il: Nope.
Both wrong? !a AND !!a?
> The proof could be expressed
> in terms of a finite number of known primes, as il seems to have
assumed
> it was, but that's not the way Richard expressed it -- he spoke
explicitly
> of the product of *all* primes.
If you view it in terms of sets, subsets of Z, my view is perfectly standard
one, it's what Ribenboim calls Euclid's proof, and as such is applicable to
the premise that Richard was assuming. Richard was the one who was
introducing
more assumptions. All I need is a finite set of primes. A maximum prime P,
and
the fact that we're in Z, gets me that. Euclid's proof just follows
immediately. Let all primes be all primes in that set, and the
connection
is made. This is why I was blathering about what all meant.
> But Richard's conclusion that P has been shown not to be the largest
> prime number is also wrong; what has been shown is that the assumption
> of a finite number of primes is wrong.
Yup, which is why I said Nope..
> The only reason to assume that
> at least one of the additional primes discovered must be larger than
> P is that we *thought* P was the largest prime, before we discovered
them.
Euclid permits me to assume {2,3,7,13,43,139,3263443} is the finite set
of primes, with maximum prime P=3263443. In what way does
{2,3,7,13,43,139,3263443}, P=3263443 violate the premise Let P be the
largest prime? In which case, surely Richard, as he originally stated his
argument should permit it too. However, _none_ of the primes that Euclid's
construction discovers is larger than P. If the premise is satisfied, but
the conclusion isn't, then there's a syllogistic error somewhere.
If Richard has simply added let all primes <=P be known to his premise
I wouldn't have jumped on it that way. However, if he had, then it would
_not_ have been Euclid's proof (and he claimed what followed was Euclid's
proof so I would have jumped on that instead), and wouldn't (on its own)
have been a proof of the infinitude of primes (as you've added another
assumption, so the proof by contradiction only disproves the _conjunction_
of the assumptions, not either one individually).
Ockham has some wise words for moments like this.
il
--
Unpatched IE vulnerability: Web Archive buffer overflow
Description: Possible automated code execution.
Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html
===
Subject: Re: Question on generation of large prime numbers
> If Richard has simply added let all primes <=P be known to his
premise
> I wouldn't have jumped on it that way.
Let all primes <=P be known.
> Ockham has some wise words for moments like this.
I was about to say don't needlessly multiply spellings, but my
dictionary
agrees with you that Ockham is an acceptable variant of Occam. I find
this deliciously ironic.
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
> If Richard has simply added let all primes <=P be known to his
premise
> I wouldn't have jumped on it that way.
> Let all primes <=P be known.
If you assume that then proof by contradiction leads to the possibilty
that that assumption is false.
I.e. the final conclusion is that
either there's a new prime >P
or there's actually a new prime Ockham has some wise words for moments like this.
> I was about to say don't needlessly multiply spellings, but my
dictionary
> agrees with you that Ockham is an acceptable variant of Occam. I find
> this deliciously ironic.
It's a town name after all, and the town's name was and is Ockham.
http://uk2.multimap.com/map/browse.cgi?X=505000&Y=155000&width=500&height=30
0&client=public&gride=&gridn=&srec=0&coordsys=gb&addr1=&addr2=&addr3=&pc=&sc
a
le=100000&advanced=&multimap.x=338&multimap.y=92
Some francoiles prefered a more italic rendering, but others,
such as the IEP (which I think contains the single most detailed biogray
of him that I've seen anywhere), list only Ockham.
il
--
Unpatched IE vulnerability: Basic Authentication URL spoofing
Description: Spoofing the URL displayed in the Address bar
Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html
===
Subject: Re: Question on generation of large prime numbers
> Well, Richard and il are both wrong here. The proof could be expressed
> in terms of a finite number of known primes, as il seems to have
assumed
> it was, but that's not the way Richard expressed it -- he spoke
explicitly
> of the product of *all* primes.
Yes.
> But Richard's conclusion that P has been shown not to be the largest
> prime number is also wrong; what has been shown is that the assumption
> of a finite number of primes is wrong. The only reason to assume that
> at least one of the additional primes discovered must be larger than
> P is that we *thought* P was the largest prime, before we discovered
them.
Hmmm. I seem to have taken a step for granted. Okay, let me make it more
explicit. We have shown that the number we thought was the largest prime
cannot be the largest prime, so there must be a bigger one. So now we find
out what that new largest prime is, and then iterate. We can do this
infinitely often (if we have the time to spare).
Is that a bit closer to the mark?
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
> Hmmm. I seem to have taken a step for granted. Okay, let me make it more
> explicit. We have shown that the number we thought was the largest prime
> cannot be the largest prime,
Yes, but not directly. You have shown that the assumption that there were
be the largest prime.
> so there must be a bigger one.
No. If you had proved *only* that the number P is not in fact the largest
prime, this would *not* prove that a larger prime exists. It would also
be possible that P is not prime after all, or that *no* largest prime
exists.
Now in fact we know that primes exist, and we have proved that there are
many, and since we're dealing with integers, from *this* it follows that
there exists a prime larger than any given number P.
In other words, you've taken a valid proof and made it invalid by falsely
inserting, as an intermediate step, a fact that you actually only know
to be true because of the final conclusion.
--
Mark Brader, Toronto Logic is logic. That's all I say.
msb@vex.net -- Oliver Wendell Holmes
===
Subject: Re: Question on generation of large prime numbers
>> Hmmm. I seem to have taken a step for granted. Okay, let me make it more
>> explicit. We have shown that the number we thought was the largest prime
>> cannot be the largest prime,
> Yes, but not directly. You have shown that the assumption that there
were
> be the largest prime.
>> so there must be a bigger one.
> No. If you had proved *only* that the number P is not in fact the
largest
> prime, this would *not* prove that a larger prime exists. It would also
> be possible that P is not prime after all, or that *no* largest prime
> exists.
But I defined P to be the largest prime. Therefore, it /is/ prime, and
therefore /either/ no largest prime exists /or/ there is a prime larger
than P, which is what I was obviously failing to say.
> Now in fact we know that primes exist, and we have proved that there are
> many, and since we're dealing with integers, from *this* it follows that
> there exists a prime larger than any given number P.
> In other words, you've taken a valid proof and made it invalid by falsely
> inserting, as an intermediate step, a fact that you actually only know
> to be true because of the final conclusion.
A mathematician cut out to be I was not.
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
> Hmmm. I seem to have taken a step for granted. Okay, let me make it
more
> explicit. We have shown that the number we thought was the largest
prime
> cannot be the largest prime,
>> Yes, but not directly. You have shown that the assumption that there
were
>> be the largest prime.
> so there must be a bigger one.
>> No. If you had proved *only* that the number P is not in fact the
largest
>> prime, this would *not* prove that a larger prime exists. It would also
>> be possible that P is not prime after all, or that *no* largest prime
>> exists.
> But I defined P to be the largest prime. Therefore, it /is/ prime, and
> therefore /either/ no largest prime exists /or/ there is a prime larger
> than P, which is what I was obviously failing to say.
Summary of Euclid's proof:
1) Suppose that there is a largest prime; call it P
2) Calculate N = (product of all primes from 2 to P, inclusive) + 1
3) N has one or more prime factors, all of which must be > P
4) #3 contradicts #1, so #1 is wrong, so there is no largest prime
You were trying to conclude that, for any prime P, N is a larger
prime. Not so. There is *some* prime larger than P, but it's not
necessarily N; it could just be one of N's factors.
Consider: P = 13; N = (2*3*5*7*11*13) + 1 = 30031
30031 is not a prime > 13, but its factors (59 and 509) are.
===
Subject: Re: Question on generation of large prime numbers
prime.
No, I wasn't. I was saying that it might be a larger prime, but that it
might be a composite of primes at least one of which is larger than P.
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
> Summary of Euclid's proof:
> 1) Suppose that there is a largest prime; call it P
> 2) Calculate N = (product of all primes from 2 to P, inclusive) + 1
Are you trying to say that the set of all primes has no primes missing?
That is vacuously true, and holds for {2,3,7}. If {2,3,7} is the set
of all primes, which we are permitted to assume according to Euclid,
then {2,3,7} is the set of all primes, bar none.
Or are you trying to say and we assume that we know all primes <=P.
> 3) N has one or more prime factors, all of which must be > P
> 4) #3 contradicts #1, so #1 is wrong, so there is no largest prime
Or the assumption in #2 was wrong.
I never thought that Euclid's proof would be hard for people to grasp.
I think I can only recommend that people look at Kummer's proof instead,
as that's even simpler.
il
--
Unpatched IE vulnerability: DNSError folder disclosure
Description: Gaining access to local security zones
Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html
===
Subject: Re: Question on generation of large prime numbers
Correct me if I'm wrong, but isn't the proof much simpler if you say
something like:
1) Assume P is the largest prime.
2) Calculate P!+1 (I.E. the product of all numbers from 1 to P, plus one)
3) Dividing that number by any number <= P will give a remainder of 1
4) Therefore, P!+1 is either prime or a multiple of a prime > P
SaSW, Willem (at stack dot nl)
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
===
Subject: Re: Question on generation of large prime numbers
> Correct me if I'm wrong, but isn't the proof much simpler if you say
> something like:
> 1) Assume P is the largest prime.
> 2) Calculate P!+1 (I.E. the product of all numbers from 1 to P, plus
one)
> 3) Dividing that number by any number <= P will give a remainder of 1
> 4) Therefore, P!+1 is either prime or a multiple of a prime > P
That's fine. (Pedantically you could change the conclusion to either prime
or a multiple of primes >P, but what you've said is correct, and
sufficient
for the proof.)
il
--
Unpatched IE vulnerability: Basic Authentication URL spoofing
Description: Spoofing the URL displayed in the Address bar
Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html
===
Subject: Re: Question on generation of large prime numbers
[Disclaimer: il Carmody is a damn sight better at mathematics than I am.
But I think I got a case. ]
>> I have looking at a few web pages dealing with the largest known
>> calculated primes and a great deal of computational time is taking
>> into searching for these numbers and verifying they are primes. I have
>> seen the Euclid's proof of infinitude primes and it occurs that me
>> that super-large prime numbers can be calculated using the following:
>> Others have already given you an example of why your idea won't
fly, but
>> I don't want you to get the idea that Euclid is broken. His proof that
>> there is no largest prime (and therefore an infinite number of primes)
is
>> as follows:
>> If there is a largest prime, P, then the number of primes is finite. We
>> can therefore form a new number, N, which is one greater than the
product
>> of all primes. Therefore, N > P (and, indeed, it would be very much
>> greater than P).
>> When divided by any prime in our list, it leaves remainder 1. Therefore,
>> /either/ it is prime /or/ it is the product of two or more primes, at
>> least one of which is greater than P.
>> In /either/ case, P has been shown not to be the largest prime number.
> Nope. Nowhere have you assumed that the list of primes known is
> complete up to P.
On the contrary, I formed N by multiplying all primes and then adding 1. To
multiply all primes together, I must first have a complete list of primes.
If the list is not complete, then I cannot perform the multiplication.
the primes are, you can't suddenly pull otherwise well-known
> features of the prime numbers out of a hat.
It's not necessary to know all the primes in order to construct the
argument. It is sufficient to know that it is /in theory/ possible to know
them, e.g. by sieving. It is only necessary to know all the primes if you
wish to perform the multiplication in real life.
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
[...]
|>> If there is a largest prime, P, then the number of primes is finite. We
|>> can therefore form a new number, N, which is one greater than the
product
|>> of all primes. Therefore, N > P (and, indeed, it would be very much
|>> greater than P).
|>>
|>> When divided by any prime in our list, it leaves remainder 1.
Therefore,
|>> /either/ it is prime /or/ it is the product of two or more primes, at
|>> least one of which is greater than P.
|>> In /either/ case, P has been shown not to be the largest prime number.
|> Nope.
Yes it has.
|> Nowhere have you assumed that the list of primes known is
|> complete up to P.
He doesn't need to assume that all the primes up to P are known for the
argument to work. He just assumes (for the sake of proof by contradiction)
that P is the largest prime.
|On the contrary, I formed N by multiplying all primes and then adding 1. To
|multiply all primes together, I must first have a complete list of primes.
|If the list is not complete, then I cannot perform the multiplication.
There are two versions of the proof. One proceeds like you just did,
starting with a (false) assumption that there is a largest prime. The
other just starts by considering an arbitrary finite set of primes. If I
remember correctly, the latter is more like Euclid's own proof. The crux is
that the N you get by adding one to the product is always divisible by a
prime not in the set. It's perhaps worth noting that you're not really
using the assumption that P is the largest prime, then, until at the end
when you contradict it.
proofs as if it was a proof by contradiction, by starting it with an
assumption that the conclusion of the proposition was false, and deducing
a contradiction in each case where the result was shown true. This is a
stylistic weakness to avoid.
In this case starting with the assumption that P is the largest prime does
not seem so bad; at least it seems more popular. It doesn't really play a
functional role, though, and it's perhaps worth doing it the other way to
make it more obvious that we're seeing as well a method for finding a prime
outside of any finite set of primes.
Keith Ramsay
===
Subject: Re: Question on generation of large prime numbers
> [Disclaimer: il Carmody is a damn sight better at mathematics than I am.
> But I think I got a case. ]
I hope you've got a hot-shot lawyer...
Your premise is:
>> If there is a largest prime, P,
and that's all, as can be deduced from the then and therefores
hereafter.
>> then the number of primes is finite. We
>> can therefore form a new number, N, which is one greater than the
product
>> of all primes. Therefore, N > P (and, indeed, it would be very much
>> greater than P).
>>
>> When divided by any prime in our list, it leaves remainder 1.
Therefore,
>> /either/ it is prime /or/ it is the product of two or more primes, at
>> least one of which is greater than P.
>> In /either/ case, P has been shown not to be the largest prime number.
>
> Nope. Nowhere have you assumed that the list of primes known is
> complete up to P.
> On the contrary, I formed N by multiplying all primes
Erm, says who? You've not defined all in this context. Nowhere do you
say anything about having a contiguous set of primes from 2 to P.
The only thing you've said is that your set of primes has a maximum
element P. Full stop. There were no further premises -- look above if
you don't believe me.
Anyway, your maximum prime P is a red herring and completely
unnecessary - why did you even introduce it in the first place?
> and then adding 1. To
> multiply all primes together, I must first have a complete list of
primes.
You must have a list of primes. What does 'complete' mean to you in
this context? If you wish to multiply together all primes between 2
and P then you must know /a priori/ all the primes between 2 and P.
How do you know that you know all those primes? If you wish to
use that as an assumption your argument you must state so. You didn't.
But why would you want to?
Let the complete list of primes be {2}.
The multiplication/addition yields 3. The factorisation oracle tells
you that that's prime.
Try again - let the complete list of primes be {2,3}
The multiplication/addition yeilds 7. The factorisation oracle tells
you that that's prime.
Try again - let the complete list of primes be {2,3,7}
Do you permit me to do the above?
Are you going to claim that I must insert 5 into that set?
If so, then why? Would Euclid insist that I inserted 5? What would be
the justification?
If not, then your argument seems to be too transient and fickle to
be able to be countered rationally.
> If the list is not complete, then I cannot perform the multiplication.
Erm??? If the assumed finite set of primes is {2,3,7}, then the
multiplication is trivial, and yields the new prime 43.
Can you not multiply 2, 3, and 7 together? Completeness is not
a clearly defined in this context. At this point we _don't know_
that 5 is prime. Unless your book of axioms says oh, you may assume
5 is prime whenever you like, but _my_ book of axioms differs
from yours in that regard.
The same for _any_ finite set of primes. Euclid's proof simply
requires that one assume there's a finite set of primes, it does
_not_ ask one to assume that this set has any other properties.
Why do you wish to give it extra unnecesary properties? In
particular, properties that rely on an external or prior prime
generation?
As I said before, and gave an example which you for some reason
snipped, if you use the Euclid method as a constructive prime
generator you very quickly end up generating primes smaller than
the previously known maximum. (On the assumption that this
generator is the only generator you have.)
> Don't forget that if you're pretending that you don't know what
> the primes are, you can't suddenly pull otherwise well-known
> features of the prime numbers out of a hat.
> It's not necessary to know all the primes in order to construct the
> argument. It is sufficient to know that it is /in theory/ possible to know
> them, e.g. by sieving. It is only necessary to know all the primes if you
> wish to perform the multiplication in real life.
That makes it sound like you've missed the entire point of Euclid's proof.
Euclid's proof shows that you never need to to _any multiplications at
all_ in real life.
As an _existance_ proof, Euclid's algorithm is non-constructive.
If you're trying to use it as a _constructive_ algorithm, why the
heck do you want/need to invoke a second constructive algorithm
(the one that you used to generate all primes up to P)?
il
--
Unpatched IE vulnerability: file-protocol proxy
Description: cross-domain scripting, cookie/data/identity
theft, command execution
Reference:
http://safecenter.net/liudieyu/WsOpenFileJPU/WsOpenFileJPU-Content.HTM
Exploit:
http://safecenter.net/liudieyu/WsOpenFileJPU/WsOpenFileJPU-MyPage.HTM
===
Subject: Re: Question on generation of large prime numbers
There appears to be some incomprehensible logic in this thread.
No-one is claiming that Euclid's method can generate all primes. But
it appears that someone IS saying that Richard is claiming this. He's
not.
Let us all agree that if we are ever to multiply all primes up to P,
we need a way of generating all of them, and knowing that they really
are all prime, and that there are none omitted. I'm sure even Euclid
himself would agree with this - so would any rules-lawyer.
So let's start. We'll keep it simple with digits we can all manage,
and work our way up.
A. Euclid says there is no largest prime, because for any given
prime P, a larger one Q can be found. (Well, not exactly that, but I
don't speak Greek, and certainly not the Greek of Euclid's time.)
B. It's not hard to see that once Q has been found it can be treated
as though it were P, and its own Q found, and the new Q can again be
treated as though it were P, and a further Q found, and so on. So if
the first Q is found, we know there will be an infinite chain of
greater and greater prime numbers.
C. Euclid then defines a number N, which is one more than the
product of all primes up to and including P. Note that Euclid never
requres the value of N to be known.
D. He states that either N (which is larger than P) is prime, or
that it has at least one prime factor M which is larger than P. Note
that Euclid never requires the value of M to be known.
E. His reasoning is that N can not be divisible by any prime number
up to and including P, as they all give a remainder of 1.
F. If N is composite, its prime factors are greater than P. There
could be just one prime factor M, in which case N is a square number.
More likely, there are two or more prime factors. Any one of these
factors will do for M, and thus for Q. Because M is larger than P, Q
is larger than P.
G. If N is not composite, it's prime (as it's greater than 1), so N
will do for Q. Because N is larger than P, Q is larger than P.
H. Because neither the value of N nor M need be known, the value of
Q need not be known, but we do know that is is prime and that it is
larger than P.
If Euclid's method is true, is should work for ANY prime number - even
small ones. So let us start at P = 5. Note, we could have started
anywhere. 5 is small enough that calculations are easy, without being
trivial.
Using a sieve of Eratosthenes, we know the primes up to and including
5 are {2,3,5}. The number N Euclid's method calculates is 31. Using
a seive, we know 31 is prime, so Euclid is correct.
Let's work with the next highest prime after 5, i.e. P = 7. N = 211,
which is prime.
Next, P = 11. N = 2311, which is prime.
Next, P = 13. N = 30031, which is NOT prime as it is 59 * 509. But
Euclid claims that either N is prime OR it has at least one prime
factor M greater than P. Either 59 or 509 will do for M. They are
both prime, and both greater than P (13).
Next, P = 17. N = 223092871, which is NOT prime as it is 19 * 97 *
277. M could be any of the three factors.
So lets's check (adding in the smaller primes):
P N N is prime? M exits? Q? Q > P?
2 3 true false Q = N = 3 true
3 7 true false Q = N = 7 true
5 31 true false Q = N = 31 true
7 211 true false Q = N = 211 true
11 2311 true false Q = N = 2311 true
13 30031 false true (59) Q = M = 59 true
17 510511 false true (19) Q = M = 19 true
So what happens when we reach the last KNOWN value of P where we also
KNOW all previous values. For any higher values of P (such as a known
mersenne prime) we will not be able to calculate N.
Well, it doesn't matter, as Euclid never required us to know all
primes smaller than P. It is sufficient to know than N is not evenly
divisible by any prime smaller than P or by P itself, simply because
of the way N is calculated (see paragras E and C).
We can then follow through paragras F, G and H to know that Euclid
was correct.
The red herring that il threw in about how do we get 5 into the
list of prime numbers is nonsense. No-one ever claimed the only way
to get the list of prime numbers was to generate N from each P by
substituting the previous N as the current P.
il has his P - (N => P) - (N => P) - (N => P) chain {2,3,7,31}.
Sure that chain doesn't include 5. Nor does it include 11, 13, 17,
19, 23 or 29. Nor 37, 41, 43 and a whole lot more. And il seems to
think that 5 therefore needs special dispensation to be allowed as a
prime number.
But il: What justification do you have for starting at 2? Why is
2 a prime number, and 5 not?
ObPuzzle: The definition of a prime number is a counting number that
has exactly two factors - 1 and itself (which is why 1 is considered
as not being prime [which is odd, given the derivation of the word
prime] - neither is it composite, as those numbers have a factor
other than 1 or itself). If we glibly change the definition from
counting number to integer, how many prime numbers are there?
What are they (if any)?
===
Subject: Re: Question on generation of large prime numbers
>> [Disclaimer: il Carmody is a damn sight better at mathematics than I
>> [am.
>> But I think I got a case. ]
> I hope you've got a hot-shot lawyer...
He might end up as just a shot lawyer at this rate...
> Your premise is:
> If there is a largest prime, P,
> and that's all, as can be deduced from the then and therefores
> hereafter.
Right. And the point of the premise (and the argument) is to take us to
reductio ad absurdum - i.e. to show that there is no largest prime.
>> Nope. Nowhere have you assumed that the list of primes known is
>> complete up to P.
>> On the contrary, I formed N by multiplying all primes
> Erm, says who? You've not defined all in this context.
Haven't I? Okay, guilty, your honour. all means every prime between 2
and
P, including 2 and P.
> Anyway, your maximum prime P is a red herring and completely
> unnecessary - why did you even introduce it in the first place?
I was merely explaining to the OP that Euclid's proof of the infinitude of
primes is no use in attempting to construct a prime.
>> and then adding 1. To
>> multiply all primes together, I must first have a complete list of
>> primes.
> You must have a list of primes. What does 'complete' mean to you in
> this context? If you wish to multiply together all primes between 2
> and P then you must know /a priori/ all the primes between 2 and P.
Well, I can get that list by starting at 2, using your rather handy
factorisation oracle, and iterating up to and including P.
> As I said before, and gave an example which you for some reason
> snipped,
Brevity, my dear chap. Mere brevity.
> if you use the Euclid method as a constructive prime
> generator you very quickly end up generating primes smaller than
> the previously known maximum. (On the assumption that this
> generator is the only generator you have.)
I wasn't trying to construct primes. I was trying to demonstrate why you
can't construct them in that way.
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
Richard Heathfield and il Carmody write:
>> If there is a largest prime, P,
> and that's all, as can be deduced from the then and therefores
hereafter.
>> then the number of primes is finite. We
>> can therefore form a new number, N, which is one greater than the
product
>> of all primes. ...
> Nope. Nowhere have you assumed that the list of primes known is
> complete up to P.
>
> On the contrary, I formed N by multiplying all primes
> Erm, says who? You've not defined all in this context. Nowhere do you
> say anything about having a contiguous set of primes from 2 to P.
il, stop blathering about defining 'all'. You misread or mis-
interpreted Richard's wording, and you've been caught.
> The only thing you've said is that your set of primes has a maximum
> element P. Full stop. There were no further premises -- look above if
> you don't believe me.
We're dealing with positive integers. If there is a largest one in a
set of them, then that set is finite, as Richard said, and if it's
finite, then a product of all its members exists. It is not necessary
to specify an algorithm to construct the set (although in this case it
would be simple to do so), nor to define 'all' in this context.
--
Mark Brader, Toronto | Professor, I think I have a counterexample.
msb@vex.net | That's all right; I have two proofs.
===
Subject: Re: Question on generation of large prime numbers
> Richard Heathfield and il Carmody write:
>> If there is a largest prime, P,
>
> and that's all, as can be deduced from the then and therefores
hereafter.
>
>> then the number of primes is finite. We
>> can therefore form a new number, N, which is one greater than the
product
>> of all primes. ...
> Nope. Nowhere have you assumed that the list of primes known is
> complete up to P.
>
> On the contrary, I formed N by multiplying all primes
>
> Erm, says who? You've not defined all in this context. Nowhere do
you
> say anything about having a contiguous set of primes from 2 to P.
> il, stop blathering about defining 'all'. You misread or mis-
> interpreted Richard's wording, and you've been caught.
I did not. He introuced concepts which were unnecessary, and I claim
harmful.
> The only thing you've said is that your set of primes has a maximum
> element P. Full stop. There were no further premises -- look above if
> you don't believe me.
> We're dealing with positive integers. If there is a largest one in a
> set of them, then that set is finite, as Richard said, and if it's
> finite, then a product of all its members exists.
Since when has that been at issue? I've even said as much myself.
You must have misread.
> It is not necessary
> to specify an algorithm to construct the set (although in this case it
> would be simple to do so), nor to define 'all' in this context.
Nor do I say it is. You must have misread.
il
--
Unpatched IE vulnerability: Click hijacking
Description: Pointing IE mouse events at non-IE/system windows
Reference:
http://safecenter.net/liudieyu/HijackClick/HijackClick-Content.HTM
Exploit: http://safecenter.net/liudieyu/HijackClick/HijackClick2-MyPage.HTM
===
Subject: Re: Question on generation of large prime numbers
> It's not necessary to know all the primes in order to construct the
> argument. It is sufficient to know that it is /in theory/ possible to know
> them, e.g. by sieving. It is only necessary to know all the primes if you
> wish to perform the multiplication in real life.
So, since you can't sieve all of them there is a (practical) largest prime!
===
Subject: Re: Question on generation of large prime numbers
> It's not necessary to know all the primes in order to construct the
> argument. It is sufficient to know that it is /in theory/ possible to
know
> them, e.g. by sieving. It is only necessary to know all the primes if
you
> wish to perform the multiplication in real life.
> So, since you can't sieve all of them there is a (practical) largest
prime!
But the (practical) largest prime is time dependant -- as time
passes and better computers and programs come into existence, the
largest know prime keeps getting larger.
With, as yet, no end in sight.
===
Subject: Re: Question on generation of large prime numbers
>> It's not necessary to know all the primes in order to construct the
>> argument. It is sufficient to know that it is /in theory/ possible to
>> know them, e.g. by sieving. It is only necessary to know all the primes
>> if you wish to perform the multiplication in real life.
> So, since you can't sieve all of them there is a (practical) largest
> prime!
Great! Please share...
--
Richard Heathfield : binary@eton.powernet.co.uk
Usenet is a strange place. - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
===
Subject: Re: Question on generation of large prime numbers
> Great! Please share...
10^80 - 3?
===
Subject: Re: Question on generation of large prime numbers
> I have looking at a few web pages dealing with the largest known
> calculated primes and a great deal of computational time is taking
> into searching for these numbers and verifying they are primes. I have
> seen the Euclid's proof of infinitude primes and it occurs that me
> that super-large prime numbers can be calculated using the following:
> p1=2 < p2=3
> A large prime number, p, can be generated using
> p = (p1 * p2) + 1 = 7
> p1=2 < p2=3 < p3=5
> p = (p1*p2*p3) + 1 = 31
> p1=2 < p2=3 < p3=5 < p4=7
> p = (p1*p2*p3*p4) + 1 = 211
> p1=2 < p2=3 < p3=5 ..... < pn=...
> p = (p1*p2*p3*p4*....*pn) + 1 = ....
> It seems to me that using the above method, super large prime numbers
> exceeding currently known largest primes can be generated rather
> quickly. For example, we calculate the first million primes, multiply
> them and just add one.
> If this is true, then calculating super-large primes should be a
> computational trivial task, shouldn't it? Then why the big fuss over
> calculating large primes?
> I suspect that I am missing fundamental here. Can super large primes
> really be calculated using the trivial method outlined above? Hope
Is 30031 a prime?
===
Subject: Re: Question on generation of large prime numbers
In sci.math, Christian Bau
I have looking at a few web pages
dealing with the largest known
>> calculated primes and a great deal of computational time is taking
>> into searching for these numbers and verifying they are primes. I have
>> seen the Euclid's proof of infinitude primes and it occurs that me
>> that super-large prime numbers can be calculated using the following:
>> p1=2 < p2=3
>> A large prime number, p, can be generated using
>> p = (p1 * p2) + 1 = 7
>> p1=2 < p2=3 < p3=5
>> p = (p1*p2*p3) + 1 = 31
>> p1=2 < p2=3 < p3=5 < p4=7
>> p = (p1*p2*p3*p4) + 1 = 211
>> p1=2 < p2=3 < p3=5 ..... < pn=...
>> p = (p1*p2*p3*p4*....*pn) + 1 = ....
>> It seems to me that using the above method, super large prime numbers
>> exceeding currently known largest primes can be generated rather
>> quickly. For example, we calculate the first million primes, multiply
>> them and just add one.
>> If this is true, then calculating super-large primes should be a
>> computational trivial task, shouldn't it? Then why the big fuss over
>> calculating large primes?
>> I suspect that I am missing fundamental here. Can super large primes
>> really be calculated using the trivial method outlined above? Hope
> Is 30031 a prime?
30031 = 2*3*5*7*11*13 + 1 = 59 * 509
Regardless of whether the number is prime or not, one always gets
a prime not in the original set, though.
As it is, RSA's algorithm requires generation of large
primes and, in order to crack it, the factorization
of a product of two large primes (the product, not the
primes, is given as part of the public key). Large prime
generation isn't all that difficult unless one gets stuck
in a desert (pick a random odd number N of the requisite
size; if it's prime, stop; otherwise add 2 to N and loop).
A known desert is just after a factorial; it's obvious
that none of N! + 2, N! + 3, ..., N! + N can be prime.
Or one can take the lcm of the numbers 1 through N, for
the same effect.
Another interesting generator is x^2 + x + 41. The first
time this fails is x = 40, as one can verify using a
simple program.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Question on generation of large prime numbers
> [...]
> p1=2 < p2=3
> A large prime number, p, can be generated using
> p = (p1 * p2) + 1 = 7
> p1=2 < p2=3 < p3=5
> p = (p1*p2*p3) + 1 = 31
> p1=2 < p2=3 < p3=5 < p4=7
> p = (p1*p2*p3*p4) + 1 = 211
> p1=2 < p2=3 < p3=5 ..... < pn=...
> p = (p1*p2*p3*p4*....*pn) + 1 = ....
> It seems to me that using the above method, super large prime numbers
> exceeding currently known largest primes can be generated rather
> quickly. For example, we calculate the first million primes, multiply
> them and just add one.
> If this is true, then calculating super-large primes should be a
> computational trivial task, shouldn't it? Then why the big fuss over
> calculating large primes?
There are at least three problems in relying on this construction
to produce primes:
Firstly in the main real-life application of large prime numbers,
namely cryptogray, the primes must be hard (in practice impossible)
to guess, and for this they must be taken from a large/random 'pool'
of possibilities. Relying on a simple recursive formula would mean
your decrypter would find it trivial to run through all practical
possibilities.
That brings us to the next problem - Values produced by the above
method very soon become impractically large, although you could
generalize it by spreading the factors between two factors, i.e.
p_{n+1} = p_1.p_3.p_4...p_m + p_2.p_5...p_n. This also increases
the pool of values, which somewhat ameliorates the first problem.
But I've saved the worst snag until last - Values obtained by your
construction, and the two-product extension of it, are not always
prime!
---------------------------------------------------------------------------
John R Ramsden (jr@adslate.com)
---------------------------------------------------------------------------
Eternity is a long time, especially towards the end.
Woody Allen
===
Subject: Re: Question on generation of large prime numbers
> it occurs that me
> that super-large prime numbers can be calculated using the following:
> p1=2 < p2=3 < p3=5 ..... < pn=...
> p = (p1*p2*p3*p4*....*pn) + 1 = ....
Alas, it usually doesn't work. 2*3*5*7*11*13+1 = 30031 = 59*509:
not prime; See
http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha102.htm
http://www.research.att.com/projects/OEIS?Anum=A014545
Subtracting one at the end is also a good try, and usually fails to
produce a prime.
--
Don Reble djr@nk.ca
===
Subject: Question on generation of large prime numbers
I have looking at a few web pages dealing with the largest known
calculated primes and a great deal of computational time is taking
into searching for these numbers and verifying they are primes. I have
seen the Euclid's proof of infinitude primes and it occurs that me
that super-large prime numbers can be calculated using the following:
p1=2 < p2=3
A large prime number, p, can be generated using
p = (p1 * p2) + 1 = 7
p1=2 < p2=3 < p3=5
p = (p1*p2*p3) + 1 = 31
p1=2 < p2=3 < p3=5 < p4=7
p = (p1*p2*p3*p4) + 1 = 211
p1=2 < p2=3 < p3=5 ..... < pn=...
p = (p1*p2*p3*p4*....*pn) + 1 = ....
It seems to me that using the above method, super large prime numbers
exceeding currently known largest primes can be generated rather
quickly. For example, we calculate the first million primes, multiply
them and just add one.
If this is true, then calculating super-large primes should be a
computational trivial task, shouldn't it? Then why the big fuss over
calculating large primes?
I suspect that I am missing fundamental here. Can super large primes
really be calculated using the trivial method outlined above? Hope
BlueBear
===
Subject: Re: Question on generation of large prime numbers
> I have looking at a few web pages dealing with the largest known
> calculated primes and a great deal of computational time is taking
> into searching for these numbers and verifying they are primes. I have
> seen the Euclid's proof of infinitude primes and it occurs that me
> that super-large prime numbers can be calculated using the following:
> p1=2 < p2=3
> A large prime number, p, can be generated using
> p = (p1 * p2) + 1 = 7
> p1=2 < p2=3 < p3=5
> p = (p1*p2*p3) + 1 = 31
> p1=2 < p2=3 < p3=5 < p4=7
> p = (p1*p2*p3*p4) + 1 = 211
> p1=2 < p2=3 < p3=5 ..... < pn=...
> p = (p1*p2*p3*p4*....*pn) + 1 = ....
> It seems to me that using the above method, super large prime numbers
> exceeding currently known largest primes can be generated rather
> quickly. For example, we calculate the first million primes, multiply
> them and just add one.
It seems that you haven't really understood Euclid's proof then. The
new number constructed does not need to be prime; the point is that it
is itself a product of primes and all of its factors must be new
primes. 2*3*5*7*11*13+1=59*509, as was mentioned earlier on this
thread. Also, 2*3*5*7-1=11*19. In either case, 59 and 509 are not in
{2,3,5,7,11,13} and 11 and 19 are not in {2,3,5,7}. So long as your
set of primes is finite, you can extend it. Therefore the full set,
which is not extensible by definition, must be infinite.
> If this is true, then calculating super-large primes should be a
> computational trivial task, shouldn't it? Then why the big fuss over
> calculating large primes?
But it's not trivial. See my examples, which are very standard and
well-known.
> I suspect that I am missing fundamental here. Can super large primes
> really be calculated using the trivial method outlined above? Hope
You're welcome.
> BlueBear
----
===
Subject: Re: Question on generation of large prime numbers
> I have looking at a few web pages dealing with the largest known
> calculated primes and a great deal of computational time is taking
> into searching for these numbers and verifying they are primes. I have
> seen the Euclid's proof of infinitude primes and it occurs that me
> that super-large prime numbers can be calculated using the following:
> p1=2 < p2=3
> A large prime number, p, can be generated using
> p = (p1 * p2) + 1 = 7
> p1=2 < p2=3 < p3=5
> p = (p1*p2*p3) + 1 = 31
> p1=2 < p2=3 < p3=5 < p4=7
> p = (p1*p2*p3*p4) + 1 = 211
> p1=2 < p2=3 < p3=5 ..... < pn=...
> p = (p1*p2*p3*p4*....*pn) + 1 = ....
> It seems to me that using the above method, super large prime numbers
> exceeding currently known largest primes can be generated rather
> quickly. For example, we calculate the first million primes, multiply
> them and just add one.
> If this is true, then calculating super-large primes should be a
> computational trivial task, shouldn't it? Then why the big fuss over
> calculating large primes?
> I suspect that I am missing fundamental here. Can super large primes
> really be calculated using the trivial method outlined above? Hope
> BlueBear
2*3*5*7*11*13 + 1 = 30031 = 59*509
===
Subject: Re: Limit at infinity
This is just to thank you for your help!
Paul
> I would like to know whether it is possible to define the limit of a
> function f as x tends to +oo, where f is defined on R^+, except on an
> infinite countable number of points of R^+ (for instance, except on
> the set of integers).
> or for instance, something like gamma(-x) ...
> I think you can safely use the standard limit definitions:
> For all P>0, there is a Q such that for all x of dom(f):
> x > Q ==> f(x) < -P
> For all e>0, there is a Q such that for all x of dom(f):
> x > Q ==> |f(x)-b| < e
> For all P>0, there is a Q such that for all x of dom(f):
> x > Q ==> f(x) > P
> I think these definitions work fine for the functions you have in mind.
> Of course the gamma(-x) would have no limit for x --> +oo
> Dirk Vdm
===
Subject: Semi-normal numbers Was: if x has normally distributed digits, does
1/x?
>The set of non-normal numbers has Lebesgue measure zero.
>If you pick a real uniformly at random from the interval [0, 1]
>then with probability 1 you have picked a normal number.
>This leads many to take the position that if some number pops up
>somewhere & there's no good reason to think it's not normal - some
>number like pi, or like the reciprocal of the number you get when
>you replace every other digit in the decimal expansion of pi with
>a zero - then it's safe to assume that the number is normal.
I started thinking about this and the same result could possibly be
extended to tell something about the reciprocals of normal numbers.
Please inform if I've made a logical error with the definition of
almost all.
---
Definition:
A normal number whose reciprocal is non-normal is called a seminormal
number. A number whose reciprocal is normal is called an inverse-
normal number.
Theorem:
Almost all real numbers are inverse-normal.
Proof:
Assume two subsets of R called R_1 and R_2 so that for each r in
R{0}, r is either in R_1 or R_2 and its reciprocal 1/r is in the
other one. This divides the reals to two subsets and creates the
bijection:
T: R_1->R_2, T(r) = 1/r
between them. Now, assume the set of all seminormal numbers in R_1
called S_1. We call the image of S_1:
U_1 = {T(s) | s in S_1}
By definition the set U_1 contains only non-normal numbers. Since
almost all reals are normal, we know that almost all numbers in R_2
don't belong in U_1.
Since there is a bijection between S_1 and U_1, we know that almost
all numbers in R_1 don't belong in S_1 either, so we have proved that
almost all numbers in R_1 aren't semi-normal. We now use the same
argument to obtain a similar result for R_2.
Since almost all numbers in R_1 union R_2 = R{0} aren't non-normal
either, the only option is that almost all numbers in R are in fact
inverse-normal.
===
Subject: Re: Semi-normal numbers Was: if x has normally distributed digits,
does 1/x?
> [...]
>Definition:
>A normal number whose reciprocal is non-normal is called a seminormal
>number. A number whose reciprocal is normal is called an inverse-
>normal number.
>Theorem:
>Almost all real numbers are inverse-normal.
>Proof: [omitted]
The reciprocal is a measure automorism on the nonzero reals. Given
that almost all reals are normal, doesn't the conclusion follow
immediately? One might need to invoke absolute continuity of measures.
--
Steen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: My research, publication announcement
>Peer review limitations:
>--reviewer subjectivity and agenda
>--constraints of scientific dogma and orthodoxy
>--by definition, *peer* reviewers are usually limited to the same
>paradigm and ilosoical traditions
> False. Counter-example: me.
I'm sure you're in a class quite by yourself.
===
Subject: Re: Please check my boolean simplification - NEW VERSION!
> To draw truth tables and calculate prime implicants,
> feel free to download from
> http://users.pandora.be/vdmoortel/dirk/Boole/boole.html
> (carful: you'll have to replace C0 with C and C1 with D)
> This is really a very useful program.
I have just uploaded a new version (1.1) of the program.
It had a little bug. Please replace.
Dirk Vdm
===
Subject: Re: Please check my boolean simplification
> in the following boolean equation simplification:
> R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
The best and my 2nd 'final' version, has only four disjunctions.
I doubt there's anything shorter.
R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
= (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B')
(C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B')
= A + C1'B' + BC1
(C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1
= C0' + A' + B'C1
R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
= (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B')
= (A' + C0' + B'C1)(A + C1'B' + BC1)
= A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A
= A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1
+ AC0'C1'B' + AC0'BC1 + C0'A + B'C1A
= A'C1'B' + A'BC1 + C0'A + B'C1A
= A'(B'C1' + BC1) + A(C0' + B'C1)
----
===
Subject: Re: Please check my boolean simplification
> in the following boolean equation simplification:
> R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
> The best and my 2nd 'final' version, has only four disjunctions.
> I doubt there's anything shorter.
> R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
> = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B')
> (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B')
> = A + C1'B' + BC1
> (C1'+C0'+A'+B')(C1+C0'+A') = C0' + A' + (C1' + B)C1
> = C0' + A' + B'C1
> R = (C1'+C0'+A'+B')(C1'+A+B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B')
> = (C1'+C0'+A'+B')(C1'+A+B)(C1+C0'+A')(C1+A+B')
> = (A' + C0' + B'C1)(A + C1'B' + BC1)
> = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A
> = A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1
> + AC0'C1'B' + AC0'BC1 + C0'A + B'C1A
> = A'C1'B' + A'BC1 + C0'A + B'C1A
> = A'(B'C1' + BC1) + A(C0' + B'C1)
Neat.
Verifying this, it helped me spot a little bug in my program.
Version 1.1 put in in place.
Dirk Vdm
===
Subject: Which series do these constants come from?
I've been disassembling an FPU emulator and as expected, it
contained a significant number of constants. A fair amount of
these are very obvious, but a quite a few are complete mysteries
to me. For all but three I figured out the (likely) relationship
between the successive terms, but as for what they are al used?
I hope someone in this group can tell me or point me to a site
that has information about mathematical series.
The constants are:
-0.005208333333333333423683514 -1/192
+0.000048828124999993646575269 1/20480 *320/3
-0.000000544956752147722229412 -1/1835008 *448/5
+0.000000006622737896066696605 1/150994944 *576/7
-0.000000000084664710789036923 -1/11811160064 *704/9
+0.000000000001118158950846881 1/big *832/11
-0.000000000000014377930269982 -1/bigger *960/13
+0.12451171875
+0.498751787292275571965858085
+0.000262287674813145674657804
+0.249050582915605093375140816
+0.498101165831210186750281632
+0.002034561246815280125422448
+0.249977256935346610686307297
+0.482158790777797965568110885
+0.798637441128639956279457873
+0.250000000000000000162630325 1/4
+0.041666666666666666576316485 1/24 /6
+0.005208333333333335212617098 1/192 /8
+0.000520833333333347420733561 1/1920 /10
+0.000043402777777744446993657 1/23040 /12
+0.000003100198412569815306638 1/322560 /14
+0.000000193762401041715562333 1/5160960 /16
+0.000000010764578285932073498 1/92897280 /18
+0.000000000538227910807861920 1/1857945600 /20
+0.000000000024464188541715309 1/408748032000 /22
+0.000000000001021270795795864 1/9809952768000 /24
+0.000000000000039701152521748 1/255058771968000 /26
+0.020833333333333334345255360 1/48 *48/1
+0.000781250000000021185310450 1/1280 *80/3
+0.000034877232141512120363463 1/28672 *112/5
+0.000001695421027976551104493 1/589824 *144/7
+0.000000086697509034958444851 1/11534336 *176/9
+0.000000004585603636012795109 1/218103808 *208/11
+0.000000000246930795093098370 1/4026531840 *240/13
+0.000000000015363636849275580 1/73014444032 *272/15
Robert
--
Robert AH Prins
prino@bigfoot.com
===
Subject: Re: Which series do these constants come from?
Robert AH Prins contained a significant number of constants. A fair amount of
> these are very obvious, but a quite a few are complete mysteries
> to me. For all but three I figured out the (likely) relationship
> between the successive terms, but as for what they are al used?
> I hope someone in this group can tell me or point me to a site
> that has information about mathematical series.
I don't have anything specific but maybe search on inverse symbolic
calculator?
> The constants are:
... [snipped, see OP]
There is this pattern:
48 = 2^4 *3
1280 = 2^8 *5
28672 = 2^12 *7
589824 = 2^16 *9
11534336 = 2^20 *11
218103808 = 2^24 *13
4026531840 = 2^28 *15
73014444032 = 2^32 *17
No doubt the software is calculating a logarithm, because these numbers are
the inverses of the coefficients of x^n in the power series for ln(1+4x).
Similarly (with alternating signs)
192 = 2^6 *3
20480 = 2^12 *5
1835008 = 2^18 *7
150994944 = 2^24 *9
11811160064 = 2^36 *11
big = 2^42 *13 (I presume)
bigger = 2^48 *15
and we are looking at a power series for ln(1+6x).
4 = 2 *2!
24 = 2^2 *3!
192 = 2^3 *4!
1920 = 2^4 * 5!
23040 = 2^5 *6!
322560 = 2^6 *7!
etc, and it's calculating an exponential.
The isolated constants in the list aren't so easy :)
===
Subject: Re: Which series do these constants come from?
>Robert AH Prins I've been disassembling an FPU
emulator and as expected, it
>> contained a significant number of constants. A fair amount of
>> these are very obvious, but a quite a few are complete mysteries
>> to me. For all but three I figured out the (likely) relationship
>> between the successive terms, but as for what they are al used?
>> I hope someone in this group can tell me or point me to a site
>> that has information about mathematical series.
>I don't have anything specific but maybe search on inverse symbolic
>calculator?
>> The constants are:
>... [snipped, see OP]
>There is this pattern:
>48 = 2^4 *3
>1280 = 2^8 *5
>28672 = 2^12 *7
>589824 = 2^16 *9
>11534336 = 2^20 *11
>218103808 = 2^24 *13
>4026531840 = 2^28 *15
>73014444032 = 2^32 *17
>No doubt the software is calculating a logarithm, because these numbers
are
>the inverses of the coefficients of x^n in the power series for ln(1+4x).
>Similarly (with alternating signs)
>192 = 2^6 *3
>20480 = 2^12 *5
>1835008 = 2^18 *7
>150994944 = 2^24 *9
>11811160064 = 2^36 *11
>big = 2^42 *13 (I presume)
>bigger = 2^48 *15
>and we are looking at a power series for ln(1+6x).
>4 = 2 *2!
>24 = 2^2 *3!
>192 = 2^3 *4!
>1920 = 2^4 * 5!
>23040 = 2^5 *6!
>322560 = 2^6 *7!
>etc, and it's calculating an exponential.
>The isolated constants in the list aren't so easy :)
a debugger is very difficult, as the emulated FPU stack and
debugger stack occupy the same storage...)
I guess it's time to trace my Carmichael & Smith and see if that
gives me any clues as to how the above series are used to calculate
logarithms or powers.
Robert
--
Robert AH Prins
prino@bigfoot.com
===
Subject: probability .........
let~ rate 0.002 of decayed tooth patients are a cancer of mouth patient.
Until first cancer of mouth patient have discovered in any dentist's
consultation room, we diagnose.
find the expected value whether we treat and you must diagnose the couple
of
patients.
---------------------------
um......i think.....answer is reciprocal of 1/500
it is right??
===
Subject: probability 2......
we cuts a two place into a wire at random.
let length of wire is 1
We got the wire of a three piece.
when we made a triangle from three wire piece,
find that probability of possibility.
----------------------------------
very difficult......um....help....me please.......
===
Subject: Re: probability 2......
> we cuts a two place into a wire at random.
> let length of wire is 1
> We got the wire of a three piece.
> when we made a triangle from three wire piece,
> find that probability of possibility.
> ----------------------------------
> very difficult......um....help....me please.......
Put the wire on a number line with left end-point at 0 and right
end-point at 1. Let c and d be the points at which the wire is
cut such that 0 < c < d < 1. Now use the fact that in a triangle
the sum of the lengths of any two sides must be greater than the
third. For example, we must have c + (d - c) > 1 - d. When
you get the appropriate inequalities between c and d, plot them
in the Cartesian plane with horizontal axes labelled c and d.
Then find the area of the solution region. This will be your
probability.
===
Subject: Re: probability 2......
> we cuts a two place into a wire at random.
> let length of wire is 1
> We got the wire of a three piece.
> when we made a triangle from three wire piece,
> find that probability of possibility.
> ----------------------------------
> very difficult......um....help....me please.......
Let say that lenghts of pieces are A,B,C. To form triangle pieces must
satisfy:
(A+B)>C
(A+C)>B
(B+C)>A
These conditions have another geometric interpretation:
A B C
0----------X-------------Y----------1
Let points X,Y be random numbers (let say that they are places where we
cut
original wire to make three pieces)
Then we may restate conditions as:
(x+y-x)>1-y <=> 2y>1 <=> y>0.5
(x+1-y)>y-x <=>2x-2y+1>0 <=> x-y+0.5>0 <=> y>x+0.5
(y-x+1-y)>x <=> 1>2x <=>x<0.5
also note that 0<=x<=1, 0<=y<=1.
Draw these in x-y plane - the area of part of square ( 0<=x<=1, 0<=y<=1)
in
which all conditions are satisfied represents your probability.
Goran
===
Subject: Re: probability 2......
thank...you....very much....
i think......if we only use 0-x-y-1
in this case, probability is 1/8
but, if we use 0-x-y-1, 0-y-x-1
in this case, probability is 1/4
which of case is right??
advice....please....sir~
===
Subject: Re: probability 2......
> thank...you....very much....
> i think......if we only use 0-x-y-1
> in this case, probability is 1/8
> but, if we use 0-x-y-1, 0-y-x-1
> in this case, probability is 1/4
> which of case is right??
> advice....please....sir~
1/4
===
Subject: Re: probability 2......
> thank...you....very much....
> i think......if we only use 0-x-y-1
> in this case, probability is 1/8
> but, if we use 0-x-y-1, 0-y-x-1
> in this case, probability is 1/4
> which of case is right??
> advice....please....sir~
> 1/4
Problem may be restated - if we have two independent uniformly distributed
variables on [0,1] variables x,y representing distances from one end of
wire, where we cut the wire - what is probabilities that pieces may form
triangle? If we consider only 0-x-y-1 then we have additional constraint
that y>x and space of all possible events are not represented by square but
upper left triangle. If we allow ywe cuts a two place into a wire at random.
>let length of wire is 1
>We got the wire of a three piece.
>when we made a triangle from three wire piece,
>find that probability of possibility.
I believe what you're saying is that the three pieces must be able to
form a genuine triangle.
WLOG, assume that the pieces are cut in increasing order of size, so the
length of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], which
also must be larger than Sqrt[a^2+b^2].
Proceed in this fashion.
Doug
===
Subject: Re: probability 2......
>WLOG, assume that the pieces are cut in increasing order of size, so the
>length of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], which
>also must be larger than Sqrt[a^2+b^2].
Yeah, I messed the details on this one, as I've found by playing around
with it some more. It's always a treat to post upon waking. :-P
Doug
===
Subject: Dream problem
This night my dream was crucially based on the
fact that sec(2x)=(sec(x)+sec(4x))/2.
Uhm, I get the nagging feeling that in that state
I'm just as lousy as a mathematician as awake
Do me a favor and compute the x for which that
equation is accidentally right before someone
gets hurt...
--
Hauke Reddmann <:-EX8
Private email:fc3a501@math.uni-hamburg.de
For our chemistry workgroup,remove math from the address
For spamming, remove anything else
===
Subject: Re: Dream problem
> This night my dream was crucially based on the
> fact that sec(2x)=(sec(x)+sec(4x))/2.
> Uhm, I get the nagging feeling that in that state
> I'm just as lousy as a mathematician as awake
> Do me a favor and compute the x for which that
> equation is accidentally right before someone
> gets hurt...
Perhaps you can plot the left function and the right function and see for
yourself before you go off and hurt someone?
===
Subject: Re: Dream problem
In sci.math, flip
<1068995840.212611@news-1.nethere.net>:
>> This night my dream was crucially based on the
>> fact that sec(2x)=(sec(x)+sec(4x))/2.
>> Uhm, I get the nagging feeling that in that state
>> I'm just as lousy as a mathematician as awake
>> Do me a favor and compute the x for which that
>> equation is accidentally right before someone
>> gets hurt...
> Perhaps you can plot the left function and the right function and see for
> yourself before you go off and hurt someone?
There are also multiple solutions -- although one of them
should jump right out and bite the OP (were the value to
have any teeth, that is -- and this particular numeral
obviously hasn't hatched any yet ).
Consider that 1 = (1 + 1)/2. Now when does sec(2x) = 1?
This hint should make solving the problem a dream...
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Dream problem
> There are also multiple solutions -- although one of them
> should jump right out and bite the OP (were the value to
> have any teeth, that is -- and this particular numeral
> obviously hasn't hatched any yet ).
Just saw that my handy plotting software (MathGV) has
sec(x) built-in. Looks like an EPR spectrum on crack
--
Hauke Reddmann <:-EX8
Private email:fc3a501@math.uni-hamburg.de
For our chemistry workgroup,remove math from the address
For spamming, remove anything else
===
Subject: Re: Dream problem
In sci.math, Hauke Reddmann
:
>> There are also multiple solutions -- although one of them
>> should jump right out and bite the OP (were the value to
>> have any teeth, that is -- and this particular numeral
>> obviously hasn't hatched any yet ).
> Just saw that my handy plotting software (MathGV) has
> sec(x) built-in. Looks like an EPR spectrum on crack
Or an EPR spectrum of someone *on* crack...
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: conjugate subgroups
> Why can't a conjugate of a subgroup be a proper subset of it?
See http://mathworld.wolfram.com/ConjugateSubgroup.html
--
Bruce Harvey
bruce@bearsoft.co.uk
The Alternative ysics Site
http://users.powernet.co.uk/bearsoft
===
Subject: Re: conjugate subgroups
>> Why can't a conjugate of a subgroup be a proper subset of it?
> See http://mathworld.wolfram.com/ConjugateSubgroup.html
Surely the question doesn't deserve an URL?
--
===
Subject: Re: conjugate subgroups
>> Why can't a conjugate of a subgroup be a proper subset of it?
> See http://mathworld.wolfram.com/ConjugateSubgroup.html
> Surely the question doesn't deserve an URL?
It doesn't have one - at any rate, if it does, that's not it.
That URL just defines conjugacy & links to some other stuff,
it doesn't address the original question.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: A potentially rewarding challenge
>For any given natural number m, the Goodstein sequence, G(m), only
>terminates if m < 4.
> G(4) terminates, so the argument is invalid. I don't have a formal
> proof, but just look at the following and it should be obvious (unless
> I made a mistake somewhere):
Bravo, indeed - I could not spot any mistake in the reasoning. If I
had bothered to apply myself to the tedious drudgery of specifically
working out the steps you did - instead of attempting to visualise
them generally - I would have, of course, held back archiving my
argument.
is not as obvious as I thought it might be); secondly, since I didn't
- fortunately - hold back my argument, for helping me bring into
better focus the issue that triggered my investigation:
As I see it, your reasoning that we can see that G(4) terminates
should, surely, be capable of expression as a formal proof sequence in
PA. Now, assuming the standard interpretation of Goodstein's argument
is valid, then, for any given natural number m, we should also,
reasonably, be able to see, similarly, that G(m) must terminate -
necessarily in 0. Thus, for any given m, we should also be able to
mirror this vision, without appeal to transfinite induction, as a
formal proof sequence in PA.
In other words, assuming that Goodstein's theorem cannot be formally
proved in PA, there should, reasonably, be some constructive
meta-proof - in Goedel's sense, i.e., one that does not appeal to
transfinite concepts - to the effect that, for any given natural
number m, there is some proof sequence in PA that implies that [G(m)]
terminates for the numeral [m].
Bhup
===
Subject: Re: A potentially rewarding challenge
> For any given natural number m, the Goodstein sequence, G(m),
> only terminates if m < 4.
Others in the thread have shown your claim is false; however,
for more detail about the structure of these sequences, you
might want to see the text file at
http://r.s.home.mindspring.com/GoodsteinSequences
which shows, incidentally, that the Goodstein sequence that
starts at 4 has (two) maximum terms equal to N = (j+1)*2^j-1,
where j=3*2^27-1, and terminates at 0 on the (2N)th term.
--r.e.s.
===
Subject: Re: A potentially rewarding challenge
> [...] for more detail about the structure of these
> sequences, you might want to see the text file at
> http://r.s.home.mindspring.com/GoodsteinSequences
> [...]
r.e.s.
sequences is, indeed, illuminating, and the graical analysis is
fascinating. I shall certainly study them carefully.
Bhup
===
Subject: Re: A potentially rewarding challenge
>
> [...] for more detail about the structure of these
> sequences, you might want to see the text file at
> http://r.s.home.mindspring.com/GoodsteinSequences
> [...]
> r.e.s.
> sequences is, indeed, illuminating, and the graical analysis is
> fascinating. I shall certainly study them carefully.
One correction to my post: The gra of the Goodstein sequence
is piecewise linear, having a plateau consisting of (N+1)/2 terms
-- not 2, of course. Rather, it's the 2nd to last, and 2nd longest
linear segment (relative to the number of terms). All the terms in
this plateau are equal to N = (j+1)*2^j-1, j=3*2^27-1.
--r.e.s.
===
Subject: Re: A potentially rewarding challenge
In response to self...
> (ii) Express m by its hereditary representation in base 2;
...
> For instance, the Goodstein sequence G(4) is given by:
> G(1, 4) = 1*(2^2) + 0*(2^1) + 0*(2^0) (base 2)
> = 4
> G(2, 4) = 1*(3^3) + 0*(3^1) + 0*(3^0) - 1
Because in hereditory representation 2 is 2^1, and therefore
bumping the base turns it into 3^1.
Sorry for any confusion.
il
--
Unpatched IE vulnerability: history.back method caching
Description: cross-domain scripting, cookie/data/identity
theft, command execution
Reference: http://safecenter.net/liudieyu/RefBack/RefBack-Content.HTM
Exploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM
===
Subject: Re: five vectors
> Hallo,
> Ich have given 5 vectors and have to find a
> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
> vector through this base.
Look up Gram-Schmidt orthogonalization.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: five vectors
> Hallo,
> Ich have given 5 vectors and have to find a
> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
> vector through this base.
> Look up Gram-Schmidt orthogonalization.
complicated
for me. Can you give me at least one really short example for the
application of this method? Please!
Karl.
[P.S. To tell the truth, I don't even understand how this method could be
helpful
for me. (:-{ ]
===
Subject: Re: five vectors
> Hallo,
> Ich have given 5 vectors and have to find a
> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
> vector through this base.
>
>
> The vectors are:
> ^^^^^^^^^^^^^^^^^^
> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);
> v3 = (0,1,3,0); v4 = (2,-1,4,-7)
> v5 = (5,-8,1,2)
> Look up Gram-Schmidt orthogonalization.
> complicated
> for me. Can you give me at least one really short example for the
> application of this method? Please!
> Karl.
> [P.S. To tell the truth, I don't even understand how this method could be
> helpful
> for me. (:-{ ]
Using Gram-Schmidt here is rather like using a sledge hammer to
crack eggs. It is way too much work for the result desired.
A much better approach is to try to determine whether each vector,
in (v1,v2,v3,v4,v5) is a linear combination of the previous ones.
This can be done pretty much by eyeball analysis.
One finds that v1 is trivially independent,
v2 is not dependent on (v1),
v3 is depentent on (v1,v2), v3 = 2*v1 - 1*v2
v4 is not dependent on (v1,v2) so not on (v1,v2,v3) either, and
v5 is dependent on (v1,v2,v3,v4), v5 = v1 + v2 + v4
Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed
as linear conbinations of basis vectors and v3 and v5 can be
expressed less trivially as linear combinations of these basis
vectors.
===
Subject: Re: five vectors
> Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed
> as linear conbinations of basis vectors and v3 and v5 can be
> expressed less trivially as linear combinations of these basis
> vectors.
Hi Virgil,
But I'm working in a R^4 room. So how can a 4D-room have a
3D-basis? Or did I misunderstand something?
===
Subject: Re: five vectors
> Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed
> as linear conbinations of basis vectors and v3 and v5 can be
> expressed less trivially as linear combinations of these basis
> vectors.
> Hi Virgil,
> But I'm working in a R^4 room. So how can a 4D-room have a
> 3D-basis? Or did I misunderstand something?
Did you mean that you want4ed a basis of the entire space, rather
than a basis of the space spanned by the 5 vectors given? If so,
that meaning was not clear to me. And such a basis is impossible
from just the 5 vectors given.
If you did want a basis of the entire space, what is wrong with
b1 = (1,0,0,0), b2 = (0,1,0,0), b3 = (0,0,1,0), b4 = (0,0,0,1)
as such a basis?
===
Subject: Re: five vectors
>> Hallo,
>> Ich have given 5 vectors and have to find a
>> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
>> vector through this base.
>> Look up Gram-Schmidt orthogonalization.
> complicated
> for me. Can you give me at least one really short example for the
> application of this method? Please!
> Karl.
> [P.S. To tell the truth, I don't even understand how this method could be
> helpful
> for me. (:-{ ]
You are given five vectors in R^4. The Gram-Schmidt algorithm allows you
to find a minimal set of vectors that spans the same subspace as the
given vectors. If your vectors happen to span the entire space, then
Gram-Schmidt gives you a basis for that space.
As a bonus, the vectors you get from Gram-Schmidt are orthonormal, which
makes it very easy to decompose any given vector as a linear combination
of the basis vectors.
The important thing is not to get bogged down in computation until you
understand what the method is doing.
The first vector is very easy. If the vector v1 is nonzero, then {v1} is
a linearly independent set, and therefore all you need to do is
normalize. Take u1 = v1 / ||v1||. That is, divide the first vector by
its Euclidean norm, and call that u1, which is a unit vector pointing in
the same direction as v1.
Next consider your second vector, v2, and its projection along u1, which
is (v2 . u1) u1. If v2 happens to be a scalar multiple of u1, this dot
product will just give you v2. If it's not a scalar multiple, then the
dot product gives you the component of v2 in the direction of u1. You
need to subtract this from v2 in order to get the orthogonal complement,
which is
w2 = v2 - (v2 . u1) u1
and then normalize, obtaining
u2 = w2 / ||w2||.
Next consider v3 and its components in the directions of u1 and u2.
Since we now have two vectors in our orthonormal set, the orthogonal
complement is
w3 = v3 - (v3 . u1) u1 - (v3 . u2) u2.
If w3 is zero, it means v3 is already in the subspace spanned by u1 and
u2, and therefore you can ignore it and move on to the next vector. If
w3 is nonzero, then you normalize it and add it to the basis you are
building:
u3 = w3 / ||w3||
and so on.
If it turns out that v5 is a linear combination of the other vectors that
you started with, then it will reduce to a linear combination in terms of
your orthonormal basis, and you can easily compute its components. Then
just substitute what each of the u_i's is in terms of the v_j's, and
you're done.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: five vectors
> Hallo,
> Ich have given 5 vectors and have to find a
> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
> vector through this base.
> Look up Gram-Schmidt orthogonalization.
> complicated
> for me. Can you give me at least one really short example for the
> application of this method? Please!
> Karl.
Take a look at
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: five vectors
You are confusing vectors with the algebra of real numbers!!!!!
Vectors are mathematical representations of real ysical entities. A
problem is meaningless unless you define the nature of the vectors.
--
Bruce Harvey
bruce@bearsoft.co.uk
The Alternative ysics Site
http://users.powernet.co.uk/bearsoft
> Hallo,
> Ich have given 5 vectors and have to find a
> base for V = L(v1, v2, v3, v4, v5). Then I must express the fifth
> vector through this base.
> I think the general approach is to try every possible linear
> combination:
> 1.) a*v1+b*v2+c*v3+d*v4 = v5
> or
> 2.) a*v1+b*v2+c*v3+d*v5 = v4
> or
> 3.) a*v1+b*v2+c*v5+d*v4 = v3
> or
> 4.) a*v1+b*v5+c*v3+d*v4 = v2
> or
> 5.) a*v5+b*v2+c*v3+d*v4 = v1
> The vectors are:
> ^^^^^^^^^^^^^^^^^^
> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);
> v3 = (0,1,3,0); v4 = (2,-1,4,-7)
> v5 = (5,-8,1,2)
> I got the following results:
> 1.) a + 2c = 1 and b - c = 1 and d = 1
> 2.) a + 2b = -3 and c - b = 1 and d = 1
> 3.) a - d = 2 and b - d = -1 and c + d = 0
> 4.) a - 3d = 2 and b + d = 0 and c + d = -1
> 5.) a + d = 0 and 2b - 3d - 1 = 0 and 2c - d - 1 = 0
> I have expected to get simple numerical values for a,b,c
> and d (like: a = 2,...) but not this. Can you tell me what my
> mistake was?
> Karl.
===
Subject: Real torsion (0,1) matrices
If A is a real nXn (0,1) matrix such that A^m = I (the identity
matrix) for some m, must A be a permutation matrix ?
===
Subject: Re: Real torsion (0,1) matrices
> If A is a real nXn (0,1) matrix such that A^m = I (the identity
> matrix) for some m, must A be a permutation matrix ?
Yes. AB = I where B = A^{m-1} has nonnegative integer entries.
are nonnegative integer linear combinations of the rows of A --
this is only possible in trivial ways.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Algebra question
Can someone explain to me what's the difference between
basis transformation and a similarity transformation.
I'm really confused ...
===
Subject: Re: Algebra question
> Can someone explain to me what's the difference between
> basis transformation and a similarity transformation.
A similarity transform of A is any transformation of the
form inv(S)*A*S with nonsingular S.
According to my Horn and Johnson, every invertible
matrix is a change-of-basis matrix, and every change-of-
basis matrix is invertible. Thus, every similarity
transform is a change of basis transform for the
same matrix.
- Randy
===
Subject: Re: Algebra question
Arnold
> Can someone explain to me what's the difference between
> basis transformation and a similarity transformation.
No doubt basis transformation refers to a square matrix which describes
a
change of basis. For any two given bases
x={x1,...,xn}
y={y1,...,yn}
the change of basis from x to y is described by a unique such matrix M.
Perforce M has a nonzero determinant and is invertible.
The only meaning of similarity transformation that I know of is this
one
from geometry: a mapping f of a Euclidean space into itself such that
[f(x),f(y)]=k[x,y] for any x,y and some real constant k>0, where [ , ]
refers to the distance between two points. If this doesn't seem to make
sense, maybe you could tell us more about the context in which you came
across the term.
LH
===
Subject: Integral of a function over [a,b]
Hello
Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be defined
in such a way that, at every irrational x in [a,b], f is continuous
and f(x) = g(x). Then, is it true that the Riemann integral of f over
[a,b] equals the integral of g? Since the set of discontinuities of f
on [a,b] is at most countable and, therefore, has measure 0, it
follows f is certainly integrable over [a,b], but I couldn't come to a
conclusion if its integral needs to equal the integral of g or if it
depends on how we define f for the rationals of [a,b].
I know that if f is Thomae's function, then the answer is yes and the
inegralas vanish, but in this case we have f(x) =0 for every
irrational, which seems to imply a loss of generality when compared to
the function f defined above.
In case the answer is yes, does it remain yes if we replace the
assumption that g is continuous by the weaker one that g is only
Riemann integrable over [a,b]?
Now, suppose in the questions above we replace the set Q cap [a,b], of
the rationals in [a,b], by an uncountable set E with measure zero?
Then, is there anything interesting we can affirm about the integral
of f?
Amanda
===
Subject: Re: Integral of a function over [a,b]
> Hello
> Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be defined
> in such a way that, at every irrational x in [a,b], f is continuous
> and f(x) = g(x). Then, is it true that the Riemann integral of f over
> [a,b] equals the integral of g? Since the set of discontinuities of f
> on [a,b] is at most countable and, therefore, has measure 0, it
> follows f is certainly integrable over [a,b], but I couldn't come to a
> conclusion if its integral needs to equal the integral of g or if it
> depends on how we define f for the rationals of [a,b].
In fact, your argument to prove that f is integrable over [a,b] is not
correct. Indeed, in order that a function defined on [a,b] is integrable
(in the sense of Riemann), the set of discontinuities must have measure
0 *and* f must be bounded. However, if f is supposed to be bounded, your
argument is correct.
On the other hand, the integrals of f and g must be equal (still
assuming that f is bounded, of course), since the function f - g is
integrable and takes the value 0 at every irrational form [a,b].
> In case the answer is yes, does it remain yes if we replace the
> assumption that g is continuous by the weaker one that g is only
> Riemann integrable over [a,b]?
Sure. I did not use the assumption g is continuous.
> Now, suppose in the questions above we replace the set Q cap [a,b], of
> the rationals in [a,b], by an uncountable set E with measure zero?
> Then, is there anything interesting we can affirm about the integral
> of f?
I don't think so.
Jose Carlos Santos
===
Subject: Re: Integral of a function over [a,b]
> Hello
> Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be defined
> in such a way that, at every irrational x in [a,b], f is continuous
> and f(x) = g(x). Then, is it true that the Riemann integral of f over
> [a,b] equals the integral of g? Since the set of discontinuities of f
> on [a,b] is at most countable and, therefore, has measure 0, it
> follows f is certainly integrable over [a,b], but I couldn't come to a
> conclusion if its integral needs to equal the integral of g or if it
> depends on how we define f for the rationals of [a,b].
> In fact, your argument to prove that f is integrable over [a,b] is not
> correct. Indeed, in order that a function defined on [a,b] is integrable
> (in the sense of Riemann), the set of discontinuities must have measure
> 0 *and* f must be bounded. However, if f is supposed to be bounded, your
> argument is correct.
Yes, sure. I forgot to say f was bounded.
> On the other hand, the integrals of f and g must be equal (still
> assuming that f is bounded, of course), since the function f - g is
> integrable and takes the value 0 at every irrational form [a,b].
How can we prove this?
Amanda
===
Subject: Re: Integral of a function over [a,b]
> On the other hand, the integrals of f and g must be equal (still
> assuming that f is bounded, of course), since the function f - g is
> integrable and takes the value 0 at every irrational form [a,b].
> How can we prove this?ffff
Thm: If f is Riemann integrable on [a,b] and = 0 a.e., then int_[a,b] f(x)
dx = 0.
Proof sketch: Let eps > 0. Then there is an open set U in [a,b] such that
m([a,b] U) = 0 and such that |f| < eps everywhere on U. Now [a,b] U is
compact, and so can be covered with finitely many disjoint open intervals,
the sum of whose lengths is < eps. The endpoints of these intervals form a
partition P of [a,b] such that the upper Riemann sum of |f| < eps*(b-a) +
eps*M (or something like that).
===
Subject: Permutation: how to detach cycles/transpositions
Let's have a permutation:
(3 5 1 2 4 6)
After detaching into cycles we have:
(1 3)(2 5 4)(6)
The same with transpositions:
(1 2 3 4) = (1 4)(1 3)(1 2)
My question is how to translate these algorithms into any
computer-understandable code. It's easy to transform it on a sheet of
paper,
but implementation seems a little bit harder for me.
--
Paweü StobiÄski
Republic of Poland
===
Subject: Re: Permutation: how to detach cycles/transpositions
def cycles(permutation):
n=len(permutation)
accountedFor=[False]*n
returnValue=[]
while True:
i=0
while i Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened. :)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Permutation: how to detach cycles/transpositions
>[something not in english!]
>You know, whenever one posts source code in a NG that is not
>comp.lang.thatlanguage or a subhierarchy of it, it would be a Very
>Nice Thing(TM) to specify in which language it is written!
It's Python. You can tell because it looks like pseudocode
but it's not.
You know, nobody ever complains about not recognizing C.
Things are gonna be different after the revolution...
>Michele
>--
>> Comments should say _why_ something is being done.
>Oh? My comments always say what _really_ should have happened. :)
>- Tore Aursand on comp.lang.perl.misc
C. Ullrich
===
Subject: Re: Permutation: how to detach cycles/transpositions
U[YAcute]ytkownik Elaine Jackson
napisaü w
> HTH
transpositions disjoining (c++):
Transposition: a class, with some methods and fields ( int* for permuation)
void Perm::detachTranspositions()
{
if (isIdent())
{
cout<0)
{
if (tmp.p[pos-1] > tmp.p[pos]) {
swap(tmp.p[pos-1],tmp.p[pos]);
X1[--index] = pos-1;
X2[index] = pos;
}
}
} while(x != tmp.getPos(x)); }
for(x=inv-1; x>=0; x--)
cout<<[< Let's have a permutation:
> (3 5 1 2 4 6)
> After detaching into cycles we have:
> (1 3)(2 5 4)(6)
I don't understand... Your cycle is already decomposed into a product of
disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).
--
Maxi
===
Subject: Re: Permutation: how to detach cycles/transpositions
> I don't understand... Your cycle is already decomposed into a product of
> disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).
By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case the
decomposition holds.
--
@+ sur le forum fran.8dais !!
===
Subject: Re: Permutation: how to detach cycles/transpositions
U[YAcute]ytkownik Julien Santini
napisaü
w wiadomo.beci
> I don't understand... Your cycle is already decomposed into a product
of
> disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).
> By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case the
> decomposition holds.
Right, sorry for lack of any description to my question.
--
Paweü StobiÄski
Republic of Poland
===
Subject: Re: diff EQ on strings, check out the math
type=text/plain;
boundary=----=_NextPart_000_001A_01C3AC3C.A7D70E10
---------------------------------------------------------------------
> Check my math!
> I've derived a differential equation for strings starting from Stokes
> Theorem to show that energy is conserved along a world-sheet. These diff
> eq's involve connection coefficients. And I'm not really sure what it all
> means yet. I would appreciate it if some who are more skilled in the art
> would take a look at this and comment.
> The math can be found at:
> http://www.sirus.com/users/mjake/diffeq.html
See link in original post.
What does it mean that ? Does this mean that the surface is a geodesic?
And what does it mean that ? Does this mean that there is no force in the
direction of time? If F is still an arbitrary field, then does this mean
that the string must be traveling in a frame where the time component is
zero?
===
Subject: Accumulation pts. in R and C ?
Hi all,
I just started Complex variables and applications by Churchill, et al.
I
am flummoxed by something already. In Churchill, they say that an
accumulation point is
...a point z0 ... of a set S if each neighborhood if z0 contains at least
one point of S distinct from z0. Why is it different from R? To wit,
every neighborhood of x0 has an infinite number of points of, say, E.
TIA,
Lurch
===
Subject: Re: Accumulation pts. in R and C ?
They are equivilent. Let me see if I can explain it on the internet. If
this helps let me know
you can just think of R for simplicity, although I don't think my argument
will use this fact.
let's say we want to check if x0 is an accumulation point of some set S.
I want to show that:
Every open neighborhood contains at least one point of S distinct from x0
is equivelent to
Every open neighborhood contains infinitely many points of S
Suppose every open neighborhood contains at least one point of S distinct
from x0. Fix an open neighborhood. (just pick any one) Pick one of the
points distinct from x0, call it x1. (again, pick one). Now, clearly
there
exists an open neighborhood of x0 that doesn't contain x1. For example, if
|x0 - x1| = a, then take the a/2 neighborhood of x0 (this is just saying
that x1 is some distance away from x0 since it is a different point, so if
we take a neighborhood that only goes out half as far, x1 is not in that
neighborhood).
Now, this smaller neighborhood must contain a point of S. call it x2. now
take an even smaller neighborhood that doesn't contain x2. this even
smaller neighborhood doesn't contain x1 or x2, but it does contain some
point of S distinct from x0. So we can keep repeating this forever. In
other words, for any n, however large n is, there is some x_n in S.
I hope that made sense. To recap, if you keep taking smaller
neighborhoods,
you keep finding points. So if every neighborhood contains one point,
every
neighborhood must contain infinitely many points, since each neighborhood
also contains every smaller neighborhood.
This might not be very rigorous, but hopefully it will get you started.
Obviously, if every neighborhood contains infinitely many points, it
contains at least one point.
Justin Van Winkle
Suppose
----- Original Message -----
===
Subject: Accumulation pts. in R and C ?
> Hi all,
> I just started Complex variables and applications by Churchill, et
al.
I
> am flummoxed by something already. In Churchill, they say that an
> accumulation point is
> ...a point z0 ... of a set S if each neighborhood if z0 contains at
least
> one point of S distinct from z0. Why is it different from R? To wit,
> every neighborhood of x0 has an infinite number of points of, say, E.
> TIA,
> Lurch
===
Subject: Re: Accumulation pts. in R and C ?
Lurch
> They are equivilent. Let me see if I can explain it on the internet. If
> this helps let me know
> you can just think of R for simplicity, although I don't think my
argument
> will use this fact.
> let's say we want to check if x0 is an accumulation point of some set S.
> I want to show that:
> Every open neighborhood contains at least one point of S distinct from x0
> is equivelent to
> Every open neighborhood contains infinitely many points of S
> Suppose every open neighborhood contains at least one point of S distinct
> from x0. Fix an open neighborhood. (just pick any one) Pick one of the
> points distinct from x0, call it x1. (again, pick one). Now, clearly
there
> exists an open neighborhood of x0 that doesn't contain x1. For example,
if
> |x0 - x1| = a, then take the a/2 neighborhood of x0 (this is just saying
> that x1 is some distance away from x0 since it is a different point, so
if
> we take a neighborhood that only goes out half as far, x1 is not in that
> neighborhood).
> Now, this smaller neighborhood must contain a point of S. call it x2.
now
> take an even smaller neighborhood that doesn't contain x2. this even
> smaller neighborhood doesn't contain x1 or x2, but it does contain some
> point of S distinct from x0. So we can keep repeating this forever. In
> other words, for any n, however large n is, there is some x_n in S.
> I hope that made sense. To recap, if you keep taking smaller
neighborhoods,
> you keep finding points. So if every neighborhood contains one point,
every
> neighborhood must contain infinitely many points, since each neighborhood
> also contains every smaller neighborhood.
> This might not be very rigorous, but hopefully it will get you started.
> Obviously, if every neighborhood contains infinitely many points, it
> contains at least one point.
> Justin Van Winkle
> Suppose
> ----- Original Message -----
===
> Subject: Accumulation pts. in R and C ?
> Hi all,
> I just started Complex variables and applications by Churchill, et
al.
> am flummoxed by something already. In Churchill, they say that an
> accumulation point is
> ...a point z0 ... of a set S if each neighborhood if z0 contains at
least
> one point of S distinct from z0. Why is it different from R? To
wit,
> every neighborhood of x0 has an infinite number of points of, say, E.
> TIA,
> Lurch
===
Subject: Re: Accumulation pts. in R and C ?
Charlie Johnson
> I just started Complex variables and applications by Churchill, et
al.
I
> am flummoxed by something already. In Churchill, they say that an
> accumulation point is
> ...a point z0 ... of a set S if each neighborhood if z0 contains at
least
> one point of S distinct from z0. Why is it different from R? To wit,
> every neighborhood of x0 has an infinite number of points of, say, E.
It is equivalent.
It is always equivalent in a metric space.
--
Maxi
===
Subject: Re: Accumulation pts. in R and C ?
How is it equivalent? If one has the usual metric in R and C, then
according to the books I am reading, they are different. C requiring only
one point and R requiring infinite amount of points. (The metric in C is
abs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)
Lurch
> Charlie Johnson
> I just started Complex variables and applications by Churchill, et
al.
> am flummoxed by something already. In Churchill, they say that an
> accumulation point is
> ...a point z0 ... of a set S if each neighborhood if z0 contains at
least
> one point of S distinct from z0. Why is it different from R? To
wit,
> every neighborhood of x0 has an infinite number of points of, say, E.
> It is equivalent.
> It is always equivalent in a metric space.
> --
> Maxi
===
Subject: Re: Accumulation pts. in R and C ?
> How is it equivalent? If one has the usual metric in R and C, then
> according to the books I am reading, they are different. C requiring
only
> one point and R requiring infinite amount of points. (The metric in C is
> abs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)
If every neighbourhood of z0 contains one point of E distict from z0, then
by considering the open balls centered in z0 of radius 1/n you see that
there is infinitely many (otherwise there would be none in some of these
disks for n sufficiently large).
--
Maxi
===
Subject: Re: Accumulation pts. in R and C ?
I don't quite get it yet, but I will work on it.
Lurch
> How is it equivalent? If one has the usual metric in R and C, then
> according to the books I am reading, they are different. C requiring
only
> one point and R requiring infinite amount of points. (The metric in C
is
> abs(z-z0) < e and the metric in R being the usual: abs(x - x0) < e.)
> If every neighbourhood of z0 contains one point of E distict from z0,
then
> by considering the open balls centered in z0 of radius 1/n you see that
> there is infinitely many (otherwise there would be none in some of these
> disks for n sufficiently large).
> --
> Maxi
===
Subject: Relational algebra help
Im having trouble understanding some questions. Some help would be
appreciated. I have read the book, but dont follow it.
Problem 1
F={B->A,A->C}
1. the non-trivial dependencies are:
B->A, A->C, B->C
is that right? I was wondering if it was a trick question.
2. Find a non-empty instance of R that satisfied every FD in F, bot
not A->B
any idea what that question MEANS. Does it want a real example? Im
confused.
3. Find an instance of R that satisfies every FD in F, bot not A-> B.
again does this mean an example? Im confused.
===
Subject: chess bishops - combinations
I understand that it is possible to express the number of ways of
having k bishops on an n*n board such that they are not attacking each
other as a combinatorial formula and i have been trying to derive the
formula without any luck, can anyone help me out ?
===
Subject: Re: chess bishops - combinations
It might be helpful to separate the board into two pieces according to the
colors of the squares, then translate the diagonals into rows and columns,
so
you can use familiar coordinate methods. An inductive proof might work.
| I understand that it is possible to express the number of ways of
| having k bishops on an n*n board such that they are not attacking each
| other as a combinatorial formula and i have been trying to derive the
| formula without any luck, can anyone help me out ?
===
Subject: Re: chess bishops - combinations
> I understand that it is possible to express the number of ways of
> having k bishops on an n*n board such that they are not attacking each
> other as a combinatorial formula and i have been trying to derive the
> formula without any luck, can anyone help me out ?
Is this right?.. Each bishop has a coordinate (r,c) (row, column). For all
the bishops, the r's are different to each other, and the c's are different
to each other.
So for k bishops the total amount of possible coordinates is how many
different ways you can take k r's out of 8, times how many different ways
you can take k c's out of 8. And that should be enough for you.. If it's
right..I think so.
--
Quaternion
===
Subject: Re: chess bishops - combinations
> I understand that it is possible to express the number of ways of
> having k bishops on an n*n board such that they are not attacking each
> other as a combinatorial formula and i have been trying to derive the
> formula without any luck, can anyone help me out ?
> Is this right?.. Each bishop has a coordinate (r,c) (row, column). For
all
> the bishops, the r's are different to each other, and the c's are
different
> to each other.
No, bishops attack on diagonals, not rows & columns.
OP; have you tried starting small? n = 2, n = 3, see whether there are
any patterns? Or try searching Sloane's On-Line Encyclopedia of Integer
Sequences for the word bishop?
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: chess bishops - combinations
>> I understand that it is possible to express the number of ways of
>> having k bishops on an n*n board such that they are not attacking each
>> other as a combinatorial formula and i have been trying to derive the
>> formula without any luck, can anyone help me out ?
>Is this right?.. Each bishop has a coordinate (r,c) (row, column). For all
>the bishops, the r's are different to each other, and the c's are
different
>to each other.
>So for k bishops the total amount of possible coordinates is how many
>different ways you can take k r's out of 8, times how many different ways
>you can take k c's out of 8. And that should be enough for you.. If it's
>right..I think so.
Thats a little too simplistic, as you can see from
http://acm.uva.es/p/v102/10237.html with 6 bishops on an 8*8 board
there are 5599888 ways and with 5 bishops on a 30*30 board there are
3127859642656 ways.
===
Subject: Re: chess bishops - combinations
> I understand that it is possible to express the number of ways of
> having k bishops on an n*n board such that they are not attacking each
> other as a combinatorial formula and i have been trying to derive the
> formula without any luck, can anyone help me out ?
>>Is this right?.. Each bishop has a coordinate (r,c) (row, column). For
all
>>the bishops, the r's are different to each other, and the c's are
>>different to each other.
>>So for k bishops the total amount of possible coordinates is how many
>>different ways you can take k r's out of 8, times how many different ways
>>you can take k c's out of 8. And that should be enough for you.. If it's
>>right..I think so.
> Thats a little too simplistic, as you can see from
> http://acm.uva.es/p/v102/10237.html with 6 bishops on an 8*8 board
> there are 5599888 ways and with 5 bishops on a 30*30 board there are
> 3127859642656 ways.
Yes I thought bishops were towers for some reason..
Different languages
--
Quaternion
===
Subject: Re: chess bishops - combinations
> I understand that it is possible to express the number of ways of
> having k bishops on an n*n board such that they are not attacking each
> other
Why would bishops attack one another? Oh--over homosexuality perhaps?
> as a combinatorial formula and i have been trying to derive the
> formula without any luck, can anyone help me out ?
--
G.C.
===
It is very good that I heard Andrei Linde speak last night at UC
Berkeley as my updated version of
http://qedcorp.com/APS/EmergentGravity.doc shows. Also a rather large
crowd showed up to hear
my talk right in the middle of the talk before mine.
Linde explained that he does not believe most of the current textbook
presentations of inflationary cosmology.
The essence of his talk is in my 2 equations II.9 and II.10 in the new
version of http://qedcorp.com/APS/EmergentGravity.doc
The key is the friction term in II.9 which is a linearization of my
II.8 near the bottom of a local minimum in the effective potential of
the vacuum coherence field in the FRW large scale limit assuming
homogeneity.
The friction term that is the root cause of chaotic inflation with
slow descent at large order parameter followed by oscillatory reheating
to make the post-inflationary Big Bang is quite simply the effect of the
connection field for parallel transport in the second application of the
covariant time derivative to the scalar field. Of course Linde has no
micro-dynamics for the emergence of the scalar field as I do nor does he
derive Einstein's gravity together with the unified exotic vacuum dark
energy/matter field in a two-way bootstrap of the never-ending
spontaneous self-organizing parallel universes splitting off into baby
universes in the sense of Andrei Sakharov's metric elasticity a
special case of P.W. Anderson's More is different.
Steve Carlip gave a interesting talk on topology change in WKB approx to
quantum gravity with a possible
explanation of why space is homogeneous on large scale consistent with
inflation.
===
Subject: Product of matrix elements?
Given two nx1 column matrices (vectors) X=(x1 x2 ... xn)^T and Z=(z1 z2
... zn)^T, all elements real numbers (T=transpose):
1) Is there a simple matrix operation to create the nx1 matrix
(x1*z1 x2*z2 ... xn*zn)^T from X and Z? I am referring to a mathematical
operation like inner product, rather than an algorithm or computer program.
2) Is there a way using matrix operations to produce the column vector
(exp(x1) exp(x2) ... exp(xn))^T (exp=exponential function) from X?
3) Related to (2), is there a way using matrix operations to produce the
nxn diagonal matrix D
(x1 0 0 ... 0)
(0 x2 0 ... 0)
( ... )
(0 0 0 ... xn)
from X? And conversely, given D as above, to write X in terms of D using
matrix operations?
===
Subject: Re: Product of matrix elements?
> Given two nx1 column matrices (vectors) X=(x1 x2 ... xn)^T and Z=(z1 z2
> ... zn)^T, all elements real numbers (T=transpose):
> 1) Is there a simple matrix operation to create the nx1 matrix
> (x1*z1 x2*z2 ... xn*zn)^T from X and Z? I am referring to a mathematical
> operation like inner product, rather than an algorithm or computer
program.
let Em be the square matrix (eij) such that emm = 1 and eij = 0
otherwise.
consider Em.Z.X, m = 1, 2, 3, ..., n
> 2) Is there a way using matrix operations to produce the column vector
> (exp(x1) exp(x2) ... exp(xn))^T (exp=exponential function) from X?
assuming exp(A) = Sum A^n/(n!) over n = 0, 1, 2, ..., infinity
you can obtain the jth entry of the sought vector:
exp(A)ej,
where A is the diagonal matrix aii = xi, aij = 0 for i not equal to j,
and ej is the jth elementary vector.
> 3) Related to (2), is there a way using matrix operations to produce the
> nxn diagonal matrix D
> (x1 0 0 ... 0)
> (0 x2 0 ... 0)
> ( ... )
> (0 0 0 ... xn)
> from X? And conversely, given D as above, to write X in terms of D using
> matrix operations?
consider X.ej
form a linear combo.
===
Subject: Proving if an inverse function to a matrix transformation exists
Hello. I am having trouble understanding how to apply the inverse function
theorem to matrix transformations. The textbook that I am using has
examples of using the inverse function theorem for ordinary R(n)->R(m)
functions but not for matrix transformations such as S(X)=X^3, where X is
in Mat(3,3) for example. Here is the problem that I need to solve and my
solution. I would greatly appreciate if someone could go over my reasoning
and point out any flaws that I have and explain to me how to solve
Problem: A = [ 0 1 0 ]
[ 0 0 1 ]
[ 1 0 0 ]
Note that A^3 = I (identity matrix). Is there a cont. differentiable
function g such that g(I)=A and (g(A))^3=A in the neighborhood of I?
My Solution:
Let f: X -> X^3 (X is a matrix)
then f(g(x)) = [g(x)]^3 = x
if X = I
f(g(I)) = I
Now use the chain rule for the derivative of f(g(I)):
[D(fog)(X)] = [Df(g(X))][Dg(X)] (by definition)
Let X = I:
[D(fog)(I)] = [Df(g(I))][Dg(I)] = [Df(A)][Dg(I)] (since g(I)=A)
Finding [Df(A)] is a bit tricky. I used the general definition of the
derivative to show that
[Df(A)]H = 3A^2H + 3AH^2
since lim {(1/H) * (f(A+H) - f(A) - (3A^2H + 3AH^2))} = 0.
Now I let H = I
so that [Df(A)]I = 3A^2 + 3A
I then went on to show that the determinant of this matrix (using A from
the problem) = 54 (is nonzero).
g(X), asked for in the problem exists.
However, I am a bit confused myself as to whether this reasoning is
correct and whether I am using the inverse function theorem correctly. I
would greatly appreciate if someone could comment on my solution and
Serge
===
Subject: Re: Axiom of Foundation (absymally stupid question)
Since we're discussing the axiom of foundation (in the textbooks I've seen,
it's called the axiom of regularity), does anyone know what the
intuitive
justification for this axiom is? I mean, all the other axioms seem pretty
natural to me, even the infamous axiom of choice. But where in world did
they come up with the axiom of regularity?
Have a tolerable existence. Eli
===
Subject: Re: Axiom of Foundation (absymally stupid question)
>Since we're discussing the axiom of foundation (in the textbooks I've seen,
>it's called the axiom of regularity), does anyone know what the
intuitive
>justification for this axiom is? I mean, all the other axioms seem pretty
>natural to me, even the infamous axiom of choice. But where in world did
>they come up with the axiom of regularity?
It was to avoid having a set x which is its only element,
and more complicated versions of this. It is easy to see
that both it and its negation are consistent.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu one: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Axiom of Foundation (absymally stupid question)
The foundation axiom rules out situations that you might call membership
loops. For example, if A1 in A2 in A3 in A1, then {A1,A2,A3} has a
nonempty
intersection with each of its elements.
| Since we're discussing the axiom of foundation (in the textbooks I've
seen,
| it's called the axiom of regularity), does anyone know what the
intuitive
| justification for this axiom is? I mean, all the other axioms seem
pretty
| natural to me, even the infamous axiom of choice. But where in world did
| they come up with the axiom of regularity?
|
| Have a tolerable existence. Eli
===
Subject: Re: Axiom of Foundation (absymally stupid question)
Cc: ecooper@mathstat.umass.edu
permission for an emailed response.
> Since we're discussing the axiom of foundation (in the textbooks I've
seen,
> it's called the axiom of regularity), does anyone know what the
intuitive
> justification for this axiom is? I mean, all the other axioms seem pretty
> natural to me, even the infamous axiom of choice. But where in world did
> they come up with the axiom of regularity?
It is not assumed because of some intuitive justification, but because
it doesn't interfere with ordinary mathematics (sets of the kind
prohibited by the axiom aren't ever needed) and it simplifies model
theory by a jillion times.
Thomas
===
Subject: Re: Axiom of Foundation (absymally stupid question)
|Since we're discussing the axiom of foundation (in the textbooks I've seen,
|it's called the axiom of regularity), does anyone know what the
intuitive
|justification for this axiom is? I mean, all the other axioms seem pretty
|natural to me, even the infamous axiom of choice. But where in world did
|they come up with the axiom of regularity?
i think the answer is that (in the presence of the usual other axioms)
it's equivalent to the statement every set belongs to some level of
the cumulative hierarchy, which is a powerful statement because it
allows you to prove theorems applying to all the sets in the universe,
by transfinite induction with respect to the level of the cumulative
hierarchy to which they belong (the levels of the hierarchy being
indexed by ordinal numbers).
the cumulative hierarchy starts out with the empty set as the bottom
level, then the power set of the empty set as the next level, and so
forth on upwards. at limit ordinals you take the union of the levels
below.
speaking as a non-specialist in set theory, what most bugged me about
the axiom of foundation when i was first trying to learn axiomatic set
theory (after already learning first-order logic) was that on the one
hand it sounds like it's trying to prevent weird loops and weird
infinite regresses of membership chains, while on the other hand the
means by which it's trying to prevent it (namely the expressive power
of axioms of first-order logic) is notoriously ineffectual at actually
preventing the existence of models with infinite regresses. but in
retrospect i guess the idea is supposed to be that that shouldn't
really bother me any more than the similar fact that first-order peano
logic is trying to prevent the same kind of infinite regresses, and
essentially fails to do so, for essentially the same reason. in
both cases you still end up with a powerful means of proving theorems
within the theory by some kind of induction even though some of the
models of the theory contain infinite regresses which naively seem
to violate the spirit of the inductive principles.
--
[e-mail address jdolan@math.ucr.edu]
===
Subject: Re: Axiom of Foundation (absymally stupid question)
>Since we're discussing the axiom of foundation (in the textbooks I've seen,
>it's called the axiom of regularity), does anyone know what the
intuitive
>justification for this axiom is? I mean, all the other axioms seem pretty
>natural to me, even the infamous axiom of choice. But where in world did
>they come up with the axiom of regularity?
The Axiom of Foundation is not really necessary - exactly 99.93% ()
of mathematics can be done without it. In fact, it even limits our
universe of discourse when we accept it. However, it is there to make
things nice. Without it, it is consistent to have pathologies such
as a set which is an element of itself, two distinct sets each of which
is a member of the other, and a sequence x of distinct sets such that
x_(n+1) is an element of x_n.
The axiom also allows an ice-cream cone construction of the class of
sets. With the axiom, there is a sequence R through the ordinals
such that a < b implies R(a) is a subset of R(b) and the universe
of sets is the well-founded sets, viz., theunion of all these sets.
(In particular, R(0) = 0 and R(a+1) = P(R(a)).) This structure
allows certain proofs in set theory; for example, the well-founded sets
provide a model for ZF.
I get most of this from Kunen.
One more question: If we didn't mind invoking the axiom of foundation,
wouldn't {a, {a,b}} suffice as a definition of the ordered pair
(a,b)? Seems to me the answer is yes.
--
Steen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Axiom of Foundation (absymally stupid question)
permission for an emailed response.
> One more question: If we didn't mind invoking the axiom of
> foundation, wouldn't {a, {a,b}} suffice as a definition of the
> ordered pair (a,b)? Seems to me the answer is yes.
Sure, but what makes {a, {a,b}} better than {{a}, {a,b}}? The key is
to prove that (a,b) = (c,d) => a=c & b=d. With your proposed
definition, the proof is a royal pain, and the resulting thingies
aren't really any simpler.
Thomas
===
Subject: Re: Axiom of Foundation (absymally stupid question)
>>One more question: If we didn't mind invoking the axiom of
>>foundation, wouldn't {a, {a,b}} suffice as a definition of the
>>ordered pair (a,b)? Seems to me the answer is yes.
>>
>Sure, but what makes {a, {a,b}} better than {{a}, {a,b}}? The key is
>to prove that (a,b) = (c,d) => a=c & b=d. With your proposed
>definition, the proof is a royal pain, and the resulting thingies
>aren't really any simpler.
Doesn't seem to like the necessary implication is more difficult than
with the usual definition.. If {a,{a,b}} = {c,{c,d}}, suppose a =
{c,d} and c = {a,b}. Then a is an element of c, which is an
element of a. This is not possible with foundation. By
contradiction, a = c and {a,b} = {c,d}, whence b = d.
I would say that my definition is in a sense simpler, since it requires
the construction of one fewer set, or the representation requires two
fewer characters. Seems to me that the real objection is that invokes
foundation unnecessarily. (Note that Halmos doesn't even mention
foundation in NST.)
Also, once an ordered pair is well-defined, one needs never refer to the
definition again.
--
Steen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Axiom of Foundation (absymally stupid question)
permission for an emailed response.
> Doesn't seem to like the necessary implication is more difficult than
> with the usual definition.. If {a,{a,b}} = {c,{c,d}}, suppose a =
> {c,d} and c = {a,b}. Then a is an element of c, which is an
> element of a. This is not possible with foundation. By
> contradiction, a = c and {a,b} = {c,d}, whence b = d.
I suppose so. (You also need to prove that {a,{a,b}} is necessarily a
doubleton, which is also a quick deduction from Foundation.)
> I would say that my definition is in a sense simpler, since it
> requires the construction of one fewer set, or the representation
> requires two fewer characters. Seems to me that the real objection is
> that invokes foundation unnecessarily. (Note that Halmos doesn't even
> mention foundation in NST.)
Ok, I suppose I grant your point then. I'd only add that I think this
objection is a very important one. As noted before, Foundation isn't
added because of some intuitive confidence, but rather because it is
known to be harmless, and it's a big help in model theory.
So that means that one must be able to develop ordinary mathematics
without using it (or else it wouldn't be so harmless, it would be
important), and since you need to show that, once you've done it, it
is no longer interesting to show that you could have used it here or
there along the way.
So invoking Foundation unnecessarily is a bad thing, but in a very
different way from (say) invoking Choice unnecessarily.
Thomas
===
Subject: Re: Axiom of Foundation (absymally stupid question)
at 06:35 PM, Elaine Jackson said:
>The whole problem is just that you're misquoting the axiom.
No he is not.
>You say: Every nonempty B contains a y
>with (B intersect y) = empty.
Not only him. Everybody who uses GBN or ZF says so as well.
>I say: Every nonempty B contains a y for which
>there is no z with z in y and z in B.
In what context do you say it?
>My axiom
What set theory is your axiom a part of? What are the other axioms?
>but it allows for the possibility that there exist
>citrus fruits that are not sets.
Then it's not part of the same set theory, it is your responsibility
to give the complete set of axioms that you are using.
>Citrus fruits that have no elements, but aren't the
>empty set, are technically called individuals.
No. There are sets theories that have individuals without having to
abandon extentionality. Again, if you wish to be taken seriously you
will need to state what set theory it is that you are using.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: Axiom of Foundation (absymally stupid question)
| at 06:35 PM, Elaine Jackson said:
|
|>The whole problem is just that you're misquoting the axiom.
|
|No he is not.
|
|>You say: Every nonempty B contains a y
|>with (B intersect y) = empty.
|
|Not only him. Everybody who uses GBN or ZF says so as well.
In a careful presentation, it would be usual to write it out in terms
of epsilon and =, and not defined terms like intersect which
technically
are not part of the language of ZFC.
I did a web search, and I see that on Mathworld the axiom is given as a
form of the axiom scheme of epsilon-induction, which again does not involve
referring directly to intersections.
|>I say: Every nonempty B contains a y for which
|>there is no z with z in y and z in B.
|
|In what context do you say it?
The point here is that speaking of intersections presupposes that the
objects in question are sets, whereas set theories with urelements
typically consider epsilon to be a relationship on the whole domain,
including both urelements and sets. So saying there doesn't exist a z
such that z in y and z in B allows for the possibility that y or B
is an urelement (like a piece of fruit). Thus the same statement of
the foundation axiom would suffice for a set theory with urelements
of this kind.
|>My axiom
|
|What set theory is your axiom a part of? What are the other axioms?
ZFU would be a familiar example of this kind of set theory.
|>but it allows for the possibility that there exist
|>citrus fruits that are not sets.
|
|Then it's not part of the same set theory, it is your responsibility
|to give the complete set of axioms that you are using.
Seymour Metz is being overly formal again.
|>Citrus fruits that have no elements, but aren't the
|>empty set, are technically called individuals.
|
|No.
But they are.
|There are sets theories that have individuals without having to
|abandon extentionality.
She didn't say this was the only sense in which the term was used. True,
another way to model urelements/atoms/individuals is to have them bear
the epsilon relation to themselves. But that requires either abandoning
or adjusting the foundation axiom.
|Again, if you wish to be taken seriously you
|will need to state what set theory it is that you are using.
The context was adequate.
Keith Ramsay
===
Subject: Vector Calculus problem
Ok, I'm trying to work on my homework and am stuck on 4.9 #10 of Vector
Analysis by Davis.
The question states By means of Stokes' theorem, find S F*dR around the
ellipse x^2+y^2=1, z=y, where F=xi+(x+y)j+(x+y+z)k.
I got the curl of F and that equalled i-j+k but I'm not really sure how to
do
the rest of the problem. Any help would be appreciated. I've wasted a lot
of
time and gotten almost nowhere.
===
Subject: Re: Vector Calculus problem
I forgot to mention that the answer is in the back of the book:
+-2pi depending upon the direction of integration.
I just can't figure out how to get that.
===
Subject: topological groups
I'm reading about topological groups and I am having trouble with the
definition of such a group. What exactly are the open sets?
Any help would be appreciated.
===
Subject: Re: topological groups
at 08:02 PM, Arthur said:
>I'm reading about topological groups and I am having trouble with the
> definition of such a group. What exactly are the open sets?
That's like asking what the open sets are in a topological space. Part
of specifying a topological group is specifying a topology; the open
sets of a topological group are the open sets of its topology.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact me. Do
not reply to spamtrap@library.lspace.org
===
Subject: Re: topological groups
> I'm reading about topological groups and I am having trouble with the
> definition of such a group. What exactly are the open sets?
Hey,
You can't talk about the open sets since a group is said topological
iff
it is continuous ( (x,y)-> x+y and x->x^(-1)are continuous); sometimes it
is
also required to be Hausdorff. But considering a group (G,+) you can put
different topologies on G so that G is a topological group (provided that G
is continuous (and sometimes, Hausdorff)). We're not definining a special
topology here.
--
Julien Santini,
France.
===
Subject: Re: topological groups
>> I'm reading about topological groups and I am having trouble with the
>> definition of such a group. What exactly are the open sets?
> Hey,
> You can't talk about the open sets since a group is said topological
> iff it is continuous ( (x,y)-> x+y and x->x^(-1)are continuous);
> sometimes it is also required to be Hausdorff. But considering a group
> (G,+) you can put different topologies on G so that G is a topological
> group (provided that G is continuous (and sometimes, Hausdorff)).
> We're not definining a special topology here.
> --
> Julien Santini,
> France.
I see.
Arthur
===
Subject: ADE Groups, Cosmology and Consciousness
Jack & Jonathan,
The *Roots of Consciousness* by Jeffrey Mishlove on the web does NOT
include
my 40 page appendix paper, Consciousness: a Hyperspace View. Jeffrey
wanted to include it but there was some difficulty with the complexity of
the figures -- if I remember correctly. Of course, it had never been in pdf
form, since the web (& home computers) were in a very primitive state in
1993. This paper has the most detailed account of what I now call ADEX
theory -- the application of the A-D-E Coxeter gras to mathematics,
ysics, and other fields such as consciousness theory. This appendix paper
was written in 1989, and published in 1993. A short summary and update of
this paper (with more ADEX examples) was titled, A Mathematical Strategy
for a Theory of Consciousness. It was written in 1994 and published in
the
book, *Toward a Science of Consciousness: the First Tucson Discussions and
Debates* (edited by Stuart R. Hameroff, Alfred W. Kaszniak, and Alwyn C.
Scott), MIT Press, 1996.
People have told me that my papers are hard to read without much more
mathematical knowledge than they possess. Jeffrey tells me that Russian
scientists have less trouble with the math since their mathematical
training
is more advanced than that of American psychologists and biologists.
Actually much of the mathematics of ADEX theory is of very recent vintage,
but the key area of mathematics is quite old -- group theory (both finite
groups and Lie groups and relationships between them). Our understanding of
these relationships depends on both algebra and geometry -- especially
hyperspace geometry -- including differential geometry and algebraic
geometry.
ysicists who study general relativity know some differential geometry,
but
they have not studied the more recently developed algebraic geometry -- and
very few ysicist (or mathematicians) have seen the great utility of the
A-D-E Coxeter gras which have been used to classify more than 20
mathematical objects. The advantage of having these classifications, is
that
the A-D-E gras provide the relationships between all these mathematical
objects. I will mention a few of these objects -- the study and
application
of which I call ADEX theory:
1. Finite reflection groups (Coxeter groups also called Weyl groups)
2. Hyperspace polytopes and thus crystallograic lattices
3. Coxeter arrangements (mirrors in reflection space)
4. Lie algebras and Lie groups (& also Kac-Moody Lie algebras)
5. Thom-Arnold catastroe bundles (useful for Jack's version of Bohm)
[BTW: Thom claimed (1975) that it models the mind-body
relationship]
[Yes, this is the aspect I want to flesh out with you.]
6. McKay groups (finite subgroups of SU(2) -- unit length quaternions)
7. Gravitational Instantons (closely related to Penrose twistors)
8. 2-d Conformal field theories (which live on hyperspace strings)
[Jack: It is interesting that O(2) macro-quantum order parameter in
ordinary space has string defects e.g. Hagen Kleinert also books on
soft condensed matter ysics and cosmic strings. Then introduce
extra space dimensions including fermi dimensions for supersymmetry to
get higher dim branes from the macro-quantum order parameters perhaps
with higher O(N) internal symmetry, hyper-complex matrix order
parameters over hyper-complex generalized space-time manifolds.
My model in http://qedcorp.com/APS/EmergentGravity.doc is only the low
energy tail of that. BTW new version shows how to go from my BIT FROM IT
Landau-Ginburg eq to Andrei Linde's specific equations for chaotic
APS-AAPT. I had discovered the friction term from my equations a year
ago not realizing their crucial role in Linde's theory of the continuous
creation of the parallel universes.]
9. Error-correcting codes (related to Jack's IT <--> BIT idea)
That too. The mind field must have error correcting codes built into it.]
10. Quantizing lattices (analog to digital transforms)
As the motto for Plato's academy said, Let no one enter here without
geometry. Today, this geometry must include hyperspace geometry --
which
dates back to the 19th century.
BTW: *Roots of Consciousness* is mentioned at the end of John McKay's very
short paper, A Rapid Introduction to ADE Theory. The URL for this paper
is:
http://math.ucr.edu/home/baez/ADE.html
This is on John Baez's very extensive website, and from the above URL you
can access 4 much longer (tutorial) papers by John Baez on the ADE related
mathematics.
Nuff said!
Saul-Paul