mm-158
I like Mathematica. Never had my computer crash.-- Bob
Day What is the best software for solving differential>
equations? Is there anything better than Matlab --> that at
least won't crash and necessitate a reboot> of my computer
when I give it a differential equation> it can't solve?
Day> http://www.bob.day.name
=I need to 'nd an algorithm
that can produce a unique non-predictable 12digit (0-9) number
for any given 12 digit number. This is to be used tocreate a
unique barcode on a ticket that cannot be predicted. It is
notrequired that the original seed number be computed from the
resultingbarcode, so some form of one-way hashing function
would be acceptable.Any help in this problem would be
appreciated.Mark.
I need to 'nd an algorithm that can
produce a unique non-predictable 12> digit (0-9) number for
any given 12 digit number. This is to be used to> create a
unique barcode on a ticket that cannot be predicted. It is
not> required that the original seed number be computed from
the resulting> barcode, so some form of one-way hashing
function would be acceptable.> Any help in this problem would
be appreciated.> Mark.1. Create a 10 element array with digits
0-9 in linear order (0 in 0,1 in 1, etc.)2. For each digit of
the 12-digit number:2a. Shuf§e the 10-element array.2b. Using
the original decimal digit as an offset into the array, lookup
the replacement digit value.
> I need to 'nd an algorithm
that can produce a unique non-predictable 12>> digit (0-9)
number for any given 12 digit number. This is to be used to>>
create a unique barcode on a ticket that cannot be predicted.
It is not>> required that the original seed number be computed
from the resulting>> barcode, so some form of one-way hashing
function would be acceptable.>> Any help in this problem would
be appreciated.>>>> Mark.1. Create a 10 element array with
digits 0-9 in linear order (0 in 0,> 1 in 1, etc.)2. For each
digit of the 12-digit number:2a. Shuf§e the 10-element array.>
2b. Using the original decimal digit as an offset into the
array, look> up the replacement digit value.You did not
specify how the array should be shuf§ed.How do you prevent
duplicates?It seems that you'll have to store a full list of
allgenerated numbers in order to verify a speci'c
bar-code,this is very awkward in most
situations.greetings,Ernst Lippe
I need to 'nd an
algorithm that can produce a unique non-predictable 12> digit
(0-9) number for any given 12 digit number. This is to be used
to> create a unique barcode on a ticket that cannot be
predicted. It is not> required that the original seed number
be computed from the resulting> barcode, so some form of
one-way hashing function would be acceptable.> Any help in
this problem would be appreciated. I've seen wiser heads
than
mine recommend a Ruby-Lackov cipher for this kind of thing.
First, you need a random function f(i) that takes a 6-digit
decimal as input and produces a 6-digit decimal output. Start
by splitting your original 12-digit decimal number into two
6-digit parts, A and B. Then perform four steps: A=A+f(B) mod
1000000 B=B+f(A) mod 1000000 A=A+f(B) mod 1000000 B=B+F(A) mod
1000000Concatenate the 'nal A and B to form the 12-digit
output. The process is reversible, so there won't be any
duplicates among the output values. f(i) should be a good
randomizing function, such as the cryptographic hash of the
concatenation of a secret key k with the operand i. While you
could convert to binary and back, if that's convenient,
it's
not necessary. If you're going to do this in decimal, you
could use the ASCII decimal digits of A or B as the input to
the hash, and take, say, the 'rst 32 bits of the
hash's
output, convert it to decimal, and use the low-order 6 digits
as the value of f. [I think Ruby-Lackov can tolerate a small
amount of bias in f. If not, I'm sure someone will post
another suggestion.]--Mike Amling
> I need to 'nd an
algorithm that can produce a unique non-predictable 12>> digit
(0-9) number for any given 12 digit number. This is to be used
to>> create a unique barcode on a ticket that cannot be
predicted. It is not>> required that the original seed number
be computed from the resulting>> barcode, so some form of
one-way hashing function would be acceptable.>> Any help in
this problem would be appreciated. I've seen wiser heads
than
mine recommend a Ruby-Lackov cipher for > this kind of thing.>
First, you need a random function f(i) that takes a 6-digit
decimal > as input and produces a 6-digit decimal output.>
Start by splitting your original 12-digit decimal number into
two > 6-digit parts, A and B. Then perform four steps:>
A=A+f(B) mod 1000000> B=B+f(A) mod 1000000> A=A+f(B) mod
1000000> B=B+F(A) mod 1000000> Concatenate the 'nal A and B
to
form the 12-digit output. The process > is reversible, so there
won't be any duplicates among the output values.> f(i)
should
be a good randomizing function, such as the cryptographic >
hash of the concatenation of a secret key k with the operand
i. While you could convert to binary and back, if that's
convenient, > it's not necessary. If you're
going to do this
in decimal, you could use > the ASCII decimal digits of A or B
as the input to the hash, and take, > say, the 'rst 32 bits
of
the hash's output, convert it to decimal, and > use the
low-order 6 digits as the value of f. [I think Ruby-Lackov can
> tolerate a small amount of bias in f. If not, I'm sure
someone will post > another suggestion.]You would have to be
careful in the selection of your hash function.All standard
hash functions have 2**n different outputs, andI don't know
any hash function that produces 10**6 outputs. An example of a
bad hash function would be to take the 'rst 20 bitsmod
1000000
of a standard cryptographical hash, because this hashfunction
is extremely biased (some values occur twice as often as
theothers). When you use a larger number of bits the bias is
reduced.Is it possible to quantify the maximum allowable bias
in yourLuby-Rackoff construction?greetings,Ernst Lippe
=As
usual, I'm missing some of the intermediatepostings, so this
is a reply to all of the partiesso far.>> I need to 'nd an
algorithm that can produce a unique non-predictable 12>
digit (0-9) number for any given 12 digit number. This is to
be used to> create a unique barcode on a ticket that cannot
be predicted. It is not> required that the original seed
number be computed from the resulting> barcode, so some form
of one-way hashing function would be acceptable.> Any help in
this problem would be appreciated.so what happens if I just
make up a 12-digitbarcode and print my own ticket? I have a
funnyfeeling that if you present the actual problemhere, you
might get other suggestions for solvingit or have possible
problems with your proposedsolution identi'ed.For example,
your use of the word ticket makesme think authentication. For
that you probablyreally want something like a ticket number
and aMAC appended to it. Or something.>> I've seen wiser
heads
than mine recommend a Ruby-Lackov cipher for >> this kind of
thing.... [snipped] ...>> ... [I think Ruby-Lackov can >>
tolerate a small amount of bias in f. If not, I'm sure
someone
will post >> another suggestion.]Depends what you mean by
tolerate. The securityproofs for Luby-Rackov certainly don't
hold up ifthere's any knowledge at all of the f()
functions.While on that subject, I'll also point out thatthe
proofs require four *independent* f()functions, not one that
is reused. You cansimulate this with f_n = hash(key, n, data)
for n = 1..4.That said, for such a small data input,
you'dprobably be safe ignoring those two nits. Problemsare
more likely to surface somewhere else.>You would have to be
careful in the selection of your hash function.>All standard
hash functions have 2**n different outputs, and>I don't know
any hash function that produces 10**6 outputs. An example of a
bad hash function would be to take the 'rst 20 bits>mod
1000000
of a standard cryptographical hash, because this hash>function
is extremely biased (some values occur twice as often as
the>others). When you use a larger number of bits the bias is
reduced.Yes, there's a regular subject here about how toget
unbiased uniform random in [0..N-1] given apseudorandom
bitstream (such as generated by ahash function), avoiding this
bias thatcreeps in when you least expect it.Greg.-- Greg
Rose232B EC8F 44C6 C853 D68F E107 E6BF CD2F 1081 A37CCrypto
Mini-FAQ: http://www.schla§y.net/crypto/faq.txtQualcomm
Australia: http://www.qualcomm.com.au
.... [snipped] ...>
>... [I think Ruby-Lackov can >tolerate a small amount of
bias in f. If not, I'm sure someone will post >another
suggestion.]> Depends what you mean by tolerate. The security>
proofs for Luby-Rackov certainly don't hold up if>
there's any
knowledge at all of the f() functions. I was thinking that,
with sample size limited to 4, the slightly biased set of
functions from which the four particular functions used are
drawn could not be distinguished from the unbiased set of
functions from which the four function should have been drawn.
Given hash output words w distributed uniformly in 0..2**32-1,
one way the code could avoid bias is by using w%1000000 when
w>=967296 (which is 2**32%1000000) and repeating with a new w
value otherwise. The probability of going on to a second w
value is about 1/4440. The probability of exhausting all four
words available from an MD5 hash even once in 10**12 barcodes
1/338. Code that just unconditionally used the 'rst w%1000000
would, on average, in 1000000 trials, produce 999775 output
values uniformly distributed in 0..999999 and 225 output
values uniformly distributed in 0..967295. I conjecture that
the number of plaintext-ciphertext pairs that an attacker
would need to take advantage of that bias would exceed the
number of barcodes the OP intends to print. In any event, I
agree that qualms about the bias can be removed by using the
uniform-in-[0..N-1] algorithms you mention. I think Benjamin
posted good code for one.> While on that subject, I'll also
point out that> the proofs require four *independent* f()>
functions, not one that is reused. You can> simulate this with
f_n = hash(key, n, data) for n = 1..4. My mistake.That said,
for such a small data input, you'd> probably be safe
ignoring
those two nits. Problems> are more likely to surface somewhere
else.>>You would have to be careful in the selection of your
hash function.>>All standard hash functions have 2**n
different outputs, and>>I don't know any hash function that
produces 10**6 outputs. >>An example of a bad hash function
would be to take the 'rst 20 bits>>mod 1000000 of a standard
cryptographical hash, because this hash>>function is extremely
biased (some values occur twice as often as the>>others). When
you use a larger number of bits the bias is reduced.> Yes,
there's a regular subject here about how to> get unbiased
uniform random in [0..N-1] given a> pseudorandom bitstream
(such as generated by a> hash function), avoiding this bias
that> creeps in when you least expect it.Greg.--Mike
Amling
I need to 'nd an algorithm that can produce a
unique non-predictable 12> digit (0-9) number for any given 12
digit number. This is to be used to> create a unique barcode on
a ticket that cannot be predicted. It is not> required that the
original seed number be computed from the resulting> barcode,
so some form of one-way hashing function would be acceptable.>
Any help in this problem would be appreciated.>The simplest way
is to encrypt the 'rst number using AES or3DES. You will have
to convert the result from binary to decimal.Any extra digits
can be thrown away. Change the key variableregularly and keep
the old ones secret.Random key variables can be generated by
rolling dice.Andrew Swallow
> I need to 'nd an algorithm
that can produce a unique non-predictable 12>> digit (0-9)
number for any given 12 digit number. This is to be used to>>
create a unique barcode on a ticket that cannot be predicted.
It is not>> required that the original seed number be computed
from the resulting>> barcode, so some form of one-way hashing
function would be acceptable.>> Any help in this problem would
be appreciated.> The simplest way is to encrypt the 'rst
number
using AES or> 3DES. You will have to convert the result from
binary to decimal.> Any extra digits can be thrown away.
Change the key variable> regularly and keep the old ones
secret.But shouldn't the bar codes be unique?Your procedure
can generate duplicate bar codes.When the bar codes must be
unique you could use thefollowing algorithm. Take a 40 bits
block cipher(you can build this with the Luby-Rackoff
construction).Now encrypt the ticket number, when the result
is>= 10**12 you re-encrypt the result as many timesas
necessary to get a value that is < 10**12.greetings,Ernst
Lippe
I need to 'nd an algorithm that can produce a
unique non-predictable12>> digit (0-9) number for any given 12
digit number. This is to be used to>> create a unique barcode
on a ticket that cannot be predicted. It is not>> required
that the original seed number be computed from the resulting>>
barcode, so some form of one-way hashing function would be
acceptable.>> Any help in this problem would be appreciated.>
The simplest way is to encrypt the 'rst number using AES or>
3DES. You will have to convert the result from binary to
decimal.> Any extra digits can be thrown away. Change the key
variable> regularly and keep the old ones secret. But
shouldn't the bar codes be unique?> Your procedure can
like AES there will be very fewduplicates. If you do not know
the key variable thenthe conversion is unpredictable.The
simplest way to ensure that the bar codes areunique is to add
a prime number to the previousvalue. Lap round when you get to
the top (or a primenumber near the top). Start at a weird
value.
I need to 'nd an algorithm that can produce a
unique non-predictable> 12> digit (0-9) number for any given
12 digit number. This is to be used to> create a unique
barcode on a ticket that cannot be predicted. It is not>
required that the original seed number be computed from the
resulting> barcode, so some form of one-way hashing function
would be acceptable.> Any help in this problem would be
appreciated.> The simplest way is to encrypt the 'rst number
using AES or>> 3DES. You will have to convert the result from
binary to decimal.>> Any extra digits can be thrown away.
Change the key variable>> regularly and keep the old ones
secret.>> But shouldn't the bar codes be unique?>> Your
grade ciphers like AES there will be very few> duplicates. I
assume that you throw away extra digits bycomputing the block
cipher output modulo 10^12.When the block cipher is
cryptographically strongthis should give a random map, in
other wordsyou can regard the output as a random number
withrange [0..10^12).So when you take the entire set of
outputs for all10^12 inputs this set should behave like a set
of10^12 random numbers that are independently drawnfrom a
uniform distribution.Now let's compute the probability that
one speci'cnumber is not present in this output set,
thisprobability is (1 - 1/n)^n and when n is largethis is
approximately equal to 1/e = 0.368 .So about 36% of the
possible outputs never occur,this means that at least 36% of
the inputs willlead to a duplicate output value.> The simplest
way to ensure that the bar codes are> unique is to add a prime
number to the previous> value. Lap round when you get to the
top (or a prime> number near the top). Start at a weird
value.Could you be more precise, I am afraid that Idon't
understand your algorithm.Anyhow, I can't see that your
algorithm wouldbe unbiased, or how it would prevent
duplicates.greetings,Ernst Lippe
to ensure that the bar codes are> unique is to add a prime
number to the previous> value. Lap round when you get to the
top (or a prime> number near the top). Start at a weird value.
Could you be more precise, I am afraid that I> don't
understand
your algorithm.> Anyhow, I can't see that your algorithm
would>
be unbiased, or how it would prevent duplicates.>Where C, Max
are prime constants and C != MaxInitialise N to a valid random
numberloop N = N + C if N > Max then N = N - Max Print Nend
loopAndrew Swallow
the bar codes are> unique is to add a prime number to the
previous> value. Lap round when you get to the top (or a
prime> number near the top). Start at a weird value. Could you
be more precise, I am afraid that I> don't understand your
algorithm.> Anyhow, I can't see that your algorithm would>
be
unbiased, or how it would prevent duplicates.> Where C, Max
are prime constants and C != Max Initialise N to a valid
random number loop> N = N + C> if N > Max then N = N - Max>
Print N> end loop> Andrew Swallow>p.s. If you want to analyse
the algorithmfor bias or duplicates just set C equal to 1.Or
in a simple case setting C = 3, Max = 5and initialise N to 25
3 1 4 2You have to stop after producing Max numbers.Andrew
Swallow
original proceduregenerates a lot of duplicates, I assume that
you couldnot 'nd any errors in the proof.>> The simplest way
to
ensure that the bar codes are>> unique is to add a prime number
to the previous>> value. Lap round when you get to the top (or
a prime>> number near the top). Start at a weird value.>>
Could you be more precise, I am afraid that I>> don't
understand your algorithm.>> Anyhow, I can't see that your
algorithm would>> be unbiased, or how it would prevent
duplicates.>> Where C, Max are prime constants and C != Max>>
Initialise N to a valid random number>> loop>> N = N + C>> if
N > Max then N = N - Max>> Print N>> end loopp.s. If you want
to analyse the algorithm> for bias or duplicates just set C
equal to 1.Or in a simple case setting C = 3, Max = 5> and
initialise N to 25 3 1 4 2You have to stop after producing Max
numbers.But this is a simple linear congruential PRNG, thatcan
be easily broken.(Also, Max is not a prime, this is not a
realproblem when C is relative prime to Max)greetings,Ernst
Lippe
original procedure> generates a lot of duplicates, I assume
that you could> not 'nd any errors in the proof.>Two separate
arguments in the same posting tend tolead to problems, best to
separate them.Since AES encryption has a 1 to 1 translation
printing theentire 128 bits would give a unique number to
eachticket. Is this consistent with your assumption thata
sample of a sample of AES+CTR output can be consideredto be
random numbers? After all they appear to beopposites. I could
believe either answer.Andrew Swallow
you snipped my proof that your original procedure>> generates
a lot of duplicates, I assume that you could>> not 'nd any
errors in the proof.> Two separate arguments in the same
posting tend to> lead to problems, best to separate them.OK.>
Since AES encryption has a 1 to 1 translation printing the>
entire 128 bits would give a unique number to each> ticket. Is
this consistent with your assumption that> a sample of a sample
of AES+CTR output can be considered> to be random numbers? As
far as we know AES behaves like a pseudo-random
permutation.When you stay suf'ciently below the birthday
limit
(for AESthat is 2^64) you cannot distinguish a pseudo-random
functionfrom a pseudo-random permutation. Because we only need
togenerate 10^12 numbers, which is slightly less than 2^40,
westay suf'ciently below the birthday limit and thereforewe
can treat the outputs as a (pseudo-) random set ofnumbers. But
if you don't believe my proofs, the easiest wayto check your
algorithm is to implement it for a smallervalue of Max and
then to check the number of duplicates.I expect that you will
be highly surprised.greetings,Ernst Lippe
you don't believe my proofs, the easiest way> to check your
algorithm is to implement it for a smaller> value of Max and
then to check the number of duplicates.> I expect that you
will be highly surprised.>I have seen true random number
generators producethe same value but it was no where near a
third of thetime. However I was not performing an in-depthsoak
test so I may not have produced suf'cientnumbers.The rate of
occurrence of duplicates is not linear,p(duplicate) =
n/10^12means that there are few at the beginning butmost of
the towards the end.What the original poster probably needs is
anencryption algorithm that is 1 to 1 over 10^12,dif'cult to
predict over short ranges and hard tobreak.Andrew Swallow
way>>
to check your algorithm is to implement it for a smaller>>
value of Max and then to check the number of duplicates.>> I
expect that you will be highly surprised.> I have seen true
random number generators produce> the same value but it was no
where near a third of the> time. However I was not performing
an in-depth> soak test so I may not have produced suf'cient>
numbers.No, you obviously did not (see below). > The rate of
occurrence of duplicates is not linear,> p(duplicate) =
n/10^12This formula is not correct when you have already found
duplicatesbecause the number of different output value that
have already beenproduced is less than n. The correct formula
is to look at theprobability that you don't have any
duplicates when you draw n out ofm different values. The
probability that you have no duplicates amongthese n values is
equal to m^n * m!/(m - n)!> What the original poster probably
needs is an> encryption algorithm that is 1 to 1 over 10^12,>
dif'cult to predict over short ranges and hard to> break.Yes,
and the algorithm that I proposed should solvethat
problem.Your algorithm has some serious problems with
duplicates. I did thesimulation I proposed with Max = 100000,
and the results are shown inthe following table:Duplicates
Inputs Outputs 0: 0 36619 1: 37030 37030 2: 36668 18334 3:
18549 6183 4: 5976 1494 5: 1375 275 6: 324 54 7: 70 10 8: 8
1The second column shows the number of inputs that give
thecorresponding number of duplicates, e.g. only 37030 of the
100000inputs will have no duplicates. The last column shows
the number ofoutputs with that number of duplicates, so 36619
possible outputs werenever produced.This corresponds well with
my expectations that the fraction of inputsthat have no
duplicates is the same as the fraction outputs that arenever
produced: for large values of Max they should both approach
1/e.of the inputs will have duplicates.greetings,Ernst
Lippe
What the original poster probably needs is
an>encryption algorithm that is 1 to 1 over 10^12,>dif'cult
to
predict over short ranges and hard to>break.Unfortunately, we
have no idea what the OP needs.I agree that this is probably
what he *wants*. Buthe didn't state very clearly what he
wanted, letalone even come close to telling us what theproblem
was.Greg.-- Greg Rose232B EC8F 44C6 C853 D68F E107 E6BF CD2F
1081 A37CCrypto Mini-FAQ:
http://www.schla§y.net/crypto/faq.txtQualcomm Australia:
http://www.qualcomm.com.au
=Who the  are you, Mr. or Ms.
anonymous reposter troll?Of course, we all know that head
hasn't learned even one tinything about either algebra or
physics or metrology in the intervening6 1/2 years, so it
might be Dense Donny himself reposting this. Henormally
reposts bits of it every day or so anyway, so the whole
thingwouldn't be any big surprise. The really sad thing is
that besidesnot learning anything in all these years, he's
forgotten how prettythat little church in the valley was.
Maybe he's just a robot with afew blown circuits.Gene
Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/=
It would have been better when it had been factually
correct. > A metallic prototype of the chosen unit of mass was
made, and is >called the Okilogram'; the French
name for
weight.It never has been a French name for weight. It is
derived from the oldGreek who used the gramme to denote a
certain amount of weight or mass(in those times the
distinction was not made) equal to 1/8 drachme. > Herein lies
the problem: > Is the kilogram a measure of weight, or a
measure of mass? OR doesn't it >really matter?It is a
measure
of mass, but the general public uses it for both weight
andmass because it is not interested in the distinction. Note
moreover thatto weigh something can both mean determine the
weight or determine themass. > The gram >and kilogram of mass
are commonly called weights, and are inconsistently >de'ned
in
various texts and dictionaries:Dictionaries are not there to
prescribe what is correct, but to describehow something is
used. When by some §uke a large percentage of thepopulation
starts to use the gramme as a unit of speed, than, for
thedictionaries, it becomes a unit of speed. > A kilogram
weighs 9.806 newtons (about 2.2 lb), on Earth. Its mass >(w/g)
is 9.806 newtons sec^2/9.806 meters = (reduces to) 1 newton
>sec^2/meter (= f/a). This connection may be rewritten, so
that (9.806 >newtons sec^2/9.806 meters) x 1 meter/sec^2 =
(reduces to) 1 newton; which >is a fundamental unit because
about the concept of fundamental unit. Bythis view 1 acc =
Mass cannot be arbitrarily, or otherwise, ... taken as
fundamental, >and force as derived, ..., because it's the
other way around!!Why is the one fundamental and the other
derived? > Better yet, for the time being at least, our
existing >foot-pound-second system - with decimals thereof -
should be retained, >and/or reinstituted. This system, where a
pint's a pound the whole world >around, has been, and still
is,
quite successful.The world is small. Only the USA and Liberia,
I think. In the UK alreadya long time a pint was not a pound,
and that was when they did use pintsand pounds. (I am not old
enough to have been in the UK when there wasa pint that also
was a pound.) In almost all of Europe there was noconnection
of the (local equivalent) of pint with the (local
equivalent)of pound. > So who *cares* if the >foot was the
length of one of some king's feet.Well, in the Netherlands
it
was not. Unless it was both a huge king and adwarf. Before
standardisation to metric there were a few hundred
differentfeet in use in the Netherlands. (O, our 'rst king is
from 1815...) > Its a more convenient >length than one ten
millionth of the distance from the Earth's equator to >one
of
its poles anyway.Convenience is in the eye of the beholder. >
[And the foot too, could be de'ned as the >distance light
travels in a second.]But it is. The inch is de'ned to be 2.54
cm exactly. The latter isde'ned in terms of light-speed and
the foot is de'ned in inches.The values are slightly
different
when you use purveyor's feet. > Finally, the decimal system
is
applicable to any numerical measure, >including the foot, the
pound, the degree of arc, and the second. The >metric systems,
with their individual names (pre'xes) for each and every
>decimal place, have no special claim to it.Indeed. Finally
something that is correct.-- dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland, +31205924131home: bovenover
215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
what a tour de force, not that I'm able to read it.> has
James
actually said what is supposed> to 'x the ring?He
hasn't even
said what's broken.Just that there exist numbers that should
be in thering, but aren't because they don't
't the de'nition
ofalgebraic integers. He hasn't been pinned down on why
theyshould be (though it has something to do with
factoring),or an example of such a number that should be but
isn't. - Randy
Given:>1.- F is a 'eld (not necessarily
Borel)>2.- u is a measure on F>3.- G is the minimal Borel
Field containing F.I really don't see what sense the
terminology > minimal Borel 'eld makes. Maybe you meant> that
G is the minimal sigma-'eld containing F?That's
what I meant.
In the material I have read, the terms BorelField, Sigma
Field, and Sigma Algebra appear to mean the same,that is a
collection of sets closed under complementation andcountable
unions. Is this not correct?>4.- v is a measure on G.>5.- v
and u agree on F.>6.- v and u are sigma-'nite on F.It can be
proved that v is the unique extension of u from F to G.
>Apparently 6.- is a suf'cient but not necessary condition
for
this>uniqueness. Can someone please indicate the necessary
condition and>outline how the proof would then proceed? Many
suf'cient condition.measure extension subject in terms of
further development of measureand probability theory? In other
words, is understanding of itimportant in terms of
understanding new material down the road? Yourhelp is always
appreciated.> ************************David C. Ullrich
>
>>Given:>>1.- F is a 'eld (not necessarily Borel)>>2.- u is a
measure on F>>3.- G is the minimal Borel Field containing F.>>
>> I really don't see what sense the terminology >> minimal
Borel 'eld makes. Maybe you meant>> that G is the minimal
sigma-'eld containing F?That's what I meant. In
the material I
have read, the terms Borel>Field, Sigma Field, and Sigma
Algebra appear to mean the same,>that is a collection of sets
closed under complementation and>countable unions. Is this not
correct?If I hadn't read the other replies I would have
simply
saidno, this is not correct - the last two are the same, but a
Borel'eld is a special case of a sigma 'eld (the
sigma 'eld
generatedby the open sets in a topological space). That's
the
standardway the terminology is used these days, but I gather
thereare people who do use the term Borel 'eld the way
you'vebeen doing - I didn't know that.>>4.- v
is a measure on
G.>>5.- v and u agree on F.>>6.- v and u are sigma-'nite on
F.>>It can be proved that v is the unique extension of u from
F to G. >>Apparently 6.- is a suf'cient but not necessary
condition for this>>uniqueness. Can someone please indicate
the necessary condition and>>outline how the proof would then
necessary and suf'cient condition.measure extension subject
in
terms of further development of measure>and probability theory?
Well, it happens a lot that the _existence_ of the extension is
used to de'ne measures, by 'rst
de'ning them on a
(non-sigma)'eld... when you do that you need the uniqueness
to
know that you've de'ned _a_ measure.>In other
words, is
understanding of it>important in terms of understanding new
material down the road? Other people may have different
opinions: If you're just learningmeasure theory my advice
would be to skim through this part asquickly as possible and
concentrate on the stuff coming up thatgets used in
applications of measures, as opposed to constructionsof
measures - if it turns out you get into something where this
isimportant there will be plenty of time to go back to the
details.>Your>help is always appreciated.>>
************************>>> David C.
Ullrich>************************David C. Ullrich
=Well, it
happens a lot that the _existence_ of the extension is > used
to de'ne measures, by 'rst de'ning
them on a (non-sigma)>
'eld... when you do that you need the uniqueness to know that
> you've de'ned _a_ measure.>In other words,
is
understanding of it>important in terms of understanding new
material down the road? Other people may have different
opinions: If you're just learning> measure theory my advice
would be to skim through this part as> quickly as possible and
concentrate on the stuff coming up that> gets used in
applications of measures, as opposed to constructions> of
measures - if it turns out you get into something where this
is> important there will be plenty of time to go back to the
details.>Your>help is always appreciated.>>
************************>> To everyone who replied, many
= >>Given:>>1.- F is a 'eld (not necessarily
Borel)>>2.- u is a measure on F>>3.- G is the minimal Borel
Field containing F. >> I really don't see what sense the
terminology >> minimal Borel 'eld makes. Maybe you meant>>
that G is the minimal sigma-'eld containing
F?>That's what I
meant. In the material I have read, the terms Borel>Field,
Sigma Field, and Sigma Algebra appear to mean the same,>that
is a collection of sets closed under complementation
and>countable unions. Is this not correct?Yes. Some authors, I
think mostly probabilists rather than analysts, use Borel
'eld
this way. See e.g. K.L. Chung in A Course in Probability
Theory. One trouble with this terminology is that it becomes
clumsy to talk about the main example of a Borel 'eld, namely
the 'eld of Borel sets. I prefer to use sigma-algebra or
sigma-'eld. >>4.- v is a measure on G.>>5.- v and u agree on
F.>>6.- v and u are sigma-'nite on F.>>It can be proved that
v
is the unique extension of u from F to G. >>Apparently 6.- is a
suf'cient but not necessary condition for this>>uniqueness.
Can
someone please indicate the necessary condition and>>outline
there _is_ a simple necessary and suf'cient condition.Me too.
Here, by the way, is an example to show that without something
like 6.- you don't have uniqueness in general. Consider the
'eld F of 'nite unions of intervals in the
reals, and the
measure u on F such that u(A) = 0 if A is a 'nite set and
u(A)
= in'nity otherwise. Of courseit is not
sigma-'nite. The
minimal sigma-'eld containing F is the
sigma-'eld B of Borel
sets. There are lots of extensions of u to B, e.g. Hausdorff
measure of any dimension d with 0 < d < 1.>measure extension
subject in terms of further development of measure>and
probability theory? In other words, is understanding of
it>important in terms of understanding new material down the
road? Your>help is always appreciated.My personal opinion is
that it's a rather specialized topic, and it's
not worth
getting too worked up about, say, the most general form ofthe
uniqueness theorem, unless you run into a case where you
really need it. Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
In the
material I have read, the terms Borel>Field, Sigma Field, and
Sigma Algebra appear to mean the same,>that is a collection of
sets closed under complementation and>countable unions. Is this
not correct?>No, as I understand the terms Sigma 'elds and
sigma algebras are indeed the same thing. The Borel 'eld is
the smallest sigma 'eld containing the topology (i.e., all
the
open sets) of a topological space. When refering to R^n, the
usually unstated topology is the usual one.-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu
In the material I
have read, the terms Borel>Field, Sigma Field, and Sigma
Algebra appear to mean the same,>that is a collection of sets
closed under complementation and>countable unions. Is this not
correct? > No, as I understand the terms Sigma 'elds and
sigma
algebras are > indeed the same thing. The Borel 'eld is the
smallest sigma 'eld > containing the topology (i.e., all the
open sets) of a topological > space. When refering to R^n, the
usually unstated topology is the usual > one.If I recall
correctly, Mr. Martinez (quoted above by Mr. Herschkorn) is
studying out of Chung's _Course_. Chung uses Borel
'eld, in a
now old-fashioned way, as a synonym for sigma-'eld (aka
sigma-algebra). The same usage can be found, for instance, in
Doob's classic book on stochastic processes.penetrating
analysis of increasing families of sigma-'elds (aka
'ltrations). Needless to say they used the term Borel
'eld.]--
A. <bkrkjj$i9ujk$1@athena.ex.ac.uk>
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= at 07:58 PM, raf@tiki-lounge.com (Ross A.
Finlayson) said:>I think it's not worthless to consider the
reals as a sequence of>points. It's worthless because it is
meaningless. You can't de'ne a
successorfunction compatible
with the ordering.>Is an in'nitesimal some ghost of a
departed
quantity, as Berkeley>suggests in his treatise damning the
foundations of the integral>calculus as in'nitesimal analysis
of what Newton called the §uents>and their §uxions?That
depends on what you mean by in'nitesimal. Bishop
Berkeley'sargument was valid as applied to
Newton's writing,
but would beinapplicable to de'nitions in terms of germs or
Nonstandard Analysis.>Is instead the in'nitesimal 1/x for
x=oo
equal to dx?The question is meaningless, unless you have
de'nitions forin'nitesimal and for 1/oo. Even
then it is
meaningless unless yourde'nition is such that 1/oo is an
in'nitesimal.>I'm still thinking that f(x)=1
for irrational x
and f(x)=0 for>rational x that f is everywhere discontinuous.
That much is correct.>What that describes is>that between each
pair of any two rationals is at least one>irrational and
between any two irrationals is at least one rational.That part
is wrong; it describes no such thing.>What I propose is that
given any>rational that the value greater than it and less
than any other>greater is irrational, There is no such number,
as several different people have shown you.>Obviously enough
then under this axiom Adding such an axiom to the standard
axioms would yield aninconsistent axiom system, and all
statements would be provable. Itwould not be an axiom system
for the real numbers.>That's like calling the
endpoints,>points adjacent to only one instead of two other
points in the>continuous sequence of the reals, of the open
interval (0,1)>irrational.I don't know what
you're trying to
say, but that sentence ismeaningless. It contains an
assumption contrary to fact.>Is the evaluation of the integral
of f(x) dx on [0,2] equal to 1? No.>I've queried before If
there are in'nite integers are there>in'nite
integers?, the
answer is not immediately necessarily, no.No. The answer is
that If P then P is always true.>because I say so.That
doesn't
constitute a proof.>Mathematics is exact. FSVO exact that
doesn't match the way you are trying to use it.>With some
level of exactitude I can note that>there are ten times as
many real numbers on [0,10) as there are on>[0,1). You can
note that Red Fuming Nitric Acid is a healthful beverage,
butfew will take your word for it. The statment that you noted
doesn'thave any meaning.>A good example comes from the
theoretical random uniform probability>distribution over the
reals.There is no such animal. There are various distinct
probabilitymeasures on the reals.>Imagine that any number
of>the real numbers is as likely to occur as a sample as any
other,I can't imagine it, because the statement has no
meanining.>Along with that there is a similar convenient
distribution over the>interval from zero to ten. Convenient to
whom?>Select ten samples, say the>sample set is {0, 1, 2, 3, 4,
5, 6, 7, 8, 9}, a representative>sample.What makes that a
representative sample? Why isn't {.01, .02, .03,.04, .05,
.06,
.07, .08, .09, .1} a representative sample.>Only one of the ten
samples, 1/10'th of the samples, is in the>interval of one
tenth length. All ten of my samples are in that interval.>That
exactly parallels the fact that>1/10 is one tenth.No, it
parallels the fact that you didn't take a random sample,
butinstead picked a speci'c sequence.>If you have a problem
with that, imagine you have a random uniform>probability
distribution over {0,1}. That casts no light on the point.>If
you don't know the answer or the question, §ip a
coin.Flipping
raf@tiki-lounge.com (Ross A. Finlayson) said:>Now then, assume
a uniform random distribution over the set {0, 1}>can be
sampled by §ipping a coin. Why would I assume that, when the
'rst sample would require anin'nite amount of
time?>Now,
consider that the point in question is showing that half of
the>reals of [0,1) are in [0,1/2), and the other half are in
[1/2,1). You keep using that expression as though it had any
meaning. You'vebeen asked repeatedly to de'ne,
in Mathematical
language, what youmean by it. So far you have refused to do
so.>Half of the real numbers in the interval [0,1) are less
than one>half,That has as much meaning as saying that half are
red and half areblue. You have not de'ned how to compare or
measure sets of reals.>Anyways it's pretty clear that it is
possible to describe the>propensity of reals What is a
propensity of reals ?>Half the reals are positive, and the
other half are negative. Meaningless.>All that might seem
patently obvious. It might. But actually it's patently
nonesense.>That's because it is.Then de'ne your
terms and
produce a proof. You can't, because you'rejust
throwing
buzzwords around.>What I want to 'gure out are things like
what is the probability>of a real being a rational, or an
algebraic irrational or a>transcendental irrational? You
can't, because you haven't
de'ned what the terms mean.
Firstde'ne them, then worry about their values in
speci'c
cases.>I have a conception that the probability>of a real
being a rational is one-half, I derive that notion from>the
consideration that the rationals and irrationals alternate
on>the real number line,You can derive any conclusion from a
false assumption. You've beenshown half a dozen times that
there can be no such alternation,because you can produce
intermediate points by taking an average.>How about this: a+b
is not de'ned,Then you're not talking about the
real numbers,
because in the realnumbers a+b is always de'ned, (a+b)/2 is
always de'ned, and if a<bthen a<(a+b)/2<b.>(a+b)/2 is not
de'ned, and it's approximately a.How can it be
approximately
anything if it's not de'ned?>In a way
that's about considering
the set of the elements of the>reals with none of the
operations de'ned on the reals, except < as>they are totally
ordered, but perhaps even not that, where density>and order
would imply something contrary to what I am trying
to>infer.You seem to be saying that if we apply the term reals
to somethingother than reals, with properties different from
those of the reals,that it's properties would be different.
That's obvious, butirrelevant; what is at issue is the
properties of the ordered 'eldknow as the reals.>I search for
modulus of precision and am unclear of its meaning.If you want
to invent terms, then you are responsible for de'ningtheir
meaning.>Similarly half of the subsets of N contain
zero,Meaningless. at 07:15 PM, raf@tiki-lounge.com (Ross A.
Finlayson) said:>There are a variety of de'nitions of
measure.K3wl! What's yours?>Consider your example about the
topological property of the>rationalsIt doesn't exist. Again
you are slinging buzzwords around. You need tolearn some
Topology before you can make any rational references
totopolgical properties.>I'm aware of logic behind the
considerations that there are many>more irrationals than
rationals. I just want to ignore it>temporarily to consider
the rationals and irrationals alternating.When you ignore
Logic then you're no longer discussing Mathematics.>If you
consider the measure of the reals [0,1] to be one, I
totally>agree.I suspect that he considers the measure of [0,1]
to be unspeci'ed. Isuspect that he considers the measure of
[0,1] to be 0, given theright measure.>Allow yourself to
consider that the measure of the rationals on>[0,1) is
1/2.Sure: use a measure in which {0} and {1} have measure 1/2
and any setnot containing 0 or 1 has measure 0.>What are the
veri'able real-world results of assuming the rationals>to be
measure zero? If you use a uniform distribution then you
don't
have to prove it;it's a standard theorem.>With what
analytical
results does it disagree?That depends on the measure you
pick.>What was the function bijecting the reals and
irrationals again? There are many.-- Shmuel (Seymour J.) Metz,
at 07:58 PM, raf@tiki-lounge.com (Ross A. Finlayson)
and f(x)=0 for>rational x that f is everywhere discontinuous.
That much is correct.What that describes is>that between each
pair of any two rationals is at least one>irrational and
between any two irrationals is at least one rational. That
part is wrong; it describes no such thing.?? Perhaps i'm
missing something.Suppose that A and B are two subsets of R,
mutually disjoint andtheir union is R. De'ne f to be 0 on A
and 1 on B. If f is everywherediscontinuous, doesn't that
imply that between any two elements ofA there is an element of
B and vice versa?>What I propose is that given any>rational
that the value greater than it and less than any other>greater
is irrational, There is no such number, as several different
people have shown you.In non-standard analysis, there might
be, however.See Alain Robert's book about NSA. Rather than
beingirrational, it would be non-standard, though.>Obviously
enough then under this axiom Adding such an axiom to the
standard axioms would yield an> inconsistent axiom system, and
all statements would be provable. It> would not be an axiom
system for the real numbers.Actually, there is an axiomatic
approach of NSA in whicha few axioms are added to ZF(C), and
in which the above suggestionmakes sense. The extra axioms are
relatively consistent w.r.t. ZF(C).Again, see Alain Robert's
book on NSA.The other poster might be interested in this
Jurjus <bkrkjj$i9ujk$1@athena.ex.ac.uk>
<3F716897.4040005@tcs.inf.tu-dresden.deqqqq>
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= at 09:08 PM, Herman Jurjus <h.jurjus@hetnet.nl>
said:>?? Perhaps i'm missing something.Slip of the
'nger; I
meant to reply to the consecutive part.>In non-standard
analysis, there might be, however.No, because you can take the
average of two nonstandard reals.>Actually, there is an
axiomatic approach of NSA in which a few>axioms are added to
ZF(C), and in which the above suggestion makes>sense.The Devil
is in the details. The nonstandard reals are a model of
thereals only in the sense that the same propositions are true
withnonstandard de'nitions of various terms. You
can't
construct anisomorphism between them. Further, even with those
axioms therewouldn't be a successor function.>The other
poster
might be interested in this approach towards NSA.He doesn't
have the background for it. He'd need to start with,
e.g.,Halmos, and work his way forward.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
What I propose is that given
any>rational that the value greater than it and less than any
other>greater is irrational, There is no such number, as
several different people have shown you.In non-standard
analysis, there might be, however.> See Alain Robert's book
about NSA. Rather than being> irrational, it would be
non-standard, though.I have yet to see any standard or
non-standard model of the reals in which there is a smallest
positive number. In the various non-standard versions, there
tend to be rather more numbers between any positive number x
and zero, there are all those y such that y/x are i'nitesimal
but positive, then all those z such that z/y is in'nitesimal
but positive, and so on ad in'nitum.
=What I propose is that
given any>rational that the value greater than it and less than
any other>greater is irrational, There is no such number, as
several different people have shown you. In non-standard
analysis, there might be, however.> See Alain Robert's book
about NSA. Rather than being> irrational, it would be
non-standard, though. I have yet to see any standard or
non-standard model of the reals in> which there is a smallest
positive number.Who says that that is meant?In nsa, there can
be a nonstandard number that can be said to be'greater than
(a
given standard rational) and less than any other (standard)
greater'That it is not unique, who cares?In nsa, though,
there
are both rational and irrational non-standard numberswith this
property. So, the statement that such values are necessarily
irrationalis not true, there.But who knows what this
poster's
intuition may eventually lead to?Herman Jurjus
<bkrkjj$i9ujk$1@athena.ex.ac.uk>
<3F716897.4040005@tcs.inf.tu-dresden.deqqqq>
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said:>Who says that that is meant?The OP has consistently
failed to de'ne his terms; Virgil's guess asto
what he meant
is as good as any.>In nsa, there can be a nonstandard number
that can be said to be>'greater than (a given standard
rational) and less than any other>(standard) greater'There
is
still no alternation of rational and irrational. Nor is
itplausible that the OP meant images of rational and
irrational under animbedding into the nonstandard reals.--
Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to
=What I propose is that given
any>rational that the value greater than it and less than any
other>greater is irrational, There is no such number, as
several different people have shown you. In non-standard
analysis, there might be, however.> See Alain Robert's book
about NSA. Rather than being> irrational, it would be
non-standard, though. I have yet to see any standard or
non-standard model of the reals in> which there is a smallest
positive number.Who says that that is meant?> In nsa, there
can be a nonstandard number that can be said to be> Ogreater
than (a given standard rational) and less than any other
(standard) > greater'> That it is not unique, who cares?Ross
wants to use non-standard numbers to con'rm his hypothesis
that there is a next real after any real, and that the
irrationals and rationals alternate on the real line or on
some non-standard real line. So he cares. On the other hand,
his hypothesis is way out in left 'eld, where he has been
stuck for months, if not years.
What I propose is that
given any>rational that the value greater than it and less
than any other>greater is irrational, There is no such number,
as several different people have shown you. In non-standard
analysis, there might be, however.> See Alain Robert's book
about NSA. Rather than being> irrational, it would be
non-standard, though. I have yet to see any standard or
non-standard model of the reals in> which there is a smallest
positive number.Who says that that is meant?> In nsa, there
can be a nonstandard number that can be said to be> Ogreater
than (a given standard rational) and less than any other
(standard) > greater'> That it is not unique, who cares?Ross
wants to use non-standard numbers to con'rm his hypothesis >
that there is a next real after any real, and that the
irrationals > and rationals alternate on the real line or on
some non-standard > real line. So he cares. On the other hand,
his hypothesis is way out > in left 'eld, where he has been
stuck for months, if not years.Actually, I have not been
claiming to be using the nonstandardnumbers. In light of the
issues that we cover and mutually understandto some extent, it
might be the case that of the set of reals that foreach that it
has a next element of opposite rationality that thatwould be in
neither the nonstandard nor classical model. Its elementswould
still be the elements of the set of real or hyperreals.About
equating density with measure in the unit interval, that
meansequating density in the unit interval to measure in the
unitinterval.In this model, say alternating analysis, or a
rationally alternatingmodel of the continuous reals, the
alternating measure of therationals in [0,1) is 1/2, as is
that of the irrationals, as is theirdensity in the unit
interval and the density of the rationals in thereals.This
doesn't disagree with the measure of the reals in the
unitinterval being equal to one. The results of classical
analysis wouldhold true. Anything that didn't hold true
would
be suspect.What's a proper subset of the rationals or
irrationals that is densein the reals, where its complement is
in'nite?The rationals and irrationals are each dense in the
reals, theirunion, the reals are continuous.Ross
Ross
wants to use non-standard numbers to con'rm his hypothesis >
that there is a next real after any real, and that the
irrationals > and rationals alternate on the real line or on
some non-standard > real line. So he cares. On the other hand,
his hypothesis is way out > in left 'eld, where he has been
stuck for months, if not years.Actually, I have not been
claiming to be using the nonstandard> numbers. In light of the
issues that we cover and mutually understand> to some extent,
it might be the case that of the set of reals that for> each
that it has a next element of opposite rationality that that>
would be in neither the nonstandard nor classical model. Its
elements> would still be the elements of the set of real or
hyperreals.As the classical model is the reals and a
non-standard is the hyperreals, you are claiming to
simultaneously be inside of and outside of some model
simultaneously, which certainly puts you outside of
rationality.About equating density with measure in the unit
interval, that means> equating density in the unit interval to
measure in the unit> interval.And how do you do equate density
to measure? It would help if you could explain what YOU mean
by dense and what you mean by measure, as the standard
meanings do not allow of equating these ideas.In this model,
say alternating analysis, or a rationally alternating> model
of the continuous reals, the alternating measure of the>
rationals in [0,1) is 1/2, as is that of the irrationals, as
is their> density in the unit interval and the density of the
rationals in the> reals.But every time you have a rational, x,
and irrational, y, supposedly next to each other, you have the
problem of (x+y)/2 between them, so they aren't really next
to
each other after all.This doesn't disagree with the measure
of
the reals in the unit> interval being equal to one.Yes it
does. For each of uncountably many irrationals, x, linearly
independent over the rationals, consider U(x) = (x+Q) / [0,1),
the intersecting ofx+Q and [0,1), where x+Q = {x+q: q in Q}.
Since each x+Q is essentially a translation of Q, each must
have the same measure, but, as they are pairwise disjoint, the
measure of [0,1) must be the sum of their separate
measures.Thus, according to your arithmetic, uncountably many
measures of 1/2 add up to 1. So your measure is
self-contradictory.> The results of classical analysis would>
hold true. Anything that didn't hold true would be
suspect.Your formulation must be suspect, then, since it
doesn' hold true.What's a proper subset of the
rationals or
irrationals that is dense> in the reals, where its complement
is in'nite?Let n be an arbirary integer grreater than 1, Z be
the set of integers, and Z[1/n] be the ring generated by
appending 1/n to Z and taking the closure under addition and
multiplication. Each such ring will be dense in the reals and
have in'nite compliment in the rationals and in the reals,
furthermore, if n is a proper factor of m, then Z[1/m] is a
proper subset of Z[1/n].Thus each of Z[1/2], Z[1/4], Z[1/8],
z[1/16], ..., is a proper subset of all its predecessors in
this sequence, but each is also dense in R.The rationals and
irrationals are each dense in the reals, their> union, the
reals are continuous.Ross, what do YOU mean by continuous? For
the standard meaning of continuous, a function may be
continuous, but not a mere set.What property does the ordered
'eld of reals have that the ordered 'eld of
rationals lacks
that makes the set of reals continuous in your sense but the
set of rationals not continuous in your sense?
= > Ross
wants to use non-standard numbers to con'rm his hypothesis >
that there is a next real after any real, and that the
irrationals > and rationals alternate on the real line or on
some non-standard > real line. So he cares. On the other hand,
his hypothesis is way out > in left 'eld, where he has been
stuck for months, if not years.Actually, I have not been
claiming to be using the nonstandard> numbers. In light of the
issues that we cover and mutually understand> to some extent,
it might be the case that of the set of reals that for> each
that it has a next element of opposite rationality that that>
would be in neither the nonstandard nor classical model. Its
elements> would still be the elements of the set of real or
hyperreals.As the classical model is the reals and a
non-standard is the > hyperreals, you are claiming to
simultaneously be inside of and > outside of some model
simultaneously, which certainly puts you > outside of
rationality.About equating density with measure in the unit
interval, that means> equating density in the unit interval to
measure in the unit> interval.And how do you do equate density
to measure? It would help if you > could explain what YOU mean
by dense and what you mean by > measure, as the standard
meanings do not allow of equating these > ideas.> In this
model, say alternating analysis, or a rationally alternating>
model of the continuous reals, the alternating measure of the>
rationals in [0,1) is 1/2, as is that of the irrationals, as is
their> density in the unit interval and the density of the
rationals in the> reals.But every time you have a rational, x,
and irrational, y, supposedly > next to each other, you have
the problem of (x+y)/2 between them, > so they aren't really
next to each other after all.> This doesn't disagree with
the
measure of the reals in the unit> interval being equal to
one.Yes it does. For each of uncountably many irrationals, x,
linearly > independent over the rationals, consider U(x) =
(x+Q) / [0,1), the > intersecting ofx+Q and [0,1), where x+Q =
{x+q: q in Q}. Since each x+Q is essentially a translation of
Q, each must have the > same measure, but, as they are
pairwise disjoint, the measure of > [0,1) must be the sum of
their separate measures.Thus, according to your arithmetic,
uncountably many measures of 1/2 > add up to 1. So your
measure is self-contradictory.The results of classical
analysis would> hold true. Anything that didn't hold true
would be suspect.Your formulation must be suspect, then, since
it doesn' hold true.What's a proper subset of
the rationals or
irrationals that is dense> in the reals, where its complement
is in'nite?Let n be an arbirary integer grreater than 1, Z be
the set of > integers, and Z[1/n] be the ring generated by
appending 1/n to Z and > taking the closure under addition and
multiplication. Each such ring > will be dense in the reals and
have in'nite compliment in the > rationals and in the reals,
furthermore, if n is a proper factor of > m, then Z[1/m] is a
proper subset of Z[1/n].Thus each of Z[1/2], Z[1/4], Z[1/8],
z[1/16], ..., is a proper > subset of all its predecessors in
this sequence, but each is also > dense in R.The rationals and
irrationals are each dense in the reals, their> union, the
reals are continuous.Ross, what do YOU mean by continuous? For
the standard meaning of continuous, a function may be >
continuous, but not a mere set.What property does the ordered
'eld of reals have that the ordered > 'eld of
rationals lacks
that makes the set of reals continuous in > your sense but the
set of rationals not continuous in your sense?I 'nd a nice
exposition on MathPages about the consideration ofalternating
rationals and irrationals, from searching for
rationalsirrationals
alternating:http://www.mathpages.com/home/kmath172.htmThis
page discusses a function continuous at all
irrationals,discontinuous at all
irrationals:http://www.math.tamu.edu/~tom.vogel/gallery/node6.
html#SECTION00024000000000000000discontinuous at the rationals
is onthere are some things to consider about that type of
function. Adescription of that is f(x)=0 for x irrational and
f(p/q)=1/q forrelatively prime p and q. I believe that that
has the same image asf(x)=0 for irrational x and f(x)=1 for
rational x with the domain ofthe irrationals, that is to say,
f(x)=0 for x irrational and f(x) isunde'ned for x rational is
nowhere continuous, or a subset of theirrationals besides the
subset of a single element would be complete. That is to say,
for the set {x}, any sequence (x, x, ...) convergesto x, an
element of {x}, and thus any singleton set is complete.A
Harvey Mudd Fun Fact: Rationals Dense but
Sparse.http://www.math.hmc.edu/funfacts/f'les/30004.3.shtmlto
an irrational. Obviously enough, I try to think of a sequence
o'rrationals that converges to a rational. How about a
rationalsequence that converges to one. Have the irrational
sequence be thevalue of any irrational minus the rational
sequence element times theirrational. For example, sum 1/2^x
converges to one, have thesequence of irrationals that
converge to a rational, zero, being pi -(sum 1/2^x) pi. Zero
is rational. That's only saying that neitherthe rationals
nor
irrationals are complete, where the reals arecomplete.I search
for rationals dense sparse, interleave rationalsirrationals,
between any two rationals, denseness rationals, andother
phrases to do with the quanti'ed density of the rationals.of
rationals and irrationals and their alternation in the reals.
Forexample, Do rational and irrational numbers
alternate?:http://mathforum.org/library/drmath/view/51573.
htmlHere is a reference to a function f(x)=0 for rational x
and f(x)=1 forirrational x:
http://www.math.rutgers.edu/courses/436/436-s00/Papers2000/
stasiak.html. Here's some discussion:
http://www.nrich.maths.org.uk/askedNRICH/edited/445.html . A
postposits the existence of sets of elements of the unit
interval of realsof measure 0.5, presumably besides x<0.5.I
read that the reals are continuous as they are complete,
anyconvergent sequence of reals converges to a real. Draw the
shortestline from zero (0, 0, ...) to one (1, 0, ...) without
lifting thepencil, each point on the line is a real, an
element of R[0,1],removing an element leaves a set
insuf'cient
to represent any point,adding an element leavse a set that can
have an element removed and besuf'cient. The constant
function
f(x)=C is continuous with thedomain of the reals, it contains
no discontinuities, it's notdiscontinuous. Its range is also
the union of the range of theconstant function for the domains
of the rationals and irrationals.About subsets of the
rationals, that was what I had in mind, p/2^q forintegers p
and q, but I am still trying to determine why it
wouldn'tcontain p/q except the prime factorization of 2^q is
always 2, 2, 2,.... Obviously enough for 'xed x there are
only
2^x many integers ythus that 0 <= y / 2^x <= 1. The set of
elements of the reals contains the same elements as the setof
hyperreals. Drawing the line from zero to one goes through
thereall the reals and all the hyperreals.About translating
each irrational by each rational and consideringtheir
intersection with the unit interval, my opinion is that it
isjust the set of irrationals of the unit interval. The sum of
arational and an irrational is always irrational. So I say that
themeasure of each in an interval is the same in this
alternation model,and where the sum of their measure is one,
the measure of each is onehalf, the measure of the irrationals
and rationals in the unitinterval are each presupposed to be
equal in the model. So I don'tthink that shows contradiction
to m([0,1))=1.Here I should note some confusion about the
measure of [0,1) vis-a-visthe measure of [0,1].About (a+b)/2
=/= a =/= b, my point is that any a and b that you
haveselected (unique numbers) are separated by in'nitely many
othernumbers. Their average is as well unique and in'nitely
distant fromeither between them in the consecutive sequence of
the continuousreals.In a way it's about reconciling
point-ness
and line-ness. A point isthe intersection of two (and
in'nitely) many lines that intersect, aline connects two (and
in'nitely many) points that are collinear. Ittakes two points
to describe a line, it takes two non-parallel(coincident)
lines to describe a point. That's obviously extravagant.
Then
there are three non-collinear points for a plane, and
threecoincident planes for a point, etcetera. The lines are
not parallel. They're correlated.Consider the set of all
Euclidean points (a_0 e1, a_1 e2, ...). Consider the set of
all lines. The points on the chords through thecenter of the
n-sphere centered at (0, 0, 0, ...) are the points
ofRxRxRx..., each point of that product besides the origin is
on onlyone of those lines.The nonstandard model contains
de'nitions of points that comprise aminimal two-point line
segment. The line described by those twopoints of the reals
goes through all the other points of the reals. There are only
points to describe the line, here the elements of a setthat
contains any and all points on a line contains as elements
onlypoints.The rationals and irrationals are each dense in the
reals, they aredisjoint and their union is the reals. Then
again, so are thealgebraics and transcendentals, or for
example the rationals p/q wherep and q are relatively prime
and q is even and that set's complementin the reals. Also a
constant function is everywhere discontinuouswith any of those
sets as its domain. This causes me angst. I derivehumor from
using the word angst. The model can't have density 1/2
foreach
of the rationals and irrationals where the same reasoning
appliesto the algebraics and transcendentals because the (set
of) algebraicsis a proper superset of the rationals with
in'nitely many propersubsets that are supersets of the
rationals. That reminds me of aboutthe algebraics and
transcendentals, with the rationals, algebraicirrationals, and
transcendentals.This throws a large cognitive wrench into the
works of determiningsome positive, 'nite measure of the
rationals in the reals. I candivide the irrationals into the
transcendentals and algebraicirrationals. That means I still
want to know that given an element,that for the next element,
what it is, or the probabilities of whatit could
be.Ross
=This page discusses a function continuous at all
irrationals,> discontinuous at all irrationals:Are you sure
about that? :-) > That means I still want to know that given
an element,> that for the next element, what it is, or the
probabilities of what> it could be.How about, just for once,
entertaining the possibility thatthere may not be a next
element after your favourite number?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
This page discusses a function continuous at all irrationals,>
discontinuous at all irrationals:Are you sure about that? :-)>
The pages discusses continuous at all irrationals,
discontinuous atall rationals, and no.> That means I still
want to know that given an element,> that for the next
element, what it is, or the probabilities of what> it could
be.How about, just for once, entertaining the possibility
that> there may not be a next element after your favourite
number?The set is only points, the set is totally ordered,
etcetera.The set is only points, the elements of the reals
each represent apoint on a line f(x)=0. The elements of the
set are not rising orfalling edges or crests or troughs of a
signal, they are not ittybitty line segments, they are points!
I suppose you could de'ne acontinuous function as one of
those
other things.Some say the real number is the limit of a
convergent sequence ofrationals as the reals are complete.
It's a point.I consider that there is not a next. I also
consider that there is. Then again I think about things like
next after Ord is zero. Nextinteger after zero is one. Next
number after zero is iota. What'sthe previous number before
zero?Anyways in talking about the reals, and how they
represent as a seteach point of fx)=0, each point is
represented explicitly.Ross
=(On RAF)> But who knows what
this poster's intuition may eventually lead to?We should all
worry :-(-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
>What I propose is that given any>rational that the value
greater than it and less than any other>greater is irrational,
There is no such number, as several different people have shown
you.In non-standard analysis, there might be, however.> See
Alain Robert's book about NSA. Rather than being>
irrational,
it would be non-standard, though.I have yet to see any
standard or non-standard model of the reals in > which there
is a smallest positive number. In the various non-standard
versions, there tend to be rather more > numbers between any
positive number x and zero, there are all those > y such that
y/x are i'nitesimal but positive, then all those z such >
that
z/y is in'nitesimal but positive, and so on ad
in'nitum.Between
any two odd integers is an even integer, between any two
evenintegers is an odd integer. The density in their union of
either isone half.Here I equate density with measure in the
unit interval. I don't care if you ignore gravity, it
won't do
you much good, I'mhere only concerned with considering a
model
where the rationals andirrationals alternate in the reals.If
there are more irrationals than rationals and rationals
andirrationals are disjoint and distinct, then, where they are
eachtotally ordered, then there necessarily would be
irrationals with norationals between them. Yet, there are
not.I'm trying to think of a function between the unit
interval's realsand irrationals. The claim is that one
exists
because the rationalsmap onto the integers and the integers
don't map to the reals, thusthat the irrationals map onto
the
reals else the reals would be aunion of two sets that don't
map onto the reals. Yet, a constructionexplicitly mapping each
element of the irrationals to each element ofthe reals is not
given. I'm also still looking for a mapping betweenR[0,1)^N
and R[0,1).I like to think that the rationals and irrationals
alternate and thatthe function f(x)=x+iota maps Q[0,1) onto
P(0,1), and f(x)=x-iota mapsQ(0,1] to P(0,1).Then again I
think the impulse function evaluates to half in'nityat zero,
and consider the Gamma function on negative integers to
havevalues of various 'nite multiples of a scalar
in'nity.Now
I'm looking at the post about mapping R <-> P. I
don'timmediately grasp vector space over a 'eld
and
linearlyindependent. You have the sequence b being a sequence
of reals eachlinearly independent over Q, and a set C of reals
of {b_0, b_1, ...}linearly independent over Q, with the initial
sequence element b_0being a rational. RQ=P, you claim that R
injects into P byf(b_n)=b_{n+1} and f(c)=c. Why do you have
braces around n+1 insteadof parentheses? Then you have F(c)=c,
for c in C. I think you meanthat c in C is not an element of
the sequence b. Then you say toextend that to all of R by
linearity over Q. So you claim a functionf(r)=p for r in R and
p in P to be de'ned for all reals. What's rfor
f(r)=pi? What's
p for f(2)? Why f and F, presumably a
shift-keyerror?http://mathworld.wolfram.com/
LinearlyIndependent.htmlhttp://mathworld.wolfram.com/
VectorSpace.htmlhttp://www.wikipedia.org/wiki/Vector_space:A
set V is a vector space over a 'eld F, if given an
operationvector addition de'ned in V, denoted v+w for all v,
w
in v, and anoperation scalar multiplication in V, denoted a*v
for all v in V and ain F, the following 10 properties hold for
all a, b, in F and u, v,and w in V:1. v+w E V2. u+(v+w) =
(u+v)+w3. v+0 = v4. v-v = 05. v+w = w+v6. a*v E V7. a*(b*v) =
(a*b)*v8. 1*v=v9. a*(v+w) = a*v + a*w10. (a+b)*v = a*v +
b*vThose each hold for V = R and F = Q. Properties 1 through 5
indicate that V is an abelian group undervector addition. The
intersection of all subspaces containing a given set of
vectorsis called their span; if no vector can be removed
without diminishingthe span, the set is called linearly
independent.So you say each element of the sequence represents
a set of vectors ora set of a vector, I'm not sure which,
and
that it is linearlyindependent over Q because removing that
vector from the set ofvectors would diminish the span of the
intersection of the subspacesof the vector space.Please neaten
that up provide a more self-contained explanation. Alsoexplain.
While you're at it show a bijection between R^N and R.Some
talk
here is about the nosntandard treatment of the reals:
thehyperreals. One thing to note is that *R, the hyperreals,
as a setcontains the same elements as R, the reals. It's
just
a different wayto consider them.About the uniform probability
distributions over intervals of reals,that's not about
making
some new de'nition of what a probabilitydistribution is.
It's
about applying the characteristics of aprobability
distribution to an in'nite population. We were talkingabout
the probability of an in'nite binary seqence having one
elementbeing on, the rest off. That probability is expressed
as n/2^n, as ndiverges to in'nity. The probability of any
possible sequence isequal to 2^n/2^n, in the limit: one. So
anyways out of those npossible sequences with one on bit and
the rest off bits, each isequally probable. The probability of
each among all possible in'nitebinary sequences is
being1/2^n,
the probability of each among allin'nite binary sequences
with
one on bit is 1/n. So a theoretical(read: thought experiment)
method to generate an element of N is toonce again §ip
in'nitely many coins. At this point it's a
crazy, orrather,
unconventional thought experiment in that the 'rst coin
tosssays whether it is oo/2 or greater or less than oo/2.
Assume it's along sequence of zeros, then it would be saying
about whether theresult is greater than or equal to oo/4,
oo/8, oo/16, etcetera. Without a method to generate a sample
from a uniform distribution overthe natural numbers, it's
still that the probability of selecting anyis 1/|N|.Of course
that's ludicrous but at the same time it allows us
toconsider
the realm of thought in concern of this issue and to thentalk
about the probability of selecting a given element of the
naturalintegers assuming a uniform probability distribution
over theintegers. At least we seem to have some agreement that
a uniformprobability distribution over an interval of the reals
exists, and asimple method to sample an element of an interval
of the reals exists.in'nitesimals, it talks about
1-in'nitesimals, 2-in'nitesimals,etcetera,
n-in'nitesimals,
with the oo-in'nitesimal being zero.Ross
<bks6sj$78u$1@pc-news.cogsci.ed.ac.uk>
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<3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net>
<3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net>
<blv35a$gpgq1$1@ID-110648.news.uni-berlin.de>
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
= at 07:09 PM, raf@tiki-lounge.com (Ross A.
Finlayson) said:>Here I equate density with measure in the
unit interval. What do you mean by measure? With the obvious
meaning, the rationalshave measure 0. As has been explained to
you.>only concerned with considering a model where the
rationals and>irrationals alternate in the reals.That's like
saying that you're concerned with a model where 1+1 >
1.It's
not a model for the reals, because it doesn't have the
propertiesof the reals.>rationals and irrationals are disjoint
and distinct,What are you trying to say?>then there necessarily
would be irrationals with no>rationals between them. Would you
care to provide a proof of that?>I like to think that the
rationals and irrationals alternateI like to think that the
nice gentleman with the map will make merich. Alas, if I do
believe it then I will lose a substantial amountof money.
It's
false.>and that>the function f(x)=x+iota maps Q[0,1) onto
P(0,1), and f(x)=x-iota>maps Q(0,1] to P(0,1).Also false.>half
in'nityMeaningless.>'nite multiples of a scalar
in'nity.Also
meaningless.>a set of a vectorMeaningless.>One thing to note
is that *R, the hyperreals, as a set>contains the same
elements as R, the reals. Incorrect.>the characteristics of a
probability distributionWhat do you believe them to be? How do
you apply them to an in'niteset?>We were talking>about the
probability of an in'nite binary seqence having one>element
being on, the rest off. No we weren't, because
you've refused
to de'ne what you mean by that.>That probability is expressed
as n/2^n, as n>diverges to in'nity. Meaningless. Assuming
that
you meant the limit of n/2^n as napproaches in'nity, that
comes
out to 0.>The probability of each among all possible
in'nite>binary sequences is being1/2^n, What is n? You seem
to
be confusing bound and unbound variables.>equally probable. Yes
0=0.>So a theoretical (read: thought experiment)Theoretical is
not the same as gedanken experiment.>method to generate an
element of N is to>once again §ip in'nitely many coins.
You
can't.>At this point it's a crazy, or>rather,
unconventional
thought experiment in that the 'rst coin>toss says whether it
is oo/2 or greater or less than oo/2.No, because oo/2 has no
meaning.>talk about the probability of selecting a given
element of the>natural integers assuming a uniform probability
distribution over>the integers. That's easy; it
doesn't
exist.>At least we seem to have some agreement that a
uniform>probability distribution over an interval of the reals
exists,No, because we don't have common understandings of
what
any of thewords mean.>simple method to sample an element of an
interval of the reals>exists.No, because a 'nite number of
coin
tosses only lets us sample a'nite
set.>in'nitesimals,And it was
clear that your understanding was fuzzy and
incorrect.Mathematics is a precise discipline; you must reason
from the axioms,de'nitions and rules of inference, not from
your preconceptions.-- Shmuel (Seymour J.) Metz, SysProg and
at 07:09 PM,
raf@tiki-lounge.com (Ross A. Finlayson) said:>Here I equate
density with measure in the unit interval. What do you mean by
measure? With the obvious meaning, the rationals> have measure
0. As has been explained to you.>only concerned with
considering a model where the rationals and>irrationals
alternate in the reals.That's like saying that
you're
concerned with a model where 1+1 > 1.> It's not a model for
the reals, because it doesn't have the properties> of the
reals.>rationals and irrationals are disjoint and
distinct,What are you trying to say?>then there necessarily
would be irrationals with no>rationals between them. Would you
care to provide a proof of that?>I like to think that the
rationals and irrationals alternateI like to think that the
nice gentleman with the map will make me> rich. Alas, if I do
believe it then I will lose a substantial amount> of money.
It's false.>and that>the function f(x)=x+iota maps Q[0,1)
onto P(0,1), and f(x)=x-iota>maps Q(0,1] to P(0,1).Also
false.>half in'nityMeaningless.>'nite multiples
of a
scalar in'nity.Also meaningless.>a set of a
vectorMeaningless.>One thing to note is that *R, the
hyperreals, as a set>contains the same elements as R, the
reals. Incorrect.>the characteristics of a probability
distributionWhat do you believe them to be? How do you apply
them to an in'nite> set?>We were talking>about the
probability of an in'nite binary seqence having one>element
being on, the rest off. No we weren't, because
you've refused
to de'ne what you mean by that.>That probability is
expressed as n/2^n, as n>diverges to in'nity. Meaningless.
Assuming that you meant the limit of n/2^n as n> approaches
in'nity, that comes out to 0.>The probability of each among
all possible in'nite>binary sequences is being1/2^n, What is
n? You seem to be confusing bound and unbound variables.>So
anyways out of those n>possible sequences with one on bit and
the rest off bits, each is>equally probable. Yes 0=0.>So a
theoretical (read: thought experiment)Theoretical is not the
same as gedanken experiment.>method to generate an element
of N is to>once again §ip in'nitely many coins. You
can't.>
>At this point it's a crazy, or>rather, unconventional
thought
experiment in that the 'rst coin>toss says whether it is oo/2
or greater or less than oo/2.No, because oo/2 has no meaning.>
>talk about the probability of selecting a given element of
the>natural integers assuming a uniform probability
distribution over>the integers. That's easy; it
doesn't
exist.>At least we seem to have some agreement that a
uniform>probability distribution over an interval of the reals
exists,No, because we don't have common understandings of
what
any of the> words mean.>simple method to sample an element
of an interval of the reals>exists.No, because a 'nite number
of coin tosses only lets us sample a> 'nite set.>
>in'nitesimals,And it was clear that your understanding was
fuzzy and incorrect.> Mathematics is a precise discipline; you
must reason from the axioms,> de'nitions and rules of
inference, not from your preconceptions.Look at a probability
distribution function. Is it everywherediscontinuous?You
don't
have to §ip in'nitely many coins to tell if a real
numberfrom
[0,1) is less than 1/2, just one.Consider x and y, independent
variables, x goes to in'nity: y/x iseffectively zero or
indeterminate, x/x is one, x is a dependentvariable of itself.
If y is a dependent variable of x, then, y/x isnot necessarily
zero or indeterminate (divergent).The probability of a binary
sequence of length n having one onelement is n/2^n, and the
sum of all the probabilities of thepossibilities is 2^n /
2^n=1. The sum from zero to in'nity of zerois zero. One half
is de'nitely a real number. The probability of abinary
sequence of length n having one on element is the same asthat
of it having one off element.Only a quarter of the subsets of
N contain both zero and one, and onlyone subset contains each
element, 1/2^n, 1 of 2^n. That has 2^(n-1)of 2^n subsets
containing zero.I'm pretty sure that the elements within the
hyperreals are the sameas those in the reals. It's just not
possible for each of theelements of the hyperreals to
distinguish which elements of the realsit is.Mathematics is a
particularly tractable subject: there are truthsthat are
agreed upon by all, that's generally exhibited by
2+2=4,it's
true. By the same token, those areas in which not all agree
arethe signposts of areas where there is room for development,
egEuclid's Fifth Postulate: Agree or Disagree.This site is
great, I could learn a lot from it:
http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml
.I'm
a mathematical ingrate. I always want more. The reals are not
going to be reordering themselves just to alternate.Have a
nice day, Ross
Between any two odd integers is an even
integer, between any two even> integers is an odd integer. The
density in their union of either is> one half.What does that
last sentence mean?Here I equate density with measure in the
unit interval.Then the integers have density zero. I don't
care if you ignore gravity, it won't do you much good,
I'm>
here only concerned with considering a model where the
rationals and> irrationals alternate in the reals.Which is
still as stupid as trying to ignore the law of gravity.If
there are more irrationals than rationals and rationals and>
irrationals are disjoint and distinct, then, where they are
each> totally ordered, then there necessarily would be
irrationals with no> rationals between them. Yet, there are
not.In the set of rationals, there are no irrationals.In the
set of irrationals, there are no rationals.In the set of
reals, there are countably many rationasl ans uncontably many
reals arranged so that between any two distinct reals there
are countably many rationals and uncountably many reals.In
fact, there is a order preserving bijection from any open
interval, (a,b), with a < b, to the set of all reals, namely
f:(a,b) -> R: x |-> (a-b)*x/[(x-a)*(x-b)], and if a and b are
rational, the mapping f carries rationals to rationals and
irrationals to irrationals. I'm trying to think of a
function
between the unit interval's reals> and irrationals. The
claim
is that one exists because the rationals> map onto the
integers and the integers don't map to the reals, thus> that
the irrationals map onto the reals else the reals would be a>
union of two sets that don't map onto the reals. Yet, a
construction> explicitly mapping each element of the
irrationals to each element of> the reals is not given. I'm
also still looking for a mapping between> R[0,1)^N and
R[0,1).A reverse bijection, from the reals to the irrationals,
was given in at least 2 versions in a prior posting in this
thread, so just take the inverse bijection of either of them.I
like to think that the rationals and irrationals alternate and
that> the function f(x)=x+iota maps Q[0,1) onto P(0,1), and
f(x)=x-iota maps> Q(0,1] to P(0,1).You may like to think a lot
of things, but that does not make them true.Then again I think
the impulse function evaluates to half in'nity> at zero, and
consider the Gamma function on negative integers to have>
values of various 'nite multiples of a scalar
in'nity.Again
you reveal that your wiring is short circuited.Now I'm
looking
at the post about mapping R <-> P. I don't> immediately
grasp
vector space over a 'eld and linearly> independent. You have
the sequence b being a sequence of reals each> linearly
independent over Q, and a set C of reals of {b_0, b_1, ...}>
linearly independent over Q, with the initial sequence element
b_0> being a rational. RQ=P, you claim that R injects into P
by> f(b_n)=b_{n+1} and f(c)=c. Why do you have braces around
n+1 instead> of parentheses? Standard newsnet notation for a
compound subscript.Then you have F(c)=c, for c in C. I think
you mean> that c in C is not an element of the sequence b.
Then you say to> extend that to all of R by linearity over Q.
So you claim a function> f(r)=p for r in R and p in P to be
de'ned for all reals. What's r> for f(r)=pi?
What's p for
f(2)? Why f and F, presumably a shift-key>
error?http://mathworld.wolfram.com/LinearlyIndependent.html>
http://mathworld.wolfram.com/VectorSpace.htmlhttp://
www.wikipedia.org/wiki/Vector_space:A set V is a vector space
over a 'eld F, if given an operation> vector addition
de'ned
in V, denoted v+w for all v, w in v, and an> operation scalar
multiplication in V, denoted a*v for all v in V and a> in F,
the following 10 properties hold for all a, b, in F and u, v,>
and w in V:1. v+w E V> 2. u+(v+w) = (u+v)+w> 3. v+0 = v> 4. v-v
= 0> 5. v+w = w+v> 6. a*v E V> 7. a*(b*v) = (a*b)*v> 8. 1*v=v>
9. a*(v+w) = a*v + a*w> 10. (a+b)*v = a*v + b*vThose each hold
for V = R and F = Q. Properties 1 through 5 indicate that V is
an abelian group under> vector addition. The intersection of
all subspaces containing a given set of vectors> is called
their span; if no vector can be removed without diminishing>
the span, the set is called linearly independent.So you say
each element of the sequence represents a set of vectors or> a
set of a vector, I'm not sure which, and that it is
linearly>
independent over Q because removing that vector from the set
of> vectors would diminish the span of the intersection of the
subspaces> of the vector space.I have this set of reals B =
{b_0,b_1,b_2, ....} whose members are are, as vectors over Q,
linearly independent and so that b_0 is a non-zero rational.
Every real in span(B) is a linear combination of 'nitely many
of the members of B. There are other reals which are linearly
independent of the span of B, the set of which I called C, and
which is a (vector) subspace of R. Given any real r, then there
are unique rationals p,q, and a unique real b in span(B) and a
unique real c in C such that r = p*b+q*c. Thus R is the direct
sum of subspaces span(B) and C.It may be proven that there is
only one Q-linear function, say f, from R to RQ and such that
f(b_0) = b_1, f(b_1) = b_2 etc., and f(c) = c for every c in
C. This function is a bijection.Please neaten that up provide
a more self-contained explanation. Also> explain. While
you're
at it show a bijection between R^N and R.The explanation is
suf'cient for those who know a little math. Learn a bit about
vector spaces and it may become clear to you, or don't and
it
won't. Some talk here is about the nosntandard treatment of
the reals: the> hyperreals. One thing to note is that *R, the
hyperreals, as a set> contains the same elements as R, the
reals. It's just a different way> to consider them.The
Robinson formulation of *R has more than a single member
corresponding to a member of R, it has uncountably many
coresponding to eanc member of R, which are in'nitesimally
close to each other, as well as some which do not correspond
to any member of R.> About the uniform probability
distributions over intervals of reals,> that's not about
making some new de'nition of what a probability> distribution
is.It is well known, to those who understand what pdf's
(probability density functions) are that there cannot be a
uniform pdf on R. Uniform pdf's require 'nite
real intervals
or 'nite sets to operate on. You cannot make one on a
countably in'nite set nor on an unbounded real interval.>
It's
about applying the characteristics of a> probability
distribution to an in'nite population. Since a uniform pdf
must have certain properties to be a pdf, there are limits on
what sets they can exist on, and N, Q and R are outside those
limits.[garbage deleted]Of course that's ludicrous but at
the
same time it allows us to> consider the realm of thought in
concern of this issue and to then> talk about the probability
of selecting a given element of the natural> integers assuming
a uniform probability distribution over the> integers. At least
we seem to have some agreement that a uniform> probability
distribution over an interval of the reals exists, and a>
simple method to sample an element of an interval of the reals
exists.We have no agreement that such a distribution exists,
because it cannot.> in'nitesimals, it talks about
1-in'nitesimals, 2-in'nitesimals,> etcetera,
n-in'nitesimals,
with the oo-in'nitesimal being zero.Your knowledge of
in'nitesimals is nil. Until you get a reasonable
understanding
of the standard reals, your hopes of understanding anything
about non-standard reals is unreal.
<vcb8yogpvht.fsf@beta13.sm.luth.se>
<vcbk780kpyu.fsf@beta13.sm.luth.se>
<1xUcb.14254$O85.6040@pd7tw1no>
<3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net>
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oftX-Terminate: SPA(GIS)
G.9adel did not show this.
He showed that for a given sound formal>system in which the
statements about natural numbers can be>formulated, there are
statements that can neither be proven nor>disproven.Google for
G.9adel number. He represented statements as naturalnumbers.
The theorems he proved about statements were also
theoremsabout the integers representing them.-- Shmuel
(Seymour J.) Metz, SysProg and JOATnot reply to
Take an n-by-n-grid, n>= 3.
Place the integers 2 to (n^2 +1) into the grid,> one DISTINCT
integer per grid-square, so that: If s(k,j) = a grid-square
(ie. an element of an> n-by-n matrix), then> (for all k and j
where n >= k >= 3 and n >= j >= 1)> s(k,j) = (any divisor >= 2
of s(k-1,j)) +> (any divisor >= 2 of s(k-2,j)),and> (for all k
and j where n >= k >= 1 and n >= j >= 3)> s(k,j) =(any divisor
>= 2 of s(k,j-1)) +> (any divisor >= 2 of s(k,j-2))>[...] an
n=3 example is:> 5 3 8> 2 6 4> 7 9 10> Is there an n=4
example??There appear to be 2 basic solutions and of course
their transposes: 5 2 7 9 3 12 6 15 8 4 10 14 11 16 13 17 5 3
8 11 2 12 4 16 7 6 10 13 9 15 14 17 11 2 13 15 3 12 6 9 14 4
16 8 5 10 7 17 11 3 14 5 2 12 4 10 13 6 16 7 15 9 8 17one
level per cell, rather than a recursive approach. The program
takes about 1 second to exhaust the 4x4 case.-jiw
Take an
n-by-n-grid, n>= 3. Place the integers 2 to (n^2 +1) into the
grid,> one DISTINCT integer per grid-square, so that: If
s(k,j) = a grid-square (ie. an element of an> n-by-n matrix),
then> (for all k and j where n >= k >= 3 and n >= j >= 1)>
s(k,j) = (any divisor >= 2 of s(k-1,j)) +> (any divisor >= 2
of s(k-2,j)),> and> (for all k and j where n >= k >= 1 and n
>= j >= 3)> s(k,j) =(any divisor >= 2 of s(k,j-1)) +> (any
divisor >= 2 of s(k,j-2))>[...] an n=3 example is:> 5 3 8> 2 6
4> 7 9 10> Is there an n=4 example??There appear to be 2
basic solutions and of course their transposes:> 5 2 7 9> 3 12
6 15> 8 4 10 14> 11 16 13 17 5 3 8 11> 2 12 4 16> 7 6 10 13> 9
15 14 17 11 2 13 15> 3 12 6 9> 14 4 16 8> 5 10 7 17 11 3 14 5>
2 12 4 10> 13 6 16 7> 15 9 8 17which a bit tediously could be
extended to 5x5 or 6x6> but not much further because it uses
nested for loops,> one level per cell, rather than a recursive
approach. > The program takes about 1 second to exhaust the 4x4
case.> -jiwAhh...So there ARE solutions after all!I was
neglecting the likelyhood of bigger integers, such as the 11
and12, being in the upper-left, perhaps.Leroy Quet
=Just
wanted to mention that Springer-Verlag's annual mathematics
booksale, the Yellow Sale, is 'nally
back.Here's the
URL:http://www.springer-ny.com/yellowsale/The book sale is
available in bookstores across North America, as wellas online
at the URL above to customers in the Americas. Springer isthe
world's leading publisher in the 'eld of
mathematics.Jason
RothSpringer-Verlag NY
=Is the polynomial
2*cos(2^k*arccos(x/2)) irreducible for each positive integer
k? If so, how is it proved?--Jim Buddenhagen------------To
reply copy jbuddenh@REMOVEtexas.net to address bar and edit
out REMOVE
=I asked:> Is the polynomial
2*cos(2^k*arccos(x/2)) irreducible for each > positive integer
k? If so, how is it proved?and giving a good deal of additional
information on the factorization of Chebeshev polynomials.
--------------------------------------------------------------
-Comments from Joe Silverman:The n'th Chebeshev polynomial
(up
to factors of 2 used by different authors)is the polynomial
F_n(x) with the property that F_n(2*cos(t)) = 2*cos(n*t).So if
we put t = arccos(x/2), then we get F_n(x) =
2*cos(n*arccos(x/2)).This is your formula.If we write cos(t) =
(e^{it} + e^{-it})/2, and for ease of notation,let z = e^{it},
then we get F_n(z+z^{-1}) = z^n + z^{-n}.This is another
characterization of F_n.The roots of F_n(x), from your formula
or from this alternative, are x = 2*cos(pi*(j+1/2)/n) for j =
0,1,2,...,n.Notice this can also be written as x = e^{it} +
e^{-it} with t = (pi*(j+1/2))/n.it's easy to see that the
splitting 'eld of F_n(x) is the real sub'eldof
the 4n'th
cyclotomic 'eld. In other words, the roots of F_n(x)generate
the 'eld K_{4n} = Q(e^{pi*i/2*n}+e^{-pi*i/2*n}).The degree of
this 'eld is [K_n:Q] = phi(4*n)/2, where phi(n) is
Euler'sphi
function. On the other hand, the degree of F_n is n. So the
conclusionis the following:Proposition: F_n(x) is irreducible
over Q if and only if phi(4*n) = 2*n.The case you're asking
about is n = 2^k, and indeed phi(4*2^k) = phi(2^{k+2}) =
2^{k+1} = 2*2^k,so your polynomials are irreducible. Further,
I think this is probably theonly case that phi(4*n) = 2*n, so
the only case that F_n is irreducible.Hope this is of some
help. Feel free to post this, if you
want.--------------------------------------------------------
Jim BuddenhagenTo reply copy jbuddenh@REMOVEtexas.net to
address bar and edit out REMOVE
asked:Is the polynomial 2*cos(2^k*arccos(x/2)) irreducible for
each> positive integer k? If so, how is it proved?and giving a
good deal of additional information on the> factorization of
Chebeshev polynomials.>
--------------------------------------------------------------
-> Comments from Joe Silverman:The n'th Chebeshev polynomial
(up to factors of 2 used by different authors)> is the
polynomial F_n(x) with the property that F_n(2*cos(t)) =
2*cos(n*t).So if we put t = arccos(x/2), then we get F_n(x) =
2*cos(n*arccos(x/2)).This is your formula.If we write cos(t) =
(e^{it} + e^{-it})/2, and for ease of notation,> let z =
e^{it}, then we get F_n(z+z^{-1}) = z^n + z^{-n}.This is
another characterization of F_n.The roots of F_n(x), from your
formula or from this alternative, are x = 2*cos(pi*(j+1/2)/n)
for j = 0,1,2,...,n.Notice this can also be written as x =
e^{it} + e^{-it} with t = (pi*(j+1/2))/n.it's easy to see
that
the splitting 'eld of F_n(x) is the real
sub'eld> of the 4n'th
cyclotomic 'eld. In other words, the roots of F_n(x)>
generate
the 'eld K_{4n} = Q(e^{pi*i/2*n}+e^{-pi*i/2*n}).The degree of
this 'eld is [K_n:Q] = phi(4*n)/2, where phi(n) is
Euler's>
phi function. On the other hand, the degree of F_n is n. So
the conclusion> is the following:Proposition: F_n(x) is
irreducible over Q if and only if phi(4*n) = 2*n.The case
you're asking about is n = 2^k, and indeed phi(4*2^k) =
phi(2^{k+2}) = 2^{k+1} = 2*2^k,so your polynomials are
irreducible. Further, I think this is probably the> only case
that phi(4*n) = 2*n, so the only case that F_n is
irreducible.Hope this is of some help. Feel free to post this,
if you want.by essentially the same argument you see that the
odd chebyshev polynomials ofprime order are irreducible if you
eliminate the trivial factor x (zero x=0)sincerelyKlaus>
--------------------------------------------------------Jim
BuddenhagenTo reply copy jbuddenh@REMOVEtexas.net to address
bar and edit out REMOVE
Is the polynomial
2*cos(2^k*arccos(x/2)) irreducible for each > positive integer
k? If so, how is it proved?and the proof is essentially
trivial: Each polynomial is the square of its predecessor less
2.This follows immediately from the identity (cos z)^2 = (1 +
cos(2*z))/2. Now for k=0,1 the polynomials are x and x^2-2 so
it is clear that for each k the leading coeff is 1, the
constant is 2 and all other coeffs are even, so Eisenstein
applies as Arturo suggests.--Jim
Buddenhagen
=http://icm.mcs.kent.edu/reports/1998/ICM-199802
-0001.pdf
=Isn't it something like Chebychev
polynomial?
Is the polynomial 2*cos(2^k*arccos(x/2))
irreducible for each >positive integer k? If so, how is it
proved?Is it a polynomial? In what? Irreducible over
what?
==
==It's not denial. I'm just very
selective about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
Arturo
Magidinmagidin@math.berkeley.edu
>Is the polynomial 2*cos(2^k*arccos(x/2)) irreducible for
each>positive integer k? If so, how is it proved?this is up to
normalization the n-th (=2^k) chebyshev polynomial (for
|x|<=2)http://mathworld.wolfram.com/
ChebyshevPolynomialoftheFirstKind.htmlits roots are related to
the n-th roots of unity, and the irreducibility to thespecial
value of phi(n)hth (and hope it is not nonsense)klausIs it a
polynomial? In what? Irreducible over
what?
==
= It's not denial. I'm just very
selective about>
what I accept as reality.> --- Calvin (Calvin and Hobbes)>
polynomial 2*cos(2^k*arccos(x/2)) irreducible for each
>positive integer k? If so, how is it proved?Is it a
polynomial? In what? Irreducible over what?> in x, irreducible
over QFor example,2*cos(2^4*arccos(x/2)) = 16 14 12 10 8 6 4 2
x - 16 x + 104 x - 352 x + 660 x - 672 x + 336 x - 64 x + 2is
irredicible.
Is the polynomial 2*cos(2^k*arccos(x/2))
irreducible for each >>positive integer k? If so, how is it
proved?>>> Is it a polynomial? In what? Irreducible over
what?>> in x, irreducible over QFor
example,2*cos(2^4*arccos(x/2)) = 16 14 12 10 8 6 4 2 >x - 16 x
+ 104 x - 352 x + 660 x - 672 x + 336 x - 64 x + 2is
irredicible.Never seen that before; but would it not be
possible to prove itirreducible by using Eisenstein's
Criterion? Looks like the leadingcoef'cient is 1, the
constant
coef'cient is 2, and all the otherterms have even
coef'cient.
If the pattern holds for arbitrary k,then there you
are.
=It's not denial. I'm just very
selective about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
>Is the polynomial 2*cos(2^k*arccos(x/2)) irreducible for
each >>positive integer k? If so, how is it proved?>Is
it a polynomial? In what? Irreducible over what?>>in x,
irreducible over Q>>For example,>>2*cos(2^4*arccos(x/2)) >16
14 12 10 8 6 4 2 >>x - 16 x + 104 x - 352 x + 660 x - 672 x +
336 x - 64 x + 2>>is irredicible.Never seen that before; but
would it not be possible to prove it> irreducible by using
Eisenstein's Criterion? Looks like the leading>
coef'cient is
1, the constant coef'cient is 2, and all the other> terms
have
even coef'cient. If the pattern holds for arbitrary k,> then
there you are.> Right you are. Let p_k(x) = cos(2^{k} *
arccos(x/2)), thenp_1(x) = x^{2} / 2 - 1andp_{k+1}(x) = 2 *
(p_k(x))^2 - 1in x^2, has constant term equal to 2 or -2 and
that 2p_k(1) = -1 and2p_k(2) = 2. As you say, there you are.
You don't even need to invokeEisenstein.Rick
Is the
polynomial 2*cos(2^k*arccos(x/2)) irreducible for each
>positive integer k? If so, how is it proved?Is it a
polynomial? In what? Irreducible over what?Yes. In x. Over Z
kuehl@inf.uni-konstanz.de
=I am trying to make Matrix
Template Library> (http://www.osl.iu.edu/research/mtl/)
interact with MatLab 6.5. If> anybody has done it, please give
me your comments on what you think> about it. I would greatly
appreciate it!IrynaI have used both. What do you mean by
having MTL Ointeract' with Matlab? Do you want
to call matlab
functions via the mex interface? Or do want to generate a C++
program that is callable from matlab?I recently did a project
where results had to be read in Matlab, but the simulation
which generates them is far too complex to run ef'ciently as
a
Matlab program. I created a class for results that allowed easy
measurments from a simulation, and automates the writing of a
corresponding mat 'le. If this is what you are trying to do,
let me know and I can send you the code (I have a version on
the web, but it needs to be updated).Regarding MTL... It's
been a long time since any updates came from MTL? In fact he
object oriented numerics websit (oon.org) is looking a bit out
of date. Is there a better reference? Has anything to replace
blitz or MTL come out ?G.S. [ See
http://www.gotw.ca/resources/clcm.htm for info about ] [
comp.lang.c++.moderated. First time posters: Do this! ]
=Can
you spell out how to do it? I can see that the general
theoremfollows easily if you can prove it where f(t) is a
linear function(byscaling, Markov property, etc.) but how do
you show easily that f(t)is approximated when f is linear? As
I mentioned, I can handle thecase f(t)=0 for all t, because
then you can use the re§ectionprincipal. Is B_t - f(t) with f
linear a BM with drift(I have heardthe term before but don't
really know what it is)?>> A while back I posted a question
about whether or not>> P[sup_{0<=t<=1}|B_t - f(t)| < d] > 0
for all d > 0 and f(t) continuous>> on [0,1] with f(0) = 0.>
>> In other words, does Brownian motion uniformly approximate
any>> continuous function(with f(0)=0) with positive
probability? Someone>> replied that it does, and this follows
from 'rst proving it for f(t)>> = 0 for all t and then
applying the Cameron-Martin Theorem. I can do>> it for f(t)=0,
but I don't seem to be able to 'nd a reference
for the>>
Cameron-Martin Theorem, though it seems to be related to
Girsanov's>> Theorem, and maybe even follows from it. Can
someone give me some help>> or lead me to a reference?>I
don't believe there is any signi'cant
difference between
the>Cameron-Martin-Girsanov theorem, the Girsanov theorem and
the Cameron-Martin>theorem. As I understand it the same
theorem was discovered independently>and is now attributed to
all three.>Rather like the Green-Gauss-Ostrogradsky
theorem.> This does not need quoting any complicated theorems.
Construct> a polygonal function h such that |f - h| < d/3 on
the interval,> bound (from below) the probability that |B_t -
h(t)| < d/3 at> all vertices, and bound the probability that B
differs by d/3> from its polygon at those vertices. The Markov
nature of B and> the boundedness and continuity of f enable
all of this to be> carried out easily.
=Could someone help
me to understand how to 'nd the minimum distancebetween a
surface (say f1(x,y,z)=c1) and a line (f2(x,y,z)=c2. ibelieve
i should be using gradients.thank you very much!
Could
someone help me to understand how to 'nd the minimum
distance>
between a surface (say f1(x,y,z)=c1) and a line (f2(x,y,z)=c2.
i> believe i should be using gradients.thank you very much!A
single equation, such as f2(x,y,z)=c2, can describe in a 3d
space a surface, possibly a plane, but not a line.The vector
parametric form for a line is g(t) = (u1 + u2*t,
v1+v2*t,w1+w2*t),where (ui,v1,w1) is a point on the line and
(u2,v2,w2) is a vector parallel to the line.If the surface and
line intersect, i.e., the distance between the surface and the
line is zero, then f1(u1+u2*t,v1+v2*t.w1+w2*t) = c1 is true
for some real value of t.If the surface and line do not
intersect and the surface has continuous gradients and no
boundary curves, then the gradient of f2 at any extremal point
(closest to or furthest from the line) must be perpendicular to
(u2,v2,w2).
A single equation, such as f2(x,y,z)=c2, can
describe in a 3d space >a surface, possibly a plane, but not a
line.I could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a
line,last time I looked.Lee Rudolph
A single equation,
such as f2(x,y,z)=c2, can describe in a 3d space >a surface,
possibly a plane, but not a line.I could *swear* that {(x,y,z)
in R^3 : x^2+y^2=0} was a line,> last time I looked.Lee
RudolphI stand corrected, but it is not a very ef'cient way
of
doing lines, and certainly not the standard way.
A single
equation, such as f2(x,y,z)=c2, can describe in a 3d space>a
surface, possibly a plane, but not a line. I could *swear*
that {(x,y,z) in R^3 : x^2+y^2=0} was a line,> last time I
looked. Lee Rudolph I stand corrected, but it is not a very
ef'cient way of doing> lines, and certainly not the standard
way.But degenerate cases come up all the time. (Well,
theoretically.) You haveto keep them in the back of your
mind.In'nitely thin cylinders, indeed.Jon Miller
A
single equation, such as f2(x,y,z)=c2, can describe in a 3d
space >a surface, possibly a plane, but not a line.I could
*swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line,> last
time I looked.Lee RudolphNote that z is unrestricted. It might
not hurt to look again.
A single equation, such as
f2(x,y,z)=c2, can describe in a 3d space >a surface, possibly
a plane, but not a line.I could *swear* that {(x,y,z) in R^3 :
x^2+y^2=0} was a line,> last time I looked.Lee RudolphNote that
z is unrestricted. It might not hurt to look again.My badMy
incredible, incredible, bad
>>A single equation, such as
f2(x,y,z)=c2, can describe in a 3d space >>a surface, possibly
a plane, but not a line.>>> I could *swear* that {(x,y,z) in
R^3 : x^2+y^2=0} was a line,>> last time I looked.>>> Lee
RudolphNote that z is unrestricted. It might not hurt to look
again.(psst: the points in that set have the property x=0,
y=0, z any real.That's a line. Think about it.)
>>A
single equation, such as f2(x,y,z)=c2, can describe in a 3d
space >>a surface, possibly a plane, but not a line.>>> I
could *swear* that {(x,y,z) in R^3 : x^2+y^2=0} was a line,>>
last time I looked.>>> Lee RudolphNote that z is
unrestricted. Oh, I do note that, I do, I do.>It might not
hurt to look again.You go 'rst!Lee Rudolph
=A single
equation, such as f2(x,y,z)=c2, can describe in a 3d space>a
surface, possibly a plane, but not a line. I could *swear*
that {(x,y,z) in R^3 : x^2+y^2=0} was a line,> last time I
looked. Lee RudolphThis actually happened.Many years ago a
problem similar to this came up in a CalcIII class I
wasteaching.I mentioned that situations like this are called
Degenerate caseA girl raised her hand and saidBut that's
what
my Father says *I* am!!Bob Pease
=Comments by Jack Sarfatti
on excerpts from:ABSTRACTEach approach to the quantum-gravity
problem originates from expertise in one or another areaI can
feel your intense interest to 'nd the mechanism of gravity
and
objects but are instead wave structures in a quantum space. Our
perception of their properties was
Oschaumkommen' of the wave
structures. (appearances.)I disagree. I agree with the
deBroglie-Bohm-Vigier pilot theory that from information
waves.IT FROM BITmatter cores /zpf < 0 that balance the
centrifugal repulsion from quantized rotation about their
centers of mass and from the repulsive self electric
charge.the pilot wave information BIT landscape it is rolling
on in a generalized gradient §ow including the 'ber space
connections or gauge potentials as in the Bohm-Aharonov effect
that is the Josephson effect in the macro-quantum case. That is
action without reaction for the micro-quantum approximation
with signal locality that applies to not apply to complex
macro-quantum systems. Einstein agreed, but nobody worked it
out.Now, it has been worked out. see QuantumMatter.com
andSpaceandMotion.comThe results are amazing. 1) All the
natural laws are found as properties of the wave structure of
the electron.2) Everything grows out of only two principles
which are properties of one thing - space.Awesome. Gravity is
the simplest piece of cake. Take a look. I would love to have
your thoughts.I do not know what you mean. Have you derived
the equations for general relativity from the information
wave? That is precisely what I have done for the giant vacuum
pilot wave along with the uni'ed dark energy/matter local
'eld.Any new proposal must be couched in mathematical
language
and must in suitable limiting cases yield the battle tested
equation of theoretical physics such asGuv =
(8piG/c^4)TuvMaxwell's equationsetc.Otherwise it is not
legitimate physics IMHO.Also there must be contact with
experimental observations both in terms of prediction and
explanation as nicely presented in David Deutsch's book The
Fabric of Reality for example in the chapters on proper
methodology in theoretical physics.Ditto for the excess verbal
baggage of less by David C. Williams below on the nature of c
in E = mc^2. The hard core of what is behind this can be found
in the book by Wheeler and Taylor's introductory text on
Einstein's relativity. The basic idea of geometrodynamics is
the block universe of 4 dimensions with time and space mixing
together in changes of perspective of uniformly moving
observers in the globally §at case without gravity to begin
with.The issue is the invariance of the 4D line element ds
under the relevant groups of local frame transformations at a
'xed spacetime event P.Given a local frame of reference with
coordinates x,y,z, tds^2 = dx^2 + dy^2 + dz^2 - c^2dt^2 =
dx'^2 + dy'^2 + dz'^2 -
c^2dt'^2Here c must be invariant under
the group as in the Lorentz transformationsdx' = (1 -
(v/c)^2)^-1/2[dx - vdt]dt' = (1 - (v/c)^2)^-1/2[t -
vx/c^2]dy'
= dydz' = dzfor a nonaccelerating frame shift at constant
velocity v along the common direction x parallel to x' of
the
two global inertial frames S and S'.One cannot describe
gravity this way if one insists on retarded causality of no
teleological future causes of past effects associated with the
ideas of destiny, fate and purpose.could use Newton's
gravity
with global special relativity to produce the three classic
tests of GR provided one introduced the
Wheeler-Feynman-Dirac-Hoyle-Narklikar trick of advanced
potentials in addition to retarded potentials. The fact that
Puthoff gets those tests as well with his variable dielectric
vacuum model is no great achievement either because
Einstein's
classical geometrodynamics goes beyond those tests, e.g.
gravimagnetism and gravity waves and black holes.With gravity
one must use LOCALLY curved spacetime in which at spacetime
point event Pds^2(P) = guv(P)dx^udx^vwith summation convention
u,v, = 0,1,2,3There are two LOCAL symmetry groups here.The
Poincare group is no good anymore. The translation subgroup
symmetry of the Poincare group is broken by the locally
variable Diff(4) curvature tensor that is the essence of
gravity. The local Lorentz group of invariant tipped light
cone structure is obeyed in the tangent 'ber space attached
to
P.The base space of the tangent bundle obeys the Diff(4) group
of LOCAL general coordinate tensor transformations xu' =
x^u'(x^u) that replaces the translation subgroup of the
globally §at Poincare group of special relativity.This is all
for zero torsion of course.All LOCAL observables must be
tensors or spinors under both groups. A spinor is a square
root of a tensor.Einstein's equivalence principle is
mathematically represented by the tetrad map eu^a(P) from
locally §at tangent space inertial coordinates a to locally
curved base space non-inertial coordinates u. The a -> a'
transformation is via the 6-parameter Lorentz group of special
relativity. The u -> u' transformation is via the
4-parameter
Diff(4) group of general relativity.dx^u = eu^a(P)dx^anab =
ea^u(P)eb^v(P)guv(P)Local elimination of the non-tensor
connection 'eld for parallel transport of tensors along world
line paths normally associated with Newtonian gravity
acceleration g for example. These are g-forces or inertial
forces from accelerating non-inertial frames like the surface
of the rotating Earth for example. They are not inhomogeneous
tidal variations in the g-force from the curvature tensor,
which are never locally eliminated although they are here on
Earth very small of order(scale of measurement) 10^-13 in
centimeters.Where nab is the §at space-time constant metric
tensor of special relativity and guv(P) is the locally
variable metric which represents a real gravity 'eld only
when
the 4th rank curvature tensor of tidal forces does not vanish.
You can have a variable guv(P) without a real gravity 'eld
from using a non-inertial local frame that is accelerating
without tidal forces. This is not physically of great interest
however.A LOCAL tensor that vanishes in one LOCAL frame
vanishes in ALL LOCAL frames at 'xed point event P for ALL
relevant symmetry groups.If tuv = 0 then tab = 0 and vice
versa.There is no such thing as a gravity force anymore in
this non-Newtonian paradigm of geometrodynamics. Newton's
gravity equations are regained in the limit of weak curvature
and slow speeds compared to c. If there that minimize their
dynamical action in the given metric 'eld guv(P)Light rays
move on null geodesics ds^2 = 0.The invariance of the speed of
light c in global special relativity is replaced by the above
remark!In general there are gravimagnetic cross terms in the
case of nonstationary metrics and this complicates what is
meant by the speed of light.For example, if there are no
gravimagnetic cross terms in a simple case with spherical
polar coordinates for a non-geodesic observer subject to spin
1 gauge forces likeds^2 = grr(P)dr^2 - gthetatheta(P)dtheta^2
+ gphiphi(P)dphi^2 - c^2gtt(P)dt^2For a light ray we have0 =
grr(P)dr^2 - gthetatheta(P)r^2dtheta^2 +
gphiphi(P)(rsin(theta))^2dphi^2 - c^2gtt(P)dt^2with the usual
convention that the LOCAL metric 'eld guv(P) is a pure
dimensionless number and r(P) is the Schwarzschild curvature
radial coordinate de'ned such that the surface area
surrounding the center in the static spherically symmetric
spacetime geometry has the Euclidean area 4pir(P)^2 .Consider
the null radial geodesic, dtheta = dphi = 00 = grr(P)dr^2 -
c^2gtt(P)dt^20 = dR^2 - c^2dT^2wheredR = grr(P)^1/2drdT =
gtt(P)^1/2dtdR and dT are physically measured space and time
intervals for the light ray using meter sticks and clocks or
radars by the non-geodesic observer for small separations
between two lightlike separated events P and P'. Small means
compared to the local radii of spacetime curvature.For example
in the static spherically symmetric Schwarzschild vacuum metric
solution ofRuv = 0for r > 2Gm/c^2gthetatheta = 1gphiphi =
1grr(P) = (1 - 2Gm/c^2r)^-1gtt(P) = (1 - 2Gm/c^2r)The black
hole event horizon is atgtt(P) = 0dR = (1 -
2Gm/c^2r)^-1/2drBut the circumference C = 2pirThe change in C
for dr isdC = 2pidrThereforedC/dR = 2pi(1 - 2Gm/c^2r)^1/2--> 0
at the event horizon.The physical radius R = integral dR is
much larger compared to physical circumference C as it would
be in §at 3D space. If one keeps R 'xed ~ h/mc ~ G*m/c^2,
the
micro-geon of Wheeler's Mass without mass Charge without
charge Spin without spin shrinks to a point in high resolution
Heisenberg microscope scattering probes with r ~ h/p for
momentum transfer p like SLAC deep inelastic electron
scattering off protons.This is a semi-classical theory without
quantum electrodynamic vacuum polarization zero point energy
density effects, however the latter were shown by Feynman and
Schwinger to obey the Poincare group in the absence of
gravity. The price for this is the obscure renormalization,
which may not be internally mathematically consistent although
its predictions are in remarkable agreement with experiments.
Indeed, the requirement of renormalization with a 'nite
number
of fudge factors or epicycles :-) has been a very useful rule
of thumb. Directly micro-quantizing Einstein's general
relativity is not renormalizable, i.e. one needs an in'nity
of
epicycle fudge factors. That tells us we have asked the wrong
question. As John A. Wheeler says:The Question is: What is The
Question?Resemblances of Wheeler's remark to
Cantor's diagonal
argument and Godel's incompleteness theorem of
self-referential spontaneous self-organization are not
accidental and random. :-) Wheeler thinks of the universe as a
self-excited circuit of observer-participators.The velocity of
light c in ordinary non-gravitating vacuum with /zpf ~ 0 is
directly measured by a variety of techniques. It is an
observable measurable property. The velocity of light in a
medium is c/n where n is the index of refraction. The physical
vacuum also has an index of refraction n(vac) that is very
close to 1 in most situations. This small variation comes
primarily from vacuum polarization zero point §uctuations of
the off mass shell or virtual electron-positron plasma
electrically neutral ionized plasma inside the vacuum. This is
vacuum.One must be careful in how to use both special and
general relativity when polarization effects in real on mass
shell media are considered. In the case of special relativity
one should, for example, consult the text books by Arnold
Sommerfeld, Panofsky and Phillips, Landau and Lifz. A real
on mass shell medium in a sense spontaneously breaks
translational symmetry on scales larger than the unit cell of
the lattice as described in more detail by P. W. Anderson in
his More is different series of papers collected in A Career
in Theoretical Physics (World Scienti'c) When one includes
all
dynamical degrees of freedom including those inside the unit
cell of the lattice, global special relativity is restored to
the propagation of light in a real medium in the usual
Paramahamsa,parties (bcc's to you and Toby) to stimulate
wider
discussion andunderstanding of the important points you have
raised towards re-evaluationand correction of fundamental
errors in scienti'c conceptualizations sincethe key departure
of Newton's work neutering science by his using
hismathematicalizations which did not incorporate properly his
own determinedreligious faith in the absolute nature of truth
as per his belief in God. Iedited you posts lightly for
clearer reading per common American Englishpunctuation etc.,
and changed one word from proof to idea relating tothe
Theosophical science quote so that your point was not mistaken
as thatquote you cited being some kind of empirical
experimental proof of theyour choice of proof over idea please
expand on this in your next posttocorrect my misunderstanding
and I will forward appropriately with apologies.Your
discussion properly considered, by those to whom it was sent,
shouldalso go a long way towards restoring, or integrating,
ethics into thescienti'c discipline of thought since the only
way out of presentconundrums in science is to replace
uncertainty about uncertainty withcertainty about the absolute
nature of truth itself and let the chipsfall where they may in
terms of how this change in principle
modi'esscienti'c
perception of the nature of reality, the various
theoryequations etc.I noticed in your writeups that you
reference C as the constant value of thespeed of light in
several places and then in one place you seem toreference it
as the velocity of light. I understand that in his
earlierworks, especially in German, Einstein used the term
velocity and thenover years he too began routinely using speed
apparently due to themathematical convention dictating that by
de'nition the value of thesquare of the direction component
of
the square of a vector property isde'ned identically equal to
one.Thus, in trying to understand, for example, what is the
true value of, say,(10mph-North)^2, the square of the velocity
of ten miles per hour in anortherly direction, the simple
(2+2=4) logic of ordinary math is thrownout the window by this
mathematical convention per this dictum and thenormal logical
value of the square of this vector
quantity(100m^2/h^2-North^2) is de'ned for all intents and
purposes as equal to(100m^2/h^2), ie, like saying because we
cannot mentally grasp orunderstand what it means (North)^2, we
de'ne it to be identical to one,ie, no meaning at all.While
this may seem a trivial point for all purposes within the
realm ofconsidering physical systems dynamics from the point
of view of currentscience since apparently Galileo's time,
ie,
the exclusively objectivenature of reality, ie, that observer
and observed are separate and allphysical systems that are
real exist independenttly of and identicallyobservable by all
observers, or they are not considered real by sciencewhen they
are not reproducible by all observers at all times, in the
caseof C^2 as the proportionality constant between the value
of the Energy ofany given system of reality under observation
and the value of the mass ofthat system, and since C itself
appears in so many equations ofelectrodynamics etc., herein
apparently lies another overlooked fundamentalmisconception of
scienti'c thought compromising the absolute nature oftruth
since C itself is not pegged to the notion of an
externalobjective reality reference frame but is pegged to the
identity of theobserver. That is to say, since C is
non-additive and its square (C^2) isthe proportionality
constant between the values of energy and mass of realphysical
systems then this mathematical convention seems evidently
themain limit of current mathematics to overcome for a full
understanding ofthe relative nature of reality, ie, the
relationship depicted in Buddhismthe universe relative to the
identity of that person, ie, the ego-centricnature of the
universe.While this view seems rightly to folks like Dr.
Sarfatti as psycho-babbleit nevertheless is the view that
offers a mathematical handle for astarting point to
conceptualize and integrate into modern physical theoriesthis
precept that there is a consciousness factor at work between
thenature of the observer and the nature of the system of
reality underobservation, ie, a relationship of mind between
observer and observedas well acknowledged by quantum physics
experiments over last decadeswhere he describes this principle
as observership.Many scientists are wrestling with this notion
of the relationship of mindbetween observer and observed, eg,
denoted in O'Leary's books asthe consciousness
factor and in
Dr. Sarfatti's post quantum physicsof consciousness theory
equations by their complexities that includethe Uncertainty
Principle mathematics and depict the operation o'ntent on the
mental 'eld of matter causing a Oback-action in
time're§ecting
back via that mental 'eld of matter in a
Ocybernetic
feedbackloop of consciousness' of predictible
Omoment of
consciousness' durationwhich corellates well with latest
experimental results in neuroscience etceven though in
Sarfatti's view, in my words, the observer is transparentie,
there is no accounting for variations between observer
identitiesand corresponding variations in observable systems
dynamics fromobserver to observed (non-reproducibility by all
observers at alltimes of all physical systems dynamics, eg,
psi-phenomena), ie, thedifferences in each observer's
Omind'
impacting the physical systemsdynamics of the
Oreality' which
each observer observes.Back-action is not temporal. It is the
direct reaction if IT back on its BIT 'eld.
IT's BIT 'eld
quantum potential Q now has sources and is not fragile, but
has macro-quantum phase rigidity of which Andrei Sakharov's
metric elasticity is an example. IT is no longer a test from
in spontaneous self-organization - the participatory universe
as a self-excited circuit. It is not to be confused with the
Wheeler-Feynman advanced potential from future to past. It is
true that when the limits of micro-quantum theory without
back-action hence with signal locality are transcended, then
one gets presponse signal nonlocality which mimics an advanced
potential effect. Back-action in my sense as quantum action +
post-quantum reaction, and advanced potentials are, of course,
not incompatible in my theory in which EPR micro-quantum
nonlocality with signal locality are as in John Cramer's
transactional generalization of the Wheeler-Feynman classical
advanced potential as 'rst noted by Costa de Beauregard back
in the 1970's. Antony Valentini shows that post-quantum
back-action that pushes the system away from sub-quantal heat
death causes signal nonlocality like what is seen by Dick
Bierman in his mind-matter presponse experiments.
Macro-quantum BEC systems in particular have post-quantum
back-action IMHO.The above is my latest attempt to explain in
words what I have renderedsince 1974 into low level
mathematical equations of a uni'ed 'eldtheory
(referencing
'eld of awareness and the human
mind'sconsciousness
orientation function of light as correction factor toadd back
this omitted true extra value of the square of the
directioncomponent of C^2 which is more than (north)^2 because
of this factthat C is non-additive and therefore as such is
fundamental, beingthe only physical not Opegged'
to the notion
of the exclusively objectivenature of reality but rather
Opegged' to the conceptualization of
the'relative' (or
Oego-centric') nature of reality a
conceptualization whichis
obvious in human daily life via common sense in all other
realms andis throughout the history of human thought a
fundamental principle asdiscussed in new age psycho-babble in
such terms as we are thecenter of our own universe and by
application of the human mindwe can in§uence and change the
nature of our universe.The above paragraph by David Williams
is New Age cargo cult pseudoscience IMHO on a par with crystal
power, orgone, spiral dynamics, monoatomic highly deformed
nuclei superconducting Ormus powder et-al. It is not even
wrong in Wolfgang Pauli's sense.Bruce DePalma is the only
person who has looked at the four equationsof my Tetron
Natural Uni'ed Field Theory and the minimal discussionof
their
meaning presented to him soon after I met him in May 1979,and
his reaction in subsequent soon talk with me was, David,
youknow, I understand your theory. OSeeing is believing,
right?' was hisresponse with a glimmer of humor behind his
eyes that made meknow that he had hit the nail on the head
with this response whichzero's in on the paradoxical
relationship of mind between seeingand believing as it relates
to all levels of human thought includingscience, religion,
spirituality, politics, etc.The above paragraph by David
Williams is New Age cargo cult pseudoscience IMHO on a par
with crystal power, orgone, spiral dynamics, monoatomic highly
deformed nuclei superconducting Ormus powder et-al. It is not
even wrong in Wolfgang Pauli's sense.Exclusive of Dr.
Sarfatti's vigorous ... criticisms,everyone else since has
either said so what? in response, or likeO'Leary have just
ignored and refused to respond to this theoryof mind, with the
single exception of Dr. Fred Wood who listenedto my 'rst-ever
public lecture on my theory, on September 10, 2001,at my alma
mater California State University at Northridge, and aftermade
a point of telling me that he would think seriously about whatI
had said (a prepared written formal read aloud lecture archived
athttp://groups.yahoo.com/group/gcsc-csun/message/6 )in the
application of my tetron thesis is beyond the mathematicsof my
education as a Bachelor of Science in Chemistry, ie, howto
understand the (tensor?) mathematical relationship betweenthis
correction factor Tetron -- the mathematical function appliedto
the square of the speed of light to correct it to its true
value ofthe square of the velocity of light oriented relative
to the identityof the observer, ie, adding back the square of
the vectorcomponent of C to overcome above described
mathematicalconvention, ie, compromise about the nature of
light and theabsolute nature of truth itself in all places
where C^2 appears inphysics equations -- as Tetron applies to
C^2, and since manyequations particularly in electromagnetism
contain C, there mustbe a similar correction factor to apply
in such equations which isalso the missing link in
understanding the consciousness factoras it relates to these
new space energy technologies and theexpected reconciliation
of their presently con§icting theories ofoperation.The above
paragraph by David Williams is New Age cargo cult
pseudoscience IMHO on a par with crystal power, orgone, spiral
dynamics, monoatomic highly deformed nuclei superconducting
Ormus powder et-al. It is not even wrong in Wolfgang Pauli's
sense.Since your theoretical approach well includes deep
backgroundunderstanding on this Eastern philosophy concept of
the mindas the creator of the universe per Vedic knowledge you
writeand feedback on my above views will help stimulate
productivethinking towards a deeper understanding of this
consciousnessfactor as it also needs to be formalized in order
to correct themind of science and fully understand what is
going on with allthese various new space energy technology
experiments andthe theories posited to try and understand
these results as wellas how some of the theories (like yours
and Joseph Newman's)have predicted and perhaps even
empowered
the successfulexperimental results that you have each obtained
by differentcon'gurations of rotating magnetic systems with
entirely differenttheoretical underpinnings yet with similar
overunity results.Consciousness IMHO is when one has
post-quantum back-action which excites the MACRO-QUANTUM BIT
BEC 'eld into self-awareness.Our minds are macroscopic
non-classical information 'elds of both memory are the
activation of unoccupied basins of attraction of the
MACRO-QUANTUM BIT LANDSCAPE when the IT system point of
sub-microtubular hydrophobically caged electric dipole
Frohlich collective modes roll into that basin.In the torsion
'eld theory developed in recent decades from theRussian
language-mind views expressed in mathematicallanguage and with
deeper consideration of many documentedpsi-phenomena
experiments in former Soviet Union, goingback to the
1950's-60's apparently, there may also be this
ideaof a
relationship of mind between observer and observed butthe
details of how this is represented in this theory are
unclearto me at my level of math education and literature
availability.But it is clear from my personal conversation
with Dr. Shipovcourtesy of Dr. Sarfatti's kind invitation
for
my informalparticipation with his group in San Francisco one
day a fewyears ago, that torsion 'eld theory also is
perplexed
by theresults of DePalma showing variation in gravitational
behaviorbetween spinning and non-spinning ball bearing drop
resultsdiscussed in earlier post.I do not believe DePalma's
claims. Gennady's beliefs in psi are not directly connected
with his torsion math. Bill Page talked about the latter at
Vigier IV.Creon Levit investigated such claims at ISSO
1999-2000 that Williams refers to here. Creon was not
impressed with any of the New Age Free Energy claims promoted
by David. But Creon can speak for himself.My hunch here on
this is that rotating objects too have avector property which
analogous to C discussed above, isbeing overlooked in its
importance as it relates to the observerbecause every rotating
system also may, similar to C, be seenin the view of its
orientation and rotational properties as relatesto the
relative identity of the observer. I think that this
otherapplication of what I talk about above will not come here
untilthis issue about C is resolved, but that there is an
importantconnection between the corrected true value of C^2 as
perabove and a deeper understanding and correction also
withtorsion 'eld theory, although the math involved is way
beyondmy education level to deal with as a language to express
theprinciples I have tried my best above to explain in my style
ofCalifornia Chemical English :-)The above paragraph by David
Williams is New Age cargo cult pseudoscience IMHO on a par
with crystal power, orgone, spiral dynamics, monoatomic highly
deformed nuclei superconducting Ormus powder et-al. It is not
even wrong in Wolfgang Pauli's sense.David Crockett Williams
661-822-3309Chartered Life UnderwriterBachelor Science
Chemistrywww.angel're.com/on/GEAR2000/vision.htmlhttp://
groups.yahoo.com/group/drums-of-peacehttp://groups.yahoo.com/
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Road Peace March ProjectUSA, California, Japan, Korea, China,
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Jerusalem,http://groups.yahoo.com/group/silk-road-to-peacehttp
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Differencewww.kucinich.usRep. Dennis
Kucinichhttp://groups.yahoo.com/group/
Kucinich-for-PresidentFor a Culture of Peace & CommunityOur
Spiritual Unity
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rianwillson.comwww.dharmawalk.orgwww.sathyasai.orgwww.tewari.or
gwww.prop1.org
=Folks,I have a couple questions. This is
homework, so please post a nudge,not a solution.1)prove that
if f,g continuous, then so are max(f,g) and min(f,g)After
drawing some graphs, this seems pretty obvious for the
singlepoint a0 -- max(f,g) has 2 cases: it equals to f or g.
Either iscontinuous. However, this question implies continuous
on R, not justat a single point. Any ideas how to approach
this?2)Let f be a function with the property that every point
ofdiscontinuity (ie the lim (x->a) f(x) exists, but is not
equal tof(x)) is a removeable discontinuity. This implies lim
(y->x)f(y)exists for all x, but f may be discontinuous at some
(even in'nitelymany) numbers x.De'ne g(x) = lim
(y->x) f(x).
Prove g is continuous.--I don't even know where to start
with
this one.-earl-
Folks, I have a couple questions. This is
homework, so please post a nudge,> not a solution. 1)prove that
if f,g continuous, then so are max(f,g) and min(f,g)> After
drawing some graphs, this seems pretty obvious for the single>
point a0 -- max(f,g) has 2 cases: it equals to f or g. Either
is> continuous. However, this question implies continuous on
R, not just> at a single point. Any ideas how to approach
this?I'm confused. If you can show that max(f,g) is
continuous
at eachpoint, then you have shown that max(f,g) is
continuous.However, you still have to prove that max(f,g) is
continuous ata point given that f and g are both continutous
at that point.Remember, the ideal of continuity is that one is
making a statementabout how control over the independent
variable allows controlover the dependent variable. For
example, if h(t) represented theheight of a balloon (h) as a
function of time (t), then tosay that h is continous is to say
that if at a given time t0,if I con'ne my attention to times
suf'ciently close to t0,then the height of the ballon is
guaranteed to be suf'cientlyclose to h(t0). Formally, the two
occurances of suf'cientlyare replaced by delta and epsilon,
respectively. Thus,if I want h is a continutous function of t,
supposeI want a guarantee that the balloon will now have
changeheight by more than 1000ft. You might respond that
thiswill be true if I con'ne my attention to a time periodof
60seconds. So you have told me that if |t-t0|<60,then
|h(t)-h(t0)|<1000. Suppose I want better--I want a
guaranteethat the balloon has not changed altitude by more
than 100 ft.Well, you might tell me that now I must con'ne my
attentionto 20 seconds. You have told me that if |t-t0|<20,
then|h(t)-h(t0)| < 100. Now, the above is the basic ideaof
continuity. In the 'rst case, I demanded what todo of an
epsilon=1000, and you told me that a delta=60 wouldsuf'ce.
Then I asked what I would need to get a guaranteefor
epsilon=100, and you told me delta=20. Note, but theway, that
if delta=20 works for epsilon=100, then delta=10also works for
epsilon=100. That is, for each epsilon youhave to produce a
delta small enough so that changes in theindependent variable
of less than delta result in a changeof less than epsilon in
the dependent variable. For checkingcontinuity, one need not
provide an optimal or largest delta--thisis part of the
abstraction. Of course, if you told me thatfor epsilon=100ft I
would need a delta of 0.0000000001sec,then this might not be
useful for practical purposes; a continuousfunction can still
jump around a lot. But for the abstractidea of continuity, you
must merely argue that for at anarbitrary point, for a given
epsilon>0, there exists a delta(perhaps a very small delta, of
no real use) such thata change in the independent variable of
less than deltaresults in a change of the dependent variable
that is*logically guaranteed* to be less than epsilon.Now back
to the problem at at. We want to guarantee thatmax(f,g)
doesn't
change too quickly. We know that fand g don't change too
quikly, so this is looking reasonable.Now we follow our nose
and try the standard openningLet a0 be a point, and epsilon>0.
[Ok, what now. Well we know that f and g are continuous.
Let's
write down what that means] So, for the particular epsilon
listed about, there exists delta_1 such that |x-a0|<delta_1
implies |f(x)-f(a0)|<epsilon [Note, we call the delta that
comes from f delta1 because we know were going need another
delta fog g, as we do now Again, for the epsilon chosen above,
there exists delta_2 such that |x-a0|<delta_2 implies
|f(x)-f(a0)|< epsilon. [Ok, so this is looking good--we've
used continuit of f and g, and it's making sense--we are
saying that we can use bounds on the independent variable to
constrain the dependent varialbe. Now we need to choose a
delta for the function max(f,g). So we write somethiing like
delta = {some function of delta_1 and delta_2} then show that
that delta works for max(f,g)]> 2)Let f be a function with the
property that every point of> discontinuity (ie the lim (x->a)
f(x) exists, but is not equal to> f(x)) is a removeable
discontinuity. This implies lim (y->x)f(y)> exists for all x,
but f may be discontinuous at some (even in'nitely> many)
numbers x. De'ne g(x) = lim (y->x) f(x). Prove g is
continuous.Try the one above again--if you still have
problems, write back.-MIke --I don't even know where to
start
with this one.> -earl->
= >1)prove that if f,g continuous,
then so are max(f,g) and min(f,g) >After drawing some graphs,
this seems pretty obvious for the single >point a0 -- max(f,g)
has 2 cases: it equals to f or g. Either is >continuous.
However, this question implies continuous on R, not >just at a
single point. Any ideas how to approach this?Show min(x,y),
max(x,y), RxR -> R are continuous.Then as f,g are continuous,
so are compositions min(f,g), max(f,g)----
1)prove that
if f,g continuous, then so are max(f,g) and min(f,g)> After
drawing some graphs, this seems pretty obvious for the single>
point a0 -- max(f,g) has 2 cases: it equals to f or g. Either
is> continuous. However, this question implies continuous on
R, not just> at a single point. Any ideas how to approach
this?> There is a slight complexity. The point may lie on both
f and g. You should try working directly from the de'nition
for
continuity andplaying around with the inequalities associated
with the min and maxfunctions.
Folks,I have a couple
questions. This is homework, so please post a nudge,> not a
solution.1)prove that if f,g continuous, then so are max(f,g)
and min(f,g)> After drawing some graphs, this seems pretty
obvious for the single> point a0 -- max(f,g) has 2 cases: it
equals to f or g. Either is> continuous. However, this
question implies continuous on R, not just> at a single point.
Any ideas how to approach this?Proving a function continuous at
any point proves it continuous at every point.> 2)Let f be a
function with the property that every point of> discontinuity
(ie the lim (x->a) f(x) exists, but is not equal to> f(x)) is
a removeable discontinuity. This implies lim (y->x)f(y)>
exists for all x, but f may be discontinuous at some (even
in'nitely> many) numbers x.De'ne g(x) = lim
(y->x) f(x). Prove
g is continuous.--I don't even know where to start with this
one.Does you de'nition of g(x) actually read g(x) = lim
(y->x)
f(x)or should it be g(x) = lim (y->x) f(y)?In the 'rst case,
g(x) = f(x) at all x, so is not continuous at discontinuities
of f.In the latter case, does g(x) = lim (y -> x) g(y), for
all x?
= >> 1)prove that if f,g continuous, then so are
max(f,g) and min(f,g) >> Any ideas how to approach this?
>Proving a function continuous at any point >proves it
continuous at every point.f(x) = 0 when x <= 0 = 1 when 0 <
xis continuous at 1. Thus it's continuous at 0?----
= >>
1)prove that if f,g continuous, then so are max(f,g) and
min(f,g)>> Any ideas how to approach this?>Proving a
function continuous at any point>proves it continuous at
every point.> f(x) = 0 when x <= 0> = 1 when 0 < x> is
continuous at 1. Thus it's continuous at 0?But 1 is not any
point, it is a speci'c point.To restate less ambiguously:If
you prove f continuous at an arbitrary point, then it is
continuous at every pointAnd to prove it at an arbitrary
point, you only need show that for an arbitrary x, g(x) =
lim_{y -> x} g(y)
=[...]>>Proving a function continuous at
any point>>proves it continuous at every point.>> f(x) = 0
when x <= 0>> = 1 when 0 < x>> is continuous at 1. Thus it's
continuous at 0?But 1 is not any point, it is a speci'c
point.To restate less ambiguously:> If you prove f continuous
at an arbitrary point, then it is > continuous at every
pointWhat is wrong with the good old English word every
?Marc
= [...]>>Proving a function continuous at any
point>>proves it continuous at every point.> f(x) = 0 when x
<= 0>> = 1 when 0 < x>> is continuous at 1. Thus it's
continuous at 0? But 1 is not any point, it is a speci'c
point. To restate less ambiguously:> If you prove f continuous
at an arbitrary point, then it is> continuous at every point
What is wrong with the good old English word every ?>Because
when you say every, many beginning students try to look at all
thepoints at once.I would say (to the beginners), Your opponent
gives you a point. Can youprove that the function is continuous
at that point? No matter what point?The catch is that when you
say any point, a single point works for acounterexample, but a
proof has to work for x no matter what x is. Forsome reason,
many beginning students have trouble with this concept.
Theythink it's a symmetric game. They have to be taught to
think logically,preferably without losing their ability to
think illogically. And, don'tforget, some have to be taught
to
think. And you're not allowed to punt(although they are, and
then they savage you on the teaching evaluationsbecause they
didn't learn anything).Jon MillerX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge:
Micro$oft
= at 01:18 PM, Virgil <vmhjr2@comcast.net>
said:>But 1 is not any point, it is a speci'c point.any point
includes every speci'c point.>To restate less ambiguously:>If
you prove f continuous at an arbitrary point, then it is
>continuous at every pointThat's just as bad. That can still
be construed as either any orevery.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
=At one point or another I read
parts of these books:* Manfredo P. do Carmo: Riemannian
Geometry> a very nice reading. good exercises> * R.W. Sharpe :
Differential Geometry. Cartan's Generalization of>
Klein's
Erlangen Program.> very broad minded. not allways accurate.> *
Kobayashi and Numizu> enciclopedic.> * J. Jost> the analytic
view. very dense.> * Spivak's pentalogy> surpsingly, some
parts of it are actually nice.Yes, now it's manufactured as
a
proper book, I'm saving up my pennies tobuy it.HTH> D.L.--
G.C.
=Can the reals be de'ned using repeated exponentative
closure?By the exponentative closure F, I de'ne F/x as the
set
of all thezeroes of all the polynomial functions with
coeffeicients ANDexponents in F. For example, the the
algebraics are the exponentativeclosure of the integers. Thus,
it can be written A=Z/x. DoesC=A/x? If not what does A/x equal?
Can C be generated byrepeatedly exponentatively closing the
integers a 'nite number oftimes? If so, how many? A countable
number of times? An uncountablenumber of times?
Can the
reals be de'ned using repeated exponentative closure?>By the
exponentative closure F, I de'ne F/x as the set of all
the>zeroes of all the polynomial functions with coeffeicients
AND>exponents in F. For example, the the algebraics are the
exponentative>closure of the integers. Thus, it can be written
A=Z/x. Does>C=A/x? If not what does A/x equal? Can C be
generated by>repeatedly exponentatively closing the integers a
'nite number of>times? If so, how many? A countable number of
times? An uncountable>number of times?The set of all that is
obtained is countable. This is becauseeach polynomial has only
a 'nite number of coef'cients andexponents, and
returns only a
'nite number of results. This is assuming that a clear
de'nition exists of x^y if x is negative and takes the value
|x|^y; more can be done.Doing this more than omega (the order
type of the integers)times yields nothing new, because there
are only a 'nite numberof arguments for each extension.--
This
address is for information only. I do not claim that these
viewsare those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
University
Can the reals be de'ned using repeated
exponentative closure?>By the exponentative closure F, I
de'ne
F/x as the set of all the>zeroes of all the polynomial
functions with coeffeicients AND>exponents in F. For example,
the the algebraics are the exponentative>closure of the
integers. Thus, it can be written A=Z/x. Does>C=A/x? If not
what does A/x equal? Can C be generated by>repeatedly
exponentatively closing the integers a 'nite number of>times?
If so, how many? A countable number of times? An
uncountable>number of times?Unless I misunderstand you, F/x is
countable if F is countable. So no,a 'nite or even a
countable
number of exponentative closures won'tdo it: the union of
countably many countable sets is countable. I don'tknow
about
an uncountable number of times.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Can the reals be de'ned
using repeated exponentative closure?>By the exponentative
closure F, I de'ne F/x as the set of all the>zeroes of all
the
polynomial functions with coeffeicients AND>exponents in F. For
example, the the algebraics are the exponentative>closure of
the integers. Thus, it can be written A=Z/x. Does>C=A/x? If
not what does A/x equal? Can C be generated by>repeatedly
exponentatively closing the integers a 'nite number of>times?
If so, how many? A countable number of times? An
uncountable>number of times?Unless I misunderstand you, F/x is
countable if F is countable. So no,> a 'nite or even a
countable number of exponentative closures won't> do it: the
union of countably many countable sets is countable. I
don't>
know about an uncountable number of times.Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel > University of British
Columbia > Vancouver, BC, Canada V6T 1Z2A further question
along those lines is what is is the set, E, ofnumbers
generated by repeated sums of products of rational powers
ofrationals? It is a subset of the algebraics, but does it
form a'eld? What can be said about the set of sums and
products ofelements of E to the power of elements of E?
A
further question along those lines is what is is the set, E,
of>numbers generated by repeated sums of products of rational
powers of>rationals? It is a subset of the algebraics, but
does it form a>'eld? I believe so: if x_1 is a member of E,
then so are its conjugates x_2, x_3, ..., x_n.Note that x_1
x_2 ... x_n is rational (being the constant term of a monic
polynomial over the rationals whose roots are x_1, x_2, ...,
x_n).And then 1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in
E.Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
A further question along
those lines is what is is the set, E, of>numbers generated by
repeated sums of products of rational powers of>rationals? It
is a subset of the algebraics, but does it form a>'eld? I
believe so: if x_1 is a member of E, then so are its
conjugates x_2, > x_3, ..., x_n.> Note that x_1 x_2 ... x_n is
rational (being the constant term of a > monic polynomial over
the rationals whose roots are x_1, x_2, ..., x_n).> And then
1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in E.Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel > University of British
Columbia > Vancouver, BC, Canada V6T 1Z2If we say that a set,
P, is exponentially closed if for all p,q in P,p^q is P, then
can the complex numbers be de'ned as the 'eld
that
isexponentially closed? Or even more simply, the set that
isexponentially closed and contains -2 and 0? If not, what are
theproperties of the smallest 'eld, S, that is exponentially
closed(cardinality, algebraic completeness, convergence
ofCauchy-Sequences)?Note: that I am only considering 'elds
with characteristic 0.
>>A further question along those
lines is what is is the set, E, of>>numbers generated by
repeated sums of products of rational powers of>>rationals? It
is a subset of the algebraics, but does it form a>>'eld? >>>
I believe so: if x_1 is a member of E, then so are its
conjugates x_2, >> x_3, ..., x_n.>> Note that x_1 x_2 ... x_n
is rational (being the constant term of a >> monic polynomial
over the rationals whose roots are x_1, x_2, ..., x_n).>> And
then 1/x_1 = x_2 ... x_n / (x_1 x_2 ... x_n) is in E.>>>
Robert Israel israel@math.ubc.ca>> Department of Mathematics
http://www.math.ubc.ca/~israel >> University of British
Columbia >> Vancouver, BC, Canada V6T 1Z2If we say that a set,
P, is exponentially closed if for all p,q in P,>p^q is P, then
can the complex numbers be de'ned as the 'eld
that
is>exponentially closed? Or even more simply, the set that
is>exponentially closed and contains -2 and 0? I assume that
your operation ^:PxP -> P is just a partial operation,not
de'ned at (0,0); so just rede'ne it to give
^(0,0) = 0.The
smallest 'eld of characteristic zero that is exponentially
closedwould be the closure of Q under the operation ^. Since ^
is a 'nitaryoperation (takes only a 'nite number
of arguments),
the closure of Qis obtained as follows:S_0 = Q;T_{n} = S_n cup
^(S_n,S_n)where ^(S_n,S_n) is the image of (S_n,S_n) in C;
that is, all numbersof the form p^q with p,q in S_n, under
that de'nition; S_{n+1} = T_{n} cup +(T_{n},T_{n}) cup -(T_n)
cup *(T_n,T_n) cup ^{-1}(T_n-{0})where +(T_{n},T_{n}) is the
sum of any two elements of T_n, -(T_n) isthe additive inverse
of any element of T_n, *(T_n,T_n) is the productof any two
elements of T_n, and ^{-1}(T_n-{0}) is the
multiplicativeinverse of any nonzero element of T_nThen the
closure of Q isS_{omega} = Union_{n<w} S_n.Going beyond
n=omega does not give you anything new, since theoperations
are all of arity strictly less than w. Or, it is easy toverify
that S_{omega} will be closed under all the operations, and
sois a 'eld closed under your exponentiation.Now, S_0 is
countable; clearly, if S_n is countable, then so are^(S_n,S_n)
and therefore T_n, and if T_n is countable, then so isS_{n+1}.
So S_{omega} is the countable union of countable sets, henceis
itself countable. Thus, it cannot equal the complex numbers.>If
not, what are the>properties of the smallest 'eld, S, that is
exponentially closed>(cardinality, algebraic completeness,
convergence of>Cauchy-Sequences)?So S=S_{omega}.Cardinality is
clearly countable, by the argument above (a standardargument of
General Algebra). You would not get convergence of
Cauchysequences: since this is countable, there is a real
number which isnot in the set, and of course that real number
is the limit of acauchy sequence of rationals, which are all
in S. I suspect algebraic completeness will also fail: if r is
an algebraicnumber such that Q(r) is not solvable by radicals,
how would you get rin S? Of course, since you also have
non-algebraic numbers in S, thisis hardly a
Oproof', more of a
OI would look at these kinds of numbersto try to settle the
question in the
negative'.
=
selective
about what I accept as reality. --- Calvin (Calvin and
Hobbes)
Arturo Magidinmagidin@math.berkeley.eduLet
f(z) = sum_k a_k*z^k be a formal power series with radius
ofconvergence 1. Put s_n = a_0 + ... + a_n and t_n = (s_0 +
... +s_n)/(n+1).Let g(z) = sum_k s_k*z^k and h(z) = sum_k
t_k*z^k.How would I show that the radius of convergence of h
and g are both 1as well? (The lim sup method doesnt seem to
work).Greg
Let f(z) = sum_k a_k*z^k be a formal power
series with radius of>convergence 1. Put s_n = a_0 + ... + a_n
and t_n = (s_0 + ... +>s_n)/(n+1).>Let g(z) = sum_k s_k*z^k and
h(z) = sum_k t_k*z^k.>How would I show that the radius of
convergence of h and g are both 1>as well? (The lim sup method
doesnt seem to work).Some hints:Note that for any epsilon > 0,
there is C such that |a_k| < C (1+epsilon)^k for all k. What
does that say about |s_k|?For the other direction, note that
a_n = s_n - s_{n-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 have a
problem in combinatorics (chapter is generating functions)
where f(x) = sum_{x>=0} a_n x^n is in C[[x]].What is C[[x]]?I
have to show that composition of two f,g in C[[x]] is again in
C[[x]] if the constant term of g is 0.Our professor de'ned
this
structure in class but I can't 'nd the
de'nition.Any help on
the de'nition of this structure is greatly appreciated.Also,
what are WEIGHTED generating functions?
I have a problem
in combinatorics (chapter is generating functions)> where f(x)
= sum_{x>=0} a_n x^n is in C[[x]]. What is C[[x]]?>Ring of
formal power series. Elements are power series in x
withcoef'cients in C, not necessarily convergent.> I have to
show that composition of two f,g in C[[x]] is again in C[[x]]>
if the constant term of g is 0. Our professor de'ned this
structure in class but I can't 'nd the>
de'nition. Any help on
the de'nition of this structure is greatly appreciated.>
Also,
what are WEIGHTED generating
functions?>
Michael A. Van OpstallPadelford
C-113opstall@math.washington.eduhttp://www.math.washington.edu
/~opstall/
I have a problem in combinatorics (chapter is
generating functions) >where f(x) = sum_{x>=0} a_n x^n is in
C[[x]].What is C[[x]]?Usually, power series on nonnegative
powers of x. Like polynomials withcoef'cients in C, except
they are allowed to have in'nitely manynonzero
monomials.
=
selective
about what I accept as reality. --- Calvin (Calvin and
Hobbes)
Arturo Magidinmagidin@math.berkeley.eduI
have a problem in combinatorics (chapter is generating
functions)> where f(x) = sum_{x>=0} a_n x^n is in C[[x]].What
is C[[x]]?I have to show that composition of two f,g in C[[x]]
is again in C[[x]]> if the constant term of g is 0.Our
professor de'ned this structure in class but I
can't 'nd the>
de'nition.Any help on the de'nition of this
structure is
greatly appreciated.Also, what are WEIGHTED generating
functions?> C[[x]] are simply formal power series, i e you do
not carefor convergence. Weighted? Just ask him (as you could
dofor the 'rst Q), may be he says: Oput some
factors at
thecoef'cients'. Hm ... there is no shame to
ask questions in
lectures ...
=I've got an array size 100 of integer values
(range: 0-5). Each plot in the array represents a 1ms time
window. I currently have these graphed as stairs in MatLab
(which looks like a bar graph). I want to make this function
smooooooth, but i'm not sure of a method to apply to it.What
steps should I take to complete this?Dustin
=Some more
corrections:The physical radius R = integral dR is much larger
compared to physical circumference C as it would be in §at 3D
space. If one keeps R 'xed ~ h/mc ~ G*m/c^2, the micro-geon
of
Wheeler's Mass without mass Charge without charge Spin
without
spin shrinks to a point in high resolution Heisenberg
microscope scattering probes with r ~ h/p for momentum
transfer p like SLAC deep inelastic electron scattering off
protons.Should be:The physical radius R = integral dR is much
larger compared to physical circumference C as it would be in
§at 3D space. If one keeps R 'xed ~ e^2/mc^2 ~ G*m/c^2
(Blackett effect), the micro-geon of Wheeler's Mass without
mass Charge without charge Spin without spin shrinks horizon.
We will see later that this is why lepto-quarks look like
probes with r ~ h/p for momentum transfer p like SLAC deep
inelastic electron scattering off protons.Note thatG(rho +
3p/c^2) = Gphro(1 + 3w) is replaced by c^2/zpf in the w = -1
exotic vacuum case that included both dark energy of negative
pressure and dark matter of positive pressure controlled by
the vacuum polarization macro-quantum coherent local order
parameter (0|e+(x)e-(x)|0) completely missed in the
Puthoff-Haisch-Rueda approach to these problems for the origin
of inertia and the origin of gravity from zero point vacuum
quantum §uctuations in the sense proposed by Andrei Sakharov
in 1967. Sakharov's metric elasticity is a special case of
P.W. Anderson's More is different generalized phase rigidity
of the macro-quantum ground state coherence from spontaneous
symmetry breaking.Imagine a dark matter exotic vacuum core of
radius R = e^2/mc^2c^2/zpfR^3/G* ~ mThe Kerr-Newmann
micro-geon picture for spatially extended IT hidden variable
is that the inner event horizon is of order classical electron
radius of ~ 1 fermi with outer event horizon out to ~ 137 fermi
= h/mc.The in-between ergosphere is ionized vacuum polarization
plasma of virtual electron-positron pairs./zpf ~ 1/Lp*^2Lp* ~
Lp^2/3(c/Ho)^1/3 ~ 1 fermi (t'Hooft-Susskind world
hologram)mass from zero point exotic vacuum energy, hadronic
mass ~ 1 GEV from Heisenberg kinetic energy of con'ned
lepto-quarks as in QCD lite bag model (Frank Wilczek).G*m^2 ~
e^2 (Blackett effect)G*m^2/hc ~ 'ne structure constantOne
must
be careful in how to use both special and general relativity
when polarization effects in real on mass shell media are
considered. In the case of special relativity one should, for
example, consult the text books by Arnold Sommerfeld, Panofsky
and Phillips, Landau and Lifz. A real on mass shell medium
in a sense spontaneously breaks translational symmetry on
scales larger than the unit cell of the lattice as described
in more detail by P. W. Anderson in his More is different
series of papers collected in A Career in Theoretical Physics
(World Scienti'c) When one includes all dynamical degrees of
freedom including those inside the unit cell of the lattice,
global special relativity is restored to the propagation of
light in a real medium in the usual situation where gravity
tidal forces are negligible.should beOne must be careful in
how to use both special and general relativity when
polarization effects in real on mass shell media are
considered. In the case of special relativity one should, for
example, consult the text books by Arnold Sommerfeld, Panofsky
and Phillips, Landau and Lifz. A real on mass shell medium
in a sense spontaneously breaks translational symmetry on
scales smaller than the unit cell of the lattice as described
in more detail by P. W. Anderson in his More is different
series of papers collected in A Career in Theoretical Physics
(World Scienti'c) When one includes all dynamical degrees of
freedom including those inside the unit cell of the lattice,
global special relativity is restored to the propagation of
light in a real medium in the usual situation where gravity
tidal forces are negligible.Some corrections:Ditto for the
excess verbal baggage of less by David C. Williams below on
the nature of c in E = mc^2.should have beenDitto for the
trite excess verbal baggage of less with more by David C.
Williams below on the nature of c in E = mc^2.could use
Newton's gravity with global special relativity to produce
the
three classic tests of GR provided one introduced the
Wheeler-Feynman-Dirac-Hoyle-Narklikar trick of advanced
potentials in addition to retarded potentials. The fact that
Puthoff gets those tests as well with his variable dielectric
vacuum model is no great achievement either because
Einstein's
classical geometrodynamics goes beyond those tests, e.g.
gravimagnetism and gravity waves and black holes.The professor
also got the basic black hole effective potential of the
Schwarzschild /zpf = 0 vacuum metric with his advanced
potential method.Interesting, but like the stochastic
electrodynamics approach to EM zero point energy and the
semi-classical attempt by Ed Jaynes not to quantize the EM
'eld, like Puthoff's PV approach et-al these
attempts at the
very best are fragmentary incomplete and fail to ask and
answer many of the really important fundamental questions
e.g.What is consciousness physically?What is the universe made
from?We are such dreams (BIT) as stuff (IT) is made
from.
=How do I do this?Express the following as a single
fraction:4/3ab - 5/6bcand(m^2 + 2)/(m^2 + m) - (m -
2)/m
How do I do this?Express the following as a single
fraction:4/3ab - 5/6bcMultiply the 'rst term by 2c/2c and the
second by a/a.and(m^2 + 2)/(m^2 + m) - (m - 2)/mNote that the
denominator of the 'rst term is m*(m+1). Use the sameapproach
described above.
=How do I do this?Express the following as
a single fraction:4/3ab - 5/6bc Multiply the 'rst term by
2c/2c and the second by a/a.I'm obviously doing it wrong but
that appears to leave two differentdenominators. I thought we
were trying to get a common one?
>How do I do
this?>>Express the following as a single fraction:>>4/3ab -
5/6bc>> Multiply the 'rst term by 2c/2c and the second by
a/a.I'm obviously doing it wrong but that appears to leave
two
different>denominators. I thought we were trying to get a
common one?>What did you get after performing the
multiplication?
=alt.math deleted because it is not
recognized by my newsreader.alt.algebra.help added because of
all the helpful people there.>>How do I do this?>>Express the
following as a single fraction:>>4/3ab - 5/6bc>> Multiply the
'rst term by 2c/2c and the second by a/a.I'm
obviously doing
it wrong but that appears to leave two different>denominators.
I thought we were trying to get a common one?Multiplying the
'rst denominator (3ab) by (2c) gives 6abc.Multiplying the
second denominator (6bc) by a gives 6abc.What values did you
get?This is only a little more abstract than doing common
denominatorswith numbers. You look for the factors of both
terms, and a commondenominator has to contain all those
factors. 3ab has a factor of 3, aand b. 6a has a factor of 2,
a factor of 3, and factor of a. For thenumeric part, 6 will
serve as common denominator for both (since it'sa multiple
of
3). So you just need to include all the other factors:a, b and
c. Hence: 6abc.You can always multiply the terms together to
get *A* commondenominator, just not the lowest one. (3ab)(6bc)
= 18abc is a commondenominator that you can use for both
fractions. - Randy
How do I do this?> Express the
following as a single fraction: 4/3ab - 5/6bc> (m^2 + 2)/(m^2
+ m) - (m - 2)/m>You do it the same way you do it for
fractions in arithematic.The general formula is derived thus
a/b + r/s = as/bs + br/bs = (as + br)/bs
= How do I do
this?> Express the following as a single fraction: 4/3ab -
5/6bc> (m^2 + 2)/(m^2 + m) - (m - 2)/m You do it the same way
you do it for fractions in arithematic.> The general formula
is derived thus> a/b + r/s = as/bs + br/bs = (as + br)/bsYep,
I understand the basic priniciple, but I just don't know how
to put itin practise with these types of fractions.