountable <99a0b764.0401221214.6a666a05@posting.google.com> <99a0b764.0401231312.226ddd00@posting.google.com> Discussion, linux)>> The intersection may be empty, a closed interval [lo, hi],>> or a semi-open interval [lo, r) or (r, hi],>> where r must be rational, but lo and hi may be rational or irrational.> That was very informative.> I hope you don't mind if I ask some stupid questions.> 1) (-1/j, 1/j) produces [0, 0]={0}.> 2) (0, 1/j) produces the empty set.> I want to understand why (1) is a singleton set and (2) is empty.> I will use a shorthand notation for the intersection.> Let w mean omega and let (-1/w,1/w) represent the intersection> of the set of intervals (-1/j, 1/j) over all j.> Define 1/w = 0 and 1/-w = 0.> Rewrite the formulas above:> 1) (1/-j, 1/j) produces (1/-w, 1/w)> 2) (0, 1/j) produces (0, 1/w)This is just nonsense. You've played around with some notation thathas nothing to do with the definition of the intersection of {(-1/j,1/j)}, then you added an arbitrary definition (1/w = 0) and youdemand to know how come things aren't correct.They're not correct because they have nothing to do with theintersection. There is no rule thatIntersection{(f(n),g(n)) | n in N} = (limit_{n->oo} f(n), limit_{n->oo} g(n)).You're just making things up.-- Jesse F. HughesWiles made somewhere around half a million dollars U.S. that I heardabout, and I know he didn't take major endorsements. --JSH on the rewards of proving Fermat's last theorem. === Subject: : Re: Rationals are Uncountable=== > That was very informative. > I hope you don't mind if I ask some stupid questions. > 1) (-1/j, 1/j) produces [0, 0]={0}. > 2) (0, 1/j) produces the empty set. > I want to understand why (1) is a singleton set and (2) is empty. >Let's first show why 2) is empty:Let X be the intersection of (0, 1/j). Assume X is not empty. Now there has to be at least one member x in X. Clearly x > 0 because none of (0, 1/j) contain negative numbers or 0.By the definition of intersection [of infinitely many sets], x must be each interval (0, 1/j), including the ones where j > 1/x.But if j > 1/x, 1/j < x, and the intervals in question would be subsets of (0, x) -- but x is not in (0, x). Now we have found intervals (0, 1/j) that don't include the member x. Therefore x cannot be in the set X. This contradicts the assumption that X is not empty; therefore X has to be the empty set, q.e.d.Now 1):Let X be the intersection of (-1/j, 1/j). Let x be a member of X. Now either x = 0 or x =/= 0.Case x = 0:this it follows that 0 is in every interval (-1/j, 1/j), and therefore 0 is in the intersection of every such interval.Case x =/= 0:Suppose x > 0. If x is in X, x has to be in each interval (-1/j, 1/j), even in the ones where j > 1/x. If j > 1/x, 1/j < x, and therefore (-1/j, 1/j) is a subset of (-x, x). But x is not in (-x, x), so we have found intervals (-1/j, 1/j) that don't include x. Therefore the intersection of every such intercal cannot contain x; thus X cannot contain any member x > 0. Similarly it can be shown that X cannot contain any member x < 0. > I will use a shorthand notation for the intersection. > Let w mean omega and let (-1/w,1/w) represent the intersection > of the set of intervals (-1/j, 1/j) over all j. > Define 1/w = 0 and 1/-w = 0. >You cannot define that 1/w = 0 and 1/-w = 0, you would be defining that w*0 = -w*0 = 1 and that is impossible in the field of real numbers.But assuming you mean that you define the symbols 1/w and 1/-w to mean 0 ... > Rewrite the formulas above: > 1) (1/-j, 1/j) produces (1/-w, 1/w) > 2) (0, 1/j) produces (0, 1/w) >... now (1/-w, 1/w) by your definition of 1/-w and 1/w means (0,0), which is the empty set. But I just showed that the intersection of (1/-j, 1/j) is {0} and not empty, so (1/-j, 1/j) does not produce(1/-w, 1/w).-Joni=== === Subject: : Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com>>Let X be the set of all rational numbers in the interval (-1,1).>If X is countable there exists a function, f(), that assigns>a natural number to every member of X.>Assume f() exists.>Use Cantor's algorithm to build sets A and B.>Assume these sets have this form:>a_n = 1/-n>b_n = 1/nFirstly, if a_1 = -1, b_1 = 1, then X is the set of rationalnumbers in [-1,1], not the set of all rational numbers in (-1,1).But they won't have this form for all n if f is surjective (specifically, if there exists m such that f(m) = 0, then not all of the intervals can have the form you dictate above), so in making the assumption that the intervals have the form that you dictate, you have already assumed that f is notsurjective (specifically, f cannot take value 0). Therefore there is no surprise if you can find an element of X which is not in the range of f (0 is one such value).>No f() that produces these sets A and B can>assign a natural number to 0.True.>No f() can assign a natural number to every rational in X.No function f which produces the sets A and B can assigna natural number to every rational in X.>Assume the sequences a_n and b_n converge to another>rational. Clearly, this rational is not in the range of f().True.>Assume A and B converge to an irrational number.>Let r be this number.>Define a new sequence, x'.>x'_1 = (x_1 - r) / r>x'_2 = (x_2 - r) / rIn particular, nearly all x'_i are irrational.>We have defined a new set X' = ( (-1-r)/r, (1-r)/r ).>f'() can not map a natural number to 0.>(remember -1 < r < 1).>Every member of X' can be mapped to X.No. X consists only of the rational elements of (-1,1) (or [-1,1]). Therefore X' = {y in ((-1-r)/r,(1-r)/r) :r(y+1) in Q}, if you are to require that every element of X' map to an element of X. Note that since r is irrational, then 0 is not an element of X'. This means that no function from N to X' can have 0 as anelement of its range.>Therefore, X did not contain every rational.Wrong. You demonstra that the function from N to X'cannot take value 0. Since 0 is not an element of X',then that is not surprising. Furthermore, you have notdemonstra that f is no surjective. There is no proof here that X is uncountable.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: Rationals are Uncountable>Assume A and B converge to an irrational number.>Let r be this number.>Define a new sequence, x'.>x'_1 = (x_1 - r) / r>x'_2 = (x_2 - r) / r> In particular, nearly all x'_i are irrational.I wondered about this.Can you give an example where (1-r)/r is irrational?Heck, I be happy to show that some (1-r)/r is rational.Does it matter if r is algebraic or transcendental?Russell- 2 many 2 count=== === Subject: : Re: Rationals are UncountableLet X be the set of rational numbers in (0,1).Let x_n be a sequence of rational numbersthat contains every member of X.I have previously described such a sequence.Construct sequences a_n and b_n usingCantor's algorithm.Are the sequences a_n and b_n finite?Assume they are finite.There exists an interval (a_i, b_i)such that x_n contains no rationalsin this interval.Assume a_n and b_n are infinite.There exists an infinite number ofrational numbers in X not in x_n.Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable <6badnbE5HvRzYIzdRVn-tw@comcast.com> Assume A and B converge to an irrational number.>>Let r be this number.>>Define a new sequence, x'.>>x'_1 = (x_1 - r) / r>>x'_2 = (x_2 - r) / rIn particular, nearly all x'_i are irrational.>I wondered about this.>Can you give an example where (1-r)/r is irrational?For ALL r, (1-r)/r is irrational iff r is irrational:(1-r)/r is irrational iff 1/r-1 is irrational iff 1/r is irrational iff r is irrational. So r = sqrt(2), orr = pi, or r = e, will do.>Heck, I be happy to show that some (1-r)/r is rational.This holds iff r is rational and nonzero.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills.=== === Subject: : Re: Rationals are UncountableLet X be the set of all rational numbers in the interval (-1,1).If X is countable there exists a function, f(), that assignsa natural number to every member of X.Assume f() exists.Use Cantor's algorithm to build sets A and B.Assume these sets have this form:a_n = 1/-nb_n = 1/nNo f() that produces these sets A and B canassign a natural number to 0.No f() can assign a natural number to every rational in X.Assume the sequences a_n and b_n converge to anotherrational. Clearly, this rational is not in the range of f().Assume A and B converge to an irrational number.Let r be this number.Define a new sequence, x'.x'_1 = (x_1 - r) / rx'_2 = (x_2 - r) / rWe have defined a new set X' = ( (-1-r)/r, (1-r)/r ).f'() can not map a natural number to 0.(remember -1 < r < 1).Every member of X' can be mapped to X.Therefore, X did not contain every rational.Russell- 2 many 2 count=== === Subject: : Re: Rationals are Uncountable> Let X be the set of all rational numbers in the interval (-1,1).> If X is countable there exists a function, f(), that assigns> a natural number to every member of X.> Assume f() exists.> Use Cantor's algorithm to build sets A and B.> Assume these sets have this form:> a_n = 1/-n> b_n = 1/nIt is not enough to show that some sequences of nes intervals have a non-empty intersection, one must show that ALL do to prove uncountability.Your construction is no more a proof that construction of a single even natural number proves all naturals are even.> No f() that produces these sets A and B can> assign a natural number to 0.> No f() can assign a natural number to every rational in X.Relevance?> Assume the sequences a_n and b_n converge to another> rational. Clearly, this rational is not in the range of f().> Assume A and B converge to an irrational number.> Let r be this number.> Define a new sequence, x'.> x'_1 = (x_1 - r) / r> x'_2 = (x_2 - r) / r> We have defined a new set X' = ( (-1-r)/r, (1-r)/r ).> f'() can not map a natural number to 0.> (remember -1 < r < 1).> Every member of X' can be mapped to X.> Therefore, X did not contain every rational.Are you trying to prove that the set of rational numbers does not contain all the rational numbers?First prove that no unicorn has his (or her) horn up your arse.=== === Subject: : Re: Rationals are Uncountable> If there is no suitable entry for a_{n+1} (for example), then the> function's range doesn't include any values in the interval (a_n,b_n).> The density of the number system was part of the hypothesis, so there> are values in that interval, and we have our contradiction.assume a function ZxN <-> R. Generalize that back to a function N <->R.So then what it does is start at zero and then goes outward from zero. How N maps to R, after N maps to R[0,1], a_1 would be zero, b_1, zero(iota and indefinitely different from zero), then x_3 is negativeiota, x_4 twice iota, x_5 twice negative iota, etcetera, thus that thesequence a is equal to (0) and b (iota).Here you'll notice I assume a mapping from the naturals to the reals,leading into concepts of the indefinitely and variously orderedsequence of real numbers as points on the real number line and thevague fugue of infinitesimality. This is where we have discussedsimilar concepts at length before.There are no values between zero and iota because that's iota'sdefinition, a minimally positive value, next in the increasingsequence of ordered real numbers after zero.It's not the square root of 2 over 99, it's not 3 divided by 4 billionzillion, it's iota.When I mentioned the function x^2, it's to get you to think of thefunction x^2 defined on the reals. How do you determine the areabounded by the function curve and the y axis, for this functiondefined and greater than or equal to zero over the reals? That'spretty simple to answer, you integrate the functions, the indefiniteintegral of x^2 dx is x^3/3. That's about as much integration as Ican handle. Anyways, evaluating that integral over the interval [a,b]is the indefinite integral x^3/3 evalua for a minus that of b, or1/3.One third: one third. Let's examine that result. Is it the sum of the area of many narrowrectangles between the curve and the axis? In a way, it is. Yet, wecan not get an exact result for any finite number of those rectangles. Instead it is said that the width of the rectangle goes to zero asthe number of rectangles goes to infinity. Is it a rational or realzero, or infinity? The points within the unit interval are densebecause they are uniformly distribu on any finite scale, andinfinumerous.Here the calculists are free to support why that is inaccurate,untrue, incomplete, semantically or syntactically flawed, and how italso gives the correct result.So then to suggest something along these lines would require arigorous foundation of iota with the characteristics I ascribe to it,and then some. Non-standard analysis may, various non-standardanalyses certainly may.So basically such a construction involves ignoring or working aroundstandard definitions of denseness for other purposes.Half the integers are even, give or take one. Half the reals arepositive, again.Ross=== === Subject: : Re: Rationals are Uncountable> There are no values between zero and iota because that's iota's> definition, a minimally positive value, next in the increasing> sequence of ordered real numbers after zero.I could define iota as the leader of the little blue people who livein Ross's walls. That would not entitle me to proceed as if iotaactually exis.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W=== === Subject: : Re: Rationals are Uncountable> There are no values between zero and iota because that's iota's> definition, a minimally positive value, next in the increasing> sequence of ordered real numbers after zero.> I could define iota as the leader of the little blue people who live> in Ross's walls. That would not entitle me to proceed as if iota> actually exis.A Google(tm) search for iota infinitesimal provides some referencesto the consideration of iota as an infinitesimal quantity,particularly with its quality of being the least real quantity.http://www.innerx.net/personal/tsmith/ surreal.htmlhttp://mathforum.org/library/drmath/view/53891. htmlhttp://www.maths.ox.ac.uk/~hardenbe/fermi96/fermi96_ main.htmlhttp://www.physicsforums.com/archive/topic/140-1. htmlhttp://thesaurus.maths.org/dictionary/map/indices/Ihttp:// www.ncc.up.pt/~mig/dic/htmlmroget/mroget_28.htmlhttp:// www.levity.com/corduroy/beckette.htmhttp://groups.google.com/ groups?as_q=iota+infinitesimal&as_ugroup=sci.*These might help us establish a context for the use of this word andits syntax, semantics, etcetera, to communicate. Largely as thosewords use the term in a similar or compatible way as I, perhaps it canhelp us discuss these amorphous concepts that are not particularlydifficult mathematics. It has a broad context within surreal numbers.Maybe the rastafarians aren't so far off from the mark: I, i, and i,or as the electrical engineers call it: j, when they speak of unity,in this context, unit-ness.In the light of this context, about sequencing the reals, I use iotato help describe the reals as a sequence of points. Then, I posit asequencing with zero, then the positive iota, then the negative iota,then the next positive value, and the next negative, ad infinitum.Heh heh heh, that's pretty good. I don't care.Ross=== === Subject: : Re: Rationals are Uncountable[garbage snipped]> Ross=== === Subject: : Re: Rationals are Uncountable> The intersection may be empty, a closed interval [lo, hi],> or a semi-open interval [lo, r) or (r, hi],> where r must be rational, but lo and hi may be rational or irrational.> That was very informative.> I hope you don't mind if I ask some stupid questions.> 1) (-1/j, 1/j) produces [0, 0]={0}.> 2) (0, 1/j) produces the empty set.In the sequence (-1/j,1/j), zero isd in every one of them, but no other, non-zero, number is in every (-1/j,1/j), so zero, and only zero, is in their intersection.In (0,1/j), neither zero nor any other number is in every one of them so the intersection ends up empty.> I want to understand why (1) is a singleton set and (2) is empty.See above.> I will use a shorthand notation for the intersection.> Let w mean omega and let (-1/w,1/w) represent the intersection> of the set of intervals (-1/j, 1/j) over all j.> Define 1/w = 0 and 1/-w = 0.A concatenation of mathematical idiocies! Let f_n = (-1/n,1/n) and f_w represent their intersection, if you must introduce unneeded notations, but 1/w is an abomination.> Rewrite the formulas above:> 1) (1/-j, 1/j) produces (1/-w, 1/w)> 2) (0, 1/j) produces (0, 1/w)You may fiddle with such non-sense if you choose, but why do you want to impose it on anyone else.You seem to take immense pleasure in designing notations to hide, rather tha reveal underlying truths. You might do well in politics. > Both formulas convert to (0,0) = empty set.> How can I assume formula (1) is not empty?Don't create misleading formulas, to start with. Mathematics can be hard enough to do right without being so creative in ways of making it harder.Your notations are not the reality. They add assumptions that are not true to make the truth obscure.Perhaps you need a good dose of Korzybski.> Using your definitions, it makes sense that> formula (1) produces [0,0].> Let's assume 1/w =/= 0.lets assume the truth, that 1/w does not mean anything at all.> 1) (1/-j, 1/j) produces (1/-w, 1/w)How? You are saying that a sequence of real numbers converges to a non-existent entity.> 2) (0, 1/j) produces (0, 1/w)Same comment.> Clearly (1/-w, 1/w) is not empty and must contain [0,0].Clearly (1/-w,1/w) is non-existent, and cannot be said to contain, or not contain, anything.> But, (0, 1/w) is not empty because 1/w is not zero.Where on the real number line do you find 1w? If you cannot place it on the real number line, then (0,1/w) is meaningless.> I don't really see how one proves (2) is empty> without also proving (1) is empty.You don't really see much of anything, do you.> Russell> - Learning is easy. Unlearning is harder. And Russell has so very much to unlearn.=== === Subject: : Re: Socrates' Nothing is Everything> In sci.logic, Garry Denke> :>> In sci.logic, Garry Denke>> <4e63857.0401220627.6600e37c@posting.google.com>:> Sadly yes, that seems clear even to me with his last post.Never mind, I could do with plenty of lessons in patience.Love and respect> Chris Hello Chris, Because every number multiplied by 0 equals 0, every number divided by>> 0 equals every number. One of those numbers is -1. Your problem, as>> others' problem here in sci.logic,sci.math,sci.physics, is _limiting_>> the solution. Don't be sad, it's only a fact. Well, you're right that x = 0/0 can solve every equation.>> You're wrong in assuming that this is even remotely useful. :-P> Concerning the sqrt(-1) = 0/0(x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.> -Doctor Garry Whilhelm Denke, 1655> solving every equation, it was of 0 use, and it made 0 sense.> Let a*x^2 + b*x + c = 0 be an equation we want to solve.> Since a*(0/0)^2 + b*(0/0) + c = 0 = 0*0/0 (obviously true if> one multiplies through by the denominator, namely 0), it is> clear that x = 0/0 is a solution of the above equation.> It is also clear that this result is of no practical value,> and violates at least one law of arithmetic.> One of those equations, of course, is a = 1, b = 0, c = 1,> as a special case. Therefore, x = 0/0 solves x^2 + 1 = 0.http://www.dalik.net/wallpaper/nothing.jpg=== === Subject: : Re: Socrates' Nothing is EverythingIn sci.logic, Garry Denke<4e63857.0401240538.3eba3f69@ posting.google.com>:>> In sci.logic, Garry Denke>> <4e63857.0401230604.60fe67d9@posting.google.com>:> In sci.logic, Garry Denke> :>> Sadly yes, that seems clear even to me with his last post. Never mind, I could do with plenty of lessons in patience. Love and respect>> ChrisHello Chris,Because every number multiplied by 0 equals 0, every number divided by> 0 equals every number. One of those numbers is -1. Your problem, as> others' problem here in sci.logic,sci.math,sci.physics, is _limiting_> the solution.Don't be sad, it's only a fact.Well, you're right that x = 0/0 can solve every equation.> You're wrong in assuming that this is even remotely useful. :-P Concerning the sqrt(-1) = 0/0 (x^2 + 1)/0 = (0)/0, Sokrates ist' Nichts Alles.>> -Doctor Garry Whilhelm Denke, 1655 solving every equation, it was of 0 use, and it made 0 sense.>> Let a*x^2 + b*x + c = 0 be an equation we want to solve.>> Since a*(0/0)^2 + b*(0/0) + c = 0 = 0*0/0 (obviously true if>> one multiplies through by the denominator, namely 0), it is>> clear that x = 0/0 is a solution of the above equation.>> It is also clear that this result is of no practical value,>> and violates at least one law of arithmetic.>> One of those equations, of course, is a = 1, b = 0, c = 1,>> as a special case. Therefore, x = 0/0 solves x^2 + 1 = 0.> http://www.dalik.net/wallpaper/nothing.jpgNothing ventured, nothing gained and you can't make anomlet [sic] without breaking eggs?Please clarify this reponse.(Also, since (0/0)^2 + 1 = 0 and (0/0)^2 - 1 = 0, itfollows that -1 = +1, leading to some practical issues.)-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Re: Socrates' Nothing is Everything> Zero is ubiquitous in that arises from the operation a-a = 0.> Zero has meaning in real experience as absence or emptiness.> In the context of physics it is rather out of place as far as the> conservation of mass/energy is concerned but lives hapily with those> That operation is only well defined for numbers. Zero, the number, is> not absence or emptiness.What you should have said is that zero does not signify absence oremptiness.What I meant was that nought, nothing, zero and the like do refer to absenceand emptinessin ordinary speech and in the practical use of arithmetic. For example, ifthere are no packets of cornflakes on a supermaket shelf the number zerowill be entered in the stock sheets..If your wallet is empty you have zero cash. Such obvuious meanings hardlywarrent comment.As to the operation a-a, if you spend all the cash you have, this expresseswhat happened in numbers. So, if you spent 2 dollars, the expression wouldbe 2-2 = 0 and if you spent ten dollars it would be 10-10 = 0. Thesestatements represent different events. Zero is a number, a token of a different> type than absence or emptiness. And I would suggest you don't> take modern physics too seriously. It is notorious for using> suggestive terminology which has little to do with the physical> meaning.Metaphysically, zero can be considered fundamental or its negation. In> spatial terms this is infinite extension in infinite dimensions. Inphysical> terms it would be an infinitely dense mass of infinite extension ie the> plenum void. In arithmetic it would be absolute infinity, that most> intractable of all entities.> This isn't metaphysics. It's barely coherent english.I was being a bit loose here. The process of negation has a dual role, thatof nullifying and that of transforming. These roles can be clarrified byconsidering four monadic operators:(1) N00: nullification(2) N01: affirmation or stasis(3) N10: negation(4) N11: plenum nullificationN00x -> 00N01x -> xN10x -> ~xN11 -> 11In its arithmetical context zero acts as type (2) with respect to addition,type (1) with respect to multiplication, type (3) with repect to subtractionand type (4) with respect to division.My point is that zero should bne trea as sui generis and recognized asdistinct from other numbers. (don't bother to tell me that zero is not anoperator)Zero is fundamental because it is bound up with symetry, particularlythe> binary symetry of positive and negative numbers. Zero is paradoxicalbecause> it exists and does not exists or, rather, is a symbol for somethingwhich is> impossible, non-existence, at least in the context of existence.> You *might* have the beginnings of a point here. I don't know if you> have the osophical sophistication to turn this into a cogent> position. (My response would be that the context of existence isn't> the right context for an analysis of the number zero.)I don't think that the term 'context of existence' is meaningful.Mathematical systems exist, if only in thought, but they exist nonetheless.The number zero cannot be analysed in the context of mathematics perse since'mathematics' is incapable of saying anything about its objects; that is therole of metamathematics.Russell's dictum is relevant here:'mathematics is the science in which we do not know what we are talkingabout, and do not care what we say about it is true'.Ideally, zero should be trea like infinity, as a necessary token fora> number but not a proper number. Infinity is unreachable by any finite> operastion on infinite numbers. Zero is unreachable too by any finite> operation other than that of a-a. This last operation, then is rather> suspect and proves to be an annoyance in forms like II[1/(n-1) x 1/(n-2)x> 1/(n-3) x ...]> What the hell is a proper number?A proper number is one which obeys all the operations of arithmetic withoutexception, as a good logical object should. Unfortunately, zero isrecalcitrant with respect to division. Infinite and infinitesimal numbersare improper and are kept out of the garden altogether. And why isn't infinity a proper> number? Surely you're familiar with the logical results about the> existence of non-standard models for, say, the ordered Peano> Arithmetic which have objects which are greater than any other object.Although infinity is referred to in analysis it is forbidden the status ofoperandin that context.> Infinity, as an object, is on the same logical status as any other> number.It is not, for the reason given above. I agree that divergent series mightbe said to have an infinite limit but this is meaningless unless one can saywhich order of infinity it tends towards. Infinity, as usually used in mathematics (like what you> described)'(as you described it)' would be a neater expression for those who careabout good English., is not an object, however. It is used as a sort of> short-hand for quantificational statements to describe the behavior of> the dependent variable (to set an upper bound, for instance) for any> given instance of the independent variable. Note that in the standard> definition of a limit, there is no mention of the word infinity. It> is all done in terms of existential quantification depending on> universal instantiation.> 'cid 'oohI know, that's why I said it wasn't a proper number or 'a number of theusual sort' if you prefer.I'm quite happy for infinite entites (not 'infinity' of course) to beregarded as proper numbers. All you need are bigger 'premises'.Tony Thomas=== === Subject: : Re: universal property of semidirect products?> |>Does the semidirect product of groups have a universal property?> |> |Well, suppose that K is an abelian group and H is group with a given> |action on K, written as k^h. If H.K is the semidirect product using> |this action, then H.K acts on K via conjugation, and we have maps> |p:H.K -> H, q:H.K -> K given by p(h,k) = h, q(h,k) = k, such that p> |is a group homomorphism, the actions of (h,k) and p(h,k) on K are the> |same, and q is a derivation; i.e. q(g1 g2) = q(g1)^g2 q(g2) for g1,> |g2 in H.K.> |> |Now suppose we are given any triple (G,p,q) such that G is a group> |with a given action on K, p:G -> H is a group homomorphism, the> |actions of g and p(g) on K are the same for all g in G, and q:G -> K> |is a derivation.> |> |Then the map f: G -> H.K given by f(g) = (p(g),q(g)) is a group> |homomorphism and the resulting diagram clearly commutes.> |> |I can't see how to make this work when K is not abelian.> hmm, apparently you decided to try to describe the _right_-universal> property of the semi-direct product, which strikes me as a somewhat> peculiar thing to do considering that the semi-direct product is most> simply characterized by its _left_-universal property. nevertheless> it sounds like you may have actually gotten something interesting (and> which would generalize pretty straightforwardly to the case where k is> check for sure.> anyway, the semi-direct product of a group h acting on a group k is> simply the homotopy colimit of the obvious functor f:h->groups> assigning the group k to the unique object of h. (the ordinary> colimit is obtained by starting with k and then universally compelling> the autofunctor of k associa with each element of h to become equal> to the identity autofunctor, whereas the homotopy colimit is obtained> by starting with k and then universally compelling the autofunctor of> k associa with each element of h to become naturally isomorphic to> the identity autofunctor, subject to a certain straightforward> coherence property of this system of natural isomorphisms.)I'm having a bit of trouble visualizing this one.h is the category with one object and with every element in the hom setinvertible..., so the obvious functor sends this object to k inside thecat grp, and sends and element of h to some automorphism of k. So where do these autofunctors arise? === === Subject: : Re: universal property of semidirect products?|> anyway, the semi-direct product of a group h acting on a group k is|> simply the homotopy colimit of the obvious functor f:h->groups|> assigning the group k to the unique object of h. (the ordinary|> colimit is obtained by starting with k and then universally compelling|> the autofunctor of k associa with each element of h to become equal|> to the identity autofunctor, whereas the homotopy colimit is obtained|> by starting with k and then universally compelling the autofunctor of|> k associa with each element of h to become naturally isomorphic to|> the identity autofunctor, subject to a certain straightforward|> coherence property of this system of natural isomorphisms.)|||I'm having a bit of trouble visualizing this one.||h is the category with one object and with every element in the hom set|invertible..., so the obvious functor sends this object to k inside the|cat grp, and sends and element of h to some automorphism of k. So where |do these autofunctors arise? automorphism = autofunctor in this context. as you poin out, thegroup h is a one-object category, but of course so is the group k;thus its automorphisms can be thought of as autofunctors.(hope i didn't misunderstand your question. also i should probablymention that my use of the terminology homotopy colimit here mightbe slightly (but only slightly) non-standard, and that in any case arigorous treatment of such homotopy colimits should take account ofthe fact that the functor f:h->groups is actually a 2-functor, or atleast of the fact that the category of groups is actually a2-category, in fact a full sub-2-category of the 2-category ofcategories.)-- [e-mail address jdolan@math.ucr.edu]=== === Subject: : MDS codes?I am struggling a bit with MDS codes and the Singleton bound. In abook (Lecture notes in control and information sciences: Topics incoding Theory, ISBN: 0387514058 (Springer-Verlag) (New York)) theSingleton bound is given as:M <= M_S(q,n,d) = q^{n-d+1},where M is the size /cardinality of the code, q is thesize/cardinality of the alphabet used, d is the minimum Hammingdistance and n is the maximum length of the codewords in the code.When M = M_S, the code is a MDS (maximum distance separable) code.Some questions that I have are:A. Given a fixed alphabet of cardinality q, can one find codes withlarge n which are MDS, albeit codes that are not quite useful. As anexample, consider the cyclic code genera by x^1 + 1. Thepolynomial will always be a factor of x^n + 1, thus a cyclic code(n,2)C can be genera that will have a minimum distance of d_min =2.B. Are there known MDS codes whose length exceeds that of Reed-Solomoncodes?(What is interesting to me is that the computational program MAGMA hasa function called MDSCode. This function gives the generator matrixof a (q+1,k) code with d_min = q - k + 2. Is this a known code?Any help, pointers to literature or suggestions will be greatlyappreciaJaco=== === Subject: : non-explicit differential equationsAre there some standard existence and/or uniqueness results for nonexplict differential equations?Actually I'm interes in the linear case, i.e. equations of the formEx' = Ax + bwith or without constant coefficients.Can anyone give me a reference? markus=== === Subject: : Re: non-explicit differential equations> Are there some standard existence and/or uniqueness results for non> explict differential equations?> Actually I'm interes in the linear case, i.e. equations of the form> Ex' = Ax + b> with or without constant coefficients.for the case of constant coefficients:if the matrix E is invertible, the equation is equivalent to x'=E^(-1)Ax +E^(-1)b, which is explicit. Otherwise, the equation IMO does not make anysense; because E does not have full rank, there are vectors that are not inthe range of E. When we have an initial condition s.t. Ax+b is one of thesevectors, there cannot be any solution. On the other hand, for an initialcondition in the range of E, an existing solution will not be uniquebecause there are infinitely many possibilities for the initial value ofx'.-- Just because you're paranoidDon't mean they're not after youreverse my forename for mail! - saibot=== === Subject: : Re: need help in understanding Torkel's ZFC comment> ...> the rules are that axiomatic systems consist of definitions, axioms,> rules of inferences and (consequently) theorems. Axiom schemas> unnecessarily (redundantly) go outside of this formalism. > ...There is some truth in this: axiom schemata are in the metalogic.> ...> Yes, the method of translating an axiom schema into a rule of> inference that I described is quite general. The only question is,> would it be better to apply this method to ZFC, i.e., to express the> axiom schemas of ZFC as rules of inference instead? I say yes and> cite Occam's Razor. I also think it is better since axioms and rules> are used differently, and calling a rule of inference an axiom> schema only clouds that distinction. (I am also not aware of anyone> having done this in ZFC before. Are you?)Axiom schemata _are_ rules of inference.-- G.C.=== === Subject: : Generating function of Repunits?I'm interes in what's the generating function of repunits?. Thatis, what's the rational function q(x) such thatq(x) = r_1 + r_2x + r_3x^2 + ....where r_n is the n-th repunit number.Thank you very much,Xan.=== === Subject: : Re: Generating function of Repunits?En el mensaje:9ce021f0.0401240740.4cb1e3d4@posting.google.com,Xan escribi.97:> I'm interes in what's the generating function of repunits?. That> is, what's the rational function q(x) such that> q(x) = r_1 + r_2x + r_3x^2 + ....> where r_n is the n-th repunit number.r_n = (10*n - 1)/9Thenq(x) = (10/9)Sum((10x)^n, n, 0, inf) - (1/9)Sum(x^n, n, 0, inf) = (10/9)(1/(1 - 10x)) - (1/9)(1/(1 - x))q(x) = 1/((x - 1)(10x - 1))-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: Generating function of Repunits?In-reply-to: xan2@ono.com (Xan)>I'm interes in what's the generating function of repunits?. That>is, what's the rational function q(x) such that>q(x) = r_1 + r_2x + r_3x^2 + ....>where r_n is the n-th repunit number.The n^th repunit is given by 10^n - 1 r_n = -------- 9So q(x), which converges for |x| < 1/10, is given by 1 1 1 q(x) = - ( ----- - --- ) 9 1-10x 1-x x = ------------ (1-10x)(1-x)Rob Johnson take out the trash before replying=== === Subject: : Re: Generating function of Repunits?Rob Johnson escribi.97:>> I'm interes in what's the generating function of repunits?. That>> is, what's the rational function q(x) such thatq(x) = r_1 + r_2x + r_3x^2 + ....where r_n is the n-th repunit number.> The n^th repunit is given by> 10^n - 1> r_n = --------> 9> So q(x), which converges for |x| < 1/10, is given by> 1 1 1> q(x) = - ( ----- - --- )> 9 1-10x 1-x> x> = ------------> (1-10x)(1-x)But the series for q(x) isq(x) = Sum(r_{n+1}*x^n, n, 0, inf) = 1/((1 - 10x)(1 - x))noSum(r_n*x^n, n, 0, inf) = x/((1 - 10x)(1 - x))that is you got.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com=== === Subject: : Re: Generating function of Repunits?In-reply-to: Ignacio Larrosa Ca.96estro , xan2@ono.com (Xan)>Rob Johnson escribi.97:> I'm interes in what's the generating function of repunits?. That> is, what's the rational function q(x) such thatq(x) = r_1 + r_2x + r_3x^2 + ....where r_n is the n-th repunit number.The n^th repunit is given by 10^n - 1>> r_n = -------->> 9So q(x), which converges for |x| < 1/10, is given by 1 1 1>> q(x) = - ( ----- - --- )>> 9 1-10x 1-x x>> = ------------>> (1-10x)(1-x)>But the series for q(x) is>q(x) = Sum(r_{n+1}*x^n, n, 0, inf) = 1/((1 - 10x)(1 - x))>no>Sum(r_n*x^n, n, 0, inf) = x/((1 - 10x)(1 - x))>that is you got.Ah, yes. What I gave was the generating function of the repunits, whichis not q(x), but oo --- n > r x --- n n=1which is x q(x). As you point out, 1 q(x) = ------------ (1-10x)(1-x)Teach me to judge a post by its title.Rob Johnson take out the trash before replying=== === Subject: : Algebra shows algebraic integer limitationFor about two years now I've been trying to explain an algebraicmethod for analyzing polynomials that relies on non-polynomialfactors. My research is a natural extension in the polynomial domainof the idea of irrationality in the integer domain. Basically, I lookat the equivalent of irrational factors with polynomials.Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he is indeed at Hamilton College: === Subject: : Re: Mathematical consistency, courageDecker put forward the quadratic(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).The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples ofnon-polynomial factors.Notice that despite not being polynomials they are algebraic integersif x is an algebraic integer because a_1(x) and a_2(x) are the tworoots ofa^2 - (x - 1)a + 7(x^2 + x).To my knowledge mathematicians have never done extensive research withnon-polynomial factors of polynomials, but instead most work in thearea has to do with finding roots to polynomials, or determining ifpolynomials with rational coefficients are reducible over rationals.The extension into the realm of non-polynomial factors has revealed aproblem with a previous understanding of mathematicians in the area ofthe ring of algebraic integers. It's easy to prove the problem.Consider that algebraically, factorizations multiply out in a certainway demonstra by(a+b)(c+d) = ac + ad + bc + bdand that's just a FACT which is not open to debate.Looking at(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) it is possible then to multiply out the factors on the left to obtain25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).Now subtracting 14 from both sides gives25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x)which reveals an imbalance as 35 is the constant left on the left,while 0 is what's on the right.A simple linear transformation helps balance the factorization, asusinga_2(x) = b_2(x) - 1, and making the substitution gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).Now multiplying out I get25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)and subtracting 14 from both sides gives25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)so as expec the constants 7 and 2 on the left are factors of 7(2)on the right.The mathematics is trivially obvious in that regard, but from thatsimple result, obvious from basic arithmetic, I have a truly profoundconclusion.That is, given that with(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)the 7 and 2 on the left are factors of 7(2) on the right, if I dividethat constant factor 7, visible as a factor of7(25x^2 + 30x + 2)from both sides then to remain in the ring of algebraic integers,there's only one way logically to do it:(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2as any other way, like factoring 7 into two other non-unit factorswill require that you have numbers not available in the ring ofalgebraic integers.For instance, assume there exists some value of x within the ring ofalgebraic integers where the factors (5a_1(x) + 7) and (5b_2(x) + 2)each have sqrt(7) as a factor (in fact, x=1 will work), then you'dhave(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2which reveals the two factors of 2 on the right, as being sqrt(7) and2/sqrt(7) on the left.Remember, algebra gives a rigid format for multiplying out, as Ipoin out before with(a+b)(c+d) = ac + ad + bc + bdso there shouldn't be debate here from people who accept algebra.Notice that works in the field of algebraic numbers, but not in thering of algebraic integers.But it turns out that it's possible to prove that even with thefactorization(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2that (5a_1(x)/7 + 1) is not in general an algebraic integer.What makes this result so surprising is that the necessary conclusionis that you cannot operate completely in the ring of algebraicintegers if you divide both sides of(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)by 7, as then *necessarily* from basic algebra, you're forced out ofthe ring as you then must have factors of 2 that are outside the ringof algebraic integers, as I demonstra with sqrt(7) as *necessarily*then the factors of 2 are sqrt(7) and 2/sqrt(7) which are not in thering of algebraic integers.Extensions of mathematical knowledge tend to produce surprisingresults, and some people can't handle a world where areas once thoughtsettled are revealed to contain new results. But my hope is that someof you are actually mathematical researchers versus seeing yourselvesas mere caretakers of dogma.For more on my research in this area and a broader overview, pleasesee my blog archives:=== === Subject: : Re: Algebra shows algebraic integer limitation> For about two years now I've been trying to explain an algebraic> method for analyzing polynomials that relies on non-polynomial> factors. My research is a natural extension in the polynomial domain> of the idea of irrationality in the integer domain. Basically, I look> at the equivalent of irrational factors with polynomials.> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.> If you wish to see his original post here are some headers which also> show that he is indeed at Hamilton College:> === Subject: : Re: Mathematical consistency, courage> Decker put forward the quadratic> (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).> The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of> non-polynomial factors.> Notice that despite not being polynomials they are algebraic integers> if x is an algebraic integer because a_1(x) and a_2(x) are the two> roots of> a^2 - (x - 1)a + 7(x^2 + x).> To my knowledge mathematicians have never done extensive research with> non-polynomial factors of polynomials, but instead most work in the> area has to do with finding roots to polynomials, or determining if> polynomials with rational coefficients are reducible over rationals.> The extension into the realm of non-polynomial factors has revealed a> problem with a previous understanding of mathematicians in the area of> the ring of algebraic integers. It's easy to prove the problem.> Consider that algebraically, factorizations multiply out in a certain> way demonstra by> (a+b)(c+d) = ac + ad + bc + bd> and that's just a FACT which is not open to debate.> Looking at> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > it is possible then to multiply out the factors on the left to obtain> 25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).> Now subtracting 14 from both sides gives> 25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x)> which reveals an imbalance as 35 is the constant left on the left,> while 0 is what's on the right.> A simple linear transformation helps balance the factorization, as> using> a_2(x) = b_2(x) - 1, and making the substitution gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).> Now multiplying out I get> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)> and subtracting 14 from both sides gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> so as expec the constants 7 and 2 on the left are factors of 7(2)> on the right.> The mathematics is trivially obvious in that regard, but from that> simple result, obvious from basic arithmetic, I have a truly profound> conclusion.> That is, given that with> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> the 7 and 2 on the left are factors of 7(2) on the right, if I divide> that constant factor 7, visible as a factor of> 7(25x^2 + 30x + 2)> from both sides then to remain in the ring of algebraic integers,> there's only one way logically to do it:> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> as any other way, like factoring 7 into two other non-unit factors> will require that you have numbers not available in the ring of> algebraic integers.> For instance, assume there exists some value of x within the ring of> algebraic integers where the factors (5a_1(x) + 7) and (5b_2(x) + 2)> each have sqrt(7) as a factor (in fact, x=1 will work), then you'd> have> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2> which reveals the two factors of 2 on the right, as being sqrt(7) and> 2/sqrt(7) on the left.> Remember, algebra gives a rigid format for multiplying out, as I> poin out before with> (a+b)(c+d) = ac + ad + bc + bd> so there shouldn't be debate here from people who accept algebra.> Notice that works in the field of algebraic numbers, but not in the> ring of algebraic integers.> But it turns out that it's possible to prove that even with the> factorization> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> that (5a_1(x)/7 + 1) is not in general an algebraic integer.> What makes this result so surprising is that the necessary conclusion> is that you cannot operate completely in the ring of algebraic> integers if you divide both sides of> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> by 7, as then *necessarily* from basic algebra, you're forced out of> the ring as you then must have factors of 2 that are outside the ring> of algebraic integers, as I demonstra with sqrt(7) as *necessarily*> then the factors of 2 are sqrt(7) and 2/sqrt(7) which are not in the> ring of algebraic integers. But recall what Rick Decker found originally:when x = 1, P(x) = P(1) = 7*57. Rick showed that P(1)/7 can be factored in the form 57 = (5*sqrt(-2) +sqrt(7))*(-5*sqrt(-2) + sqrt(7)),that is, there is a factorization in which all thecoefficients are algebraic integers. You appear to be claiming this is impossible. Howdo you resolve this? > Extensions of mathematical knowledge tend to produce surprising> results, and some people can't handle a world where areas once thought> settled are revealed to contain new results. But my hope is that some> of you are actually mathematical researchers versus seeing yourselves> as mere caretakers of dogma.> For more on my research in this area and a broader overview, please> see my blog archives:> === === Subject: : Re: Algebra shows algebraic integer limitation Discussion, linux)> If you wish to see his original post here are some headers which> also show that he is indeed at Hamilton College:Is there any particular reason you're fascina with Decker'saffiliation?For about a month or more, you've mentioned Hamilton College everytime you post about him (as if his affiliation is difficultinformation to come by, or perhaps that the fact that he's at HamiltonCollege ought to make a difference in how one interprets the disputebetween you two). Once, you even helpfully poin out that, if onewants to learn more about Hamilton College, he could use a searchengine. That's crucial technical advice, that is, and very helpful tothe literally thousands, perhaps millions, of us that chose to learnabout Hamilton College because Rick Decker disagreed with JSH. James S. Harris reinvents the Slashdot effect.But *now*, you've given even more vital and obscure information. Itturns out (and I didn't know this *at all*) that one can learn one's*now*, no matter how Rick Decker denies it, we know he really *is*affilia with Hamilton College. Really, I didn't believe it untilpresen with this irrefutable evidence.Now, can you remind me? What, exactly, am I to do with this newinformation? Why ought it matter to me that Rick Decker is affiliawith Hamilton College? It must be staggeringly important information,since Google reports that jstevh@msn.com has mentioned HamiltonCollege 31 times -- to be fair, one of those mentions was when youfailed to snip Decker's .sig.So *why* ought we care about Rick's employer? Aside from the factthat you know who it is? This reminds me of how proud you are torealize that Nora Baron might be a pseudonym -- you began tellingeveryone of this fact (as well as helpfully defining palindrome)right after someone else mentioned it's a palindrome.-- What I've learned is that [mathematicians are] the gatekeepers, andseem to have almost absolute power when it comes to mathematics. -- , on All I Really Ever Needed to Know I Learned in /Ghostbusters/.=== === Subject: : Re: Algebra shows algebraic integer limitation> For about two years now I've been trying to explain an algebraic> method for analyzing polynomials that relies on non-polynomial> factors. My research is a natural extension in the polynomial domain> of the idea of irrationality in the integer domain. Basically, I look> at the equivalent of irrational factors with polynomials.I thought you had died and I went to heaven, but here you are to prove me wrong.Bob Kolker=== === Subject: : Re: Algebra shows algebraic integer limitation[snip]> But it turns out that it's possible to prove that even with the> factorization> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> that (5a_1(x)/7 + 1) is not in general an algebraic integer.> What makes this result so surprisingSurprising? To whom?> is that the necessary conclusion> is that you cannot operate completely in the ring of algebraic> integers if you divide both sides of> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> by 7,That is *not* a necessary conclusion.[snip error-ridden pseudo-math]--There are two things you must never attempt to prove: the unprovable -- andthe obvious.----http://www.crbond.com=== === Subject: : Re: Practical problems with correlation dimension> Hallo!> I want to calculate the correlation dimension of a time series.> Hope somebody can help!> KarlYou can't do that without a sonic screwdriver. Sorry.O'Ryan Wells.=== === Subject: : Re: Practical problems with correlation dimensiontargeting@OMCL.mil,I stand on what I've pos.So do you :-]K. P. Collins> , apparently tired of ranting to> those as cognitively diverse as he, chose to annoy newsgroups> typically free of idiocy like his, with:> } > Hallo!> } } > I want to calculate the correlation dimension of a time serie.> } } > What I have done> } > I calcula the correlation integral C(r) (number of point having a> } > distance smaller than r) for different embedding dimensions. Taking> } > the slopes of the curve of log C(r) against log r for the different> } > embedding dimensions and plotting them against the embedding dimension> } > should result in a limes of the slopes: the correlation dimension.> } } > My problem> } > Which slope shall I take?> } } > In examples I saw in text books there is a nice limit of the slopes> } > with higher embedding dimensions. In my data I do not know which slope> } > I should take because the slope of the curve varies. If I take the> } > slope at a certain value of log r I can not get a limes.> } } > My curves (log C(r) against log r) can be seen in> } > http://karlknoblich.4t.com/korrdim.jpg> } } } > What to do? Does anybody knows such data and how to handle it?> } } > Hope somebody can help!> } } > Karl> }> } What I will say has not yet been accep by others,> That's because you're about to prove to everyone not already aware of> the fact that you're a complete moron. Watch:> } You're missing some crucial data that cross-correlates> } your 'time' series to the cerebellar topology.> I couldn't possibly add anything to prove the point better.> Please keep your imaginary mathematical pollution confined to your> consciousness groups, although alt.usenet.kooks could use your> input.=== === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et alI sense another meltdown coming. Once again the hopelessness of yourposition is becoming apparent to you, as the truth of your opponents'arguments becomes harder for you to ignore. No doubt we can expectanother obscenity-strewn drunken rant in the near future, followed bya pseudo-light-hear attempt to pretend it didn't happen...-- rs, === === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et al> Now I can show you over and over again *basic* algebra, but for many> of you mathematics is the antithesis of what it was to those who came> before you, as to them it was a refuge from the often crazy world of> opinions, beliefs, religion, and just plain nuttiness which most> people take for gran.> For them logic and preciseness in mathematics made it beautiful.> Your generation clearly hates that in mathematics.contained either logic or preciseness?=== === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et al> Now I can show you over and over again *basic* algebra, but for many> of you mathematics is the antithesis of what it was to those who came> before you, as to them it was a refuge from the often crazy world of> opinions, beliefs, religion, and just plain nuttiness which most> people take for gran.Then along came James along with all those circle squarers, cube duplicators and angle trisectors, to save math from all those people who say such things cannot be done.> For them logic and preciseness in mathematics made it beautiful.> Your generation clearly hates that in mathematics.Logic and preciseness are just what are missing in JSH's mathematical paradies. We could use a lot more of them. in hhis work.> I have outlined a position that follows from algebra basic enough to> explain to a kid in school, but most of you have refused to accept it> because you don't *like* it!!! I recall recently how JSH rejec a proof of mine only because he did not like it. Since it turns out Dedekind, one of JSH's nominal heroes of old, proved the tough part (that relatively prime implies coprime in the ring of algebraic integers) he is hardly in a strong moral postion to talk about rejecting what one doesn't like.> What does liking have to do with it?Thank you for asking. Can you now answer why you are so quick to reject what you do not like?> What in the hell is wrong with you people?What is wrong with you, JSH, that you accuse us of your own crimes. > Do you despise the mathematicians who came before you? The one's who> actually had principles?Whatever principles you have clearly do not extend into the world of mathematics, so why do you care?> Do you hate Gauss? Despise Dedekind? Spit on Archimedes? Feel> disgust for Newton? Blast Euler?No. I actually respect what they have done, rather than, like you, merely reeling their names off periodically.> What is WRONG WITH YOUR PEOPLE?I don't have any PEOPLE.> Mathematics is about logic, and accepting what follows mathematically> because it is true. Not because it makes you feel good. Not because> it makes you feel smart, or because you can brag to people about what> you know, but because it's TRUE.Then what are you, who want in for all the wrong reasons, playing at mathematics for. It will never make you rich, or famous, or right when you are actually wrong.> If you have decided to continue to turn your back on the spirit of> mathematics, to spit on the legacy of so many who came before you, to> try and destroy the value in the search for truth, then you are just> debased animals.I guess that makes JSH a debased animal, unlike all of us undebased animals who like math for its own sake.> Pretenders to a grand tradition, who have no hope, no freedom, no> place to turn to, as God in Heaven despises you for despising the> greatest thing of all: Truth.JSH keeps demanding truth, and rejecting it when it appears because it isn't what he really wants. But he does give me a good excuse to vent my spleen without endangering any innocent bystanders.=== === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et al> Now I can show you over and over again *basic* algebra, [snip] Who's Madigin? === === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et al> === Subject: : JSH: Problem with Decker, Madigin, Ullrich et al>Message-id: <3c65f87.0401231512.34efec63@posting.google.com>>Pretenders to a grand tradition, who have no hope, no freedom, no>place to turn to, as God in Heaven despises you for despising the>greatest thing of all: Truth.>Hitting the bottle early this weekend? I thought you usually post these drinkenrants on Sunday.--=== === Subject: : Re: JSH: Problem with Decker, Madigin, Ullrich et al> Now I can show you over and over again *basic* algebra, but for many> of you mathematics is the antithesis of what it was to those who came> before you, as to them it was a refuge from the often crazy world of> opinions, beliefs, religion, and just plain nuttiness which most> people take for gran.I agree with you that most people are nuts. If you were a nut, what kind would you be? I'd be a chestnut. Hard to crack but yummy inside.> Your generation clearly hates that in mathematics.Talkin' bout mah gen-eration . . . hope I die before I get old!> What does liking have to do with it?What's love got to do . . . got to do with it? > Do you hate Gauss? Despise Dedekind? Spit on Archimedes? Feel> disgust for Newton? Blast Euler?Fun game. Hate Hausdorff? Blame Banach? Abhor Abel? Frown on Fermat? Diss Descartes? Punish Poisson?> What is WRONG WITH YOUR PEOPLE?My people? What kind of people are those?> Pretenders to a grand tradition, who have no hope, no freedom, no> place to turn to, as God in Heaven despises you for despising the> greatest thing of all: Truth.Praise Glory Hallelujah, I am feelin' it. I say, Praise Glory I am FEELIN' it. Preach On Brother James!!=== === Subject: : Re: Topological genus of the human body Discussion, linux)> The Prophet Daniel Grubb known to the wise as grubb@math.niu.edu, opened the Book of Words, and read unto the people:>If we treat the human body as a topological surface, male and female, what> is its genus? How many holes do we have in our bodies; what counts as>a hole?>>This may depend on whether your mouth is open or closed.> Mathematics made difficult had a chapter on topology, which drew a> distinction between a 'Quiet Man' (closed orifices, so a surface of> genus 0) and a 'Loud Man' (constantly flatulent and speaking, so, with> an oversimplified view of the digestive system, a surface of genus> 1). They made the rather disturbing point that two loud men could be> interlocked, which is probably best not visualized.But a so-called closed orifice isn't closed in the relevanttopological sense, right? After all, a continuous function turns aclosed mouth into an open mouth, right?At least, I don't recall tearing or ripping in order to open mymouth. Seems to me that topology oughtn't distinguish between openand closed mouths.-- This confused and outraged many Matrix fans, who'd already spent hourson the web explaining that man and computers could never really livein such a state of harmony and mutual benefit. -- http://www.pointlesswasteoftime.com=== === Subject: : Re: JSH: Push for Java> This is _Python_, an extremely fabulous language,You might want to take a look at the following linkon another language called whitespace: http://compsoc.dur.ac.uk/whitespace/ Hale=== === Subject: : Re: JSH: Push for Java> === Subject: : Re: JSH: Push for Java>Message-id: This is _Python_, an extremely fabulous language,>You might want to take a look at the following link>on another language called whitespace:> http://compsoc.dur.ac.uk/whitespace/> HaleYou Python bashers can laugh all you want, but David Ullrich is right aboutPython being a fabulous language. It is the perfect fit for what _I_ need in aprogramming language and that is all that matters to _me_.I, for one, appreciate guys like David Ullrich promoting their favoritelanguages, perl, Python, Java, Rexx, UBASIC, Cygwin, .NET SDK and have downloaded and installed all of them.I'll be the judge of whether a language is suitable for my needs. Just tell mewhat's available and where to get it. I'm not interes in such things as thesignificance of whitespace, whether or not the language is self-hosting or itssuitability to large projects. Those issues can be discussed in languagegroups.In sci.math, when the question is I'm stuck because Excel only has 18 digits,what I want to hear is which languages support unlimi precision integers,not which languages are suitable for writing compilers.--=== === Subject: : Re: JSH: Push for Java Discussion, linux)>> === Subject: : Re: JSH: Push for Java>>Message-id: >This is _Python_, an extremely fabulous language,>>You might want to take a look at the following link>>on another language called whitespace: http://compsoc.dur.ac.uk/whitespace/>> Hale> You Python bashers can laugh all you want, but David Ullrich is> right about Python being a fabulous language. It is the perfect fit> for what _I_ need in a programming language and that is all that> matters to _me_.No idea whether Python rocks or sucks, personally, but Bill'sreference to whitespace was just funny.-- It has been shown that no man can sit down to write without a very profounddesign. Thus to authors in general trouble is spared. A novelist, for example,need have no care of his moral. It is there -- that is to say, it is somewhere-- and the moral and the critics can take care of themselves. --E.A. Poe=== === Subject: : Re: JSH: Push for Java> === Subject: : Re: JSH: Push for Java>Message-id: <87n08d3w2h.fsf@phiwumbda.org>> === Subject: : Re: JSH: Push for Java>Message-id: This is _Python_, an extremely fabulous language,>>You might want to take a look at the following link>on another language called whitespace: http://compsoc.dur.ac.uk/whitespace/HaleYou Python bashers can laugh all you want, but David Ullrich is>> right about Python being a fabulous language. It is the perfect fit>> for what _I_ need in a programming language and that is all that>> matters to _me_.>No idea whether Python rocks or sucks, personally, but Bill's>reference to whitespace was just funny.Yeah, I enjoyed it also.Just wan to use this as a point to butt into this tiresome discussion and insert my 2 cents.>-- >It has been shown that no man can sit down to write without a very profound>design. Thus to authors in general trouble is spared. A novelist, for>example,>need have no care of his moral. It is there -- that is to say, it is>somewhere>-- and the moral and the critics can take care of themselves. --E.A. Poe--=== === Subject: : Re: JSH: Push for Java permission for an emailed response.> But you haven't _mentioned_ any problems that are going to come> up when one attempts to use Python in a large system! Let's see: ever changing semantics. Variable scoping rules that arebroken. Maintains a data/program distinction. Hard to parse. Syntaxonly easy if you already know an Algol-like language. Can't capturecontinuations. Can't create lexical closures.These come up in any system, but they are worse and worse the largerthe system, because the weight of working around them multiplies.Moreover, as with statue comparison, I could point out that David isbetter sculp in little ways than the crappy metal thing outside myoffice, but that would even miss the point. Correct the technicalproblems in the metal thing, and David is *still* much better. Comingup with a clear statement of exactly why David is amazingly morebeautiful is famously hard. > Ok, I lied, you did mention one. But it says _nothing_ about why,> say, Python 1.5.2 is not suitable for large systems. Python 1.5.2 will not be maintained for very long. People will notcontinue to be using it, it will stop being available. Bugs in itdon't get fixed. Python (with version abstrac), by contrast, doesnot maintain backwards compatibility well at all. Consider the case of C, where old-style K&R C is *still* suppor,despite being obsole by ANSI C quite a few years ago now. Thomas=== === Subject: : Re: JSH: Push for Java>> But you haven't _mentioned_ any problems that are going to come>> up when one attempts to use Python in a large system! >Let's see: ever changing semantics. Variable scoping rules that are>broken. Maintains a data/program distinction. Hard to parse. Syntax>only easy if you already know an Algol-like language. Can't capture>continuations. Can't create lexical closures.>These come up in any system, but they are worse and worse the larger>the system, because the weight of working around them multiplies.>Moreover, as with statue comparison, I could point out that David is>better sculp in little ways than the crappy metal thing outside my>office, but that would even miss the point. Correct the technical>problems in the metal thing, and David is *still* much better. Coming>up with a clear statement of exactly why David is amazingly more>beautiful is famously hard. >> Ok, I lied, you did mention one. But it says _nothing_ about why,>> say, Python 1.5.2 is not suitable for large systems. >Python 1.5.2 will not be maintained for very long. People will not>continue to be using it, it will stop being available. If it's not available when someone starts a new large projectthat's irrelevant, because there's no reason the new projectneeds to use 1.5.2. The problem was supposed to be stability -if someone already has a large project using, say, 1.5.2,and he's concerned about stability there's no reason hecan't just keep using 1.5.2 - it's going to remain availableto him until he decides to delete it.>Bugs in it>don't get fixed. Python (with version abstrac), by contrast, does>not maintain backwards compatibility well at all. >Consider the case of C, where old-style K&R C is *still* suppor,>despite being obsole by ANSI C quite a few years ago now. >Thomas************************=== === Subject: : Re: JSH: Push for Java permission for an emailed response.> If it's not available when someone starts a new large project> that's irrelevant, because there's no reason the new project> needs to use 1.5.2. The problem was supposed to be stability -> if someone already has a large project using, say, 1.5.2,> and he's concerned about stability there's no reason he> can't just keep using 1.5.2 - it's going to remain available> to him until he decides to delete it.My point is that any version has bugs, always will. Not that this isPython's fault; it's a big complex interpreter. And there's reallyonly one Python implementation out there.So if you stick with version 1.5.2, you're stuck with the bugs it'sgot, forever.If you upgrade to get bug fixes, you also get language changes to copewith. By contrast, if you stick with old-style C, you can upgrade yourcompiler, and old-style C, being a stable thing, doesn't change. Evenwhen compilers star supporting ANSI C, it's still extremely commonfor them to continue to support old-style C, so that upgrading yourcompiler doesn't force you to use a new version of the language.Thomas=== === Subject: : Re: JSH: Push for Java permission for an emailed response.X-Tom-Swiftie: I'm having deja-vu, Tom said again quotes me saying:> 'I don't think you can prove programming skill by saying crazy things> like Python is an extremely fabulous language. Sorry.'I did not intend this as disrespectful. I apologize for theimpression I gave with it. I believe I was correct in my statementthat to label Python extremely fabulous does demonstrate a lack ofunderstanding of the actual width of the spectrum of programminglanguage fabulousness. But I did not mean to say that saying Python is extremely fabulous isa disproof of programming skill: it is surely not. I said theconverse; that saying such things does not prove programming languageskill. But what I said was a little too casual, and in no way did Ihad that effect.Thomas=== === Subject: : Re: JSH: Push for Java> quotes me saying:>> 'I don't think you can prove programming skill by saying crazy things>> like Python is an extremely fabulous language. Sorry.'>I did not intend this as disrespectful. Ah, I guess I misunderstood the wording, sorry.> I apologize for the>impression I gave with it. I believe I was correct in my statement>that to label Python extremely fabulous does demonstrate a lack of>understanding of the actual width of the spectrum of programming>language fabulousness. Could be. The other day you asked me whether I thought therewas a reason Macsysma(sp?) used Lisp. I was at www.python.orgjust now, looking for something else, and I noticed this at thetop of the page:'Python has been an important part of Google since the beginning, and remains so as the system grows and evolves. Today dozens of Google engineers use Python, and we're looking for more people with skills in this language. said Peter Norvig, director of search quality at Google, Inc.'I wonder whether there's a reason they use Python there.(I'm _temp_ to wonder whether Google counts as alarge system but I'm not going to, because no, ofcourse probably parts of Google use other languagesBut I did not mean to say that saying Python is extremely fabulous is>a disproof of programming skill: it is surely not. I said the>converse; that saying such things does not prove programming language>skill. But what I said was a little too casual, and in no way did I>had that effect.>Thomas************************=== === Subject: : Re: JSH: Push for Java permission for an emailed response.> By far the most significant of your complaints about Python, and the> _only_ one you offered for quite a while, was that it's not suitable> for large systems. It's as if I said Michelangelo's David is a much better sculpturethan the bad statue outside my office, and you demanded I give exactquantified descriptions of exactly why. Frankly, anyone who thinksthe statue outside my office is as good as David is crazy. I can't easily tell you why. I can give statements about how David isfamously wonderful, exhibits power and beauty and grace, and thestatue outside my office is a dorky small metal thing dropped there totry and make a boring garden a little nicer. But that doesn't proveit. I never tendered an offer of proof, and expressing judgment callsabout the quality of programming languages is not like asserting amathematical theorem. I offered my judgment: that to claim thatPython is extremely fabulous demonstrates a lack of awareness ofwhere it really falls in the fabulousness scale of programminglanguages. Sure, it's way better than C or Perl. To people who areused to C and Perl, Python can indeed *seem* extremely fabulous. Butthat's like comparing a decent statue with the crappy one outside myoffice, and saying it's extremely fabulous. With language likethat, what words are left for David?> A lot of this has been interesting. But if I've been showing a lack> of programming expertise, I do think that you've been showing> exactly the sort of [can't come up with the right adjective]> debating style that William has been commenting on.I freely admit that I am entirely unable to come up with a proof or astraightforward objecting standard by which people with only marginalexperience could judge which programming languages are better.This is very different from math, where in principle any proof can bewritten so that someone with no experience or understanding can checkit. Thomas=== === Subject: : Re: JSH: Push for Java > I said that it *would* have been a nightmare in C and C++, and that I > suspec it would have been similar in Scheme or Lisp. But I do not > know enough about the latter two to be sure about it. Anyhow, the > multiplexer is about 300 lines of Python code. > Good grief, that's not even a small program! > Did I say it was large? > I thought it was offered as an example of a large system written in > Python.Nope. It was offered as an example in a question to you what you consideredlarge. What *do* you consider large? > Nobody disputes that Python can run tiny programs well. Yes, of course. Have you looked at the multimedia authoring system Ioffered? Is that large? And please, do not evade the question.Note: the numbers of lines of code does not define the largeness of thesystem. In that case I can offer you a package with 30936 lines of codefor primality proving. In C, Fortran and assembly (about 30 differentmachines). But that is one I would consider quite small.-- === === Subject: : Re: JSH: Push for Java permission for an emailed response.> Nope. It was offered as an example in a question to you what you considered> large. What *do* you consider large?The example I already gave was of a good optimizing compiler. GCC isan example of a large system. The core source is about 400K lines ofsource. That includes the C front end, but not the other ones. (Ada,C++, Fortran, Java, Objective C, and Pascal in GCC 3.3.2.) A typicalbackend in GCC is twenty to sixty thousand lines more.Other examples of large systems are a production OS kernel (say Linuxor the NetBSD kernel). The X window system is another example.In all these cases, you can see the effects of bad language problems(in those cases, C). They are filled with ad-hoc techniques forworking around things; the need to invent whole new languages forparts of the system because C is inflexible and can't be adap tosuit (GCC has RTL and md files; X has the whole horrid resourcesystem, and more; BSD kernel have a special config file syntax thatusers have to deal with; etc). None of these problems are serious insmaller systems, but all of them make GCC hacking (or kernel, or Xinternals) a pain.Moreover, these problems are not so apparent when you are writing Cprograms an order of magnitude or so smaller. They are also oftentotal surprises to people who are not familiar with systems of thissort. Failure to take into account the problems that come up in suchcases is entirely typical of people with less experience.> Yes, of course. Have you looked at the multimedia authoring system I> offered? Is that large? And please, do not evade the question.No, I haven't looked at it. I'm about to go out of town for a week,and ious argument has a low priority.Thomas=== === Subject: : Re: A problem I hope you can give me some advice in resolving the following>> question, which is a Number Theory problem. I've tried and tried but>> can't figure out the right way to solve it.>> The problem is:>> Let q(x,y) be a primitive (that is, with (a,b,c)=1) quadratic form>> ax^2+bxy+cy^2,>> and n a positive integer.>> Show the existence of two integers s,t , with (s,t)=1, such that>> (q(s,t),n)=1.>> Use Chinese remainder theorem to reduce to case n = p^r, p prime.>> Notice that is equivalent to n = p.>> Then one of (1,0), (0,1) and (1,1) works.> You mean, these are the values to assign to s and t?> I thought that the condition (a,b)=1 was defined just for a>1 and b>1,> not for a, b taking values 0 and-or 1.> And, the condition a,b,c coprime seems not necessary to me, if I solve> the problem in this way you told me.> Anyway, thank you for your advice!I'm not sure I see what you are getting at.I first say that one can always reduce to the case n = p prime.Now if gcd(a,b,c) = 1, at least one of the threevalues q(1,0) = a, q(0,1) = c and q(1,1) = a + b + c isnot a multiple of p.-- === === Subject: : Re: p-adic zero of a polynomial in extension Q_p(zeta)>> I wouldn't use Hensel. I'd note that as long as r > = 2,>> (1 + c lambda^r)^p = 1 + p c lambda^r (mod lambda^{p+r}).>> Take b and write it as 1 + p c_2 lambda^2>> for some integer c_2 in Q(zeta_p).>> Now b/(1 + c_2 lambda^2) = 1 (mod lambda^{p+2}) so>> write b/(1 + c_2 lambda^2)^p = 1 + p c_3 lambda^3.>> Keep going: for each j >= 2,>> b/(1 + c_j lambda^j)^p = 1 + p c_{j+1} lambda^{j+1}.>> Then b = a^p where a = product_{j=2}^infinity(1 + c_j lambda^j).> Thanks a lot!> One typo puzzled me a bit, but now I understand your solution.I'm sure that if one is careful with the p-adic log, onecan use that as well.-- === === Subject: : Re: Skeptickal Inquirer UFO>> MOON before Venus, as in one small step, then whatever is possible.>Crank Information> http://www.google.com/search?q=guth+++site%3Awww.crank.netBrad Guth is a complete moron.... We even have a link to his silly Venus site over athttp://spaceprojects.tk along with other stupidity on the net...SpaceProjects.tk=== === Subject: : Re: Nothing's Zero (0)> OK, time for the engineering hat.An equation represents, say, the stress on a bridge. An architect> has to decide whether to use concrete (rocks), paper, or scissors --> erm, I mean, iron -- for a certain critical subspar. He solves> an equation and gets the answer 0/0. Which does he use, and why? Paper, because his paper covered rock (limestone) cut by iron. Uh huh. Please show me a (non-model) bridge made out of paper.> http://www.bennington.com/chamber/Bridges/images/paper_ mill.jpg> PAPER MILL BRIDGE> and follow to first street on left, Murphy Road (.5 miles) Bridge may> be seen from Rt. 67. This bridge, originally built in 1889 by Charles> F. Sears, was completely reconstruc in 2000.> DIMENSIONS: 125.5 feet long, 14.25 feet wide, 8.67 feet high at truss,> 11.17 feet high at center. This bridge was built by the son of the> Silk Bridge builder.> Want to see a Silk Bridge?> Arrgh. What I meant was one using paper in its construction as the> primary structural element (as opposed to wooden beams, steel girders,> concrete, etc.)O. 0.=== === Subject: : Re: Nothing's Zero (0)In sci.math, Garry Denke<4e63857.0401241014.55d3c52f@ posting.google.com>:>> OK, time for the engineering hat. An equation represents, say, the stress on a bridge. An architect>> has to decide whether to use concrete (rocks), paper, or scissors -->> erm, I mean, iron -- for a certain critical subspar. He solves>> an equation and gets the answer 0/0. Which does he use, and why?Paper, because his paper covered rock (limestone) cut by iron.Uh huh.Please show me a (non-model) bridge made out of paper. http://www.bennington.com/chamber/Bridges/images/paper_ mill.jpg PAPER MILL BRIDGE and follow to first street on left, Murphy Road (.5 miles) Bridge may>> be seen from Rt. 67. This bridge, originally built in 1889 by Charles>> F. Sears, was completely reconstruc in 2000. DIMENSIONS: 125.5 feet long, 14.25 feet wide, 8.67 feet high at truss,>> 11.17 feet high at center. This bridge was built by the son of the>> Silk Bridge builder. Want to see a Silk Bridge?>> Arrgh. What I meant was one using paper in its construction as the>> primary structural element (as opposed to wooden beams, steel girders,>> concrete, etc.)> O. 0.Erm....yeah.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.=== === Subject: : Resolvable SpaceA space S is resolvable when S is union two disjoint dense sets.Are all spaces without open singletons resolvable?I think not. Is there an example why not?A space S is equally resolvable when S is union two disjoint equinumerous dense sets.Are all resovable spaces equally resovable?No, any finite multipoint resovable space with odd number of pts iscounterexample. Are there any Hausdorff or infinite counterexamples?-- theoremsinfinite topological group G, proper dense subgroup D ==> G equally resovableproofPick e not in D. e+D is dense and equinumerous and disjoint to D.Now divvy up G - (D / e+D) putting equal many in D and e+D.infinite S, disjoint dense D,E, |D| = |E| or max(|D|,|E|) <= |S - D/E| ==> S equally resolvableProof uses method similar to that used in first theorem.----=== === Subject: : Re: Resolvable Space> A space S is resolvable when S is union two disjoint dense sets.> Are all spaces without open singletons resolvable?> I think not. Is there an example why not?An irresolvable T_1 space (X,t) with no isola points: Let X be aninfinite set. Let u be a free ultrafilter on X. Let t be the topologyon X consistint of the elements of u and the empty set.An irresovable Hausdorff space (X,t) with no isola points: Let X bean infinite set. Let t_0 be a Hausdorff topology on X with no isolapoints. Let T be the set of all topologies on X that extend t_0 andcontain no singletons. By Zorn's lemma, T has maximal elements. Let tbe a maximal element of T.> A space S is equally resolvable when S is union> two disjoint equinumerous dense sets.> Are all resovable spaces equally resovable?> No, any finite multipoint resovable space with odd number of pts is> counterexample. Are there any Hausdorff or infinite counterexamples?No. Let X be an infinite topological space, say |X| = m. Suppose X isthe union of two disjoint dense sets A and B of different sizes, say|A| = m > |B|. Partition A into m disjoint sets A_i, each of size m.Assume for a contradiction that none of the sets AA_i is dense in X.Since B is dense in X, the closure of AA_i does not contain B. Choosea point b_i in B which is not in the closure of AA_i. Since |B| < m,there are two distinct indices i and j such that b_i = b_j. But thenb_i is not in the closure of A, a contradiction. Thus, for some i, Xis the union of the two disjoint sets AA_i and BuA_i, both of size m.=== === Subject: : Re: Binary Quadratic Forms> I'm looking for primes of the form> a^2 - ab + b^2, where b is congruent to 0 (mod 3) and a is not congruent to> 0 (mod 3)All positive primes of the form 6n+1 can be so represen, and no others.Let a = x and b = 3y, so the form becomes x^2 - 3xy + 9y^2. This hasdiscriminant 3^2 - 4* 9 = -27. The discriminant -27 has class number 1, and soone genus class and one class per genus. Modulo 4*27 =108 the numbersrelatively prime to 108 represen by the form x^2 - 3xy + 9y^2 are exactlythe numbers congruent to 1 modulo 6. So every prime congruent to 1 modulo 6 isrepresen by any primitive form with discriminant -27 (primitive means thegcd of the coefficients is 1).Someone else mentioned Cox. See also:Harvey Cohn, A Classical Introduction to Algebraic Numbers and Class Fields,especially chapter 14.Duncan Buell, Binary Quadratic Forms, chapter 4, section 4.Paulo Ribenboim, My Numbers, My Friends, chapter 6, section 12.G. B. Mathews, Theory of Numbers.Any thorough treatment of binary quadratic forms will probably have enough toanswer your question.John Robertson === Subject: : Re: factoring===>Don McDonald schreef in bericht>> factoring : what are the factors of 48?>> 24.01.04 00:04>Those are some of its divisors, but not all of them!I think you've critiqued a time-stamp here.willy === === Subject: : Re: factoring>>Don McDonald schreef in bericht> factoring : what are the factors of 48?> 24.01.04 00:04>>Those are some of its divisors, but not all of them!>I think you've critiqued a time-stamp here.>willy What are the factors of 48?1, 2, 3, 4, 6, 8, 12, 16, 24, 48.Take the sum of those factors, other than 48 (as in the definition ofa perfect number; 6 is a perfect number as it is the sum of 1, 2 and3, its factors other than itself.)sum of 1+2+3+4+6+8+12+16+24 = 7676 factors as 2*2*19: factors 1, 2, 4, 19, 38 (and 76); factor-sum1+2+4+19+38 =64.Factors of 64: 1, 2, 4, 8, 16, 32: sum = 63;and so on.What happens to these series? Do any come to a defined end (if so,how?) Do some regenerate a previous term and hence loop infinitely(obviosly 6 does!)Do any NOT terminate and NOT repeat. If so why?A pleasant diversion for a Sunday afternoon (?)Steve B.=== === Subject: : Re: factoring>Don McDonald schreef in bericht>> factoring : what are the factors of 48?>> 24.01.04 00:04>>Those are some of its divisors, but not all of them!>>I think you've critiqued a time-stamp here.>>willy > What are the factors of 48?> 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.> Take the sum of those factors, other than 48 (as in the definition of> a perfect number; 6 is a perfect number as it is the sum of 1, 2 and> 3, its factors other than itself.)> sum of 1+2+3+4+6+8+12+16+24 = 76> 76 factors as 2*2*19: factors 1, 2, 4, 19, 38 (and 76); factor-sum> 1+2+4+19+38 =64.> Factors of 64: 1, 2, 4, 8, 16, 32: sum = 63;> and so on.> What happens to these series? Do any come to a defined end (if so,> how?) Do some regenerate a previous term and hence loop infinitely> (obviosly 6 does!)> Do any NOT terminate and NOT repeat. If so why?> A pleasant diversion for a Sunday afternoon (?)> Steve B.Sounds a bit like drawing a mandelbrot set :)An exercise from a programming book that I thought might be an interestingdiversion for a sunday afternoon and spent months and months playing with. My very amateurish efforts are here http://tubs.hypermart.net/warwick/fractals.htmlSome much better examples are herehttp://www.fractalus.com/gumbycat/cheersWarwick=== === Subject: : JSH: Past posters, consider factsOne of the things that's interesting to do in light of the Deckerexample is to think back over past posters.The Decker example, of course, is(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), and using a_2(x) = b_2(x) - 1, gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)where now the factors of 7(2) on the right are visible in thefactorization on the left.Several posters have argued for some time that in a case like that ofthe Decker example it must be true that the factors(5a_1(x) + 7) and (5b_2(x) + 2)have some varying factors in common with 7, but they've noticeablybeen vague about explicitly showing what is required from theirposition, which I've remedied by considering what if each had a factorof sqrt(7).Then, the algebra is simple enough as then you'd have(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2where the problem now is that 2/sqrt(7) is not an algebraic integer.That is, the factorization (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2is not in the ring of algebraic integers.Now then, Decker apparently picked this example because at x=1, youcan easily evaluate it to get a result that happens to be in the ringof algebraic integers, but that's even more fascinating as a reason,as consider (x^2 + 2x)/3is NOT in the ring of algebraic integers though for x=1 you get aresult in that ring.That is, to give the expression you need to be in another ring, likebe in the field of algebraic numbers, where it just so happens thatfor a particular value you get an algebraic integer.To the extent that I've seen any posters even consider such problemsat all, they've claimed that there's a shifting factor, like considerf(x) some factor of 3, where you have (x^2 + 2x)/f(x).However, in algebra, if you do have a general factor, then you canjust divide it out, like if you have (x^2 + 3x + 2)/(x+1) you get x+2,which shows that it is a factor in the ring of algebraic integers.Notice no violations of what is in the ring of algebraic integers.But if you have (x^2 + 3x + 2)/3, then you get x^2/3 + x + 2/3, andnotice you have two elements not in the ring of algebraic integers ingeneral.Similarly with(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2you have 2/sqrt(7), which is not in the ring of algebraic integers.It doesn't matter if (5b_2(x) + 2)/sqrt(7) is true for some particularvalue, as in general the expression is outside of the ring, just likex^2/3 + x + 2/3.Algebraically special cases don't matter. So if someone were to cometo you to claim that x^2/3 + x + 2/3 *was* in the ring of algebraicintegers because it is at x=1, they would be wrong, as mathematicallythe special case is irrelevant, as the elements x^2/3 is not ingeneral an algebraic integers, and 2/3 is a constant that is not.The problem I've had in trying to assess motivations for people likeRick Decker, Dik Winter, , David Ullrich, Nora Baron,and C. Bond, among others is in figuring out whether or not they canhonestly be mistaken with such basic algebraic principles or if infact they're deliberately engaging in fraud.Whatever their reasons, what's not open to debate is the fact that ifthe two factors (5a_1(x) + 7) and (5b_2(x) + 2) each have sqrt(7) as afactor then you necessarily must have(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2which is not a factorization within the ring of algebraic integers.Remember, from *algebra* if you have(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)and you divide both sides by 7, dividing each factor on the left bysqrt(7) then you get(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2which is NOT in the ring of algebraic integers.Now some of you may feel that denying that result, or just ignoringit, to give people you may feel are part of your group the benefit ofthe doubt is ok. You may even feel it's your duty to protect peoplelike Decker or Magidin, but remember, mathematics is not just thepeople, it's also the knowledge.What if you were faced with arguments over the existence of sqrt(2),or sqrt(-1)?Imagine if as a group mathematicians had risen to protect the statusquo, and rejec sqrt(-1)?What if it didn't happen only because society needed sqrt(-1) in thereal world for real world applications?Aren't you then providing an indictment of pure math by showing thatmathematicians can't be trus to follow mathematics itself if theysee a personal interest in protecting their group by avoiding certainresults?Physics needed sqrt(-1) but I don't know if it will ever needfactorization into non-polynomial factors, though from past history,it's quite possible that sometime in the future it will be part ofapplied mathematics.Still what you are faced with today is a clear test of your value ofmathematics or social groups, a test of whether or not you will tellthe truth, or try to protect Decker, Winter or Nora Baron from it.=== === Subject: : Re: JSH: Past posters, consider factsOkay, just so I get this...A ring 'garantees' that if you add A and B, and A and B are on thatring, the result A+B is also on that ring. A ring also 'garantees'that A-B is on that ring (because every element has an additiveinverse), so substraction is covered too. Multiplication is also'covered' (A*B is on the ring as well).Division is however not 'garanteed'. Example: 1/7 is not an algebraicinteger even though 1 and 7 are algebraic integers... correct so far?Suppose f(x), g(x) and h(x) are functions whose result are algebraicintegers if x is an algebraic integer. So if h(x) is an algebraicinteger than 7*h(x) is as well. Now suppose:f(x)*g(x) = 7*h(x)The right side of this equation is obviously dividable by 7. Now fromwhat I've read, Mr. Harris seems to think that therefore either f(x)or g(x) must also be dividable by 7. My question is: why? I mean, itis easily seen in the following example that this doesn't have to bethe case:Suppose f(x) = sqrt(7)*x, g(x) = -sqrt(7)*x, h(x) = -(x^2). Neitherf(x)/7 or g(x)/7 is necessarily an algebraic integer (with x analgebraic integer), but (f(x)*g(x))/7 is.Why try to prove that a ring is incomplete with one of the things thata ring doesn't allow? I mean, from posts on this group I understandthat Mr.Harris has been at it a couple of years. How can you miss sucha thing? Or have I missed a key argument in Mr.Harris's argument wherehe explains why in his case one of the factors must be dividable by 7and stay in the ring of algebraic integers?=== === Subject: : Re: JSH: Past posters, consider facts> To the extent that I've seen any posters even consider such problems> at all, they've claimed that there's a shifting factor, like consider> f(x) some factor of 3, where you have> (x^2 + 2x)/f(x).> However, in algebra, if you do have a general factor, then you can> just divide it out, like if you have (x^2 + 3x + 2)/(x+1) you get x+2,> which shows that it is a factor in the ring of algebraic integers.> Notice no violations of what is in the ring of algebraic integers.> But if you have (x^2 + 3x + 2)/3, then you get x^2/3 + x + 2/3, and> notice you have two elements not in the ring of algebraic integers in> general.In the ring of integers, 2 is a factor of x(x+1) for every value of x,but of course 2 doesn't always divide x, or always divide x+1. I don'tget the point of all this. What are you trying to show?=== === Subject: : Re: JSH: Past posters, consider facts> One of the things that's interesting to do in light of the Decker> example is to think back over past posters.Past posters in these threads, other than JSH, have sought truth, regardless of consequences.JSH, in his past posts, sought personal aggrandizement, regardless of truth.=== === Subject: : Re: JSH: Questioning my conclusions >>Here's some identifying to find Decker's post, which also shows that>>he is indeed at Hamilton College:> Good to have that confirmed, my paychecks say the same thing.>>I just don't accept that mushing position as it's not mathematics.> Regardless of whether you accept it or not, it's correct, as> a number of respondants have poin out.> It's not algebraically correct.> Here is algebra, so that you can see what it looks like Decker.> Originally a slight modification to your example gives> (5a_1(x) + 7)(5b_2(x)) + 2) = 7(25x^2 + 30x + 2),> and the question is how to divide both sides by 7 and remain in the> ring of algebraic integers. That's important. It's not just about> dividing by 7, but can you do so and remain in the ring of algebraic> integers?> the left have sqrt(7) as a factor, then you have> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2.> The problem though is that 2/sqrt(7) is not in the ring of algebraic> integers.> However, algebra covers more inclusive rings, so I can display that> and readers can see that now you're in a ring which has 2/sqrt(7).> The algebra doesn't mind.> Now then, even if you now say, hey, but wait, (5b_2(x) + 2)/sqrt(7)> will give a result--that is, if you do the operations, work it out and> get a value--for *some* x, or even some family of x's that IS in the> ring of algebraic integers, I say, ok, but here and now, whether it> works or not for some x, you need 2/sqrt(7), which is not available in> the ring of algebraic integers.> That's algebra Decker.> Mushing everything together in the hope that somehow, someway it will> all work out isn't algebra Decker. Algebraically there's one answer,> and it's not debatable, as algebra isn't a democratic process.> Algebraically, to try and remain in the ring of algebraic integers,> your only option is to have> (5a_1(x)/7 + 1)(5b_2(x)/7 + 2) = 25x^2 + 30x + 2> and I know that it bothers you because it's possible to show that> 5a_1(x)/7 in general is not an algebraic integers, but sorry Decker,> that's what logically follows from the algebra.> The simple fact is that you can't even present any other outcome> without having members that aren't in the ring of algebraic integers,> like I showed with 2/sqrt(7) and it's basic algebra.> The conclusion that follows algebraically is that you can't divide 7> from both sides and guarantee that you're still in the ring of> algebraic integers. Of course Rick Decker showed that P(1)/7 = 57 = (-5*sqrt(-2) + sqrt(7))*(5*sqrt(-2) + sqrt(7)),a factorization in which ALL the coefficients are algebraicintegers. You appear to be saying this is impossible. Howdo you resolve this? Consider another polynomial: R(x) = 7*(50*x^2 - 10*x + 2), which can be written as R(x) = 7*(25*(2*x^2 - x) + 5*(3*x - 1) + 7)If this is factored in the form R(x) = (5 a_1(x) + 7)*(5 a_2(x) + 7),then a_1(x) and a_2(x) are roots of a^2 - (3*x - 1)*a + 7*(2*x^2 - x).Evaluating this at x = 0 gives a^2 + a, which has roots a_1(0) = 0, a_2(0) = -1.Evaluating it at x = 1 gives a^2 - 2*a + 7.The roots are: (2 +/- sqrt(-24))/2 = 1 +/- sqrt(-6).The two roots a_1(1) and a_2(1) are algebraic integers such that a_1(1)*a_2(1) = 7. Therefore R(1)/7 = 42 = (5 + a_2(1))*(5 + a_1(1)).Explicitly, R(1)/7 = 42 = (5 + (1 - sqrt(-6)))*(5 + (1 + sqrt(-6))).Check the arithmetic for yourself.Again, this is a perfectly good factorization of R(1)/7 as a polynomial in the number 5 in which all the coefficients are algebraic integers. Again, you say this kind of factorization is impossible. Please explain.> You're forced out, as the algebra clearly shows. See above. In neither case are you forced out ofthe ring of algebraic integers to obtain a factorization.This is not a question of someone denying algebra. This is a question of someone denying *arithmetic*. > === === Subject: : Re: JSH: Questioning my conclusions >>Here's some identifying to find Decker's post, which also shows that>>he is indeed at Hamilton College:> Good to have that confirmed, my paychecks say the same thing.>>I just don't accept that mushing position as it's not mathematics.> Regardless of whether you accept it or not, it's correct, as> a number of respondants have poin out.> It's not algebraically correct.> Here is algebra, so that you can see what it looks like Decker.> Originally a slight modification to your example gives> (5a_1(x) + 7)(5b_2(x)) + 2) = 7(25x^2 + 30x + 2),> and the question is how to divide both sides by 7 and remain in the> ring of algebraic integers. That's important. It's not just about> dividing by 7, but can you do so and remain in the ring of algebraic> integers?> the left have sqrt(7) as a factor, then you have> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2.> The problem though is that 2/sqrt(7) is not in the ring of algebraic> integers.> However, algebra covers more inclusive rings, so I can display that> and readers can see that now you're in a ring which has 2/sqrt(7).> The algebra doesn't mind.> Now then, even if you now say, hey, but wait, (5b_2(x) + 2)/sqrt(7)> will give a result--that is, if you do the operations, work it out and> get a value--for *some* x, or even some family of x's that IS in the> ring of algebraic integers, I say, ok, but here and now, whether it> works or not for some x, you need 2/sqrt(7), which is not available in> the ring of algebraic integers.> That's algebra Decker.> Mushing everything together in the hope that somehow, someway it will> all work out isn't algebra Decker. Algebraically there's one answer,> and it's not debatable, as algebra isn't a democratic process.> Algebraically, to try and remain in the ring of algebraic integers,> your only option is to have> (5a_1(x)/7 + 1)(5b_2(x)/7 + 2) = 25x^2 + 30x + 2That should be(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2and what's amazing to me now is how simple it all actually is.(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)there's ONE WAY to multiply it out, which reveals that the 7 and 2seen on the left are the factors of 7(2) on the right.So despite the complexity of non-polynomial factors, the decidingfactor is arithmetic!!!Here only numbers in the ring of algebraic integers are allowed ifyou're to stay in that ring, and that leaves only one way to go(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2ignoring unit factors as trivial.The algebra is rigid here. There's just no other way it can be. > and I know that it bothers you because it's possible to show that> 5a_1(x)/7 in general is not an algebraic integers, but sorry Decker,> that's what logically follows from the algebra.One can consider this a test of your ability to accept what logicallyfollows, versus what you may wish to be true.Physicists faced it with quantum mechanics. Maybe mathematicians areweaker minded than physicists though.> The simple fact is that you can't even present any other outcome> without having members that aren't in the ring of algebraic integers,> like I showed with 2/sqrt(7) and it's basic algebra.> The conclusion that follows algebraically is that you can't divide 7> from both sides and guarantee that you're still in the ring of> algebraic integers.> You're forced out, as the algebra clearly shows.And I'd think that *some* mathematicians would actually acceptalgebra!!!=== === Subject: : Re: JSH: Questioning my conclusions> And I'd think that *some* mathematicians would actually accept> algebra!!!We have. That's why we attack your proofs.=== === Subject: : Re: JSH: Questioning my conclusions[snip error-ridden equation]> That should be> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> and what's amazing to me now is how simple it all actually is.If it's so simple, how come you got it wrong? Hahahahah.......Oops!--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com=== === Subject: : Re: JSH: Questioning my conclusions> Regardless of whether you accept it or not, it's correct, as> a number of respondants have poin out.> It's not algebraically correct.Based on past history, JSH's judgement of what is or is not algebraically correct is unreliable in the extreme. Particularly at anything beyond high school levels.=== === Subject: : Arcsin X Maclauryn seriesHey, I'm trying to see how the Maclauryn Series for the arcsin xfunction is derived. I don't see the pattern of the derivities setequal to zero. I've seen it given on Eric Weistand's MathematicalEncyclopedia, but I wasn't able to realize the pattern behind eachcoefficient. Any and all help is apprecia. Thanks.=== === Subject: : Re: Arcsin X Maclauryn series> Hey, I'm trying to see how the Maclauryn Series for the arcsin x> function is derived. I don't see the pattern of the derivities set> equal to zero. I've seen it given on Eric Weistand's Mathematical> Encyclopedia, but I wasn't able to realize the pattern behind each> coefficient. Any and all help is apprecia. Thanks.Another way is:Find the series for sin(x), and then do a reversion of series on that series.To learn more about reversions of series, see, for example, http://mathworld.wolfram.com/SeriesReversion.html=== === Subject: : Re: Arcsin X Maclauryn series> Hey, I'm trying to see how the Maclauryn Series for the arcsin x> function is derived. I don't see the pattern of theUse the development of the derivative of arcsin.=== === Subject: : Re: Arcsin X Maclauryn series> Hey, I'm trying to see how the Maclauryn Series for the arcsin x> function is derived. I don't see the pattern of the> Use the development of the derivative of arcsin.Yes, (1-x^2)^(-1/2) is a binomial series, then integrate term-by-term.=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityExpires: 28 daysYou have made it quite plain in your note that you do not have the>foggiest>> idea of what the experiment of which you are talking, so that last>> sentence of yours is totally without foundation, and merely expresses>> wishful thinking on your part.It was actually a most beautiful observation which convincingly>vindica>> the source>> independence of the speed of photons, and as a bonus confirmed that the>> relativistic expression for the Doppler shift is correct, even at speeds>in>> excess of 99% of that of light.Sorry chum, you don't have a leg to stand on.FranzSummary of your position to date:>Mistake number 1. No it does not. .>> which when>> accellera to close to c, decays to become one going forward>Mistake number 2. The decay angular disteibution is isotropic in the CM of>the pi0>> at>> Exactly c, and the other opposite direction at exactly c. That only>> two photons were produced,>That is so. It is an extremely well known property of the pi0 that its>major decay mode, by a very long chalk, is into 2 photons.>> and both measured so perfectly>Mistake number 3. Not measured perfectly, whatever that may mean.>The actual experimental uncertainties are quo properly in the paper, as>is the case in all good papers describing experimental observations.You mean a statistical approach is used?You know what one can do with statistics Franz. Almost anything!>> I find>> unlikely:---- that they are both exactly on the axis of travel of the>> pion I find bull.>Mistake number 3. They entered two detectors placed on the axis defined by>the direction of>travel of the pi0. The detectors subtended small but finite solid angles at>the decay position. That is discussed in the paper.>> And before you waffle anymore, note again that it was the ENERGIES>> which were looked at , NOT the VELOCITIES!!>Mistake number 4>On the contrary, both the energies and the speeds were measured. Please>read the paper. The times between the birth and death of the photons were>measured. The distances between the birth and death points were known. The>speeds could thus be obtained by simple arithmetic division. The energies>were measured by the use of well-calibra photon calorimeters.How do you know they were the same photons?What is a photon anyway?>> I took my daughter to a circus, and she cried when a bloke climbed a>> rope seemingly just hanging in the air, and vanished! She thought he>> must have died, but you need not cry: just realise that you are seeing>> illusions. (helped by a lot of positive thinking, if your>> job/reputation depends on these results)>There is no physics in that final paragraph of yours, but I leave it in, to>testify to your feebleness.>Franz>Franz HeymannHenri Wilson. www.users.bigpond.com/hewn/index.htm=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity> Germaine to this 'dicussion' is the fact that contrary to all> claims> that experiments have been performed which showed 'actually'> that light was independent of source motion, apparently only one has> been> confirmed, AND TURNS OUT TO BE A CROCK!> Turs out, did it not, that the photon speed was not really> measured-- just an increased energy no which was not allowed to> an increased velocity.You have quite obviously not read anything about that expariment. The> times> between the birth and the death of the two photons were measured with> extreme precision. Pray tell....could it be that the stop watch which timed said time oftravel of photon from point of emission to detector was star by alight signal from said point of emission? (Important note to DHR's: ifthe photon IS travelling >c, the watch will be self-correc by timeof light signal activating timing of event)These times agreed with the assertion that the> speed> of> the photons were independent of the speed of the source, in spite of> the> fact that the source moved at around 99% of the speed of light.> Additionally, the energies of the forward and backward photons were> measured. These turned out to be in agreement with what was expec> as> a> result of the relativistic Doppler effect.It is feeble for Relativity to claim photon speed is independent ofsource, while simultaneously claiming source velocity affects energyof photon.> My daughter was present at the circus: the rope man didn't die> What I strongly suspect, is that the photon emit, and the remnant> velocity after the decay!Your srtong suspicion is unfounded. There was no remnant of whatever> the> photons themselves.see your self-contradiction below> If it is the pi0 decay experiment of which you speak, the> following> apply:> They were most certainly not travelling at zero speed when they> decayed.> They decayed in flight. This is borne out by the fact that the> forward> going photons were shif up in energy in full accordance with> the> relativistic Doppler effect and the backward going photons were> shif> downwards in energy by the expec amount, for the same reason.Stand in the back of a moving vehicle, and throw two stones, oneforward and one back. The forward one is shif up in energy, and therearwards one back---AND nothing to do with Relativity>Uncle Al (and others) have ridiculed me previously for considering twophotons at once: some unispiring arguement about two photons NEVERbeing able to be looked at at once. Was he wrong?> On the contrary, that experiment is but a tiny weapon in the> armoury> of> physicists. The really abundant evidence in favour of SR comes> from> the> accelerators> designed in accordance with the principles of relativistic> mechanics> and> relativistic> kinematics as accurately as can be measured.> All the rest involve> tweaked clocks and wavelength.(you- Twaddle)I submit that any clock using or activa by light is very suspect.If light was in no way a factor of the time measurement which was usedto calculate the velocities, I will apologise.> You have no comprehension of what _causes_ Doppler. It is ALWAYS the> result of a past accelleration (decelleration). So if a Doppler> shift> is observed, then the photon suffered said accelleration DUE TO the> MOTION of its SOURCE.On the contrary. I know quite precisely how the Doppler effect> occurs.> In> the experiment under consideration, it arose from the fact that the> source> wa smoving a a speed very close to that of light at the moment when> the> two> photons were born.>> No comparison was done between the photon and source (remaining> bits )> right?There were no remaining bits> Summary of your position to date:Mistake number 1. No it does not. You clearly imply that after emission of two photons from pi0, thereis no residule other than in the photons!!!which when> accellera to close to c, decays to become one going forwardMistake number 2. The decay angular disteibution is isotropic in the CM> of> the pi0at> Exactly c, and the other opposite direction at exactly c. That only> two photons were produced,That is so. It is an extremely well known property of the pi0 that its> major decay mode, by a very long chalk, is into 2 photons.> There were no remaining bitsSee any discrency in the above??>and both measured so perfectlyMistake number 3. Not measured perfectly, whatever that may mean.> The actual experimental uncertainties are quo properly in the paper, as> is the case in all good papers describing experimental observations.I find> unlikely:---- that they are both exactly on the axis of travel of the> pion I find bull.Mistake number 3.> The astute reader will notice that I cannot count. An astute reader will also note that your position changed somewhat asto what occurred between time of decay, and arrival of (both) photonsat detectors. > They entered two detectors placed on the axis defined by> the direction of> travel of the pi0. The detectors subtended small but finite solid angles> at> the decay position. That is discussed in the paper.And before you waffle anymore, note again that it was the ENERGIES> which were looked at , NOT the VELOCITIES!!Mistake number 4> On the contrary, both the energies and the speeds were measured. Please> read the paper. The times between the birth and death of the photons were> measured. The distances between the birth and death points were known.> The> speeds could thus be obtained by simple arithmetic division. The energies> were measured by the use of well-calibra photon calorimeters.The one true process for deciding which object is faster, is TOCOMPARE THEM.To tell which is the faster runner, one need know1 That they were at the same place at the same moment2 Be able to observe them at the finish line- see which crosses firstNo clocks are needed; the distance doesn't have to be knownI challenge you, or any other DHR, to point to an experiment wherethis was done.(hint: I can give two easy examples for said comparisons, which woulddestroy the myth of c irrevocably)=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity>message> Germaine to this 'dicussion' is the fact that contrary toall> claims> that experiments have been performed which showed 'actually'> that> light was independent of source motion, apparently only onehas> been> confirmed, AND TURNS OUT TO BE A CROCK!> Turs out, did it not, that the photon speed was not really> measured-- just an increased energy no which was notallowed to> an increased velocity.> You have quite obviously not read anything about that expariment.The> times> between the birth and the death of the two photons were measuredwith> extreme precision.> Pray tell....could it be that the stop watch which timed said time of> travel of photon from point of emission to detector was star by a> light signal from said point of emission?Yes. (Important note to DHR's: if> the photon IS travelling >c, the watch will be self-correc by time> of light signal activating timing of event)Nonsense.Only one clock was used in the Alvager et al experiment.It measured the flight times for both the forward and backward goingphotons. You have little clue of what you are talkinng about. So to which one'smotion was it self-correc, whatever that might mean.Why do you think yo are entitled to criticise an experiment when you have soobviously not read the relevant paper?> These times agreed with the assertion that the> speed> of> the photons were independent of the speed of the source, in spiteof> the> fact that the source moved at around 99% of the speed of light.> Additionally, the energies of the forward and backward photonswere> measured. These turned out to be in agreement with what wasexpec> as> a> result of the relativistic Doppler effect.> It is feeble for Relativity to claim photon speed is independent of> source, while simultaneously claiming source velocity affects energy> of photon.No. You are wrong. Photons exist with energies anywhere from as near zeroas you please, up to that of the most energetic photon found in cosmic rays.They all travel at a speed c.> My daughter was present at the circus: the rope man didn't die> What I strongly suspect, is that the photon emit, and the remnantfor velocity after the decay!> Your srtong suspicion is unfounded. There was no remnant ofwhateverfor> the> photons themselves.> see your self-contradiction belowThere is no self-contradiction in anything I have said in this thread. Yousimply cannot folow what I am saying, because of your wilful ignorance.> If it is the pi0 decay experiment of which you speak, the> following> apply:> They were most certainly not travelling at zero speed whenthey> decayed.> They decayed in flight. This is borne out by the fact thatthe> forward> going photons were shif up in energy in full accordancewith> the> relativistic Doppler effect and the backward going photonswere> shif> downwards in energy by the expec amount, for the samereason.> Stand in the back of a moving vehicle, and throw two stones, one> forward and one back. The forward one is shif up in energy, and the> rearwards one back---AND nothing to do with RelativityNon-relativistic and relativistic kinematics predict differing relationshipsbetween the energies of the forward and backward going stones. Thedifference becomes very marked at speeds comparable to that of light. Atspeeds like 99.9% of that of light, the difference is quite gross. And itis as predic by relativity, not by pre-relativity physics.I cannot for the life of me understand why you insists batting on a losingwicket.> Uncle Al (and others) have ridiculed me previously for considering two> photons at once: some unispiring arguement about two photons NEVER> being able to be looked at at once. Was he wrong?I have not the slightest doubt that you have, as is usually the case, got itwrong. What *is* impossible is to look at the same photon twice.> On the contrary, that experiment is but a tiny weapon in the> armoury> of> physicists. The really abundant evidence in favour of SRcomes> from> the> accelerators> designed in accordance with the principles of relativistic> mechanics> and> relativistic> kinematics as accurately as can be measured.> All the rest involve> tweaked clocks and wavelength.> (you- Twaddle)> I submit that any clock using or activa by light is very suspect.You have no clue about what you are talking about.> If light was in no way a factor of the time measurement which was used> to calculate the velocities, I will apologise.I don't care a blind damn whether you choose to apologise or not.I am not conducting a debate with you. I am merely pointing out what anunimpressive fool you are.> You have no comprehension of what _causes_ Doppler. It is ALWAYSthe> result of a past accelleration (decelleration). So if a Doppler> shift> is observed, then the photon suffered said accelleration DUE TOthe> MOTION of its SOURCE.> On the contrary. I know quite precisely how the Doppler effect> occurs.> In> the experiment under consideration, it arose from the fact thatthe> source> wa smoving a a speed very close to that of light at the momentwhen> the> two> photons were born.> No comparison was done between the photon and source (remaining> bits )> right?> There were no remaining bits> Summary of your position to date:Mistake number 1. No it does not.> You clearly imply that after emission of two photons from pi0, there> is no residule other than in the photons!!!Absolutely so. That does not in any way whatsoever imply that the pi0consists of 2 photons, as you put it.A positronium atom also leaves only a number of photons when itsconstituents annihilate by interacting with one another. That does not meanthat the electron and the positron consist of photons only.Nature is chock full of creation and annihilation events.> which when> accellera to close to c, decays to become one going forwardMistake number 2. The decay angular disteibution is isotropic in theCM> of> the pi0at> Exactly c, and the other opposite direction at exactly c. That only> two photons were produced,That is so. It is an extremely well known property of the pi0 thatits> major decay mode, by a very long chalk, is into 2 photons.> There were no remaining bitsSee any discrency in the above??Not at all. I am only once more reminded of your essential ignorance of atopic which you have the cheek to criticise.> and both measured so perfectlyMistake number 3. Not measured perfectly, whatever that may mean.> The actual experimental uncertainties are quo properly in thepaper, as> is the case in all good papers describing experimental observations.I find> unlikely:---- that they are both exactly on the axis of travel ofthe> pion I find bull.Mistake number 3.The astute reader will notice that I cannot count.> An astute reader will also note that your position changed somewhat as> to what occurred between time of decay, and arrival of (both) photons> at detectors.Not at all. All that happened during those times was that the two photonseach travelled at a speed c between their common birthplace and thepositions at which their arrival times were measured.> They entered two detectors placed on the axis defined by> the direction of> travel of the pi0. The detectors subtended small but finite solidangles> at> the decay position. That is discussed in the paper.And before you waffle anymore, note again that it was the ENERGIES> which were looked at , NOT the VELOCITIES!!Mistake number 4> On the contrary, both the energies and the speeds were measured.Please> read the paper. The times between the birth and death of the photonswere> measured. The distances between the birth and death points wereknown.> The> speeds could thus be obtained by simple arithmetic division. Theenergies> were measured by the use of well-calibra photon calorimeters.> The one true process for deciding which object is faster, is TO> COMPARE THEM.> To tell which is the faster runner, one need know> 1 That they were at the same place at the same moment> 2 Be able to observe them at the finish line- see which crosses first> No clocks are needed; the distance doesn't have to be knownBalls. Physicists have intelligence and there are those who can measuresimultaneously the flight times of two photons on different paths, whetheryou like that fact or not. In principle it is dead easy. In practice youhave to be as good as the physicists who actually did the experiment, if youare to get results as good as theirs.> I challenge you, or any other DHR, to point to an experiment where> this was done.Your request is irrelevant.> (hint: I can give two easy examples for said comparisons, which would> destroy the myth of c irrevocably)Rubbish. If you had an idea, you would have trot it out.Franz=== === Subject: : Re: Final Rout of Synchronization Clocks in Relativity message> Pray tell....could it be that the stop watch which timed said time of> travel of photon from point of emission to detector was star by a> light signal from said point of emission?> Yes.> (Important note to DHR's: if> the photon IS travelling >c, the watch will be self-correc by time> of light signal activating timing of event)> Nonsense.> Only one clock was used in the Alvager et al experiment.Very handy! You can lok at two photons at once with the same clock,but NOT the same one twice> It measured the flight times for both the forward and backward going> photons.> You have little clue of what you are talkinng about. So to which one's> motion was it self-correc, whatever that might mean.I could point to synchronising clocks in Relativity on threadsgalore. The photons going in opposite directions BOTH SEE THE OTHERCLOCK RUN SLOW, (remember?)> These times agreed with the assertion that the> speed> of> the photons were independent of the speed of the source, in spite> of> the> fact that the source moved at around 99% of the speed of light.> Additionally, the energies of the forward and backward photons> were> measured. These turned out to be in agreement with what was> expec> as> a> result of the relativistic Doppler effect.Oh goody! Next time I hear a train whistle, I can use your Dopplerclock to know how fast the train is going, and when it will arrive!And all this without knowing the whistle tone when train stationary!It is feeble for Relativity to claim photon speed is independent of> source, while simultaneously claiming source velocity affects energy> of photon.> No. You are wrong. Photons exist with energies anywhere from as near zero> as you please, up to that of the most energetic photon found in cosmic rays.> They all travel at a speed c.Prediction: Cosmic ray luminescent tracks as studied by the arraysbeing built in Argentina will be produced in a time indicating abillions (?) magnitudeWhat I strongly suspect, is that the photon emit, and the remnant> for> velocity after the decay!> Your srtong suspicion is unfounded. There was no remnant of> whatever> for> the> photons themselves.see your self-contradiction below> There is no self-contradiction in anything I have said in this thread. You> simply cannot folow what I am saying, because of your wilful ignorance.> If it is the pi0 decay experiment of which you speak, the> following> apply:> They were most certainly not travelling at zero speed when> they> decayed.> They decayed in flight. This is borne out by the fact that> the> forward> going photons were shif up in energy in full accordance> with> the> relativistic Doppler effect and the backward going photons> were> shif> downwards in energy by the expec amount, for the same> reason.Stand in the back of a moving vehicle, and throw two stones, one> forward and one back. The forward one is shif up in energy, and the> rearwards one back---AND nothing to do with Relativity> Non-relativistic and relativistic kinematics predict differing relationships> between the energies of the forward and backward going stones. The> difference becomes very marked at speeds comparable to that of light. At> speeds like 99.9% of that of light, the difference is quite gross. And it> is as predic by relativity, not by pre-relativity physics.> Uncle Al (and others) have ridiculed me previously for considering two> photons at once: some unispiring arguement about two photons NEVER> being able to be looked at at once. Was he wrong?> I have not the slightest doubt that you have, as is usually the case, got it> wrong. What *is* impossible is to look at the same photon twice.> I submit that any clock using or activa by light is very suspect.> You have no clue about what you are talking about.see below: the runners crossing the finishing line provided theinformation for the stop watch to start. As the watch knew c, and nowthe runner has arrived, it couldn't get ANY OTHER RESULT than c forthe runner> If light was in no way a factor of the time measurement which was used> to calculate the velocities, I will apologise.> I don't care a blind damn whether you choose to apologise or not.> I am not conducting a debate with you. I am merely pointing out what an> unimpressive fool you are.> You have no comprehension of what _causes_ Doppler. It is ALWAYS> the> result of a past accelleration (decelleration). So if a Doppler> shift> is observed, then the photon suffered said accelleration DUE TO> the> MOTION of its SOURCE.> On the contrary. I know quite precisely how the Doppler effect> occurs.> In> the experiment under consideration, it arose from the fact that> the> source> wa smoving a a speed very close to that of light at the moment> when> the> two> photons were born.> No comparison was done between the photon and source (remaining> bits )> right?> There were no remaining bits> Summary of your position to date:Mistake number 1. No it does not.You clearly imply that after emission of two photons from pi0, there> is no residule other than in the photons!!!> Absolutely so. That does not in any way whatsoever imply that the pi0> consists of 2 photons, as you put it.This is smacking now of dishonesty! Read again- A piO consists of aunderstand?> A positronium atom also leaves only a number of photons when its> constituents annihilate by interacting with one another. That does not mean> that the electron and the positron consist of photons only.> Nature is chock full of creation and annihilation events.Once again: your explanation of the piO event suggests that ALL thematter of the piO was accoun for. Then it goes to most.........> which when> accellera to close to c, decays to become one going forwardMistake number 2. The decay angular disteibution is isotropic in the> CM> of> the pi0> at> Exactly c, and the other opposite direction at exactly c. That only> two photons were produced,That is so. It is an extremely well known property of the pi0 that> its> major decay mode, by a very long chalk, is into 2 photons.Isotropic: in all directions. So these two photons were travellingin ALL directions ?> An astute reader will also note that your position changed somewhat as> to what occurred between time of decay, and arrival of (both) photons> at detectors.> Not at all. All that happened during those times was that the two photons> each travelled at a speed c between their common birthplace and the> positions at which their arrival times were measured. They entered two detectors placed on the axis defined by> the direction of> travel of the pi0. The detectors subtended small but finite solid> angles> at> the decay position. That is discussed in the paper.> And before you waffle anymore, note again that it was the ENERGIES> which were looked at , NOT the VELOCITIES!!Mistake number 4> On the contrary, both the energies and the speeds were measured.> Please> read the paper. The times between the birth and death of the photons> were> measured. The distances between the birth and death points were> known.> The> speeds could thus be obtained by simple arithmetic division. The> energies> were measured by the use of well-calibra photon calorimeters.The one true process for deciding which object is faster, is TO> COMPARE THEM.> To tell which is the faster runner, one need know> 1 That they were at the same place at the same moment> 2 Be able to observe them at the finish line- see which crosses first> No clocks are needed; the distance doesn't have to be known> Balls. Physicists have intelligence and there are those who can measure> simultaneously the flight times of two photons on different paths, whether> you like that fact or not. In principle it is dead easy. In practice you> have to be as good as the physicists who actually did the experiment, if you> are to get results as good as theirs.All we need now is One with the balls to do a COMPARISON OF FLIGHT> I challenge you, or any other DHR, to point to an experiment where> this was done.> Your request is irrelevant.What's this? Another EISTEIN? (God has made me an authority etc)> (hint: I can give two easy examples for said comparisons, which would> destroy the myth of c irrevocably)> Rubbish. If you had an idea, you would have trot it out.receiever), simultaneously send another pulse. Compare their arrivaltimes! Note well: this does NOT mean timing- just position of spots ona moving photographic plate will do. Agreement of time of emission ofsignals is done by signal exchange between (satellite) and rockettarget.With the train moving, fire a pulse from the same position. Vectoraddition of trains motion to photons will result in a miss in thetrain's direction (for a fast train, in the order of a foot at target100 mile distant).This experiment may be impractical due to vibration, the accuracy oflasers, and delays in triggering the pulse from the time of arrivalof train at emission point, due to technical limitations--- whichmight let you off the hook for now!PS: I back race horses, not trotters. They are much more likely to winon their abilityJim G=== === Subject: : Re: Final Rout of Synchronization Clocks in RelativityMessage-Id: <1074940740.91613.0@demeter.uk.clara.net>> c' * cos(b') = (c + v') - v * cos(a')>>and you should see that c' = c>The vertical component of A is not quite v'.>>The vertical component at S is v*sin(a) while at A it>>is -v*sin(a). The horizontal component adds to the>>speed of the light at s and subtracts at A and they are>>equal.I'm afraid I find all that a bit hard to follow. It is too difficult to>draw>> such things properly here.>>I understand, I tried to do a sketch for the a' and b' angles>and it was hopeless, you can't get the resolution at all. The>key point is simply that the motions are symetrical so their>contribution cancels out and the relative speed of the light>incident on the mirror is then exactly c. If I get time I'll>draw it with Paint later.> Yes. Putting complica diagrams on a webpage is the only satisfactoryway.> Trouble is it all takes TIME!This should explain the way I calcula v' better thanthe ASCII sketch.http://www.briar.demon.co.uk/Henri/speed.gifRemember this is in the non-rotating lab frame whilethe discussion below is in the co-rotating table frame.The v*cos(a) adds on emission and subtracts at thenext reflection and the cos(b) term is identical atboth ends so the speed of incidence on the mirror mustbe c.>If the table is rotating anti-clockwise, the apparent>curvature of the paths in the rotating frame will be>clockwise for both rays but you need to be careful which>way you draw it because the rays progress round the table>in opposite directions. As a result, one ray gets nearer>to the centre when half way between the mirrors while the>other gets further away. This didn't take long:>>http://www.briar.demon.co.uk/Henri/paths.gif> That's good.> Two questions:> How can you assume that both rays hit the same point on each mirror?> Like I said, the apparatus rotates slightly while the light is in transit.> Therefore the point where each ray hits will vary with rotation speed.This is drawn in the co-rotating frame in which the mirrors arenot rotating.> Also, how can you say that the curves are identical and circular? I don'tthink> that is right.They are not necessarily circular but any deviation is too smallto show with Paint. The direction of the path turns at a constantrate as the light moves along path because the table rotates ata constant angular velocity.The curves needn't be identical but a straight line path must bethe shortest so for very small v the length of the path must berela to v by L = L_0 + k*v^2There cannot be a linear term such as L = L_0 + k'*v + k*v^2because for sufficiently small v, |k*v^2| can be made smaller than|k'*v| and for the right sign of v the path would be shorter thana straight line, hence k'=0For sufficiently small v we can also ignore v^3 and higher terms so L = L_0 + k*v^2is valid and this is symmetrical about v=0, hence the path lengthsmust be identical. If the paths lengths are identical, the angleturned by the table in the time taken is the same so the anglesby which the paths deviate from the straight path when they hitthe mirrors are also identical.>> Thus they will be displaced sideways in opposite directions when>> the meet and cause movement of the fringes.>>Sideways displacement doesn't cause fringe shift since you>are moving parallel to the wavefront.> I'm sure it will have some kind of effect - the angle between the twobeams> varies with rotational speed.http://www.briar.demon.co.uk/Henri/paths.gifThe diagram shows a square layout and at the corners the lightwould turn through 90 degrees just as an example. Note the greenlines indicating the angle between the tangents to the rays atthe mirror. It should be clear both rays have moved in the samedirection (clockwise) and since the magnitude is identical, theangle between the rays remains 90 degrees. What this shows isthat the angle between the rays stays the same for small v sincethey rotate the same amount in the same direction.> In Ritzian theory, light gains a velocity when reflec from amoving mirror.> If that is not well known, it is now - because I said it.>>Sure, but you forgot the source was moving.Moving AND rotating.>>Yes, but it is the moving part that affects the speed and>your diagram even shows the light from the source moving>at c+v, but you seem to have forgotten the effect of the>+v when considering the relative speed of the light>incident on mirror A.> Actually I am reaching the conclusion that source dependency or not makelittle> difference to the operation of a sagnac. The travel time of the lightbetween> mirrors is barely affec by c+v when v is extremely small.The travel time is barely affec but more importantly it isincreased by exactly the same amount for both rays since itdepends on v^2, not v, so there should be _no_ path difference.However, the observed fringe shift is exactly as if the speedwas c-v and c+v in the co-rotating frame.> The standard analysis for source dependency in a ring gyro, using arotating> frame, is wrong.> It should treat the device as a 'four mirror interferometer with aninfinite> number of mirrors' rather than a fibre optic internal reflection thing -if you> know what I mean.I think of it as sending the light round the inside of amirrored cylinder. The cylinder, source and detector allco-rotate so then it is obvious that the path length isjust the circumference and the speed is c relative to themirror in both directions. All reflections are grazingincidence and the rotation between reflections isnegligible. The speed in the lab frame is then simplyc+v and c-v for the two rays and the movement of thedetector exactly cancels the speed difference.>> The curvature is opposite for the two different rays. You are missingthis>> point.>>See the diagram. Curvature alters the path length as v^2>(the path is shortest when v=0) and increases both paths>equally so does not produce a path length difference. It>is easiest to see in the fibre case which as you said is>equivalent to an infinite series of mirrors so both paths>are just the circumference.> But you have assumed both rays strike the mirrors at the same points. Thatis> plainly wrong.See above, because of the symmetry it must be correct.> But the displacement is in opposite directions for the two beams.>>Draw out the paths, you'll find it is the same. It's a bit>>surprising.The path length might be the same but the angle of incidence at the>eyepiece is>> different - as is the sideways displacement.>>Fringes are cused by the relative angle between the beams>which stays the same as both rays rotate the same way. The>beam has to be wide enough to fall on the eyepiece so>sideways movement doesn't have an effect, the actual shift>is very small and much less than the size of the beam.> but the sideways movement is also indicative of an angular differencebetween> the two beams.No, see the green lines on the diagram. The rays turn the sameway by the same amount so the angle between them is unaffec.I think your 'infinite mirrors' approach is the simplest tosee what happens but that method, the inertial frame approachand the co-rotating frame we worked through here all give thesame result, there should be no fringe shift if the speed wassource-dependent while in fact the shift is what is expecif the speed of the light was c+v/c-v in the co-rotating frameor always c in the lab frame. Ritz sugges the idea in 1908and Sagnac performed the experiment around 1913. De Sitter hadalready poin out that this was incompatible with observationsof eclipsing binary stars, both they and the Sagnac experimentrule out the Ritzian (source speed dependent) model.George=== === Subject: : graph question boundary=----=_NextPart_000_0037_01C3E294.45685B10------------- -------------------------------------------------------- charset=iso-8859-1I have a graph question. What would the graph of this function look like:x = V4 - y^2and what would this look likex = -V4 - y^2I can't figure out how I could type these into the TI-83 or maple.Thanks in advance for your help.Mark=== === Subject: : graph questionI have a graph question. What would the graph of this function look like: x = V4 - y^2 (V is square root symbol) and what would this look like x = -V4 - y^2 I can't figure out how I could type these into the TI-83 or maple. Thanks in advance for your help. Mark === Subject: : re:graph question===Your 2 expressions are simply the positive and negative halves of acircle.x^2+y^2=4.Specifically a circle of radius 2 centered at the origin.----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : re:graph questionThank you, I was able to figure that out, but which equation goes withwhich graph(to the right of the y-axis)y|o| o| o| o|----o-------x| o| o| o|o(to the left of the y-axis) o|y o | o | o | o | o |o------|----------x o | o | o | o | o |(sorry for the lame semi-circles, but I drew them so you can understandwhat I'm talking about)Thanks!Mark> Your 2 expressions are simply the positive and negative halves of a> circle.> x^2+y^2=4.> Specifically a circle of radius 2 centered at the origin.> ----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet News==----> http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups> ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =---=== === Subject: : A tiny puzzle in number theoryIs it always true that:a) there are distinct integers x,y,z >0 so that x+y+z always dividesx^n+y^n+z^n as n is any integer >0 ?b) there are distinct integers x,y,z >0 so that x+y+z always dividesx^n+y^n-z^n as n is any integer >0 ?=== === Subject: : Re: A tiny puzzle in number theory> Is it always true that:> a) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n+z^n as n is any integer >0 ?> b) there are distinct integers x,y,z >0 so that x+y+z always divides> x^n+y^n-z^n as n is any integer >0 ?make |x+y+z|<=1 ?-- Unpatched IE vulnerability: mhtml wecerr CAB flipDescription: Delivery and installation of an executableReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/48.html=== === Subject: : Re: Inverse Function Theorem--stronger result, better proof> Shouldn't this be the strong form? Theorem: Suppose y is a differentiable> function in a neighborhood of p and Dy/Dx is invertible and continuous at> p. Then y maps a neighborhood U of p bijectively onto a neighborhood V of> y(p), with y^(-1) differentiable in V and continuous at y(p).I believe this is true, and I think it follows from the standard proofs> quite easily (although I haven't checked this super carefully).> Yes, looking at the standard proofs it does seem to follow quite easily.> Looking quickly at the standard proof it seems the standard proof involving> the contraction lemma doesn't really seem meaningfully to use Dy/Dx being> continuous at any place other than p. I don't know why they didn't claim to> prove the stronger result, guess they weren't interes in it. So I suppose> my result is not as interesting as I first thought, and is mainly just the> observation that local injectivity is a little easier to prove than local> homeomorphism onto an open set.By the way, you can prove injectivity using just the mean value theorem. Let's assume WLOG that Df(p) = I, the identity (I'm using f instead of y). Write f = (f1, ..., fn). Then if we're in a convex neighborhood U of p, the MVT shows f(b) - f(a) = (grad(f1)(p1)*(b-a), ..., grad(fn)(pn)*(b-a)),where p1, ..., pn are on [a,b] and * denotes the dot product. Let T be the linear transformation whose matrix has rows equal to the gradients above. We can write f(b) - f(a) = (b-a) + (T-I)(b-a).The continuity of df at p shows that for each j, grad(fj)(pj) is close to the jth row of I if U is small. Hence ||T-I|| < 1/2 in such a nbhd. So we'll have |f(b) - f(a)| >= |b-a|/2 for a, b close to p and we're done.=== === Subject: : Re: Inverse Function Theorem--stronger result, better proof> It's trivial that the map is continuous throughout the neighborhood> (differentiability implies continuity). What I am not assuming is that the> derivative is continuous throughout the neighborhood.Yes, sorry, should have thought a bit longer before replying.=== === Subject: : Re: Action Device Tragedy> [snip]> 17 years ago, in 1986, when I was 17 year old, I ran away from one of> the most prestigious engineering college in India, VJTI Bombay.> [snip]> You misspelled expelled for incompetence and mental disorder. For a> crazy person to be ostracized rather than worshipped in India is> saying something truly vile. Why are you trolling your disease> through professional newsgroups?> Professional newsgroup? And what are you doing here? In almost every> reply you comment on my country, not on me or my post. What> professionalism you are showing?He is but mad north-north-west: when the wind is southerlyhe knows a hawk from a handsaw.=== === Subject: : Action Device: Destination Alpha Centauri ALook, until I plan my future life, I can not build this Action Device.Alpha Centauri A is 4.35 ly away. So, please tell me, I amaccelerating my bike in space at the rate of 1g i.e. 9.8 m/s^2. Howmuch time it will take to cover 2.18 lightyears distance i.e. when Ihave covered half the road?Please Note: Velocity Does Not Matter....Thanks!-Abhi.=== === Subject: : Re: Action Device: Destination Alpha Centauri AContent-transfer-encoding: 8bit> Look, until I plan my future life, I can not build this Action Device.I am precognitive -- I predic you would continue to post about it!-=-=-=-=-=== === Subject: : Re: GCD of multiple numbers>If you have a bunch of numbers and want to find the GCD of them all, what's >the best way to do it?>The method that suggests itself is to keep a running GCD of the first 1, 2, >3, etc. by computing >GCD(x1,x2,x3,...,xn) = GCD(x1,GCD(x2,GCD(...,GCD(xn-1,xn)...)))>or reverse order.>And then for each one you use Euclid's Algorithm or binary GCD as >appropriate.>But I can't stop thinking that there's a better way. I don't know if the following is any better (probably not) than what hasalready been described, but it is a little different and hence may be of someinterest. Suppose n=3 above, then gcd(x1,x2,x3) can be compu as follows:gcd(x1,x2,x3)=repeat repeat tx1=x2-floor(x2/x1)*x1; tx2=x3-floor(x3/x1)*x1; x3=x1 x1=tx1; x2=tx2 if x1>x2 then swap(x1,x2) until x1=0 x1=x2until x2=0gcd=x3 Finding a,b,c s.t. a*x1+b*x2+c*x3=g only requires a bit more computation. rich=== === Subject: : Coprimality: One-Person Game (& question)[I apologize if this post appears twice. Google is giving me sh**..]I was playing around with this today, and found it somewhat addicting. A 1-person game/ puzzle (with a score), which involves a little math.First, start with a n-by-n grid drawn on paper.I suggest a grid of at least n=5, and (for serious fun) perhaps an nof at least 10.So, you start by writing 1 in any of the grid's n^2 squares.You then write the 2 adjacent to the 1, the 3 adjacent to the 2, the 4adjacent to the 3, etc, in a chain of increasing integers, in suchaway that (ideally...) all n^2 integers are placed into the grid, oneinteger (and ONLY one integer) per square.(By adjacent, I mean immediately next to in the direction of up,down, left, or right, but *not* diagonally.)And, oh by the way, the integers are to be placed so that EVERY pairof adjacent squares contains two integers which are COPRIME with eachother...(Again, adjacent is as defined above.)So, you place the integers into the grid as far as you can until yourpath cannot be continued for whatever reason.And your score is the last integer you were able to write into asquare.Oh,...no erasing after you write down an integer!(Well, perhaps a *little* erasing can be allowed...)Here are my best, as of now, n=6 game, for examples:21 22 25 26 27 *20 23 24 * 28 2919 18 1 2 3 ** 17 6 5 4 *15 16 7 8 9 *14 13 12 11 10 *(My score = 29)And better,1 2 3 34 27 266 5 4 33 28 257 8 9 32 29 2412 11 10 31 30 2313 14 17 18 * 22* 15 16 19 20 21(My score: 34)(I viola the no-erasing rule a little bit above...)(And I apologize if I missed that any 2 adjacent integers in myexamples have a GCD which is greater than 1.)(And, double-check every time you write down an integer so as to makesure that it is coprime with each of its neighbors, I suggest!{unless playing on a computer, which should be checking forcoprimality for you}) I had intended long ago to post this game idea, which is based uponthe puzzle/math problem at:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off &threadm=b4be2fdf.0211011659.79913415%40posting.google.com& rnum=22&prev=(I may have actually already mentioned this exact idea, but morelikely a 2-player variation. And, anyway, I have a question regardingit, so as to justify a new thread.)Question:Let, for an n-by-n game, the average score, taken over ALL possiblegames, be:a(n).(By possible games, I mean all games following the rules and beingplayed to the highest possible score, without any regards tostrategy.)What is a(n) asymptotical towards??Leroy Quet=== === Subject: : derivatives being in L^1Fix y and defineg_y (x) = [f(x + y) - f(y)] / [e^(ix) - 1].How do you show thatINT |g'_y (x)| < infty iff INT |f ''(x)| < infty ?I see how to do this if you can let x --> 0 (and think of g_y as a sort ofNewton quotient for f at y, but, I don't see how to show this w/o that. Cansomeone lead me down the right path?thanx=== === Subject: : Re: derivatives being in L^1> Fix y and define> g_y (x) = [f(x + y) - f(y)] / [e^(ix) - 1].> How do you show that> INT |g'_y (x)| < infty iff INT |f ''(x)| < infty ?> I see how to do this if you can let x --> 0 (and think of g_y as a sort of> Newton quotient for f at y, but, I don't see how to show this w/o that. Can> someone lead me down the right path?Come on. You're asking for free advice, how about some hypotheses? What kind of function is f? Where is it defined. Is it infinitely differentiable, two derivatives, what? What are the limits of integration?=== === Subject: : Re: derivatives being in L^1I think f is in L^2(0,2 pi), and the integration is performed over (0,2 pi).And yes, f' and f'' (at least) are defined...> Fix y and defineg_y (x) = [f(x + y) - f(y)] / [e^(ix) - 1].How do you show thatINT |g'_y (x)| < infty iff INT |f ''(x)| < infty ?I see how to do this if you can let x --> 0 (and think of g_y as a sortof> Newton quotient for f at y, but, I don't see how to show this w/o that.Can> someone lead me down the right path?> Come on. You're asking for free advice, how about some hypotheses? What> kind of function is f? Where is it defined. Is it infinitely> differentiable, two derivatives, what? What are the limits of integration?=== === Subject: : Re: A Certain Random Maze-Generating AlgorithmBy the way, it occured to me that you can have two or more people,or two or more computer algorithms/stategies, creating these mazes.Just take turns drawing the walls by the rules.(As a game:Perhaps 2+ human players can take turns drawing walls so as to forma maze, then {after duplicating the finished maze} race each otherin trying to solving it.)Leroy Quet(PS: this is *not* a duplicate of the other reply I pos today.)> maze-generating program (in BASIC).> Even though the resulting mazes LOOKED interesting,> the mazes' paths would practically always just pass through> the centers of the mazes...> Very simple to solve, each of these mazes were...> The basic (basic...snicker) idea behind the algorithm is:> start with a rectangular lattice of vertexes, and have the> vertexes along the outside wall of the maze each be occupied.> Then the program, on each loop, simply randomly picks an occupied> vertex, and then draws a line segment (equal to distance between> adjacent {above/below/right/left} vertexes in the maze) to a random> adjacent unoccupied vertex, if such a vertex exists, recording> the unoccupied vertex as now occupied.> (Again, line segments are all vertical and horizontal and of> constant-length.)> (And, I must stress, the program never connec two unoccupied> vertexes, nor did it connect two occupied vertexes.)> The program is complete when all the vertexes are occupied,> ie have at least one line-segment connec to each of them.> (And, of course, there are 2 doors in the maze's outer-wall,> diametrically opposed, for entrance-into and exit-out-of the maze.)> But I believe that we can improve the mazes' difficulty by> using a variable v, where 0 <= v <= 1.> v is the probability that the program, instead of simply> picking an occupied vertex to connect to at-random,> will connect to the most recently conver-to-occupied vertex> in the maze, if this vertex has at least one currently> unoccupied neighbor.> And (1-v) is the probability that the program will just> choose from the list of already-occupied vertexes at-random,> without regards to when these vertexes became occupied, of course.> So, in the original program I had essentially v = 0, which leads> to through-the-center paths (using a predictable> uniformly-random-number generator).> On the other extreme, v=1 would make certain the path passes> along the maze's outer wall, at least for some part of the path.> So, somewhere between v=0 and v=1 is a value for v which leads> to the most interesting mazes.> By interesting, I mean: the mazes are at greatest difficulty> and unpredictability in regards to their solution-paths.> Also, interesting refers to the aesthetic appeal of the mazes.> But in either case, the definition of interesting is subjective.> I am unable currently to do any actual programming on my computer.> If anyone has a desire to make such a random maze-generator> program, which v(or v's) would you suggest??> Perhaps v can even vary as the maze is drawn....> Any comments?> Leroy > Quet=== === Subject: : Re: A Certain Random Maze-Generating AlgorithmI was impressed with the maze-description site you gave a link to.Interesting.I had some other items to add to this discussion.Regarding my wall-building algorithm specifically, but notnecessarily:First, perhaps we could simply get rid of the outer walls at thestart.We would have a theoretically infinite grid.But the maze would be genera from 2 physically-close occupiedseed vertexes.So, after the algorithm has run a certain amount of time,perhaps being shut off by a human operator,then we should have a blob-shaped maze, where we know not for surewhere the entrance and exit are.The goal, of course, would be to find a way to pass through the blob,which is topologically 2 intertwined trees.For extra weirdness, perhaps we could have 3 or more seeds,giving us a maze with 3 or more entrances/exits.Or we could have a walled maze, without an exit or entrance door,plus an internal seed.Then the goal of this maze would be to find the closed loop within themaze.(Hmmm...this sounds fun, perhaps! Did I steal this idea??) Another possiblity, with any of the algorithms, is to involve mathsomehow,instead of using a pseudo-random number generator.Using my original wall-building algorithm:So, we could allow, after labelling the vertexes from 1 to...n^2(or as vectors {1,1},{1,2},...{n,n}), say,have each wall drawn such that the number of thedrawn-to unoccupied vertex is some function of,or meets some condition based upon, the number of the drawn-fromoccupied vertex.(Such as the drawn-to integer should have a value which is the closestapproximation to a logarithm of an integer than any of the other0,1,or 2 unoccupied neighboring vertexes.)SOME mathematics SHOULD lead to some interesting-looking mazes,theoretically.Leroy Quet>Even though the resulting mazes LOOKED interesting,>the mazes' paths would practically always just pass through>the centers of the mazes...>Very simple to solve, each of these mazes were...> In a wall adding algorithm where you pick a random cell each time, the> Maze will tend to grow inward from the four boundary walls at equal rates,> always finishing up in the center. The passage layout will roughly be a> bunch of passages radiating from the center to the edges, meaning the> solution indeed always goes through the center. Passage carved versions of> this algorithm have the same issue, where all passages will radiate from> the first cell visi.> This is very similar to Prim's Algorithm. The one subtle difference is> Prim's keeps track of frontier cells, or unoccupied cells adjacent to> an occupied cell, and chooses randomly among the frontier cells. Prim's> Algorithm results in Mazes that are similar, but look more random. Your> algorithm tends to smooth away differences in the wave of occupied cells> that move toward the center, like throwing sand on a rough surface. For> example, have a smooth wall with a one unoccupied cell indentation inward,> and there's triple the chance that unoccupied cell will get visi, since> it's adjacent to three occupied cells. Prim's Algorithm however tends to> accentuate and grow any differences, like a crystal in a solution. For> example, have a smooth wall with a one occupied cell indentation outward,> and there's triple the chance that occupied cell will get extended, since> it's adjacent to three frontier cells.>(And, I must stress, the program never connec two unoccupied>vertexes, nor did it connect two occupied vertexes.)> Adding a wall segment between two unoccupied vertices would create a loop.> Adding a wall segment between two occupied vertices would create an> inaccessible section or make the Maze unsolvable (unless there are already> loops, in which case it may reconnect a detached wall). Hence your Mazes> are perfect, or form a minimal spanning tree over the cells, with exactly> one solution and one path between any pair of points.>But I believe that we can improve the mazes' difficulty by>using a variable v, where 0 <= v <= 1.>v is the probability that the program, instead of simply>picking an occupied vertex to connect to at-random,>will connect to the most recently conver-to-occupied vertex>in the maze, if this vertex has at least one currently>unoccupied neighbor.> What you describe here is the Growing Tree Algorithm. Growing Tree is a> generalized algorithm that grows the passages or walls in the Maze like> other algorithms, where all occupied cells are appended to a list. You have> total freedom how to select the next cell from the list to try to grow> from. Always select the most recent cell (i.e. v=1) and this turns into the> Recursive Backtracking Algorithm. Always select a random cell (i.e. v=0)> and this turns into your algorithm. You can do things like select among the> n most recent cells, or always select the most recent cell but a certain> percentage of the time select a random cell, or have a variable v like you> have, for different textures.>On the other extreme, v=1 would make certain the path passes>along the maze's outer wall, at least for some part of the path.> When v=1, your algorithm becomes very similar to a wall added version of> the Recursive Backtracking Algorithm. The only difference is that when you> can no longer grow a wall in any direction, you go to a random vertex,> instead of following the stack back to the first available vertex. This> indeed has a potential problem in that the solution path may follow the> boundary wall. In general wall added Maze algorithms are harder to get> right than passage carved ones. If v is too low, all paths go through the> center. If v is too high, the solution goes around the outside edge.> The Mazes are indeed simple, but this is actually one of the faster> algorithms if implemen right. Have a list of all occupied vertices,> and a mapping from each cell coordinate pair to its index in the list. When> a new cell is visi, add it to the end of the list. This allows picking> a random occupied vertex in constant time. If an occupied vertex being> considered has no unoccupied neighbors, remove it from the list (by swapping> it with the end). This works regardless of the value of v and is just as> fast at the start and at the end of the Maze creation process.>So, somewhere between v=0 and v=1 is a value for v which leads>to the most interesting mazes.>Also, interesting refers to the aesthetic appeal of the mazes.>But in either case, the definition of interesting is subjective.> Having v near 1, or the Recursive Backtracking Algorithm, works well for> passage carved Mazes, and will result in long convolu solutions.> Interesting is indeed subjective, however I can say for the best Mazes,> create it by carving passages, and have a v at least 0.5> All the algorithms mentioned: Recursive Backtracking, Prim's Algorithm, and> Growing Tree, and many other types (including perfect Mazes that don't> grow the Maze like a tree, and methods of non-perfect Mazes with loops) are> described at http://www.astrolog.org/labyrnth/algrithm.htm. That page also> links to a free program called Daedalus that implements each algorithm,> in both passage carved and wall added versions, where you can watch the> Maze get drawn if you like. :)> O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O* O*O*O*O*O*O*O> * Walter D. Cruiser1 Pullen :) ! Astara@msn.com *> O Get lost in my Labyrinth Web page: http://www.astrolog.org/labyrnth.htm O> *O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O*O *O*O*O*O*O*O*=== === Subject: : Re: A Certain Random Maze-Generating Algorithm> I had some other items to add to this discussion.> Regarding my wall-building algorithm specifically, but not> necessarily:> First, perhaps we could simply get rid of the outer walls at the> start.> We would have a theoretically infinite grid.> But the maze would be genera from 2 physically-close occupied> seed vertexes.> So, after the algorithm has run a certain amount of time,> perhaps being shut off by a human operator,> then we should have a blob-shaped maze, where we know not for sure> where the entrance and exit are.> The goal, of course, would be to find a way to pass through the blob,> which is topologically 2 intertwined trees.After I contempla this a little, I realized there is a big problem.The blob-maze is not guaranteed to have a solution!This is because the walls being drawn from one of the two seeds maycompletely surround those walls being drawn from the other seed.So we would have then, in that case (which could actually be verycommon), one path leading into the maze, connecting to a loop, andhave no exit.I could not come up with a simple algorithm-fix for this problem.As an interesting (perhaps) question itself, what wouldsci.math/rec.puzzle-repliers suggest for the *simplest* fix whichGUARANTEES a path *through* the blob?? A simple solution: Just say that the maze is solved, if not onlysomeone finds a path through the maze, but if the solver's pathreconnects with itself somewhere within the maze (without just doing aU-turn, obviously).;)Also, a problem (which is a result of the maze's imperfect boundry):it would be more difficult to occupy ALL of the internal vertexes witha SIMPLE algorithm.I suggest that after some point, the algorithm only work in occupyingthe unoccupied INTERNAL vertexes.> For extra weirdness, perhaps we could have 3 or more seeds,> giving us a maze with 3 or more entrances/exits.> Or we could have a walled maze, without an exit or entrance door,> plus an internal seed.> Then the goal of this maze would be to find the closed loop within the> maze.> (Hmmm...this sounds fun, perhaps! Did I steal this idea??)> . .WHERE did I see the closed-loop maze??And regarding the v-variable (which is referred to in my originalpost), perhaps it would be interesting to build mazes where valternates {depending on if the order of the move is odd or even}between very low (possibly 0) and very close to 1 (possibly 1). (Inthe v = ~1 case, the wall would perhaps be joined to the the vertexoccupied *2* moves back, instead of 1 move back.)Leroy Quet=== === Subject: : Re: Recommend a Book?> Hi all,> I am a graduate - well 4 years ago anyway. I havnt been using maths toany> real standard and I don't want to forget the knowledge.> I wonder if anyone could recommend a good reference book - I would likethe> book to contain details/tutorials etc on topics such as calculus, Fourier,> ODE/PDE - matrix ops... I know these topics are wide ranging - so I wonder> does a good book covering these topics decently exist?> Thanks..> CLAn Advanced Calculus book would cover quite a bit of material, i.e. Singleand Multi-variable Calculus, Fourier series, Linear Algebra, and a littleReal analysis.Lurch=== === Subject: : Re: Recommend a Book?> I wonder if anyone could recommend a good reference book - I would like the> book to contain details/tutorials etc on topics such as calculus, Fourier,> ODE/PDE - matrix ops... I know these topics are wide ranging - so I wonder> does a good book covering these topics decently exist?The mix of topics you are asking about (with the addition of basic useof complex functions and tensors) seem to be standard in the booksused in undergraduate 'mathematical methods for physics' courses. Ifyou browse through a few of these, you should be able to findsomething in a style that suits your needs.