mm-108
Please explain what you meant by Even Homer nods.
Is it a nod > of assent?You betray your lack of a classical
education by that question.Homer, of Iliad and Odyssey fame,
was considered the ultimate poet and storyteller, and the
ideal model for all those who follow him, but he did make
some minor errors in his two great works. Those noting the
errors usually also note that, as great as he was, even Homer
nods (Homer dormitat).
Im not versed in classics. Classic
rock, sure, Ive probably heardit and may know the band,
classic tragedies, comedies, and epics, lessso. La commedia
es la tragedia. I think the rst book with morethan a hundred
pages I read was a childrens Popeye novel that hadscenes that
reminded me of Scylla and Charybdis. The Dungeons andDragons
Monster Manuals and Deities and Demigods summarized
someaspects of the pantheons members, mythical beasts, and
otherfantastic creatures, such as their hit points,
immunities, etcetera. I centaur has 4d8 hit points, or four
hit dice, 4d-8d. Heh,Sahaugin (sa-HOO-jen). I have 74 hit
points, immunity to drizzle, andin some venues, a power st.
Also, I y, and lasers shoot out of myeyes.Homer, take a bow.
I cant string my own damn bow, it has to go in apress, it has
more than a 70 pound draw weight, 35% let-off, speedcams.
Twang, it pings. It puts some arrows _through_ the
eternaltarget. Im proud of
it.http://www.pagebypagebooks.com/Homer_Butler_Tr/The_Odyssey
Studies/Greek/Homer/http://www.perseus.tufts.edu/cache/
perscoll_Greco-Roman.htmlDid you ever read Philip Jose
Farmers Riverworld Series? It hadUlysses, Odysseus, but it
wasnt actually Ulysses, it was Loga. Mythof Sisyphus,
indeed.Even Homer plauds. (...)So anyways were talking about
normal numbers. The number normal tobase two appears to have a
binary zero-density of one-half. Theexpression n!/ (n/2)!^2
2^n evaluates asymptotically to, what was thatagain, sqrt (
pi/ 2 n)? I thought it would be 1/2. Now Im hearingthat
almost all irrationals are absolutely normal. Splitting
thedifference between almost all and almost none is around
one half.Ross
Splitting the> difference between almost
all and almost none is around one half.RossSplitting the
difference between almost all and almost none in a real
interval is almost anything in between.Unless some highly
specic splitting mechanism is guaranteed.
Each player
takes turns coming up with algorithmic steps, one step at> a
time in sequence, each step giving a rule to transform an
integer> into another.(such as: multiply m by greatest prime
p, where p divides m and is < sqrt(|m|). If no such prime
exists, leave m unchanged.)> (or such as: m =
oor(|m|/d(|m|)), where d(m) is number of positive> divisors
of m.)> (or rules may involve meta-transformations, such as
sending players to> previous rules, depending somehow upon
the value of the integer when> reaching the step.)Anyway,
players are encouraged to be creative when inventing rules!>
Each step is capable of transforming any integer, always
giving an> integer as output.After a predetermined number of
steps have been created (the same> number for each player), a
random integer is generated somehow.Players then try to guess
what the output intger will be.The algorithm is run using the
random start-integer. The winner of the game is the player who
comes closest to guessing the> (needs work....)Of course,
there should be a time limit between when the randominteger
is picked (after the list of rules is completed) and when
theplayers must submit their predictions as to what will be
the output ofthe algorithm.(Otherwise, each player could just
go through the algorithm manually,each player perhaps
predicting the output exactly...)Leroy Quet
Let m be a
xed positive integer.Let x be a xed positive real.Let
n(m,x) be the highest positive integer such that1/m + 1/(m+1)
+ 1/(m+2) + ...+ 1/n(m,x) <= x.> A lot of questions can be
asked about this.But what I am wondering now is, what
is:E(m,x) = x - (1/m + 1/(m+1) + 1/(m+2) + ...+
1/n(m,x))asymptotical to?All this is highly related to the
topic
off&threadm=agt133%24ek%241%40nntp.itservices.ubc.ca&rnum=1&
prev Now, m does not have to be an integer.(But the number
of terms in the sum is an integer. n(y,x) = y + integer.)So,
if we plot w = E(y,x), for xed x, what does the graph look
like?(My copy of Mathematica got fried.)Is it continuous, for
By the way, I might as well add
this:>exp(sum{p=primes} 1/p^x) = >>sum{k=1 to oo}
A(k)/k^x,>where A(k) = product{p=primes} 1/(a(p,k))! ,>>and
where each a(p,k) is a nonnegative integer such that>>p^a(p,k)
is the highest power of the prime p which divides k.>>(I
think...)I think so, too. In other words the coefcient of
n^{-x} in > exp(sum_p p^{-x}) is 1/prod_i k_i! if n=prod_i
p_i^{k_i}.However, returning to the OP, today I made a few
other calculations> and found out that *formally*sum_p p^{-x}
= sum_{k=1}^{infty} mu(n)/n log(zeta(nx)).There do not *seem*
to be any issues suggesting that this equality> should not
hold also in the analytical sense, but what interests me>
most is that as I had guessed initially this function once
again seems> to be independent of zeta in the sense that it
cannot be expressed> in terms of zeta by means of (a nite
number of) algebraic> operations, composition with elementary
functions and/or derivation.> Am I right?> MicheleI forgot to
mention this
webpage:http://mathworld.wolfram.com/PrimeZetaFunction.htmlIt
seems to have a lot of info on what is known as the prime zeta
function.Yes, this sum is well-known, but still
=Ive seen one cataloged, as
it were; ifIfind the book, Ill get the authors. > Has a
single polygon that tiles the plane only aperiodically been
found,> or proved not to exist, or is this still an open
question?--Dec.2000 WAND Chairman Paul ONeill, reelected
to Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=Suppose
H is a hash function with N-bit output. Dene h_n to be H
restricted to n-bit input (1 <= n <= N), and take h_n(x) to
be the rst n bits of H(x). (1) Referring to n-bit strings by
the integers they represent in binary, let p(n) be the
proportion of values 0, ..., 2^n-1*missing* from the range of
h_n, i.e.,p(n) = ( #duplicates among h_n(0), ..., h_n(2^n-1) )
/ 2^n.Are there theoretical values to be expected for the
p(n)?(2) When H is SHA-1 (N = 160), Ifind the valuesp(2^8) =
90/256 ~ 35%p(2^16) = 24100/65536 ~ 38%.What is a reasonable
value to expect for
p(2^160)?BCgNrSjVlRhxQXcdZzWpGFLtDswMkHvqPnTKJmbf
(1)
Referring to n-bit strings by the integers they represent >in
binary, let p(n) be the proportion of values 0, ...,
2^n-1>*missing* from the range of h_n, >Are there theoretical
values to be expected for the p(n)?1/e ~=
0.36787944117144232159http://www.cs.berkeley.edu/~daw/
my-posts/hash-iteration
=| >(1) Referring to n-bit strings
by the integers they represent | >in binary, let p(n) be the
proportion of values 0, ..., 2^n-1| >*missing* from the range
of h_n, | >Are there theoretical values to be expected for the
p(n)?| | 1/e ~= 0.36787944117144232159| |
http://www.cs.berkeley.edu/~daw/my-posts/hash-iterationhttp:/
/www.cs.berkeley.edu/~daw/my-posts/random-mapsfrom which I
think the following is clear:If |X| < oo, and f: X -> X is a
random function such that for all x,y in X, P( f(x) = y ) =
1/|X|, then E |X - f(X)| / |X| = (1 - 1/|X|)^|X|.In the
present context, X is the set of n-bit stringsand h_n is
apparently to be treated like h, giving us E p(n) = E |X -
h_n(X)| / |X| ~ 1/e.But why can we treat h_n (a given,
non-random function-- say the restriction of SHA-1) as though
its random? E.g., why and in what sense do we have P( h_n(x)
= y ) = 1/2^n?
why can we treat h_n (a given,
non-random function> -- say the restriction of SHA-1) as
though its random? If we take SHA1, by internal contruction
for small number of bits,the output is S1 = S0 ^
ENC(paddedMessage, S0)where S0 is a xed constant (160 bits),
andC = ENC(K,P) is a cipher with K 512 bits, P and C 160
bits.On the assumption that the cipher is perfect [any
distinct Krandomly selects a mapping from P to C among
(2^160)!], you candemonstrate properties on duplicates in
h_n.Without more precision than H is a hash function with
N-bit outputyou cant. For example, consider the following
hash function,dened on same input and output domain as SHA1-
for input up to 158 bits, output is the input, followed by
just enough bits at 0 to reach 158 bits, followed by two bits
at 1- for input of 159 bits or more, the output is that of
SHA1, with the last bit set to 0.This pathological hash
function enjoys excellent second premimageresistance (and
thus is perfectly suitable to protect the integrityof a le).
But p(n)=0 for all n<=158, and p(160) >> 1/e.A less
pathological example is H = SHA(SHA1(M)); it experimentalyhas
properties similar to SHA1 (including
experimentalindistinguishibility from random, and preimage
resistance),but p(160) >> 1/e. Fran.8dois GrieuOriginal
problem quated below> Suppose H is a hash function with N-bit
output. Dene h_n > to be H restricted to n-bit input (1 <= n
<= N), and take > h_n(x) to be the rst n bits of H(x). >
(1) Referring to n-bit strings by the integers they represent
> in binary, let p(n) be the proportion of values 0, ...,
2^n-1> *missing* from the range of h_n, i.e.,> p(n) = (
#duplicates among h_n(0), ..., h_n(2^n-1) ) / 2^n.> Are there
theoretical values to be expected for the p(n)?(2) When H is
SHA-1 (N = 160), Ifind the values> p(8) = 90/256 ~ 35%> p(16)
= 24100/65536 ~ 38%.> What is a reasonable value to expect for
p(160)?
>why can we treat h_n (a given, non-random
function>>-- say the restriction of SHA-1) as though its
random? > If we take SHA1, by internal contruction for small
number of bits,> the output is S1 = S0 ^ ENC(paddedMessage,
S0) The ^ is more like a +, although, since its 5 separate
32-bit additions, its not really + either.> where S0 is a
xed constant (160 bits), and> C = ENC(K,P) is a cipher with
K 512 bits, P and C 160 bits.> On the assumption that the
cipher is perfect [any distinct K> randomly selects a mapping
from P to C among (2^160)!], you can> demonstrate properties
on duplicates in h_n.Without more precision than H is a hash
function with N-bit output> you cant. For example, consider
the following hash function,> dened on same input and output
domain as SHA1> - for input up to 158 bits, output is> the
input, > followed by just enough bits at 0 to reach 158
bits,> followed by two bits at 1> - for input of 159 bits or
more, the output is that of SHA1,> with the last bit set to
0.This pathological hash function enjoys excellent second
premimage> resistance (and thus is perfectly suitable to
protect the integrity> of a le). But p(n)=0 for all n<=158,
and p(160) >> 1/e.A less pathological example is H =
SHA(SHA1(M)); it experimentaly> has properties similar to
SHA1 (including experimental> indistinguishibility from
random, and preimage resistance),> but p(160) >> 1/e. This
also applies to H=SHA1(SHA1(M)).--Mike Amling
| >(1)
Referring to n-bit strings by the integers they represent > |
>in binary, let p(n) be the proportion of values 0, ...,
2^n-1> | >*missing* from the range of h_n, > | >Are there
theoretical values to be expected for the p(n)?> | > | 1/e ~=
0.36787944117144232159> | > |
http://www.cs.berkeley.edu/~daw/my-posts/hash-iterationhttp:/
/www.cs.berkeley.edu/~daw/my-posts/random-maps> from which I
think the following is clear:If |X| < oo, and f: X -> X is a
random function > such that for all x,y in X, P( f(x) = y ) =
1/|X|, > then E |X - f(X)| / |X| = (1 - 1/|X|)^|X|.In the
present context, X is the set of n-bit strings> and h_n is
apparently to be treated like h, giving us > E p(n) = E |X -
h_n(X)| / |X| ~ 1/e.But why can we treat h_n (a given,
non-random function> -- say the restriction of SHA-1) as
though its random? Because no distinguisher has been found
that can differentiate between SHA-1 output (from unknown
input) and random output?> E.g., why and in what sense do we
have > P( h_n(x) = y ) = 1/2^n? The measured probabilities of
SHA-1 outputs (truncated to n bits) are statistically
indistinguishable from 2**-n.--Mike Amling
= [referring to
David Wagners posting]| > I think the following is clear:In
this little theorem, X is supposed to be non-empty:| > If [0
<] |X| < oo, and f: X -> X is a random function | > such that
for all x,y in X, P( f(x) = y ) = 1/|X|, | > then E |X - f(X)|
/ |X| = (1 - 1/|X|)^|X|.| > In the present context, X is the
set of n-bit strings| > and h_n is apparently to be treated
like h, giving us | > E p(n) = E |X - h_n(X)| / |X| ~ 1/e.| >
| > But why can we treat h_n (a given, non-random function| >
-- say the restriction of SHA-1) as though its random? | |
Because no distinguisher has been found that can
differentiate | between SHA-1 output (from unknown input) and
random output?That has to be relevant, but its just not clear
how. The trouble is that the theorem does not refer to
randomness of inputs or outputs -- its the *mapping* thats
random.E.g., in the 1-bit case with h_1 = SHA1 =
(restricted)SHA-1,the expression P( SHA1(0) = 0 ) = 1/2 makes
no sense, because SHA1 is not a random mapping.| > E.g., why
and in what sense do we have | > P( h_n(x) = y ) = 1/2^n?| |
The measured probabilities of SHA-1 outputs (truncated to n
bits) | are statistically indistinguishable from 2**-n.Again,
this 1/2^n result does not refer to randomness of the inputs
or outputs, but to randomness of the mapping (and SHA1is a
known xed mapping). To make the issue clearer, consider the
simple 1-bit case ...There are 4 possible mappings {0,1} ->
{0,1}, and the above theorem applies when these mappings are
equally likely: mapping probability-------------------
-----------f1: 0 -> 0, 1 -> 0 1/4f2: 0 -> 0, 1 -> 1 1/4f3: 0
-> 1, 1 -> 0 1/4f4: 0 -> 1, 1 -> 1 1/4For the random function
f dened by this distribution,P( f(0) = 0 ) = P( {f1, f2} ) =
1/2, andE p(1) = E |{0,1} - f({0,1})| / 2 = (1/2)/2 = 1/4.In
the n-bit case, there are 2^n equally-likely mappings,
exactly one of which corresponds to SHA-l. So something more
is needed to explain the behavior of a specic one of these
mappings.
=Yes. Im using the random oracle model; replace
the (xed)hash function by a random function, do your
computations assumingyoure dealing with a random function,
and then hope that yourcomputations carry over to the real,
xed hash function.This is only a heuristic -- but it seems
to be a darn effective one.
=[typo corrections]Suppose H is
a hash function with N-bit output. Dene h_n to be H
restricted to n-bit input (1 <= n <= N), and take h_n(x) to
be the rst n bits of H(x). (1) Referring to n-bit strings by
the integers they represent in binary, let p(n) be the
proportion of values 0, ..., 2^n-1*missing* from the range of
h_n, i.e.,p(n) = ( #duplicates among h_n(0), ..., h_n(2^n-1) )
/ 2^n.Are there theoretical values to be expected for the
p(n)?(2) When H is SHA-1 (N = 160), Ifind the valuesp(8) =
90/256 ~ 35%p(16) = 24100/65536 ~ 38%.What is a reasonable
value to expect for
p(160)?BCgNrSjVlRhxQXcdZzWpGFLtDswMkHvqPnTKJmbf
What is
the family of real -> real analytic functions f(x), such
there are an innite number of fs,not necessarily
continuous.As to determine f(x) for all positive x, just
dene the interval from>0 to 1 arbitrarily (but I suggest all
of it be > 0 to avoid undenedf(x)s at any x), and then get
the rest of f(x), for positive x,recursively.(Duh, but must
What is the family of real
-> real analytic functions f(x), such that:> f(x+1) = f(x) +
1/f(x)> for all real x?>> Even if there is no analytic f,
there are an innite number of fs,> not necessarily
continuous.>> As to determine f(x) for all positive x, just
dene the interval from>0 to 1 arbitrarily (but I suggest all
of it be > 0 to avoid undened> f(x)s at any x), and then get
the rest of f(x), for positive x,> recursively.>Would you work
that out? Ive much doubt.See my reply to Herman Rubin where I
extend his method.There I show f(x) = +-oo for all x. Am I
correct?
What is the family of real -> real analytic
functions f(x), such that:> f(x+1) = f(x) + 1/f(x)> for
all real x?>> Even if there is no analytic f, there are an
innite number of fs,> not necessarily continuous.>> As to
determine f(x) for all positive x, just dene the interval
from>0 to 1 arbitrarily (but I suggest all of it be > 0 to
avoid undened> f(x)s at any x), and then get the rest of
f(x), for positive x,> recursively.>> Would you work that
out? Ive much doubt.> See my reply to Herman Rubin where I
extend his method.> There I show f(x) = +-oo for all x. Am I
correct?Ah, I was thinking only in terms of the function
being denedjust for *positive* real x, not for all real x.My
mistake.If the relation need only be true for all x > some
constant X, then weshould be able to get a dened (but not
analytic) function f, atleast when f is positive for all x >
What is the family of real -> real
analytic functions f(x), such that:>f(x+1) = f(x) +
1/f(x)>for all real x?>Ah, I was thinking only in terms of
the function being dened>just for *positive* real x, not for
all real x.>If the relation need only be true for all x > some
constant X, then we>should be able to get a dened (but not
analytic) function f, at>least when f is positive for all x >
X.It might be interesting to try for solutions that are
analytic in somedomain. f(x) = 0 produces a pole at x+1. On
the other hand, f(x) = 2is likely to produce a branch point
at x-1.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
Is a
limit point compact subspace of a Hausdorff space
necessarilyclosed?>> I suppose that a limit point compact
subspace is a subspace such that> each sequence has a
convergent subsequence. The usual expression for> that is
sequencially compact.>The denition of a limit point compact
space is such that every innitesubset of that space has a
limit point. Sequential compactness is only anequivalent
denition when the space is metrizable.> Well since I cant
seem to prove this and since if this is true, theproof> would
look nothing like the proof that a compact subspace of a
Hausdorff> space is closed, I am trying to come up with a
counterexample.>> Let S be the rst uncountable ordinal and
consider the set [0,S] of all> ordinals smaller or equal than
S, endowed with the topology generated by> all open intervals
]a,b[. Then [0,S[ is sequentially compact, but it is> not a
closed subset.>> However, if you add the hypothesis that your
space X is not only Hausdorff> but also rst countable, then
it is true that any sequentially compact> subspace Y is a
closed subset. If this was not the case, you could take a>
point p that belongs to the closure of Y but not to Y and a
sequence y_1,> y_2, ... of points of Y that converges to y.
Since Y is sequentially> compact, then some subsequence
converges to some y in Y. But then the> subsequence
converges to two distinct points, y and y, and this
isimpossible,> since X is Hausdorff.>> There is an example of
a limit point compact space that isnt compact inmy> Munkres
book which is :>> Let Y consist of two points and give Y the
topology consisting of Y andthe> empty set. Then X = Z_+ x Y
is limit point compact but not compact.(Z_+> is the positive
integers).>> Can I use this to show a counterexample?>> No.
Your space is obviously non-Hausforff.> Jose Carlos
Santos
Is a limit point compact subspace of a Hausdorff
space necessarilyclosed?>>I suppose that a limit point
compact subspace is a subspace such that>>each sequence has a
convergent subsequence. The usual expression for>>that is
sequencially compact.> The denition of a limit point
compact space is such that every innite> subset of that
space has a limit point. Sequential compactness is only an>
equivalent denition when the space is metrizable.OK. But
since, obviously sequentially compact countably compact
limit point compactmy previous counter-example is still a
counter-example.Jose Carlos Santos
> Is a limit point
compact subspace of a Hausdorff space necessarilyclosed?>>
Whats a limit point compact space?>A space is said to be
limit point compact if every innite subset of thespace has a
limit point.> Let Y consist of two points and give Y the
topology consisting of Y andthe> empty set. Then X = Z_+ x Y
is limit point compact but not compact.(Z_+> is the positive
integers).>> Can I use this to show a counterexample? I guess
the question is does Xsit> inside of a Hausdorff space making
X open? I dont think it doesbecause X> itself is not
Hausdorff since the open sets consist of unions of {n} x
Yand> thus if Y = {a,b}, then 1 x a and 1 x b do not have
disjointneighborhoods.>> No, subspaces of Hausdorff spaces
are Hausdorff.
Its not much easier to dene a method to
get a random (and in this> context implicitly of a uniform
distribution) real number from the> unit interval, but the
probability of it being less than one half is> one half.In
this case you appear to be dening the probability of
choosing a number in any given subinterval of the unit
interval as being the length of that subinterval. It can be
shown that such a measure satises the necessary properties
of probability functions, at least for intervals and nite
unions of them. For anything else, there are problems to deal
with before you can speak of probability as casually as you
do.
I dont have a method to select a random integer
uniformly, but for> any such method that there ever could be
the probability of its result> being an even integer is one
half.If you mean a method for choosing one object from a
countably innite set in such a way that every member of that
set has an equal chance of being the one chosen, it is quite
impossible, so your probability of its being an even integer
is mathematically undened.
=The ARC Centre of Excellence
for Mathematics and Statistics ofComplex Systems (MASCOS)
announces the availability of a ResearchFellowship at the
University of Queensland within the Disciplineof
Mathematics.The successful applicant will work closely with
Dr Phil Pollett,Director (Queensland) of MASCOS, and will be
associated with oneor more of the following projects: o The
Construction and Classication of Markov Chains o Stochastic
Models for Complex Biological Systems o Strong and Geometric
Ergodicity in Markov ChainsAn appointment will be made at the
level of Postdoctoral Fellow,Research Fellow or Senior
Research Fellow (depending on quali-cations and experience).
Applicants should hold a PhD in any areaof Stochastic
Processes and have a demonstrated record of re-search
achievement. An appointment at the level of Research Fel-low
and Senior Research Fellow will require a substantial body
ofpublished work, and a proven capacity for self-directed
research.Further details:
http://www.maths.uq.edu.au/~pkp/Fellowship.html______________
__________________________________________________Dr Philip K
Pollett, Director (Qld) ARC Centre of Excellence
forMathematics and Statistics of Complex Systems
www.complex.org.au
==
Let us say we have a loop of chain (or
we have a necklace) of m^2links/beads, each link/bead colored
with one of n colors.Now, we take a m-by-m grid, over which we
can place the chain so thateach square contains exactly one
link.(m is even.)How many ways can we expect to be able place
the chain so that nosimilar colors are in any immediately
adjacent (up/left/down/right)squares?(Of course, the exact
number varies according to how the links arecolored. I guess
I am just wishing for some analysis on this. Perhapsthe
number is exactly determined or approximated asymptotically
basedon some statistic of how the chain is colored.)And what
would the number be in the range of, if the chain need not
bea loop?(m may be either odd or even, in this case.)this
question was directly inspired by this puzzle from way
&rnum=13&prev=I wonder if I can get any idea if any such maze
has multiple solutions(after rotation/reection), and if so,
what would the magnitude ofthis number of solutions
>>James Harris posted a
new version of his proof of a core >error on October 12. He
made a simple algebra mistake>which he has since
acknowledged. >> The interesting thing about this is that,
when the algebra>mistake is corrected, it leads immediately
to a proof that>him main conclusions are wrong.>>...>>JSH
will ignore your post.>Gib,>> I dont think he has ignored
it. However he has none >too courageously avoided replying to
it. >>...>>The distinction is too subtle for me
;-)>>GibGib, Oh, I doubt that. Avoid and ignore are
quite > distinct. You might avoid a polar bear if one showed
up> in your neighborhood, but you probably wouldnt ignore >
it. OK, good example :-) On the other hand, if Im trying to
get your attention by, say, standing next to you and uttering
your name, and you do not respond, we would commonly say that
you were ignoring me.GibGib
=Gib Bogle
frontthe shed door: ^ > I dont think he has ignored it.
However he has none ^ >too courageously avoided replying to
it. ^ >>The distinction is too subtle for me ;-)^ > Oh, I
doubt that. Avoid and ignore are quite ^ > distinct. You
might avoid a polar bear if one showed up^ > in your
neighborhood, but you probably wouldnt ignore ^ > it. ^ OK,
good example :-) On the other hand, if Im trying to get your
^ attention by, say, standing next to you and uttering your
name, and you ^ do not respond, we would commonly say that
you were ignoring me.Both true, and not inconsistent with
each other. That is because ofthe distinction Nora has
unaccountably introduced between thedistinction between
avoid and ignore (as per the polar bearexample) and the
distinction between avoid replying to andignore (as per
the original JSH example). If a polar bear showed upin your
neighbourhood, you would be well advised not to ignore it;
butif it spoke to you, you would be very well advised indeed
not to avoidreplying to it.Andy--Hell! - dont worry about
old raving Dave Ullrich ...Basically hes a sociopath who
cant see a red ragwithout regarding it as a personal
insult.Bill Taylor, sci.math
=Evaluate the summation 1/1 +
2/2 + 3/4 + 4/8 + ... + n/2^n-1 ???
Evaluate the
summation 1/1 + 2/2 + 3/4 + 4/8 + ... + n/2^n-1 ???Did the
instructor specify that you were supposed to post _all_
thehomework questions to sci.math? (And are you going to
haveinternet access when you take the nal exam?) Just
curious...************************David C.
Ullrich
==
Evaluate the summation 1/1 + 2/2 + 3/4 + 4/8 +
... + n/2^n-1 ???Did the instructor specify that you were
supposed to post _all_ the> homework questions to sci.math?
(And are you going to have> internet access when you take the
nal exam?) Just curious...> ************************David C.
UllrichAsshole im just asking for some help what are forum
for !
Asshole im just asking for some help what are forum
for !Maybe thats what *you* should be asking. What are the
forums for?And anyone that was going to do your homework
problems for you probablywont now that youve called us all
assholes.Doug
Evaluate the summation 1/1 + 2/2 + 3/4 +
4/8 + ... + n/2^n-1 ???Did the instructor specify that you
were supposed to post _all_ the> homework questions to
sci.math? (And are you going to have> internet access when
you take the nal exam?) Just curious...>
************************David C. UllrichYes exactly
Evaluate the summation 1/1 + 2/2 + 3/4 + 4/8 + ... + n/2^n-1
???You might mean 1/1+2/2+3/4+...+n/2^(n-1) .The answer is 4
- (2 + n)/2^(n-1) .Can you explain to the instructor how you
got it?
Evaluate the summation 1/1 + 2/2 + 3/4 + 4/8 +
... + n/2^n-1 ???You might mean 1/1+2/2+3/4+...+n/2^(n-1)
.The answer is 4 - (2 + n)/2^(n-1) .Can you explain to the
instructor how you got it?> I try it but i cant get the same
answer!!
Evaluate the summation 1/1 + 2/2 + 3/4 + 4/8 +
... + n/2^n-1 ???
=Evaluate the innite sum 1/1 + 2/2 + 3/4
+ 4/8 + ... .?????
Evaluate the innite sum 1/1 + 2/2 +
3/4 + 4/8 + ... .?????I would have, but not if you are going
to keep posting the same messageover and over again.
Sorry.Doug
Evaluate the innite sum 1/1 + 2/2 + 3/4 +
4/8 + ... .?????Draw a rectangle of 2*1, width 2 and height
1. Over it, aligned to right,draw a rectangle of 1*1 (yes, a
square is a rectangle). Over it, draw otherrectangle of
(1/2)*1, aligned to right, over it other of (1/4)*1and so
on...The sum of the areas of all of them is ...?Now, continue
to down the left vertical sides of the rectangles until
thebasis of the rst rectangle, and sum the areas, gouping
that are in thesame vertical:1 + 2(1/2) + 3(1/4) + 4(1/8) +
...that is your series.That is the method follow by Oresme
(1350 a.c.) to sum that series and othersimilars.-- Ignacio
Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
Evaluate
the innite sum 1/1 + 2/2 + 3/4 + 4/8 + ... .?????Whats in
it for me?particular for x = 1/2.
=I am certain this is
well-known, but it does not look familiar. (It ismost
probably known in a simpler form.)Let us say we have a
polygon P which is inscribed in a unit circle(each vertex of
P coincides with the circle).P is an n-gon which has the
known interior angles{a(1),a(2),a(3),...a(n)}.(So, sum{k=1 to
n} a(k) = (n-2) pi radians)So, am I right in assuming that the
area of P is:sum{k=1 to n} cos(b(k)) sin(b(k)),where b(k)
=(1/2) sum{j=1 to n} a(j) c(j,k),where c(j,k) = (-1)^(j+k) if
j <= k;and c(j,k) = (-1)^(1+j+k) if j > k.A related question:
If a(k) = 2 pi (1 - 2/(k+r)) radians, for k <= n-1,and a(n) =
(n-2) pi - sum{k=1 to n-1} a(k), so that the polygoncontains
the angles, one angle per m-gon, of the regular (1+r)-gon
tothe regular (n+r-1)-gon, plus one extra angle (to make
things workout);then, what is this polygon named? Harmonic
Let us say we have a polygon
P which is inscribed in a unit circle> (each vertex of P
coincides with the circle).> P is an n-gon which has the
known interior angles> {a(1),a(2),a(3),...a(n)}.> (So,
sum{k=1 to n} a(k) = (n-2) pi radians)So, am I right in
assuming that the area of P is:sum{k=1 to n} cos(b(k))
sin(b(k)),where b(k) =(1/2) sum{j=1 to n} a(j) c(j,k),where
c(j,k) = (-1)^(j+k) if j <= k;> and c(j,k) = (-1)^(1+j+k) if
j > k.> I do not think this is correct, but I may be
mistaken. I took thecase that P is a square. Then a(1) = a(2)
= a(3) = a(4) = pi / 2.I also got that b(1) = b(3) = 0 and
b(2) = b(4) = pi / 2For every b(k), either sin(b(k)) = 0 or
cos(b(k)) = 0 sosum {k=1 to n} cos(b(k)) sin(b(k)) = 0Your
formulation looks very interesting, though. Id like very
muchto see your reasoning behind it. Its quite possible that
it can becorrected.In any case, I think I know a slightly
easier formulation of the areaof P(1/2) sum{k=1 to n} cos(pi
- a(k)/2 + a(k+1)/2 )where we take a(n+1) = a(1).In case
youre wondering, this is a sum over the areas of
thetriangles formed by the middle of the circle and each one
of the sidesof the polygon P.A related question: If a(k) = 2
pi (1 - 2/(k+r)) radians, for k <= n-1> ,> and a(n) = (n-2)
pi - sum{k=1 to n-1} a(k), so that the polygon> contains the
angles, one angle per m-gon, of the regular (1+r)-gon to> the
regular (n+r-1)-gon, plus one extra angle (to make things
work> out);You must choose r rather carefully so that sum{k=1
to n-1} a(k) < (n-2) 2 pisum_k (1-2/(k+r)) < (n-2)2 > sum_k 2
/ (k+r)1 > sum_k 1 / (k+r)> then, what is this polygon named?
Harmonic Polygon??Never heard of it. I wouldnt call it the
Harmonic Polygon because itonly has a nite number of angles.
Why are you interested? Perhapsthe application will suggest a
=Well, the below MAY
be true for all integers n, n >= 3, but this wouldbe a
coincidence.For I only actually found, if I did not err even
in this regard, thatthe below is true (true) for ODD
does not look familiar. (It is> most probably known in a
simpler form.)> Let us say we have a polygon P which is
inscribed in a unit circle> (each vertex of P coincides with
the circle).> P is an n-gon which has the known interior
angles> {a(1),a(2),a(3),...a(n)}.> (So, sum{k=1 to n} a(k) =
(n-2) pi radians)So, am I right in assuming that the area of
P is:sum{k=1 to n} cos(b(k)) sin(b(k)),where b(k) =(1/2)
sum{j=1 to n} a(j) c(j,k),where c(j,k) = (-1)^(j+k) if j <=
k;> and c(j,k) = (-1)^(1+j+k) if j > k.> A related question:
If a(k) = 2 pi (1 - 2/(k+r)) radians, for k <= n-1> ,> and
a(n) = (n-2) pi - sum{k=1 to n-1} a(k), so that the polygon>
contains the angles, one angle per m-gon, of the regular
(1+r)-gon to> the regular (n+r-1)-gon, plus one extra angle
(to make things work> out);> then, what is this polygon
1/1 + 2/2
+ 3/4 + 4/8 + ... + n/2n-1.> AND> 1/1 + 2/2 + 3/4
+ 4/8 + ...> Comment on notation:> By usual newsnet
interpretation of math symbols,> 1/2n-1 reduces to -1/2.>
Does your n/2n-1 mean the same as n/2^(n-1),> where ^ means
raised to the power of, and the parentheses are >
necessary?> Re the problem:> Note f_n(x) = 1 + x + x^2
+ ...+ x^n = (1-x^(n+1))/(1-n)> and g(x) = 1 + x + x^2 +
... = 1/(1-x) near x = 1/2.> Their derivatives with respect
to x evaluated at x = 1/2 are> what you want.> I have to
calculate using the geometric formula for sequences...> and
n/2n-1 is not n/2^(n-1) its n/2n-1In the absence of any
parentheses, multiplications and divisions are > always given
priority over additions and subtractions, soeither n/2n-1 =
((n/2)*n) - 1 = n^2/2 - 1, or n/2n-1 = (n/(2*n)) - 1 = -1/2,
and in neither case does it t into your sequence of> 1 + 1/2
+ 1/4 + 1/8 + ... 1/2^0 + 1/2^2 + 1/2^2 + 1/2^3 + ...> by
ane stretch of the imaginationSorry the sequences are
separated1/1 + 2/2 + 3/4 + 4/8 + ... + n/2^n-1.and 1/1 + 2/2
+ 3/4 + 4/8 + ...are two separated sequences!!
1/1
+ 2/2 + 3/4 + 4/8 + ... + n/2n-1.> AND>
1/1 + 2/2 + 3/4 + 4/8 + ...> Comment on notation: By usual newsnet interpretation of math symbols,> 1/2n-1
reduces to -1/2.> Does your n/2n-1 mean the same as
n/2^(n-1),> where ^ means raised to the power of, and the
parentheses are > necessary?> Re the problem:>
Note f_n(x) = 1 + x + x^2 + ...+ x^n = (1-x^(n+1))/(1-n)>
and g(x) = 1 + x + x^2 + ... = 1/(1-x) near x = 1/2.>
Their derivatives with respect to x evaluated at x = 1/2 are what you want.> I have to calculate using the geometric
formula for sequences...> and n/2n-1 is not n/2^(n-1) its
n/2n-1> In the absence of any parentheses, multiplications
and divisions are > always given priority over additions and
subtractions, so> either n/2n-1 = ((n/2)*n) - 1 = n^2/2 - 1,
> or n/2n-1 = (n/(2*n)) - 1 = -1/2, > and in neither case
does it t into your sequence of> 1 + 1/2 + 1/4 + 1/8 + ...
1/2^0 + 1/2^2 + 1/2^2 + 1/2^3 + ...> by any stretch of the
imagination> Sorry the sequences are separated> 1/1 + 2/2 +
3/4 + 4/8 + ... + n/2^n-1.and > 1/1 + 2/2 + 3/4 + 4/8 +
...are two separated sequences!!The rst terminates and the
second does not, but the second should contain the rst, as
it contains all of the terms of the rst and more besides.The
last term of the rst sequence is still not correctly
denoted.It should be n/2^(n-1) if n-1 is supposed to be the
exponent and 2 the base of a power.The point is that the
series whose general term is k*x^(k-1), with x = 1/2, is
related to the series whose general term is x^k by being,
term by term, the derivative of x^k with respect to x.And 1 +
x + x^2 + ... + x^n = (1-x^(n+1))/(1-x), unless x = 1,and1 + x
+ x^2 + ... = 1/(1-x), for x^2 < 1.The sercret of solving the
problem is in these hints.
=participant,> but this link has
the stuff:>> http://www.math.dartmouth.edu/~euler/for, for
sure.-- Karl M. Bunday Christ has set us free. Galatians
5:1Learn in Freedom (TM) http://learninfreedom.org/
The
number of ways of distributing m identical balls into n urns
is> C(m+n-1,n-1), where C(n,r) denotes n choose r.> Could
somebody outline a proof for this?The proof I recall is hard
to outline without giving it away; theres alarge aha!
factor. It involves taking m red balls and n-1 green
balls,and lining them up. Notice how the green balls divide
the red balls intogroups...> Also, is there a more general>
result where we have conditions like p_i <= x_i <= q_i, and 0
< p_i,q_i <=m, that is the number of balls in the i^{th} urn
has> arbitrary lower and upper bounds.There may be, but I
dont see a way to generalize the proof Ive
suggested.Brian
I really need some help on this
question. Ive not been studying matrices> for more than a
couple of weeks and I am getting in a muddle. Let lamda = L
and let epsilon = e. Im is the identity matrixThe elementary
matrices E(r,s;L) and P(r,s) are dened as follows:E(r,s;L) =
Im + Le(r,s)P(r,s) = Im - e(r,r) - e(s,s) + e(r,s)
+e(s,r)Where r does not = se(r,s)e(u,t) = { e(r,t) s=u and 0
where s doesnt = uIf the indices i, r, s are all different
nd the elementary matrix X such> thatP(i,r)E(r,s;L) =
XP(i,r)Your notation makes no sense to me - sorry. Ill
assume these are theusual elementary matrices:E(r, s; L) is
the identity matrix I with its s-th row relaced by thesum of
the s-th row of I and L times the r-th row of I.
ConsequentlyE(r, s; L)*A is A with the s-th row of A replaced
by the sum of thes-th row of A and L times the r-th row of
A.P(r, s) is I with the r-th and s-th rows interchanged. So
P(r, s)*A isA with its r-th and s-th rows interchanged.Now,
note P(i, r)*P(i, r) = I so you have no choice about X:X =
X*P(i, r)*P(i. r) = P(i, r)*E(r, s; L)*P(i, r).So, the
question is whether P(i, r)*E(r, s; L)*P(i, r) is elementary.
[ Yes, it is E(i, s; L) providing neither i nor r is equal to
s ]-- Paul SperryColumbia, SC (USA)
=i have the following
problem: i need tofind efciently, in linear time,the longest
contiguoussubsequence in a sequence of ones and zeroes in
which number of 0s - numberof 1sis 2 mod 5. The only
algorithm i can think of is is quadratically timerelated to
the size of thesequence.
i have the following problem: i
need tofind efciently, in linear time,>the longest
contiguous>subsequence in a sequence of ones and zeroes in
which number of 0s - number>of 1s>is 2 mod 5. The only
algorithm i can think of is is quadratically time>related to
the size of the>sequence.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
> I was wondering if I could
get some help on this problem.>> We know that multiplication
of integers is commutative; i.e. ab=ba for> all pairs of
integers a and b. Prove that, for every natural number n,>
the product of n integers is independant of the order of the
factors.>Induct over n using your axiom (ab=ba) as your base
case
> I was wondering if I could get some help on this
problem.>> We know that multiplication of integers is
commutative; i.e. ab=ba for> all pairs of integers a and b.
Prove that, for every natural number n,> the product of n
integers is independant of the order of the factors.>Induct
over n using your axiom (ab=ba) as your base caseAnd assuming
complete associativity of multiplication!
=Let
phi=(1+sqrt(5))/2 and Q(X)= X^9 - (4*phi^2)X^7 + a(3)X^6 +
... +a(9) .It is known that all roots x_1,x_2,...,x_9 of Q(X)
are real, andmoreover that Q(X) > 0 for all x in ( phi/3 ,
infty) .Its possible tofind the roots of Q(X) ? How
?
look athttp://www.library.cornell.edu/nr/bookcpdf/c9-5.pdfp
375Roland> Let phi=(1+sqrt(5))/2 and> Q(X)= X^9 -
(4*phi^2)X^7 + a(3)X^6 + ... +a(9) .> It is known that all
roots x_1,x_2,...,x_9 of Q(X) are real, and> moreover that
Q(X) > 0 for all x in ( phi/3 , infty) .> Its possible to
nd the roots of Q(X) ? How ?>
sum 1/1 + 2/2 + 3/4 + 4/8 + ... .?????>Okay, I evaluated
it... now what should I do?adam
>Evaluate the innite sum
1/1 + 2/2 + 3/4 + 4/8 + ... .?????>Okay, I evaluated it... now
what should I do?adam
>Evaluate the innite sum 1/1 + 2/2
+ 3/4 + 4/8 + ... .?????>Okay, I evaluated it... now what
should I do?adami WANT TO KNOW HOW????
i WANT TO KNOW
HOW????Fantastic. Doug
You can deduce easily that must
be 4 inection points: before the maximum,> between maximum
and minimum and after the minimum.... the root x = - 1, ..>
the roots 2 +/- sqrt(3); May be a typo ... 3 inection
points. __ Homework problem ?
> You can deduce easily
that must be 4 inection points: before the>> maximum,
between maximum and minimum and after the minimum.... the>>
root x = - 1, .. the roots 2 +/- sqrt(3);>> May be a typo ...
3 inection points. __ Homework problem ?Yes, it is a typo.
There is three inection points.-- Ignacio Larrosa
Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
I have 2
quick questions...> 1. Sketch the graph of the following
function showing vertical and> horizontal asymptotes and
relative extrema. Label your x-axis in> terms of a and your
y-axis in terms of 1/a.> f(x)=(a-x)/(x^2+a^2) where a is a
constant, a greater than 0.> I canfind out the 2 stationary
points, the HA, and that their are no> VAs. I dont get how
tofind other points to make that graphs> however, or
inection points etc.2.Sketch the graph of the following:>> I
canfind that its an odd function. The VA is x=-2. I cant
gure out the relative> extrema or IPs or HA for some reason
though.Scott Eliasonthe function in the second question
should be... f(x) = (5(x^2 + 6)^(.5))> ---------------->
x+2sorry about the mistake.scott eliason
> 2.Sketch the
graph of the following:>> I canfind that its an odd
function. The VA is x=-2. I cant>> gure out the relative>>
extrema or IPs or HA for some reason though.>> Scott
Eliason>> the function in the second question should be...>>
f(x) = (5(x^2 + 6)^(.5))>> ---------------->> x+2> sorry
about the mistake.> scott eliasonAs I said before, that isnt
an odd function, becaus f(-x) =/= f(x).In order to sketch the
graph:Study the domain and vertical asymptotes or other
discontinuities. Alsohorizontals or oblicues. Study
intersections of the graph with axis andasymptotes (onlu HA
and OA, no VA, of course). With that, place the graph atthe
correct side of axis and asymptotes in each interval.Get the
rst derivative. Get the critical points with f(x) = 0.
Studyintervals of increasing and decreasing with the sign of
the derivative,using critical and discontinuity points. This
says you if the criticalpoints are maximum or minimums.The
same with second derivative: solve f(x) = 0, study
sign(f(x)), asbefore with f(x). With that, determine what of
the roots of f(x) areinection points.Sketch all result as
soon as you determine it. The graph will be sketchingalone
..Ignacio Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
=Let A be
the real matrix / 0 a b c A= / a 0 d e b d 0 f / c e f 0 /
where a^2+b^2+c^2+d^2+e^2+f^2 = 6 .Consider that r_1 =< r_2
=< r_3 =< r_4 are eigenvalues of A.It is known that r_4= 10.
Its true that all eigenvalues are integer numbers ?If A is
non-singular, which is the inverse matrix A^{-1}
?
> b d 0 f / > c e f 0 / >where a^2+b^2+c^2+d^2+e^2+f^2 = 6
.>Consider that r_1 =< r_2 =< r_3 =< r_4 are eigenvalues of
A.>It is known that r_4= 10. >Its true that all eigenvalues
are integer numbers ?>If A is non-singular, which is the
inverse matrix A^{-1} ?>
==It cant happen that r_4 = 10.
The characteristic polynomial ofA is t^4 - (a^2+b^2+...+f^2)
t^2 + c1 t + c0 for some c1 and c0.So youd need r1 + r2 + r3
+ r4 = 0 and -6 = r1 r2 + r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3
r4 = 1/2 ((r1 + r2 + r3 + r4)^2 - r1^2 - r2^2 - r3^2 -
r4^2)i.e. r1^2 + r2^2 + r3^2 + r4^2 = 12. In particular r4 <=
2 sqrt(3). Suppose we remove the requirement r_4=10. The only
ways to get 12 as a sum of four squares are 1^2 + 1^2 + 1^2 +
3^2 and 0^2 + 2^2 + 2^2 + 2^2, so the only integer solutions
tor1+r2+r3+r4=0 and r1^2+r2^2+r3^2+r4^2=12 with
r1<=r2<=r3<=r4are [-1,-1,-1,3] and [-3,1,1,1]. It is possible
to get theseeigenvalues, e.g. with a=b=c=d=e=f=1 or
a=b=c=d=e=f=-1 respectively.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Let A be the real matrix> /
0 a b c > A= / a 0 d e > b d 0 f / > c e f 0 / >where
a^2+b^2+c^2+d^2+e^2+f^2 = 6 .>Consider that r_1 =< r_2 =< r_3
=< r_4 are eigenvalues of A.>It is known that r_4= 10. >Its
true that all eigenvalues are integer numbers ?>If A is
non-singular, which is the inverse matrix A^{-1} ?>
==It
cant happen that r_4 = 10. The characteristic polynomial of>
A is t^4 - (a^2+b^2+...+f^2) t^2 + c1 t + c0 for some c1 and
c0.> So youd need r1 + r2 + r3 + r4 = 0 and > -6 = r1 r2 +
r1 r3 + r1 r4 + r2 r3 + r2 r4 + r3 r4 > = 1/2 ((r1 + r2 + r3
+ r4)^2 - r1^2 - r2^2 - r3^2 - r4^2)> i.e. r1^2 + r2^2 + r3^2
+ r4^2 = 12. In particular > r4 <= 2 sqrt(3). Suppose we
remove the requirement r_4=10. The only ways to > get 12 as a
sum of four squares are 1^2 + 1^2 + 1^2 + 3^2 and > 0^2 + 2^2
+ 2^2 + 2^2, so the only integer solutions to> r1+r2+r3+r4=0
and r1^2+r2^2+r3^2+r4^2=12 with r1<=r2<=r3<=r4> are
[-1,-1,-1,3] and [-3,1,1,1]. It is possible to get these>
eigenvalues, e.g. with a=b=c=d=e=f=1 or a=b=c=d=e=f=-1
respectively.Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel > University of
British Columbia > Vancouver, BC, Canada V6T 1Z2Robert Israel
,According to your important remarks, conditioninstead of
r_4=10 we must have ,e.g. x_4=3 .
Let A be the real
matrix> / 0 a b c > A= / a 0 d e > b d 0 f / > c e f 0 / >
where a^2+b^2+c^2+d^2+e^2+f^2 = 6 .> Consider that r_1 =< r_2
=< r_3 =< r_4 are eigenvalues of A.> It is known that r_4= 10.
> Its true that all eigenvalues are integer numbers ?> If A
is non-singular, which is the inverse matrix A^{-1} ?Matlabs
Maple engine says this:A =[ 0, a, b, c][ a, 0, d, e][ b, d, 0,
f][ c, e, f, 0]det(A)
=a^2*f^2-2*a*d*c*f-2*a*e*b*f+b^2*e^2-2*b*e*c*d+c^2*d^2The
inverse of A was a huge unreadable thing involvingfractions
with the expression for det(A) in the denominator.This is
more readable:det(A)*inv(A) =[ 2*d*e*f, f*(a*f-b*e-c*d),
-e*(a*f-b*e+c*d), -d*(a*f+b*e-c*d)][ f*(a*f-b*e-c*d),
2*b*c*f, -c*(a*f+b*e-c*d), -b*(a*f-b*e+c*d)][
-e*(a*f-b*e+c*d), -c*(a*f+b*e-c*d), 2*a*c*e,
a*(a*f-b*e-c*d)][ -d*(a*f+b*e-c*d), -b*(a*f-b*e+c*d),
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9G24cL25908;
= There was pretty hard to harm this
conjecturein some more easy way as it did lately prof. A.
Wiles ... Anyhow some special selection of FLT rules
andproperties seems to extend some procedure anddismiss some
unknown value just to 0 ... What is already true once so
called trivialsolution exists for one of values considered
tobe 0 but so on not the natural value and notthe sort of
intigers denied rst by Fermat... Details very soon once the
cheeck procedurewill be nished with my friend prof.
W.Heiermann Be patient, Yours Ro
= There was pretty hard
to harm this conjecture> in some more easy way as it did
lately prof. A. Wiles ...> Anyhow some special selection of
FLT rules and> properties seems to extend some procedure and>
dismiss some unknown value just to 0 ...> What is already true
once so called trivial> solution exists for one of values
considered to> be 0 but so on not the natural value and not>
the sort of intigers denied rst by Fermat...> Details very
soon once the cheeck procedure> will be nished with my
friend prof. W.Heiermann> Be patient, Yours RoWill the prof
be checking the grammar and spelling as wellas the maths?
Just wondering.
= > There was pretty hard to harm this
conjecture > in some more easy way as it did lately prof. A.
Wiles ... > Anyhow some special selection of FLT rules and >
properties seems to extend some procedure and > dismiss some
unknown value just to 0 ... > What is already true once so
called trivial > solution exists for one of values considered
to > be 0 but so on not the natural value and not > the sort
of intigers denied rst by Fermat... > Details very soon
once the cheeck procedure > will be nished with my friend
prof. W.Heiermann > Be patient, > Will the prof be checking
the grammar and spelling as well > as the maths? Just
wondering.I was wondering in what language the proof would be
written.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
(from approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9G3kTe32625;
of two intersecting circles? If> there are two circles with a
radius of 30 feet and the center of the two> circles were 50
feet apart from each other, then there would be 10 foot> of
overlap. What would the surface area of these circles be?You
asked this question earlier in the thread overlapping
circles.The answer I gave you there for the area of the
overlap wasUse formula (12)
at<http://mathworld.wolfram.com/
Circle-CircleIntersection.html>.If this answer is not
adequate for your purpose, please tell us why.David
=Chris>
How do you calculate the surface area of two intersecting
circles?> If there are two circles with a radius of 30 feet
and the center of thetwo circles were 50 feet apart > from
each other, then there would be 10foot of overlap. What would
the surface area of> these circles be? I got as far as
A=pi*r^2.> For the two circles in this question that would be
5,654 Sq Ft.> So to answer this question, it would be 5,654 Sq
Ft minus overlap area?> How do you calculate the overlap
area?The overlap consists of two chunks (lunae, I think
theyre called) each ofwhich is in the shape of a pie slice
minus a triangle. See it now?LH
=Where could Ifind a proof
for following:f : R^n -> R+ If f is strictly convex in a
convex region Dthen the Newtons iteration x_k+1 = x_k -
inv(f(x_k)) f(x_k)^Tconverges to the minimum of f from any
starting point in D.Ive been trying to apply various known
convergence conditions, but they seem a bit overkill for this
and I havent reallybeen able to use the convexity of f for
anything.Is this too trivial a result or am I mistaken in my
assumption?
Where could Ifind a proof for following:>>f
: R^n -> R+ >If f is strictly convex in a convex region
D>then the Newtons iteration>> x_k+1 = x_k - inv(f(x_k))
f(x_k)^T>>converges to the minimum of f from any starting
point in D.You cantfind a proof of this anywhere, because
its not true.To start, f need not _have_ a minimum in D. But
even if itdoes Newtons method need not converge to it. This
isshown by a standard example:Lets take n = 1 for
simplicity. Youre using Newton tofindthe zero of g = f.
Saying f is strictly convex says that gis strictly
increasing. Concoct a strictly increasing functiong such that
g(-1) = -1, g(1) = 1, g(-1) = g(1) = 1/2. Thenif you apply
Newtons method starting at x = 1 you getthe sequence 1, -1,
1, -1, ..., which does not convergeto the zero of g.>Ive
been trying to apply various known convergence conditions,
>but they seem a bit overkill for this and I havent
really>been able to use the convexity of f for anything.>Is
this too trivial a result or am I mistaken in my >assumption?
>************************David C. Ullrich
Let p_k be
the kth prime number.> i.e. P_1=2, p_2=3, p_3=5,....Then is
it true that (p_n + p_1)/2 >= (p_1+p_2+...+p_n)/n for all
natural > number n?Probably. Its certainly true for n up to
10000. And numerically it seems the difference between the
two sides is asymptotic to cn for someconstant c, 0.25 < c <
0.3.Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T
1Z2
>>It could be a constant vector, but my guess is that the
equation is>>actuallyr = -k r/|r|^3>>Well of
course that was my guess as well, which is why I>didnt
decide right away that k was a constant vector...Well,
of course it has many closed-form solutions. Its>>not
possible to express the general solution in closed>>form,
is it?>> Not as far as I know (for r as a function of
last 300 years> or so, so it would be pretty remarkable if
you got an> answer in a newsgroup in a couple of days.>> -
RandyIm a hoping and optimist boy!! :-)Marco
There is
no need for more food. People go hungry not because there is>
not enough food, but because they do not have enough money to
buy the> food of which there is plenty.Countries like USA
keep the price of food high to protect their ownagriculture.
If the price of food was lower, the less developed
countrieswould be able to compete on the market and thus
produce more food also fortheir own people.
There is no
need for more food. People go hungry not because there is>
not enough food, but because they do not have enough money to
buy the> food of which there is plenty.Countries like USA keep
the price of food high to protect their own> agriculture. If
the price of food was lower, the less developed countries>
would be able to compete on the market and thus produce more
food also for> their own people.Food does not reach people
who need it for logistic reasons, not becauseof any lack of
money or any conspiracy on the part of developed nations.The
nations that have the fastest growing populations also tend
to havethe least advanced agriculture.The solution is not to
produce and distribute ever increasing amounts offood; it is
to instead reduce overpopulation. The world populationcannot
increase innitely, and the larger it gets, the lower
thestandard of living will be for everyone. It makes more
sense tomaintain smaller populations with higher standards of
living. Developednations tend naturally in this direction, in
most cases. Developingnations tend to have unacceptably
natural rates of increase, and so theytend to have periodic
famines and very low standards of living.--
There is no
need for more food. People go hungry not because there is>
not enough food, but because they do not have enough money to
buy the> food of which there is plenty.It really has nothing
to due with money. Famines are caused due to lackof the
political infrastructure needed to distribute the food. A
fewyears ago, a Nobel prize in Economics was awarded for
research in thisarea - there has never been a famine in a
democratic country, independentof monetary considerations.>
Countries like USA keep the price of food high to protect
their own> agriculture. If the price of food was lower, the
less developed countries> would be able to compete on the
market and thus produce more food also for> their own
people.Thats somehwat true. The USA pays more for sugar than
any other nationowing to our tariffs. Your implication is that
developing nations somehowneed to compete on the agricultural
world market. Thats less of a concernthan producing what
their nation requires. Keep in mind that the
privateorganizations (e.g., CARE) export raw agricultural
products to developingnations so they can monitize the aid.
That is, some country might receivetons of soybeans so they
can resell it on the open market for hard cash.Food does not
reach people who need it for logistic reasons, not because>
of any lack of money or any conspiracy on the part of
developed nations.> The nations that have the fastest growing
populations also tend to have> the least advanced
agriculture.I am unaware of any evidence suggesting that the
fastest growing populationsare the least advanced in terms of
agriculture. What are some examples andwhat implications do
you intend to draw from this assertion?The solution is not to
produce and distribute ever increasing amounts of> food; it is
to instead reduce overpopulation. The world population> cannot
increase innitely, and the larger it gets, the lower the>
standard of living will be for everyone. It makes more sense
to> maintain smaller populations with higher standards of
living. Developed> nations tend naturally in this direction,
in most cases. Developing> nations tend to have unacceptably
natural rates of increase, and so they> tend to have periodic
famines and very low standards of living.You seem to view the
world as some nite set of resources ever taxed bya growing
population. Yet we continue to discover new resources - not
necessarily agricultural or natural (i.e., oil), but more
tailored to thepresent economy. The economic history of the
US for example, exhibits anincreasing standard of living
despite its increasing population.Can you reasonably say that
100 years ago, the US had a better standardof living than we
have today? The population was lower after all.
I am
unaware of any evidence suggesting that the fastest growing
populations> are the least advanced in terms of agriculture.
What are some examples and> what implications do you intend
to draw from this assertion?Subsaharan Africa comes to
mind.The implication is that the same stupidity that prevents
a country frombeing skilled in agriculture also prevents it
from controlling its ownpopulation. Usually things like
disease will maintain equilibrium, butwhen more advanced
countries help, the diseases go down, but the rateof natural
increase goes up, and the production of food doesnt
change.So eventually lots of people being to starve.
Eventually starvationrestores equilibrium, at least
temporarily, but its not very elegant.> You seem to view the
world as some nite set of resources ever taxed by> a growing
population.Yes.> Yet we continue to discover new resources -
not necessarily agricultural> or natural (i.e., oil), but
more tailored to the present economy.That will not continue
forever, and its best not to increase thepopulation until
you are sure that you can support the increase.> The economic
history of the US for example, exhibits an> increasing
standard of living despite its increasing population.The
United States has an educated population (including women,
which iscritical).> Can you reasonably say that 100 years
ago, the US had a better standard> of living than we have
today? The population was lower after all.Technology has
outpaced population growth in the U.S. That wontcontinue
forever, nor is it transferable to other countries.--
=<...>> The implication is that the same stupidity that
prevents a country from> being skilled in agriculture also
prevents it from controlling its own> population.Wrong. In
many underdeveloped countries, having many children is
theonly possible insurance for old age support, and that
keeps the birthrate high. Thats why education helps. A good
profession provides enoughsupport even with less children.--
Lassi
In many underdeveloped countries, having many
children is the> only possible insurance for old age support,
and that keeps the birth> rate high.Thats true when mortality
is high. But as mortality is reduced(usually with the help of
outsiders with technology), that same highbirth rate rapidly
increases the population and exhausts the food.> Thats why
education helps. A good profession provides enough> support
even with less children.And women who are educated are less
likely to be willing to spend theirlives reproducing as
well.--
You seem to view the world as some nite set of
resources ever taxed by>a growing population. Yet we continue
to discover new resources - not >necessarily agricultural or
natural (i.e., oil), but more tailored to the>present
economy. The economic history of the US for example, exhibits
an>increasing standard of living despite its increasing
population.>Can you reasonably say that 100 years ago, the US
had a better standard>of living than we have today? The
population was lower after all.The world *is* a nite set of
resources...although improvements in technology increase the
usefulness and valueof those resources.Hence, population can
cause problems if it increases faster thantechnology keeps
up... and, given that there do exist people who havelow
standards of living, apparently population does tend to
increasefaster than improvements in technology in most
places.Unfortunately, we dont have any more empty continents
with whichother people could get the jump on the game the way
the Europeaninhabitants of the New World did.John
Savardhttp://home.ecn.ab.ca/~jsavard/index.html
=Were
getting pretty far from math or crypt...>> There is no need
for more food. People go hungry not because there is> not
enough food, but because they do not have enough money to buy
the> food of which there is plenty.One good way to keep the
famines happening is to send food for free.Local producers go
bust, and the famines keep on repeating...> The solution is
not to produce and distribute ever increasing amounts of>
food; it is to instead reduce overpopulation.The best known
way to reduce birth rate is to teach women to read. Thesocial
impact is enormous.-- Lassi
The best known way to reduce
birth rate is to teach women to read. The> social impact is
enormous.Agreed. Equal education for women in general reduces
birth rates andimproves the standard of living. So does a good
general level ofeducation for both sexes.--
= > The best
known way to reduce birth rate is to teach women to read.
Well, rst they will have to be considered as human beings,
not ascommodity items, as they customarily are in Moslem
countries and India.Yep, this is no politically correct, but
it is nevertheless true.
<Ddtjb.80$en5.60@reader1.news.jippii.net>
<h2ssovoks77lgfguub16s3ubd42mjile13@4ax.com The best
known way to reduce birth rate is to teach women to read.>>
Well, rst they will have to be considered as human beings,
not as> commodity items, as they customarily are in Moslem
countries and India.> Yep, this is no politically correct,
but it is nevertheless true.And what do you think about how
the women are considered in western(i.e. not majorily
Moslem) countries, when you watch TV series,advertisings,
magazines, etc ? Not much better, as they sometimes
arerepresented as geese, often as sexual objects. If theyre
not blondes withbig breasts, they are of no value.Still not
politically correct, but also true.-- -----je crosspost sur
fr rec moto pour ce triste mod.8ele dintol.8erance. [...]PS
:D.8esol.8e mon logiciel de news ne permet pas les follow up
et je nenchangerai certainement pas pour vous etre
agr.8eable.-+- CC in Guide du Neuneu Usenet - Bien congurer
son incomp.8etence -+-
And what do you think about how
the women are considered in western> (i.e. not majorily
Moslem) countries, when you watch TV series,> advertisings,
magazines, etc ? Not much better, as they sometimes are>
represented as geese, often as sexual objects.Orders of
magnitude better, actually. Ask any woman who has lived
inboth societies.--
=Were getting pretty far from math or
crypt...>> There is no need for more food. People go
hungry not because there is> not enough food, but because
they do not have enough money to buy the> food of which
there is plenty.One good way to keep the famines happening is
to send food for free.> Local producers go bust, and the
famines keep on repeating...> The solution is not to produce
and distribute ever increasing amounts of> food; it is to
instead reduce overpopulation.The best known way to reduce
birth rate is to teach women to read. The> social impact is
enormous.With any luck, when the present pope dies his
successor will be lessopposed to articial methods of birth
control.-- G.C.
With any luck, when the present pope
dies his successor will be less> opposed to articial methods
of birth control.You still have to teach and convince people
to use it.--
With any luck, when the present pope dies
his successor will be less> opposed to articial methods of
birth control.You still have to teach and convince people to
use it.Education is crucial in all matters relating to
poverty I suspect.-- G.C.
<Ddtjb.80$en5.60@reader1.news.jippii.net>
<h2ssovoks77lgfguub16s3ubd42mjile13@4ax.com>
<3F8E898D.3C24B60F@ieee.orgasm-research.invalid> Were
getting pretty far from math or crypt...>There is
no need for more food. People go hungry not because there
is>not enough food, but because they do not have enough money
to buy the>food of which there is plenty. One good way to
keep the famines happening is to send food for free.>> Local
producers go bust, and the famines keep on repeating...
The solution is not to produce and distribute ever increasing
amounts of>> food; it is to instead reduce overpopulation.>> The best known way to reduce birth rate is to teach women
to read. The>> social impact is enormous.>> With any luck,
when the present pope dies his successor will be less>
opposed to articial methods of birth control.The thread
simply wasnt controversial enough for you? Why not bringup,
say, skinhead abortions and gun control laws for baby
seals?Take it elsewhere, willya?-- When I am grown to mans
estateI shall be very proud and great,and tell the other
girls and boysnot to meddle with my toys. --Robert Louis
Stevenson
There is no need for more food. People go
hungry not because there is> not enough food, but because
they do not have enough money to buy the> food of which there
is plenty.Countries like USA keep the price of food high to
protect their own> agriculture. If the price of food was
lower, the less developed countries> would be able to compete
on the market and thus produce more food also for> their own
people.An ancient Chinese monarch was dining when he was
toldthat the people didnt have enough rice to eat. Why dont
they eat meat instead, the monarch exclaimed.M. K. Shen
=--
KindlyKonrad-------------------------------------------------
places;their souls be chased by demons in Gehenna from one
room toanother for all eternity and more.Sleep - thing used
by ineffective people as a substitute for coffeeAmbition - a
poor excuse for not having enough sence to be
lazy---------------------------------------------------
=
Heres the existing function I have for rotating a 3D point
(x,y,z) byangleX, Y, Z around point px,py,pz (its actually
in a 3D vectorclass). Is there a simple way of adding 4D
rotation to this? Does itnot involve Quaternion numbers at
all anyway? /** ** This method rotates this 3D Vector around
the X/Y/Z axis byamounts ** angleX/angleY/angleZ around the
points centerX/centerY/centerY */ public void rotate3D(double
angleX, double angleY, double angleZ, double centerX, double
centerY, double centerZ){ double sinX, cosX; double sinY,
cosY; double sinZ, cosZ; double tempX, tempY, tempZ; sinX =
Math.sin(angleX); cosX = Math.cos(angleX); sinY =
Math.sin(angleY); cosY = Math.cos(angleY); sinZ =
Math.sin(angleZ); cosZ = Math.cos(angleZ); tempX = x -
centerX; tempY = y - centerY; tempZ = z - centerZ; double xy
= tempY*cosX - tempZ*sinX; double xz = tempY*sinX +
tempZ*cosX; double xx = tempX; tempX = xx; tempY = xy; tempZ
= xz; double yx = tempX*cosY + tempZ*sinY; double yz =
-tempX*sinY + tempZ*cosY; tempX = yx; tempZ = yz; double zx =
tempX*cosZ - tempY*sinZ; double zy = tempX*sinZ + tempY*cosZ;
tempX = zx; tempY = zy; tempX += centerX; tempY += centerY;
tempZ += centerZ; x = tempX; y = tempY; z = tempZ; }> Ive
add c.g.a to the newsgroups as its probably a more
appropriate forum> for this.> I have written a program to
display Quaternion Julia set fractals. If> you dont know
what these are then I think it is sufciant to know> that
they are 4D object (in Quaternion space - I guess that is a>
unique type of 4D space, with its own set of rules?). I
can display> this object in 3D (using raytracing), ignoring
the 4th dimension (I> can of course set the 4th to any
number but it remains static for> each 3D image). I have
programmed this in such a way that I can> rotate this in 3
Dimensions, ie rotate the view cube to form a> different
view - so I would like to be able to rotate the (this is>
gonna sound silly) 4D cube to form another view.I would not
think of them as quaternions as I dont think its helpful
in> this case: you have a 4D object you would like to display
in 3D by chosing> a 3D slice through it. You already have
rotations in 3D. What you need> is rotations involving the
4th dimension, i.e. the 4th axis.Consider a similar situation
in 3D, applied to a sphere, the earth say. If> you ignore/x
the north-south axis you would have the rotation in the>
plane of the equator, the normal rotation of the earth. Then
combine these> with rotations taking the poles towards points
on the equator, i.e. in> planes passing through the poles, to
get the full set of 3D rotations.Your situation is similar
but in 4D. You already have three dimensional> rotations. To
generate the full set of 4D rotations you need to combine>
these with rotations from the 4th axis to all possible axes
in 3D, e.g. in> planes through this fourth axis.It should be
possible to devise an interface for this, such a sphere,>
which you click on to indicate the 3D axis which will be
paired with the> 4th axis to generate the rotation into the
fourth dimension. Rotations in> 4D have six degrees of
freedom, the three DOF youve already implemented +> the
three DOF of the rotations into 4D.John
More like 2300
years. Euclid proved there is an innitude of primes.Naah,
actually, he never mentions innite at all.He just does:
Given any [nite] set of primes, to construct a primenot in
the set.
More like 2300 years. Euclid proved there is an
innitude of primes.Naah, actually, he never mentions innite
at all.He just does: Given any [nite] set of primes, to
construct a prime> not in the set.Does the usual proof
actually construct a prime?
More like 2300 years.
Euclid proved there is an innitude of primes.> Naah,
actually, he never mentions innite at all.> He just does:
Given any [nite] set of primes, to construct a prime> not in
the set.Does the usual proof actually construct a prime?No,
but it shows that such a prime must exist.
More like
2300 years. Euclid proved there is an innitude of primes.>
Naah, actually, he never mentions innite at all.> He just
does: Given any [nite] set of primes, to construct a prime not in the set.> Does the usual proof actually construct a
prime?No, but it shows that such a prime must exist.Since
nobody else sketched the argument, I will.Let p1, p2, p3,
..., pn be the rst n primes. Considerthe number
(p1*p2*...*pn)+1. This number has a remainderof 1 when
divided by any prime up to pn. Therefore eitherthe number is
prime, or if it is composite then its primefactors are all
larger than pn.Thus given the rst n primes where n is
nite,there exists another prime which is larger than allof
them.If p1, p2, ... pn are any n primes (not the rst n),then
youve proved as Virgil says that there exists anotherprime
not in this list, but its not necessarilya larger one. For
instance, (3*5*13)+1 has 2 and 7 asprime factors. Still,
since that means no nite list ofprimes is complete it is
sufcient to prove there areinnitely many primes. -
Randy
More like 2300 years. Euclid proved there is an
innitude of primes.Naah, actually, he never mentions innite
at all.He just does: Given any [nite] set of primes, to
construct a prime> not in the set.Thus he proved that the
number of primes is not nite, though perhaps he did not
express his result in quite those terms.
More like 2300
years. Euclid proved there is an innitude of primes.Naah,
actually, he never mentions innite at all.He just does:
Given any [nite] set of primes, to construct a prime> not in
the set.Does he prove unique factorisation of the
integers?Isnt that also necessary?Phil-- Unpatched IE
vulnerability: Web Archive buffer overowDescription:
Possible automated code execution.Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html=
>> More like 2300 years. Euclid proved there is an
innitude of primes. Naah, actually, he never mentions
innite at all. He just does: Given any [nite] set of
primes, to construct a prime>> not in the set.> Does he prove
unique factorisation of the integers?> Isnt that also
necessary?No, because the key step is tofind a number that is
not divisible by anyof the primes in the given (nite) set.
Either this number is prime, orit has (at least one) prime
divisor not in the set. Unique factorizationis not needed.--
Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
We can understand the proof that the set
of primes has no largest member. >Thus, we may conclude that
some innite primes exist.We may? No, We
PrettyDoug
We can understand the proof that the
set of>>primes has no largest member. Thus, we may conclude
that some innite>>primes exist.>> Nonsense. If you
conclude that, I conclude that you dont know much>>
mathematics. Assuming youre talking about the usual
denition of>> primes, primes are integers. There are no
innite integers.>>Doesnt the compactness theorem for FOL
imply that there are>non-archimedean models of the
integers?>> Its not entirely clear to me what a
non-archimedean model> of the integers would be. Yes, the
compactness theorem> shows that there are non-standard, very
large, models of> the _theory of_ the integers. These models
include things> which might reasonably be called innite.>>
That does not imply that there exists an innite integer,
however.>>There are. They are, of course, nonstandard.>>A
nonstandard prime is not a _prime_.>>Maybe we should start
with the question of how many legs a>dog has if you call the
tail a leg...>All the constructions that Im aware of and
understand (that leaves out>>ultralters) create nonstandard
integers. Construction is in quotes>>because they arent
really constructions -- theyre proofs of
existence.>>However, if you start with the standard rst
order axioms of integers, and>>nd a model with innite
elements, how can you not call the elements of>>that model
integers? They t all the axioms of integers, but were
going>>to call them monsters? No, theyre integers. >>Huh?
_Any_ object whatever is an element of some model of
the>theory of the integers. By your criterion _everything_ is
an integer.Well, it might be true that people who spend half
their lives working withnonstandard models of the integers
get into the habit of calling thesethings just integers. No
problem with that - just like algebraic numebrtheorists often
refer to algebraic integers just as integers.There is no
largest prime - therefore there is an innite prime, then
themost appropriate and helpful answer is surely `Nonsense
or maybe even*Nonsense* (could even be a case for NONSENSE!)
To start discussinginnite primes in nonstandard models of
the integers in the companyof such a poster does not seem
very helpful.Derek Holt.>>Since theyre not in the>>standard
model, theyre obviously nonstandard, but theyre still
integers.>>As you say, it does not imply that there exists
an innite integer. I>>suspect that there are nonstandard
models without innite integers -- some>>numbers are innite,
but all integers are nite.>>Jon
Miller>>************************>>David C. Ullrich
N^* - N>> Excuse me (again) for being stupid, but is that
expression>> meaningful?>> [I ask because I know exactly
nothing about a subject called>> non-standard analysis.]Its
intended to mean the set difference, which I would write N*
N.The tricky part is that although N* is an internal set in
NSA, N is not.> Both N and (N* N) are external sets, which
means that in some sense> they are not quite meaningful sets
in NSA.But then, its common to keep leaping between internal
reasoning and> external reasoning in NSA.This is the whole
point of using NSA :-)-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
However, if you start with the standard rst order axioms of
integers,> andfind a model with innite elements, how can you
not call the elements> of> that model integers?No.> They t
all the axioms of integers,No. Not the second order induction
axiom (a subset of N containing 1and the successor of each of
its elements is N).> but were going> to call them
monstersNo, only the largest sporadic simple group is called
a monster.We call them nonstandard integers.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
=Proof the following proposition or give a
counterexample:For all homoeomorphisms (not sure about the
spelling, I mean an openand closed bijective mapping) between
two open subsets of (of perhapstwo different) euclidean vector
spaces, the dimensions of the twoeuclidean spaces are
identical.Is this perhaps a famous theorem (at least in nite
dimensions, whatabout dimensions of innite cardinality?)
whose name I dont know?
Proof the following proposition
or give a counterexample:>>For all homoeomorphisms (not sure
about the spelling, I mean an open>and closed bijective
mapping) between two open subsets of (of perhaps>two
different) euclidean vector spaces, the dimensions of the
two>euclidean spaces are identical.>The nite case has been
answered already>Is this perhaps a famous theorem (at least in
nite dimensions, what>about dimensions of innite
cardinality?) whose name I dont know?>As to the innite
case: one might consider two innite versions of R^n:1.
R^kappa, for innite cardinals kappa, with the product
topology2. ell_2(kappa), the set of square-summable sequences
of length kappa, In either case the answer is much easier than
that for R^n: kappa is the topological weight of any
nonemptyopen set; so. yes, the dimensions have to equal in
that case as well.KP-- E-MAIL: K.P.Hart@EWI.TUDelft.NL PAPER:
Faculty EWIPHONE: +31-15-2784572 TU DelftFAX: +31-15-2786178
Postbus 5031URL: http://aw.twi.tudelft.nl/~hart 2600 GA Delft
the Netherlands
Proof the following proposition or give a
counterexample:For all homoeomorphisms (not sure about the
spelling, I mean an open> and closed bijective mapping)
between two open subsets of (of perhaps> two different)
euclidean vector spaces, the dimensions of the two> euclidean
spaces are identical.Is this perhaps a famous theorem (at
least in nite dimensions, what> about dimensions of innite
cardinality?) whose name I dont know? As G.A. Edgar noted,
its a version of Brouwers Invariance of Domain Theorem:
Theorem: If X can be imbedded in R^n as an open subset, then
h(X) is an open subset of R^n for _any_ imbedding h: X -->
R^n. Your result is a corollary. If U subseteq R^m and V
subseteq R^n are open sets and h: U --> V is a homeomorphism
between them, it follows that m = n. (Otherwise, say m < n.
Then R^m sits inside R^n as a coordinate hyperplane which has
empty interior. This shows that U can be imbedded in R^n as
something other than an open subset. But h imbeds it as V,
which _is_ open -- contradiction.) A proof of Brouwers
theorem requires some machinery -- just what machinery
dependson what kinds of things you already know ...
=
Theorem: If X can be imbedded in R^n as an open subset, then>
h(X) is an open subset of R^n for _any_ imbedding> h: X -->
R^n. Your result is a corollary. If U subseteq R^m and V
subseteq R^n> are open sets and h: U --> V is a homeomorphism
between them, it> follows that m = n. (Otherwise, say m < n.
Then R^m sits inside> R^n as a coordinate hyperplane which
has empty interior. This shows> that U can be imbedded in R^n
as something other than an open subset.> But h imbeds it as V,
out already. A proof of Brouwers theorem requires some
machinery -- just what> machinery dependson what kinds of
things you already know ...Concerning algebraic topology, I
know what homotopy is, but am not(yet) familiar with
homology. I doubt that homotopy is of any usebecause e.g.
open balls are contractible and thus have trivialhomotopy.Are
there any measure-theoretic invariants that can be used
tocharacterize full-dimensionality of sets? That is, is the
propertymeasure zero preserved under homeomorphisms?My
original intention was to prove that the dimension of a
manifold iswell-dened (i.e. given two atlases, they map two
the same R^n). Arethere any other ways to do this?
>>
Theorem: If X can be imbedded in R^n as an open subset,
then>> h(X) is an open subset of R^n for _any_ imbedding>> h:
X --> R^n. Your result is a corollary. If U subseteq R^m
and V subseteq R^n>> are open sets and h: U --> V is a
homeomorphism between them, it>> follows that m = n.
(Otherwise, say m < n. Then R^m sits inside>> R^n as a
coordinate hyperplane which has empty interior. This shows>>
that U can be imbedded in R^n as something other than an open
subset.>> But h imbeds it as V, which _is_ open --
>> A proof of Brouwers theorem requires some machinery --
just what>> machinery dependson what kinds of things you
already know ...>Concerning algebraic topology, I know what
homotopy is, but am not>(yet) familiar with homology. I doubt
that homotopy is of any use>because e.g. open balls are
contractible and thus have trivial>homotopy.>Are there any
measure-theoretic invariants that can be used to>characterize
full-dimensionality of sets? That is, is the property>measure
zero preserved under homeomorphisms?>My original intention
was to prove that the dimension of a manifold is>well-dened
(i.e. given two atlases, they map two the same R^n).
Are>there any other ways to do this?You could also use the
Lebesgue dimension (a.k.a. covering dimensionor topological
dimension). This is invariant under homeomorphisms
anddim(R^n) = n. This isnt a measure-theoretic dimension,
though.You can take a look at Hurewicz & Wallmans Dimension
Theory formore information.John
A proof of Brouwers
theorem requires some machinery -- just what> machinery
dependson what kinds of things you already know
...>Concerning algebraic topology, I know what homotopy is,
but am not>(yet) familiar with homology. I doubt that
homotopy is of any use>because e.g. open balls are
contractible and thus have trivial>homotopy.The usual proofs
condisider the open ball with the center point removed(or the
one point compactications). This set is homotopic to asphere,
which is not contractible. This allows a proof that open
setsin R^2 and open sets in higher dimensions are not
homeomorphic usingfundamental (rst homotopy) groups.If you
know about higher homotopy groups, this will also give the
generalresult. However, the higher homotopy groups are not so
easy to calculate.Typically, the invariance theorem is proved
using homology theory.>Are there any measure-theoretic
invariants that can be used to>characterize
full-dimensionality of sets? That is, is the property>measure
zero preserved under homeomorphisms?No, it is not. There is a
homeomorphism from the real line to itselfwhich takes a set
of measure zero to one of positive measure. Essentially,the
Cantor ternary set is stretched to a set of positive
measure.>My original intention was to prove that the
dimension of a manifold is>well-dened (i.e. given two
atlases, they map two the same R^n). Are>there any other ways
to do this?As another poster suggested, you can use the
Lebesgue covering dimension.The problem is that proving the
dim(R^n)=n is again not easy. I belive itrequires the fact
that a sphere is not a retract of the closed ball, which is
also proved using homology (usually).---Dan Grubb
...>>My
original intention was to prove that the dimension of a
manifold is>>well-dened (i.e. given two atlases, they map
two the same R^n). Are>>there any other ways to do this?>>As
another poster suggested, you can use the Lebesgue covering
dimension.>The problem is that proving the dim(R^n)=n is
again not easy. I belive it>requires the fact that a sphere
is not a retract of the closed ball, which >is also proved
using homology (usually).Thats true, but there are also
direct geometric proofs that thesphere is not a retract of
the closed ball. There was a recentsci.math thread on that
topic (Non-contractibility of S^n, orsomething like
that).John Mitchell
Are there any measure-theoretic
invariants that can be used to> characterize
full-dimensionality of sets? That is, is the property>
measure zero preserved under homeomorphisms?Not if measure
means Lebesgue measure.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
Proof the following proposition or give a counterexample:For
all homoeomorphisms (not sure about the spelling, I mean an
open> and closed bijective mapping) between two open subsets
of (of perhaps> two different) euclidean vector spaces, the
dimensions of the two> euclidean spaces are identical.Is this
perhaps a famous theorem (at least in nite dimensions, what>
about dimensions of innite cardinality?) whose name I dont
know?Invariance of the domain ... proved by Brouwer, I
believe.When it says euclidean vector spaces, it means
nite-dimensional.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
Proof the
following proposition or give a counterexample:For all
homoeomorphisms (not sure about the spelling,Usually
homeomorphism.> I mean an open> and closed bijective mapping)
between two open subsets of (of perhaps> two different)
euclidean vector spaces, the dimensions of the two> euclidean
spaces are identical.In the local homology groups of an open
subset of R^n vanish exceptin dimension n. (See texts on
algebraic topology).> Is this perhaps a famous
theoremRaasonably; it hasnt anyones name AFAIK.> (at least
in nite dimensions, what> about dimensions of innite
cardinality?)Dunno about that.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
= at 02:23 PM, Robin Chapman
homology groups of an open subset of R^n vanish except>in
dimension n. (See texts on algebraic topology).Whats H_1 of
an annulus?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot
at 02:23 PM, Robin
the local homology groups of an open subset of R^n vanish
except>>in dimension n. (See texts on algebraic
topology).Whats H_1 of an annulus?Its isomorphic to Z. Now
is that a homework question?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
=Ive been reading his posts and I have a hard time
following them lately.Here are some suggestions/questions I
have for him.1. What is a key variable? Last I heard, there
was no such thing as a keyvariable, however, I think you
might mean independent variables. You shouldknow about this
term.2. Use the right terminology. Your biggest problem is
that you usenonstandard terms and expect everyone to know
what youre talking about.Just because we dont know what
youre talking about doesnt mean werestupid, as you
insinuate.3. Cut the personal attacks; they are irrelevent to
the math and make youappear childish.David Moran
Ive
been reading his posts and I have a hard time following them
lately.> Here are some suggestions/questions I have for
him.1. What is a key variable? Last I heard, there was no
such thing as a key> variable, however, I think you might
mean independent variables. You should> know about this
term.2. Use the right terminology. Your biggest problem is
that you use> nonstandard terms and expect everyone to know
what youre talking about.> Just because we dont know what
youre talking about doesnt mean were> stupid, as you
insinuate.3. Cut the personal attacks; they are irrelevent to
the math and make you> appear childish.If they thought there
was a chance that their suggestions would be taken seriously,
I imagine others could add quite a few more.Gib
Ive been
reading his posts and I have a hard time following them
lately.> Here are some suggestions/questions I have for him.>
1. What is a key variable? Last I heard, there was no such
thing as a key> variable, however, I think you might mean
independent variables. You should> know about this term.> 2.
Use the right terminology. Your biggest problem is that you
use> nonstandard terms and expect everyone to know what
youre talking about.> Just because we dont know what youre
talking about doesnt mean were> stupid, as you insinuate.>
3. Cut the personal attacks; they are irrelevent to the math
and make you> appear childish.If they thought there was a
chance that their suggestions would be taken > seriously, I
imagine others could add quite a few more.Like:4. Dont post
when drunk or when in full-blown delusionalmode.5. Talk to
somebody in your local community health services. -
Randy
Ive been reading his posts and I have a hard time
following them lately.> Here are some suggestions/questions I
have for him.> 1. What is a key variable? Last I heard, there
was no such thing as a key> variable, however, I think you
might mean independent variables. You should> know about this
term.> 2. Use the right terminology. Your biggest problem is
that you use> nonstandard terms and expect everyone to know
what youre talking about.> Just because we dont know what
youre talking about doesnt mean were> stupid, as you
insinuate.> 3. Cut the personal attacks; they are irrelevent
to the math and make you> appear childish.If they thought
there was a chance that their suggestions would be taken >
seriously, I imagine others could add quite a few more.Gib4.
Cut down on the conspiracy speculations (actually, get rid of
them completely).---J K Hauglandhttp://www.neutreeko.com
Ive been reading his posts and I have a hard time following
them lately.> Here are some suggestions/questions I have for
him.>> 1. What is a key variable? Last I heard, there was no
such thing as akey> variable, however, I think you might mean
independent variables. Youshould> know about this term.>> 2.
Use the right terminology. Your biggest problem is that you
use> nonstandard terms and expect everyone to know what
youre talking about.> Just because we dont know what youre
talking about doesnt mean were> stupid, as you insinuate.>>
3. Cut the personal attacks; they are irrelevent to the math
and make you> appear childish.>> David Moran>>One other thing
is dont use the word should. That word has nomathematical
relevance. If you need the word, then you need to analyze
thesituation.
=its just a matter of tense, since shall is
the same as must,being forms of to be. in other words,should
is cognate with yet to be shown, to be --or not to be! > 2.
Use the right terminology. Your biggest problem is that you
use > One other thing is dont use the word should. That word
has no> mathematical relevance. If you need the word, then you
need to analyze the> situation.--Dec.2000 WAND Chairman Paul
ONeill, reelected to Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=How do
I do the following problem:Let f(x) = x^2 - 6x + 7. By
completing the square, or otherwise, determinethe unique
number a such that f(x) is a one-to-one function over the
domain(-?,a] and over the domain [a,?).For this value of a,
nd the inverses of the functions f1 and f2 denedby:f1(x) =
x^2 - 6x + 7, x?af2(x) = x^2 - 6x +7, x?aTIA.
How do I
do the following problem:Let f(x) = x^2 - 6x + 7. By
completing the square, or otherwise, determine> the unique
number a such that f(x) is a one-to-one function over the
domain> (-?,a] and over the domain [a,?).For this value of a,
nd the inverses of the functions f1 and f2 dened> by:f1(x) =
x^2 - 6x + 7, x?af2(x) = x^2 - 6x +7, x?a> TIA.f(x) = (x-3)^2
- 2Let g be the inverse of fy = (g(y) - 3)^2 - 2 y+2 = (g(y)
- 3)^2+/- sqrt(y+2) = g(y) - 33 +/- sqrt(y+2) = g(y)So g(y) =
3 + sqrt(y) and g(y) = 3-sqrt(y) are both inverses of fg(y) =
3+sqrt(y) >= 3 and g(y) = 3-sqrt(y) <= 3So g(y) = 3+sqrt(y)
is the inverse of f(x) for x >= 3and g(y) = 3 - sqrt(y) is
the inverse of f(x) for x <= 3~ Chris
How do I do the
following problem:Let f(x) = x^2 - 6x + 7. By completing the
square, or otherwise, determine> the unique number a such
that f(x) is a one-to-one function over the domain> (-?,a]
and over the domain [a,?).For this value of a,find the
inverses of the functions f1 and f2 dened> by:f1(x) = x^2 -
6x + 7, x?af2(x) = x^2 - 6x +7, x?a> TIA.What have you done
so far? Did you complete the square in x^2-6x+7 ? What did
you get?
> What have you done so far? Did you complete
the square in x^2-6x+7 ?> What did you get?In completing the
square I got (x-3)^2 - 2.My problem with the question is that
Im unsure exactly what it is askingand how a relates to the
inverses of the two functions.
> What have you done so
far? Did you complete the square in x^2-6x+7 ?> What did you
get?In completing the square I got (x-3)^2 - 2.My problem
with the question is that Im unsure exactly what it is
asking> and how a relates to the inverses of the two
functions.Look at the graph of any function. The inverse
function,if it exists, is created by swapping the x and y
axes,by reecting the graph through an imaginary
45-degreeline at y=x.A function cant be multi-valued: if you
draw a verticalline, it should only intersect the graph at
most once.So for a function to be invertible, any HORIZONTAL
lineshould only intersect the graph at most once.That means
you need to restrict your function to domainover which this
property holds. If you look at the wholegraph of a quadratic,
it has a symmetry. Every horizontalline that intersects it,
intersects it twice. Flip itaround and youd have a curve
with two y-values for everyx which is not a valid
function.Thats the purpose of the a: to restrict the domain
to aplace where you can ip the curve sideways and not havea
multi-valued curve. - Randy
> What have you done so
far? Did you complete the square in x^2-6x+7 ?> What did
you get?> In completing the square I got (x-3)^2 - 2.> My
problem with the question is that Im unsure exactly what it
is asking> and how a relates to the inverses of the two
functions.Look at the graph of any function. The inverse
function,> if it exists, is created by swapping the x and y
axes,> by reecting the graph through an imaginary 45-degree>
line at y=x.A function cant be multi-valued: if you draw a
vertical> line, it should only intersect the graph at most
once.> So for a function to be invertible, any HORIZONTAL
line> should only intersect the graph at most once.That means
you need to restrict your function to domain> over which this
property holds. If you look at the whole> graph of a
quadratic, it has a symmetry. Every horizontal> line that
intersects it, intersects it twice. With one important
exception, the horizontal through the vertex.> Flip it>
around and youd have a curve with two y-values for every> x
which is not a valid function.Thats the purpose of the a: to
restrict the domain to a> place where you can ip the curve
sideways and not have> a multi-valued curve. -
Randy
complete the square in x^2-6x+7 ?> What did you get?In
completing the square I got (x-3)^2 - 2.My problem with the
question is that Im unsure exactly what it is asking> and
how a relates to the inverses of the two functions.>For
some functions, like f(x) = m*x+b with m <> 0, there is only
one x for any y, so a uniquely dened inverse, hereg(x) =
(x-b)/m, can be found. For many y = f(x) functions there is
sometimes more than one x for some values of x, so that there
is no unique inverse. In these cases it is sometimes possible
to split up the domain (set of inputs) for the function into
pieces, possibly intervals, on which there is only only one x
for each y, and thenfind a separate inverse on each piece. For
example f(x) = x^2 has no total inverse, but the peice of f(x)
= x^2 for which x >=0, has inverse g(x) = sqrt(x), and the
piece for which x <= 0 has inverse h(x) = - sqrt(x). Consider
that equation y = (x-3)^ - 2 is a parabola with vertex at
(3,-2) and vertical line of symmetry {(x,y) | x = 3 }.It
might help to sketch this graph, manually or on a graphic
calculator. Have you done this?
> What have you done so
far? Did you complete the square in x^2-6x+7 ?>> What did you
get?>> In completing the square I got (x-3)^2 - 2.>> My
problem with the question is that Im unsure exactly what it
is> asking and how a relates to the inverses of the two
functions.-- Ignacio Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
=Certain
symbols didnt come out right.> How do I do the following
problem:>> Let f(x) = x^2 - 6x + 7. By completing the square,
or otherwise, determine> the unique number a such that f(x) is
a one-to-one function over thedomain> (-?,a] and over the
domain [a,?).? is innity> For this value of a,find the
inverses of the functions f1 and f2 dened> by:>> f1(x) = x^2
- 6x + 7, x?a? is less than or equal to> f2(x) = x^2 - 6x +7,
x?a? is greater than or equal to> TIA.>>
When one has
intuited an innite proof, which cant be> communicated to
someone lacking this sort of intuition, one can> nevertheless
use language to sort of nudge other people with> necessary
intuitions into intuiting the proof construction.Reminds me
of something--The rst evening Zen master Momataki arrived at
Rain Mountain Monastery the monks requested their new master
show them the way. When Momataki assented with a nod and
pointed to the rising moon all but two of the assembled
looked at his nger.I like the anecdote, but to my lay self
it doesnt sound much like a way of doing maths ;)
[.snip.]>>My point in using the example is to show that
certain posters are full>>of it, as they *have* been
attacking rather basic algebra, especially>>with one
particularly annoying tactic.>>Its not such a complicated
thing in analysis when you have terms>>independent of a
particular variable to set that variable to 0 to>>clear it
out, but these posters attacked that basic technique
and>>claimed that m=0 was a special case!!!> Your basic
technique seems to be a mess. Whenever people ask you>> what
the term is, instead of answering you just start new>>
threads. You claim something changes, when people ask you
what and>> how, you dont answer, you just start new threads
and repeat the same>> claim.>>Thats a rather interesting
falsehood.Really? Lets see:>A key expression has long been
g_1 = a_1 x + uf, where, though not>seen, certain variables
are dependent on m, such that some posters>want me to write
g_1(m) = a_1(m) x + uf, and at m=0, g_1 = uf, as uf>is the
independent term.>>So it is not true that I dont answer what
the independent term is.[changes], you just start new threads
and repeat the same claim.What changed?I repeatedly explain.
I have P(m) with a factor that is f^2 andanother that is g_1,
where g_1 varies with P(m), but has uf as anindependent term
i.e. a term that doesnt change as m changes.Given that P(m)
also has an independent term that is u^2 f^2(3x + uf),I now
move to P(m)/f^2, where I know that the independent terms
haveto have changed, as in one case you have uf, but with
P(m)/f^2, theindependent term is now u^2(3x + uf), so theres
been a clear andvisible change. The f had to divide off, so
now the independent termfor that factor is u. That is, it was
uf, now it has to be u, whenyou divide P(m) by f^2.Given the
condition that f is coprime to 3, x, and I toss in u, justin
case, its clear that u^2(3x + uf) is coprime to f.So now,
from the terms *independent of m* I know that dividing P(m)
byf^2 means that f is divided off of uf, which as an
independent term,is not affected by the value of m, so then
in general g_1 has f as afactor.>> I went to the Hong Kong
forum you were recommending. I noted that as>> soon as a
rather basic question about your basic technique was>> asked,
you ran away like a frightened little kid. No wonder you>>
havent been posting there any more.>>I last posted there a
few days ago, which doesnt mean I wont post>there
again.>>The poster doesnt give a link so Ill give one for
readers who wish>to see the full discussion on the Hong Kong
website, where theres an>additional advantage as they allow
use of LaTeX.>>See
http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782>> Let
me quote it here, maybe youll have the courage to actually
answer> As there are continued postings it might help to
consider the>> alternative.> So, consider again $P(m)=g_1 g_2
g_3$, $g_1 = a_1 x + uf$ where its>> established that $uf$ is
independent of $m$ as determined by setting>> $m=0$, note
$P(0)=u^2f^2(3x + uf)$.> Now suppose that instead of $f$ as a
factor $g_1$ had $sqrt{f}$ when>> $m=5$, then youd have>
$g_1/sqrt{f} = a_1/sqrt{f} x + usqrt{f}$> so the independent
term has changed.> You were asked which function had an
independent term that changed,>> what it changed from, and
what it changed to.>>Oh, I remember that, as some poster kept
going on and on asking the>same question after I
answered.Really? Where was the answer where you said WHAT
changed, and from> what to what? You replied, but you never
->answered<-. As usual.In reality, I repeatedly explain, and
the posters like this one just*claim* I dont, and people
seem to believe him, which I think he seesas approval of his
lying. Again, I have P(m) with a factor that isf^2 and
another that is g_1, where g_1 varies with P(m), but has uf
asan independent term i.e. a term that doesnt change as m
changes.Given that P(m) also has an independent term that is
u^2 f^2(3x + uf),I now move to P(m)/f^2, where I know that
the independent terms haveto have changed, as in one case you
have uf, but with P(m)/f^2, theindependent term is now u^2(3x
+ uf), so theres been a clear andvisible change.Given the
condition that f is coprime to 3, x, and I toss in u, justin
case, its clear that u^2(3x + uf) is coprime to f.So now,
from the terms *independent of m* I know that dividing P(m)
byf^2 means that f is divided off of uf, which as an
independent term,is not affected by the value of m, so then
in general g_1 has f as afactor.James Harris
= [.snip.]> I
have P(m) with a factor that is f^2 and>another that is g_1,
where g_1 varies with P(m), but has uf as an>independent term
i.e. a term that doesnt change as m changes.>>Given that P(m)
also has an independent term that is u^2 f^2(3x + uf),>I now
move to P(m)/f^2, where I know that the independent terms
have>to have changed, as in one case you have uf, but with
P(m)/f^2, the>independent term is now u^2(3x + uf), so
theres been a clear and>visible change. The f had to divide
off, so now the independent term>for that factor is u. That
is, it was uf, now it has to be u, when>you divide P(m) by
f^2.>>Given the condition that f is coprime to 3, x, and I
toss in u, just>in case, its clear that u^2(3x + uf) is
coprime to f.>>So now, from the terms *independent of m* I
know that dividing P(m) by>f^2 means that f is divided off of
uf, which as an independent term,>is not affected by the value
of m, so then in general g_1 has f as a>factor.I have P(m) =
m^2 + 3m + 2, with a factor that is 2, and another whichis
g_1 = (m+2), where g_1 varies with P(m), but has 2 as
anindependent term i.e. a term that doesnt change as m
changes.Given that P(m) also has an independent term that is
2, I now move toP(m)/2, where I know that the independent
terms have to have changed,as in one case you have 2, but
with P(m)/2 the independent term is now1, so theres been a
clear and visible change. The 2 had to divideoff, so now the
independent term for that factor is 1. That is, it was2, now
it has to be 1, when you divide P(m) by 2.Given the condition
that 2 is coprime to 1, it is clear that 1 iscoprime to 2.So
now, from the terms *independent of m* I know that dividing
P(m) by2 means that 2 is divided off 2, which as an
independent term is notaffected by the value of m, so then in
general g_1 has 2 as a factor.[Here, P(m) = (m+2)(m+1), g_1(m)
= m+2, g_2(m) = m+1].So, turns out that for every integer
value of m, m+2 is a multiple of2.<Monty
Burns>Excellent...</Monty Burns>The error lies in the nal
paragraph, as it always has. I know thatdividing P(m) by f^2
means that f is divided of uf is an unjustied(and in fact
false) assertion. The claim So then in general g_1 has fas a
factor is unjustied, and in general
false.
a reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
gures few readers can criticize. A great many people are
staggered to this extent, that they imagine there must be the
indenite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
[.snip.]> I have P(m) with a factor that is f^2 and>another
that is g_1, where g_1 varies with P(m), but has uf as
an>independent term i.e. a term that doesnt change as m
changes.>>Given that P(m) also has an independent term that
is u^2 f^2(3x + uf),>I now move to P(m)/f^2, where I know
that the independent terms have>to have changed, as in one
case you have uf, but with P(m)/f^2, the>independent term is
now u^2(3x + uf), so theres been a clear and>visible change.
The f had to divide off, so now the independent term>for that
factor is u. That is, it was uf, now it has to be u, when>you
divide P(m) by f^2.>>Given the condition that f is coprime to
3, x, and I toss in u, just>in case, its clear that u^2(3x +
uf) is coprime to f.>>So now, from the terms *independent of
m* I know that dividing P(m) by>f^2 means that f is divided
off of uf, which as an independent term,>is not affected by
the value of m, so then in general g_1 has f as a>factor.I
have P(m) = m^2 + 3m + 2, with a factor that is 2, and
another which> is g_1 = (m+2), where g_1 varies with P(m),
but has 2 as an> independent term i.e. a term that doesnt
change as m changes.Heres a fascinating case of Arturo
Magidin clearly lying by trying touse something true only in
the ring of *integers*.In the actual proof that I use you can
see f^2 as a factor asP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3
- 3(-1+mf^2 )x u^2 + u^3 f)yet here you see the poster using
an example where 2 is NOT a visiblefactor.Look further and
see dramatic desperation. > Given that P(m) also has an
independent term that is 2, I now move to> P(m)/2, where I
know that the independent terms have to have changed,> as in
one case you have 2, but with P(m)/2 the independent term is
now> 1, so theres been a clear and visible change. The 2 had
to divide> off, so now the independent term for that factor is
1. That is, it was> 2, now it has to be 1, when you divide
P(m) by 2.Given the condition that 2 is coprime to 1, it is
clear that 1 is> coprime to 2.So now, from the terms
*independent of m* I know that dividing P(m) by> 2 means that
2 is divided off 2, which as an independent term is not>
affected by the value of m, so then in general g_1 has 2 as a
factor.> [Here, P(m) = (m+2)(m+1), g_1(m) = m+2, g_2(m) =
m+1].So, turns out that for every integer value of m, m+2 is
a multiple of> 2.<Monty Burns>> Excellent...> </Monty
Burns>And yes, Arturo Magidin is a graduate of Berkeley
Universitys Ph.Dprogram in mathematics, which tells me a lot
about what math people*actually* learn there.Clearly Berekeley
is NOT the place to get a math degree, as shown byArturo
Magidin.> The error lies in the nal paragraph, as it always
has. I know that> dividing P(m) by f^2 means that f is
divided of uf is an unjustied> (and in fact false)
assertion. The claim So then in general g_1 has f> as a
factor is unjustied, and in general false.The error is in
Arturo Magidin using a result only valid in the ringof
integers, as what he claims isnt even true in the ring
ofalgebraic integers.Hes rather pathetic, now isnt he?>
Arturo Magidin> magidin@math.berkeley.eduAnd whats
particularly pathetic is his emphasizing his
Berkeleyeducation to the shame of that institution, the place
that gave thisanti-mathematician his degree.I think their
entire math department should be shut down.James Harris
And yes, Arturo Magidin is a graduate of Berkeley
Universitys Ph.D> program in mathematics, which tells me a
lot about what math people> *actually* learn there.And yes,
James Harris is an idiot who *still* doesnt know its not
calledBerkeley University in spite of having been told this
before.> Clearly Berekeley is NOT the place to get a math
degree, as shown by> Arturo Magidin.Clearly Vanderbilt is NOT
the place to learn to get your facts straight,as shown by
James Harris.-- Wayne Brown (HPCC #1104) | When your tails
in a crack, you improvisefwbrown@bellsouth.net | if youre
good enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
>> [.snip.] I have P(m) with a factor
that is f^2 and>>another that is g_1, where g_1 varies with
P(m), but has uf as an>>independent term i.e. a term that
doesnt change as m changes.>>Given that P(m) also has an
independent term that is u^2 f^2(3x + uf),>>I now move to
P(m)/f^2, where I know that the independent terms have>>to
have changed, as in one case you have uf, but with P(m)/f^2,
the>>independent term is now u^2(3x + uf), so theres been a
clear and>>visible change. The f had to divide off, so now
the independent term>>for that factor is u. That is, it was
uf, now it has to be u, when>>you divide P(m) by
f^2.>>Given the condition that f is coprime to 3, x, and I
toss in u, just>>in case, its clear that u^2(3x + uf) is
coprime to f.>>So now, from the terms *independent of m* I
know that dividing P(m) by>>f^2 means that f is divided off
of uf, which as an independent term,>>is not affected by the
value of m, so then in general g_1 has f as a>>factor. I
have P(m) = m^2 + 3m + 2, with a factor that is 2, and another
which>> is g_1 = (m+2), where g_1 varies with P(m), but has 2
as an>> independent term i.e. a term that doesnt change as m
changes.>>Heres a fascinating case of Arturo Magidin clearly
lyingHeres a fascinating case of James Harris engaging in ad
hominemsbecause his math fails him.> by trying to>use
something true only in the ring of *integers*.Nowhere in your
presentation do you use anything about the fact thatyou are in
the ring of algebraic integers. >In the actual proof that I
use you can see f^2 as a factor as>>P(m) = f^2((m^3 f^4 -
3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)>>yet here you
see the poster using an example where 2 is NOT a
visible>factor.So what? 2 is a factor of P(m) = m^2+3m+2 for
every integer. Do youdeny that?Nowhere in your argument do
you requre the factor to be visible. Youonly require it to be
a factor.>Look further and see dramatic desperation.No need;
you demonstrate the desperation quite thoroghly.>> Given that
P(m) also has an independent term that is 2, I now move to>>
P(m)/2, where I know that the independent terms have to have
changed,>> as in one case you have 2, but with P(m)/2 the
independent term is now>> 1, so theres been a clear and
visible change. The 2 had to divide>> off, so now the
independent term for that factor is 1. That is, it was>> 2,
now it has to be 1, when you divide P(m) by 2. Given the
condition that 2 is coprime to 1, it is clear that 1 is>>
coprime to 2. So now, from the terms *independent of m*
I know that dividing P(m) by>> 2 means that 2 is divided off
2, which as an independent term is not>> affected by the
value of m, so then in general g_1 has 2 as a factor.>
[Here, P(m) = (m+2)(m+1), g_1(m) = m+2, g_2(m) = m+1].
So, turns out that for every integer value of m, m+2 is a
multiple of>> 2. <Monty Burns> Excellent...>> </Monty
Burns>And yes, Arturo Magidin is a graduate of Berkeley
Universitys Ph.D>program in mathematics, which tells me a
lot about what math people>*actually* learn there.I do not
see anything that addresses the math. Why?>Clearly Berekeley
is NOT the place to get a math degree, as shown by>Arturo
Magidin.I do not see anything that addresses the math?>> The
error lies in the nal paragraph, as it always has. I know
that>> dividing P(m) by f^2 means that f is divided of uf is
an unjustied>> (and in fact false) assertion. The claim So
then in general g_1 has f>> as a factor is unjustied, and in
general false.>>The error is in Arturo Magidin using a result
only valid in the ring>of integers, as what he claims isnt
even true in the ring of>algebraic integers.What is this
mystery result?For every integer value of m, P(m) is
divisible by 2 in the ring ontegers and in the ring of
algebraic integers. Your argument onlyrelies on integer
values of m, so there is no difference between thefalse
argument above and your false argument. [more desperate ad
hominems
deleted]
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A mans capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
gures few readers can criticize. A great many people are
staggered to this extent, that they imagine there must be the
indenite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
think their entire math department should be shut down.Are you
like this with the people at work, too? Im curious?
I
think their entire math department should be shut down.Are
you like this with the people at work, too? Im
curious?Mathematicians have proven to be...weak opponents.Im
bored.Do I have to challenge your physical lives before any of
you rise to the challenge?Dont worry, why should I
bother?Youre going to die anyway.James Harris
Mathematicians have proven to be...weak opponents.>> Im
bored.>> Do I have to challenge your physical lives before
any of you rise to the challenge?>> Dont worry, why should I
bother?>> Youre going to die anyway.>> James HarrisWatch out
everyone! Hes getting ready to blow!!!--A fool and his proof
are soon refuted.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
> I think their
entire math department should be shut down.>> Are you like
this with the people at work, too? Im curious?>>
Mathematicians have proven to be...weak opponents.>> Im
bored.>> Do I have to challenge your physical lives before
any of you rise to thechallenge?>> Dont worry, why should I
bother?>> Youre going to die anyway.> James HarrisYou are
so immature that it is scary. If you would harm people
overmathematics, there is something SERIOUSLY wrong with
you.David Moran
> I think their entire math
department should be shut down.>> Are you like this with
the people at work, too? Im curious?>> Mathematicians have
proven to be...weak opponents.>> Im bored.>> Do I have to
challenge your physical lives before any of you rise to the>
challenge?>> Dont worry, why should I bother?>> Youre going
to die anyway.> James HarrisYou are so immature that it is
scary. If you would harm people over> mathematics, there is
something SERIOUSLY wrong with you.David MoranYou dont
understand as *I* wont physically harm mathematicians.Why
bother? Youre going to die anyway.James Harris
[.snip.]>My point in using the example is to show that
certain posters are full>of it, as they *have* been attacking
rather basic algebra, especially>with one particularly
annoying tactic.>>Its not such a complicated thing in
analysis when you have terms>independent of a particular
variable to set that variable to 0 to>clear it out, but these
posters attacked that basic technique and>claimed that m=0 was
a special case!!!> Your basic technique seems to be a mess.
Whenever people ask you> what the term is, instead of
answering you just start new> threads. You claim something
changes, when people ask you what and> how, you dont answer,
you just start new threads and repeat the same>
claim.>>Thats a rather interesting falsehood. Really?
Lets see:A key expression has long been g_1 = a_1 x +
uf, where, though not>>seen, certain variables are dependent
on m, such that some posters>>want me to write g_1(m) =
a_1(m) x + uf, and at m=0, g_1 = uf, as uf>>is the
independent term.>>So it is not true that I dont answer
what the independent term is. [changes], you just start
new threads and repeat the same claim. What changed?>>I
repeatedly explain. No, you repeatedly repeat. You do not
explain the point where peopleask.> I have P(m) with a factor
that is f^2 and>another that is g_1, where g_1 varies with
P(m), but has uf as an>independent term i.e. a term that
doesnt change as m changes.Meaning, g_1(0) = uf. Yes.>Given
that P(m) also has an independent term that is u^2 f^2(3x +
uf),Meaning, P(0) = u^2 f^2(3x+uf); yes.>I now move to
P(m)/f^2, where I know that the independent terms have>to
have changed, as in one case you have uf, but with P(m)/f^2,
the>independent term is now u^2(3x + uf), so theres been a
clear and>visible change.What a waste of space. You mean:
P(0) = u^2 f^2(3x+uf) and P(0)/f^2 = u^2(3x+uf).Yes.> The f
had to divide off, so now the independent term>for that
factor is u. That is, it was uf, now it has to be u, when>you
divide P(m) by f^2.>>Given the condition that f is coprime to
3, x, and I toss in u, just>in case, its clear that u^2(3x +
uf) is coprime to f.>>So now, from the terms *independent of
m* I know that dividing P(m) by>f^2 means that f is divided
off of uf, which as an independent term,>is not affected by
the value of m, so then in general g_1 has f as a>factor.And
that last part is what you neither explain nor do I follow;
norhave I seen anybody who claims to follow. What you need to
explain isSo then in general g_1 has f as a factor.Thats the
part I dont get.In addition, in the Hong Kong forum this is
exactly what you
said:http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782(
message from 11/10/03-23:18:06) So, consider again $P(m)=g_1
g_2 g_3$, $g_1 = a_1 x + uf$ where its established that $uf$
is independent of $m$ as determined by setting $m=0$, note
$P(0)=u^2f^2(3x + uf)$. Now suppose that instead of $f$ as a
factor $g_1$ had $sqrt{f}$ when $m=5$, then youd have
$g_1/sqrt{f} = a_1/sqrt{f} x + usqrt{f}$ so the independent
term has changed.And I ask: what independent term changed?
What did it change from, andwhat did it change to, assuming
that it was the case that instead of fas a factor g_1 had
sqrt(f) when m=5?You continue: Now suppose that at $m=7$ the
factor is $f^{2/5}$, then again the independent term will
change.So I ask again: what independent will term changed,
assuming that at m=7the factor is f^{2/5}? The independent
term of g_1? Of g_1/f? Well then, the independent term would
clearly NOT be independent of $m$ if its value changed that
way depending on whether $m=0$, $m=5$, or $m=7$. Which is a
direct contradiction.So you are claiming that if g_1(5) was
divisible by sqrt(f) andg_1(5)/sqrt(f) was coprime to f, then
the independent term wouldchange. You claim this by looking at
g_1(5)/sqrt(f). But you had never looked at g_1(m)/sqrt(f)
before. So I am stillwondering just what it is you think you
accomjplished by looking atg_1(5)/sqrt(f), and what it was
you were comparing it to so you couldsay so the independent
term has changed.You dont explain, you just repeat over and
over the same thing.
reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
gures few readers can criticize. A great many people are
staggered to this extent, that they imagine there must be the
indenite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
> Well readers, did anyone see whats wrong with this
posters position? > He suggests > f_0(m).b^3 +
f_1(m).b^2 + f_2(m).b + f_3(m), > where the f_i(m)
are algebraic integer functions of m and f_0(0) = 1, >
f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0? > But the
poster fails to point out that his position requires that for
> certain values of m and another important variable I call
f, that his > f_0(m) *must* have f as a factor, so its
basically a step function. > Is that the right word
step function? > I think not. I have no idea what you mean
with that term. Can you explain > what you mean? > Your
position requires that f_0(m) either be a unit or have f as
its > only non-unit factor, depending on certain conditions
as m varies from > 0 to positive innity.Yes. And so what? >
The condition is that if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4
- 3m^2 f^2 + 3m) > is irreducible over Q, by your position,
f_0(m) must have f as its > only non-unit factor.That is
correct in part. It must also have f as its only non-unit
factorin some cases when that polynomial *is* reducible. So
what? > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) > P(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf) > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2
f^2 + 3m). > R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f > R(m) = (b_1 x + u)(b_2 x +
u)(b_3 x + uf) > Here reducibility over rationals of the
cubic dening the as > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 -
3m^2 f^2 + 3m) > has been given as an issue, and the posters
position requires that his > f_0(m) vary such that its 1 if
that cubic is reducible over > rationals, and f, if its
irreducible over rationals. > See my other post. It is a
fallacy. With m=1 and f=2, the polynomial > in a is
reducible. The polynomial I come up with is: > 2b^3 + (25 +
sqrt(105))/2.b^2 + (-7 + sqrt(105))/2.b - 14. > That looks
ok as one of the roots is 1 as required, as one of the bs >
is 1. > If it is the correct polynomial, then *each* of the
coefcients has 2 > as a factor, but NOT in the ring of
algebraic integers, as the middle > coefcients are
arbitrarily excluded by the denition of algebraic > integers
from having 2 as a factor.I am *talking* about the ring of
algebraic integers. There are indeedrings where 2 is a
divisor of those two factors. But that is somethingcompletely
different. Such rings much more units. So lets say thatthere
is a ring (the object ring) where (25 + sqrt(105))/4 is in.
Why is(25 - sqrt(105))/4 *not* in that ring? What is the
essential differencebetween the two? > Those who doubt that
assessment can instead believe in the step > function, which
somehow varies the leading coefcient from f to 1 > based on
whether or not > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2
+ 3m) > is reducible over Q. > Note that for the example
above, it reduces to a linear and a > quadratic.Yes, and
f_0(m) is 2 in that case. > Now theres no *mathematical* way
to have a function f_0(m) which will > vary in such an extreme
manner depending on whether or not some cubic > is reducible
over Q or not, that is consistent with the other >
mathematical facts here.Why? I have been thinking about it.
It is a fact that f_0(m) is f when*all three* of the as
share a factor with f in the algebraic integers.In the case
above the algebraic integer factors shared are
2,(-sqrt(7)+sqrt(15))/2 and (-sqrt(7)-sqrt(15))/2. All three
non-unitsin the algebraic integers. (Note the case f=3 is
special as in thatcase the constant term of the polynomial is
divisible by f^3, and indeedall three of the as are divisible
by f, so f_0(m) = 1, I will not restatethe restriction f != 3
again, and also not the trivial restriction f != 1.) >
Depends on the roots a1 to a3, which two you divide by f, how
the resulting > bs can be written as a quotient of an
algebraic integer and a normal > integer (in lowest form),
and perhaps a bit more. It is *not* possible to > give a
general formula. But I have given the algorithm through which
you > can calculate that number for each m. > Thats
nonsense. Its not necessary to give the expression
explicitly > as given that only f is being divided off, it
follows that only f can > be the non-unit factor, if there is
one.You are not entirely right here. f_0(m) can also be any
multiple of f andany multiple of 1. But when the polynomial
is reduced to primitive withalgebraic integer coefcients, it
is indeed either 1 or f. > Remember, this poster is trying to
cast doubt on the factorization > (b_1 x + u)(b_2 x + u)(b_3
x + uf) > which comes from the factorization > (a_1 x +
uf)(a_2 x + uf)(a_3 x + uf) > where > (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf) = > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) > and the as are given by > a^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) > which is a
monic with integer coefcients.... > which works when m=1,
f=sqrt(2), so if you wish to believe this poster > you need
to believe in that odd function which varies from being a >
unit to having f as a factor dependent on the reducibility of
> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).You state
the wrong condition again. f_0(m) = f when the as all
threeshare a non-unit factor with f, it is 1 otherwise. All
three as *can*share a non-unit factor with f even when the
polynomial is reducible.But lets try it otherwise, to see
why this is so. Note that the productof the factors shared is
always f^2. > This poster either is unwilling to concede to
the math, or is > incapable of realizing the aw in his
position.Clearly this poster is unwilling to concede to the
math and incapable ofrealising the aw in his position. > The
reality is that a function with the behavior he needs doesnt
> exist in this context, as his f_0(m) cant vary in the way
he needs as > m goes from 0 to positive innity, where it is
a unit for certain > values, but switches to having f as its
only non-unit factor for > others.*Why* does such a function
not exist in this context? That is a prettystrange argument.
Such a function exists independent of context. > What Im
showing here is a outlandish consequence of the posters >
position, since this poster and others have continued to
argue against > the mathematical argument which shows that
the leading coefcient is > always monic as in fact two of
the as have a factor that is f, as > shown by their terms
independent of m, for all m.The leading coefcient is *not*
always 1 when you want algebraic integercoefcients. And you
need algebraic integer coefcients to be ableto tell whether
the bs are algebraic integers or not. You can make
theleading coefcient always 1 (trivially so), but in that
case you do notalways have all coefcients algebraic
integers. > You see, m=0 is NOT a special case.Very, very
special. Because it has the property that in that caseexactly
two of the as are divisible by f for every f, f vanishesalso
from the polynomial in a, so the two as that are 0 are
triviallydivisible by f. When m != 0 it can be shown that
when there areexactly two of the as divisible by f, that the
polynomial in a isreducible. (The other way around is false.)
This has been provenmany times, by various posters, both in
general and by specicexamples. Also, when m != 0, f does not
vanish from the polynomialin a.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
...> Well
readers, did anyone see whats wrong with this posters
position?> He suggests > f_0(m).b^3 +
f_1(m).b^2 + f_2(m).b + f_3(m),> where the f_i(m)
are algebraic integer functions of m and f_0(0) = 1,>
f_1(0) = -3, f_2(0) = 0 and f_3(0) = 0?> But the poster
fails to point out that his position requires that for>
certain values of m and another important variable I call f,
that his> f_0(m) *must* have f as a factor, so its
basically a step function.> Is that the right word step
function?> I think not. I have no idea what you mean with
that term. Can you explain> what you mean?> Your position
requires that f_0(m) either be a unit or have f as its> only
non-unit factor, depending on certain conditions as m varies
from> 0 to positive innity.Yes. And so what?Readers should
note that the poster admits that his position requiresa
function that is either a unit or has f as its only non-unit
factordepending on certain conditions as m varies from 0 to
positiveinnity. A condition follows. > The condition is
that if> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 +
3m)> is irreducible over Q, by your position, f_0(m) must
have f as its> only non-unit factor.That is correct in
part. It must also have f as its only non-unit factor> in
some cases when that polynomial *is* reducible. So
what?However, do you accept that also for *certain* values of
f, like ifthat polynomial is reducible into all linear factors
over rationals,that function would be monic?Do you accept that
as true?> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f)> P(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2
f^2 + 3m).> R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f> R(m) = (b_1 x + u)(b_2 x +
u)(b_3 x + uf)> Here reducibility over rationals of the
cubic dening the as> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4
- 3m^2 f^2 + 3m)> has been given as an issue, and the
posters position requires that his> f_0(m) vary such that
its 1 if that cubic is reducible over> rationals, and f,
if its irreducible over rationals.> See my other post.
It is a fallacy. With m=1 and f=2, the polynomial> in a is
reducible. The polynomial I come up with is:> 2b^3 + (25 +
sqrt(105))/2.b^2 + (-7 + sqrt(105))/2.b - 14.> That looks
ok as one of the roots is 1 as required, as one of the bs>
is 1.> If it is the correct polynomial, then *each* of the
coefcients has 2> as a factor, but NOT in the ring of
algebraic integers, as the middle> coefcients are
arbitrarily excluded by the denition of algebraic>
integers from having 2 as a factor.I am *talking* about the
ring of algebraic integers. There are indeed> rings where 2
is a divisor of those two factors. But that is something>
completely different. Such rings much more units. So lets
say that> there is a ring (the object ring) where (25 +
sqrt(105))/4 is in. Why is> (25 - sqrt(105))/4 *not* in that
ring? What is the essential difference> between the two?Your
intuition, or guesses about the math are not a concern.Im in
the process of showing readers just how illogical your
positionis by getting you to admit to certain facts that
would follow fromyour position.So far its established that
you believe that a function is forced tohave f as its only
non-unit factor dependent in some way on thereducibility over
rationals of the cubica^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2
f^2 + 3m)where further Im establishing that you understand
that if that cubicis reducible over Q into linear factors
that your hypotheticalfunction, called f_0(m) by you, would
have to be monic.> Those who doubt that assessment can
instead believe in the step> function, which somehow varies
the leading coefcient from f to 1> based on whether or not a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)> is
reducible over Q.> Note that for the example above, it
reduces to a linear and a> quadratic.Yes, and f_0(m) is 2
in that case.Here readers should note the posters belief
that the cubic deningthe bs now has a leading term that is
2, which is the value of f,which is consistent with previous
positions. > Now theres no *mathematical* way to have a
function f_0(m) which will> vary in such an extreme manner
depending on whether or not some cubic> is reducible over Q
or not, that is consistent with the other> mathematical
facts here.Why? I have been thinking about it. It is a fact
that f_0(m) is f when> *all three* of the as share a factor
with f in the algebraic integers.> In the case above the
algebraic integer factors shared are 2,>
(-sqrt(7)+sqrt(15))/2 and (-sqrt(7)-sqrt(15))/2. All three
non-units> in the algebraic integers. (Note the case f=3 is
special as in that> case the constant term of the polynomial
is divisible by f^3, and indeed> all three of the as are
divisible by f, so f_0(m) = 1, I will not restate> the
restriction f != 3 again, and also not the trivial
restriction f != 1.)The poster is apparently now trying to
come up with a reason for howthis function can jump between 1
and f, in such a peculiar manner.However, the reality is that
the ring of algebraic integers does nothave numbers required
to keep such oddities from occurring.> Depends on the roots
a1 to a3, which two you divide by f, how the resulting> bs
can be written as a quotient of an algebraic integer and a
normal> integer (in lowest form), and perhaps a bit more.
It is *not* possible to> give a general formula. But I have
given the algorithm through which you> can calculate that
number for each m.> Thats nonsense. Its not necessary to
give the expression explicitly> as given that only f is
being divided off, it follows that only f can> be the
non-unit factor, if there is one.You are not entirely right
here. f_0(m) can also be any multiple of f and> any multiple
of 1. But when the polynomial is reduced to primitive with>
algebraic integer coefcients, it is indeed either 1 or
f.Thats not true, and is yet another basic error from this
poster.In fact, only f itself is available as a non-unit
factor for theleading coefcient of the bs by his
position.That he does not know that is just another
indication of limitedmathematical understanding from this
poster.> Remember, this poster is trying to cast doubt on
the factorization> (b_1 x + u)(b_2 x + u)(b_3 x + uf)>
which comes from the factorization> (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)> where> (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf) = > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x
u^2 + u^3 f)> and the as are given by> a^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)> which is a
monic with integer coefcients.> ...> which works when m=1,
f=sqrt(2), so if you wish to believe this poster> you need
to believe in that odd function which varies from being a>
unit to having f as a factor dependent on the reducibility
of> a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).You
state the wrong condition again. f_0(m) = f when the as all
three> share a non-unit factor with f, it is 1 otherwise. All
three as *can*> share a non-unit factor with f even when the
polynomial is reducible.> But lets try it otherwise, to see
why this is so. Note that the product> of the factors shared
is always f^2.Ive stated the strong condition, which is that
if the cubic isirreducible over Q, your claims would require
that the leading term ofthe cubic dening the bs is
non-monic with f as its only non-unitfactor, which is
easier.So far the poster has never given anything like a
rational explanationfor why hed suppose itd matter for some
cubics that are reducible,while not for others, as I can give
m=1, f=sqrt(2), to show a cubicthat is reducible in the same
way as his example, where its clearthat only two of the as
have algebraic integers, and the cubicdening the bs isb^3 +
(1+sqrt(2))b^2 + (sqrt(2)-1)b - 1.Basically the posters
position requires a rather odd mathematics withan
unexplainable quirkiness.That kind of mathematics sounds like
how people are, not how a logicaland consistent system
operates.I remind of Ptolemys circles.> This poster either
is unwilling to concede to the math, or is> incapable of
realizing the aw in his position.Clearly this poster is
unwilling to concede to the math and incapable of> realising
the aw in his position.The irrationality of believing in a
function which jumps in valuesbased on reducibility over
rationals is so striking that Ifind itrather interesting that
Dik T. Winter is willing to keep debatingabout it.Then again,
its kind of interesting to see how long such a poster cango
on with such a public display.> The reality is that a
function with the behavior he needs doesnt> exist in this
context, as his f_0(m) cant vary in the way he needs as> m
goes from 0 to positive innity, where it is a unit for
certain> values, but switches to having f as its only
non-unit factor for> others.*Why* does such a function not
exist in this context? That is a pretty> strange argument.
Such a function exists independent of context.You *can* have
step functions which behave in surprising ways.However, the
mathematics here does not allow for the behavior Dik T.Winter
claims.mathematical logic.In claiming its validity he has to
rely on the aw with thedenition of algebraic integers,
which is like trying to walk on abroken leg.> What Im
showing here is a outlandish consequence of the posters>
position, since this poster and others have continued to
argue against> the mathematical argument which shows that
the leading coefcient is> always monic as in fact two of
the as have a factor that is f, as> shown by their terms
independent of m, for all m.The leading coefcient is *not*
always 1 when you want algebraic integer> coefcients. And
you need algebraic integer coefcients to be able> to tell
whether the bs are algebraic integers or not. You can make
the> leading coefcient always 1 (trivially so), but in that
case you do not> always have all coefcients algebraic
integers.Here the poster relies on the broken leg.> You
see, m=0 is NOT a special case.Very, very special. Because it
has the property that in that case> exactly two of the as are
divisible by f for every f, f vanishes> also from the
polynomial in a, so the two as that are 0 are trivially>
divisible by f. When m != 0 it can be shown that when there
are> exactly two of the as divisible by f, that the
polynomial in a is> reducible. (The other way around is
false.) This has been proven> many times, by various posters,
both in general and by specic> examples. Also, when m != 0, f
does not vanish from the polynomial> in a.Whats established
is that Dik T. Winter believes in this odd andinconsistent
function which goes from being a unit to having only f asits
non-unit factor in a way that he cant even explain.Hes
given an example which he believes has 2 as a leading
coefcientwhen in fact all of the coefcients have 2 as a
factor, in a properring.James Harris
=... > Your position
requires that f_0(m) either be a unit or have f as its >
only non-unit factor, depending on certain conditions as m
varies from > 0 to positive innity. > Yes. And so what?
> Readers should note that the poster admits that his
position requires > a function that is either a unit or has f
as its only non-unit factor > depending on certain conditions
as m varies from 0 to positive > innity. A condition
follows.Yes, I still see no problem with such a function.  The condition is that if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3
f^4 - 3m^2 f^2 + 3m) > is irreducible over Q, by your
position, f_0(m) must have f as its > only non-unit factor.
> That is correct in part. It must also have f as its only
non-unit factor > in some cases when that polynomial *is*
reducible. So what? > However, do you accept that also for
*certain* values of f, like if > that polynomial is reducible
into all linear factors over rationals, > that function would
be monic? > Do you accept that as true?Yes, why not? As I
have stated: when all as have common factors with fthe
leading term of the resulting polynomial will be f. If it
reducesto three linear terms over Q, trivially only two as
can have common factorsof f, as f is a prime in the integral
part of Q (Z). The problem is whenthere are *not* three
linear factors. Because in that case not all asare integers,
but at least two of them are algebraic integers. And, whilef
is prime in the integers, it is not prime in the algebraic
integers. > I am *talking* about the ring of algebraic
integers. There are indeed > rings where 2 is a divisor of
those two factors. But that is something > completely
different. Such rings much more units. So lets say that >
there is a ring (the object ring) where (25 + sqrt(105))/4 is
in. Why is > (25 - sqrt(105))/4 *not* in that ring? What is
the essential difference > between the two? > Your
intuition, or guesses about the math are not a concern.I have
no intuition in the question, not guesses. That is why I pose
aquestion. > Im in the process of showing readers just how
illogical your position > is by getting you to admit to
certain facts that would follow from > your position. > So
far its established that you believe that a function is
forced to > have f as its only non-unit factor dependent in
some way on the > reducibility over rationals of the cubic  a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) >
where further Im establishing that you understand that if
that cubic > is reducible over Q into linear factors that
your hypothetical > function, called f_0(m) by you, would
have to be monic.Yes. I still see no problem. There are
stranger functions in mathematics. > Those who doubt that
assessment can instead believe in the step > function,
which somehow varies the leading coefcient from f to 1 >
based on whether or not > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4
- 3m^2 f^2 + 3m) > is reducible over Q. > Note that for
the example above, it reduces to a linear and a >
quadratic. > Yes, and f_0(m) is 2 in that case. > Here
readers should note the posters belief that the cubic
dening > the bs now has a leading term that is 2, which is
the value of f, > which is consistent with previous
*mathematical* way to have a function f_0(m) which will >
vary in such an extreme manner depending on whether or not
some cubic > is reducible over Q or not, that is consistent
with the other > mathematical facts here. > Why? I have
been thinking about it. It is a fact that f_0(m) is f when >
*all three* of the as share a factor with f in the algebraic
integers. > In the case above the algebraic integer factors
shared are 2, > (-sqrt(7)+sqrt(15))/2 and
(-sqrt(7)-sqrt(15))/2. All three non-units > in the algebraic
integers. (Note the case f=3 is special as in that > case the
constant term of the polynomial is divisible by f^3, and
indeed > all three of the as are divisible by f, so f_0(m) =
1, I will not restate > the restriction f != 3 again, and also
not the trivial restriction f != 1.) > The poster is
apparently now trying to come up with a reason for how > this
function can jump between 1 and f, in such a peculiar
manner.There are enough functions that jump up and down. So
what? Her we haveanother function. > However, the reality is
that the ring of algebraic integers does not > have numbers
required to keep such oddities from occurring.The ring of
algebraic integers does indeed not have numbers jumping up
anddown. But we were talking about a function. > Thats
nonsense. Its not necessary to give the expression
explicitly > as given that only f is being divided off, it
follows that only f can > be the non-unit factor, if there
is one. > You are not entirely right here. f_0(m) can also
be any multiple of f and > any multiple of 1. But when the
polynomial is reduced to primitive with > algebraic integer
coefcients, it is indeed either 1 or f. > Thats not true,
and is yet another basic error from this poster. > In fact,
only f itself is available as a non-unit factor for the >
leading coefcient of the bs by his position.Eh? Given the
polynomial (x - 1) which has the root x = 1, so the
leadingcoefcient is 1. (2x - 2) has the same root, as has
(3x - 3). Youprimitive in the paragraph above. > That he does
not know that is just another indication of limited >
mathematical understanding from this poster.That he does not
know that is just another indication of limitedreading
capabilities from this poster. > You state the wrong
condition again. f_0(m) = f when the as all three > share a
non-unit factor with f, it is 1 otherwise. All three as
*can* > share a non-unit factor with f even when the
polynomial is reducible. > But lets try it otherwise, to see
why this is so. Note that the product > of the factors shared
is always f^2. > Ive stated the strong condition, which is
that if the cubic is > irreducible over Q, your claims would
require that the leading term of > the cubic dening the bs
is non-monic with f as its only non-unit > factor, which is
easier. > So far the poster has never given anything like a
rational explanation > for why hed suppose itd matter for
some cubics that are reducible, > while not for others, as I
can give m=1, f=sqrt(2), to show a cubic > that is reducible
in the same way as his example, where its clear > that only
two of the as have algebraic integers, and the cubic >
dening the bs isI thought all the time that f was integral?
Even prime? > b^3 + (1+sqrt(2))b^2 + (sqrt(2)-1)b - 1. >
Basically the posters position requires a rather odd
mathematics with > an unexplainable quirkiness.What is odd
about it, what is a quirkiness? > This poster either is
unwilling to concede to the math, or is > incapable of
realizing the aw in his position. > Clearly this poster is
unwilling to concede to the math and incapable of > realising
the aw in his position. > The irrationality of believing
in a function which jumps in values > based on reducibility
over rationals is so striking that Ifind it > rather
interesting that Dik T. Winter is willing to keep debating >
about it.You believe such a function does not exist? Please
sleep on. > The reality is that a function with the
behavior he needs doesnt > exist in this context, as his
f_0(m) cant vary in the way he needs as > m goes from 0 to
positive innity, where it is a unit for certain > values,
but switches to having f as its only non-unit factor for >
others. > *Why* does such a function not exist in this
context? That is a pretty > strange argument. Such a function
exists independent of context. > You *can* have step
functions which behave in surprising ways. > However, the
mathematics here does not allow for the behavior Dik T. >
Winter claims.Can you *please* tell us why such a function is
not allowed her? > In claiming its validity he has to rely on
the aw with the > denition of algebraic integers, which is
like trying to walk on a > broken leg.The broken leg is
yours. You just *want* that exactly two of the asare not
co-prime to f. When it is shown that such is not possible
inthe algebraic integers you just go away. > The leading
coefcient is *not* always 1 when you want algebraic integer
> coefcients. And you need algebraic integer coefcients to
be able > to tell whether the bs are algebraic integers or
not. You can make the > leading coefcient always 1
(trivially so), but in that case you do not > always have all
coefcients algebraic integers. > Here the poster relies on
the broken leg.Did I not show that is not possible in the
algebraic integers, and justbecause you *whish* it to be true
you are trying to go to another ring,Where is the broken leg?
> Very, very special. Because it has the property that in
that case > exactly two of the as are divisible by f for
every f, f vanishes > also from the polynomial in a, so the
two as that are 0 are trivially > divisible by f. When m !=
0 it can be shown that when there are > exactly two of the
as divisible by f, that the polynomial in a is > reducible.
(The other way around is false.) This has been proven > many
times, by various posters, both in general and by specic >
examples. Also, when m != 0, f does not vanish from the
polynomial > in a. > Whats established is that Dik T.
Winter believes in this odd and > inconsistent function which
goes from being a unit to having only f as > its non-unit
factor in a way that he cant even explain.You state
inconsisten. Inconsistent with what exactly? > Hes given an
example which he believes has 2 as a leading coefcient >
when in fact all of the coefcients have 2 as a factor, in a
proper > ring.Ah. So you do not falsify my statements at all.
What you imply here isthat I was correct, but you wish to move
to another ring. Fine. Pleasestate the denition of the ring
*such that it can be determined whethera particular number
belongs to the ring or not*. As long as you have notdone the
latter, you have no ring.So much verbiage by James to simply
state: you are correct, but you arenot working in the ring
where I am working.-- dik t. winter, cwi, kruislaan 413, 1098
sj amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland; http://www.cwi.nl/~dik/
...>
Your position requires that f_0(m) either be a unit or have f
as its> only non-unit factor, depending on certain
conditions as m varies from> 0 to positive innity.>
Yes. And so what?> Readers should note that the poster
admits that his position requires> a function that is
either a unit or has f as its only non-unit factor>
depending on certain conditions as m varies from 0 to
positive> innity. A condition follows.Yes, I still see no
problem with such a function.Then give a *single* function in
ALL of mathematics which behaves asyou wish.Youre making up
some wacky mathematics.For readers who havent kept up, this
poster needs a function which isa unit whena^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)is fully
reducible into all linear factors over rationals, which has
acomplex (and so far unexplained) behavior when its
reducible into alinear and a quadratic, where sometimes its
unit or else it has onlyf as a non-unit factor, which would
always have f as a non-unit factorifa^3 + 3(-1+mf^2)a^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m)is irreducible over rationals. The condition is that if> a^3 + 3(-1+mf^2)a^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m)> is irreducible over Q, by
your position, f_0(m) must have f as its> only non-unit
factor.> That is correct in part. It must also have f as
its only non-unit factor> in some cases when that
polynomial *is* reducible. So what?> However, do you accept
that also for *certain* values of f, like if> that
polynomial is reducible into all linear factors over
rationals,> that function would be monic?> Do you accept
that as true?Yes, why not? As I have stated: when all as
have common factors with f> the leading term of the resulting
polynomial will be f. If it reduces> to three linear terms
over Q, trivially only two as can have common factors> of f,
as f is a prime in the integral part of Q (Z). The problem is
when> there are *not* three linear factors. Because in that
case not all as> are integers, but at least two of them are
algebraic integers. And, while> f is prime in the integers,
it is not prime in the algebraic integers.Huh? It looks like
the poster conceded that if a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4
- 3m^2 f^2 + 3m)is reducible over rationals into all linear
factors that his positionis bogus.No reason is given. Then
the poster basically babbles at the end.Yes, f is prime in
the integers and not prime in algebraic integers,but that has
no relevance here.Its kind of interesting to watch this
person jump, dance and dovarious gyrations as he tries to
shoehorn mathematics.Lets see what else interesting comes
from him.James Harris
=... > Yes. And so what? >
Readers should note that the poster admits that his position
requires > a function that is either a unit or has f as its
only non-unit factor > depending on certain conditions as m
varies from 0 to positive > innity. A condition follows.  Yes, I still see no problem with such a function. > Then
give a *single* function in ALL of mathematics which behaves
as > you wish. > Youre making up some wacky
mathematics.Oh, come on. Do you know the Moebius function?
mu(n) = 1 if n = 1 0 if n is divisible by a square (-1)^r if
n = the product of r distinct primes.Wacky enough? But
indeed, not the function we seek, see below. > For readers
who havent kept up, this poster needs a function which is >
a unit when > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2
+ 3m) > is fully reducible into all linear factors over
rationals, which has a > complex (and so far unexplained)
behavior when its reducible into a > linear and a quadratic,
where sometimes its unit or else it has only > f as a
non-unit factor, which would always have f as a non-unit
factor > if > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 +
3m) > is irreducible over rationals.f_0(m,f) = f if f != 3
and all three roots of the polynomial are not co-prime to f =
1 otherwise. > However, do you accept that also for
*certain* values of f, like if > that polynomial is
reducible into all linear factors over rationals, > that
function would be monic? > Do you accept that as true?
> Yes, why not? As I have stated: when all as have common
factors with f > the leading term of the resulting polynomial
will be f. If it reduces > to three linear terms over Q,
trivially only two as can have common factors > of f, as f
is a prime in the integral part of Q (Z). The problem is when
> there are *not* three linear factors. Because in that case
not all as > are integers, but at least two of them are
algebraic integers. And, while > f is prime in the integers,
it is not prime in the algebraic integers. > Huh? It looks
like the poster conceded that if > a^3 + 3(-1+mf^2)a^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m) > is reducible over rationals
into all linear factors that his position > is bogus.What is
the contradiction with my position? > No reason is given.
Then the poster basically babbles at the end.A pity you cant
follow it. If the cubic splits in linear terms over Q,the
three roots are integer. Because the product of the roots is
amultiple of f^2, with f prime, at most two of the as are
co-prime to f(note again: the as are integer in this case).
So f_0(m) = 1 in thiscase. > Yes, f is prime in the integers
and not prime in algebraic integers, > but that has no
relevance here.Well, that is crucial, you know. If *not* all
three are integer, but atleast two of them are algebraic
integer it is possible that all *three*are not co-prime to f,
just because f is not a prime in the algebraicintegers.
Capisce? > Its kind of interesting to watch this person
jump, dance and do > various gyrations as he tries to
shoehorn mathematics.I see gyrations, but not by me I am
afraid.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland; http://www.cwi.nl/~dik/
... Yes. And so what?> Readers should note that the
poster admits that his position requires> a function that
is either a unit or has f as its only non-unit factor>
depending on certain conditions as m varies from 0 to
positive> innity. A condition follows.> Yes, I still
see no problem with such a function.> Then give a *single*
function in ALL of mathematics which behaves as> you wish. Youre making up some wacky mathematics.Oh, come on. Do you
know the Moebius function?> mu(n) = 1 if n = 1> 0 if n is
divisible by a square> (-1)^r if n = the product of r
distinct primes.> Wacky enough? But indeed, not the function
we seek, see below.> Ive found a simpler refutation of this
posters position.Remember I haveP(m) = f^2((m^3 f^4 - 3m^2
f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Variables: m, f, x,
u E Ring of Algebraic integersand the factorization P(m) =
(a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) Variables: a_1,
a_2, a_3, roots of cubic dened as follows. Cubic: a^3 +
3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)The poster wishes
for a variable function which has the property ofbeing a
factor of f, so Ill use w_1(m) w_2(m) w_3(m) = f^2, where
a_1has w_1(m) as a factor for all integer m.Then a_1(m) x +
uf must have w_1(m) as a factor, so dividing
throughgivesa_1(m)/w_1(m) + uf/w_1(m)where uf/w_1(m) cant be
an algebraic integer for all integer m.It might help for me to
put in actual numbers for u and f, which I cando as the
variables are independent of each other, so let u=2, f=7,then
itsa_1(m)/w_1(m) + 14/w_1(m)and clearly, if w_1(m) varies with
m, then 14/w_1(m) is not analgebraic integer for all integer
m.For those who STILL need help, consider that if you had
14/w_1(m) = r(m)introducing r(m) for the result of the
division, thenw_1(m) r(m) = 14sow_1(m) r(m) - 14 = 0which
would force zeroes for m.That is, you cant have algebraic
integer functions, that is functionsthat give algebraic
integer results, and not have only certain valuesof m that
would work.That is, you can have something like 2m+ 7 = 21,
that works for aparticular value of m, but you cant have
functions in algebraicintegers that will multiply to give 14
for all integer m.To get such functions, you have to go
outside the ring into a eld.That refutes the position of Dik
T. Winter, and note that as Ivepointed out that poster
clearly either has limited mathematicalability, or hes been
lying now for some time.James Harris
...>b^3 +
3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),> This is
not correct. Note that a1/f = b1, or a1 = f*b1.> Putting
this into the equation above that the as> satisfy and
simplifying, one obtains> f*b1^3 + 3*(-1 +
m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m).> Clearly
b2 is a root of the same polynomial. Since > b3 = a3, b3
satises a different equation:> b3^3 +
3*(-1+mf^2)b3^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0.>
Um, Nora Baron do you accept that there exists a SINGLE cubic
which> has the bs as its roots?> Nora Baron *only*
state that the single cubic you give is not one which> has
the bs as its roots.> Um, do *you* accept that there
exists a SINGLE cubic which has the bs> as its roots?>
Yes, I do, and I think everyone does. (x-b1)(x-b2)(x-b3)
serves just ne.> It is a cubic in b, but as the bs depend
on m, it is not necessarily a> cubic in m.The question is to
the solution for the bs, just like earlier in the> post a
solution for the as is given as they are roots of the cubic(***) a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).>
This entire argument can be shortened even further thanI
thought. Look at the polynomial (***) in a just above.We all
agree that it is correct. Harris says that at least one of
the roots is divisible by f. So let a = f*c (where c is an
A.I.) and substitute that into (***): f^3*c^3 +
3(-1+mf^2)f^2*c^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). Factor out
f^2: f*c^3 + 3(-1+mf^2)c^2 - (m^3 f^4 - 3m^2 f^2 + 3m).Let m
= 1 and f = 5 and set the expression to zero: 5*c^3 + 72*c^2
- 553 = 0Its nonmonic, primitive, and irreducible.
Thereforec is NOT an A.I. Therefore NO root of {***} is
divisible by f = 5. No need to worry about whether you need
one or twocubics for the bs. No complications. Totally
directand simple, and totally irrefutable. Harris, I bet you
remember that scene in The Matrix where Cary-Ann Moss puts a
gun right against the head of an Agent, and says: Dodge
The question is to the solution
for the bs, just like earlier in the> post a solution for
the as is given as they are roots of the cubic>(***) a^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).> This entire
argument can be shortened even further than> I thought. Look
at the polynomial (***) in a just above.> We all agree that
it is correct. Harris says that at least > one of the roots
is divisible by f. So let a = f*c (where > c is an A.I.) and
substitute that into (***): f^3*c^3 + 3(-1+mf^2)f^2*c^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m).> Factor out f^2: f*c^3 +
3(-1+mf^2)c^2 - (m^3 f^4 - 3m^2 f^2 + 3m).Let m = 1 and f = 5
and set the expression to zero: 5*c^3 + 72*c^2 - 553 = 0Its
nonmonic, primitive, and irreducible. Therefore> c is NOT an
A.I. Therefore NO root of {***} is > divisible by f = 5.Aha!
Core error! Youve found one of the missing
numbers.Obviously, c isnt in the algebraic integers, butit
SHOULD BE!!!(As a matter of fact, I think that this is what
Jamesshould test is: If it would make his argument work
forsome x to be an algebraic integer, then it should beone.
And by James logic, he has therefore proved that it is. So
when you show that it isnt, thats a contradiction in core.)
- Randy
The question is to the solution for the bs,
just like earlier in the> post a solution for the as is
given as they are roots of the cubic>(***) a^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).> This entire
argument can be shortened even further than> I thought. Look
at the polynomial (***) in a just above.> We all agree that
it is correct. Harris says that at least > one of the roots
is divisible by f. So let a = f*c (where > c is an A.I.) and
substitute that into (***):> f^3*c^3 + 3(-1+mf^2)f^2*c^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m).> Factor out f^2:> f*c^3 +
3(-1+mf^2)c^2 - (m^3 f^4 - 3m^2 f^2 + 3m).> Let m = 1 and f =
5 and set the expression to zero:> 5*c^3 + 72*c^2 - 553 = 0>
Its nonmonic, primitive, and irreducible. Therefore> c is
NOT an A.I. Therefore NO root of {***} is > divisible by f =
5.Aha! Core error! Youve found one of the missing numbers.>
Obviously, c isnt in the algebraic integers, but> it SHOULD
BE!!!(As a matter of fact, I think that this is what James>
should test is: If it would make his argument work for> some
x to be an algebraic integer, then it should be> one. And by
James logic, he has therefore proved that > it is. So when
you show that it isnt, thats a > contradiction in core.)>
Randy, That is his general modus operandi. In this
casehowever the polynomial {***} is his own creation. It is a
somewhat Procrustean stretch to have somethingthat you have
used as the basis for a proof turn out tobe exactly the
basis for a DISproof of exactly the samething. How can you
attack your own polynomial? He has not responded to my posts
for several days. No doubt he will maintain that I cheat and
lie,and I must be worse at it than Arturo and Dik Winter
because he is on speaking terms with them. Youcan see for
yourself in my post that it is chock fullof cheating and
lying. As for the idea that c should be an algebraic integer:
he has described the algebraic integers asincomplete. I
think this is based on Einsteinsdescription of quantum
theory as incomplete - Einstein believed (without proof)
that local hiddenvariables must be present which would
resolve problemslike quantum entanglement and eliminate the
uncertaintyprinciple. Physics would be determinate rather
thanprobabilistic on the small scale. However later
developments in theory (Bells theorem) and experiments
indicated thatEinstein was wrong [though some physicists and
a largernumber of cranks still disagree]. But for Harriss
situation, this harking back to Einsteinwould be gilding the
lily. Harriss result, claiming that cSHOULD be an algebraic
integer or MUST be one, does not lead one to think that the
algebraic integers are incomplete. It simply at out
contradicts a known theorem. There are reallyonly three
possibilities: 1. The known theorem is wrong [but Harris has
accepted it as correct some time ago; it goes back to Gauss].
2. Mathematics is inconsistent [Harris rejects that idea]. 3.
Harriss argument is wrong. There is no room for any other
choices. I have my own opinion on this which you can no doubt
guess. Nora B.> - Randy
=... > The reason is that posters
like this one have worked to convince > readers of
mathematically false things, and I can show that their >
assertions would force the cubic that has the bs for roots
to be > *selectively* non-monic, where if they are correct,
the cubic has to > be non-monic if > a^3 + 3(-1+mf^2)a^2 -
f^2(m^3 f^4 - 3m^2 f^2 + 3m). > is irreducible over Q, and
monic if its not. > Lessee whether that is true. Lets
take m = 1, f = 2. The polynomial > becomes: > a^3 + 9.a^2 -
28, > this one is reducible: (a + 2)(a^2 + 7a - 14).
Nevertheless, every > polynomial with integer coefcients
that denes the bs is not monic, > because the bs are not
all algebraic integers. So your claim is false.Let me
elaborate. With m=1, f=2 the polynomial in a is written as
above.The three roots are (sqrt meaning the positive square
root of an integer): a1 = -2 a2 = [-7 + sqrt(105)]/2 =
sqrt(7).r2 a3 = [-7 - sqrt(105)]/2 = sqrt(7).r3where r2 and
r3 are: r2 = [-sqrt(7) + sqrt(15)]/2 r3 = [-sqrt(7) -
sqrt(15)]/2.Now r2 and r3 are algebraic integers (they are
roots of the polynomialx^4 - 11.x^2 + 4, the other two roots
have the sign of sqrt(7) inverted),moreover, r2.r3 = 2, so
they are both divisors of 2. a1 shares thefactor 2 with
f(=2), a2 shares the factor r2 with f and a3 shares thefactor
r3 with f. Now b1 (= a1/f) and b3 (= a3) are algebraic
integers,b2 (= a2/f) is not. So there is no monic cubic
polynomial with integercoefcients that has all three of b1
to b3 as root, because that wouldmake b2 an algebraic
integer. > Yes. Do you want a proof?About the proof that
every algebraic number can be written as the quotientof an
algebaric integer and a normal integer. I *think* it can be
madestronger (here follows a conjecture, WLT): Lets have
P(x) an irreducible, primitive polynomial with integer
coefcients. Let a be the leading coefcient of the
polynomial and let b be a root.Then: a.b is an algebraic
integer (that is easy) and (a.b,a) = 1 in the algebraic
integers (that is hard).I will think about it.-- dik t.
winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
...> The reason
is that posters like this one have worked to convince>
readers of mathematically false things, and I can show that
their> assertions would force the cubic that has the bs
for roots to be> *selectively* non-monic, where if they
are correct, the cubic has to> be non-monic if> a^3 +
3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).> is
irreducible over Q, and monic if its not.> Lessee whether
that is true. Lets take m = 1, f = 2. The polynomial>
becomes:> a^3 + 9.a^2 - 28,> this one is reducible: (a +
2)(a^2 + 7a - 14). Nevertheless, every> polynomial with
integer coefcients that denes the bs is not monic,>
because the bs are not all algebraic integers. So your claim
is false.Now readers whove kept up realize that my point is
that the ring ofalgebraic integers is awed because of the
denition of algebraicinteger, so you should be surprised at
this poster attempting to relyon the denition of algebraic
integers.But besides that please pay attention to the full
claim, whichrequires that the polynomial--a cubic--that has
the bs as roots havea leading coefcient that switches from
being monic to having f asits only non-unit factor depending
on the cubic dening the as beingreducible over
rationals.The poster requires a complex relationship where
sometimes if thecubic for the as is reducible into a linear
and a quadratic the cubicis monic, while in others, its not,
while the posters position wouldrequire that its never monic
if the cubic dening the as isirreducible over Q.Let me
elaborate. With m=1, f=2 the polynomial in a is written as
above.> The three roots are (sqrt meaning the positive square
root of an integer):> a1 = -2> a2 = [-7 + sqrt(105)]/2 =
sqrt(7).r2> a3 = [-7 - sqrt(105)]/2 = sqrt(7).r3> where r2
and r3 are:> r2 = [-sqrt(7) + sqrt(15)]/2> r3 = [-sqrt(7) -
sqrt(15)]/2.> Now r2 and r3 are algebraic integers (they are
roots of the polynomial> x^4 - 11.x^2 + 4, the other two
roots have the sign of sqrt(7) inverted),> moreover, r2.r3 =
2, so they are both divisors of 2. a1 shares the> factor 2
with f(=2), a2 shares the factor r2 with f and a3 shares the>
factor r3 with f. Now b1 (= a1/f) and b3 (= a3) are algebraic
integers,> b2 (= a2/f) is not. So there is no monic cubic
polynomial with integer> coefcients that has all three of b1
to b3 as root, because that would> make b2 an algebraic
integer. > Yes. Do you want a proof?Thats the posters on
question from the prior post. Hes replying tohimself in what
follows, I guess, unless hes ignoring his ownquestion!!! >
About the proof that every algebraic number can be written as
the quotient> of an algebaric integer and a normal integer. I
*think* it can be made> stronger (here follows a conjecture,
WLT):> Lets have P(x) an irreducible, primitive polynomial
with integer> coefcients. Let a be the leading coefcient of
the polynomial and> let b be a root.> Then:> a.b is an
algebraic integer (that is easy) and> (a.b,a) = 1 in the
algebraic integers (that is hard).> I will think about it.I
remind of the position this poster needs, which is that the
cubicdening the bs has a leading coefcient that varies
from being aunit to having f as its only non-unit factor
dependent, on the cubicdening the as, where supposedly
reducibility over rationals is thebig deal.When pushed the
poster starts making claims about proving thatotherwise the
coefcients arent algebraic integers, when my pointfor some
time has been that the denition of algebraic integers
isawed.The correct answer is that the cubic dening the bs
is always monic,and for certain values of m and f, its
coefcients are pushed out ofthe ring of algebraic integers
because the denition is awed byexcluding certain numbers,
which you can see need to be included, oryou get
contradictions.James Harris
=... > The correct answer is
that the cubic dening the bs is always monic, > and for
certain values of m and f, its coefcients are pushed out of
> the ring of algebraic integers because the denition is
awed by > excluding certain numbers, which you can see need
to be included, or > you get contradictions.As I have written
already multiple times, it is trivially monic in otherrings,
the cubic (x - b1)(x - b2)(x - b3) serves just ne. But
youwant to go to a ring where your divisibility claims work.
It includesthe ring of algebraic integers. But which of the
two numbers (-sqrt(7) + sqrt(15))/2 and (-sqrt(7) -
sqrt(15))is a unit in that ring? And why precisely that
number? How can wedetermine whether a number is in that ring
or not? As long as youdo not give a clear indication on how
to determine the numbers inthat ring it is not possible to
know whether the division requirementsyou need are indeed in
that ring.You have also not yet shown what the contradiction
is.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland, +31205924131home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/
...> The
correct answer is that the cubic dening the bs is always
monic,> and for certain values of m and f, its coefcients
are pushed out of> the ring of algebraic integers because the
denition is awed by> excluding certain numbers, which you
can see need to be included, or> you get contradictions.>>As
I have written already multiple times, it is trivially monic
in other>rings, the cubic (x - b1)(x - b2)(x - b3) serves
just ne. But you>want to go to a ring where your
divisibility claims work. It includes>the ring of algebraic
integers. But which of the two numbers> (-sqrt(7) +
sqrt(15))/2 and (-sqrt(7) - sqrt(15))>is a unit in that ring?
And why precisely that number?James situation gets a bit more
complicated than that; I explained itto him, but as he
usually does with uncomfortable truths, he justignored it and
started repeating his claim again.Assume that
(-sqrt(7)+sqrt(15))/2 is a unit in the ring. Then so isits
square,(22 -2sqrt(105))/4 = (11-sqrt(105))/2.This element is
a root of x^2 - 11x + 4.So if this is a unit, that means that
the ring contains one root ofthe irreducible quadratic
polynomial 4x^2 - 11x + 1.James has never told us whether, if
an irreducible polynomial has aroot in his ring, does it have
all roots in his ring? Because heclaims that his denition
xes the algebraic integers, by allowingnumbers that had been
excluded, it seems reasonable to assume thatit does. But of
course he has never come out and said it straight.So, since
it has a root of 4x^2-11x+1, should it not contain the
otherroot as well? The other root is the inverse of
(11+sqrt(105))/2. Butthis is the square of
(-sqrt(7)-sqrt(15))/2; so (-sqrt(7)-sqrt(15))/2is also a unit
in the ring. Which would mean they BOTH are units inthe ring,
which would mean that their product, 2 is a unit, which
weknow cannot be true.Ah, so it must be that
(-sqrt(7)+sqrt(15))/2 is not a unit in thering. So it must be
that it is (-sqrt(7)-sqrt(15))/2 that is aunit. So its square,
(11+sqrt(105))/2 is a unit; so the ring containsa root of 4x^2
- 11x + 1; so it contains both roots; so it containsthe
inverse of (11-sqrt(105))/2. So (-sqrt(7)+sqrt(15))/2 is
aunit. Oops.Which leaves us with two possibilities:(A) The
Object ring does not satisfy the properties James
claimseither: in that ring, neither (-sqrt(7)-sqrt(15))/2
nor(-sqrt(7)+sqrt(15))/2 are units;or(B) In the Object ring,
this marvellous construction that xes thering of algebraic
integers by allowing numbers that should be in,there are
polynomias with integer coefcients, irreducible over Q,which
have ->SOME<- roots in the ring but not all of them.And in
case (B), how do we know which one is there, as you
ask?
reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A mans capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
gures few readers can criticize. A great many people are
staggered to this extent, that they imagine there must be the
indenite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
>As I have written already multiple times, it is trivially
monic in other >rings, the cubic (x - b1)(x - b2)(x - b3)
serves just ne. But you >want to go to a ring where your
divisibility claims work. It includes >the ring of algebraic
integers. But which of the two numbers > (-sqrt(7) +
sqrt(15))/2 and (-sqrt(7) - sqrt(15)) >is a unit in that
ring? And why precisely that number?... > (B) In the Object
ring, this marvellous construction that xes the > ring of
algebraic integers by allowing numbers that should be in, >
there are polynomias with integer coefcients, irreducible
over Q, > which have ->SOME<- roots in the ring but not all
of them. > And in case (B), how do we know which one is
there, as you ask?I think (B) is indeed precisely the case.
James wants all those numbersin his ring such that the cubics
in the bs are monic with numbers inhis rings for all possible
polynomials P(m). So when we look at thepolynomial in a for
m=1 and f=2 we get (in the algebraic integers)three roots
that are obviously not coprime to f. To get at his claimthat
exactly *one* of the as is coprime to f he has to make one
ofthose common factors (that I have written above) units.
They are bothroots of an irreducible polynomial with integer
coefcients of degree 4.Actually sometime ago he made a
remark that suggested this.The main problem is that he has to
make this choice for each and everypolynomial that comes up.
Now I am not sure, but it may be possibleto have a ring where
only a single root of a quadratic is in his ring,I very much
doubt that a choice can be made across *all* polynomialshe
comes up with without getting conicts. He has to show that
aconsistent choice can be made.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
answer is that the cubic dening the bs is always monic,>
and for certain values of m and f, its coefcients are pushed
out of> the ring of algebraic integers because the denition
is awed by> excluding certain numbers, which you can see
need to be included, or> you get contradictions.>> James
HarrisThere is no aw in the denition of algebraic integers
and no contradiction in any results obtained byvalid
operations. There is an error in your argument.--A fool and
his proof are soon refuted.--Democracy: The triumph of
popularity over principle.--http://www.crbond.com
=Given a
kxk diagonal matrix, under what conditions we can ensure the
matrix is positive denite?Alos, if the matrix is not
diagonal with all the elements are positive real number, how
can we ensure it is positive denite?Willy
=Given a kxk
diagonal matrix, under what conditions we can ensure the
matrix is positive > denite?The matrix is positive denite
if and only if all entries are >0(this is easy to see, make
use of the canonical basis to proof this)Alos, if the matrix
is not diagonal with all the elements are positive real
number, how > can we ensure it is positive denite?> There is
a very useful subdeterminant criterion for symmetric
(resp.hermitian) matrices (elements need not all be >0 to use
this test):check the number in the upper left corner. For the
matrix to bepositive denite, it must be positive. In the
same way, check thedeterminant of the 2x2-matrix in the upper
left, then thecorresponding 3x3-matrix, etc. If all these
determinants are positive(including the last determinant
which is that of the whole matrix),then the matrix is pos.
def..The converse is also true.It may save much computational
work if you permute the rows andcolumns in the matrix before
applying the criterion so thatcalculation of the determinants
is as easy as possible.
>Given a kxk diagonal matrix,
under what conditions we can ensure the matrix>is positive
>denite?>>Alos, if the matrix is not diagonal with all the
elements are positive real>number, how >can we ensure it is
positive denite?>Willy>>If youre looking to check,
look up Sylvesters criterion.There are many other special
cases. For example, diagonally dominant matriceswith positive
diagonal entries.--irascible since 1957
=Can anyone please
point me toward an implementation (in any language,
evenpseudo ones) of Knuths spectral test?I have a random
generator that I need to test, and I can not
ndimplementations of the test...(I have found implementation
specially geared for congrual tests, but thisis not what I
need as my generator is not congrual).
=Note subject
change.In sci.math, Justin
Davis<jkd3@duke.edu><5bfec9e3.0310150545.5bdf3dfd@
<rphenry@home.com> <ZjVib.33914$La.10104@fed1read02>:>>
[snip for brevity]> Does it matter? Given any nite digit
sequence, pi embeds it.>> Somewhere. :-)>> (At least, such is
my understanding.) Im afraid I dont understand your
understanding.> I can try to be more precise, if it
will help. Consider a sequence D_0, D_1, ... D_{k-1},
for some>> integer k > 0, where D_i is in the set of digits>>
{0,1,2,3,4,5,6,7,8,9}. Then consider the sequence P_0,>> P_1,
... where P_i = oor(pi * 10^i) mod 10. Therefore>> P_0 = 3,
P_1 = 1, P_2 = 4, P_3 = 1, P_4 = 5, P_5 = 9,>> etc., or more
compactly and familiarly written pi = 3.14159...
The idea is that there exists an N such that P_N = D_0,>>
P_{N+1} = D_1, P_{N+2} = D_2, ... , P_{N+k-1} = D_{k-1},>>
given the sequence D_0 to D_{k-1}. I hesitate to call it
a theorem as Ive no idea how to>> prove it. (Im not even
sure Im formulating it correctly,>> although I suspect it is
true.) Its obvious pi is irrational, but so are the>>
numbers 0.121121112111121111121111112.... and>>
0.345334455333444555333344445555... . Theres>> no 6 in
either one. If you dont like base 10 you can of course
modify the>> digit denitions appropriately. :-)I think the
property youre looking for is normality. Pi would be> normal
in base 10 if the above were true. Look>
http://pi314.at/math/normal.html for a more in-depth
notehttp://mathworld.wolfram.com/NormalNumber.html(an obvious
search for those familiar with
mathworld.wolfram.com).Evidently the problems not quite as
cut and dried as I thought, andlooks like an open question.--
#191, ewill3@earthlink.netIts still legal to go
.sigless.
=Regarding Grahams number,here is a diagram that
Ive seen on the web- / 1) 3^^^^3 | | 2) 3^^...1)...^^3 [where
there are 1) = 3^^^^3up-arrows] | | 3) 3^^...2)...^^3 [where
there are 2) up-arrows] | . 64 levels < . | . | . | 63)
3^^...62)...^^3 | 64) 3^^...63)...^^3 <--- Grahams #(-the
process of so and so many arrows is far from
beingmathematicallyprecise and needs to be
reworked;however,it lends itself to beingspaciallycompact.
:-) )Labelling sections and bays one section
________________^_______ | / | 2 bays |
______________^_________ | / | bay 1 bay 2 | ___^___
______^________ | / / > one section / 1) 3^^^^3 | | 2)
3^..1)..^3 | | 3) 3^..2)..^3 | 64 levels< . | | . | | . | 64)
3^..63)..^3 /where a bay (OL) includes all levels and a
section is all bays andtheir levels.Using this mechanism, the
Conway-Guy expression a -> b -> ..... x -> y -> zcan be dened
in terms of Knuth up-arrows, e.g. 2->3->3->4 8 bays
__________________________________^__________________________
__________/ bay 1 bay 2 bay 3 4-7 bay 8 __^___
_________^__________ __________^_________ __^__ _____^______/
/ / / / / 1) 2^3 = 8 / 1) 2^3 / / 1) 2^3 | 2) 2^..1)..^3 | 2)
2^..1)..^3 | | 2)2^..1)..^3 | 3) 2^..2)..^3 | . | | .8
levels< . | . | 4 | . | . | . | bays| . 8) 2^..7)..^3 levels<
. |here | . = 2->3->8->2 | . | | . = 2->3->2->3 ..2^........^3
levels<.....< . = 2->3->3->3 | | . ..2^........^3 = 2->3->8->3
bays = 2->3->2->4
_____________________________________________________________
____^_____/ / 1) 2^3 / 1) 2^3 / / 1) 2^3 | . | . | | .8
levels< . | . | | . | . | . | | . 8) 2^..7)..^3 levels< . | |
. | . | | . ..2^........^3 levels<.....< . | | .
..2^........^3 =2->3->3->4the above can be called 2 section
levelscrunch the diagram down to this;dropping bays # 1 and 3
--- 7 bays ___________^_____________ / 1) 2^3 / / 1) 2^3 . | |
. . | | . 8)2^..7)..^3<..< . levels | | . ..2^.....^3 =
2->3->2->4 bays ___________________^_____ /1) 2^3 / / 1) 2^3
. | | . . | | . 8)2^..7)..^3<..< . levels| | . ..2^.....^3 =
2->3->3->4Following and extending the concept---> another try
at 9->9->9->9--- 387420488 section bays>
_______________________________________________^_____________
________________________________________________ > / > /
387420488 bays > |
_______________________^_____________________ > |/ 1)
9^9=387420489 / / 1) 9^9 > | 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | . | | . > | . | | . > |387420489) 9^..387420488)..^9<..<
. > | levels| | . > | ..9^........^9 > | bays > /387420488<
______________________________________^______ > | section|/ .
> | levels | . / / 387420488 bays > | | . | |
_______________________^_____________________ > | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 > | |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 > | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | . | | .
> | | . | | . | | . | | . > | | . | | . | |387420489)
9^..387420488)..^9<..< . > | |387420489)
9^..387420488)..^9<..< . | | levels| | . > | | levels | | . |
| ..9^........^9 > | ..9^.........^9<...< bays > | section
levels | | ___________________________________^_________ > |
| |/ . > | | | . > | | |
___________________________________^_________ > | | |/ 1) 9^9
/ / 1) 9^9 > | | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 >
387420488< | | . | | . > levels of| | | . | | . > section | |
|387420489) 9^..387420488)..^9<..< . > levels | | | levels | |
. > | ..9^........^9 > | section bays > |
_____________________________________________________________
_____________________________________^_________ > |/ . > | .
> |
_____________________________________________________________
_____________________________________^_________ > |/ /
387420488 bays > | |
_______________________^_____________________ > | |/ 1)
9^9=387420489 / / 1) 9^9 > | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 > | | . | | . > | | . | | . > | |387420489)
9^..387420488)..^9<..< . > | | levels| | . > | |
..9^........^9 > | | bays > 387420488<
______________________________________^______ > section|/ . >
levels | . / / 387420488 bays > | . | |
_______________________^_____________________ > |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 > |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 > | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | . | | . >
| . | | . | | . | | . > | . | | . | |387420489)
9^..387420488)..^9<..< . > |387420489) 9^..387420488)..^9<..<
. | | levels| | . > | levels | | . | | ..9^........^9 >
..9^.........^9<...< bays > section levels | |
___________________________________^_________ > | |/ . > | |
. > | | ___________________________________^_________ > | |/
1) 9^9 / / 1) 9^9 > | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 > | |
. | | . > | | . | | . > | |387420489) 9^..387420488)..^9<..< .
> | | levels | | . > ..9^........^9 > levels of > 387420488
bays of section bays section levels >
_____________________________________________________________
__^___________________________________________________ |> /
v> v> 387420488 section bays v>
_____________________________________________________________
__^____________________________________________ | | v> / | |
v> / 387420488 bays | | v> |
_______________________^_____________________ | | v> |/ 1)
9^9=387420489 / / 1) 9^9 | | v> | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | v> | . | | . | | v> | . | | . | | v>
|387420489) 9^..387420488)..^9<..< . | | v> | levels| | . | |
v> | ..9^........^9 | | v> | bays | | v> 387420488<
______________________________________^______ | | v>
section|/ . | | v> levels | . / / 387420488 bays | | v> | . |
| _______________________^_____________________ | | v> |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | | v> |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | v> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | . | |
. | | v> | . | | . | | . | | . | | v> | . | | . | |387420489)
9^..387420488)..^9<..< . | | v> |387420489)
9^..387420488)..^9<..< . | | levels| | . | | v> | levels | |
. | | ..9^........^9 | | v> ..9^.........^9<...< bays | | v>
section levels | |
___________________________________^_________ | | v> | |/ . |
| v> | | . | | v> | | . | | v> | |
___________________________________^_________ | | v> | |/ 1)
9^9 / / 1) 9^9 | | v> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
v> | | . | | . | | v> | | . | | . | | v> | |387420489)
9^..387420488)..^9<..< . | | v> | | levels | | . | | v>
..9^........^9 | | v> section bays | | v>
_____________________________________________________________
_____________________________________^_________ | | v> / . |
| v> . >....>> . | | >
_____________________________________________________________
_____________________________________^_________ | |> / /
387420488 bays | |> |
_______________________^_____________________ | |> |/ 1)
9^9=387420489 / / 1) 9^9 | |> | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | |> | . | | . | |> | . | | . | |> |387420489)
9^..387420488)..^9<..< . | |> | levels| | . | |> |
..9^........^9 | |> | bays | |> 387420488<
______________________________________^______ | |> section|/
. | |> levels | . / / 387420488 bays | |> | . | |
_______________________^_____________________ | |> |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | |> |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | |> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | . | | .
| |> | . | | . | | . | | . | |> | . | | . | |387420489)
9^..387420488)..^9<..< . | |> |387420489)
9^..387420488)..^9<..< . | | levels| | . | |> | levels | | .
| | ..9^........^9 | |> ..9^.........^9<...< bays | |>
section levels | |
___________________________________^_________ | |> | |/ . |
|> | | . | |> | | . | |> | |
___________________________________^_________ | |> | |/ 1)
9^9 / / 1) 9^9 | |> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |>
| | . | | . | |> | | . | | . | |> | |387420489)
9^..387420488)..^9<..< . | |> | | levels | | . | |>
..9^........^9 / /> bays of section bays >
_____________________________________________________________
__________________________________________^__________________
_____________________________________________________________
____________________________________________________ |> /
387420488 section bays |>
_____________________________________________________________
__^____________________________________________ |> / |> /
387420488 bays | > |
_______________________^_____________________ |> |/ 1)
9^9=387420489 / / 1) 9^9 / / 387420488 section bays |> | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | |
_____________________________________________________________
__^____________________________________________ | > | . | | .
| |/ | > | . | | . | | / 387420488 bays |> |387420489)
9^..387420488)..^9<..< . | | |
_______________________^_____________________ |> | levels| |
. | | |/ 1) 9^9=387420489 / / 1) 9^9 |> | ..9^........^9 | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | bays | | | . | | . |>
/387420488< ______________________________________^______ | |
| . | | . |> | section|/ . | | |387420489)
9^..387420488)..^9<..< . |> | levels | . / / 387420488 bays |
| | levels| | . |> | | . | |
_______________________^_____________________ | | |
..9^........^9 |> | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | | | bays |> | |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9
| | 2) 9^..1)..^9 | |387420488<
______________________________________^______ |> | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | . | | . | | section|/ . |> |
| . | | . | | . | | . | | levels | . / / 387420488 bays |> | |
. | | . | |387420489) 9^..387420488)..^9<..< . | | | . | |
_______________________^_____________________ |> |
|387420489) 9^..387420488)..^9<..< . | | levels| | . | | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 |> | | levels | | . | | ..9^........^9 | | |/ 1) 9^9 /
/ 1) 9^9 | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> |
..9^.........^9<...< bays | | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . | | . |> | section levels | |
___________________________________^_________ | | | . | | . |
| . | | . |> | | |/ . | | | . | | . | |387420489)
9^..387420488)..^9<..< . |> | | | . | | |387420489)
9^..387420488)..^9<..< . | | levels| | . |> | | | . | | |
levels | | . | | ..9^........^9 |> | | |
___________________________________^_________ | |
..9^.........^9<...< bays |> | | |/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |> | | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | | |/ . |> 387420488< | | . |
| . | | | | . |> levels of| | | . | | . | | | | . |> section |
| |387420489) 9^..387420488)..^9<..< . | | | |
___________________________________^_________ |> levels | | |
levels | | . | | | |/ 1) 9^9 / / 1) 9^9 |> | ..9^........^9 |
| | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | section bays | | |
| . | | . |> |
_____________________________________________________________
_____________________________________^_________ | | | | . | |
. |> |/ . | | | |387420489) 9^..387420488)..^9<..< . |> | . |
| | | levels | | . |> | . | | ..9^........^9 |> |
_____________________________________________________________
_____________________________________^_________ | | section
bays |> |/ / 387420488 bays | |
_____________________________________________________________
_____________________________________^_________ |> | |
_______________________^_____________________ | |/ . |> | |/
1) 9^9=387420489 / / 1) 9^9 | | . |> | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . |> | | . | | . | |
_____________________________________________________________
_____________________________________^_________ |> | | . | |
. | |/ / 387420488 bays |> | |387420489)
9^..387420488)..^9<..< . | | |
_______________________^_____________________ |> | | levels|
| . | | |/ 1) 9^9=387420489 / / 1) 9^9 |> | | ..9^........^9
| | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | | bays | | | . | |
. |> |387420488< ______________________________________^______
| | | . | | . |> | section|/ . | | |387420489)
9^..387420488)..^9<..< . |> levels | . / / 387420488 bays | |
| levels| | . |> | . | |
_______________________^_____________________ | | |
..9^........^9 |> |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | | | bays |> |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 |
| 2) 9^..1)..^9 | |387420488<
______________________________________^______ |> | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | . | | . | | section|/ . |> |
. | | . | | . | | . | | levels | . / / 387420488 bays |> | . |
| . | |387420489) 9^..387420488)..^9<..< . | | | . | |
_______________________^_____________________ |> |387420489)
9^..387420488)..^9<..< . | | levels| | . | | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 |> | levels | | . | | ..9^........^9 | | |/ 1) 9^9 / /
1) 9^9 | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |>
..9^.........^9<...< bays | | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . | | . |> section levels | |
___________________________________^_________ | | | . | | . |
| . | | . |> | |/ . | | | . | | . | |387420489)
9^..387420488)..^9<..< . |> | | . | | |387420489)
9^..387420488)..^9<..< . | | levels| | . |> | | . | | |
levels | | . | | ..9^........^9 |> | |
___________________________________^_________ | |
..9^.........^9<...< bays |> | |/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |> | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | | |/ . |> | | . | | . | | |
| . |> | | . | | . | | | | . |> | |387420489)
9^..387420488)..^9<..< . | | | |
___________________________________^_________ |> | | levels |
| . | | | |/ 1) 9^9 / / 1) 9^9 |> ..9^........^9 | | | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 |> levels of section bays<....<
| | . | | . |> | | | | . | | . |> | | | |387420489)
9^..387420488)..^9<..< . | > | | | | levels | | . |>
..9^........^9 |> bays of bays > section bays |>
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
__________________________________________^_________ |> / .
|> . |> . |>
_____________________________________________________________
_____________________________________________________________
_____________________________________________________________
__________________________________________^_________ |> / |
8> 387420488 section bays > levels>
_____________________________________________________________
__^____________________________________________ |of bays> /
|of sect.> / 387420488 bays |bays> |
_______________________^_____________________ |> |/ 1)
9^9=387420489 / / 1) 9^9 / / 387420488 section bays |> | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | |
_____________________________________________________________
__^____________________________________________ | > | . | | .
| |/ | > | . | | . | | / 387420488 bays |> |387420489)
9^..387420488)..^9<..< . | | |
_______________________^_____________________ |> | levels| |
. | | |/ 1) 9^9=387420489 / / 1) 9^9 |> | ..9^........^9 | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | bays | | | . | | . |>
/387420488< ______________________________________^______ | |
| . | | . |> | section|/ . | | |387420489)
9^..387420488)..^9<..< . |> | levels | . / / 387420488 bays |
| | levels| | . |> | | . | |
_______________________^_____________________ | | |
..9^........^9 |> | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | | | bays |> | |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9
| | 2) 9^..1)..^9 | |387420488<
______________________________________^______ |> | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | . | | . | | section|/ . |> |
| . | | . | | . | | . | | levels | . / / 387420488 bays |> | |
. | | . | |387420489) 9^..387420488)..^9<..< . | | | . | |
_______________________^_____________________ |> |
|387420489) 9^..387420488)..^9<..< . | | levels| | . | | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 |> | | levels | | . | | ..9^........^9 | | |/ 1) 9^9 /
/ 1) 9^9 | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> |
..9^.........^9<...< bays | | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . | | . |> | section levels | |
___________________________________^_________ | | | . | | . |
| . | | . |> | | |/ . | | | . | | . | |387420489)
9^..387420488)..^9<..< . |> | | | . | | |387420489)
9^..387420488)..^9<..< . | | levels| | . |> | | | . | | |
levels | | . | | ..9^........^9 |> | | |
___________________________________^_________ | |
..9^.........^9<...< bays |> | | |/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |> | | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | | |/ . |> 387420488< | | . |
| . | | | | . |> levels of| | | . | | . | | | | . |> section |
| |387420489) 9^..387420488)..^9<..< . | | | |
___________________________________^_________ |> levels | | |
levels | | . | | | |/ 1) 9^9 / / 1) 9^9 |> | ..9^........^9 |
| | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | section bays | | |
| . | | . |> |
_____________________________________________________________
_____________________________________^_________ | | | | . | |
. |> |/ . | | | |387420489) 9^..387420488)..^9<..< . |> | . |
| | | levels | | . |> | . | | ..9^........^9 |> |
_____________________________________________________________
_____________________________________^_________ | | section
bays |> |/ / 387420488 bays | |
_____________________________________________________________
_____________________________________^_________ |> | |
_______________________^_____________________ | |/ . |> | |/
1) 9^9=387420489 / / 1) 9^9 | | . |> | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . |> | | . | | . | |
_____________________________________________________________
_____________________________________^_________ |> | | . | |
. | |/ / 387420488 bays |> | |387420489)
9^..387420488)..^9<..< . | | |
_______________________^_____________________ |> | | levels|
| . | | |/ 1) 9^9=387420489 / / 1) 9^9 |> | | ..9^........^9
| | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |> | | bays | | | . | |
. |> |387420488< ______________________________________^______
| | | . | | . |> | section|/ . | | |387420489)
9^..387420488)..^9<..< . |> levels | . / / 387420488 bays | |
| levels| | . |> | . | |
_______________________^_____________________ | | |
..9^........^9 |> |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 | | | bays |> |/ 1) 9^9 / / 1) 9^9 | | 2) 9^..1)..^9 |
| 2) 9^..1)..^9 | |387420488<
______________________________________^______ |> | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | . | | . | | section|/ . |> |
. | | . | | . | | . | | levels | . / / 387420488 bays |> | . |
| . | |387420489) 9^..387420488)..^9<..< . | | | . | |
_______________________^_____________________ |> |387420489)
9^..387420488)..^9<..< . | | levels| | . | | |
______________________________________^______ | |/ 1) 9^9 / /
1) 9^9 |> | levels | | . | | ..9^........^9 | | |/ 1) 9^9 / /
1) 9^9 | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |>
..9^.........^9<...< bays | | | 2) 9^..1)..^9 | | 2)
9^..1)..^9 | | . | | . |> section levels | |
___________________________________^_________ | | | . | | . |
| . | | . |> | |/ . | | | . | | . | |387420489)
9^..387420488)..^9<..< . |> | | . | | |387420489)
9^..387420488)..^9<..< . | | levels| | . |> | | . | | |
levels | | . | | ..9^........^9 |> | |
___________________________________^_________ | |
..9^.........^9<...< bays |> | |/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |> | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 | | | |/ . |> | | . | | . | | |
| . |> | | . | | . | | | | . |> | |387420489)
9^..387420488)..^9<..< . | | | |
___________________________________^_________ |> | | levels |
| . | | | |/ 1) 9^9 / / 1) 9^9 |> ..9^........^9 | | | | 2)
9^..1)..^9 | | 2) 9^..1)..^9 |> levels of section bays<....<
| | . | | . |> | | | | . | | . |> | | | |387420489)
9^..387420488)..^9<..< . |> | | | | levels | | . |>
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
h9GFkWI12344;
=Malcolm generally covered within a position
of both the second and third raised to indicate a v and the
rst touching the fourth at a midway position whilst the fth
is aligned to 180 degrees to the second and third yes a st
full of digits as descibed you have placed a victory sign
what is then held high in front of the world to see Peter,
Bloody inventor of ings ><pre>>Are there any mathematicians
out there who are concerned about the>implications of the
concept of Constructivism for maths? I must admit that>what
little I know about it gives cause for a modicum of alarm. I
would>very much like to discuss this matter with someone who
is familiar with the>limitations of oating point numbers,
particularly the theory of>signicant digits. >></pre>
Exposing the war for prot game is like turning>on the tap of
a high pressure bull faucet.> Tom Potter
http://tompotter.us>> The fact of the matter is that the
Germans are perhaps the> most moral, intelligent and
productive folks on the planet,> and they have been demonized
because they had to turn> to a political [Nazi] party pledged
to control the Bolshevik> instigators, terrorists, murderers
and assassins, who were> creating death and destruction all
over the world in the 1900s,> committed to going after the
Bolsheviks terrorists. -- Tom>Tom, in case you are leaving or
have left for Germany,the land of your dreams with its most
intelligent peoplelet me offer you some info that might save
your ass fromembarrassment and perhaps even more severe
consequences.your age (except perhaps todays youngsters
below 25) donot have any tolerance for what you are
preaching.== During my years in DE, there was the case where
aguy stopped at the autobahn and helped two woman to
changetheir at tire. They offered him 20 bucks for his
service, buthe brushed it away with the remark: ich bin doch
kein Judeand drove off. 3 weeks latter he got hauled into
court for havingmade disparaging remarks against Jews.==
Dont make any of your propaganda neither in
neighboringSwitzerland. They have their Artikel 261. Due to
this law a doctorsits in jail because he railed against the
Jewish method ofSchaechten = butchering cattle, kosher
style.== Again from Germany here is a story off the wires
from
today:http://www.reuters.com/newsArticle.jhtml?type=
oddlyEnoughNews&storyID=3621259BERLIN (Reuters) - A German
man who taught his dog Adolf to giveGermanys anti-Nazi laws
for a series of incidents in recent years, aBerlin court said
on Wednesday. Germany has strict laws banning themustache and
military tunic, said he didnt understand what thefuss was
about. The trial is set for Thursday. Thein is accused
ofshouting the Nazi battlecry Sieg Heil in front of Berlin
police and ofcity in separate incidents in 2002. (Additional
reporting by Erik Kirschbaum)All your stories you were
telling at our NGs cyber party here, aboutthe Germans
greatness, aint gonna be an effective defense
inDeutschland...........ahahahahah..........ahahahaEnjoy the
beer while youre over there. Not WWII
politics.ahahahaha..........ahahahahahanson
Exposing the
war for prot game is like turning>on the tap of a high
pressure bull faucet.> Tom Potter http://tompotter.us>>
The fact of the matter is that the Germans are perhaps the>
most moral, intelligent and productive folks on the planet,>
and they have been demonized because they had to turn> to a
political [Nazi] party pledged to control the Bolshevik>
instigators, terrorists, murderers and assassins, who were>
creating death and destruction all over the world in the
1900s,> committed to going after the Bolsheviks terrorists.
-- Tom>> Tom, in case you are leaving or have left for
Germany,> the land of your dreams with its most intelligent
people> let me offer you some info that might save your ass
from> embarrassment and perhaps even more severe
consequences.> your age (except perhaps todays youngsters
below 25) do> not have any tolerance for what you are
preaching.== During my years in DE, there was the case where
a> guy stopped at the autobahn and helped two woman to
change> their at tire. They offered him 20 bucks for his
service, but> he brushed it away with the remark: ich bin
doch kein Jude> and drove off. 3 weeks latter he got hauled
into court for having> made disparaging remarks against
Jews.== Dont make any of your propaganda neither in
neighboring> Switzerland. They have their Artikel 261. Due to
this law a doctor> sits in jail because he railed against the
Jewish method of> Schaechten = butchering cattle, kosher
style.== Again from Germany here is a story off the wires
from today:>
http://www.reuters.com/newsArticle.jhtml?type=oddlyEnoughNews
&storyID=3621259> BERLIN (Reuters) - A German man who taught
his dog Adolf to give> Germanys anti-Nazi laws for a series
of incidents in recent years, a> Berlin court said on
Wednesday. Germany has strict laws banning the> mustache and
military tunic, said he didnt understand what the> fuss was
about. The trial is set for Thursday. Thein is accused of>
shouting the Nazi battlecry Sieg Heil in front of Berlin
police and of> city in separate incidents in 2002.
(Additional reporting by Erik Kirschbaum)All your stories you
were telling at our NGs cyber party here, about> the Germans
greatness, aint gonna be an effective defense in>
Deutschland...........ahahahahah..........ahahaha> Enjoy the
beer while youre over there. Not WWII politics.>
ahahahaha..........ahahahahahansonhanson makes a good point
about Defeated nations having the dogma of the victors
enforced on them, regardless of the truth or value of the
dogma.He also makes a good point about the fact that harsh
laws have been forced on the Germans, inhibiting their free
speech.Of course, the same thing happened when the Bolsheviks
co-opted the government of Russia, from the native Russians.
The rst law they passed made criticism of Jews punishable by
ring squad.Intelligent, rational, moral folks, who believe in
free speech, justice, equity, and the brotherhood of man, and
who are concerned about making the world a better place to
live in for ALL people should learn from these examples.If
folks are inhibited by brainwashing, and harsh laws, from
criticizing certain classes of people, it gives carte blanche
to that class to do as they please.Can you imagine what kind
of world it would be if folks were jailed, or even executed,
for criticizing Microsoft?????--Tom Potter
> Modern
socialism is a balance of public and private>> enterprise
with an adiquate social safety net.>>That means stealing from
one and giving to other via taxes.>Taxation is Theft>>
Stupidity alert!>Parki-pooh, you do not have to precede any
of your intellectually>>empty posts with any alert!. Everyone
knows that you are stupid.>>The poster is obviously a
taxpayer, hence a contributor to your>>salary, since you are
on the public payroll, as a college professor.>> I see
somebody is too dumb to know the difference between public>>
universities and private
ones.>>ahahahah.......ahahahahha.......>lie, twist it any
way you want, Parki-pooh, the lparker@emory.edu :>You are
feeding off the public trough, which the OP, me and millions
>of other taxpayers ll up for you.....thats why your
hangout has... >--- .....to comply with new **Federal**
regulations, Emory University>has developed a set of **Cost
Accounting** Standard Policies. >--- EMORY UNIVERSITY
RECEIVES $225,000 GRANTHaliburton received a multi-billion
dollar contract from the government. Does that make all its
employees public employees?>FROM U.S. DEPARTMENT OF ENERGY
......etc., etc....>Do I need to post more such tid bits and
their urls, Parki-pooh?>ahahahahha........ahahahansonPerhaps
you need to learn how to read.
> Modern socialism is a
balance of public and private>> enterprise with an adiquate
social safety net.>>That means stealing from one and giving
to other via taxes.>Taxation is Theft>> Stupidity
alert!>Parki-pooh, you do not have to precede any of your
intellectually>>empty posts with any alert!. Everyone knows
that you are stupid.>>The poster is obviously a taxpayer,
hence a contributor to your>>salary, since you are on the
public payroll, as a college professor.>> I see somebody is
too dumb to know the difference between public>> universities
and private ones.>ahahahah.......ahahahahha.......>lie,
twist it any way you want, Parki-pooh, the lparker@emory.edu
:>..... where as an associate professor for the last 25
years, ..... >you are feeding off the public trough, which
the OP, me and millions >of other taxpayers ll up for
you.....thats why your hangout has... >--- .....to comply
with new **Federal** regulations, Emory University>has
developed a set of **Cost Accounting** Standard Policies.
>--- EMORY UNIVERSITY RECEIVES $225,000 GRANT>FROM U.S.
DEPARTMENT OF ENERGY ......etc., etc....>Do I need to post
more such tid bits and their urls,
Parki-pooh?>ahahahahha........ahahahanson>> Haliburton
received a multi-billion dollar contract from the >
government. Does that make all its employees public
employees?> Perhaps you need to learn how to read.>
Parki-pooh, see, you are twisting and spinning again, but it
doesntdo you no good. We are not talking about them, Parki.
We aretalking about you cussing at taxpayers who fund your
salary.If you are denying that, its probably due to the fact
that you do know this and that you are not delivering enough
for paymentreceived, and feel guilty..........your guilt
shows Parki, it
shows........ahahahahaha.......ahahahahanson
Modern
socialism is a balance of public and private> enterprise with
an adiquate social safety net.>>That means stealing from one
and giving to other via taxes.>>Taxation is Theft>>
Stupidity alert!>Parki-pooh, you do not have to precede any
of your intellectually>empty posts with any alert!. Everyone
knows that you are stupid.>The poster is obviously a
taxpayer, hence a contributor to your>salary, since you are
on the public payroll, as a college professor.>> I see
somebody is too dumb to know the difference between public>
universities and private
ones.>ahahahah.......ahahahahha.......>>lie, twist it any
way you want, Parki-pooh, the lparker@emory.edu :>>.....
where as an associate professor for the last 25 years, .....
>>you are feeding off the public trough, which the OP, me and
millions >>of other taxpayers ll up for you.....thats why
your hangout has... >>--- .....to comply with new **Federal**
regulations, Emory University>>has developed a set of **Cost
Accounting** Standard Policies. >>--- EMORY UNIVERSITY
RECEIVES $225,000 GRANT>>FROM U.S. DEPARTMENT OF ENERGY
......etc., etc....>>Do I need to post more such tid bits and
their urls, Parki-pooh?>>ahahahahha........ahahahanson>>
Haliburton received a multi-billion dollar contract from the
>> government. Does that make all its employees public
employees?>> Perhaps you need to learn how to read.>Parki-pooh, see, you are twisting and spinning again, but it
doesnt>do you no good. We are not talking about them, Parki.
We are>talking about you cussing at taxpayers who fund your
salary.Well, that obviously doesnt include you.>If you are
denying that, its probably due to the fact that you do >know
this and that you are not delivering enough for
payment>received, and feel guilty..........your guilt shows
Parki, it shows........>ahahahahaha.......ahahahahanson
Modern socialism is a balance of public and private>
enterprise with an adiquate social safety net.>>That means
stealing from one and giving to other via taxes.>>Taxation is
Theft>> Stupidity alert!>Parki-pooh, you do not have to
precede any of your intellectually>empty posts with any
alert!. Everyone knows that you are stupid.>The poster is
obviously a taxpayer, hence a contributor to your>salary,
since you are on the public payroll, as a college
professor.>> I see somebody is too dumb to know the
difference between public> universities and private
ones.>ahahahah.......ahahahahha.......>>lie, twist it any
way you want, Parki-pooh, the lparker@emory.edu :>>.....
where as an associate professor for the last 25 years, .....
>>you are feeding off the public trough, which the OP, me and
millions >>of other taxpayers ll up for you.....thats why
your hangout has... >>--- .....to comply with new **Federal**
regulations, Emory University>>has developed a set of **Cost
Accounting** Standard Policies. >>--- EMORY UNIVERSITY
RECEIVES $225,000 GRANT>>FROM U.S. DEPARTMENT OF ENERGY
......etc., etc....>>Do I need to post more such tid bits and
their urls, Parki-pooh?>>ahahahahha........ahahahanson>>
Haliburton received a multi-billion dollar contract from the
>> government. Does that make all its employees public
employees?>> Perhaps you need to learn how to
read.>>Parki-pooh, see, you are twisting and spinning again,
but it doesnt>do you no good. We are not talking about them,
Parki. We are>talking about you cussing at taxpayers who fund
your salary.Well, that obviously doesnt include you.>
AHAHAHHAHA........ahahahaha........taxpayers DO fund your
salary. It wasnt so hard, wasnt it, Parki?Take care dude,
laughs,hahahahaha........ahahahanson>If you are denying that,
its probably due to the fact that you do >know this and that
you are not delivering enough for payment>received, and feel
guilty..........your guilt shows Parki, it
shows........>ahahahahaha.......ahahahahanson
=[snip...]>
ahahahah.......ahahahahha.......> lie, twist it any way you
want, Parki-pooh, the lparker@emory.edu :> You are feeding
off the public trough, which the OP, me and millions > of
other taxpayers ll up for you.....thats why your hangout
has... > --- .....to comply with new **Federal** regulations,
Emory University> has developed a set of **Cost Accounting**
Standard Policies. > --- EMORY UNIVERSITY RECEIVES $225,000
GRANT> FROM U.S. DEPARTMENT OF ENERGY ......etc., etc....> Do
I need to post more such tid bits and their urls, Parki-pooh?>
ahahahahha........ahahahansonPS: Or are you so lthy rich,
that you can afford to sit there, doing> tale,
[snip...]Hanson. Hanson. For Gods sake,
man....Hanson...reach around behind your skull;feel with your
ngertips. Find the smallnode, the bump? Thats the RESET
button,Hanson. PUSH IT, Hanson, PUSH IT. Mark (I probably am
in over my head, but.... :-)
[snip...]>
ahahahah.......ahahahahha.......> lie, twist it any way you
want, Parki-pooh, the lparker@emory.edu :> You are feeding
off the public trough, which the OP, me and millions> of
other taxpayers ll up for you.....thats why your hangout
has...> --- .....to comply with new **Federal** regulations,
Emory University> has developed a set of **Cost Accounting**
Standard Policies.> --- EMORY UNIVERSITY RECEIVES $225,000
GRANT> FROM U.S. DEPARTMENT OF ENERGY ......etc., etc....> Do
I need to post more such tid bits and their urls, Parki-pooh?>
ahahahahha........ahahahanson>> PS: Or are you so lthy rich,
that you can afford to sit there, doing> tale,
[snip...]>> Hanson. Hanson. For Gods sake, man....>
Hanson...reach around behind your skull;> feel with your
ngertips. Find the small> node, the bump? Thats the RESET
button,> Hanson. PUSH IT, Hanson, PUSH IT.>Awe, .....your are
plagiarizing out of Ripleys. Shame on you. Besides,
Parki-pooh already did it . He broke it. Now what
?awawawawa......hahahanson>> Mark (I probably am in over my
head, but.... :-)
[snip...]>
ahahahah.......ahahahahha.......> lie, twist it any way you
want, Parki-pooh, the lparker@emory.edu :> You are feeding
off the public trough, which the OP, me and millions> of
other taxpayers ll up for you.....thats why your hangout
has...> --- .....to comply with new **Federal**
regulations, Emory University> has developed a set of
**Cost Accounting** Standard Policies.> --- EMORY
UNIVERSITY RECEIVES $225,000 GRANT> FROM U.S. DEPARTMENT OF
ENERGY ......etc., etc....> Do I need to post more such tid
bits and their urls, Parki-pooh?>
ahahahahha........ahahahanson>> PS: Or are you so lthy
rich, that you can afford to sit there, doing> tale,
[snip...]>> Hanson. Hanson. For Gods sake, man....>
Hanson...reach around behind your skull;> feel with your
ngertips. Find the small> node, the bump? Thats the RESET
button,> Hanson. PUSH IT, Hanson, PUSH IT.>> Awe, .....your
are plagiarizing out of Ripleys. > Shame on you. Besides,
Parki-pooh already did it . > He broke it. Now what ?>
awawawawa......hahahansonI thought, and still do think, that
theidea was an original.So burst my bubble, already --
Whos/whatsRipleys (Believe It or Die?) and what
didParki-pooh already do thats related to theRESET idea?>
Mark (I probably am in over my head, but.... :-) Mark (I have
not followed this thread, nor intend to but for this small
intrusion :-)
> <SNIP> -->> Intelligence is a constant.
Population grows.>>Irrwahn Grausewitz is living proof of his
sig.Its you who is a living prove of my sig, as it turned
out you are not > even able to read my post, despite
understand its contents. Now go > trolling somewhere else.
Fup2 set accordingly.Ah yes, I almost forgot: *PLONK*Note
That Irrwahn Grausewitzs sig reveals a lot about him.He
apparently thinks [sic] that he is of superior intelligence,
and that this makes him superior to the folks who he thinks
[sic] are of lesser intelligence, even though many, if not
most of the things that make life worth while are produced by
the very folks that Irrwahn Grausewitz thinks [sic] that he is
superior to.that he shouldnt play newgroup policeman, or if
he has an irresistable compulsion to play policemen, that he
should go after the real culprits, rather than bushwhack
posters who respond to posts.--Tom Potter
Im trying to
work out how to calculate the> distance from an N-dimensional
point to an> N-dimensional hull.Ive looked around for an
answer, but most> are specic to a certain dimension.I need
this for a machine learning method Im trying> to implement,
which replaces hyperrectangles in> Neareast Neighbour with
Generalisation with convex> hulls instead. Im currently
using qhull for creating> convex hulls.Any help would be much
appreciated, as Ive> spent the last few days trying to come
up with> a method. I have some ideas, but Im not sure about>
them and would take alot of time to implement.> SO I thought
Id check rst ;)This is a convex program:find the
coordinates of a linear function that is nonnegative on each
of the points of the hull and negative on the query point (a
collection of linear inequality constraints) maximizing the
distance from the query point to the hyperplane of zeros of
the linear function (convex quadratic objective function). So
you should be able to handle this with any convex programming
package, but unless N is very small there is not going to be
much else to do.If youre looking at this in the context of
machine learning, you should also look at Support Vector
Machines, which typically involve similar convex programs for
the distance between two hulls.-- David Eppstein
http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine,
School of Information & Computer Science
=Apparently, you do
->NOT<- want to take me up on the offer. Readerswho have seen
you complain for years about my blocking you and howif I
wasnt things would go swimmingly, might wonder what exactly
areyou afraid of, when I offer to simply step out of the way
if youlljust say the word.>>Theres actually a rather small
group of posters, including this>>poster, a poster that goes
by Nora Baron, Arturo Magidin, and Dik>>Winter, who make
rather remarkably dumb mistakes throughout their>>posts, yet
somehow theyve managed to convince much of the
readership>>that they are competent at mathematics.
Strange, that you have never been able to substantiate any of
these>> rather remarkably dumb mistakes.>>For months now this
poster has been working to cast doubt on a rather>simple
argument, where Ifind terms independent of a variable I
call>m, by challenging my use of m=0 to clear it out.This is,
as you like to say, a fascinating lie.While I have pointed out
that your use of independent term isnonstandard for functions
which are not polynomials, nobody hasobjected to you dening
them to be the value when m=0. So yourclaim that I have been
working to cast doubt [...] by challenging myuse of m=0 to
clear it out is a lie.In fact, everyone agrees that what you
do with m=0 is correct. Whatnobody seems to agree with you
about is the conclusions you derivefrom your analysis of the
m=0 case. Namely, you claim that becausewhen m=0, you have
g_1(0) and g_2(0) multiples of f and g_3(0) coprimeto f, then
it follows that g_1(m) and g_2(m) are always multiples of ffor
every value of m, and g_3(m) is always coprime to f for
everymultiple of m.You lie when you claim that people are
challenging the antecedent ofthe implication. They are
challenging the implication. Challenging theimplication means
saying that the antecedent is true, BUT theconsequent is
not.The challenge is not to what happens when m=0; the
challenge is toyour claim that what happens when m=0 MUST
ALSO HAPPEN WHEN m<>0.>However, its rather a basic thing
that if terms are independent of a>variable, then it doesnt
matter WHAT that variables value is, which>is why theyre
independent.This is a misrepresentation of the issue, born
out of your lack ofunderstanding of the criticisms leveled at
your argument.>Thats a fascinatingly dumb thing to question
for MONTHS and in fact>shows that either Arturo Magidin is
remarkably dumb, or is, as I>believe, a well practiced liar
who knows how to appear convincing.False dichotomy. Other
possibilities:(a) You have misunderstood what I have said,
whether deliberately or through incompetence.(b) You are
remarkably dumb and have made mistakes you are unable to
recognize.(c) Both (a) and (b).(d) Other.
[.snip.]
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A mans capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
gures few readers can criticize. A great many people are
staggered to this extent, that they imagine there must be the
indenite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus
de
Morgan
Magidinmagidin@math.berkeley.edu
=Apparently, you do
->NOT<- want to take me up on the offer. Readers> who have
seen you complain for years about my blocking you and how> if
I wasnt things would go swimmingly, might wonder what exactly
are> you afraid of, when I offer to simply step out of the way
if youll> just say the word.>>Theres actually a rather small
group of posters, including this>>poster, a poster that goes
by Nora Baron, Arturo Magidin, and Dik>>Winter, who make
rather remarkably dumb mistakes throughout their>>posts, yet
somehow theyve managed to convince much of the
readership>>that they are competent at mathematics.> Strange,
that you have never been able to substantiate any of these>>
rather remarkably dumb mistakes.>>For months now this poster
has been working to cast doubt on a rather>simple argument,
where Ifind terms independent of a variable I call>m, by
challenging my use of m=0 to clear it out.This is, as you
like to say, a fascinating lie.Readers who want to know the
truth need only read further down, to seejust how Arturo
Magidin operates.Read below...> While I have pointed out that
your use of independent term is> nonstandard for functions
which are not polynomials, nobody has> objected to you
dening them to be the value when m=0. So your> claim that I
have been working to cast doubt [...] by challenging my> use
of m=0 to clear it out is a lie.In fact, everyone agrees that
what you do with m=0 is correct. What> nobody seems to agree
with you about is the conclusions you derive> from your
analysis of the m=0 case. Namely, you claim that because>
when m=0, you have g_1(0) and g_2(0) multiples of f and
g_3(0) coprime> to f, then it follows that g_1(m) and g_2(m)
are always multiples of f> for every value of m, and g_3(m)
is always coprime to f for every> multiple of m.Here readers
can see the poster focusing on m=0, but obviously ifterms are
independent of the value of m, then ms value just
doesntmatter. > You lie when you claim that people are
challenging the antecedent of> the implication. They are
challenging the implication. Challenging the> implication
means saying that the antecedent is true, BUT the> consequent
is not.The challenge is not to what happens when m=0; the
challenge is to> your claim that what happens when m=0 MUST
ALSO HAPPEN WHEN m<>0.>However, its rather a basic thing
that if terms are independent of a>variable, then it doesnt
matter WHAT that variables value is, which>is why theyre
independent.This is a misrepresentation of the issue, born
out of your lack of> understanding of the criticisms leveled
at your argument.>Thats a fascinatingly dumb thing to
question for MONTHS and in fact>shows that either Arturo
Magidin is remarkably dumb, or is, as I>believe, a well
practiced liar who knows how to appear convincing.False
dichotomy. Other possibilities:(a) You have misunderstood
what I have said, whether deliberately or> through
incompetence.(b) You are remarkably dumb and have made
mistakes you are unable to> recognize.(c) Both (a) and
(b).(d) Other. [.snip.]All thats important is that values
independent of m are independentof m.Its a simple technique
tofind values independent of m by settingm=0.James
Harris
>> Apparently, you do ->NOT<- want to take me up
on the offer. Readers>> who have seen you complain for years
about my blocking you and how>> if I wasnt things would go
swimmingly, might wonder what exactly are>> you afraid of,
when I offer to simply step out of the way if youll>> just
say the word.>>Theres actually a rather small group of
posters, including this>poster, a poster that goes by Nora
Baron, Arturo Magidin, and Dik>Winter, who make rather
remarkably dumb mistakes throughout their>posts, yet somehow
theyve managed to convince much of the readership>that they
are competent at mathematics.> Strange, that you have never
been able to substantiate any of these> rather remarkably dumb
mistakes.>>For months now this poster has been working to
cast doubt on a rather>>simple argument, where Ifind terms
independent of a variable I call>>m, by challenging my use of
m=0 to clear it out. This is, as you like to say, a
fascinating lie.> Readers who want to know the truth need
only read further down, to see> just how Arturo Magidin
operates.We can see very clearly how *you* operate. Youre
afraid to acceptArturos offer to stop replying to you,
because you know that theattention that he (and Nora and Dik
and others) give you with theirreplies is the only thing that
lends any semblance of credibility toyou at all.-- Wayne Brown
(HPCC #1104) | When your tails in a crack, you
improvisefwbrown@bellsouth.net | if youre good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) = -1
-- Euler | -- John Myers Myers, Silverlock
=no, dont say
it ... its too obvious ... youll ruin your mind! oh, heck;
the medium is the message. yeeha. > attention that he (and
Nora and Dik and others) give you with their> replies is the
only thing that lends any semblance of credibility to--les
ducs dEnron!
<snip>:> What is the false implication?>
::> <snip>> : The false implication is that b is a non-unit
factor of 2 like> : sqrt(2), or 2^{1/3}, when in fact,
appropriately, its a factor of 1.> : Unfortunately the
poster deleted out pertinent information, so Ill> : make the
effort to put it back in so that it makes sense to readers> :
without forcing them to go back to previous posts.> : I
gave the example of xy = 2, where x and y are algebraic
integers,> : where x=2a and y=b, so b is an algebraic
integer, but a is not, but> : it is a member of what I call
the object ring.> : The object ring is a commutative ring
that includes all numbers such> : that -1 and 1 are the only
members that are both a unit and an> : integer, where no
non-unit member is a factor of any two integers that> : are
coprime.> : Notice that denition for the object ring
excludes possibilities like> : a=1/2 as then 2 would be a
unit since 2(1/2) = 1.> : So, in fact, in the object ring,
ab=1 and b IS a unit, but because a> : is inappropriately
excluded from the ring of algebraic integers by the> :
denition of algebraic integers as roots of monic polynomials
with> : integer coefcients, you have the false implication
that b is some> : other type factor of 2.> : Now then, say
you follow algebra, and then youfind that you have 2 as> : a
factor of x, well youve been pushed out of the ring of
algebraic> : integers where x does NOT have 2 as a factor,
even though x=2a.> :> :> : And you can see several problems
that popped up by that exclusion as b> :> :> : is not a unit
in the ring of algebraic integers, when it should be,> :> :>
: and x does NOT have 2 as a factor in the ring, though
x=2a.> :> ::> :> So, what is the problem? I havent seen
anything that looks like > :> :> a contradiction, only an
assertion that something is not a unit> :> :> which should be
according to some unstated principle. Whats the> :> :>
principle?> :> : Given that xy=2, with x=2a, y=b, with b an
algebraic integer, x an> :> : algebraic integer, while a is
not, it *should* be that y does not> :> : share non-unit
factors with 2, since x *should* have 2 as a factor,> :> :
but in the ring of algebraic integers, because the denition>
:> : arbitrarily excludes a, neither of those is the case.>
::> What is behind the shoulds, i.e. what are the bad
consequences of > :> excluding a? Before you were talking
about being able to prove two > :> contradictory results.>
::> Mike> : The problem occurs because algebra does NOT
recognize the arbitrary> : denition, so using algebra you
canfind yourself in situations where> : you prove that x has
2 as a factor, but then again, because of this> : arbitrary
denition, it does not, in the ring of algebraic integers.>
: Since the assumption is that algebraic integers is an
appropriate> : ring, it appears that youve proven two
contradictory things:What do you mean by an appropriate
ring?One where you cant appear to prove contradictory
things. > : 1. x has 2 as a factor> : 2. x does NOT have 2
as a factorBut arent 1. and 2. actually:1. x has 2 as a
factor in the object ringIll remind of the example of 2 and
6 in the ring of evens to helpreaders with context, as there
2 is not a factor of 6 because 3 isnteven.Now then, with my
example, you have x=2a, where x is an algebraicinteger, but
as a is not, it doesnt have 2 as a factor, in the ringof
algebraic integers, which falsely implies, given that xy=2,
where yis an algebraic integer that x and y have non-unit
factors of 2 in thesame sense as like if you have sqrt(2) as
a factor for both.But, in fact, *neither* has what you might
call non-trivial factors of2 in the ring of algebraic
integers, which I have to say as triviallythey have
themselves as factors.That is, given that xy=2, x is
trivially a factor of 2, in the ring ofalgebraic integers, as
is y, when in fact x has a factor that IS 2, anon-trivial
factor, in an appropriate ring, that is, one which doesnot
allow contradictions.Its worth noting that proving that x
doesnt have a factor of 2, inthe ring of algebraic integers,
requires going to the eld ofalgebraic numbers, which is
itself, of course, not awed.> 2. x does NOT have 2 as a
factor in the algebraic integersAnd proving that requires
going to the eld of algebraic numbers,which hasnt really
been discussed at this point.So then *in the ring of
algebraic integer* you can appear to prove*both* things, and
only gure out that youve been pushed out of thering of
algebraic integers, by going to the eld of
algebraicnumbers.> These dont seem contradictory. MikeI keep
talking about using mathematical logic as you can NOT just
relyon your intuition.If you use mathematical logic then I
can simply take you through eachstep of the proof, and then
once you see what has to bemathematically, then you can begin
to tailor your intuition to thereality.Months back--over a
year ago--I even talked about quantum mechanics intrying to
give an analogy, as an example of an advance in the eldthat
forced its adherenents--there physicists--to rely less on
theirintuition.But as a ccontrary group, mathematicians may
simply be refusing to dowhat physicists had to do--rely less
on intuition.What Ifind troubling is that even when
repeatedly directed to rely onmathematical logic,
mathematicians seem almost childishly unwilling todepend on
what should be the very basis for their discipline.James
Harris
Im not following. You let x = 2a where x is an
algebraic integer> and a is not. So x does not have a factor
of 2 in the algebraic> integers. What is the contradictory
proof that x *does* have a factor > of 2 in the algebraic
integers?JSH does not answer questions.Gib
=it does, if you
permit a change of tense of shall;that is to say,its a
conjecture that remains to be proven ... surely,I shall ...
at least, I certainly could! Who you callin Inscrute,
mister? > The fundamental problem is that should is not a
word with a generally > accepted meaning in mathematics. JSH
follows his own inscrutable rules, --Dec.2000 WAND Chairman
Paul ONeill, reelected to Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=( also
posted to
sci.research.careers,comp.programming,sci.physics,
sci.physics.computational.uid-dynamics,sci.math.num-analysis
)I was wondering if anyone could give me feedback on this
topic. I am34, have a scientic background, with Bachelors
Degrees in Physicsand Math, graduate work in Physics and
Optics culminating with an MSin Optics. I have worked as an
engineer, and then as a programmer forthe last 5 years.
However, it has been comercial programming (usingMicrosoft
technologies such as Vsual C++, ASP, etc) and while I
amthankful to have a job in the current economic climate, I
am alsounhappy with the kind of work I am doing. I would like
to do some sortof scientic programming, but I have not had
much success applying tosuch jobs. They either require an
existing security clearance or askfor tons of prior scientic
programming experience.So my questions are:1) What do you
think my chances are of getting such a scienticprogramming
job? Would you expect a big difference when the
economyimproves, hopefully in a few years?2) Any suggestions
of specic places where I can apply for arelatively entry
level scientic programming position?2) Are scientic
programming jobs more secure than your typicalcommercial
position? I have heard that being over 40 and a
commercialprogrammer is a risky place to be, because of age
discrimination andgeneral outsourcing. Are there many
scientic programmers who so thattill they retire?If I cant
nd such a job, I am considering getting an MBA, but Ireally
think I would not enjoy that type of work nearly as much.-Sam
Banerjee
( also posted to
sci.research.careers,comp.programming,sci.physics,
sci.physics.computational.uid-dynamics,sci.math.num-analysis
> )> I was wondering if anyone could give me feedback on
this topic. I am> 34, have a scientic background, with
Bachelors Degrees in Physics> and Math, graduate work in
Physics and Optics culminating with an MS> in Optics. I have
worked as an engineer, and then as a programmer for> the last
5 years. However, it has been comercial programming (using>
Microsoft technologies such as Vsual C++, ASP, etc) and while
I am> thankful to have a job in the current economic climate,
I am also> unhappy with the kind of work I am doing. I would
like to do some sort> of scientic programming, but I have
not had much success applying to> such jobs. They either
require an existing security clearance or ask> for tons of
prior scientic programming experience.>> So my questions
are:>> 1) What do you think my chances are of getting such a
scientic> programming job? Would you expect a big difference
when the economy> improves, hopefully in a few years?>> 2) Any
suggestions of specic places where I can apply for a>
relatively entry level scientic programming position?>> 2)
Are scientic programming jobs more secure than your typical>
commercial position? I have heard that being over 40 and a
commercial> programmer is a risky place to be, because of age
discrimination and> general outsourcing. Are there many
scientic programmers who so that> till they retire?>> If I
cantfind such a job, I am considering getting an MBA, but I>
really think I would not enjoy that type of work nearly as
much.> -Sam BanerjeeYou didnt mention where you have tried
tofind work, but one of the major employers of scientic
programmers are companies involvedin test equipment and
medical equipment. There are positions at all levels in such
companies, and you mightfind a niche in one ofthem.--There
are two things you must never attempt to prove: the
unproveable -- and the obvious.--Democracy: The triumph of
popularity over principle.--http://www.crbond.com
( also
posted to
sci.research.careers,comp.programming,sci.physics,
sci.physics.computational.uid-dynamics,sci.math.num-analysis
> )> I was wondering if anyone could give me feedback on this
topic. I am> 34, have a scientic background, with Bachelors
Degrees in Physics> and Math, graduate work in Physics and
Optics culminating with an MS> in Optics. I have worked as an
engineer, and then as a programmer for> the last 5 years.
However, it has been comercial programming (using> Microsoft
technologies such as Vsual C++, ASP, etc) and while I am>
thankful to have a job in the current economic climate, I am
also> unhappy with the kind of work I am doing. I would like
to do some sort> of scientic programming, but I have not had
much success applying to> such jobs. They either require an
existing security clearance or ask> for tons of prior
scientic programming experience.So my questions are:1) What
do you think my chances are of getting such a scientic>
programming job? Would you expect a big difference when the
economy> improves, hopefully in a few years?2) Any
suggestions of specic places where I can apply for a>
relatively entry level scientic programming position?2) Are
scientic programming jobs more secure than your typical>
commercial position? I have heard that being over 40 and a
commercial> programmer is a risky place to be, because of age
discrimination and> general outsourcing. Are there many
scientic programmers who so that> till they retire?If I
cantfind such a job, I am considering getting an MBA, but I>
really think I would not enjoy that type of work nearly as
much.My guess, FWIW, is that people usually get into
scientic programming by being involved with a research
project in a university. Maybe your best bet would be to seek
work in the eld of optics, where you have already established
your credentials.Gib
>could some expert recommend a
Liapunov function that might help>control this plant
represented by a nonlinear differential equation:>>y(t) =
x(t) * ( 1 - F(s) * (|x(t)|^2) )>>here F(s) = K/(1+sb),
represents a 1st order low pass lter>(convolution with
Kexp(-bt)) with time constant 1/b, and K is a>small positive
number, so y is nearly x but not exactly; both b and>K are
unknown although I have a good idea of their value within
a>factor of 2.>>( By F(s) * (|x|^2) is meant the square of
x(t), x(t)^2, is low pass>ltered by F(s).)>>I need a control
law x = C[u, y],for positive u>0, such that the>tracking error
|y-u| is small.>I think what you said was dx/dt = x - F x^3 Is
this what you mean? If so, you get dz/dt = (-z + F )/2with z =
1/x^2This equation should be much easier to work
with.
Evaluate the innite sum 1/1 + 2/2 + 3/4 + 4/8
+ ... .?????>> Okay, I evaluated it... now what should I
do?>> adam>>with a preposition? do is just add up all those
numbers.Ill get you started:1 + 2 = 33 + 3/4 = 15/415/4 + 4/8
= 17/4etcadam
=The is a physics problem, but Im hoping
someone here might be able to help.Im trying tofind the
tangential velocity of a body (mass: .127kg), yingaround in
a circle from a string of unknown length. The radius is the
circleis .43m, and using an astrolabe, I found the angle to
be 58 degrees.With v = wr, my problem right now is trying to
nd the angular velocity (w)to mutliply by the radius. I
converted the 58 degrees to radian to get 1.012rad, but Im
not sure what else to do. The time for the body to complete
afull circle is not given, and were supposed to be able to
get this usingthe information listed above.Please help!
The is a physics problem, but Im hoping someone here might
be able to help.> Im trying tofind the tangential velocity
of a body (mass: .127kg), ying> around in a circle from a
string of unknown length. The radius is the circle> is .43m,
and using an astrolabe, I found the angle to be 58
degrees.The angle of *what* is 58 deg? And it sounds like
regardless of thedetail of the string setup, the object
ultimately moves in acircle of radius 0.43m. So just measure
the time it takes forthe object to go around the circle, and
the tangential velocityis then simply (0.43m*2pi)/T.If you
cant measure the time, then try to describe the
problembetter. One end of the string is apparently attached
to theobject. What is the other end attached to? Is that
mountpointmoving? If so, how is it moving? And where does the
58 degree anglet in to all that?-- Rich Carreiro
rlcarr@animato.arlington.ma.us
> The is a physics
problem, but Im hoping someone here might be able to help.>
Im trying tofind the tangential velocity of a body (mass:
.127kg), ying> around in a circle from a string of unknown
length. The radius is the circle> is .43m, and using an
astrolabe, I found the angle to be 58 degrees.>> The angle of
*what* is 58 deg? And it sounds like regardless of the> detail
of the string setup, the object ultimately moves in a> circle
of radius 0.43m. So just measure the time it takes for> the
object to go around the circle, and the tangential velocity>
is then simply (0.43m*2pi)/T.>> If you cant measure the
time, then try to describe the problem> better. One end of
the string is apparently attached to the> object. What is the
other end attached to? Is that mountpoint> moving? If so, how
is it moving? And where does the 58 degree angle> t in to
all that?Already answered on sci.physics and
sci.physics.relativity.Dirk Vdm
=answer.> The is a
physics problem, but Im hoping someone here might be able
tohelp.> Im trying tofind the tangential velocity of a
body (mass: .127kg),ying> around in a circle from a string
of unknown length. The radius is thecircle> is .43m, and
using an astrolabe, I found the angle to be 58 degrees.>> The
angle of *what* is 58 deg? And it sounds like regardless of
the> detail of the string setup, the object ultimately moves
in a> circle of radius 0.43m. So just measure the time it
takes for> the object to go around the circle, and the
tangential velocity> is then simply (0.43m*2pi)/T.>> If you
cant measure the time, then try to describe the problem>
better. One end of the string is apparently attached to the>
object. What is the other end attached to? Is that
mountpoint> moving? If so, how is it moving? And where does
the 58 degree angle> t in to all that?>> Already answered on
sci.physics and sci.physics.relativity.>> Dirk Vdm>>
Of
course its not. But it seems to me that his skepticism
regarding>whether that proof _can_ be formalized must in fact
be skeptism>regarding whether its possible to state all the
necessary axioms>and deduce the theorem without any appeal to
the way things>look.As I have stressed in my other post, Im
*not* skeptic about the factthat that proof can be
formalized!Im simply puzzled thinking of how it would be
done, since IMHO it isa proof that indeed *does* make heavily
appeal to the way things look,and in this sense it is so
simple that often one canfind it inpopularization books.(BTW)
Also, since you mentioned axiomatic euclidean geometry, I
justdont know anything about it[*], but I dont know wether
it ispossible in its formalism to dene/talk about
continuoustransformations, that are indeed the basis of that
proof.So, if I am right in my assumption above, while the
object of theproof can be reasonably dened in axiomatic
euclidean geometry,either the theorem cant be proved in that
teory or there exists analternative proof that does not use
continuity arguments.[*] This means: so I may well be
wrong!Michele-- > Comments should say _why_ something is
being done.Oh? My comments always say what _really_ should
have happened. :)- Tore Aursand on
comp.lang.perl.misc
=Could someone please give me a proof
for distributive law of the crossproduct:a x (b + c) = a x b
+ a x cBut theres a catch! I am assuming the we have dened
the crossproduct as (|a||b|sinO)n, etc. Using the above proof
(that is, takingthe above as true) THEN I will use the above
truth to derive therectangular form of the cross product.
So I need the above provedwithout resorting to the
rectangular form/defn of the cross product.Thanx in
advanceMB
=Could someone please give me a proof for
distributive law of the cross> product:a x (b + c) = a x b +
a x cBut theres a catch! I am assuming the we have dened
the cross> product as (|a||b|sinO)n, etc. Using the above
proof (that is, taking> the above as true) THEN I will use
the above truth to derive the> rectangular form of the
cross product. So I need the above proved> without resorting
to the rectangular form/defn of the cross product.> Proof it
for the canonical unit vectors e1, e2, e3. This is easy,there
are only a few cases to consider.Then set a = a1*e1 + a2*e2 +
a3*e3 where a1, a2, a3 are the coodinatesof a. The same for b
and c and see what you can get. Probably it isuseful to do the
case rst where a is one of the canonical unitvectors and only
b and c are arbitrary.
=If we visualize a cubic graph (with
2n vertices and 3n edges) as anetwork of water pipes, and if
water is owing through the pipes in aclosed system, then the
ows through each of the edges must satisfythe condition that
the ows into any vertex are equal to the owsout. This leads
to a set of 2n equations in 3n unknowns which must besatised
by any ow. At least one of these 2n equations is impliedby
the rest, since the sum of all the equations is 0, thus the
rank ofthe system of equations is at most 2n-1. Ive tested
all bridgefreeregular cubic graphs up to n=7 and it appears
that for these graphs,the rank is always exactly equal to
2n-1. Is this true in general, orare there counterexamples,
or is it an open question?
If we visualize a cubic graph
(with 2n vertices and 3n edges) as a> network of water pipes,
and if water is owing through the pipes in a> closed system,
then the ows through each of the edges must satisfy> the
condition that the ows into any vertex are equal to the
ows> out. This leads to a set of 2n equations in 3n unknowns
which must be> satised by any ow. At least one of these 2n
equations is implied> by the rest, since the sum of all the
equations is 0, thus the rank of> the system of equations is
at most 2n-1. Ive tested all bridgefree> regular cubic
graphs up to n=7 and it appears that for these graphs,> the
rank is always exactly equal to 2n-1. Is this true in
general, or> are there counterexamples, or is it an open
question?Form a cycle basis of the graph: take a spanning
tree (2n-1) edges) and form n+1 cycles by using each
remaining edge together with the path connecting its
endpoints in the tree.You can form a ow (circulation) in the
graph by choosing for each cycle a ow that is zero elsewhere,
and adding these cycles.Conversely any ow is the sum of ows
on the cycles in a unique way determined by the ow values on
the non-tree edges.So the space of ows has dimension exactly
n+1, as a subspace of the 3n-dimensional space of independent
ow amounts on edges, and the system of ow conditions has
rank exactly 2n-1.This can all be viewed as doing basic
homology theory on the graph...-- David Eppstein
http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine,
School of Information & Computer Science
=James Harris
[...]-- G.C.
=What is the interpretation on z_1*z_2 = ln
z_1 + ln z_2on a complex plane? Is it true thatfor any ln z_1
and any ln z_2 exists ln z_1*z_2 such thatln z_1*z_2 = ln z_1
+ ln z_2 ?If not, what the rst order logic statement is
correct, then?
What is the interpretation of>ln z_1*z_2 =
ln z_1 + ln z_2>on a complex plane? Is it true that>for any ln
z_1 and any ln z_2 exists ln z_1*z_2 such that>ln z_1*z_2 = ln
z_1 + ln z_2 ?In this case that does happen to be true,
because different branchesof ln(z) differ by a multiple of 2
pi i. Or you could choose, say, values of ln z_1 and ln(z_1
z_2) and get a value of ln z_2.On the other hand, a statement
such as ln z^2 = 2 ln z is trickier: every value of the right
side is a value of the left side, but not every value of the
left side is a value of the right side. IMHO it wouldbe
better to writeln (z_1 z_2) = ln(z_1) + ln(z_2) + 2 pi i n
for some integer nwhich is true no matter what branches you
choose for all the logarithms.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
What is the interpretation
of>>ln z_1*z_2 = ln z_1 + ln z_2>>on a complex plane? Is it
true that>>for any ln z_1 and any ln z_2 exists ln z_1*z_2
such that>>ln z_1*z_2 = ln z_1 + ln z_2 ?>> In this case that
does happen to be true, because different branches> of ln(z)
differ by a multiple of 2 pi i. Or you could choose, say,>
values of ln z_1 and ln(z_1 z_2) and get a value of ln z_2.>
On the other hand, a statement such as ln z^2 = 2 ln z is
trickier:> every value of the right side is a value of the
left side, but not every> value of the left side is a value
of the right side. IMHO it would> be better to write>> ln
(z_1 z_2) = ln(z_1) + ln(z_2) + 2 pi i n for some integer n>>
which is true no matter what branches you choose for all the
logarithms.Could the identityln (z_1 z_2) = ln(z_1) +
ln(z_2)be somehow interpreted in terms of Riemann Surfaces,
without referring tobranches and multivalued functions?
Could the identity>> ln (z_1 z_2) = ln(z_1) + ln(z_2)>> be
somehow interpreted in terms of Riemann Surfaces, without
referring to> branches and multivalued functions?Or to put it
bluntly, can I add Riemann Surfaces?
=I left the quotes that
are not just silly, but it has to be saidthat a)Bucky *did*
use at least one mathematicians work, Coxeters, and b)_S_
is not a textbook of math, and contains no proofs. as I
said,nothing more difcult than deploying the pythagorean
theorem -- andhere are two things that are even useful,
sinceI realized that they could be
magnied:http://www.rwgrayprojects.com/synergetics/plates/gs
/plate02.htmlhttp://www.rwgrayprojects.com/synergetics/plates
/gs/plate01.htmlthe words about co-ordinate systems are ne,
but, then,Bucky was also entirely correct about the
importanceof the IVM, the dual of the net of packed rhombical
dodecahedra;yes, trigonal or hexagonal tiling of hte plain is
equivalentto doing it with tetragona (skwares), but it misses
the point. as for Synergetics co-ordinates,Bucky didnt ever
nd what he was looking for, as faras i can tell, and Nelson
(so far) has only come-upwith a kind of homogenous
co-ordination --which is awfully obfuscated in his Wolframite
layout, ifthats what it is. (actually,trigonal co-ordinates
are inherently homogenous in the plain,unless you just ignore
one of the sets of lines, andthink of the as oblique
(cartesian) ones.) > (1)
http://www.rwgrayprojects.com/synergetics/synergetics.html>
(2) http://www.mathforum.org/epigone/geometry-research> (3)
http://library.wolfram.com/infocenter/MathSource/600> (4)
http://mathworld.wolfram.com/SynergeticsCoordinates.html >
fundamental structure of the plane is hexagonal) that sound
wonderful > but have no meaning. In fact its a mere matter
of convenience whether > we use orthogonal or hexagonal
coordinates - neither is intrinsically > better than the
other; its just that some coordinate-systems are better >
suited to some problems than others. Ive used dozens of
different > I agree with you that Fuller was under no
obligation to study earlier > work. But if he had done so,
his writings might have been much more > ...Fuller gives
plenty of evidence of not knowing what a proof is, and > As I
said, Fullers writings on geometry are meaningful and
probably > usually correct. But let me assure you that you
would have gained so > much more by reading geometry from a
more sensible source. --UN HYDROGEN (sic; Methanex (TM)
reformanteurs) ECONOMIE?...La Troi Phases dExploitation de
la Protocols des Grises de Kyoto:(FOSSILISATION
[McCainanites?] (TM/sic))/BORE/GUSH/NADIR @
http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/
bushb.htm (content partiale, below): 17 -- LATTEMPTER de
COUP DETAT, 3/30/81
=oops; let me see if I can get those
excerpts seperatedby whitespace. but heres the best one by
JHC-- and very true, ye olde Bucky Witches --which Nelson
cleverly put at the end:As I said, Fullers writings on
geometry are meaningful and probably usually correct. But let
me assure you that you would have gained so much more by
reading geometry from a more sensible source. >
http://www.rwgrayprojects.com/synergetics/plates/gs/plate01.
html > Bucky didnt everfind what he was looking for, as far>
as i can tell, and Nelson (so far) has only come-up> with a
kind of homogenous co-ordination --> which is awfully
obfuscated in his Wolframite layout, if> thats what it is.
(actually,> trigonal co-ordinates are inherently homogenous
in the plain, > (1)
http://www.rwgrayprojects.com/synergetics/synergetics.html >
(4) http://mathworld.wolfram.com/SynergeticsCoordinates.html fundamental structure of the plane is hexagonal) that sound
wonderful > but have no meaning. In fact its a mere matter
of convenience whether > we use orthogonal or hexagonal
coordinates - neither is intrinsically > better than the
other; its just that some coordinate-systems are better >
suited to some problems than others. Ive used dozens of
different > I agree with you that Fuller was under no
obligation to study earlier > work. But if he had done so,
his writings might have been much more > ...Fuller gives
plenty of evidence of not knowing what a proof is, and
--Dec.2000 WAND Chairman Paul ONeill, reelected to Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
oops;
let me see if I can get those excerpts seperated> by
whitespace. but heres the best one by JHC> -- and very true,
ye olde Bucky Witches --> which Nelson cleverly put at the
end:> As I said, Fullers writings on geometry are meaningful
and probably> usually correct. But let me assure you that you
would have gained so> much more by reading geometry from a
more sensible source.I suppose that Synergetics is internally
consistent.It is signicant ( to me, at least) that nobody I
ever met who thinksthat Synergetics is a Mathematic
Masterpiece, knows any Mathematics whateverbeyond CalcI, if
that.Although I have a small sample of his audience, I fell
safe to state that heis probably a guru with Mensa-type
numerologists.RJ Pease
=Bob, can I call you, Dickhead?
seriously,it isnt difcult to read, if you just peruse itby
the pictures that appeal to you, andignore the run-on
language. you can just as well begin with _S2_, ifyou canfind
a copy at your library, as wellas _S1_, if you dont want use
the combined online thing. but, yes, you are referring to
whom I call Bucky Witches,wich is what a gaggle of them
called themselves,when I met them at a local modelling
workshopput on by Amy Edmundson. oh, ****;search for *her*
book, which is also online,for a fabulously concise
accounting of the math.Bucky didnt fall for numerology at
all, althoughhe came close with the Scheherezade
numbers,which you canfind in _S_. > Although I have a small
sample of his audience, I fell safe to state that he> is
probably a guru with Mensa-type numerologists.--les ducs
dEnron!
=as someone whos actually read _S_, but who left
both volumesat a library, since its really quite simple--
nothing more difcult than the pythagorean theorem is needed
--and I can always get the online edition, I have to saytaht
Conways comments are primarily irrelavent prattle (butIll
review them, when I get the chance). one can take issue with
a lot of things (as I just did,below), as I do with his
so-called a-political practice, andsome of his speculative
prehistory, or with his cute waywith English, but I dont
care. he actaully has made some dyscoveriesthat are
incredibly important (or just very basic; hey).thus quoth:
discussion of Buckminster Fullers works
<GEODESIC@LISTSERV.BUFFALO.EDU>in other words, if youre
really into it,get an elementary book on numbertheory, orits
basically hopeless. truly fascinating,once you can stand
it!--UN HYDROGEN (sic; Methanex (TM) reformanteurs)
ECONOMIE?...La Troi Phases dExploitation de la Protocols des
Grises de Kyoto:(FOSSILISATION [McCainanites?]
(TM/sic))/BORE/GUSH/NADIR @
http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/
bushb.htm (content partiale, below): 17 -- LATTEMPTER de
COUP DETAT, 3/30/81 > <opinions of Conway, full of hearty
common sense, snipped>
Here are some quotes of John
Horton Conway about R. Buckminster Fullers > books
Synergetics: an exploration in the geometry of thinking, and
> Synergetics 2 (see 1), from geometry-research (see 2). All
of the quotes > are completely out of context; you cant tell
what he was replying to.> Do mathematicians follow JHC like he
said some people follow RBF?Its nice to see that JSH isnt
the only outsider persecuted byestablishment
mathematicians.
=Clifford J. Nelson> Here are some quotes
of John Horton Conway about R. Buckminster Fullers> books
Synergetics: an exploration in the geometry of thinking, and>
Synergetics 2 (see 1), from geometry-research (see 2). All of
the quotes> are completely out of context; you cant tell
what he was replying to.>> Ill bet a lot of people wont
read the Mathematica notebook> SynergeticsApplication7 (see
3), or see anything new or worthwhile in> the mathworld entry
on synergetics coordinates (see 4) because of the> answer to
the question below.No argument there.This Synergetics cult,
and others like it, might have fewer followers if themasses
were better educated about simple stuff like coordinates
andtesselations. On the other hand, there will always be
zealots, and peoplewho, owing to vanity, subscribe to
whatever would-be super-theory is aucourant.The word
cybernetics came massively into fashion in the 50s; maybe
thesetwo coinages were attempts to cash in on it.Enough of
this crap:)LH
=[SNIP - mersennes]> I also wonder about y^n
+ 1 and about y! - 1 or y! + 1 being prime.> And eh, another
thing I wonder is if there have been experiments wiht>
representing primes in a different system than the decimal
system (for> instance a kind of system where each digit is
2*previous_digit^2 in value.> And lastly I also wonder if
there have been any theories on correlations> between prime
number dispersion (I cant think of the right word) and>
fractals, or any complex iteration model.By similar
arguments, if y^n + 1 is prime, then y = 2 and n = 2^k. The>
numbers 2^2^k + 1 are Fermat numbers, and only ve of them
are proved> primes, for k = 0, 1, 2, 3 and 4. For k = 5 to 25
or more (tha last a big> number), they are composite.Y need
not be 2.Daniel Heuers recent 1372930^2^17+1 is prime (and
huge, the 5th largest in the world presently at 804474
digits)These are called Generalised Fermat Numbers, of just
GFNs.http://primes.utm.edu/top20/page.php?id=3Phil--
Unpatched IE vulnerability: history.back method
cachingDescription: cross-domain scripting,
cookie/data/identity theft, command executionExploit:
http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM
=Phil Carmody <thefatphil_demunged@yahoo.co.uk> escribi.97
1 and about y! - 1 or y! + 1 being prime.> And eh, another
thing I wonder is if there have been experiments wiht>
representing primes in a different system than the decimal
system (for> instance a kind of system where each digit is
2*previous_digit^2 invalue.> And lastly I also wonder if
there have been any theories oncorrelations> between prime
number dispersion (I cant think of the right word) and>
fractals, or any complex iteration model.>> By similar
arguments, if y^n + 1 is prime, then y = 2 and n = 2^k. The>
numbers 2^2^k + 1 are Fermat numbers, and only ve of them
are proved> primes, for k = 0, 1, 2, 3 and 4. For k = 5 to 25
or more (tha last abig> number), they are composite.>> Y need
not be 2.>> Daniel Heuers recent 1372930^2^17+1 is prime>
(and huge, the 5th largest in the world presently at 804474
digits)>> These are called Generalised Fermat Numbers, of
just GFNs.>> http://primes.utm.edu/top20/page.php?id=3>>
Phil>Of course, I mismatch the statements. The correct
statement is:2^n + 1 is an odd prime
> n = 2^kI apologize
...-- Ignacio Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
[SNIP -
mersennes]>> By similar arguments, if y^n + 1 is prime, then
y = 2 and n = 2^k. The>> numbers 2^2^k + 1 are Fermat
numbers, and only ve of them are proved>> primes, for k = 0,
1, 2, 3 and 4. For k = 5 to 25 or more (tha last a big>>
number), they are composite.>Y need not be 2.>Daniel Heuers
recent 1372930^2^17+1 is prime >(and huge, the 5th largest in
the world presently at 804474 digits)>These are called
Generalised Fermat Numbers, of just GFNs.Or for smaller
examples, try 6^2+1, 6^(2^2)+1, 10^2+1, 14^2+1,20^2+1,
20^(2^2)+1. Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
I also
wonder about y^n + 1 and about y! - 1 or y! + 1 being
prime.>and snoozing 4! + 1 = 25; 5! + 1 = 121; 6! + 1 =
721are compositey! - 1 however is more interesting.
I
will answer your questions below up here.> I dont like the
idea of using a real value in the vector elds.> I believe
that they should be unsigned magnitudes.> To impose a
sub-sign seems incoherent to me.> I admit that the math still
works out but its like buildng the reals> from the reals.> I
agree there are some odd discontinuities as a path passes
between> the thirds of the plane (three-signed).> If for
example you look at a graph of the unit circle in the>
three-signed domain for one third of the plot each sign has
zero> component in the reduced form:star(theta)> ^> |. .> | .
.> | . .> | . .> | . .> | . .> | . . > | ............. >
------------------------------------>theta> 0 2pi/3 pi 4pi/3
2piThis is a crude graph of the star component of the unit
circle versus> theta.> The star pole is at theta equals
zero.Still, even with these discontinuities the math matches
the complex> numbers for sum and product.So I guess the
four-signed product doesnt ring any bells?> I still havent
managed the transform with all of its triginometry.> I can
understand the desire to work in vectors. The philosophy of
what> we are doing is still polysigned numbers.Just an
observation: these arent discontinuities in the sense used
in calculus. The derivative, however, has two
discontinuities.-- Will Twentyman
=160 years ago today, on
October 16, 1843, mathematician W. R. Hamiltoninvented
quarternions, also known as hypercomplex numbers. He beganby
postulating a second imaginary, j, whose square is -1, but
which isnot equal to i. He then proved:1) that the product of
i x j must be a third imaginary, k, whosesquare is also -12)
multiplication of quarternions is not commutative (e.g., i x
j = -j x i).
W. R. Hamilton invented... hypercomplex
numbersBeing currently enrolled for a course in Complex
Analysis, i strongly opose to that theory. For once, id like
to know it all without a next bigger idea awaiting. :)--
KindlyKonrad-------------------------------------------------
places;their souls be chased by demons in Gehenna from one
room toanother for all eternity and more.Sleep - thing used
by ineffective people as a substitute for coffeeAmbition - a
poor excuse for not having enough sence to be
lazy---------------------------------------------------
160 years ago today, on October 16, 1843, mathematician W. R.
Hamilton> invented quarternions, also known as hypercomplex
numbers.Who actually uses these numbers?I mean, ordinary
complex numbers are abundant, even in engineering.
Buthypercomplex? Are they related to Minkowski space, i.e.
special relativity(simply because the dimension is four). Are
they used in theoretical physicsotherwise?I seem to remember
some eight-dimensional numbers too (Cayley?), but thenyoud
lose either the associative or distributive law, I forget
which. Arethey of any use?And thats it. Apparently, you
cannot go into higher dimension, and stillretain familiar
properties. What is the signicance of that
fact?-Michael.
160 years ago today, on October 16,
1843, mathematician W. R. Hamilton>>invented quarternions,
also known as hypercomplex numbers.Who actually uses
these numbers?>>I mean, ordinary complex numbers are
abundant, even in engineering. But>hypercomplex? Are they
related to Minkowski space, i.e. special relativity>(simply
because the dimension is four). Are they used in theoretical
physics>otherwise?>>I seem to remember some eight-dimensional
numbers too (Cayley?), but then>youd lose either the
associative or distributive law, I forget which. Are>they of
any use?>>And thats it. Apparently, you cannot go into
higher dimension, and still>retain familiar properties.
What is the signicance of that fact?>>-Michael.>Arent
quaternions used in number theory? (A sincere question - I
dont really know.) Plus they provide a canonical example of
a skew eld.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
Arent quaternions used in
number theory? (A sincere question - I dont >really know.)
Plus they provide a canonical example of a skew eld.They can
be used to prove that every positive integer is a sum of four
squares (see e.g. Herstein, Topics in Algebra, sec. 7.4).On
the other hand, Lagrange proved that theorem long before
quaternionswere invented.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
160 years ago today, on
October 16, 1843, mathematician W. R. Hamilton> invented
quarternions, also known as hypercomplex numbers.Who actually
uses these numbers?I mean, ordinary complex numbers are
abundant, even in engineering. But> hypercomplex? Are they
related to Minkowski space, i.e. special relativity> (simply
because the dimension is four). Are they used in theoretical
physics> otherwise?I probed that question a bit when I rst
learned about hypercomplexnumbers while I was an
undergraduate. A classmate of mine, who hadknown about
hypercomplex numbers for awhile, said that WernerHeisenburg
had originally used them in his matrix formulation ofwas
wrong, and Heisenburg proceeded to redo his matrix
formulationusing ordinary imaginary numbers. Since then, as
far as we coulddetermine, nobody had used them for
anything.The bottom line is that theres no guarantee a
mathematical conceptwill prove useful just because we can
conceive it. But Im not awarethat ordinary complex numbers
had any apparent practical use when theywere rst formulated,
either. I think Heisenburg had the right idea,even though his
theory turned out to be wrong: If youfind
somethingunexplained, at least consider what would happen if
hypercomplexnumbers were involved and see whether this would
explain anything. Itmay not. But, then again, it may!Bob
> 160 years ago today, on October 16, 1843, mathematician W.
R. Hamilton> invented quarternions, also known as
hypercomplex numbers.> Who actually uses these numbers?> I
mean, ordinary complex numbers are abundant, even in
engineering. But> hypercomplex? Are they related to Minkowski
space, i.e. special relativity> (simply because the dimension
is four). Are they used in theoretical physics> otherwise?>
=Michael Jrgensen> 160 years ago today, on October 16,
1843, mathematician W. R. Hamilton> invented quarternions,
also known as hypercomplex numbers.>> Who actually uses these
numbers?>> I mean, ordinary complex numbers are abundant, even
in engineering. But> hypercomplex? Are they related to
Minkowski space, i.e. special relativity> (simply because the
dimension is four). Are they used in theoreticalphysics>
otherwise?Physicists tend to use spinors instead of
quaternions, i.e. pairs of complexnumbers instead of
quadruples of real numbers. Quaternions are used insoftware
for rendering, if that counts for anything.>> I seem to
remember some eight-dimensional numbers too (Cayley?), but
then> youd lose either the associative or distributive law,
I forget which. Are> they of any use?Cayleys octonians,
which are not associative. They are the prototype
ofalternative algebras, which have developed a little in
recent times....LH
Im looking for a reference for the
proof that the innite seriesAs for a reference, the proof
appears in many elementarycalculus texts, and in almost any
text covering Fourierseries.There is an entirely elementary
proof of this relation,which I suspect is well known but Ive
never seen it.Factor (1+x)^N - (1-x)^N, for N an odd integer,
asproduct ( w^k(1+x) - w^{-k}(1-x) ),
k=-(N-1)/2..(N-1)/2where w = exp( i pi / N )Thenfind the
coefcient of x^3 for that expression, andcompare it to the
same coefcient using the Binomial Theoremfor the rst
expression, and take a limit.
Thats a different it. If
you apply Parseval to the function x on> [-pi,pi] you get the
sum of 1/n^2. If you use the convergence> of the Fourier
series for x^2 you also get the sum of 1/n^2 > coming
out.
=I have a concept about trying to map 3-D coordinates
to the plane. My idea is to divide the plane into three
sections, radially about theorigin. Thus the x axis is to be
represented in the area swept by theangle [0, 2/3 pi], the y
axis in the area swept by the angle [2/3 pi,4/3 pi], and the
z-axis in the area of [4/3 pi, 2 pi], as an example.Then I
gure to bisect each angle and represent positive values
inone section of the angle and negative values in the other
section. x-/y+ / x+----0-------y- / z- / z+Then a three
dimensional point x, y, z, is one to three points on
theplane. For example (0, 0, 0) is mapped to (0, 0), non-zero
(x, y, z)is mapped to polar coordinates ( r(x), theta(x)), (
r(y), theta(y)),and (r(z), theta(z)).A part of that is that
where x is real then for positive x it ismapped to (x, pi/6)
for positve x and (x, pi/2) for negative x.Connecting the
points of non-zero t(x, y, z) would form a triangle. Then
something like a 3-d polygon would be the interconnection of
eachof the points mapped into the 2-d space.I got this notion
and it is mostly that, Im trying to gure out ameaningful way
to map n-d coordinates to the plane, in this case wheren=3. It
might only be useful where for n that a regular polygon of
2nsides tiles in the plane.Im hoping that you can tell me
something about mathematical enquiryand practice into
representing 3-d or generally n-d objects fully in2-d.Ross
F.
=Im not so sure what you mean,butwhy not make positive
& negative sectors,opposite from each other?> I have a
concept about trying to map 3-D coordinates to the plane. My
idea is to divide the plane into three sections, radially
about the> origin. Thus the x axis is to be represented in
the area swept by the> angle [0, 2/3 pi], the y axis in the
area swept by the angle [2/3 pi,> 4/3 pi], and the z-axis in
the area of [4/3 pi, 2 pi], as an example.Then I gure to
bisect each angle and represent positive values in> one
section of the angle and negative values in the other
section.> x-/> y+ / x+> ----0-------> y- / z-> / z+Then a
three dimensional point x, y, z, is one to three points on
the> plane. For example (0, 0, 0) is mapped to (0, 0),
non-zero (x, y, z)> is mapped to polar coordinates ( r(x),
theta(x)), ( r(y), theta(y)),> and (r(z),
theta(z)).--Dec.2000 WAND Chairman Paul ONeill, reelected
to Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=Youre
not mapping 3-d to 2-d, but rather space to either the
powerset of the plane or the cartesian product of three
copies of theplane. Ive never heard of anything like this
being done, and I donot think that further inquiry would
benecial.> Connecting the points of non-zero t(x, y, z)
would form a triangle. You will not always get a triangle. If
two coordinates are 0, thenyou will get a line segment.> Then
something like a 3-d polygon would be the interconnection of
each> of the points mapped into the 2-d space.Im not sure if
I understand what this means. Heres what I think itmeans. We
have a point (x,y,z) in 3-d. You map this point to thethree
points (x,theta(x)), (y,theta(y)), and (z,theta(z)). Then
drawthe three line segments connecting these three points.
The problemis that in general, these line segments contain
points that are notmapped to by any point in 3-d. So it is
rather meaningless (anduntrue) to say that this is a 3-d
polygon. But, admittedly, I dontbelieve I understand what
you meant. ~ Chris
=Im not quite sure if what I had in
mind was any use either, but atthe time it seemed to be some
kind of intellectually soothing insightabout itself: mapping
3-d points onto a plane.In mapping the 3-d point to the 3 2-d
points, they may coincide, asyou say, for (x, y, z) if any two
points equal zero than there wouldonly be two points to
represent it on the plane. The 3-d origin isuniquely
represented by the 2-d origin.In drawing the lines among the
3 points that represent one point,basically the idea is that
that triangle, or rather the set of itsvertices, represents
one 3-d point. Consider (1, 1, 1). It would mapto polar (1,
pi/6), (1, 5pi/6) and (1, 9 pi/6). Adding 2pi/3 to eachof
those angles would yield the points on the plane which map
back to(1, 1, 1). For the coordinates that represent (2, 1,
1), the sameaddition to each polar coordinates angle would
represent (1, 2,1). Ina way that change would represent a
change from the 3-d coordinaterepresenting a 2x1x1
rectangular solid to a 1x2x1 rectangular solid,for width,
length and depth swapping width and length. Thats quitemore
directly done by exchanging the x and y coordinates.I was
trying to think how something like a polyhedron in 3
dimensionwould be represented. It could be a list of its
vertices in 3-d, or alist of the triples of 2-d coordinates
for 2-d.I noticed on the little ASCII diagram after I sent it
that it lookslike a perspective drawing of the x, y, and x
axes except the labelsare different, with the positive axis
being orthogonal to the negativeaxis, and x+ || (parallel to)
y-, y+ || z-, and z+ || x-.Going back to 2 3-d points mapped
into the 2-d points, each wouldrepresent a set of one, two,
or three points. Connect each pair ofpoints, some may cross,
thus forming multiple polygons.A large problem with this is
plotting 2-d points wth z considered tobe zero into the
construction. For example, a straight line segmentfrom (1, 0)
to (-1, 0) would be plotted as two adjoining line segmentsfrom
(1, theta(x+)) to the origin and from there to (1, theta(x-)).
What concerns me more are what would end up being plotted as
separatedline segments. For example the 2-d point (1, 1, 0,
...)would beplotted as the set of two points (1, theta(x+))
and (1, theta(y+)),instead of just one point. Im not saying
that this is a problemspace, just that it might be
interesting.This might be something where making a bunch of
these little plots andseeing how they represent things that
have utility without beingmutlilated would be. As well, I
should draw regular polygons of theplane onto it and see what
they could represent if they were 3-dpoints represented on the
plane this way.The equilateral triangle with its center at the
origin and to havedistance from its center to each vertex
being equal to one would havevertexes representing (1, 1, 1).
The lines of the triangle also passthrough some (-x,
theta(-x)), (-y, theta(-y)), and (-z, theta(-z)),where x, y,
and z are equal and the length of the segment from theorigin
to the midpoint of the side of the equilateral triangle
withdistance from the center(centroid) to the vertex of 1.
Here it iskeen to have a protracter, which I have misplaced.
The trianglebreaks into six 30-60-90 triangles, with
hypotenuse of length one. Its been a while since I have
thought of this kind of thing. Thelines of the equilateral
triangle would pass through (1/ 2,theta(x-)), etcetera,
(-1/2, -1/2, -1/2).Anyways, how about drawing a box from 3-d
with corners of +-1, thathas volume 8 and area 6 x 4. I gure
to determine where each pointof te line segments connecting
the vertices is marked on the 2-d plot. On each of the six
radial rays the point at distance from the origin1 is marked,
as is that of 1/2, 1/4, and each point between the originand
1, the cube symmetric about the origin with faces parallel to
thexy, yz, and zx planes plots in the 2-d plane as a set of
three linesegments from (1, theta(x+)) to (1, theta(y-)),
from (1, theta(y+)) to(1, theta(z-)), and from (1, theta(x+))
to (1, theta(x-)), each minusthe origin. A problem here is
recovering the cube from the plot,there may well be other 3-d
polyhedra that have the same plot, forexample all polyhedra
with vertices (+-1, +-1, +-1) with no vertex (x,y, z) with
|x| (absolute value), |y|, or |z| greater than 1.Im trying
to gure out a tractable sphere in cartesian coordinatesto
get a plot. A sphere centered on the origin x^2+y+2+z^2=1 as
anexample would perhaps also have the same plot as the cube
above.Ill think about this some more and try and gure out
if there isanything useful about it, it might be a waste of
time but I tend totrust my instincts.Heh, cubing the
sphere.Besides real coordinates x, y, z, Im also considering
complexcoordinates x, y, z. For example where the angle is
between that ofx+ and x-, at distance r from the origin, on
the x+ side it would be0-ri and on the x- side it would be
0+ri.Im not too worried about this. Ross
=Im trying to
solve a puzzle. Its a cube with parts that move NOT like
aRubics cube. I need a method to tell me if certain
positions are reachablethrough the allowed manipulations from
a starting conguration. Here is asimple sub problem.At one
point along the way, Ive got four corners oriented like
this:A BD CI can x any corner, and rotate the other three.
For example, xing A Ican reach:A DC BthenA CB DthenA BD Cthe
original.After ddling with this for a while I decided that I
can not reach:A BC DThere seem to be two sets of
congurations, one reachable fromA BD Cand one reachable
fromA BC DIt seems there must be a property, e.g. left vs
right, that is different forthe two groups and conserved by
the manipulation. How can I dene andcompute such a property
for a given conguration? How do I know it will beconserved
by the manipuation?Ive tried looking at displacements from
starting position in terms ofrotations and/or horizontal and
vertical ips. I dontfind anything easy.If I learn how to
dene simple symmetry properties, I will be able to do sofor
larger problems within the puzzle. That will save me
considerable timeddling in an attempt to achieve the
impossible.John
Im trying to solve a puzzle. Its a
cube with parts that move NOT like a> Rubics cube. I need a
method to tell me if certain positions are reachable> through
the allowed manipulations from a starting conguration. Here
is a> simple sub problem.At one point along the way, Ive got
four corners oriented like this:> A B> D CIf this is a
physical square object, why cant you perform A BD C->D AC
B?Do the faces have some kind of orientation?> I can x any
corner, and rotate the other three. For example, xing A I>
can reach:> A D> C Bwhich, if the above rotation is
permitted, is D B A Cso you can swap A and D. Therefore you
can swap any neighbouring corners. Therefore you can swap C
and D.> then> A C> B Dthen> A B> D C> the original.After
ddling with this for a while I decided that I can not
reach:> A B> C DWhats wrong with xing D:A B D C->C AD B=A
BC DPhil-- Unpatched IE vulnerability: Notepad
popupsDescription: Opening popup windows without
scriptingReference:
http://computerbytesman.com/security/
notepadpopups.htmFollowup:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0308/55.html==
= If this is a physical square object, why cant you
perform> A B> D C> ->> D A> C B> ?>> PhilBecause of
relationships between these corners and other parts of the
faceand other faces of the cube.You are right, if I could
perform the 90 degree rotation, then I cold reachany
conguration.P.S.Im using Outlook Express to read and reply
to this newgroup atnews.ucr.edu. It wouldnt let my response
include more of the originalmessage than new text that I
added! Ever heard of such a thing? DidOutlook express do
that, or did my news server do that. What is a goodnewsreader
for Windows?
Im trying to solve a puzzle. Its a cube
with parts that move NOT like a>Rubics cube. I need a method
to tell me if certain positions are reachable>through the
allowed manipulations from a starting conguration. Here is
a>simple sub problem.>At one point along the way, Ive got
four corners oriented like this:>A B>D C>I can x any corner,
and rotate the other three. For example, xing A I>can
reach:>A D>C B...>There seem to be two sets of congurations,
one reachable from>A B>D C>and one reachable from>A B>C D>It
seems there must be a property, e.g. left vs right, that is
different for>the two groups and conserved by the
manipulation. How can I dene and>compute such a property for
a given conguration? How do I know it will be>conserved by
the manipuation?Yes. A permutation is odd if it is the
product of an oddnumber of transpositions, and even if it is
the product of aneven number of transpositions. Your
rotations are 3-cycles, whichare products of two
transpositions and thus even, e.g.AB AB ADDC to CD to CB.The
product of even permutations is even, so you can only
performeven permutations, and not odd permutations.Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
=Robert,original text in my
replies.Now, when I attempt to apply this idea to larger
problems, Ive got morequestions.My transpositions were of
objects in adjacent positions. However, yourstatement holds
even if the transpositions are not adjacent, true?My list had
four objects in four positions, and also position 1 was
adjacentto position 4. Your statement holds true regardless
the size of the listand regardless of whether the rst and
last positions are adjacent, true?Is there a way that can I
compute the oddness or evenness of a congurationwithout
having to actually transpose it back to the desired position
andcounting the number of transpositions needed? For example
if I call (ABCD)even, then how can I know in one step that
(DCBA) is also even?
My transpositions were of objects in
adjacent positions. However, your>statement holds even if the
transpositions are not adjacent, true? True>My list had four
objects in four positions, and also position 1 was
adjacent>to position 4. Your statement holds true regardless
the size of the list>and regardless of whether the rst and
last positions are adjacent, true?True>Is there a way that
can I compute the oddness or evenness of a
conguration>without having to actually transpose it back to
the desired position and>counting the number of
transpositions needed? For example if I call (ABCD)>even,
then how can I know in one step that (DCBA) is also even?Use
disjoint cycle notation. For example, the permutation taking
ABCDto DCBA consists of the cycles [AD][BC] so its even. Or
the permutation taking ABCDEFG to DBACGFE is [ADC][B][EG][F]
(i.e. A goes to D, which goes to C, which goes back to A, B
goes to itself, E goes to G which goes to E, F goes to
itself). Any permutationcan be written as a product of
disjoint cycles; an n-cycle is odd ifn is even and even if n
is odd. So [ADC][B][EG][F] is even*even*odd*even= odd.Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Im trying to solve a
puzzle. Its a cube with parts that move NOT like a> Rubics
cube. I need a method to tell me if certain positions are>
reachable> through the allowed manipulations from a starting
conguration. Here is> a simple sub problem.At one point
along the way, Ive got four corners oriented like this:> A
B> D CI can x any corner, and rotate the other three. For
example, xing A I> can reach:> A D> C Bthen> A C> B Dthen> A
B> D C> the original.After ddling with this for a while I
decided that I can not reach:> A B> C DThere seem to be two
sets of congurations, one reachable from> A B> D C> and one
reachable from> A B> C DIt seems there must be a property,
e.g. left vs right, that is different> for> the two groups
and conserved by the manipulation. How can I dene and>
compute such a property for a given conguration? How do I
know it will> be conserved by the manipuation?Ive tried
looking at displacements from starting position in terms of>
rotations and/or horizontal and vertical ips. I dontfind
anything> easy. If I learn how to dene simple symmetry
properties, I will be able> to do so> for larger problems
within the puzzle. That will save me considerable> time
ddling in an attempt to achieve the impossible.John--
Transformations of the square can be represented by the
subgroup of S_4 (the symmetric group on four letters)
generate by 3-cycles. The question is, what is the index of
this subgroup. In particular, if the index is 1, you have all
of S_4 so every arrangement is reachable. The order of S_4 is
24 and there are 8 3-cycles. The subgroup must contain these
8 and the identity. The subgroup must contain these 8 members
and the identity so its order must be at least 9.. Since the
order must divide 24, it must be either 12 or 24 so the index
must be 1 or 2.
Furthermore(2314)(1423)=(4213)(3124)(1423)=(2143)(2314)(1342)
=(3412)(3124)(1342)=(4132)so the order is at least 9+4=13 so
the order is 24, the index is 1, and every arrangement is
reachable.Have a tolerable existence. Eli
Transformations
of the square can be represented by the subgroup of S_4 (the
>symmetric group on four letters) generate by 3-cycles. The
question is, >what is the index of this subgroup. In
particular, if the index is 1, you >have all of S_4 so every
arrangement is reachable. The order of S_4 is 24 >and there
are 8 3-cycles. The subgroup must contain these 8 and the
>identity. The subgroup must contain these 8 members and the
identity so >its order must be at least 9.. Since the order
must divide 24, it must be >either 12 or 24 so the index must
be 1 or 2. Furthermore>(2314)(1423)=(4213) This is a
3-cycle>(3124)(1423)=(2143)>(2314)(1342)=(3412)>(3124)(1342)=
(4132)>so the order is at least 9+4=13 so the order is 24,
the index is 1, and >every arrangement is reachable.Sorry,
the order is 12. Its the alternating group A_4. Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israelUniversity of British
ColumbiaVancouver, BC, Canada V6T 1Z2
(2314)(1423)=(4213)> This is a
3-cycle>>(3124)(1423)=(2143)>>(2314)(1342)=(3412)>>(3124)(
1342)=(4132)>>so the order is at least 9+4=13 so the order
is 24, the index is 1, and>>every arrangement is reachable.>
Sorry, the order is 12. Its the alternating group A_4.Yeah,
I should have realized that. I guess I have an itchy trigger
nger.Have a tolerable existence. Eli
=Has anyone here used
the book The Best Test Preparation for the GREin in
Mathematics? Its published by the the Research and
EducationAssociation.I was trying a few of the practice tests
just go give myself an ideaof the test and feeling quite
annoyed. They seem to ask a lot of veryobscure things that I
never covered in any calculus/linearalgebra/abstract
algebra/etc class Ive ever taken.Several reviews Ive seen
gave the same opinion that I had of the book(not too great
practice for the test itself).Whats the best book you would
recommend to prepare for the test?
=I have been using The
Princeton Reviews GRE Math test book. Once again,it is a
little harder than what is really on the test (from what I
have beentold), but I think that it will serve as a good
practice book. If you cando those problems, then the test
should be easy. If I recall correctly, thetest consists of
over 50% Calculus(Analysis) problems. Review Calc.
andAnalysis!Lurch> Has anyone here used the book The Best
Test Preparation for the GRE> in in Mathematics? Its
published by the the Research and Education> Association.>> I
was trying a few of the practice tests just go give myself an
idea> of the test and feeling quite annoyed. They seem to ask
a lot of very> obscure things that I never covered in any
calculus/linear> algebra/abstract algebra/etc class Ive ever
taken.>> Several reviews Ive seen gave the same opinion that
I had of the book> (not too great practice for the test
itself).>> Whats the best book you would recommend to
prepare for the test?
Has anyone here used the book The
Best Test Preparation for the GRE> in in Mathematics? Its
published by the the Research and Education> Association.I
was trying a few of the practice tests just go give myself an
idea> of the test and feeling quite annoyed. They seem to ask
a lot of very> obscure things that I never covered in any
calculus/linear> algebra/abstract algebra/etc class Ive ever
taken.Several reviews Ive seen gave the same opinion that I
had of the book> (not too great practice for the test
itself).Whats the best book you would recommend to prepare
for the test?I was very pleased with the book by the
Princeton Review. Idsay that the practice tests and strategy
gave me at leastanother 100 points on each section. It just
really helpedto get a feeling for how fast I should expect to
be moving.I still blew the timing in the math test, spending
way toomuch time on some early questions that I thought just
a*little* more calculation could help with. But the
strategyhelped there too: I knew that guessing was a good
strategyon the subject tests, if you could condently
eliminate all but 3 or fewer choices. So I zipped through the
last third of the test as fast as possible, eliminating a
fewchoices on each question and then guessing. This
strategygot me a 990. - Randy
=i need a small result to
prove something, and it seems so simple and idont know why i
cant get itsup (xn + yn) <= sup xn + sup ynfor all real
sequences xn and yn. i dont understand, this seems
i need a small
result to prove something, and it seems so simple and i> dont
know why i cant get itsup (xn + yn) <= sup xn + sup ynfor all
real sequences xn and yn. i dont understand, this seems so>
xn so xa + ya <= sup xn + ya. Taking the supremum over a
yields, sup (xn + yn) = sup (xa + ya) <= sup (sup xn + ya) =
sup xn + sup ya = sup xn + sup yn.Have a tolerable existence.
Eli
=Type in half a cup in teaspoonsor sqrt(3) * sqrt(5) *
sqrt(7)in Google.comPretty cool.Discovered it when typing in
a phone number.
Type in half a cup in teaspoons> or
sqrt(3) * sqrt(5) * sqrt(7)in Google.com> Pretty cool.Nah. It
doesnt do symbolic calculations ;-)Gib
Type in half a
cup in teaspoons> or sqrt(3) * sqrt(5) * sqrt(7)in
Google.com> Pretty cool.Discovered it when typing in a phone
number.Excellent. It appears to be a copy of the unix units
utilityIt even has the geeks favorite unit: furlongs per
fortnight:1 (furlong per fortnight) = 3.23383197 [Times]
10-25 megaParsecs per minuteGoodfind.Dale.
> It even has
the geeks favorite unit:>> furlongs per fortnight:>> 1
(furlong per fortnight) = 3.23383197 [Times] 10-25
megaParsecs per minute>> Goodfind.>> Dale.Useful when they
have intergalactic horse racing. <g>
Could someone
please give me a proof for distributive law of the
cross>product:>>a x (b + c) = a x b + a x c>>But theres a
catch! I am assuming the we have dened the cross>product as
(|a||b|sinO)n, etc. Using the above proof (that is,
taking>the above as true) THEN I will use the above truth
to derive the>rectangular form of the cross product. So I
need the above proved>without resorting to the rectangular
form/defn of the cross product.>>Thanx in advance>Use the
denition of the cross product: That is if v = q X pthen the
ith component of v is:v_i = eps_ijk q_j p_kwhere eps_ijk is
the totally antisymetric tensor. That is,eps_123 = eps_231 =
eps_312 = 1eps_132 = eps_321 = eps_213 = -1and all others are
zeroThe ith component of the cross product on the LHS of
yourequation is:(a X (b + c))_i= eps_ijk (a_j) ((b + c)_k)=
eps_ijk (a_j) (b_k + c_k)= ( eps_ijk (a_j) (b_k) ) + (
eps_ijk (a_j) (c_k) )= (a X b)_i + (a X c)_i<--> a X (b + c)
= (a X b) + (a X c) hope that helps,adam
=The idea was to
derive the distributive law for the cross productassuming we
dene the cross product as a x b = (|a||b|sinO)n, forvectors
a, b, n, etc.The thing I am actually trying to do that many
textbooks dont botherto do is to derive one denition from
the other. You see, onetextbook says we dene the cross
product as <a2b3 - a3b3, ..., ...>and then derives |axb| =
|a||b|sinO and so we have the magnitude ofthe cross product.
It then goes onto to explain why the cross productis
perpendicular to these 2 vectors (e.g. (axb).b = 0, etc). BUT
itdoes not explain how we DERIVE the need for the RIGHT-HAND
RULE to getthe direction of n, and here is where I am
stuck.Now the other way to do it, is to take the
cross-product to be denedasa x b = (|a||b|sinO)n and then
try to get it into <a2b3 - a3b3, ...,...> form. Easy PROVIDED
you accept that the cross-product isdistributive! But I wont
just assume - I would like a proof!Perhaps you know such a
proof?Thnx in advanceMB>Could someone please give me a
proof for distributive law of the cross>product:>>a x (b + c)
= a x b + a x c>>But theres a catch! I am assuming the we
have dened the cross>product as (|a||b|sinO)n, etc. Using
the above proof (that is, taking>the above as true) THEN I
will use the above truth to derive the>rectangular form of
the cross product. So I need the above proved>without
resorting to the rectangular form/defn of the cross
product.>>Thanx in advance>> Use the denition of the cross
product: That is if v = q X p> then the ith component of v
is:v_i = eps_ijk q_j p_kwhere eps_ijk is the totally
antisymetric tensor. That is,eps_123 = eps_231 = eps_312 = 1>
eps_132 = eps_321 = eps_213 = -1> and all others are zero> The
ith component of the cross product on the LHS of your>
equation is:(a X (b + c))_i= eps_ijk (a_j) ((b + c)_k)=
eps_ijk (a_j) (b_k + c_k)= ( eps_ijk (a_j) (b_k) ) + (
eps_ijk (a_j) (c_k) )= (a X b)_i + (a X c)_i<--> a X (b + c)
= (a X b) + (a X c) hope that helps,adam
=It has been a
long time since I have used a matrix, and I dontremember
quite how to do it. I am not even sure that it is the
rightapproach to solving this problem. Please read the
problem, and then ifa matrix is the way to solve it and there
is a web reference (I havesearched, but am confused) that will
help, please point me there, andif there is a better way of
going at this rather than a matrix, I aminterested in that
too.Problem: Our company voted on 15 movies to screen for a
Halloween lmfestival, one movie each day. Each employee
voted on their top 5 picksand noted which day they will be
able to attend. Their top pick isworth 7 points, second pick
is worth 6 points, third pick is worth 5points, 4th=4 points,
5th=3 points. Any day they plan on attendingtheir votes count
towards the tally on that day, and any day theyarent
attending their votes dont count. We are going to screen
5movies (Monday-Friday) and the goal is to maximize the total
pointsfor the week, with the assumption that if we do that
then the mostpossible people will be satised.That may sound
like a problem out of a textbook, but it is really whatwe are
doing, and while most normal people would just show the top
5movies voted for (and not even bother weighting the scale
nor breakingit down to who is there each day), well Im just
not that normal. Ihave to make things complicated and desire
to have the theoreticalmaximum happy points. When I realized
my rst solution was notmaximized, it drove me to seek a more
appropriate mathematical modelto solve the problem.Ill give
just the top 7 movies to make it simpler:(A, B, C, D, E, F,
G)Monday = (37 29 22 22 11 3 14)Tuesday = (45 29 27 21 17 14
26)Wednesday = (51 39 40 21 20 11 20)Thursday = (46 22 28 15
18 16 20)Friday = (29 19 12 9 7 6 10)My rst solution was to
nd the largest number total (51 onWednesday for movie A) and
then remove all other choices for Wednesdayand movie A and
look for the next largest number (Movie B on Tuesday =29
points).Then I realized if I moved Movie A to Tuesday it
would reduce thepoints by 6 but moving movie B to Wednesday
increased it by 10 points,a net increase of 4 points. So
thats where I stand, 144 points.Can anyone point me in the
direction of solving this problem thecorrect way?Eric
Wikman
Problem: Our company voted on 15 movies to screen
for a Halloween lm>festival, one movie each day. Each
employee voted on their top 5 picks>and noted which day they
will be able to attend. Their top pick is>worth 7 points,
second pick is worth 6 points, third pick is worth 5>points,
4th=4 points, 5th=3 points. Any day they plan on
attending>their votes count towards the tally on that day,
and any day they>arent attending their votes dont count. We
are going to screen 5>movies (Monday-Friday) and the goal is
to maximize the total points>for the week, with the
assumption that if we do that then the most>possible people
will be satised.This is called an Assignment Problem, and
there are standard algorithmsfor solving it. You can use
linear programming: there are more specialized algorithms
that are more efcient, but that shouldnt bea consideration
here as your problem is not that big. The linear
programmingformulation is as follows. You have variables
x_{ij} for each movie iand each day j: x_{ij} = 1 in the
solution means you show movie i on day j. Let c_{ij} be the
number of points for movie i on day j.maximize sum_i sum_j
c_{ij} x_{ij}subject to sum_i x_{ij} = 1 for each j (i.e.
exactly one movie is shown on each day) sum_j x_{ij} <= 1 for
each i(i.e. each movie is shown on at most one day)all x_{ij}
>= 0The Integrality Theorem for network ow problems
guarantees that in a basic solution (which is what standard
linear programming softwarends) all variables will be 0 or
1.Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Problem: Our company voted
on 15 movies to screen for a Halloween lm>festival, one
movie each day. Each employee voted on their top 5 picks>and
noted which day they will be able to attend. Their top pick
is>worth 7 points, second pick is worth 6 points, third pick
is worth 5>points, 4th=4 points, 5th=3 points. Any day they
plan on attending>their votes count towards the tally on that
day, and any day they>arent attending their votes dont
count. We are going to screen 5>movies (Monday-Friday) and
the goal is to maximize the total points>for the week, with
the assumption that if we do that then the most>possible
people will be satised.>>That may sound like a problem out
of a textbook, but it is really what>we are doing, and while
most normal people would just show the top 5>movies voted for
(and not even bother weighting the scale nor breaking>it down
to who is there each day), well Im just not that normal.
I>have to make things complicated and desire to have the
theoretical>maximum happy points. When I realized my rst
solution was not>maximized, it drove me to seek a more
appropriate mathematical model>to solve the problem.>>Ill
give just the top 7 movies to make it simpler:>(A, B, C, D,
E, F, G)>Monday = (37 29 22 22 11 3 14)>Tuesday = (45 29 27
21 17 14 26)>Wednesday = (51 39 40 21 20 11 20)>Thursday =
(46 22 28 15 18 16 20)>Friday = (29 19 12 9 7 6 10)Sounds
like a combinatorial optimization problem, which can
sometimesbe approximated by a linear programming problem. You
have an objectivefunction (total score for the festival?)
which appears to be linear,which you are trying to maximize,
you have some constraints (eachmovie shown once, only one
movie per day).You have 75 variables (one for movie A on
Monday, one for movie A onTuesday, ..., one for movie B on
Monday,.... ), each of which can be 0or 1. Set up as a linear
program, you would relax the requirement sothat the variables
were required to be between 0 and 1. Then you roundat the
end.Excel has a built in solver which, correctly set up, can
solve someoptimization problems. I know it does LPs, and it
may do integerproblems. - Randy
=i want know why number 1
does not contain in the prime number.if 1 in prime, what
trouble is happen??
i want know why number 1 does not
contain in the prime number.if 1 in prime, what trouble is
happen??We lose the Fundamental Theorem of Arithmetic: that
every integer hasa unique prime factorization except for the
order of the factors. If 1is not prime then, for example, we
can factor 6 as 2*3. If we DO count1 as a prime then 6 can be
factored in an innite number of ways,such as 2*3 = 1*2*3 =
1*1*2*3 and so on. It just makes things
moretidy.Darren
=from math2050@yahoo.co.kr:>>i want know
why number 1 does not contain in the prime number.>>if 1 in
prime, what trouble is happen??>The factors of 1 are only
1.The ONLY factor of 1 is 1. A prime number has only the
factors 1 and itself. A composite number hasfactors of 1,
itself, and at least one other prime number. Since 1 has only
1 as a factor, it is not prime and not composite. To say that
one has 1 and 1 as its factors is not productive, because
anynumber may be multiplied by 1 any number of times and will
still produce 1 as aresult. We do not do this when we are
listing or factoring a number to showits prime factors. G
C
i want know why number 1 does not contain in the prime
number.if 1 in prime, what trouble is happen??The trouble is
that, if 1 is a prime, then positive integers cant be
factored into primes uniquely because you can include any
number one 1s in the list of factors. For example, if 1 is
not a prime, then the only way to write 4 as a product of
primes is 2*2. However, if 1 is prime then
4=2*2=1*2*2=1*1*2*2=1*1*1*2*2=. . . Have a tolerable
existence. Eli--
i want know why number 1 does not
contain in the prime number.if 1 in prime, what trouble is
happen??Fundamental theorem of arithmetic (unique
factorization) fails to hold, forone.--Bill Barksdale
=Now
maybe some of you understand the test.The Universe
periodically tests species. Humanity has a version
ofsentience, so She tested you in mathematics.Remember?
Mathematics is the universal language.I told you the Universe
has a sense of humor.Were having fun. Please continue.James
Harris
=I was going to not post (just read, sheepishly),
butit hasnt been resolved about TheRealHSJ@MSN.com;was it
always him, or does he have to personalities, ordid he beat
the impostor out of it? (Ferrys posting reminds meof that
pentacostal wingnut that Rumsfeld just appointed;what they
dont note is thatthe military chaplaincy, even the Catholic
ones,are largely converted to that style of preaching,Holy
Rollers or Amy Semple McPherson e.g.) oh, I see;the real HSJ
would never make a blatant suicide-note header;would it?> Now
maybe some of you understand the test.--les ducs dEnron!
I was going to not post (just read, sheepishly), but> it
hasnt been resolved about TheRealHSJ@MSN.com;> was it always
him, or does he have to personalities, or> did he beat the
impostor out of it?> (Ferrys posting reminds me> of that
pentacostal wingnut that Rumsfeld just appointed;> what they
dont note is that> the military chaplaincy, even the
Catholic ones,> are largely converted to that style of
preaching,> Holy Rollers or Amy Semple McPherson e.g.)And you
never ask questionsWith God on your sideGib
Now maybe
some of you understand the test.Very few, no doubt.> The
Universe periodically tests species. Humanity has a version
of> sentience, so She tested you in mathematics.Remember?
Mathematics is the universal language.I told you the Universe
has a sense of humor.Were having fun. Please continue.I had
another dream last night. It may have been inspired bythe
promo I saw for the *trailer* of LotR 3, but Frodo could bea
symbol of you.I was Sam, standing with Frodo at the Crack of
Doom. We wereboth just standing there, sort of waiting
around, confused. Iwas thinking, well, this is the part where
he takes Ring andcasts it into the re, but then decides to
wield it himself,but Gollum bites off his nger. None of this
happened though.We were just standing there. Wheres the Ring,
Frodo? I thoughtyou had it, Sam -- youre the best man. What?
Oh no, we didntleave it back in the Shire, did we?So we
walked out onto the slope of Mt. Doom. The entire base ofthe
mountain was swirling with orcs. It was beautiful,
actually,as if we were the vortex. We looked up to see
Gwaihir (king ofthe eagles), coming to our rescue, but as he
swooped toward ushe turned into an F-16. Wind. Noise. When we
looked aroundagain, the swirling mass of orcs were Islamic
pilgrims. We werein Mecca, standing on the very axis about
which the Muslim worldrotates.I thought Id better leave. I
got separated from Frodo. I wanderedabout, and ended up
studying the Koran with a teacher of some kind(white robes).
I was bored. So he taught me to dance instead.Iss tay la na
bude lay la . . . -- some sort of dancing and singing.I felt
like I was getting into the spirit of the place.When I went
outside, Frodo was up on a soapbox. But now he lookedlike
Uncle Sam meets G. I. Joe. He was haranguing the crowd
abouthow they lived in a dictatorship, that democracy was the
only way toprosperity and human dignity, that women were being
oppressed, andI tried to get to him to make him stop
(especially about the Allahpart), but I couldnt. Then I woke
up.
Now maybe some of you understand the test.>> The
Universe periodically tests species. Humanity has a version
of> sentience, so She tested you in mathematics.>> Remember?
Mathematics is the universal language.>> I told you the
Universe has a sense of humor.>> Were having fun. Please
continue.>> James HarrisLets see if I get this straight.
You, James Harris, are the Universe.We, the readers of this
newsgroup, are the species under test. Is thatright? (Please
seek professional help.)--There are two things you must never
attempt to prove: the unproveable --and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
Now maybe some of you
understand the test.>>The Universe periodically tests
species. Humanity has a version of>sentience, so She tested
you in mathematics.>>Remember? Mathematics is the universal
language.>>I told you the Universe has a sense of
humor.>>Were having fun. Please continue.If you dont like
it when people say youre crazy then stopsaying things that
make you sound like a raving lunatic.>James
Harris************************David C. Ullrich
Now maybe
some of you understand the test.The Universe periodically
tests species. Humanity has a version of> sentience, so She
tested you in mathematics.Remember? Mathematics is the
universal language.And you are a broken record, with bits of
mathematics mixed together incorrectly, and forever
repeating.
Now maybe some of you understand the test.>>
The Universe periodically tests species. Humanity has a
version of> sentience, so She tested you in mathematics.>>
Remember? Mathematics is the universal language.>> I told you
the Universe has a sense of humor.>> Were having fun. Please
continue.> James HarrisAnd the point of this is....?
>>Now maybe some of you understand the test.>>The Universe
periodically tests species. Humanity has a version
of>>sentience, so She tested you in mathematics.>>Remember?
Mathematics is the universal language.>>I told you the
Universe has a sense of humor.>>Were having fun. Please
continue.>>James Harris> And the point of this is....?To
keep attention focussed on JSH?Gib
> <snip>> So, lets
not be complimentary. However:> Dr. David Ullrich has argued
that at least one is a head and at> least one is a tail are
_the same_ statement. What does that mean to> you?Blimey! Did
he really?What he actually said was that surely, even I, Eldon
Moritz, is notdumb enough to think changing those two words
makes a difference.Is he dumb enough? Right now, hes non
committal. With this questionhe goes into the obfuscation
mode rather easily.> To me it means that with a ip of HH, we
had, Two coins were ipped> and at least one is a head. What
are the chances for two heads? and> with TT we had Two coins
were ipped and at least one is a tail.> What are the chances
for two tails?> With HT, or TH, either statement was true, was
equally likely, and one> was made.> Mathematically speaking
they would be the same statement, and the same> question.
They would have the same mathematics and the same answer.>
When they are the same, they would have correct answer 1/2.>
To have answer other than 1/2, they would have different
answers.Sorry, I cant be complimentary about your clarity.>
I cant make head nor tail of your arhuments :-(Suppose that
we ipped umpteen thousand times, we should getumpteen/4 HH
and umpteen/r TT. So we would have the heads
statementumpteen/4 and the tails statement umpteen/4.Then
there are u/4 HTs and u/4 THs. Each one of these ips was
recreating a rst time event. On a rst time event, with HT,
or TH,the generator of the statement would not know whether
to favor heads,or tails. Therefore there should be another
u/4 for the heads andanother u/4 for the tails for the HTs
and THs.If you altered all the HTs and THs to say heads
every time, or tailsevery time then the answer to the favored
color would be 1/3, howeverthe unfavored color would only have
the double, therefore when oneside answers 1/3, the other side
answers 1. The only place where theywould both be equal is at
zero prejudice, or 1/2.Dr. Ullrich cant make up his mind
whether he wishes to be wrong now,or six months ago. Thats a
dilemma:~)Eldon
If you altered all the HTs and THs to
say heads every time, or tails> every timeAs before, I still
cant make head nor tail of your arguments.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
<snip>> Dr. David Ullrich has argued that at least
one is a head and at>> least one is a tail are _the same_
statement. What does that mean to>> you?>>Blimey! Did he
really?Of course not.I beg your pardon, Doctor Ullrich. Its
in the archives, they can lookit up.I dug the following out
of the archives:##################3<from the
archives>###############this>> is 1/3. B. Two coins were
ipped. At least one is a head. What are the>> chances for two
heads? Many mathematicians do not understand thatthis>> is
1/2.You switched from tails to heads here. Presumably even
_you_ dontthink that that makes a difference; presumably you
would also sayA. Two coins were ipped. Given that there is
at least one tail,what are the chances for two tails? Many
laymen do not understand that this is 1/3. B. Two coins were
ipped. At least one is a tail. What are thechances for two
tails? Many mathematicians do not understand that thisis
1/2.##############<end of archives>###################I had
dened A. and B. You redened them as A and B
becausepresumably even me doesnt think that makes a
difference. You wereright, I dont, but you were also arguing
that at least one is atail and at least one is a head are,
mathematically speaking, thesame statement.We agreed that
they are, we agree that they are, my argument stands,and your
apology is accepted:)Eldon>> To me it means that with a ip of
HH, we had, Two coins were ipped>> and at least one is a
head. What are the chances for two heads? and>> with TT we
had Two coins were ipped and at least one is a tail.>> What
are the chances for two tails?> With HT, or TH, either
statement was true, was equally likely, and one>> was made.>
Mathematically speaking they would be the same statement, and
the same>> question. They would have the same mathematics and
the same answer.> When they are the same, they would have
correct answer 1/2.>> To have answer other than 1/2, they
would have different answers.>>Sorry, I cant be
complimentary about your clarity.>I cant make head nor tail
of your arhuments :-(> My clarity is suspect, but my argument
is correct. When two coins areipped and they land HT, or TH,
then either statement is true, eitherthe heads or the tails
statement. Either could be made and eithershould have the
same answer. When they do, its a half. You could skewthe
odds toward heads and away from tails, or vise versa. Have
extremeprejudice toward heads; always make the heads
statement when itstrue, only make the tails statement when
the heads statement isnttrue, then the probability for HH is
1/3 on the heads statement andthe probability for two TT with
the tails statement is 1.(it couldonly be made when the heads
statement wasnt possible, that would beTT)To answer one
question 1/3, you must answer the other question 1.
Whentheyre the same, when the are equally likely at HT and
TH, then theyboth answer 1/2. Thats about as clear as I can
make it.Eldon> ************************David C. Ullrich
<snip>> Dr. David Ullrich has argued that at least one is
a head and at> least one is a tail are _the same_ statement.
What does that mean to> you?>>Blimey! Did he really?
Of course not.>I beg your pardon, Doctor Ullrich. Its in the
archives, they can look>it up.>>I dug the following out of the
archives:Good for you. Youd have a better point, and look a
little lessridiculous, if you supported your assertion that I
said thosetwo statements were the same statement with
something inthe archives that actually shows me saying that!
I _dont_say that in the quote
below.>##################3<from the
archives>###############>Counter-intuitive mathematical
results>> A. Two coins were ipped. Given that there is at
least one tail,>what> are the chances for two tails? Many
laymen do not understand that>this> is 1/3.> B. Two coins
were ipped. At least one is a head. What are the> chances
for two heads? Many mathematicians do not understand
that>this> is 1/2.>>You switched from tails to heads here.
Presumably even _you_ dont>think that that makes a
difference; presumably you would also say>>A. Two coins were
ipped. Given that there is at least one tail,>what are the
chances for two tails? Many laymen do not understand >that
this is 1/3.>B. Two coins were ipped. At least one is a
tail. What are the>chances for two tails? Many mathematicians
do not understand that this>is 1/2.>##############<end of
archives>###################Yes, I said the above. And
nowhere above does it show me sayingthat at least one is a
head and at least one is a tail are thesame statement.>I had
dened A. and B. You redened them as A and B
because>presumably even me doesnt think that makes a
difference. You were>right, I dont, but you were also
arguing that at least one is a>tail and at least one is a
head are, mathematically speaking, the>same statement.I was
arguing that? Only in your imagination (or to be fair,
yourloose/imprecise interpretation of what I said.)(HINT:
Saying that it doesnt make any difference whether we
talkabout one problem or another problem says that if we know
theanswer to one we then know the answer to the other. It
doesnot say that the _statements_ in the problems are the
_same_.Duh.)>We agreed that they are, we agree that they are,
my argument stands,>and your apology is accepted:)Youre a
funny guy, we all agree on that.>Eldon>> To me it means
that with a ip of HH, we had, Two coins were ipped> and at
least one is a head. What are the chances for two heads? and>
with TT we had Two coins were ipped and at least one is a
tail.> What are the chances for two tails?> With HT, or TH,
either statement was true, was equally likely, and one> was
made.> Mathematically speaking they would be the same
statement, and the same> question. They would have the same
mathematics and the same answer.> When they are the same,
they would have correct answer 1/2.> To have answer other
than 1/2, they would have different answers.>>Sorry, I
cant be complimentary about your clarity.>>I cant make head
nor tail of your arhuments :-(>>My clarity is suspect, but
my argument is correct. When two coins are>ipped and they
land HT, or TH, then either statement is true, either>the
heads or the tails statement. Either could be made and
either>should have the same answer. When they do, its a
half. You could skew>the odds toward heads and away from
tails, or vise versa. Have extreme>prejudice toward heads;
always make the heads statement when its>true, only make the
tails statement when the heads statement isnt>true, then the
probability for HH is 1/3 on the heads statement and>the
probability for two TT with the tails statement is 1.(it
could>only be made when the heads statement wasnt possible,
that would be>TT)>To answer one question 1/3, you must answer
the other question 1. When>theyre the same, when the are
equally likely at HT and TH, then they>both answer 1/2.
Thats about as clear as I can make it.>>Eldon>
************************ David C.
Ullrich************************David C. Ullrich
> <snip>>
Dr. David Ullrich has argued that at least one is a head and
at> least one is a tail are _the same_ statement. What does
that mean to> you?>>Blimey! Did he really?> Of course
not.>I beg your pardon, Doctor Ullrich. Its in the archives,
they can look>it up.>>I dug the following out of the
archives:Good for you. Youd have a better point, and look a
little less> ridiculous, if you supported your assertion that
I said those> two statements were the same statement with
something in> the archives that actually shows me saying
that! I _dont_> say that in the quote
below.>##################3<from the
archives>###############>Counter-intuitive mathematical
results>> A. Two coins were ipped. Given that there is at
least one tail,> what> are the chances for two tails? Many
laymen do not understand that> this> is 1/3.> B. Two coins
were ipped. At least one is a head. What are the> chances
for two heads? Many mathematicians do not understand that>
this> is 1/2.>>You switched from tails to heads here.
Presumably even _you_ dont>think that that makes a
difference; presumably you would also say>>A. Two coins were
ipped. Given that there is at least one tail,>what are the
chances for two tails? Many laymen do not understand >that
this is 1/3.>B. Two coins were ipped. At least one is a
tail. What are the>chances for two tails? Many mathematicians
do not understand that this>is 1/2.>##############<end of
archives>###################Yes, I said the above. And
nowhere above does it show me saying> that at least one is a
head and at least one is a tail are the> same statement.>I
had dened A. and B. You redened them as A and B
because>presumably even me doesnt think that makes a
difference. You were>right, I dont, but you were also
arguing that at least one is a>tail and at least one is a
head are, mathematically speaking, the>same statement.> We
have statement B:Two coins were ipped and at least one is a
head.We have statement B:Two coins were ipped and at least
one is atail.What do you argue? What do you say now?I say
that mathematically speaking, B and B are the same
statement.That changing from heads to tails doesnt change
the statement, thequestion, or the answer.What do you say
now? Do you now say that changing from heads to tailschanges
the question?Changing from heads to tails either alters the
question, or itdoesnt. What does Doctor Ullrich
say?Eldon
ipped and at least one is a head.> We have statement B:Two
coins were ipped and at least one is a> tail.>> What do you
argue? What do you say now?> I say that mathematically
speaking, B and B are the same statement.> That changing
from heads to tails doesnt change the statement, the>
question, or the answer.>> What do you say now? Do you now
say that changing from heads to tails> changes the question?>
Changing from heads to tails either alters the question, or
it> doesnt. What does Doctor Ullrich say?>> EldonI cant
speak for Dr. Ullrich, but I can ask if you see any
distinction between two phrases which are identical, and
twophrases which are equivalent? To most analysts,
equivalence holds within limits, identity does not.--There
are two things you must never attempt to prove: the
unprovable -- and the obvious.--Democracy: The triumph of
popularity over
principle.--http://www.crbond.com
statement B:Two coins were ipped and at least one is a
head.>> We have statement B:Two coins were ipped and at
least one is a>> tail.>> What do you argue? What do you say
now?>> I say that mathematically speaking, B and B are the
same statement.It depends on what question youre going to
ask. Youve taken half theproblem statement but havent said
whether youre changing the otherhalf.If you make a swap of
heads and tails THROUGHOUT THE ENTIRE QUESTION,then they are
equivalent questions.To whit: With a weighted coin which
falls on heads with probability0.75, what is the probability
of three heads in a row?Same question as: With a weighted
coin which falls on tails withprobability 0.75, what is the
probability of three tails in a row?But that does not mean
that a weighted coin which falls on tails withprobability
0.75 is identical to a weighted coin which falls onheads with
probability 0.75. Here is a different question:With a weighted
coin which falls on tails with probability 0.75, whatis the
probability of three heads in a row?See how that works?
Change the labelling in the whole question: samequestion.
Change the labelling in half the question, might be the
sameor might be different. - Randy
statement B:Two coins were ipped and at least one is a
head.>> We have statement B:Two coins were ipped and at
least one is a>> tail.>> What do you argue? What do you say
now?>> I say that mathematically speaking, B and B are the
same statement.It depends on what question youre going to
ask. Youve taken half the> problem statement but havent
said whether youre changing the other> half.If you make a
swap of heads and tails THROUGHOUT THE ENTIRE QUESTION,> then
they are equivalent questions.To whit: With a weighted coin
which falls on heads with probability> 0.75, what is the
probability of three heads in a row?Same question as: With a
weighted coin which falls on tails with> probability 0.75,
what is the probability of three tails in a row?But that does
not mean that a weighted coin which falls on tails with>
probability 0.75 is identical to a weighted coin which falls
on> heads with probability 0.75. Here is a different
question:With a weighted coin which falls on tails with
probability 0.75, what> is the probability of three heads in
a row?See how that works? Change the labelling in the whole
question: same> question. Change the labelling in half the
question, might be the same> or might be different. - RandyWe
have coins of equal density. Heads and tails are equally
likely. Iwas trying to get Ullrichs opinion nailed down. He
seems to bewafing.You and I seem to agree.When we shipped HH
and said, Two coins were ipped and at least oneis a head.
What are the chances for two heads? or we ipped TT, andsaid
Two coins were ipped and at least one is a tail. What are
thechances for two tails? We would have the same question
with the sameanswer.Likewise, with a HT, or a TH ip, either
statement would be true.Either question would have the same
answer. For one of them to haveanswer other than 1/2, they
would have to have different answers.Eldon
 We have statement B:Two coins were ipped and at least
one is a head.>> We have statement B:Two coins were ipped
and at least one is a>> tail.>> What do you argue?
What do you say now?>> I say that mathematically speaking,
B and B are the same statement.> It depends on what question
youre going to ask. Youve taken half the> problem statement
but havent said whether youre changing the other> half.> If
you make a swap of heads and tails THROUGHOUT THE ENTIRE
QUESTION,> then they are equivalent questions.> To whit: With
a weighted coin which falls on heads with probability> 0.75,
what is the probability of three heads in a row?> Same
question as: With a weighted coin which falls on tails with>
probability 0.75, what is the probability of three tails in a
row?> But that does not mean that a weighted coin which falls
on tails with> probability 0.75 is identical to a weighted
coin which falls on> heads with probability 0.75. Here is a
different question:> With a weighted coin which falls on
tails with probability 0.75, what> is the probability of
three heads in a row?> See how that works? Change the
labelling in the whole question: same> question. Change the
labelling in half the question, might be the same> or might
be different.> - RandyWe have coins of equal density. Heads
and tails are equally likely. I> was trying to get Ullrichs
opinion nailed down.No, he is not.
You are misinterpreting his quite clear statement.He is
trying to say your misinterpretation is not his
originalstatement. The difference is quite clear when the two
arelaid side by side, as you did in your quote.You and I seem
to agree.I mostly avoid your threads because your silly
obsessionwith this particular probability problem and
yourpeculiar reading of it. So no, I would not say we agree.>
When we ipped HH and said, Two coins were ipped and at least
one> is a head. What are the chances for two heads? or we
ipped TT, and> said Two coins were ipped and at least one
is a tail. What are the> chances for two tails? We would have
the same question with the same> answer.No. The question does
not include what was ipped. Take thestatements in quotes,
i.e.Two coins were ipped and at least one is a head. What
are the chances for two heads?andTwo coins were ipped and at
least one is a tail. What are thechances for two tails? And I
will say these are the same question given a faircoin. Youve
just relabelled heads and tails.When you add that there is a
particular outcome given, itssilly to ask about probability
since theres only onepoint in the sample space. The
probability is 1 in bothof those cases.> Likewise, with a HT,
or a TH ip, either statement would be true.Huh? The statement
includes a question: What are thechances for two heads
(tails)? How can that questionbe true?> Either question would
have the same answer. For one of them to have> answer other
than 1/2, they would have to have different answers.No, they
have the same answer of 1/3.Case 1: I tell you two coins were
ipped and one of thecoins is a head. Therefore you know that
I have eitherHH, TH, or HT with equal probability. I ask you
what isthe probability that I got HH? The answer is 1/3.Case
2: I tell you two coins were ipped and one of thecoins is a
head. Therefore you know that I have eitherTT, HT, or TH with
equal probability. I ask you what isthe probability that I got
TT? The answer is 1/3.You will see that in exchanging H and T
in thetwo paragraphs throughout, I have not changed
themeaning or the interpretation. There is nothing likea
requirement that the probability be 1/2. Your nalstatement
is coming from somewhere outside of probabilitytheory. -
Randy
were ipped and at least one is a head.>> We have statement
B:Two coins were ipped and at least one is a>> tail.> What do you argue? What do you say now?>> I say that
mathematically speaking, B and B are the same statement.>
It depends on what question youre going to ask. Youve taken
half the> problem statement but havent said whether youre
changing the other> half.> If you make a swap of heads
and tails THROUGHOUT THE ENTIRE QUESTION,> then they are
equivalent questions.> To whit: With a weighted coin which
falls on heads with probability> 0.75, what is the
probability of three heads in a row?> Same question as:
With a weighted coin which falls on tails with> probability
0.75, what is the probability of three tails in a row?> But
that does not mean that a weighted coin which falls on tails
with> probability 0.75 is identical to a weighted coin
which falls on> heads with probability 0.75. Here is a
different question:> With a weighted coin which falls on
tails with probability 0.75, what> is the probability of
three heads in a row?> See how that works? Change the
labelling in the whole question: same> question. Change the
labelling in half the question, might be the same> or might
be different.> - Randy> We have coins of equal density.
Heads and tails are equally likely. I> was trying to get
Ullrichs opinion nailed down. No, he
is not. You are misinterpreting his quite clear statement.> He
is trying to say your misinterpretation is not his original>
statement. The difference is quite clear when the two are>
laid side by side, as you did in your quote.>He chastised me
for thinking that changing from heads to tails wouldchange
that question. I didnt think it would and he didnt
either.Now he says he wasnt arguing that it wouldnt.. > You and I seem to
agree.I mostly avoid your threads
because your silly obsession> with this particular probability
problem and your> peculiar reading of it. So no, I would not
say we agree.>We agree that changing from heads to tails does
not alter thequestion, or the answer. Or we dont agree? I say
that it doesnt.Surely you agree with that? > When we ipped
HH and said, Two coins were ipped and at least one> is a
head. What are the chances for two heads? or we ipped TT,
and> said Two coins were ipped and at least one is a tail.
What are the> chances for two tails? We would have the same
question with the same> answer.No. The question does not
include what was ipped. Take the> statements in quotes,
i.e.Two coins were ipped. It says that right in the
question. WEREFLIPPED, past tense. Were talking about a
historical event. All weknow about the event is what we read
in the statement.Two coins were ipped and at least one is a
head. What > are the chances for two heads?andTwo coins were
ipped and at least one is a tail. What are the> chances for
two tails? And I will say these are the same question given a
fair> coin. Youve just relabelled heads and tails.>When they
landed HH the rst statement was true, when they landed
TT,the second statement was true. At HT, or TH, either
statement wouldhave been true.Consider the second statement.
We know that the coins landed TT, orTH, or HT. Suppose that
they landed TH. The other statement would havebeen true. If
it had been made, would we have had a differentquestion? >
When you add that there is a particular outcome given, its>
silly to ask about probability since theres only one> point
in the sample space. The probability is 1 in both> of those
cases.> This is a probability question. The coins were ipped
once. We can doprobability because we can re iterate the ip,
umpteen thousands oftimes. The tricky part is that we are re
iterating a rst time ip.We ip umpteen thousand times and
each one was a rst time ip.With HH, the heads statement was
made, with TT, the tails statementwas made, and with HT, or
TH, one, or the other. Either would havebeen true, and would
have been the same statement. OR, changing thecolor would
have changed the statement.> Likewise, with a HT, or a TH
ip, either statement would be true.Huh? The statement
includes a question: What are the> chances for two heads
(tails)? How can that question> be true?> Thats the point.
Does changing from heads to tails alter thequestion.> Either
question would have the same answer. For one of them to have>
answer other than 1/2, they would have to have different
answers.No, they have the same answer of 1/3.>Thats why I
have this strange obsession. There are always
qualiedpeople like you who will argue with me, and dont
seem to want tounderstand my argument. > Case 1: I tell you
two coins were ipped and one of the> coins is a head.
Therefore you know that I have either> HH, TH, or HT with
equal probability. I ask you what is> the probability that I
got HH? The answer is 1/3.> I know that they landed HH, or
HT, or TH. With HH, you made the headsstatement with
probability one, but you made the tails statement
withprobability 1/2.There are now three possible outcomes.
They are no longer equallylikely outcomes.> Case 2: I tell
you two coins were ipped and one of the> coins is a head.
Therefore you know that I have either> TT, HT, or TH with
equal probability. I ask you what is> the probability that I
got TT? The answer is 1/3.>I know that you have TT, HT, or
TH. They are no longer equallyprobably. Try it. You cant do
it.Flip the coins but dont look. You cant make either
statement, priorto inspection. After inspection you can make
one statement, or theother. With HH, you can only make the
heads statement. With TT, youcan only make the tails
statement.With HT, or TH, you can make one, or the other, but
not both. If youmake both, then I know that the probability is
one that the two coinsare different. The four are equally
probable, prior to inspection.After inspection, you can only
make one statement, or the other. Afterthe statement, the
three that are left are not equally probable. > You will see
that in exchanging H and T in the> two paragraphs throughout,
I have not changed the> meaning or the interpretation. There
is nothing like> a requirement that the probability be 1/2.
Your nal> statement is coming from somewhere outside of
probability> theory.You ipped the coins and got the answer
wrong. Thats what keeps meobsessed. My reading of the
question is strange? My reading may bestrange, but you get
the wrong answer, and you ipped the coins.Eldon - Randy
=I
am continuing, in a way, the thread
FUNctions/Continued-FractionPuzzle, but not crossposting it
to rec.puzzles this time.(For the focus is now not the
original puzzle, but something moregeneral.)(What I have to
add is below copy/pasted relevant parts of theprevious
thread.)>>...>>...>>For all real x > 1, and for some
function of x, y(x);>where each y is a real, y =y(x), based
on x:>>it is so that:>>f([x; x^2, x^3, x^4,...,x^m]) >[y;
y^2, y^3, y^4,...,y^m],>>for EVERY positive integer
m;>>where:>>[x; x^2, x^3, x^4,...,x^m] is the
continued-fraction>> 1>x + ------------------ ;> 1> x^2 +
--------------> 1> x^3 + ---------> ....> + 1/x^m>>and [y;
y^2, y^3, y^4,...,y^m] is also a
continued-fraction(obviously);>>and [x; x^2, x^3,
x^4,...,x^m] converges to X;>>and f(w) is a real -> real
function, such that>>f(X) exists and is nite nonzero.>So, what are the possible f(w)s, given all of the
conditionsabove??> First, by the way, f(X) is the
(1st) derivative of f(w) at w =X, in>> case this is not
obvious. I should mention that f can equate to an
innite number offunctions>> if it need not be analytic. If
it need by analytic, however,there are>> a nite number of
possible functions that can equal f(w).>> (So,find the set of
analytic f(w)s.)>>... ....>> ...the solution: I
get that the only possible analytic f is: f(w) = w.
Proof:> limit{m -> oo} (x/y)^(2m-1) = 1. So, x
must = y. And, consequently, f(w) must = w.> * earlier
result at:>>
rnum=4&prev>I highly suspect my PROOF is far from the
simplest.>Is there a PROOF which is any simpler, even
trivial?>>(Perhaps my result itself, that f(w) = w is the
ONLY analytic>function, is wrong.)>I believe that the result
above can be generalized.Let g_k(x) be a family of real->
real functions, wherelimit{k->oo} g_k(x)/g_k(y) = 1ONLY if x
= y, but the limit exists in any case,and where g(x) > 1 and
is dened for all x in a given domain D.f is as above in the
copied message (lots of conditions placed uponit).And g and f
are so that, for all x in D, the continued fraction[g_1(x);
g_2(x), g_3(x), g_4(x),..., g_m(x)]converges to X(x).And,
analogous to above,the derivative of f(w) exists at all real
w (just to be on the safeside),and is always nite and
nonzero.Now, y(x) = y is a real-> real function such that:
f([g_1(x); g_2(x), g_3(x), g_4(x),..., g_m(x)]) =[g_1(y);
g_2(y), g_3(y), g_4(y),..., g_m(y)]for ALL positive integers
m,and f and g are such that y(x) is in D for all x in D.(And
did I mention that D contains an interval of positive
length?)So, if f is analytic (so that f(w) is determined
completely by itsevaluation within some possibly-nite domain
for w), then I amguessing that it *must* be so thatf(w) =
w.Because...limit{m -> oo} g_m(x) g_{m-1}(x)/(g_m(y)
g_{m-1}(y)) = 1.And sinceg_m() is nonzero and nite,and
g_m(x)/g_m(y) is not 1 if x is not y(nor is ever -1 since g
is positive),then x = y, and then again f(w) = w.(I MUSt have
missed something here, most likely a condition, if Isimply did
= I have recently
nished teaching myself basic group/ring/eldtheory using
I.N. Hersteins Abstract Algebra. I am seeking
thename/publisher/possibly ISBN of a good textbook of Galois
theory. Itshould be one which lends itself well to being
self-taught; and itshould be well steeped in theory (I like
textbooks where a givensection gives the basics of something
but the problems proceed toexpound upon that thing in great
detail with numerous otherresults...) Money is not an issue
since I consider this a toppriority investment however I lack
access to a university bookstoreand so it would be good if the
textbook can be ordered and shipped via SamP.S. Links to
online tutorials are also good, although they do notseem to
exist for Galois theory, or if they do they are rare and
wellhidden...
=Not sure if it is still in print, consider
Algebraic Extensions of Fields by Paul J. McCarthy.
= I
have recently nished teaching myself basic group/ring/eld>
theory using I.N. Hersteins Abstract Algebra. I am seeking
the> name/publisher/possibly ISBN of a good textbook of
Galois theory. It> should be one which lends itself well to
being self-taught; and it> should be well steeped in theory
(I like textbooks where a given> section gives the basics of
something but the problems proceed to> expound upon that
thing in great detail with numerous other> results...) Money
is not an issue since I consider this a top> priority
investment however I lack access to a university bookstore>
and so it would be good if the textbook can be ordered and
shipped viaOthers have suggested Ian Stewarts book, and I
agree with them. I just wantto inform you that the most well
steeped in theory of all textbooks on Galoistheory is Harold
M. Edwards book Galois theory.Jose Carlos Santos
P.S.
Links to online tutorials are also good, although they do
not> seem to exist for Galois theory, or if they do they are
rare and well>
hidden...http://www.maths.tcd.ie/~dwilkins/Courses/311/ has
good online noteson Galois Theory.
= I have recently
nished teaching myself basic group/ring/eld> theory using
I.N. Hersteins Abstract Algebra. I am seeking the>
name/publisher/possibly ISBN of a good textbook of Galois
theory. It> should be one which lends itself well to being
self-taught; and it> should be well steeped in theory (I like
textbooks where a given> section gives the basics of something
but the problems proceed to> expound upon that thing in great
detail with numerous other> results...) Money is not an issue
since I consider this a top> priority investment however I
lack access to a university bookstore> and so it would be
good if the textbook can be ordered and shipped via> SamP.S.
Links to online tutorials are also good, although they do
not> seem to exist for Galois theory, or if they do they are
rare and well> hidden...My experience (in the U.S.) is that
one can often examine a book byrequesting an interlibrary
loan. And if you want online materials,try a search engine
such as www.alltheweb.com with a compound searchterm such
asgalois lecture notesand you are sure tofind something.David
Ames
I have recently nished teaching myself basic
group/ring/eld> theory using I.N. Hersteins Abstract
Algebra. I am seeking the> name/publisher/possibly ISBN of a
good textbook of Galois theory. Ian Stewart, Galois Theory,
published by Chapman and Hall.--
I have recently
nished teaching myself basic group/ring/eld> theory using
I.N. Hersteins Abstract Algebra. I am seeking the>
name/publisher/possibly ISBN of a good textbook of Galois
theory.> Ian Stewart, Galois Theory, published by Chapman and
Hall.Many ciphers use calculations in GF(2^8). I would like to
understand thosecalculations better. Would this book be a
total overkill for that matter, orwould it be *the* book to
buy?