mm-361 
==
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||<epsilon}and
let V_epsilon = exp(U_epsilon)|| . || is the hilbert-schmidt
normi.e. ||A|| = [ sum (from k,l = 1 to n) of |A_(kl)|^2 ]
^(1/2)Consider A_0 an element of G, a matrix group. Then for
any A_0 in G,consider the neighborhood (A_0)V_epsilon of A_0
in M_n(C).Why is it true that A is in (A_0)V_epsilon if and
only if ((A_0)^{-1})(A) isin V_epsilon?I guess I don't
understand how to consider neighborhoods of a matrix. Iassume
that we think of them by identifying an n x n matrix by itsn^2
-tuple in C(the complex numbers). Then, if we think in this way
ofcourse U_epsilon is open, and V_epsilon is open (right? exp
is an openmapping?). Therefore, (A_0)V_epsilon is a
neighborhood of A_0 becauseV_epsilon is a neighborhood about
the identity matrix, and the identitymatrix is mapped to A_0
when you multiply V_epsilon by A_0.But, now when we pick a
matrix A in the neighborhood (A_0)V_epsilon, this isequivalent
to ((A_0)^{-1})(A) is in V_epsilon? The question is, how can
Ithink of this now in terms of n^2-tuples? i.e., if you
identify A_0 as ann^2-tuple, then (A_0)^{-1} is what? You take
the inverse of the matrix,then write the inverse of the matrix
as an n^2-tuple. But now it is notclear to me that
((A_0)^{-1})(A) is in V_epsilon. I am sure I am thinkingabout
this in the wrong way. I would ASSUME that what you really
want to dois, if you identify A as an n^2-tuple, then you
would want to multiply eachentry of A by the inverse of each
entry of A_0.(i.e., if A_0 = (1,2,3,4) and A = (3,4,5,6), then
(A_0)^{-1} =(1,1/2,1/3,1/4) and multiply accordingly)Could
someone clarify for me what is going on?Thanks a
bundle,Moshe
===
Subject: An Unusual Generating FunctionLet 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, leta(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 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.>><Nitpicking To get any 3n+2 you actually _need_ one 20
(or 4, 7, ... 20s).> 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,
96779, 96905, 97011, 97105, 97156, 97206,97370, 97488, 97540,
97545, 97616, 97687, 97693, 97755, 97809, 97911,98010, 98041,
98255, 98304, 98339, 98400, 98591, 98811, 98857, 99055,99092,
99174, 99290, 99436, 99795, 99886, 99895, 100067,
100218,100240, 100418, 100513, 100532, 100617, 100842, 100848,
100907,100930, 100931, 100934, 100935, 101004, 101126, 101455,
101492,101559, 101713, 101745, 101783, 101788, 102001, 102570,
102572,102870, 102914, 103068, 103245, 103517, 103676, 103761,
103794,103836, 103869, 103908, 104004, 104245, 104286, 104455,
104595,104662, 104941, 104959, 104991, 105065, 105085, 105166,
105217,105259, 105332, 105553, 105861, 105977, 106168, 106262,
106337,106356, 106426, 106543, 106544, 106550, 106686, 106796,
107188,107230, 107264, 107655, 107779, 107799, 107911, 107941,
108131,108170, 108181, 108202, 108268, 108348, 108882, 108940,
109101,109168, 109295, 109480, 109524, 109870, 110177, 110188,
110309,110416, 110531, 110538, 110906, 110938, 111051, 111085,
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.<tongue in cheek>Aha! But that depends on the
convention that 0/0 = 0. What ifsomeone adopts the convention
that 0/0 = 1?</tongue in cheek>> So are you using some
different definition of real numbeone 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: <bururj$l9l$1@lust.ihug.co.nz>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 otheor 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 integefor 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 predecessothat 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 predecessothat 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> <HrvG26.Ms4@cwi.nl>
<O7OdnYSerPBP1ZLdRVn-hQ@comcast.com>
<877jzkxzdf.fsf@phiwumbda.org>
<jaqdnfu-6OAOAJLdRVn-jg@comcast.com>
<OuSdnWCa2ai95I3dRVn-ug@comcast.com>
<z8WdnQXNSaT4SI3dRVn-tA@comcast.com>
<tOudnUHvipzvWYzdRVn-jw@comcast.com>
<HLmdnX3URMpEgY_dRVn-sw@comcast.com> Discussion, linux)For
intersections of any finite family of nes sets, meaning each>
is>> a proper subset of all its predecessothat 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 predecessothat 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>
<HrvG26.Ms4@cwi.nl> <O7OdnYSerPBP1ZLdRVn-hQ@comcast.com>
<877jzkxzdf.fsf@phiwumbda.org>
<jaqdnfu-6OAOAJLdRVn-jg@comcast.com>
<OuSdnWCa2ai95I3dRVn-ug@comcast.com>
<z8WdnQXNSaT4SI3dRVn-tA@comcast.com>
<tOudnUHvipzvWYzdRVn-jw@comcast.com>
<HLmdnX3URMpEgY_dRVn-sw@comcast.com>
<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> <HrvG26.Ms4@cwi.nl>
<O7OdnYSerPBP1ZLdRVn-hQ@comcast.com>
<877jzkxzdf.fsf@phiwumbda.org>
<jaqdnfu-6OAOAJLdRVn-jg@comcast.com>
<OuSdnWCa2ai95I3dRVn-ug@comcast.com>
<z8WdnQXNSaT4SI3dRVn-tA@comcast.com>
<tOudnUHvipzvWYzdRVn-jw@comcast.com>
<HLmdnX3URMpEgY_dRVn-sw@comcast.com>
<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 integeand (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 integeand 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
integeand 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,employeFBI, 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
integethesecond 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 integeexcept that this definitionis
not concerned with the sign, so in the integethe 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 <waderameyxiii@comcast.remove13.net>  [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
answere: 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
<waderameyxiii@comcast.remove13.net>
[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 answere: 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> numbeand 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 numbe2 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 yeaand 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 logicSubject: 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:
<GspQb.254424$ts4.213189@pd7tw3no>>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,
countememory 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 powewhich
termdominates the sum? What happens after you take the n:th
root of thesum of powers of n?
===
Subject: Re: Isoperimetric
ZeppJim Ferry <corklebath@hotmail.com> a .8ecrit dans le
message deJim Ferry <corklebath@hotmail.com> 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 <paris63@pacbel.net>>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 numbewhich are irrational simply because
they haveinfinite continued fractions, inf { n * (nx -
INT(nx)) } = 0.Rob Johnson <rob@trash.whim.org>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* integeyes. Allow zero in the set, no. Allow
negative integeit 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<pawlosko5@aol.com><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<rdecker@hamilton.edu><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.> <OT> 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
>> </OT>> 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<willem@stack.nl><slrnc14oth.24tb.willem@toad.stack.nl>:
> ) 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 <roottop@hotmail.com> 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]<http://www.cartoonbank.com/assets/1/40967_m.gif>
===

Subject: Re: The Lost Proof of Fermat<another proof dele>Keep
working at it, Slappy. You'll prove it someday.
===
Subject: Re:
The Lost Proof of Fermat<snip nonsenseAha! so we can't have>
(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 Fermat<proof dele>Now
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
<n.karjanto@math.utwente.nl>>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 <rob@trash.whim.org>take out the
trash before replying
===
Subject: Re: Rationals are Uncountable
<yKGdnTJavO2eS5Dd4p2dnA@comcast.com>
<bukt3q$l4n$1@mozo.cc.purdue.edu>
<nKydnYyvmbotY5DdRVn-jw@comcast.com>
<bul34g$mf5$2@mozo.cc.purdue.edu>
<tPudnYKsV6t4gpPd4p2dnA@comcast.com>
<bum5rh0pp8@drn.newsguy.com>
<YcqdnUOsPqPpZJPdRVn-sQ@comcast.com>
<bunj5b01dor@drn.newsguy.com>
<rqudnWYwFKq3BpLd4p2dnA@comcast.com>
<fab2a28f.0401220738.409c7044@posting.google.com>
<99a0b764.0401221214.6a666a05@posting.google.com>
<99a0b764.0401231312.226ddd00@posting.google.com>
<npOdnfWSt_qMJozdRVn-sA@comcast.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>
<butbju$iqh$1@bunyip.cc.uq.edu.au>
<O-2dndZPoM2D3Y_d4p2dnA@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>
<garrydenke@catholic.org<4e63857.0401230604.60fe67d9@
posting.google.com>:>> In sci.logic, Garry Denke>>
<garrydenke@catholic.org>
<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<garrydenke@catholic.org><4e63857.0401240538.3eba3f69@
posting.google.com>:>> In sci.logic, Garry Denke>>
<garrydenke@catholic.org>
<4e63857.0401230604.60fe67d9@posting.google.com>:> In
sci.logic, Garry Denke>
<garrydenke@catholic.org<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>> 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 dollathe 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 thesevectothere 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
<xan2@ono.com> 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 <rob@trash.whim.org>take out the trash
before replying
===
Subject: Re: Generating function of
Repunits?Rob Johnson <rob@trash.whim.org> 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 <ilarrosaQUITARMAYUSCULAS@mundo-r.com>,
xan2@ono.com (Xan)>Rob Johnson <rob@trash.whim.org>
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
<rob@trash.whim.org>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 numbebut 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 integeas 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.