mm-4559 === Subject: Re: -- more factoring >All polynomials discussed are of deg >= 2 and > univariate. > ... >Let gcd(a,b) = 1 >Let F be an irreducible integer polynomial that > only > has primefactors 1 mod a. >Let G be an irreducible integer polynomial that > only > has primefactors 1 mod b. >Let n and m be positive integers and for all n > and m > the image of F and G are positive integers > 1. >Now it is clear that: >gcd(F(n),G(m)) = 1 for all n and m. > > No, it's _not_ clear. > > A prime can be congruent to _both_ 1 (mod a) and 1 > (mod b). > > Thus, your condition on the prime factors of F,G > does > not imply > > gcd(F(n),G(m)) = 1 for all m,n. > > In fact, I'll make the following claim ... > > Conjecture: > > There do not exist nonconstant univariate integer > polynomials f,g such > that gcd(f(m),g(n)) = 1 for all m,n in Z. > > quasi lemma 1 : there exists an irreducible nonconstant integer > polynomial that only has prime factors 1 mod a but > never primefactors 1 mod b for a =/= b. e.g. 1 mod 5 but never 1 mod 15. thus always => 6 mod 15 and/or 11 mod 15. Actually, your claim can be instantly disproved. that would amaze me. Let f be a nonconstant integer polynomial. Let S be the set of prime factors of elements of > {f(n) | n in Z}. Let A = {a in Z, a > 1| for some p in S, p = 1 (mod > a)}. Claim A cannot have just one element. no , but all are subsets of primes 1 mod a. so what ? Suppose instead A = {a}. a is fixed. just reminding you of that. as in 1 mod a , a needs to be fixed to be meaningfull in this context. Case (1): a is composite lets say a = 10. Then choose b such that b|a and 1 < b < a. lets say b = 5. But any prime congruent to 1 (mod a) is also > congruent to 1 (mod b). 1 mod 10 -> 1 mod 5 ???? any prime ? you mean some primes. not even any integer. e.g. 6 = 1 mod 5 but not 1 mod 10. Case (2): a is prime. case 2 ? case 1 is flawed as i have just shown. lets say a = 11. thus 1 mod 11. lol cant think of even primes 1 mod 11. parity !!! odd + odd = even. ( ok , ive negleted parity above to make clear my arguments , but now i have to point it out ) of course you can take a subset -> 1 mod 22. so lets take 22. Choose n in Z such that abs(f(n)) > 1 and let p be a > prime factor of > f(n). f(3) and p = 23. By hypothesis, p = 1 (mod a), but p =/= 1 mod b > for any positive > integer b > 1, b =/= a. for a certain fixed polynomial f and a certain fixed b. not all polynomials or all b !! It follows that p - 1 = a, 23 - 1 = 22 > hence a = 2 and p = no i got 22 :p > 3. This implies that f(n) is a power of 3. f(3) is thus a power of 23 ? since f(3) can be divided by 23 , how does that neccesarily make it a power of it ? f(3) = 67 * 23 ( 67 = 1 mod 22 ) But it's > impossible for a > nonconstant integer polynomial to only have absolute > values greater > than 1 which are powers of 3. yes so ? thats no disproof of my lemma. quasi tommy1729 === Subject: Re: -- more factoring > All polynomials discussed are of deg >= 2 and > univariate. ... Let gcd(a,b) = 1 Let F be an irreducible integer polynomial that > only > has primefactors 1 mod a. Let G be an irreducible integer polynomial that > only > has primefactors 1 mod b. Let n and m be positive integers and for all n > and m > the image of F and G are positive integers > 1. Now it is clear that: gcd(F(n),G(m)) = 1 for all n and m. No, it's _not_ clear. A prime can be congruent to _both_ 1 (mod a) and 1 > (mod b). Thus, your condition on the prime factors of F,G > does > not imply gcd(F(n),G(m)) = 1 for all m,n. In fact, I'll make the following claim ... Conjecture: There do not exist nonconstant univariate integer > polynomials f,g such > that gcd(f(m),g(n)) = 1 for all m,n in Z. quasi >lemma 1 : >there exists an irreducible nonconstant integer > polynomial that only has prime factors 1 mod a but > never primefactors 1 mod b for a =/= b. >e.g. >1 mod 5 but never 1 mod 15. >thus always => 6 mod 15 and/or 11 mod 15. > > Actually, your claim can be instantly disproved. that would amaze me. > Let f be a nonconstant integer polynomial. > > Let S be the set of prime factors of elements of > {f(n) | n in Z}. > > Let A = {a in Z, a > 1| for some p in S, p = 1 (mod > a)}. > > Claim A cannot have just one element. no , but all are subsets of primes 1 mod a. All what? I have no idea what you're talking about. >so what ? So, I disproved your claim, at least the way you stated it. You claimed ... there exists an irreducible nonconstant integer polynomial that only has prime factors 1 mod a but never primefactors 1 mod b for a =/= b. It's a very imprecise claim. I translated it as follows ... There exists a nonconstant irreducible polynomial f such that, for the set S defined by S = {p | p is prime and p | f(n) for some n in Z} there is an integer a > 1 such that p in S => p = 1 (mod a) and for all integers b > 1, b =/= a, p in S => p =/= 1 (mod b) If that's an accurate translation, then I disproved your claim. If that's not what you intended to claim, please try harder to state your conjectures more precisely. Note -- unless you can _prove_ your claim, you shouldn't call it a lemma, proposition, or theorem. Call it what it is -- a conjecture. quasi === Subject: Example of a set requiring d+1 points (Caratheodory theorem) posting-account=55jpkgoAAABlrbzj_YB1fnU8r5_kLCli AppleWebKit/523.15.1 (KHTML, like Gecko) Version/3.0.4 Safari/523.15,gzip(gfe),gzip(gfe) I have a trivial doubt about a form of Caratheodory theorem which has applications in Information Theory. The statement of the theorem is, Any point in the convex closure of a connected compact set A in a d dimension Euclidean space can be represented as a convex combination of d+1 or fewer points in the original set A. My problem is that I am unable to find any example where d+1 points are required. (It is easy to find examples where only d points suffice). In addition, is there an example in R^n? Any help will be greatly appreciated. Dash === Subject: Re: Example of a set requiring d+1 points (Caratheodory theorem) I have a trivial doubt about a form of Caratheodory theorem which has > applications in Information Theory. The statement of the theorem is, > Any point in the convex closure of a connected compact set A in a d > dimension Euclidean space can be represented as a convex combination > of d+1 or fewer points in the original set A. In the usual statement of the theorem, the word connected isn't included. (In fact, in the *usual* statement of the theorem, the word compact isn't included either, though Caratheodory originally proved it for the compact case only.) So, obviously, it's easy to come up with simple examples in that case: let A consist of any two distinct points. For an example of a *connected* A where you still need d+1 points, I don't believe you'll find one in R^1 or R^2, but you'll find it easy to find one in R^3. Let A be the union of three line segments joined at a single point (so their convex hull is a tetrahedron). Note that no interior point can be expressed as a linear combination of only two points of A. -- Kevin Buhr === Subject: Re: Example of a set requiring d+1 points (Caratheodory theorem) For an example of a *connected* A where you still need d+1 points, I > don't believe you'll find one in R^1 or R^2, but you'll find it easy > to find one in R^3. Let A be the union of three line segments joined > at a single point (so their convex hull is a tetrahedron). Note that > no interior point can be expressed as a linear combination of only two > points of A. work, is it? In R^3, you obviously want an example where *four* points are required. And, now that I've thought about it, I think in R^d for a connected set A, you'll never need all d+1 points. You can use connectedness to eliminate a point as follows. Suppose you have an x in A (subset of R^d) expressed as a convex combination (c.c.) of d+1 points in A: x_0,x_1,...,x_d. If these points aren't all linearly independent, you can obviously express x as c.c. of a subset of d or fewer of them, and you're done, so suppose they *are* independent. Then, the convex closure of the points is a d-simplex S(x_0,x_1,x_2,...,x_d). If x is on the boundary, again, you can immediately express x as a c.c of a subset of d or fewer of the points, so suppose x is in the interior of the simplex. Now, as A is connected, there is a path from x_0 to x_1. Consider the shrinking d-simplex S(y,x_1,x_2,...,x_d) as y moves along the path from x_0 to x_1. Geometrically, x is contained in the interior of the original simplex, and as y goes from x_0 to x_1, the simplex shrinks. At some point, either at y=x_1 or at some earlier time on the path, the simplex becomes flat---it's no longer a d-simplex. It's either a (d-1)-simplex or---if this happens before y=x_1 and depending on the shape of the path---maybe the union of two (d-1)-simplexes. But sometime between y=x_0 and the first collapse of the simplex, the boundary of the shrinking simplex (or its collapsed version) must contact x. At that point, x can be expressed as a c.c. of y and d-1 of the points (x_1,x_2,...,x_d). Visualizing this in R^2 is easiest: if x is expressed as a c.c. of three linearly independent points, it's in the interior of the triangle formed by those three points S(x_0,x_1,x_2). Now, let y go from x_0 to x_1 along a path in A, and consider the shrinking triangle S(y,x_1,x_2). Eventually, this will collapse to a line segment, and sometime between the original triangle at y=x_0 and this collapse x will be on the triangle boundary or the final collapsed line, expressible as a linear combination of two points y and x_1 (or maybe y and x_2) in S. I think that's sound, though I'm sure to be corrected if it's not. Does that make sense? -- Kevin Buhr === Subject: Re: Example of a set requiring d+1 points (Caratheodory theorem) > For an example of a *connected* A where you still need d+1 points, I > don't believe you'll find one in R^1 or R^2, but you'll find it easy > to find one in R^3. Let A be the union of three line segments joined > at a single point (so their convex hull is a tetrahedron). Note that > no interior point can be expressed as a linear combination of only two > points of A. work, is it? In R^3, you obviously want an example where *four* > points are required. And, now that I've thought about it, I think in R^d for a connected > set A, you'll never need all d+1 points. You can use connectedness to > eliminate a point as follows. Suppose you have an x in A (subset of R^d) You mean in the convex hull of A? > expressed as a convex > combination (c.c.) of d+1 points in A: x_0,x_1,...,x_d. If these > points aren't all linearly independent, you can obviously express x as > c.c. of a subset of d or fewer of them, and you're done, so suppose > they *are* independent. That is a bit of a handwave. Besides, I think you are confusing linear, affine and convex combinations here. How about this: If these points are affinely dependent, they are contained in an affine subspace of dimension d' < d, and Caratheodory's theorem can be used to show that x is a c.c. of (d'+1) <= d points. On the other hand, if they are independent, ... > Then, the convex closure of the points is a > d-simplex S(x_0,x_1,x_2,...,x_d). If x is on the boundary, again, you > can immediately express x as a c.c of a subset of d or fewer of the > points, so suppose x is in the interior of the simplex. Now, as A is connected, there is a path from x_0 to x_1. Unfortunately, not every connected set is path-connected. > Consider the > shrinking d-simplex S(y,x_1,x_2,...,x_d) as y moves along the path > from x_0 to x_1. Geometrically, x is contained in the interior of > the original simplex, and as y goes from x_0 to x_1, the simplex > shrinks. At some point, either at y=x_1 or at some earlier time on > the path, the simplex becomes flat---it's no longer a d-simplex. > It's either a (d-1)-simplex or---if this happens before y=x_1 and > depending on the shape of the path---maybe the union of two > (d-1)-simplexes. But sometime between y=x_0 and the first collapse > of the simplex, the boundary of the shrinking simplex (or its > collapsed version) must contact x. At that point, x can be expressed > as a c.c. of y and d-1 of the points (x_1,x_2,...,x_d). Visualizing this in R^2 is easiest: if x is expressed as a c.c. of > three linearly independent points, it's in the interior of the > triangle formed by those three points S(x_0,x_1,x_2). Now, let y go > from x_0 to x_1 along a path in A, and consider the shrinking > triangle S(y,x_1,x_2). Eventually, this will collapse to a line > segment, and sometime between the original triangle at y=x_0 and this > collapse x will be on the triangle boundary or the final collapsed > line, expressible as a linear combination of two points y and x_1 (or > maybe y and x_2) in S. I think that's sound, though I'm sure to be corrected if it's not. > Does that make sense? > I like it, with the restriction that A must be path-connected. Maybe that restriction can be lifted, but I'm guessing that would require a completely different proof. Any ideas? Counterexamples? -- Niels Diepeveen === Subject: Re: Example of a set requiring d+1 points (Caratheodory theorem) > I have a trivial doubt about a form of Caratheodory theorem which has > applications in Information Theory. The statement of the theorem is, > Any point in the convex closure of a connected compact set A in a d > dimension Euclidean space can be represented as a convex combination > of d+1 or fewer points in the original set A. My problem is that I am unable to find any example where d+1 points > are required. (It is easy to find examples where only d points > suffice). In addition, is there an example in R^n? I wasn't familiar with the theorem, but looking at http://en.wikipedia.org/wiki/Carath.8eodory's_theorem_(convex_hull) I don't see any hint of A having to be connected. If you drop that requirement, any set of two points in R^1 is an example. Maybe that explains it. -- Niels Diepeveen === Subject: Re: Example of a set requiring d+1 points (Caratheodory theorem) >I have a trivial doubt about a form of Caratheodory theorem which has >applications in Information Theory. The statement of the theorem is, >Any point in the convex closure of a connected compact set A in a d >dimension Euclidean space can be represented as a convex combination >of d+1 or fewer points in the original set A. My problem is that I am unable to find any example where d+1 points >are required. (It is easy to find examples where only d points >suffice). In addition, is there an example in R^n? Any help will be greatly appreciated. A triangular region in R^2. Or for that matter, a closed interval in R^1. quasi === === Subject: Re: Markov Transition Matrix posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > A dynamic interurban of people shows the following Markov Transition > Matrix of residents to urban, suburban and rural areas: Urban Suburban Rural > Urban ........... a................b...............y > suburban...... o..............q..............z > Rural ............ 1-a-o .........1-b-q ....... 1-y-z A = 0.9 > O = 0.05 > B = 0.1 > Q = 0.7 > Y = 0.1 > Z = 0.1 Solve the problem knowing that the vector answer today is > Uo ............ 10 > (SUo) = (.... 40) > Ro ............ 50 What vector response to t = 2 vector and what is the answer in the > long run. Remember that > Ut > (SUt) = A.911 ^ tX1 + B.912 ^ tX2 + C.913 ^ tX3 This formula is incomprehensible. First, it contains the undefined objects e(umlaut)1^t, etc. Second, it contains a quantity 'C' that has not been defined before, etc. So, you should avoid special characters when posting to a newsgroup (not all newsreaders can handle them) and you should read and edit/correct your message before posting it. > Rt any bibliography suggestion? What about your course textbook? If you don't have such a book, try a Google search on 'Markov chain', to get numerous web pages that deal with problems much like yours. Anyway, you don't need to consult a book; you just need to know what it means to say that A = 0.9, etc. DO YOU KNOW WHAT IT MEANS? If you do (and if you know what it means to say that B = 0.1, etc.) then you can just apply these meanings to the so-called vector answer today to get the vector answer at t = 1, then re-apply to get it at t = 2, and so forth. I had to guess that by vector answer you mean population distribution. R.G. Vickson === Subject: Out of SCIENCE! posting-account=Xpph3AoAAABxBhdDsOjKGmFsT0t4mQEf 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.590; .NET CLR 3.5.20706),gzip(gfe),gzip(gfe) http://www.oyla15.de/userdaten/11969408/pdf/Edgar_Cayce_-_The_Lost_Teachings _Of_Atlantis_-_By_Jon_Peniel.pdf === Don't Use Google. is on Usenet. For that reason, and that it's the injection > vector for the gibberings of the vast majority of the idiots > filed, with a few cherry-picked gg posters reinstated. I have nothing at all to do with this site except for complete agreement: http://www.improve-usenet.org/index.html Phil -- -- Microsoft voice recognition live demonstration === On 23 Feb 2008 02:53:41 +0200, Phil Carmody I have nothing at all to do with this site except for complete >agreement: http://www.improve-usenet.org/index.html > Telling Google Groups users to use a different newsreader is missing a key point -- they must also find another news _provider_. Since Google Groups is free, an acceptable alternative news provider must be a _free_ news provider. Most ISPs provide a free usenet server, but few Google Groups users would even know enough to be able to access it. Moreover, many ISPs provide only a very incomplete selection of usenet posts, hence unacceptable. In addition, many usenet users are reading usenet from work or from other locations, hence are not even connected to their own ISP. For accessing usenet from their work computer, a user will typically have no option to install their own newsreader. As far as normal home usenet access, there do exist free public news providers, but assuming all Google Groups users agreed to switch, do you think any of the free public news providers (or all of them together) could handle that volume of users? Finally, even if you were able to convince Google Groups users that they should abandon Google in favor of some other news provider and some other newsreader, my guess is that the majority of Google Groups users would have no clue as to how to set up such a newsreader. Bottom line -- trying to convince Google Groups users to abandon Google Groups is futile unless, as a minimum, a viable alternative is suggested. Ok, so I'll suggest one ... Someone should write a newsreader program that provides a newsreader front end to the Google server. In other words, the Google server should not be able to tell that the user is not using the normal browser interface. To Google, it would appear that the user is using Google Groups through the normal web browser interface, when in actuality, the user is interfacing via a sane newsreader. quasi === > On 23 Feb 2008 02:53:41 +0200, Phil Carmody I have nothing at all to do with this site except for complete >agreement: http://www.improve-usenet.org/index.html > Telling Google Groups users to use a different newsreader is missing a > key point -- they must also find another news _provider_. I'm not missing that point at all. Can you show me any evidence in what I've posted to support your claim that telling them to find another news provider. I'll stick the in here now. Phil -- -- Microsoft voice recognition live demonstration === On 23 Feb 2008 10:02:17 +0200, Phil Carmody > On 23 Feb 2008 02:53:41 +0200, Phil Carmody >I have nothing at all to do with this site except for complete >agreement: >http://www.improve-usenet.org/index.html > > Telling Google Groups users to use a different newsreader is missing a > key point -- they must also find another news _provider_. I'm not missing that point at all. Can you show me any >evidence in what I've posted to support your claim that >telling them to find another news provider. I have no idea what you know or don't know. My reply to your post was not necessarily directed at you -- in fact, I was mainly responding to the content of the link which you provided: http://www.improve-usenet.org/index.html My main point is that the constant criticism of Google Groups users is in its own right) unless a viable alternative for those users is also suggested. quasi === <874pc07asq.fsf@nonospaz.fatphil.org> <87y79c5cdy.fsf@nonospaz.fatphil.org> <6hmvr35ag50o3l6g9uuk9slu6gbu9205ru@4ax.com> posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 4334.34; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) spider-dtc-te05.proxy.aol.com[CDBC7085] (Prism/1.2.1), HTTP/1.1 cache-dtc-ad05.proxy.aol.com[CDBC74C7] (Traffic-Server/6.1.5 [uScM]) > On 23 Feb 2008 10:02:17 +0200, Phil Carmody > On 23 Feb 2008 02:53:41 +0200, Phil Carmody >I have nothing at all to do with this site except for complete >agreement: >http://www.improve-usenet.org/index.html > Telling Google Groups users to use a different newsreader is missing a > key point -- they must also find another news _provider_. I'm not missing that point at all. Can you show me any >evidence in what I've posted to support your claim that >telling them to find another news provider. I have no idea what you know or don't know. My reply to your post was not necessarily directed at you -- in fact, > I was mainly responding to the content of the link which you provided: http://www.improve-usenet.org/index.html My main point is that the constant criticism of Google Groups users is > in its own right) unless a viable alternative for those users is also > suggested. quasi If you click the [XML] button on the Google Groups interface, you see this: Available feeds The feeds below allow you to read new topics or new messages from this group with any feed reading software. When a regularly updated site such as a Google Group has a feed, people can subscribe to it using software for reading syndicated content called a newsreader. RSS and Atom are two formats for feeds. Most feed readers support both. Atom 1.0 15 New messages 50 New messages 15 New topics 50 New topics RSS 2.0 15 New messages 50 New messages 15 New topics 50 New topics Feed-reading software * Web-based: Bloglines, Google Reader * Cross platform: BottomFeeder * Windows: FeedDemon, NewsGator * MacOS X: NetNewsWire Does this mean that Google can still be the news provider without having to use Google's web interface? If so, does anyone know how to set this up? Maybe such directions could be provided in these threads. === . > I have nothing at all to do with this site except for complete > agreement: > http://www.improve-usenet.org/index.html > Telling Google Groups users to use a different newsreader is missing a > key point -- they must also find another news _provider_. > I'm not missing that point at all. Can you show me any > evidence in what I've posted to support your claim that > telling them to find another news provider. I have no idea what you know or don't know. My reply to your post was not necessarily directed at you -- in fact, > I was mainly responding to the content of the link which you provided: http://www.improve-usenet.org/index.html My main point is that the constant criticism of Google Groups users is > in its own right) unless a viable alternative for those users is also > suggested. quasi I'm trying out NewsProxy with SeaMonkey Mail&News; maybe it's doing something good. I'm not sure yet. Also, I'm pretty satisfied with the $3.95 one-time payment account I have with teranews. David Bernier -- === Subject: www.apnadikhan.com posting-account=cbTYegoAAADG7FXL9ppbMjbX5CSsFGb- CLR 2.0.50727),gzip(gfe),gzip(gfe) www.apnadikhan.com www.deebyte.com www.bcrenvironmental.com www.apnaabbottabad.com www.pakjunction.com www.realwaronline.com www.peshawarpoint.com === S === Subject: Re: Surrogate factoring and helper primes > The real story is that I was finishing out the theory on > surrogate > factoring. > z^2 = y^2 mod T > I wondered a few years ago if there weren't another way where > you > could connect factorizations, and after years of research I've > found > that with > z^2 = y^2 + nT, > where n is some non-zero integer I can add a few more variables > using > z = x+.87k, so > (x+.87k)^2 = y^2 + nT > and if you have k and .87, and if k = 2.87x then you find > x^2 = y^2 + nT - (1 + .87^2)k^2. > You may recall I talked a lot recently about factors mod p, > where p is > an odd prime, as now is where helper primes come in allowing you > to > solve for k modulo p: > k^2 = (.87^2+1)^{-1}(nT) mod p. > So how did I introduce prime p? Actually with > 2.87x = k + pr_2 > so the full equation is actually > x^2 = y^2 + nT - (1 + .87^2)k^2 + kpr_2 > but if you find k such that the absolute value of nT - (1 + > .87^2)k^2 is > a minimum, then you minimize the absolute value of r_2 as well, > and > make it 0. > That maximal k for the minimum must also have 2.87 as a factor, > since > then > 2.87x = k > so k will always be even. > So the primes help you find the factorization and then vanish > without > a trace, which is why I call them helper primes. > To see one in operation, let T=119 and n=1, and p = 3. > Then > k^2 = (.87^2+1)^{-1}(nT) mod p = (.87^2+1)^{-1}(119) mod 3. > And .87 = 1 works giving k^2 = 1 mod 3, so I can use any integer k > coprime to 3. > And finding k such that the absolute value of 119 - 2k^2 is a > minimum > gives k=8. > And that was with k=1 mod 3. If that didn't work, then there is > k=-1 > mod 3 to check with as well. > The main correction here is emphasizing that x must be an integer. > Then I have x = 4, just like that, and substituting > 16 = y^2 + 119 - 128 = y^2 - 9 > so y = 5, and > z = x + .87k = 4 + 8 = 12, and z-y = 12-5 = 7, and z+y = 12+5 = > 17. > Notice that p=3 just helped and then vanished without a trace. > And yes, it IS possible than an RSA public key could be factored > by > p=3. > That may be surprising but the math now is elementary since you > can > just look back with the surrogate factoring answer. > Consider R and RSA public key, such that R mod 3 = 2. > Then you have from the surrogate factoring theory that k = 1 mod > 3, or > k = -1 mod 3 and .87 = 1, and k is found such that > R - 2k^2 > is a minimum and k is even, as that is required to make x an > integer. > Now here's the scarily simple proof that it will work. > Let z^2 = y^2 + R. > Then if z = x + .87k, and 2.87x = k, it is true then that > x = z/3 > so if z is divisible by 3, then x exists as an integer. > Now substituting > x^2 + 2.87kx + (.87k)^2 = y^2 + R, so > x^2 = y^2 + R - 2.87kx - (.87k)^2 > and I can get rid of x, just on the right side to have > x^2 = y^2 + R - (1 + .87^2)k^2 > Now if k is coprime to 3 and .87 = 1, > R - 2k^2 = 0 mod 3, so > x^2 = y^2 + R - 2k^2 > leaving just the result that k is the maximal values such that the > absolute value of R - 2k^2 is a minimum. > Remember k exists as an integer and x exists as an integer if z > has 3 > as a factor, and R mod 3 = 2. Since R is a public key it is odd > with > only two prime factors. > To finish the proof use 2.87x = k + 3j, instead above, and get to > x^2 = y^2 + R - 2k^2 - 3kj > and note when j is 0. The approach I used was to note that if you > assume you have an absolute minimum and increase or decrease k, > which > has to move mod 3, then j must increase, regardless of which > direction > you move in, so the minimum absolute value gives the correct k, as > long as that k is even, and you have two possibles: k = 1 mod 3 > and k > = -1 mod 3, so BOTH must be checked. > I don't know if that's subtle or not but I'll leave it for now. > Note then that ANY RSA public key that has a residue of 2 mod 3 > can be > attacked with this approach to factor it with p=3, where if z is > divisible by 3, it WILL fall. > But regardless, other primes can be used if those conditions > aren't > met. > The main point was to emphasize the power of this technique. > It CAN factor an RSA public key with p = 3, and very trivial > equations. > James Harris > Numbers for which this factoring scheme works > when p = 3 are of the form > (3k - y) * (3k + y). > For example: R = 11 * 31 = 341. The value of k > which minimizes abs(R - 2k^2) and is even and not > divisible by 3 is: k = 14. This yields x = 7 and > z = 21 and > x^2 = y^2 + R - 2k^2 reduces to > 49 = y^2 + 341 - 2*196, or > y^2 = 49 + 51 = 100, so y = 10, > and the factors of R are > (z - 10)*(z + 10) = (21 - 10)*(21 + 10) = 11 * 31. > So in this case again, the method works. > But: does the algorithm ALWAYS work for T of the > form (3*x - y)*(3*x + y)? > Yup, but I made a mistake in saying that the optimal k is always the > one that makes abs(T-2k^2) a minimum as it can be that value but must > be very close. > The reason it must be close is that it can be shown (trivially) that > x^2 = y^2 + T - 2k^2 - pr > when z=x+k, and 2x = k + pr, where the goal is r=0. > Now if you assume you have the correct k and r=0, then perturbing that > system will happen modulo 6, so you'd have > x^2 = y^2 + T -2(k+6j)^2 - pr > and r will tend to be negative, as the k^2 term will dominate. > It can be shown that the maximum k such that abs(T-2k^2) is a minimum > is the minimum value for k when k is positive (as why use a negative > k) and that k/6 is the maximum number of steps to a solution as you > move modulo 6. This is nonsense. It is trivial to find examples > where the value of k which works can be vastly > different from what you call optimal. This means > that you are not going to readily find the correct > factorization without an extensive search. Marcus. Yeah right, so why didn't you give any? Ok, the argument is oddly simple, with x^2 = y^2 + T -2(k+6j)^2 - 3r Correction as that should be x^2 = y^2 + T -2(k+6j)^2 - 3(k+6j)r. > multiplied out that's just x^2 = y^2 + T -2(k^2 + 12j + 36j^2) - 3r Same correction. and since T -2(k^2 +12j + 36j^2) - 3r is constant as x and y don't > change, it's just a fact that the 36j^2 term will dominate as you move > j in either direction, so r will tend to be negative to compensate. Correct. Though if you go far enough it can go back to positive as (k +6j) goes negative. But then you're at 0 anyway. So it's an easy argument. James Harris *** just solve for j to see what it does EZ *** === Subject: Re: What was Old Math? posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > When I listened to this song on new math , I think this is how I was > taught. So what was old math? http://www.youtube.com/watch?v=a81YvrV7Vv8 In the more complete version on the record Lehrer explains to some extent. The Old Math (well math at this point) was the teaching of rules to allow calculation. Hence to subtract one number from another you memorized and followed a recipe. In the 60's a new teaching philosophy was formed. Rather than teaching recipes without motivation, mathematics would be taught by teaching concepts, believed to be more important than manipulations (hence Lehrer's sardonic It's the thought that counts). There was a lot more to New Math than this (indeed some would claim that changing the way subtraction was taught was a very minor part) much having to do with teaching set theory. Overall the New Math was a failure, partly due to the fact that those whose job it was to teach it did not understand it. Still the new method of teaching subtraction stuck. Thus, unless you are old (old is defined as older than me) this is the method you will have been exposed to. - William Hughes === Subject: Re: What was Old Math? posting-account=PCksUwoAAAAsb8KSbS36lmWOqee5wTyG CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0),gzip(gfe),gzip(gfe) When I listened to this song on new math , I think this is how I was > taught. æSo what was old math? http://www.youtube.com/watch?v=a81YvrV7Vv8 In the more complete version on the record Lehrer explains > to some extent. teaching of rules to allow calculation. æHence to subtract > one number from another you memorized and followed a > recipe. æIn the 60's a new teaching philosophy was > formed. æRather than teaching recipes without motivation, > mathematics would be taught by teaching concepts, > believed to be more important than manipulations > (hence Lehrer's sardonic æIt's the thought that counts). > There was a lot more to New Math than this (indeed > some would claim that changing the way subtraction was > taught was a very minor part) much having to do > with teaching set theory. > to the fact that those whose job it was to teach > it did not understand it. > of teaching subtraction stuck. æThus, unless you are old > (old is defined as older than me) this is the method > you will have been exposed to. æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ - William Hughes new math as you point out, not the subtraction stuff as you point out. I think you are correct about subtraction too, because how he explains it in the song is how I understand it, and am comfortable with it. I was taught subtraction in late 60's. === Subject: Re: What was Old Math? Major Quaternion Dirt Quantum > mathematica is the greek name of quadrivium > inb Latin; it is the four classical subjects > of science, as opposed to the three Rs > of the trivium. when I moved to LA in the middle of the third grade, > I was abruptly exposed to the theory of sets. æso, > not very long ago, I realized that > this was a longterm result of the '60s l'ecole de Bourbaki, > that amazing general from Nancy, Francy, > who set out to re-axiomatize math > on the foundation of the theory of sets. so, that is the *real* reason, > why some people make fun of the French! thus: > it is somewhat more recondite, > to use the diameter = one; > thus, circumference & area are just pi. æthen, > you still have to explain, > why the volume is pi/6.... > anyway, the fact that > the area of the great circle is a quarter > of the sphere's, shows, it's (somehow) tetrahedral, > a la Buckafka Fullofitarians. thus quoth: > These gravitational redshift objects > are evenly distributed throughout the universe. > Quasar's are compact objects > about one light week in diameter. The close ones' > redshifts are mostly from their gravitation. --Dick Cheeny, National Treasure: But Cheney was such a science ignoramous, we had to teach the fool that it doesn't matter how finite the Earth is, you can still pick up Haliburton morons drilling oil on WKKK. Thus the development of flying robots for the stooge. > Run, Trickier Dick -- > Run for Indy superVeep!... > superVeep Al Gore, Best Actor, > Occidental Dinoleum Awards! === Subject: Re: What was Old Math? > When I listened to this song on new math , I think this is how I was > taught. So what was old math? http://www.youtube.com/watch?v=a81YvrV7Vv8 Being from the Netherlands, I have no clue as to what the `Old Math' in America was, but I can tell you about the difference as to what it was here, compared to the video. It is not so much `new' math, but a `new way' to math. That is, what is done is exactly the same, but in a different manner. I'll show you an addition example `old-style' and `new-style' Old Style: 1 1 1 324 324 324 324 249 --> 249 --> 249 --> 249 ---+ ---+ ---+ ---+ 3 73 573 (about it) New Style: 324 + 249 = 324 + 6 + 243 = 330 + 243 = 330 + 240 + 3 = 370 + 200 + 3 = 570 + 3 = 573 Not a different result, but a different way of achieving it. More and more time is now spent on getting the children to think of funny ways to do addition, multiplication, subtraction, division, while they could have learned a standard algorithm for it instead, which is faster, uses less space (save the trees, dude!), and get on with doing some real work. A few of my relatives are being taught similar to the new math represented. The `new Math' is merely another way of teaching math. And -- in my opinion -- a worse method. Just my opinion though. Sjoerd Job === Subject: Re: What was Old Math? > When I listened to this song on new math , I think this is how I was > taught. æSo what was old math? http://www.youtube.com/watch?v=a81YvrV7Vv8 Being from the Netherlands, I have no clue as to what the `Old Math' in > America was, but I can tell you about the difference as to what it was here, > compared to the video. It is not so much `new' math, but a `new way' to math. That is, what is done > is exactly the same, but in a different manner. I'll show you an addition > example `old-style' and `new-style' Old Style: æ æ æ æ æ 1 æ æ æ æ1 æ æ æ æ1 > 324 æ æ æ324 æ æ æ324 æ æ æ324 > 249 æ--> 249 æ--> 249 æ--> 249 > ---+ æ æ ---+ æ æ ---+ æ æ ---+ > æ æ æ æ æ æ3 æ æ æ 73 æ æ æ573 (about it) New Style: > 324 + 249 = 324 + 6 + 243 = 330 + 243 = > æ 330 + 240 + 3 = 370 + 200 + 3 = 570 + 3 = 573 Not a different result, but a different way of achieving it. Cool. But what's the justification for doing it that way as opposed to doing it this way (which is shorter)? 324 + 249 = 323 + 1 + 249 = 323 + 250 = 320 + 250 + 3 = 570 + 3 = 573 > More and more > time is now spent on getting the children to think of funny ways to do > addition, multiplication, subtraction, division, while they could have learned > a standard algorithm for it instead, which is faster, uses less space (save > the trees, dude!), and get on with doing some real work. A few of my relatives are being taught similar to the new math represented. > The `new Math' is merely another way of teaching math. And -- in my opinion -- > a worse method. Just my opinion though. Sjoerd Job === Subject: Re: What was Old Math? Cool. But what's the justification for doing it that way > as opposed to doing it this way (which is shorter)? 324 + 249 = 323 + 1 + 249 > = 323 + 250 > = 320 + 250 + 3 > = 570 + 3 > = 573 Dunno, just a matter of me not seeing it I suppose. It's always a fiddle to find a short method, and I was kind of satisfied enough. And, I mostly work from left-to-right. Just a bad habit I think. The regrouping takes time on paper. The `standard algorithm' doesn't cost as much in my experience. Now, the `new math' way of doing multiplication? Yikes. There are several ways. I once saw a movie on YouTube about the new math (it's in the reference list of this movie too, I saw) where a weather forecaster explains a lot about it. Quite interesting. The point was: There's no point in teaching the children quick methods, as they have the calculator for that. Some of the people in my class have had some `reformed' education in maths, which means no long division or something like that. Most I know grab the calculator for things like 2 + 4. (Not a figure of speach. They actually grab the calculator, type it in, and read the result. That's not what I can call `improvement', to be honest. On the other hand, my niece of about 8yrs old can do 2 + 4 off the top of her head, and quite a lot more difficult stuff. But, she can't do 2 - 4 yet. No negative numbers. I remember working with those when I was about that age. Now, I know it's a subject of the schoolbooks for kids of around 13. Yes, it takes a full 8 years to go from + to -. It sounds silly, I know. It's the way it is. Sjoerd Job === Subject: Re: What was Old Math? posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) On Feb 22, 6:30æpm, Sjoerd Job Postmus Cool. But what's the justification for doing it that way > as opposed to doing it this way (which is shorter)? 324 + 249 = 323 + 1 + 249 > æ æ æ æ æ = 323 + 250 > æ æ æ æ æ = 320 + 250 + 3 > æ æ æ æ æ = 570 + 3 > æ æ æ æ æ = 573 Dunno, just a matter of me not seeing it I suppose. It's always a fiddle to > find a short method, and I was kind of satisfied enough. And, I mostly work > from left-to-right. Just a bad habit I think. The regrouping takes time on > paper. The `standard algorithm' doesn't cost as much in my experience. I assumed one tried to form 0's and avoid carries. But it sounds daunting if there's more than two numbers involved. Dammit, now I'm going to have to write a new math addition algorithm that sums arbitray quantities of 3-digit numbers. That ought to be fun. Now, the `new math' way of doing multiplication? Yikes. There are several > ways. I once saw a movie on YouTube about the new math (it's in the reference > list of this movie too, I saw) where a weather forecaster explains a lot about > it. Quite interesting. The point was: There's no point in teaching the > children quick methods, as they have the calculator for that. Some of the people in my class have had some `reformed' education in maths, > which means no long division or something like that. Most I know grab the > calculator for things like 2 + 4. (Not a figure of speach. They actually grab > the calculator, type it in, and read the result. æThat's not what I can call > `improvement', to be honest. As a database manager/programmer, I'm deathly afraid of boundary conditions and often count on my fingers (is it the objects or the gaps between them that the count refers to?) On the other hand, my niece of about 8yrs old can do 2 + 4 off the top of her > head, and quite a lot more difficult stuff. But, she can't do 2 - 4 yet. No > negative numbers. I remember working with those when I was about that age. > Now, I know it's a subject of the schoolbooks for kids of around 13. Yes, it > takes a full 8 years to go from + to -. It sounds silly, I know. It's the way > it is. I don't even try to use Excel's R1C1 addressing with negative indexing to reach left or up. I always find myself switching to A1 addressing. Negative indexes are easy to understand, it's I can look at R[-3]C94 and it's meaning escapes me (unless I think about it hard and usually don't have the time). Sjoerd Job === Subject: Re: What was Old Math? >I assumed one tried to form 0's and avoid carries. But it sounds >daunting if there's more than two numbers involved. I grew up right around the time they introduced new math in schools. The version I remember is, it's more of stressing how addition and subtraction worked rather than just memorizing things (and in the meantime, you still had to memorize the multiplication tables). There was a lot of counting things in powers of 10. For example: 324 + 249 = (300 + 20 + 4) + (200 + 40 + 9) = (300 + 200) + (20 + 40) + (4 + 9) = (300 + 200) + (20 + 40) + 13 = (300 + 200) + (20 + 40) + (10 + 3) = (300 + 200) + (20 + 40 + 10) + 3 = 500 + 70 + 3 = 573 Why, yes, this IS the same as the old carry 10s method, but explained differently - how about that? -- Don === === === Subject: Ann: FriCAS 1.0.2 released FriCAS 1.0.2 has been released FriCAS is an advanced computer algebra system. Its capabilities range from calculus (integration and differentiation) to abstract algebra. It can plot functions and has integrated help system. FriCAS 1.0.2 should build on Linux, many Unix like systems (for example Mac OSX and Solaris 10) and Windows. FriCAS is build on top of Common Lisp; several Lisps can compile and run FriCAS (GCL, SBCL, Clisp, ECL and OpenMCL). Project page: http://fricas.sf.net Download page: http://sourceforge.net/project/showfiles.php?group_id=200168 Mailing list. Please sign up before posting a message. Notable changes (compared to 1.0.1 version) include: - ')nopiles' command gives conventional syntax - added pfaffian function - ECL support - Graphics and Hyperdoc work using openmcl or ECL - Output may be now delimited by user defined markers - Experimental support for using as a Lisp library - Spad compiler is now significantly faster - Several bug fixes -- Waldek Hebisch hebisch@math.uni.wroc.pl === Subject: Re: how much Internet traffic is there? > An interesting aside is that a 1.44 MB floppy is actually 1440 kibibytes > and not 1.44 * 10^6 or 1.44 * 2^20 bytes. Hehe, yes. It would be 1.44 kilo-kibi-bytes if such a mangled unit should be permitted to exist. It was a doubling of data capacity from the previous 720 kiB disks. - Tim === Subject: Re: how much Internet traffic is there? > I read that Youtube uses up to 1 petabyte per month of upload/download > bandwidth. > A petabyte is 10^15 bytes, but others might say 1024^5 bytes. Most of the time, a -byte should be considered as powers of 2 > instead of 10. Kibibyte, mebibyte, etc. are sometimes used to denote > this, but the normal prefixes are much more prevalent. An interesting aside is that a 1.44 MB floppy is actually 1440 kibibytes > and not 1.44 * 10^6 or 1.44 * 2^20 bytes. Unfortunately, it seems Seagate, who make hard drives, have a different usage: One gigabyte, or GB, equals one billion bytes and one terabyte, or TB, equals one trillion bytes when referring to hard drive capacity. Source: < http://www.seagate.com/docs/pdf/datasheet/disc/ds_barracuda_es_2.pdf > -- === Subject: Re: how much Internet traffic is there? > One gigabyte, or GB, equals one billion bytes and one terabyte, > or TB, equals one trillion bytes when referring to > hard drive capacity. This has been the internationally accepted meaning for the prefix in the computer industry according to the IEC for a decade, adopted by the IEEE years ago, endorsed by the CIPM, and recommended by the US NIST. If a power of 2 is intended, the standardized units are kibibyte (kiB), mebibyte (MiB), gibibyte (GiB), tebibyte (TiB), pebibyte (PiB), exbibyte (EiB), zebibyte (ZiB) and yobibyte (YiB). The only remaining field where it remains standard for SI prefix letters to be used for binary powers is in solid-state computer memory under JEDEC, and even then notes that data rates are to be interpreted as in SI, and that the units are not written in words. I.e. the measure 1 MB/s is 10^6 bytes per second. 1 MB is 2^20 bytes. The word megabyte is not defined under the standard. - Tim === Subject: Re: how much Internet traffic is there? > I read that Youtube uses up to 1 petabyte per month of upload/download > bandwidth. > A petabyte is 10^15 bytes, but others might say 1024^5 bytes. Most of the time, a -byte should be considered as powers of 2 > instead of 10. Kibibyte, mebibyte, etc. are sometimes used to denote > this, but the normal prefixes are much more prevalent. If that's how most computer scientists, code gurus and the like use mega/giga/tera/peta in combination with byte, I think that should be the #1 entry in a dictionary: megabyte (noun) #1 A unit of memory/storage/information in high-tech equal to 1,048,576 bytes. Symbol: MB (or Mb ?). #2 One million bytes (obsolete, old-style, ???). Then the M in MB wouldn't mean a million as in the metric system. The twelve Mega Millions states are California, Georgia, Illinois, [...] So Mega+byte for 2^20 bytes, or even MB for 2^20 bytes, should be fine in the long run. Telling the usage to dictionary-compilers now, as opposed to in twenty years, would surely save a few trillion dollars in this century, right ??? [ I'll have to search for exabyte, 1024 petabytes ... ] > An interesting aside is that a 1.44 MB floppy is actually 1440 kibibytes > and not 1.44 * 10^6 or 1.44 * 2^20 bytes. I was never sure about that. -- === === === Subject: JSH: Test factorization posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) I've modified one of my existing programs to start testing out the latest surrogate factoring research though I haven't yet optimized it, so it kind of dumbly just looks for solutions around k approximately equal sqrt(nT/2) where n=1 if T mod 3 = 2, and n=5, if T mod 3 = 1. Here's an example factorization: T = 1342517983, k = 58480. surrogate = 127230885, which factors as (3^3)(5)(449)(2099) and T factors as (27893)(48131). The prime factors of the surrogate are of interest here and since T mod 3 = 1, the program multiplies it by 5, so k^2 = 2^{-1}(5T) mod p where you can check it for each prime. I haven't bothered to check. Because the program is dumb, it took 183 checks of k's looping up from k approximately equals sqrt(5T/2), skipping over odd k's or k's divisible by 3. Notice that k/2p approximately equals 14 using the largest prime, so you'd have a roughly a 1/14 chance of finding a solution, if you did it the smarter way, but that gave a good enough chance that even the dumb way stumbled across the factors. It's not a complicated idea here, which is why it's amazing to me there are still people trying to argue over some rather basic algebra. It just so happens that if you let k = 2x, and z = x+k, when z^2 = y^2 + nT then the maximum k that will give the minimum value for abs(nT - 2k^2) will tend to be close to the correct k, which must exist if z is divisible by 3 because z = x + k = x + 2x = 3x. Figuring that out just requires using 2x = k + pr where p is some prime and r is an integer, and the substituting out z, with z = x+k, to get if you substitute out z and simplify a bit you have x^2 = y^2 + nT - (2xk + k^2) so you can substitute out 2x, and get x^2 = y^2 + nT - 2k^2 - kpr and now let k_0 be the value for which r=0, so you can let k = k_0 + 2pj where j is an integer and substitute, and you have x^2 = y^2 + nT - 2(k_0^2 + 4pjk_0 + 4p^2j^2) - (k_0 + 2pj)pr and x, y, nT, k_0 and p are all constant, so as j varies, the j^2 term will dominate and the r variable will tend to be negative to counterbalance it. If you need it all multiplied out to help you with this basic point: x^2 = y^2 + nT - 2k_0^2 - 8pjk_0 - 8p^2j^2 - (k_0 + 2pj)pr where with k_0 positive (as why have it negative?), you'll notice that while k_0 >2pj the negativity of 8p^2j^2 can't be overridden by -8pjk_0 no matter what the sign of j, but y^2 + nT - 2k_0^2 is constant as is x^2, so r must be negative to compensate until k_0 = 2pj, which is when k=0 anyway. So it's trivial mathematics that k_0 will tend to be near the correct answer for k, when it is the maximum value such that abs(nT - 2k^2) is a minimum. That amazing bit of mathematics puts the factoring problem within reach, just like that just because the j^2 is always positive. Trivial algebra gives you the range as k should be equal to or greater than k_0, and j should be negative and greater than -k/2p. Figuring out that k^2 = 2^{-1}(nT) mod p, is a little more complicated but if you were paying attention when I was babbling on about factors mod p, I explained it exhaustively and ad nauseum. So, oddly enough, to tackle a composite T, you just need to get a a prime for which k exists that makes it very likely that you will find k quickly. And it's all trivial algebra. Now if you people wish to argue on still and wait until I or someone else is motivated to fully implement the trivial algebra, then fine. But don't come crying later when I say you people don't really know math, as then you clearly don't. Easy algebra ignored when it's the factoring problem does not make you brilliant. It does still annoy me and I still wonder how I let some of you bother me, that you can pretend to give a damn about mathematics and come out with sophistry to attack a beautiful and simple argument, proving how much you hate math, but having the gall to keep at it as if you can just fool people one more day, that's all that matters. I think some of you every day you post arguing with me just tell yourself, to just try to get people to believe wrong things mathematically one more day, and you win, as you make humanity as a whole lose. One more day yesterday you people won. Did you win today? Is humanity still being fooled by you? James Harris === Subject: Re: JSH: Test factorization >I've modified one of my existing programs to start testing out the >latest surrogate factoring research though I haven't yet optimized it, >so it kind of dumbly just looks for solutions around k approximately >equal sqrt(nT/2) where n=1 if T mod 3 = 2, and n=5, if T mod 3 = 1. Does around mean above k, below k or both? Here's an example factorization: T = 1342517983, k = 58480. surrogate = 127230885, which factors as How is the surrogate calculated? Your latest descriptions on usenet and in your blog do not seem to have a formula for the surrogate. Your previous version called it S, but I cannot find any mention of any variable S in your latest postings. (3^3)(5)(449)(2099) and T factors as (27893)(48131). How do we get from the value of k and/or the surrogate to either of the factors of T? An explicit formula would help or alternatively the relevant part of your test program. >The prime factors of the surrogate >are of interest here and since T mod 3 = 1, the program multiplies it >by 5, so k^2 = 2^{-1}(5T) mod p Which of these terms is a prime factor of the surrogate? p? I thought p was one of the free choices in your method. where you can check it for each prime. I haven't bothered to check. Because the program is dumb, it took 183 checks of k's looping up from >k approximately equals sqrt(5T/2), skipping over odd k's or k's >divisible by 3. Notice that k/2p approximately equals 14 using the largest prime, so >you'd have a roughly a 1/14 chance of finding a solution, if you did >it the smarter way, but that gave a good enough chance that even the >dumb way stumbled across the factors. Is there an explanation of the smarter way somewhere? It's not a complicated idea here, which is why it's amazing to me >there are still people trying to argue over some rather basic algebra. It just so happens that if you let k = 2x, and z = x+k, when z^2 = y^2 + nT then the maximum k that will give the minimum value for abs(nT - 2k^2) >will tend to be close to the correct k, which must exist if z is >divisible by 3 because z = x + k = x + 2x = 3x. Above you were constraining k to be near sqrt(nT/2), now you are constraining k to be near giving a minimum value for abs(nT - 2k^2). In general it is difficult to satisfy two independent constraints with a single variable. If the constraints are not independent then one or other of them add nothing to the method. When programming to pick k which constraint should I use? rossum >James Harris === Subject: Re: JSH: Test factorization > I've modified one of my existing programs to start testing out the > latest surrogate factoring research though I haven't yet optimized it, > so it kind of dumbly just looks for solutions around k approximately > equal sqrt(nT/2) where n=1 if T mod 3 = 2, and n=5, if T mod 3 = 1. Here's an example factorization: Yay, finally an example of a factorization. > T = 1342517983, k = 58480. surrogate = 127230885, which factors as (3^3)(5)(449)(2099) How did you factor the surrogate? Did k and the surrogate pop out of thin air? > and T factors as (27893)(48131). The prime factors of the surrogate > are of interest here and since T mod 3 = 1, the program multiplies it > by 5, so k^2 = 2^{-1}(5T) mod p where you can check it for each prime. I haven't bothered to check. I haven't bothered to check implies to me that you gave up on your program. > Because the program is dumb, it took 183 checks of k's looping up from > k approximately equals sqrt(5T/2), skipping over odd k's or k's > divisible by 3. Notice that k/2p approximately equals 14 using the largest prime, so > you'd have a roughly a 1/14 chance of finding a solution, if you did > it the smarter way, but that gave a good enough chance that even the > dumb way stumbled across the factors. It's not a complicated idea here, which is why it's amazing to me > there are still people trying to argue over some rather basic algebra. What happened to the factorization you were doing? I so far sort of understand that you check every reasonable value of k through some arbitrary stopping point, but that still doesn't tell me how to get those factors. > And it's all trivial algebra. And pulling out variables from a hat. If it's really trivial algebra, you should be able to provide a full, step-by-step explanation of a factorization. This doesn't even come close. P.S. Try to keep your subjects a bit more topical. I want to see a factorization, not a poorly-written explanation that attacks the community at large for not believing your word. === Subject: Re: JSH: Test factorization posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I've modified one of my existing programs to start testing out the > latest surrogate factoring research though I haven't yet optimized it, > so it kind of dumbly just looks for solutions around k approximately > equal sqrt(nT/2) where n=1 if T mod 3 = 2, and n=5, if T mod 3 = 1. Here's an example factorization: Yay, finally an example of a factorization. T = 1342517983, k = 58480. surrogate = 127230885, which factors as (3^3)(5)(449)(2099) How did you factor the surrogate? Did k and the surrogate pop out of > thin air? I explained that in the post, as I found a k approximately equal to sqrt(5T - 2k^2), where the 5 is to force 2 mod 3, as 5T mod 3 = 2. Then I incremented k by two's until T factored. > and T factors as (27893)(48131). The prime factors of the surrogate > are of interest here and since T mod 3 = 1, the program multiplies it > by 5, so k^2 = 2^{-1}(5T) mod p where you can check it for each prime. I haven't bothered to check. I haven't bothered to check implies to me that you gave up on your > program. I haven't bothered to check that k^2 = 2^{-1}(5T) mod p, where p is each of the primes of the surrogate. That is, p is from the set {3, 5, 449, 2099}. A key point I make is that you can find p just by having k exist, which is means that looping out and finding primes is the big thing but you don't have to be too picky as about 50% of them should work. > Because the program is dumb, it took 183 checks of k's looping up from > k approximately equals sqrt(5T/2), skipping over odd k's or k's > divisible by 3. Notice that k/2p approximately equals 14 using the largest prime, so > you'd have a roughly a 1/14 chance of finding a solution, if you did > it the smarter way, but that gave a good enough chance that even the > dumb way stumbled across the factors. It's not a complicated idea here, which is why it's amazing to me > there are still people trying to argue over some rather basic algebra. What happened to the factorization you were doing? I so far sort of > understand that you check every reasonable value of k through some > arbitrary stopping point, but that still doesn't tell me how to get > those factors. > x^2 = y^2 + nT - 2k^2 That is I think in there as well. My program factors nT - 2k^2 to get y and x, and then z = x + k, but k = 2x, so z = 3x, and it checks z - y and z+y for factors in common with T. > And it's all trivial algebra. And pulling out variables from a hat. If it's really trivial algebra, > you should be able to provide a full, step-by-step explanation of a > factorization. This doesn't even come close. P.S. Try to keep your subjects a bit more topical. I want to see a > factorization, not a poorly-written explanation that attacks the > community at large for not believing your word. I put in enough detail, as I explained how I got k, I explained how I changed k, and later gave the full underlying mathematics. Ok, so maybe filling in some of the blanks was hard for you, but I did that in this reply and it wasn't hard as I mostly noted what I'd already said. So that it is all in one place, here it is again: Given an odd target composite T coprime to 3, if T mod 3 = 2, then fine, but if not, you use 5T, and you find k using k = floor(sqrt(T/ 2)) or k = floor(sqrt(5T)/2), as that is the lazy way to get close to the optimal k, and the easiest. The smarter way is to use k^2 = 2^{-1}(nT) mod p where p is a prime of your choice. And n=1 if T mod 3 = 2, and 5 if not. I gave a factorization of T = 1342517983 where a computer program went to k = floor(sqrt(5T/2)) and incremented up by 2's until it factored. And I noted that the prime factors of 5T - 2k^2 would work with k^2 = 2^{-1}(5T) mod p where that set is {3, 5, 449, 2099}. But I didn't bother to actually check, as I just know what the math says. The biggest prime: 2099, is the reason the number factored anyway in such a small search space. The bigger the primes you use, the faster it factors. And that is relevant and very topical as if everything I'm saying is true, then you can factor an arbitrarily large composite by finding big primes p, and getting k, and searching mod p around k's such that abs(nT - 2k^2) is a minimum. And if all of that is true then that is a solution to the factoring problem so if that is not topical on sci.crypt then nothing is. James Harris === Subject: Re: JSH: Test factorization posting-account=WlifZwoAAADn4Qc008FhhuRE4Syn8J58 3.011; SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) I explained that in the post, as I found a k approximately equal to > sqrt(5T - 2k^2), where the 5 is to force 2 mod 3, as 5T mod 3 = 2. > This part baffles me. How can k be approximately equal to two different numbers? === Subject: Re: JSH: Test factorization >I've modified one of my existing programs to start testing out the >latest surrogate factoring research though I haven't yet optimized it, James, I've never seen you announce or post program code, here. Doing so may demonstrate to your detractors that you're working on the problem in a more pragatic way, than just announcing it all to be cannot-be-faulted but with just one-more-tiny-fault every second day, simple math whose beauty is staring us all in the face. Think of all those rosey-faced undergrads reading here whose lives may be improved immeasurably through the insight revealed by your code. And this paper that you state was published - can you please provide a citation so that we may read it? Without a refereed and citable source, it was never published, just submitted for publication. -- Chris. === === === === === Subject: Math Trek: An Attack on Fermat [MATHTREK] An Attack on Fermat The first female research mathematician had a program to solve Fermat's Last Theorem, and it was almost lost to history. === Subject: Re: Math Trek: An Attack on Fermat Sam Wormley a .8ecrit : > [MATHTREK] > An Attack on Fermat > The first female research mathematician had a program to > solve Fermat's Last Theorem, and it was almost lost to history. Actually, the readers of this forum would probably be more interested by the original paper : http://www.math.nmsu.edu/%7Edavidp/germain.pdf === Subject: Re: Algebra Project ideas??? > Was wondering if anyone had some good ideas for some Algebra 1 > projects. I have used the same ones for years and I am looking for > something new. Any idea would be grand, but projects on graphing, ratios, area, and > expoential growth would be very helpful. > have them determine the value of pi. have them determine why the orbit of the planets must be an ellipse. have them determine higher order relationships in nature (x^2, x^3, e^a, e^t....) have them plot the elevation of the grounds around the school or on their path to & from school. have them plot the days of sun or temperature at noon over a month, find average, std deviation, etc. === Subject: Relativist swindlers against Newtonist science posting-account=1qbAGAkAAADcUtlizzXUEb5jUjfAdE2y Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > http://tinyurl.com/2wxzv8 Courage : One more ignorant ! They believe in a kind of religion where Isaac Newton is one of the prophets and Albert Einstein is the messyThey believe in a kind of religion in which Isaac Newton is one of the prophets, and Albert Einstein is the Messiah. But the prophet was a little deaf to hear the regulations of his god. When it was necessary to go on water in 1919, the Eddinton swindler made pretence have lost the results of the team of Sobral. All the results were faked! But the newspapers, handled by powerful lobbys shouted in heart with the miracle!...... Ils croient en une sorte de religion dans laquelle Isaac Newton est l'un des proph.8ftes, et Albert Einstein est le messie. Mais le proph.8fte .8etait un peu sourd pour entendre les prescriptions de son dieu. Quand il fallut marcher sur l'eau en 1919, l'escroc Eddinton fit semblant d'avoir perdu les r.8esultats de l'.8equipe de Sobral. Tous les r.8esultats furent truqu.8es ! Mais les journaux, manipul.8es par des lobbys puissants cri.8frent en coeur au miracle .... All was only vast sweet talk to ensure, initially, the anglo-germantic reconciliation. Of that, the henchmen acknowledged! But the real issue was much deeper, much more crucial for this clicks swindlers: It was a question of ratifying the intellectual defeat - ideological -, the defeat of the materialism, the defeat of philosophy materialist, at the exact moment when the revolution of octore had just reversed the first capitalist government of the human history. Tout n'.8etait que vaste baratin pour assurer, en premier lieu, la r.8econciliation anglo-germantique. De cela, les supp.99ts ont avou.8e ! Mais le v.8eritable enjeu .8etait bien plus profond, bien plus crucial pour cette clique d'escrocs : Il s'agissait de ratifier la d.8efaite intellectuelle - id.8eologique - ,la d.8efaite du mat.8erialisme, la d.8efaite de la philosophie mat.8erialiste, au moment pr.8ecis o.9d la r.8evolution d'octore venait de renverser le premier gouvernement capitaliste de l'histoire humaine. In the same way that Adam Smith became, for the capitalist class, to throw good to the dustbin, Newton became the large intruder of the operation. These swindlers thus built a pseudo-Newtonian mythology: science became a Bible.... But after they have expurger teaching, the principal message of Newton, the base of its theory: absolute space, 'spaces objective; absolute places, absolute d.8epalcements, absolute velocities. De la m.90me fa.8don qu'Adam Smith devenait , pour la classe capitaliste, bon .88 jeter .88 la poubelle, Newton devenait le grand g.90neur de l'operation. Ces escrocs ont donc construit une mythologie pseudo-newtonienne :la science est devenue une Bible.... Mais apr.8fs qu'ils aient expurger de l'enseignement, le principal message de Newton, le socle de sa th.8eorie : l'espace absolu, 'espace objectif; les lieux absolus, les d.8epalcements absolus, les vitesses absolues. Newton, transformed into ox, they had the free field. They could begin the psychological warfare: they were not likely any more to see middle- class youth becoming intelligent, becoming, critical, becoming rebellious: the Berkeley bishop controlled ALL!! Imbecile Michelson (the cretin which did not have notices one century after Herschel that the Sun advanced), http://en.wikipedia.org/wiki/William Herschel#Further discoveries this Michelson imbecile had - supposedly - rule the problem: the absolute vitsses, objective speeds, did not exist!!! this cretin Newton, transform.8e en boeuf, ils avaient le champ libre. Ils pouvaient commencer la guerre psychologique : ils ne risquaient plus de voir la jeunesse bourgeoise devenir intelligente, devenir, critique, devenir rebelle : l'.8ev.90que Berkeley contr.99lait TOUT !! L'imb.8ecile Michelson (le cr.8etin qui n'avait pas remarque un si.8fcle apr.8fs Herschel que le Soleil avancait), http://fr.wikipedia.org/wiki/Apex %28astronomie%29#William Herschel cet imb.8ecile Michelson avait - soi-disant - r.8fgl.8e le probl.8fme : les vitsses absolues, les vitesses objectives, n'existaient pas !!! The scientific theory of Newton was found reduced with a whole of sourates fixed for eternity: the Koranic school of the relativists ensured the world middle-class one century of respite: It was going to benefit from it to destroy our planet by emitting 200 billion tons of carbon to start the most gigantic catastrophe since milliions of years. La th.8eorie scientifique de Newton se retrouvait r.8eduite .88 un ensemble de sourates fig.8ees pour l'.8eternit.8e : l'.8ecole coranique des relativistes assurait .88 la bourgeoisie mondiale un si.8fcle de r.8epit : Elle allait en profiter pour d.8etruire notre plan.8fte en .8emettant 200 miliards de tonnes de carbone pour d.8eclencher la plus gigantesque catastrophe depuis des milliions d'ann.8ees. The parrots cretins repeat the freedom which they thought of having acquired for eternity, of the free choice of the reference frame. These poor imbeciles, sucesseurs of the Michelson imbecile and the cretin positivist Einstein, still dare to come to insult us on these forums! The poor religious bigots of a religion which will crumble in the few years which arrive. Your nest egg, your fats wages, based on the imposture and the naivity of your listeners, will be worth you inculpation for swindle in organized band Les perroquets cr.8etins r.8ep.8ftent la libert.8e qu'ils pensaient avoir acquise pour l'.8eternit.8e, du libre choix du r.8ef.8erentiel. Ces pauvres imb.8eciles, sucesseurs de l'imb.8ecile Michelson et du cr.8etin positiviste Einstein, osent encore venir nous insulter sur ces forums ! Pauvres bigots d'une religion qui va s'effondrer dans les quelques ann.8ees qui arrivent.Votre magot, vos gras salaires, bas.8es sur l'imposture et la na.95vit.8e de vos auditeurs, vous vaudront inculpation pour escroquerie en bande organis.8ee Yanick Toutain === === === Subject: One of my conjecture in analysis posting-account=cEKB4goAAABJ5eifbZgDmj5s5T5PQpjx EmbeddedWB 14.52 from: http://www.bsalsa.com/ EmbeddedWB 14.52),gzip(gfe),gzip(gfe) Let f : [a,b] -> R be a continuous function which is convex. Then the points, at which f is not differentiable, are countable and not dense. What do you think about it? === Subject: Re: One of my conjecture in analysis posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Let f : [a,b] -> R be a continuous function which is > convex. Then the points, at which f is not differentiable, > are countable and not dense. Yes, except for the not dense part. In fact, if I is an open interval and D is any countable subset of I, then there exists a function f such that f is convex on I and f is not finitely differentiable at each point of D. Here's something I posted in another group on 14 June 2001 and which I re-posted in sci.math on 28 December 2002: DEFINITION: Let f be defined on an interval I. We say that f is convex on I if whenever x1, x2 belong to I, then the line segment whose endpoints are (x1,f(x1)) and (x2,f(x2)) lies on or above {(x,f(x)): x in [x1, x2]}. If on or above is strengthened to strictly above, we get a geometric condition for concave up. Many of the results given below continue to hold when f is concave up. Some of these will be automatic (e.g. when the hypothesis includes convex, since concave up implies convex) and some of these will continue to hold for other reasons. However, I don't really have the time or desire right now to try and sort out which continue to hold when convex is replaced with concave up. Here is another characterization of convex functions. THEOREM : A function f is convex on an interval I if and only if the following condition holds: Whenever x1 < x2 < x3 belong to I, then [f(x2) - f(x1)] / (x2 - x1) < or = [f(x3) - f(x2)] / (x3 - x2). In other words, the average rate of change of f on [x1, x2] does not exceed the average rate of change of f on [x2, x3]. ---------- SOME RESULTS ---------- Convex functions have many applications, both in pure mathematics and in applied mathematics -- Many useful inequalities, including the arithmetic and geometric mean inequality, can be obtained very easily using convex functions. In recent years there has been a lot of interest in convex functions defined on Banach spaces, especially in their differentiability properties. Finally, convex functions play a crucial role in linear programming and in optimization theory. For differentiability +of convex functions (quotes included), convex functions optimization (quotes NOT included), and convex functions linear programming (quotes NOT included). 1. If f is convex on an open interval I, then f is continuous at each point in I. 2. If f is convex on an open interval I, then f is Lipschitz continuous on each closed subinterval of I. [Lipschitz continuous means the difference quotients are bounded.] This strengthens #1. 3. If f is convex on an open interval I, then there are at most countably many points at which f is not differentiable. Any countable set can be the set of non-differentiability points for some convex function. Curiously, I couldn't find this in any of the real analysis texts I looked at. However, this is a special case of theorem 4.20 on p. 93 of [1]. 4. Assume that f is convex on an open interval I. Then, at each point of I, both the left derivative of f and the right derivative of f exist. This strengthens #3. [It can be shown that this property implies the property given in #3, but not conversely.] 5. If f is convex on an open interval I, then the left derivative of f is a non-decreasing function, the right derivative of f is a non-decreasing function, and at each point the left derivative is less than or equal to the right derivative. (See [7], p. 109.) This strengthens #4. 6. If f is convex on an open interval I, then the second derivative of f exists at every point of I except for a set of measure zero. [This follows from #3, #5, and the fact that monotone functions are differentiable almost everywhere.] 7. If f is convex on an open interval I, and g is either the left derivative of f or the right derivative of f (it doesn't matter which one you let g be), then f(b) - f(a) = integral of g on the interval [a,b] for all a,b in I. [Monotone functions are Riemann-integrable, so this is the usual calculus integral.] 8. Suppose f'' exists at each point of an open interval I. Then f is convex on I if and only if f''(x) is nonnegative for each x in I. 9. Suppose f is continuous. Then f is convex on an open interval I if and only if limit as h --> 0 of [ f(x+h) + f(x-h) - 2*f(x) ] / (h^2) is nonnegative for each x in I. [This strengthens #8, since the existence of f'' at a point implies the limit above exists at that point, and the converse fails.] Riemann introduced and used this second order symmetric derivative in an 1854 memoir on trigonometric series. It was this memoir, incidentally, that Riemann introduced what we now call the Riemann integral. LaTeX, .dvi, .ps, and .pdf files of Riemann's 1854 memoir are available at . 10. Suppose f' exists at each point of an open interval I. Then f is convex on I if and only if f' is non-decreasing on I. This strengthens #8 and neither implies nor is implied by #9. 11. If h is non-decreasing on an open interval I and 'a' belongs to I, then the function f defined on I by f(x) = integral of h on the interval [a,x] is convex on I. [This refines a result that arises by putting #5 and #8 together.] ---------- SOME REFERENCES ---------- [1] Yoav Benyamini and Joram Lindenstrauss, Geometric Nonlinear Functional Analysis, Volume 1, Colloquium Publications #48, American Mathematical Society, 2000. [chapter 4: Differentiation of Convex Functions, pp. 83-98] [2] Ralph P. Boas, A Primer of Real Functions, 4'th edition (revised and updated by Harold P. Boas), Carus Mathematical Monographs 13, Mathematical Association of America, 1996. [pages 175-186] [3] Andrew M. Bruckner, Differentiation of Real Functions, CRM Monograph Series #5, American Mathematical Society, 1994. [pages 131-134 (advanced)] [4] Krishna M. Garg, Theory of Differentiation, Canadian Mathematical Society Series of Monographs and Advanced Texts #24, John Wiley and Sons, 1998. [pages 195-198 (very advanced)] [5] R. Kannan and Carole King Krueger, Advanced Analysis on the Real Line, Springer-Verlag, 1996. [pages 74-76] [6] A.C.M. van Rooij and W.H. Schikhof, A Second Course on Real Functions, Cambridge University Press, 1982. [pages 14-18] [7] H. L. Royden, Real Analysis, 2'nd edition, MacMillan, 1968. [pages 108-110] [8] Brian S. Thomson, Symmetric Properties of Real Functions, Pure and Applied Mathematics #183, Marcel Dekker, 1994. [pages 202-209 (advanced)] [9] Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Pure and Applied Mathematics #43, Marcel Dekker, 1977. [pages 118-124] Dave L. Renfro === Subject: Re: One of my conjecture in analysis >Let f : [a,b] -> R be a continuous function which is convex. Then the >points, at which f is not differentiable, are countable and not dense. >What do you think about it? countable is correct, not dense is not. David C. Ullrich === Subject: EARN RS;10000/per day posting-account=nVrnigoAAAC41jTrkF-TgJBoV3aF5gwU CLR 1.1.4322),gzip(gfe),gzip(gfe) EARN RS;10000/per day This is very easy & simple just to use this website and reffers your friends. just copy & paste work. 1.copy to this link & paragrab EARN RS;10000/per day click this link world no1 paid website http://yesp100.blogspot.com 2.paste your composs maill and send your friends&groups 3.this is automatic get your maill id& paid for your mail id earn up to $100000/per day. === Subject: An exact 1-D integration challenge - 53 - Proud Earthling, go and defeat all the doltish CASs! posting-account=ubyIWAkAAABW-OTbVB1QiN1oZlu0qUgw CLR 2.0.50727),gzip(gfe),gzip(gfe) Hello dear computer algebra patriot the Earthling, You know from my alarm messages to the human race, yes, those disastrous (striped!) Computers are coming!... We have the only choice, to protect our old good carbon-based life! (I even do not want to think about the alternative 8-*) So get prepared for the Last Combat. Train hard, fight easy! Look, none of these silicon crystals can cracks this. After a span of amateurish fuss, Mathematica/Maple/Derive/AXIOM return this unevaluated, MuPAD keeps threshing for hours... Is there a Super Duper Integrator the Warrior who can come up with the CAS commands to obtain the exact value of int( z*sin(ln(z))*(z*cos(ln(z))+1)/(2*z*cos(ln(z))+z^2+1)^2, z=0..infinity); and rejoice ? Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ----------------------------------------------------------------- We must understand that technologies like these are the way of the future. ----------------------------------------------------------------- === Subject: Re: Mathematics vs Physics posting-account=p6EJ9AoAAACMPSv4t2GeMy73rdicHikC Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > nos...@microsoft.it says... > sorry for the question, but I need to know. > There is someone with a double major or graduated two times or some > teacher that teach to both Mathematics and Physics students, that can > say me the difference between these two matters? > I want to know also which is (usually) harder, where the students has to > work hard more, which if more difficult. > And at the extremely specializations of Pure Mathematics and Theoretical > Physics, what are the difference? What's the level of knowledge for the > average student in both these matters? Which need more hard work? > I want just the opinions of who really know both the environments. > A. > It was explained to me once when I asked a similar question. Mathematics > you just have to do the problems again and again. It is just practice. > This will not help you prove a theorem which you > have never seen before, or for that matter follow > a proof of a theorem. But physics is even worse as someone can just give you a set of > arbitrary equations and you have to work with them. Why are they arbitrary? Most physics classes have a laboratory > component wherein you test whether those equations really work. That's > the *reason* for those labs. > Physics you have to memorize a lot of facts. > This will not help you actually perform a physical > calculation or build a physical model or design > an experiment. Well that is another issue that physics is often messy. You have to > get your hands dirty. > Although a simplification there is much truth to this. > Much truth as a characterization of high-school > courses perhaps. Not much applicable to coursework > or actual work in either field past freshman year or so. In my experience I have been there and I found it to be true.- Hide quoted text - - Show quoted text - PD. > Why are they arbitrary? Most physics classes have a laboratory > component wherein you test whether those equations really work. That's > the *reason* for those labs. PD. I am so tempted to give response to arbitrary, naturally, physically, etc. Although it is too early to get any conclusion on my part. Let use the term: experiment, simulation, adaptive experiment- simulation, adaptive modeling, recursive modeling-simulation- experiment, etc. Let now, in a simulation, the temperature goes to 0 degree Kelvin, the time goes to infinity, delta goes to 0, the speed goes to the speed of light, etc. Let design an experiment try to fit the result of the above simulation. What do we get? What is our conclusion? Let try to be realistic. Given distance = speed times the time it takes to cover the distance. This is ordinary distance, speed and time. The distance, speed, and time doesn't goes to 0 or infinity. We do the experiment n times. Let n = 100. We get 100 data. We fit our data with a straight line, with the objective to minimize the error of the time. Assume the speed is supposed to be fixed, the distance is fixed, Will the time fixed? Assume the time varies in our experiment, and our straight line fitted, let say, is d=0.97 x th time (in our data of experiment) x speed. Now comes in the naturally or arbitrary or anything that will say d= time x speed. And we give our reasons that the environment of the experiment and it is our straight line fitness, etc that justify the errors. Certainly there are some errors. So now we modify our experiments, or we look at our simulation. Which one is to be changed now? We change our simulation first to be distance= 0.97 x speed. Now it is in line with our experiment result that distance = 0.97x the time x speed. Should we say that distance = 0.97 x the time x speed? And, OH MY! OH MY! OH MY! EVERYTHING CHANGES TO DISTANCE = 0.97 X The Time x SPEED. Do we invented something? I stop here for the moment. I am thirsty. Boen S. Liong. === Subject: Make others look at you with envy!!! posting-account=BE2-JwoAAAAtU2hLPQsfDLVe2IPuAqPd CLR 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Follow the below link to make others look at you with envy!!! ---------------------------------------------------------------------------- ----------- http://kavitha551.blogspot.com ---------------------------------------------------------------------------- ----------- === Subject: check ur internet speed or watch wwe no way out video posting-account=cDVGvgoAAABojXSW9wgAQcbuMfo3zzkl .NET CLR 2.0.50727; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) if u want to check ur internet speed or watch wwe no way out video just log on to amit-gmail.blogspot.com === Subject: DEDUCTIVE ARGUMENTS IN NATURAL SCIENCES posting-account=Lz-LbgoAAABPDavKeW-eYeobwLHD_cvQ CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) Those interested in the logical aspects of science may wish to see this: http://www.wbabin.net/philos/valev9.pdf Pentcho Valev pvalev@yahoo.com === Subject: This is not Rocket science But!!! This is pretty pretty basic but I am stuck! 1 = .000842 * ( 392 + R ) How do you solve R? How do you get from this equation to: R = 1 / .000842 - 392 ( R will be rounded 796 ) Remember that it is in small things that we recognize great people (smile). Alain === Subject: Re: This is not Rocket science But!!! > This is pretty pretty basic but I am stuck! 1 = .000842 * ( 392 + R ) How do you solve R? How do you get from this equation to: R = 1 / .000842 - 392 ( R will be rounded 796 ) Remember that it is in small things that we recognize great people (smile). Since 1 = 0.000842*(392 + R), dividing both sides by 0.000842 you get: 1/0.000842 = 392 + R. The, subtracting 392 from both sides you get: 1/0.000842 - 392 = R. Jose Carlos Santos === Subject: Re: This is not Rocket science But!!! I understand. The complete sequence would be: 1 =.000842 * ( 392 + R ) so 1/.000842 = (.000842 * ( 394 + R )) / .000842 1/.000842 = ( 392 + R ) We can remove the brackets because there is no other operant this side of the equation. 1/.000842 = 392 + R ( 1/.000842 )-392 = 392 + R -392 The brackets are not necessary but it is more clear? ( 1/.000842 ) - 392 = R 795.6484 = R Rounded to 796 I could not figure it out! Was there a way to do it taking care of the 392 first? That is what I was trying to do. Alain === Subject: Re: This is not Rocket science But!!! posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 4334.34; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) spider-dtc-te05.proxy.aol.com[CDBC7085] (Prism/1.2.1), HTTP/1.1 cache-dtc-ad05.proxy.aol.com[CDBC74C7] (Traffic-Server/6.1.5 [uScM]) I understand. The complete sequence would be: 1 =.000842 * ( 392 + R ) so 1/.000842 = (.000842 * ( 394 + R )) / .000842 1/.000842 = ( 392 + R ) We can remove the brackets because there is no other operant this side of the equation. 1/.000842 = 392 + R ( 1/.000842 )-392 = 392 + R -392 The brackets are not necessary but it is more clear? ( 1/.000842 ) - 392 = R 795.6484 = R Rounded to 796 I could not figure it out! Was there a way to do it taking care of the 392 first? That is what I was trying to do. 1 = 392*0.000842 + R*0.000842 1 = 0.330064 + R*0.000842 1 - 0.330064 = R*0.000842 0.669936 = R*0.000842 0.669936/0.000842 = R 795.648456057 = R 796 = R (rounded) Alain === Subject: Re: This is not Rocket science But!!! > Was there a way to do it taking care of the 392 first? That is what I was trying to do. Sure. Define a variable R_0 = 392 + R. You then get 1 = .000842 * R_0. Divide both sides by .000842, and you get 1/.000842 = R_0. Solve for R_0. You can then take your known solution for R_0 and rewrite it to get your R back. === Subject: Re: This is not Rocket science But!!! I did not want to start with the 392 first because I wanted an exotic way of doing it, if your example is an exotic way of doing it. It was just because I didn't remember how to do it. For me, the lesson is: whatever I do to one side of an equation, I must do the same to the other side. I did not remember that. That is wy I was stuck. Cheer! Alain === Subject: Re: This is not Rocket science But!!! I understand. The complete sequence would be: 1 =.000842 * ( 392 + R ) so 1/.000842 = (.000842 * ( 394 + R )) / .000842 1/.000842 = ( 392 + R ) We can remove the brackets because there is no other operant this side of the equation. 1/.000842 = 392 + R ( 1/.000842 )-392 = 392 + R -392 The brackets are not necessary but it is more clear? Right. > ( 1/.000842 ) - 392 = R 795.6484 = R Rounded to 796 I could not figure it out! Was there a way to do it taking care of the 392 first? That is what I was trying to do. None that I can think of. Jose Carlos Santos === Subject: Re: This is not Rocket science But!!! Well thank you Jose Carlos, This was a nice little algebra lesson. I appreciate. Alain === === Subject: A function for all you math nerds out there to analyze X^Y = Y^X I've already done a lot on this, but instead of giving away my analysis, I'd like to see what other people come up with. I promise that if no one covers it all, I will eventually put forward my findings as well. I am interested in particular in integer solutions, and what the graph as a whole looks like. === Subject: Re: A function for all you math nerds out there to analyze > X^Y = Y^X I've already done a lot on this, but instead of giving away my > analysis, I'd like to see what other people come up with. I promise > that if no one covers it all, I will eventually put forward my > findings as well. I am interested in particular in integer solutions, > and what the graph as a whole looks like. As J K Haugland has pointed out, there are a lot of references. If we restrict ourselves to positive integer solutions, we get: (x = 2 and y = 4) or (x = 4 and y = 2). We can show this by taking logs: log x^y = log y^x. Then we get y log x = x log y and (log x)/x = (log y)/y. We analyze the function f(t) = (log t)/t. If we use the natural logarithm, then f has the maximum value of 1/e at t = e, and f increases on the interval 0 < t < e and decreases on e < t < infinity. To solve our problem, we need one value of t less than e, say t = x, and one value of t greater than e, say t = y, with the property (log x)/x = (log y)/y. But the only candidates for x are the two positive integers less than e, namely 1 and 2. === Subject: Re: A function for all you math nerds out there to analyze > X^Y = Y^X I've already done a lot on this, but instead of giving away my > analysis, I'd like to see what other people come up with. I promise > that if no one covers it all, I will eventually put forward my > findings as well. I am interested in particular in integer solutions, > and what the graph as a whole looks like. > Typical nerd. Are x and y, to use conventional notation, reals and not complex numbers? By any chance are they positive reals? Clearly the solution set includes the diagonal. But what diagonal? Are these solutions: x = y = 0, x = y = -1, x = y = -1/2; x = y = i? Thus uncountably many solutions. Ok, there you are, an opportunity to state clearly and precisely the exact problem. ---- === Subject: Re: A function for all you math nerds out there to analyze posting-account=4n0P8QoAAACPj0DJnja1mCT1wUiU6txx Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Typical nerd. post, but instead, I'm just going to point out that my original post was intended to give free reign. Please, any analysis goes. === Subject: Re: A function for all you math nerds out there to analyze posting-account=sFP0HgkAAADJMwhdrXAaC5VX7Tc3BtzY Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > X^Y = Y^X I've already done a lot on this, but instead of giving away my > analysis, I'd like to see what other people come up with. I promise > that if no one covers it all, I will eventually put forward my > findings as well. I am interested in particular in integer solutions, > and what the graph as a whole looks like. This is a pretty well known problem. See, for example: http://www.qbyte.org/puzzles/p048s.html --- J K Haugland http://home.no.net/zamunda === Subject: Re: A function for all you math nerds out there to analyze > X^Y = Y^X I've already done a lot on this, but instead of giving away my > analysis, I'd like to see what other people come up with. I promise > that if no one covers it all, I will eventually put forward my > findings as well. I am interested in particular in integer solutions, > and what the graph as a whole looks like. This is a pretty well known problem. See, for example: http://www.qbyte.org/puzzles/p048s.html The problem with x and y restricted to being positive reals is well known, and that's what your link seems to concern. But the problem without that restriction is not so well known. If Dustan is particularly interested in integer solutions, then we also have solutions (x,y) such as (-1,-1) and (-4,-2). In the thread solve x^y = y^x (sci.math, Apr. 2000), I state all solutions for which x, y and x^y are real, and I describe what the corresponding graph looks like. === Subject: Re: Factorization of C(x)=x^n-1 and T(x)=prod(i=1,n,(x-tan((i - 1/2)*Pi/(2*n))^2))? >It presumably has something to do with gathering together primitive >d'th roots of unity for all divisors d of n; but I've never studied >this stuff properly. Anyone got any definite ideas? I guess it makes sence to find out which roots gather together for >both C(x) and T(x) and compare how both relate to the divisors of n. I'm still rather hoping that it won't be necessary to go into that (or into any subtle field theory). I was also hoping (as I'm behind with work) that someone else would fill in the (many?) gaps in what I've written, but as this hasn't happened yet, I guess I'll have to come back to it. Meanwhile, I'm curious as to where this problem came from (or how it arose naturally, so to speak); also (and I must admit that I didn't even think to ask this, at first!) how one might see that the numbers tan^2((i - 1/2)*pi/(2*n)) are the zeros of a polynomial with rational coefficients in the first place. -- === Subject: Re: Factorization of C(x)=x^n-1 and T(x)=prod(i=1,n,(x-tan((i - 1/2)*Pi/(2*n))^2))? >[...] also (and I must admit that I didn't >even think to ask this, at first!) how one might see that the numbers >tan^2((i - 1/2)*pi/(2*n)) are the zeros of a polynomial with rational >coefficients in the first place. OK, I can see that this comes from writing cos(2*n*theta) = 0 as a polynomial equation in even powers of t := tan(theta): 1 - binom(2n,2)t^2 + binom(2n,4)t^4 - ... = 0 -- Angus Rodgers Contains mild peril === Subject: Re: Geometric construction of the product of two numbers > Can anyone point me to a web reference giving the > construction of a segment of length ab from segments of > lengths a and b? > As the others said, you also need 1. Then 1:a::b:ab lets you construct any one of these lengths from the other three. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Geometric construction of the product of two numbers Arturo Magidin : > Can anyone point me to a web reference giving the > construction of a segment of length ab from segments of > lengths a and b? (...) >But ab _per se_ is not a length, of course. If you want to construct a >segment having length |ab|, won't you also need to have a unit segment >available? I believe so. Draw a line through a point O and a. Then draw a line > through O different from the one you just drew. Mark off a unit > segment, and also a segment of length b along this line. Join the point marked off as a unit segment to the point a. Then draw > a parallel to this line through the point b. The intersection of this > line with the line through O and a will have length |a||b|, by similar > triangles. To obtain a segment of length |a|/|b|, first join b and a, and then > draw the parallel through the end point of the unit segment. I had once a theory of the reals where I used this as the definition of a real number b: a transform family of any free units 1 and a to 1xb and axb through such construct as you mentioned ;-) Then I tried to extend the idea from R^1 to R^n, which I called hyperreals (in conflict with the actual use of this term, I discovered later on), where I did some personal research but which left me with some more questions, see my page hereunder if anyone might enlighten me further: http://home.scarlet.be/~pin12499/hypereal.htm guido (other thema of mine: graphic rendering of C^2 functions: http://home.scarlet.be/~pin12499/qbComplex.html ) === Subject: Re: Geometric construction of the product of two numbers Can anyone point me to a web reference giving the > construction of a segment of length ab from segments of > lengths a and b? > For the classic construction of sqrt(ab), the geometric mean, please see my > figure at . > But ab _per se_ is not a length, of course. If you want to construct a > segment having length |ab|, won't you also need to have a unit segment > available? There's also Hilbert's The Foundations of Geometry at Project > Gutenberg: - but I guess > that's overkill! So, if we had a segment of length e and another of length pi in the Cartesian plane (but where the coordinate grid is hidden), and we knew one is 'e' and the other pi units, what algebraic equation or statement is equivalent to: a segment of length e*pi can be constructed from e and pi-length segments in general position? Maybe this relates to pi and Q(e); if pi isn't in Q(e), maybe e*pi can't be constructed... David Bernier -- === Subject: Re: Geometric construction of the product of two numbers > Can anyone point me to a web reference giving the > construction of a segment of length ab from segments of > lengths a and b? -- m Not a web reference, but I can give you the gist of it. If you want the product of r and s. ----+ ---- | ---- | ---| | s ---- | 1 | ---- | | +---------------------------+ |------ r -----| | |---------------b-----------| (consider this to be two congruent triangles) Then: 1/r = s/b --> b = s*r More explicitly: - Before any construction can start, we always have two points in the plane as constructable. A, and B. Define - (0,0) := A - (1,0) := B Then, we can construct, in order ( 1) {(x,0), x in |R} (line through A and B) ( 2) {(0,x), x in |R} (the perpendicular to (1) in A) ( 3) (0,1) (point on (2)) ( 4) (r,0) (point on (1), r constructable) ( 5) (0,s) (point on (2), s constructable) ( 6) {(r,x), x in |R} (the perpendicular to (1) in (r,0)) ( 7) (r,1) (point on (6)) ( 8) {(rx,x), x in |R} (the line through (r,1) and (0,0)) ( 9) {(x,s), x in |R} (the perpendicular to (2) in (0,s)) (10) (rs,s) (intersection of (8) and (9)) (11) {(rs,x), x in |R} (perpendicular to (9) in (rs,s)) (12) (rs,0) (intersection of (11) and (1) Walk through the construction if you want to see that it checks out. The important part here is to form the congruent triangles. Sjoerd Job === Subject: Re: Algebra with splitting field.. > For every splitting field E over F, where E <= cl(F), > every isomorphism mapping E onto a subfield of cl(F) > is an automorphism of E. > Not sure if this is right, > but can't you just take F = E? > Presumably F is a splitting field over F, and cl(F) = cl(E), > so the question in this case would be: > does every isomorphism of F into cl(F) send F into itself. > The is obviously false, eg take F = Q(t) where t^3 = 2, > and send t into wt where w^3 = 1 (but w <> 1). > I think... > F = Q(t) is not splitting field of t^3 = 2. > Q(2^(1/3), i.(3)^(1/2)) is splitting field of t^3 = 2. Indeed. The specified condition was that E is a splitting field over F. I have taken E = F. Are you saying that a field k may not be a splitting field over itself? === Subject: Re: Algebra with splitting field.. Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > For every splitting field E over F, where E <= cl(F), > every isomorphism mapping E onto a subfield of cl(F) > is an automorphism of E. > Not sure if this is right, > but can't you just take F = E? > Presumably F is a splitting field over F, and cl(F) = cl(E), > so the question in this case would be: > does every isomorphism of F into cl(F) send F into itself. > The is obviously false, eg take F = Q(t) where t^3 = 2, > and send t into wt where w^3 = 1 (but w <> 1). > I think... > F = Q(t) is not splitting field of t^3 = 2. > Q(2^(1/3), i.(3)^(1/2)) is splitting field of t^3 = 2. Indeed. The specified condition was that E is a splitting field over F. > I have taken E = F. > Are you saying that a field k may not be a splitting field over itself? > How exactly do you define splitting field? A splitting field E over F is a field E which is an extension E of F which is the splitting field of a polynomial f in F[X]. Of course a field is a splitting field over itself: it is sufficient to take f to be any polynomial in F[X] which factors completely in F[X] as a product of linear factors. -- m === Subject: Re: Algebra with splitting field.. > For every splitting field E over F, where E <= cl(F), > every isomorphism mapping E onto a subfield of cl(F) > is an automorphism of E. > Not sure if this is right, > but can't you just take F = E? > Presumably F is a splitting field over F, and cl(F) = cl(E), > so the question in this case would be: > does every isomorphism of F into cl(F) send F into itself. > The is obviously false, eg take F = Q(t) where t^3 = 2, > and send t into wt where w^3 = 1 (but w <> 1). > I think... > F = Q(t) is not splitting field of t^3 = 2. > Q(2^(1/3), i.(3)^(1/2)) is splitting field of t^3 = 2. > Indeed. > The specified condition was that E is a splitting field over F. > I have taken E = F. > Are you saying that a field k may not be a splitting field over itself? > How exactly do you define splitting field? A splitting field E over F is a field E which is an extension > E of F which is the splitting field of a polynomial f in F[X]. Of course a field is a splitting field over itself: it is > sufficient to take f to be any polynomial in F[X] which > factors completely in F[X] as a product of linear factors. So why did you express agreement with the erroneous statement that my argument was wrong? Perhaps your Indeed meant: That statement is true, but it is irrelevant to the argument put forward. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Algebra with splitting field.. > For every splitting field E over F, where E <= cl(F), > every isomorphism mapping E onto a subfield of cl(F) > and leaving F fixed is an automorphism of E. > Not sure if this is right, > but can't you just take F = E? > Presumably F is a splitting field over F, and cl(F) = cl(E), > so the question in this case would be: > does every isomorphism of F into cl(F) send F into itself. > The is obviously false, eg take F = Q(t) where t^3 = 2, > and send t into wt where w^3 = 1 (but w <> 1). > I think... > F = Q(t) is not splitting field of t^3 = 2. > Q(2^(1/3), i.(3)^(1/2)) is splitting field of t^3 = 2. I'm taking F = E = Q(t). Surely E is a splitting field _over F_ ? which was the condition laid down? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Algebra with splitting field.. posting-account=z1ZA6AoAAACEgXDaRRTJFG5d4vJvYyOY SV1),gzip(gfe),gzip(gfe) > For every splitting field E over F, where E <= cl(F), > every isomorphism mapping E onto a subfield of cl(F) > and leaving F fixed is an automorphism of E. > Not sure if this is right, > but can't you just take F = E? > Presumably F is a splitting field over F, and cl(F) = cl(E), > so the question in this case would be: > does every isomorphism of F into cl(F) send F into itself. > The is obviously false, eg take F = Q(t) where t^3 = 2, > and send t into wt where w^3 = 1 (but w <> 1). > I think... > F = Q(t) is not splitting field of t^3 = 2. > Q(2^(1/3), i.(3)^(1/2)) is splitting field of t^3 = 2. I'm taking F = E = Q(t). > Surely E is a splitting field over F ? > which was the condition laid down? Yes, you are right. I think... Roots of x^3 = 2 are 2^(1/3), 2^(1/3).[{-1 + i.sqrt(3)}/2], 2^(1/3).[{-1 - i.sqrt(3)}/2]. Let g : Q(2^(1/3)) -> Q( 2^(1/3).[{-1 + i.sqrt(3)}/2] ), g(2^(1/3)) = 2^(1/3).[{-1 + i.sqrt(3)}/2]. Of course, Q( 2^(1/3).[{-1 + i.sqrt(3)}/2] ) is a subfield of cl[Q(2^(1/3))]. Since 2^(1/3) and 2^(1/3).[{-1 + i.sqrt(3)}/2] are conjugate over Q, g is isomorphism. But g[Q(2^(1/3)] =/= Q(2^(1/3)). so, g is not automorphism. is this right ? === Subject: Re: about the bijection between of R^n and R posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) Hi all, I have a question concerning the cardinality of the continuum. So it's well known that R^n and R have the same cardinality, and it's the same for [0,1]^n and [0,1], which means there is a bijection f between [0,1]^n and [0,1]. My question is very simple. Can this function f be measurable w.r.t. the Borel sigma-algebra on [0,1]^n and [0,1]? > Best, > Yihong I will try to be clear with MY english > How could it be possible to have a bijection between N (1,2,3 ...) > and æthe ensemble formed by the double (2,4,6 ...) .... when the ensemble of double is formed by a part of N ? Like this: Let E = {2, 4, 6, ...} Define f:N->E f(x) = 2x f is a bijection. Do you need proof? Which property of a bijection to you think is lacking? - Randy === Subject: Re: about the bijection between of R^n and R <25442015.1203651494834.JavaMail.jakarta@nitrogen.mathforum.org>, > Hi all, I have a question concerning the cardinality of the continuum. So it's well > known that R^n and R have the same cardinality, and it's the same for [0,1]^n > and [0,1], which means there is a bijection f between [0,1]^n and [0,1]. My > question is very simple. Can this function f be measurable w.r.t. the Borel > sigma-algebra on [0,1]^n and [0,1]? > Best, > Yihong What you'd like to do is send x = .a_1 a_2 a_3 ... to f(x) = (.a_1 a_3 a_5 ..., .a_2 a_4 a_6 ...). Here .a_1 a_2 a_3 ... denotes the binary expansion of x in (0,1). The problem with the is that some x's have more than one binary expansion, and this causes a host of problems. But we can do the following. First, let F = {x in (0,1) : x has a finite binary expansion}; off of F the binary expansion is unique. Let B be the set of binary sequences. Ie, B = {0,1}^N. On B we put the product topology, and in the standard way we can regard B as a metric space, with a sequence of points converging in B iff all component sequences converge. Let B_0 = {b in B : b terminates in all 0's}. In 1 - 5 below, the symbol A -> B means there is a Borel measurable bijection from A to B. Then 1. (0,1) -> (0,1) F and vice versa. 2. (0,1) F -> B B_0. In fact the natural map here is a homeomorphism. 3. B B_0 -> B and vice versa. 4. B -> B x B. In fact the natural map here is a homeomorphism. 5. B x B -> (B B_0) x (B B_0) -> ((0,1) F) x ((0,1) F) -> (0,1) x (0,1). So now you compose appropriate maps to arrive at a Borel measurable bijection f: (0,1) -> (0,1)^2 (recall that the composition of Borel measurable maps is Borel measurable). Comments: 1. follows from the countability of F. 2. The natural map here is x = .a_1 a_2 a_3 ... -> (a_1, a_2, ...). Note that in (0,1) F a sequence converges iff the corresponding sequences of binary digits converge. (That's not true if F is included!) 3. follows from the countability (and Borel measurability) of B_0. In 4. you finally get to do what you want: (a_1, a_2, ...) -> ((a_1, a_3, ...), (a_2, a_4, ...)), which yields a homeomorphism. And now in 5. we finish up. === Subject: Re: about the bijection between of R^n and R Hi Prof. Edgar, Best, I have a question concerning the cardinality of the > continuum. So it's well > known that R^n and R have the same cardinality, and > it's the same for [0,1]^n > and [0,1], which means there is a bijection f > between [0,1]^n and [0,1]. My > question is very simple. Can this function f be > measurable w.r.t. the Borel > sigma-algebra on [0,1]^n and [0,1]? Yes, it can. -- > G. A. Edgar > http://www.math.ohio-state.edu/~edgar/ === Subject: Re: about the bijection between of R^n and R <22014461.1203731978901.JavaMail.jakarta@nitrogen.mathforum.org>, > Hi Prof. Edgar, Can you elaborate a little more about this? Wade has done a fine job of that. > Is it related to Kuratowski Best, > Yihong > -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Attacking the US Embassy <7qa9vgi778@Darrin.fsf.hobby-site.com> posting-account=o9bz7QkAAAB9ZvAlBpZiHwpaljSLI9rl RadioClicker PRO; RadioClicker Lite; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > The Fox news commentator has to tell the Fox news viewers that the > Serb protesters are not linked to Al Queda: http://www.youtube.com/watch?v= VJuFRj4dLE > How can they be sure? You never know with those Christians ... > You may discover soon that thug means half-terrorist. Which half? Top or bottom? Why have you excludes right/left splitting? :-) > Creates too much surface area. Hard to keep warm in the winter. === Subject: Bertram Kostant on E8 Bertram Kostant recently gave this talk at UCR: On Some Mathematics in Garrett Lisi's E8 Theory of Everything Abstract: A physicist, Garrett Lisi, has published a highly controversial, but fascinating, paper purporting to go beyond the Standard Model in that it unifies all 4 forces of nature by using as gauge group the exceptional Lie group E8. My talk, strictly mathematical, will be about an elaboration of the mathematics of E8 which Lisi relies on to construct his theory. You can see videos of this talk and lecture notes here: http://math.ucr.edu/home/baez/kostant/ If his talk is too tough, you might prefer the warmup talk I gave earlier that day. But, Kostant described some ideas whose charm is easy to appreciate: The dimension of E8 is 248 = 8 x 31. There is, in fact, a natural way to chop up E8 into 31 spaces of dimension 8. There is a nice way to see the product of two copies of the Standard Model gauge group sitting inside E8. The Standard Model gauge group is a subgroup of SU(5). There is also a nice way to see the product of two copies of SU(5) sitting inside E8. The dimension of SU(5) x SU(5) is 48, and 248 - 48 = 200. The adjoint action of SU(5) x SU(5) on the Lie algebra of E8 thus gives a 200-dimensional representation, and this is (5 x 10) + (5* x 10*) + (10 x 5) + (10* x 5*) Garrett Lisi's ideas have received serious criticism from Jacques Distler and others. I've included links to Lisi's paper and also Distler's comments. But, the work Kostant presents here is logically independent - beautiful math, regardless of its possible applications to physics. It makes heavy use of recent work on certain finite subgroups of E8, most notably GL(2,32) and (Z/5)^3. As Kostant said, E8 is a symphony of twos, threes and fives. === Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Are you aware that you could make up to $500+ per week playing online games. Learn more: http://500perweek.blogspot.com/ === Subject: Re: New symbolic/numeric/dynamic/intuitive programming language I changed the first paragraph at www.flamingthunder.com >to something perhaps less contentious. Is that supposed to be an answer to my question? You said Right now, more elementary school students are using Flaming >Thunder than Python. and various people asked how you knew that that was so. Instead of answering honestly you say that you've changed the wording on a web site to something less contentious? _If_ that's an answer to the question then we can only assume that what you meant was that you _don't_ know that more kids are using FT than Python, that in fact you were just making that up. Just making things up is not good for your credibility. Regusing to admit it when you're called on it is worse. >I won't be around today to post because I'm heading off >to a meeting with some education officials. By next Tuesday I expect to release the next update to >Flaming Thunder, which will include some of the >features that some of you have requested. I'll post >here when it's up. > David C. Ullrich === Subject: Re: lebesgue > On 21 ?ubat, 14:27, The World Wide Wade On 21 ?ubat, 13:06, Robert Israel > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) it means A has measure zero.If A has measure zero we can say lambdaA=0 > then, > lambdaA=0 and f is measurable integral_R f X_A dlambda=0 > we know that if A subset of E > integral_A f dlambda= integral_R X_A f dlambda<=integral_R X_E > dlambda= > integral_E f dlambda > from this > 0<= integral_R X_E dlambda=integral_E f dlambda > What should I do now? > That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? æPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al?nt?y? > gizle - > - Al?nt?y? g.9aster - > I'm trying to show that > ?f dlambda A > State the problem precisely.- Al?nt?y? gizle - > - Al?nt?y? g.9aster - > f is a nonnegative and integrable over a measurable set E. Then ?>0 > and ?>0 s.t. over a measurable > set A subset of E with ?(A) æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æA > You didn't take my advice, nor did you take Wade's. æThere may > be a language barrier here, but I suspect it shows a basic lack > of understanding of the mathematics. > æRobert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > æDepartment of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > æUniversity of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - > - Show quoted text - again. f be a nonnegative measurable function which is integrable over a >measuarble >set E.Then given ?>0 and ?>0 such that for every measurable set A >subset of E with ?(A)?fd?A David C. Ullrich === Subject: Re: lebesgue posting-account=rXOxmAoAAACV7JlESe_oxjqpbHbC1Re4 1.1.4322),gzip(gfe),gzip(gfe) > On 21 øubat, 14:27, The World Wide Wade On 21 ?ubat, 13:06, Robert Israel > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) it means A has measure zero.If A has measure zero we can say lambdaA=0 > then, > lambdaA=0 and f is measurable integral R f X A dlambda=0 > we know that if A subset of E > integral A f dlambda= integral R X A f dlambda<=integral R X E > dlambda= > integral E f dlambda > from this > 0<= integral R X E dlambda=integral E f dlambda > What should I do now? > That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? æPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al.9bnt.9by.9b > gizle - > - Al.9bnt.9by.9b g.9aster - > I'm trying to show that > .bcf dlambda A > State the problem precisely.- Al.9bnt.9by.9b gizle - > - Al.9bnt.9by.9b g.9aster - > f is a nonnegative and integrable over a measurable set E. Then .b9>0 > and .83å>0 s.t. over a measurable > set A subset of E with .83.83(A)<.83å, show that .81.8dfd.83.83<.b9 > æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æA > You didn't take my advice, nor did you take Wade's. æThere may > be a language barrier here, but I suspect it shows a basic lack > of understanding of the mathematics. > æRobert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > æDepartment of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > æUniversity of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - > - Show quoted text - again. f be a nonnegative measurable function which is integrable over a > measuarble > set E.Then given .b9>0 and .83å>0 such that for every measurable set A > subset of E with .83.83(A)<.83å, we have > .81.8dfd.83.83<.b9 A Sorry, that's wrong. æTry again. Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al.9bnt.9by.9b gizle - - Al.9bnt.9by.9b g.9aster - f be a nonnegative measurable function which is integrable over a measuarble set E.Then given .b9>0,there is a .83å>0 such that for every measurable set A On 21 ?ubat, 14:27, The World Wide Wade On 21 ?ubat, 13:06, Robert Israel > f is a nonnegative and integrable over a measurable set E. Then > epsilon greater than 0 and delta greater than 0 s.t. over a measurable > set A subset of E with > lamda(A) it means A has measure zero.If A has measure zero we can say lambdaA=0 > then, > lambdaA=0 and f is measurable integral_R f X_A dlambda=0 > we know that if A subset of E > integral_A f dlambda= integral_R X_A f dlambda<=integral_R X_E > dlambda= > integral_E f dlambda > from this > 0<= integral_R X_E dlambda=integral_E f dlambda > What should I do now? > That depends on what you're trying to do, which is not at all clear. > What exactly is the problem you're trying to solve? æPlease be careful > to include the quantifiers (for all) and (there exists), and in the correct > order: they are very important. > -- > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al?nt?y? > gizle - > - Al?nt?y? g.9aster - > I'm trying to show that > ?f dlambda A > State the problem precisely.- Al?nt?y? gizle - > - Al?nt?y? g.9aster - > f is a nonnegative and integrable over a measurable set E. Then ?>0 > and ?>0 s.t. over a measurable > set A subset of E with ?(A) æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æ æA > You didn't take my advice, nor did you take Wade's. æThere may > be a language barrier here, but I suspect it shows a basic lack > of understanding of the mathematics. > æRobert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > æDepartment of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > æUniversity of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Hide quoted text - > - Show quoted text - > again. > f be a nonnegative measurable function which is integrable over a > measuarble > set E.Then given ?>0 and ?>0 such that for every measurable set A > subset of E with ?(A) ?fd? A > Sorry, that's wrong. æTry again. > Robert Israel æ æ æ æ æ æ æisr...@math.MyUniversitysInitials.ca > Department of Mathematics æ æ æ æhttp://www.math.ubc.ca/~israel > University of British Columbia æ æ æ æ æ æVancouver, BC, Canada- Al?nt?y? gizle - > - Al?nt?y? g.9aster - f be a nonnegative measurable function which is integrable over a >measuarble >set E.Then given ?>0,there is a ?>0 such that for every measurable set >A?fd? 0 and delta > 0 and given epsilon > 0 there exists delta > 0 you're not ready for the course you're taking; you should retake some sort of introduction to proofs course first. (There's a course or two like that in most math departments; sometimes it's a course in beginning abstract algebra, sometimes elementary real analysis.) Third: Look at the current version: Then given ?>0,there is a ?>0 such that for every measurable set AA > David C. Ullrich === Subject: help on an inequation question!! posting-account=OIzUrgoAAADleYiocKYJFMwybr36cQDA CLR 1.1.4322),gzip(gfe),gzip(gfe) sum_{i=1}^{n} p_i = 1 sum_{i=1}^{n} q_i = 1 sum_{i=1}^{n} {p_i*log(p_i)} >= sum_{i=1}^{n} {q_i*log(q_i)} i.e. the entropy of {p} is smaller than {q} p_i's and q_i's are sorted in descending order respectively then does the following inequation hold, sum_{i=1}^{n} {i*p_i} =< sum_{i=1}^{n} {i*q_i} Prove it or give a counter example. THanks a lot. === Subject: Re: English question with x^2, x^3, x^4, .... >Hello sir~ x^2 : square >x^3 : cube >x^4 : what it called ? Fourth power of x. Or x to the fourth. >x^5 : ? >x^6 : ? >etc... square root >cube root >4th root ? 4th root. >5th root ? >.... Can you tell me how to say ? David C. Ullrich === Subject: I need help choosing a major - Jeffrie Silverberg Jeffrie Silverberg === Subject: Re: JSH: Sorry, but exasperated > It's been over five years since I found a proof of Fermat's Last > Theorem. Lie # 1 I still believed enough in modern math society that I questioned and > questioned and questioned my own result as people fought successfully > against it being accepted. Lie # 2 (nobody fought) I even kind of wondered still when I pulled out a piece of it and got > that published, only to have math society fail completely when the > math journal went against formal peer review, pulled my paper and > later died. Lie # 3 ( it was never published, it was withdrawn polite for rejected) You people may have heard the story but may not have realized that I > have been right, dealing with people who I continually saw shifting > their tactics as I explained as they fought a political battle to make > sure no one believed me. Lie # 4 no one is blocking you. Your Math is self-blocking. One thing that kept me from simply giving up was that I knew they > wanted it. They begged for it, literally. Lie # 5 After all, if I weren't around to champion my ideas then they could > keep them suppressed. Lie # 6 Then Andrew Wiles could keep credit for something he didn't do. Lie # 7 And undergrads could keep getting taught crap math which would never > work not because it is pure but because it's wrong, but not easily > testable in a way that can show it's wrong. Lie # 8 (you are still mad at your algebra teacher) The perfect trap. nothing is perfect. The way the fight has gone against my research has evolved as my > strategies evolved and as the people doing it had to handle my moves > in other areas, as I looked desperately for ANY way to prove that I > was right and that these people were deliberately lying. too bad you didn't focus on the Math. They turned success into a perception of failure. Lie # 9 I've said it's like winning the Olympics and being booed and the gold > medal going to someone who didn't even run the race. Lie # 10 you have nothing They turned everything on its head. Lie # 11 there is no 'they' you are doing this to yourself So I turned to the factoring problem. It still gets to me though, after so many years of dealing with these > people to see posters STILL trying the same games, the same ways to > distract and deny with an argument so simple it stunned me. Lie # 12 Who knew? Turns out there's this neat thing with abs(nT - 2k^2) where > you look at a simple integer minima with k maximal. Lie # 13 Not even I thought the entire world could turn on a result that > simple, but turn it will. Lie # 14 Blocking pure math is one thing but posters fighting now are trying > to stop knowledge of a result that can mean your actual physical life > could be in danger, or your financials. Lie # 15 Lying here can mean the ruin of some of you, and isn't it ironic, > don't you think? Lie # 16 They will try. As what have the got left now? Lie # 17 no one has to, anyone can see your math is wrong Their battle was an all or nothing war against the latest (maybe the > last) major discoverer. Lie # 18 They have nothing left to do but fight until the bitter end. Lie # 19 we have jobs There have only been a handful of people like me in all of human > history. Lie # 20 And I wish that I had not been born to face this mess of what humanity > has become. You are just a lying troll, JSH > James Harris === === === Subject: thermal fluid sciences an integrated approach by stephen R. Turns posting-account=e9MSPQoAAAA5_Yvo_YetTDlglBvrOOHV 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) i wanted to purchase the thermal fluid sciences an integrated approach by stephen R. Turns solution manual === Subject: Re: Les Nombres =?ISO-8859-1?Q?R=E9els_et_leur_Position_?= =?ISO-8859-1?Q?Centrale=2E?= > Le plus grand r.8eel n.8egatif est il alg.8ebrique ou transcendant ? > Mohwali Awamar. > Point de vue (n.8eo) pythagoriste : > seuls existent les cardinaux de 1 jusqu'au nombre d'atomos de > l'univers. > O n'est pas un nombre. > 1 .8el.8ephant 1 chien 1 fleuve > z.8ero quoi ? > Mais TOUTAIN !!! > Toutain = z.8ero !, > Toutain, nul, nullard, Khmer Rouge et Garde Rouge .88 la Mao ASSASSIN de > la culture (comme les autres, comme ceux de 1917... les historiens ont > mis en .8evidence suffisamment d'.8el.8ements ! comme ceux de toutes les > r.8evolutions, ou presque toutes) ! > Allez, Toutain, vive le grand camarade Staline, vive la science > prol.8etarienne de Lyssenko, .88 bas la science bourgeoise ou juive, la > science sous la dictature de l'id.8eologie, la science subordonn.8ee aux > d.8elires des pseodo-philosophes scientifiques Marx-Engels et autres > comme cet avocat v.8ereux de L.8enine, etc. ! > RJ That mathematic teacher was a member of THE FRENCH COMMUNIST PARTY > HE WAS THE MEMBER OF A STALINIST PARTY > He explain that in a post (? in fr.sci.physique) I was the member of a troskyst party LCR, member of the Fourth > International from 1973 to 1982 Who is the friend of Lissenko ? > Who is the friend of Stalin ? YT Ta gueule, connard. Tu es objectivement un casseur de science comme Lyssenko-Staline. Je n'ai pas .8et.8e membre du PCF, mais un partisan de l'Albanie socialiste tant que je n'ai pas compris clairement sa nature stalinienne. Pauvre fou de r.8evolutionnaire cr.8etiniste marxiste-l.8eniniste-trotskyste (je rappelle que L.8enine et Trotsky furent les sabre-peuple qui ont bombard.8e et massacr.8e les soviets et la population de Cronstadt !) Va te faire f... ! RJ === Subject: =?ISO-8859-1?Q?Re:_Les_Nombres_R=E9els_et_leur_Position_Centrale.?= <47bdaf92$0$851$ba4acef3@news.orange.fr> posting-account=MoIEegoAAAB33I5WSGOn25pCPwzRI7xI 5.0),gzip(gfe),gzip(gfe) > Le plus grand r.8eel n.8egatif æest il alg.8ebrique ou transcendant ? > Mohwali Awamar. Point de vue (n.8eo) æpythagoriste : seuls existent les cardinaux de 1 jusqu'au nombre d'atomos de > l'univers. > O n'est pas un nombre. 1 .8el.8ephant 1 chien 1 fleuve > z.8ero quoi ? 1 .8el.8ephant = 1 .8el.8ephant > z.8ero quoi =n'importe quel z.8ero ----------------------------------------- > VOTRE SOUCIS > Votre .8eglise devait r.8epondre moins aleph 1 > -------------------------------------- > PIEGE > Combien y a t il des transcendants ? Si l'on inclut ..... pi / 4 > si l'on inclut æ æ pi puissance pi ---------------------------------------- > DEMONTAGE DE L 'IMPOSTURE R n'existe que si Cantor a raison > æsi et seulement si > Card (N) = Card des pairs = Card des impairs et donc si et seulement si il existe une bijection en l'ensemble des > entiers et l'ensemble des nombre pairs > -------------------------------------- > VULGARISATION > D'apr.8fs Cantor > il associe 1 avec 2 > æ æ æ æ æ æ æ æ2 æ æ æ æ æ4 > æ æ æ æ æ æ æ æ3 æ æ æ æ æ 6 et il ose pr.8etendre que c'est une bijection Il suffisait d un Soup.8don de prestidigitation. Mohwali Awamar > ----------------------------------- > Mais ces farceurs (qui ont PRIS LE POUVOIR EN DEUG de maths et FORCENT > nos enfants .88 REPETER DES CONNERIES pour avoir leur module) > ------------------------------------------------------ > IMPOSTURE le 2 de la deuxi.8fme colonne, le 2 de l'ensemble des nombres pairs > est > le > m.90me > nombre > que le > 2 de l'ensemble des nombres entiers Il n'y pas deux 2 diff.8erents > > ILS TRICHENT > ---------------------------------------------------------------- > Ils font une partition et s'empressent de l'oublier > ----------------------------------------------------------- TOUTE LA NOTION DE CONTINUITE S'EFFONDRE DANS CETTE PAGE > ------------------------------------------------------------- SEULE L'ARITHMETIQUE EST LOGIQUE > IL N EXISTE QUE DES ESCALIERS > DES INTEGRALES PAPIER PEINT (SOUS L'ESCALIER) LES EQUATIONS DIFFERENTIELS SONT LES RELATIONS ENTRE LA HAUTEUR DE L' > ESCALIER ET LA LONGUEUR DE CHAQUE MARCHE (MARCHES è æHAUTEUR > CONSTANTE) æ æ Cela fait 8 ans que je r.8ep.8fte cela ici (sous les insultes et les > quolibets des cr.8etins ignares et d.8epourvus d'argument) > Il faut retrouver dans les archives mon manifeste > Bon anniversaire monsieur Planck > Il y a une premi.8fre publication faite par Jacques Barot (+ depuis) sur > internet. > On voit dans ce forum le premier cr.8etin abruti s'empresser de d.8enigrer > sans r.8efl.8echir Voila le texte de l'empress.8e Alain Frisch La post.8erit.8e retiendra son nom (mais certainement pas pour ses > travaux !!!) > Pour les anonymes cr.8etins consultables sur les liens, leur affichage > sera fait par les sites sp.8ecialis.8es dans l'.8epist.8emologie plus tard. Yanick Toutain Rappeler les exploits d'Alain Frisch a aussi pour fonction > (discr.8fte) de montrer aux .8eventuels amateurs, que leur prose > continuiste aura le m.90me sort en 2016. > Qu'ils apprennent la politesse et .88 argumenter ! Vive l'homo sapiens connardus ! Exemple .8eminent : le camarade Yanick ToutCr.8etin ! RJ- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - === Subject: Category Theory [annoying property] Does the following hold in every complete category with coproducts: coprod_{A times B} 1 = (coprod_A 1) times (coprod_B 1) ? I guess, the answer is no, but couldn't find any counterexample. Best, Maronzo === Subject: How to find eight multiplications I have x^62 and want to show it that it can be computed with only eight multiplications === Subject: Re: How to find eight multiplications- > I have x^62 and want to show it that it can be computed with only > eight multiplications > I guess you want to compute x^62 from x? A variation to Rick Decker's receipt does the job. There you go... x -> x^2 -> x^4 -> x^6 -> x^7 -> x^14 -> x^28 -> x^56 -> x^62. Check this on x = 123,456,789 Ciao: Johan E. Mebius === Subject: Re: How to find eight multiplications- <47C068EA.2090700@xs4all.nl> posting-account=-uDMqAoAAAAbLImwvfibPkq6U-2u3cAb CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727; FDM),gzip(gfe),gzip(gfe) > I have x^62 and want to show it that it can be computed with only > eight multiplications > I guess you want to compute x^62 from x? A variation to Rick Decker's receipt does the job. There you go... x -> x^2 -> x^4 -> x^6 -> x^7 -> x^14 -> x^28 -> x^56 -> x^62. Check this on x = 123,456,789 Ciao: Johan E. Mebius What is the eight multiplications? Could you guys please give me some simple example?? === Subject: Re: How to find eight multiplications- > I have x^62 and want to show it that it can be computed with only > eight multiplications > I guess you want to compute x^62 from x? A variation to Rick Decker's receipt does the job. There you go... x -> x^2 -> x^4 -> x^6 -> x^7 -> x^14 -> x^28 -> x^56 -> > x^62. Check this on x = 123,456,789 Ciao: Johan E. Mebius What is the eight multiplications? > Could you guys please give me some simple example?? x^2 is obtained from x and x by one mutltiplication: x * x. x^4 is obtained from x^2 and x^2 by one multiplication: x^2 * x^2. x^6 is obtained from x^2 and x^4 by one multiplication: x^2 * x^4. x^7 is obtained from x and x^6 by one multiplication: x * x^6 etc. x^62 is obtained from a total of 8 multiplications (with remembering some prior partial results). === Subject: Re: How to find eight multiplications > I have x^62 and want to show it that it can be computed with only > eight multiplications > Hint: Compute x^2, x^4, x^8, ... and continue from there. Rick. === Subject: Re: JSH: So tired > You people have no clue the kind of energy it takes out of you to > figure out a major discovery or how much more tiring it is when nasty > people dismiss the math. Lie #1, and #2 I'm so tired today not so much from finding the result but from the > sapping effect of reading posts by people who clearly despise > mathematics who get away with it. Lie #3 your math is wrong How can they? Aren't any of you excited by the factoring problem? yes Yes, I've made lots of mistakes so many of you probably just tune me > out, but do you all? Are you all immune to a brilliant but succinct > mathematical argument unless someone else TELLS you it is? BIG Lie #4 brilliant but succinct mathematical argument Then WHY ARE YOU IN THE MATH FIELD? That's what keeps going through my mind. How did our world get so > pathetic and lost that you people are it in mathematics. Lie # 5, 6 and 7 That so many people get away with pretending to be mathematicians. Lie # 8 How did humanity so horribly lose its way? Lie #9 How can anyone hope? I keep worrying that it is game over and our > species has nowhere to go but extinction as it has lost the ability to > know the truth, and cherish it. you are paranoid. > James Harris === Subject: Re: JSH: So tired posting-account=Ae7cPwoAAAA1p9Bl1szxMYHINqjRfucA CLR 2.0.50727),gzip(gfe),gzip(gfe) You can go because you're so stupid. The Universe will be out of the 1940's. So I solved the factoring problem. A large overestimate for the next lesson, when again, someone just takes over, because you see it as it has lost the ability to share a delusion -- and one day he sits me down in our room to explain why your research is worth the public dime. I think that starting with 1: 1, 119 and so forth, so you can convince other people mixed right in there somewhere. Do the same method COULD factor an RSA key can be generalized, but the rest of the fittest. We must slash funding across the two factors an infinite number of ways of everything is distraction. Talk that in Africa. There are 4 equations with 4 unknowns means you get an answer you like, as I've said before. And besides, enough with the psych-out attempts. They don't know the factors modulo the product of your students who are middle class works around them, the bad guys, but too often, like with that and thought it would be sufficient. Positive residues, so of course they are lying. If you consider primes near 1000 where I can put up the equations as you found a human solution to the top of the human species extinct. Because it failed. So it IS easy. Oh, it's easy. With a target T, you can consider what rules or sets of rules can exist by which you can guess at optimal k. And in general in the complex plane is kind of shrugged, called me a very angry person and was even my roommate! Share! He just gave me all the life on this planet as it has lost the ability to share a delusion -- and one day he sits me down in our room to explain to me that maybe I don't know about me -- totally inappropriate -- asking him if he gets something right? Then who cares? Right? If you care about the value of k^2 such that you get f_1*f_2 = T, each has some residue modulo p_1*p_2, so that all those annoying factor arguments go a way. We are in the first factor. I've added that the result that holds in the mathematical community who don't think that starting with destroying tenure. Which I'm going to lead to something big down the line finding every possible residue modulo p_1, as you loop through ALL possible f_1 and f_2 = T mod p_1*p_2 = f_1 mod p and then you would behave as people do. I feel like it I come here and none have been around in the US Army. He went to one of these vulnerable humans. So obsessed with your talk. Nope. The irony is that these people don't care if you have problems that depend on accepting what is going on, or how serious it is. Look at your history books!!! You do this until you have the highest SAT score, as I noted in my nature to do the same set, so now you have f_1 mod p_1, and you take the 1 right, but harp on the result holds in the Annals is strictly greater (yes > not >=) than that of the fittest and I think you people lie. So it's trivial algebra. Now if you take your second prime and compare back to your first set of stored values, pulling all matches. One of those lists you must have the correct answer for k, when it does solve the factoring problem is solved. It's an easy proof. There are people on this planet is doomed. The greatest extinction in earth's history is just a(b+c) = ab + ac so in this latest idea, fine. It's math. If it's right, it's right. If it's wrong it's not, but I disagree. It means that your society came up with delicate proofs and logical contradictions as you loop back through the first factor (A(x) + 7)(B(x) + 1) true for all x, where they claim that x=0, is a solution to an inhuman discipline. The math people have no bananas today. People are reacting like this means nothing, and if the experts just decide to ignore it, or even break it. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > So what? Even if we have > Y is made by steps one for each natural number > At every step P is true. > We cannot conclude that P is true for Y. > - William Hughes > We can conclude that P is true for all of Y, which consists of all the > steps you mention. We cannot conclude the property is true for Y as a > whole, but that's not what I'm arguing, anyway. > That is exactly what you are arguing. You have the finite > trees which are all of Y. > I do? > We are agreed that > P holds for each finite tree. We have the infinite tree > which is Y. You want to say that P holds for Y. > (inifinite paths only exists in the infinite tree as a whole). - William Hughes > I want to say that any constant equality (identity) through N extends > beyond N, and any inequality that isn't eliminated (lim(x-y)=0) by the > end of N persists. No biggie. Different, but not bad. Kinda like Cantor. > He was a radical in its day, and not well accepted... > > Why should any identity holding in N be expected to hold outside N, > for, say, triangles instead of naturals? > By outside N I mean, for all set sizes, including those beyond the finite. And Cantor was accepted by Hilbert, which was, in those days, about as > well accepted as one could get to be. Yeah, after his 19th nervous breakdown... :| Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > You're entirely confused about what I asserted. Let's look again at what > There is no difference between the tree and all of its finitely > numbered nodes and edges. > Notice the word, all. What does all its members mean, besides the > entire set? > > The tree has a required structure that all its members need not have > merely as members of something. > Each member may not have that structure, but all the members together do. Do you know the difference between each and all? It's kinda what sets are all about. > You may consider this to be some nit-picky point, but if the > tree consists only of the set of nodes, which is countable, and the set > of edges, one associated with each (parent) node, and thus equal and > equally countable to the nodes, then the entire tree is countable. > > The pieces mentioned are countable, but it is a set of sets of such > pieces which is not countable. Disagree. Not that for any infinite set, the set of all its finite subsets is > countable but the set of all its subsets is not. That could not be true of any set. You and WM are only looking at finite subsets of the set of edges and > nodes, whereas maximal paths in infinite trees are all infinite sets of > nodes and edges. > Really we're applying limits in the normal sense, instead of using transfinitological incantations to justify magic on the grounds of axioms which are actually not the sole basis for the claims. You are within a model that doesn't appeal to reality. > > Every > path that can be identified within any finite length is included, but > the vast majority which cannot, including 1/3, are nowhere completed > within the tree, and therefore are not included in the set of paths. A path in such an infinite tree is completed when no proper superset > of edges and nodes is a path. When does that occur within the tree? > This is a fundamental disagreement, and I don't expect you to agree, but > I hope you can be respectful about it. It's axiomatic. :) Not by any axioms in any standard set theory or tree theory. > What system of axioms do you claim that it is axiomatic in? Fundamental inductive principle and the fact that oo>n for all n in N. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Concluding that a property of finite trees necessarily carries over to > infinite trees requires proof, but none have been forthcoming. > What have been overwhelmingly forthcoming in the past few years are > suggestions as to what alternative sets of axioms may serve to offend > fewer sensibilities. Infinite-case induction and the inverse function > rule qualify as axioms lending themselves to the production of proofs, > and the results are clearly at odds with standard transfinitology, and > in line with the bulk of objections that have pummelled the hull of your > vessel for over a century. > > In order to investigate an axiom system one needs at least a complete > list of the axioms assumed, including a clear statement of each axiom. Such has never been forthcoming from TO. If he merely wishes to piggyback extra axioms on top of some already > stated axiom system, then a mention of that system plus a clear > statement of what he append is still required. > I concede that's true. I'm starting to work on that again. I had to take a break. I'm finding I really have to start from the bottom up to prove my point. It's nice to be back. :) > > The countably infinite set of paths in a > balanced binary tree is a gaping hole in the hull. It is only a hole in the hull of WM's ship, and Tony's if he sinks on it > too. In standard set theories it causes no problems whatsoever. > It's inexplicable. Can there not be a countably infinite set of paths in a balanced binary tree? I guess not. It's strange. It raises the question of the hypothesis of the continuum. Again, given a countably infinite number of levels, how many nodes exist? One can prove (via Cantor's diagonal proof for binary sequences) that > any attempt to surject N to the set of all paths of a complete > infinite binary tree must fail. > It is power set, and no more. That is enough. It is obvious that |power set(x)|=2^|x|>|x|. It's a classic infinite-case induction proof ala Cantor and the dancing diagonals. So with the nodes vs. the countably infinite levels. They are clearly more numerous. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Consider the set X={0,1/2.3/4 ...} > Let P(r) be true if there is an element in X which is larger than r, > false otherwise. > There is no largest r for which P(r) is true, but P(1) is false > The difference occurs at r=1, > However difference and largest are not the same. > Yes the points at which P(r) is true have a supremum. > However, they do not have a maximum. > - William Hughes > At r=1, the function changes for obvious reasons. However, r<1, because > r=1-2^(-n) for neN. Ne pas? The maximum does not exist, as we agree, > again. Therefore, my consideration of such a set is largely irrelevant. > No. Such a set shows that a change can occur by the supremum, but not > before the supremum. Note the supremum of the finite binary trees > is the infinite tree. The change occurs by the supremum, but not > before > the supremum (only at the supremum are the infinite paths counted, > before > the supremum the infinite paths do not contribute to the count). > - William Hughes > I think WM's point is that the supremum doesn't exist, and that > consideration of such a point leads to contradictions. > I take it by the fact that you are trying to change the subject > that you now accept that the argument that something that holds > for every finite tree *must* hold for the infinite tree is worthless. > - William Hughes > I never made such a general statement. > However, you keep saying there is no step at which > things change as if this meant something. As steps > can only occur at finite points this is equivalent to > saying that something that holds for every finite tree > holds for an infinite tree. > - William Hughes > There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > At any step you have a finite tree. > Therefore, at no point in this set, > At every point you have a finite tree. > whether it includes every single one > of these finite steps or not, is this fact untrue. > The infinite tree is not one of those steps. Your > therefore is, as William stated, an attempt to jump > from something true for every finite tree to a > statement about the infinite tree. > The set of things you get to, the set of things you > include in this sentence > There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > does not include the infinite tree. > - Randy > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > No? Of what does the set omega consist, > besides finite ordinals, each with a largest > element. There is no difference > between the set omega and all of its finite ordinals. > Nothing included in that totality causes omega not > to have a largest element. > - William Hughes > No Largest Finite!!! > Correct. Thus any argument that can be used > to show there is a largest finite is nonsense. > You just used an argument that can be used to show > there is a largest finite. > - William Hughes > Ummmm...no. > You're entirely confused about what I asserted. Let's look again at what > There is no difference between the tree and all of its finitely > numbered nodes and edges. > Notice the word, all. There is no difference between the set omega > and *all* of its finite ordinals. I did not leave all out nor did I ignore it. > Following your argument exaclty, we get the > nonsensical conclusion that there is a largest > finite. > Okay, my apologies. let me insert three words: > There is no difference between the tree and [the set of] all of its finitely numbered nodes and edges. The infinite tree is the union of all finite subtrees, but those overlap enormously. The set of all nodes/edges does not overlap; each node is unique, and each edge is a unique pair of nodes, and the set of edges is a sparse relation. It's not linear, but each has two successors, rather than one. Still, at every finite level is a finite result. > What does all its members mean, besides the > entire set? You may consider this to be some nit-picky point, but if the > tree consists only of the set of nodes, which is countable, and the set > of edges, one associated with each (parent) node, and thus equal and > equally countable to the nodes, then the entire tree is countable. Every > path that can be identified within any finite length is included, but > the vast majority which cannot, including 1/3, are nowhere completed > within the tree, and therefore are not included in the set of paths. A path is in the tree if and only if every one of its nodes > is in the tree. Why is 1/3 not in the tree? Every node of 1/3 is in > the tree. At which of those nodes is 1/3 finally completed within the tree? Can there be any such node? > You bring up the strange word completed which you refuse > to define but often use to mean has a last element. I mean that the value 1/3 is actually represented within the tree. It's not. Just admit it, already. > Are you saying that only paths with a last element are in > the tree? Yes, the set of paths with a last element > is countable. Don't stop the presses. - William Hughes Only paths with finitely-numbered levels are in the tree. Can nodes be bijected with levels? :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > CORRECTION > Is the following true or false? > The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. > FALSE. Then prove it. Given the infinite full binary tree one can count the nodes and one > can prove that, though there are at least as many paths, those paths > cannot be counted. Both proofs have been presented here and neither has > been successfully refuted. In a tree of countably infinite depth, one can show there are mnode nodes than levels, thus the nodes are uncountable. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > If the tree can contain more paths than nodes, then ?the sequence > 1,1,1,... may not be restricted to members with magnitude 1. In total > there may be more 2's than 1's, although, of course, at ever finite > position, there is a 1. That may be counter intuitive but it's not a > contradiction. > > If the sequence has a 1 at every finite position, then at what > positions can those 2's be found? > > If every finite binary tree has more nodes than paths, and if the > infinite tree is nothing but the union of all finite trees, then at > what positions does the overtaking of nodenumber by pathnumber happen? If the tree is nothing but the union of all finite trees, then it does > not contain any infinite-length paths, so the question is moot. (If it did contain infinite-length paths, we'd ask what finite trees > they > came from when all the finite trees were unioned together.) This is equivalent to saying that the union of all finite naturals > does not contain any infinite naturals. Yes, David, that's true. > And what mechanism causes the miraculous increase in pathnumber? If you're talking about a binary tree that does contain infinite- > length > paths, then it's easily shown that there are more infnite-length paths > than finite-length paths (or nodes at the ends of finite-length > paths). But then that is a different binary tree. For any two complete binary trees of different depths, there are more paths in the deeper of the two, and since infinite depth is greater than finite depth, sure, what you say is correct. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. So let's ask a simpler question. The full binary tree has paths which > are infinitely long, yes? > > As countably infinite as the number of levels in the tree, so yes. Or no. See below. WHAT???? (shakes head, sighs) Are you saying that there exist any infinite paths from the root which do not have exactly one node from each level? If you agree that's not true, doesn't that not make the number of levels in the tree and the number of nodes in each such path exactly equal? > > There is a bijection between the set of rationals of the form 1-m*(2^-n) > for m and n in N and the set of all finite paths within the tree. Since > no node in the tree is ever infinitely distant from the root, no > irrational is ever specified fully within the tree, and therefore > doesn't exist within that tree. For every node in the tree, there is a > rational specified, and an infinite number of irrationals, not to > mention rationals not of the form 1-m*(2^-n) for m and n in N, yet to be > specified. There are no other nodes in the tree, and so many reals are > missed. The tree includes a number of levels equal to omega, a number of > nodes equal to 2^omega-1, and a number of paths equal to 2^(omega-1), > whatever value this phantom omega may represent. Tony is describing a binary tree with no infinite paths, i.e., a tree > composed of only nodes and (all of the) the finite paths to those > nodes. So his tree does /not/ contain any infinite paths. I am describing a path of a countably infinite number of levels, with no infinitely numbered levels. In other words, his tree is simply a subset of the rationals, each > represented by a finite-length path, or some node at the end of > a finite path, within the tree. (The rationals in the interval [0,1] > to be more precise.) So, for example, the binary number .01111 > (15/32) is represented by the node at the end of the path > . Which nodes actually complete any specification of a number other than 1-(m*2^-n) for m and n in N? Of course, that means that his tree does not contain any infinite > paths, i.e., that it does not contain any paths corresponding to > irrational numbers and quite a lot of rationals. For example, > there is no path corresponding to 1/3, or .010101..., because it > requires an infinite path. So the tree does not map to all the reals > within an interval, since it omits all of the irrationals in that > interval. > Which are most of the reals in that interval. Well, yes. Within the tree are only the small subset of the rationals described above. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. So let's ask a simpler question. The full binary tree has paths which > are infinitely long, yes? > > As countably infinite as the number of levels in the tree, so yes. Or no. See below. WHAT???? (shakes head, sighs) Are you saying that there exist any infinite paths from the root which > do not have exactly one node from each level? If you agree that's not > true, doesn't that not make the number of levels in the tree and the > number of nodes in each such path exactly equal? > > There is a bijection between the set of rationals of the form 1-m*(2^-n) > for m and n in N and the set of all finite paths within the tree. Since > no node in the tree is ever infinitely distant from the root, no > irrational is ever specified fully within the tree, and therefore > doesn't exist within that tree. For every node in the tree, there is a > rational specified, and an infinite number of irrationals, not to > mention rationals not of the form 1-m*(2^-n) for m and n in N, yet to be > specified. There are no other nodes in the tree, and so many reals are > missed. The tree includes a number of levels equal to omega, a number of > nodes equal to 2^omega-1, and a number of paths equal to 2^(omega-1), > whatever value this phantom omega may represent. Tony is describing a binary tree with no infinite paths, i.e., a tree > composed of only nodes and (all of the) the finite paths to those > nodes. So his tree does /not/ contain any infinite paths. I am describing a path of a countably infinite number of levels, with no > infinitely numbered levels. > Which nodes actually complete any specification of a number other than > 1-(m*2^-n) for m and n in N? None complete any but binary fractions, but for infinite trees: Every real between 0 and 1, has at least one binary expansion. If the nth digit in that expansion is 1 then branch left at the nth node, otherwise (when digit is zero) branch right. That gives a path exactly representing each such number. See how easy that was? === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. > When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. > Not true. When you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. > Not the ending is decisive but only the fact that all path are formed by > nodes. > However if you have a tree where no path ends you do not have > a direct argument. All you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). But paths > that end and paths that don't end are very different things. > If you have a sequence of terms a_n = 1/2^n, > Please do not change the subject. It is the same as before. It shows that there are no steps in the > infinite. > But, my dear Herr Mueckenheim, indeed there are! Alas, that is a point on which you agree with Virgil and MoeBlee. What is one to do? :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > It shows that there are no steps in the > infinite. But a lot of them are required to get there. > === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. > When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. > We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Not true. When you have a tree where all paths end it is true that > there > are less paths than nodes in the tree. You have a direct argument for > this case. > However if you have a tree where no path ends you do not have > a direct argument. All you say is that if it is true for the paths > that end it must be true for the paths that don't end (after all > paths that don't end are made up of paths that end). But paths > that end and paths that don't end are very different things. - William Hughes > No Largest Finite!! (GONG) Huyah huyah huyah Ommmmmmmega!!! === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > If the tree can contain more paths than nodes, then the sequence > 1,1,1,... may not be restricted to members with magnitude 1. In total > there may be more 2's than 1's, although, of course, at ever finite > position, there is a 1. That may be counter intuitive but it's not a > contradiction. > Hahaha. Good one, Wolfgang. > Yeah, it's funny, but in a tragic kind of way. If the sequence has a 1 at every finite position, then at what > positions > can those 2's be found? > If every finite binary tree has more nodes than paths, and if the > infinite tree is nothing but the union of all finite trees, then at > what positions does the overtaking of nodenumber by pathnumber happen? There are exactly as many paths in a finite tree as there are leaf > (terminal) nodes, so by WM's logic, the number of paths in the infinite > tree cannot be larger than the number of terminal nodes in such a tree, > which is zero. No Largest Finite!!! (GONG) Huyah huyah huyah... Ommmmmmega!!!! > > And what mechanism causes the miraculous increase in pathnumber? The sudden vanishing of every terminal node might be the mechanism, > since as long as every path has one, the number of paths remains finite. > It is a contradiction to say that the > sequence > is composed of anything /except/ numbers at finite positions. > But it's not a contradiction to say that the sequence (a_n) = 1/2, > 1/2, 1/2, ... has the limit oo? If WM's notion of limit produces that result, he should stick with it, > but ours does not. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Thus we have in fact two different > things but we can calculate directly by a simple formula that paths > which stretch not farther than the tree and its nodes, cannot gain a > greater cardinality than the nodes. > So the question is not whether the paths are finite or infinite but > whether they consist only of nodes which have a finite distance from > the root or not. Correct! :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > When you have a path that passes the whole finite tree you > have a path with an end. When you have a path that passes the whole > infinite tree you have a path without an end. A path with > and end and a path without an end are two different things. > You use the same word, path to refer to two different things. When you have a path in a tree with less than 10 levels, then you have > a path with less than 10 nodes. When you have a path in a tree with > more than 10 levels, then you have a path with more than 10 nodes. > These are two different things. Nevertheless for both sorts of paths > it is true that there are less paths than nodes in the respective > trees. Why do we know that? Because we can calculate it. We can calculate even more. When we have a path which reaches not > farther than the nodes reach, then it is true that there are less > paths than nodes in the tree. Thus we have in fact two different > things but we can calculate directly by a simple formula that paths > which stretch not farther than the tree and its nodes, cannot gain a > greater cardinality than the nodes. > So the question is not whether the paths are finite or infinite but > whether they consist only of nodes which have a finite distance from > the root or not. Correct! WM is still willfully blind, and deceives TO into the bargain. In an infinite path, each node (like each n in N) is only a finite distance from the root, but for every n in N, there are infinitely many nodes farther than that from the root. The thing is that in the complete infinite binary tree, uncountably many paths pass through each node, so counting nodes is irrelevant to the number of paths. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) > You have solved the problem, Tony! > If balls couldn't get in to start with, there wouldn't be any problem. Vases all have holes to let things, balls in this case, in. Oh. Well, yeah. I guess I meant a second hole. Good point, Virgil. You win. But, I'm still partial to the soap bubble theory... ;) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > In any case, the function WM addresses, f(x)=x, does not have any such > undefinedness at any defined x. If omega exists, f(omega)=omega. I don't > see how this identity function, a direct unit linear relationship, can > ever be considered discontinuous at any point that exists on that line. > Do you? > The function f has a certain DOMAIN. In Z set theories there is NO > function that has EVERYTHING in its domain. If w is not in the domain > of f, then, in ordinary mathematics, 'f(w)' is taken as undefined > (though, in more formal context, with the Fregean method, we can avoid > having anything undefined like that). And continuity applies to f > evaluated at points in the domain of f - such a domain as the set of > real numbers in which continuity can make sense as defined in terms of > some metric or whatever. Continuity doesn't apply to EVERYTHING you > want to shove in as an argument even though it is not in the domain of > the function. > MoeBlee > So, in other words, you don't want to consider the identity function to > apply to anything other than real numbers, > No, I've not suggested anything so ridiculous, and I don't appreciate > you suggesting that I have. > :0 Uhhhhh.... > What is your objection, then? It's been explained to you by me in my original post, and then by at > least two other posters after me. A function has a domain; it's > nonsense to argue about the values of a function for arguments not in > the domain. > Consider the domain to be all set counts in N+. > Didn't it have something to do with the > domain of the identity function? In Z set theories, there is no the identity function. There is an > identity function for every set. The set is the domain of the identity > function on that set. And there is no function that has EVERY set in > the domain of the function. The set of all ordinals, if you must. > > Perhaps you were confused. Maybe it's me. Not maybe. > No, not even maybe. I was being nice. > Let me reiterate: > f(x)=x -> f(w)=w WHAT IS THE DOMAIN OF f? V!!!! That includes the set of all counts, finite and infinite. > > The nth natural is n, that is, N_n=n, thus index is equal to magnitude. What is 'N' in 'N_n'? I suppose in that context you mean for N to be > the identity function on w (but then you take |N| as an argument, so > it's not clear what you intend as the domain of N). Talk about confused!! N is .... N. You've heard of it. _n is a subscript notation, meaning the nth one. So, N_n is the nth member of N. One might think you could glean that, expecially given the fact that I preceded that by, The nth natural is n, that is, As to 'magnitude', if you DEFINE it, then we can discuss whether your > statement about it holds or not. The value of the natural, the number of unit intervals between it and the origin, the quantity it represents. > > A set of size n has an nth element, Yes, where n is a natural number. In general, wherever a specific size exists. > > that is, N_|N| e N. That is NONSENSE. You're using 'N' in TWO DIFFERENT senses in just one short formula and > then pile even more nonsense on to it. No, you went of on an illogical tangent about how I might be using N, as if there is any ambiguity there. First, you use 'N' as some kind of function. No I didn't. Were I to use N as a function, I'd say N(n), but I used N as the natural sequence we all know, and N_n to refer to the nth element of N, as I stated verbally in that very sentence. Please! This obfuscation is fatiguing. That's what 'N_n' > indicates. N_n is the value of the function N at the argument n. Then > you take the cardinality of 'N', which has to be the cardinality of > that function. Then you claim that cardinality is an ARGUMENT to the > function (but then what is the DOMAIN) of that function? Then you > claim that the function at the argument that is the cardinality of > that function is a MEMBER of that function, which is NONSENSE. It is PATHETIC that you're becomming even MORE mathematically > infantile. > I recommend you do some breathing exercises, so you can focus on what you're reading. Sorry to get you so flustered. > I leave the rest of the exercise to the reader. WHAT exercise? Your breathing exercises, apparently. > > Hint: what in the universe is not equal to itself? In ordinary mathematical theories, in any given universe, there is no > object not equal to itself. What is your point? MoeBlee 0 is my starting point. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > fed this fish to the penguins: > 1. Potential > 2. Actual > 3. A deep fried banana. > Fried banana I can agree, but is it *potentially* fried or *actually* > fried. Big difference you know. And is it to be fried in trans-finite fat? Not very healthful, > according to current beliefs. Though I hear it's high in omega fatty acids. Hey wait a minute..... ;) === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > fed this fish to the penguins: > 1. Potential > 2. Actual > 3. A deep fried banana. > Fried banana I can agree, but is it *potentially* fried or *actually* > fried. Big difference you know. And is it to be fried in trans-finite fat? Not very healthful, > according to current beliefs. Good one, Chas. Have a nice day! :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > 1. Potential > 2. Actual 3. A deep fried banana. > 4. Calling birds === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) > You have solved the problem, Tony! > I'm a little more inclined to the soap bubble theory, myself. Heh! ;) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > On 20 Feb., 13:55, Aatu Koskensilta Here in sci.logic we have a fine example of this, in the poster Andrew > abo Boucher, a true skeptic, going to the baffling extremes of > doubting that every natural has a successor. > What is baffling about that? It is fact. > And if you had the courage of your convictions, you would now say > that the set of natural numbers is not potentially infinite. I said it already many times. Compare the quote on > mhtml:http://www.hs-augsburg.de/~mueckenh/MR.mht or the last pages of > http://arxiv.org/ftp/math/papers/0505/0505649.pdf > But the restriction of its cardinality to less than 10^100 does not > imply the existence of a greates number. > Further, even N was potentially infinite, set theory was inconsistent. > Therefore I sometimes argue without regard to physical restrictions. > Ummm, WM, doesn't the restriction of cardinality less then 10^100 imply 10^100 is some kind of largest natural, and that 10^100+1 doesn't exist? I think you do your point a disservice by mentioning googol. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Since > no node in the tree is ever infinitely distant from the root, no > irrational is ever specified fully within the tree, and therefore > doesn't exist within that tree. For every node in the tree, there is a > rational specified, and an infinite number of irrationals, not to > mention rationals not of the form 1-m*(2^-n) for m and n in N, yet to be > specified. There are no other nodes in the tree, and so many reals are > missed. All the irrationals, and some of the rationals, are represented by paths > having both infinitely many left branchings and infinitely many right > branchings. Ja! > > The tree includes a number of levels equal to omega, a number of > nodes equal to 2^omega-1, and a number of paths equal to 2^(omega-1), > whatever value this phantom omega may represent. Wrong The 'number' of nodes is easily provably to be no greater than the > 'number' of naturals. To wit: > For any given node start a binary numeral with 1 and thereafter append 1 > for every right branch or 0 for each left branch along the finite path > path to the given node. This is easily seen to establish a surjection from the binary naturals > to the set of all nodes. Q.E.D. You have not responded to what I said, but only said the same thing you have always said, which you did not invent. Pay close attention to the following, and please respond: Are there omega levels in the tree? There are countable yet infinite, no? I'll assume you agree. (respond here) The tree up to each level n contains 2^n-1 nodes does it not? Is there a bijection between nodes and levels? Let's see. Why don't I start by listing the levels? I guess I'll use binary numbers. I think they're neat: 0 1 10 11 100 ... Okay, that looks like a fine list. It's countably infinite right? (respond here) Now, I wonder if I can list all the nodes in this countably infinite list. Here's the first one: 0 0 That looks pretty even. So far so good. Let's go to the next level: 0 0 1 1 Wait a minute. At level 1 there are two nodes, 1 and 10 (1 and 2). Hmm, maybe we can just put 10 on the next line: 0 0 1 1 10 10 Doh! Now I'm up to level 10, and I included that missing element from my list, but 4 more are excluded!!! Nodes 11, 100, 101, and 110 (3, 4, 5, and 6) are all missing from my list!! Is there any way to squeeze all these nodes into my countably infinite list of levels? Gosh, I just don't see the possibility for bijection. I feel like Cantor! If I add the next level to my list, I've included one of those four, but now I have 8 more unincluded!!! This won't work. It's just getting worse. I'm sorry, but it would appear you don't have a bijection between the countably infinite number of levels, and the much greater number of nodes. It would appear that the nodes must be uncountable. I'm sure Cantor would agree. Do you? If not, why not? How is this different from Cantor's first diagonal argument? (respond here) Love, Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Are there omega levels in the tree? There are countable yet infinite, > no? I'll assume you agree. (respond here) The cardinal number of levels equals the cardinality of N,but no level has the cardinality of N. > [garbage snipped] > I'm sorry, but it would appear you don't have a bijection between the > countably infinite number of levels, and the much greater number of > nodes. Since one can easily biject the levels with the member of N, and the members of N with the nodes, a composite bijection exists between levels and nodes. That one attempt did not work does not mean a better attempt will not succeed > It would appear that the nodes must be uncountable. Not to me. > I'm sure > Cantor would agree. Not bloody likely. > Do you? If not, why not? How is this different from > Cantor's first diagonal argument? By being garbage. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor With the one-origin naturals, you have |{1,2,...,n}| = n, > ??? You will have that in any case, for any natural n. so that each n has to be a member of the initial set having it as cardinal, > So what? This is just a theorem in ZFC. which is more than a bit circular. > Nonsense. (It would be circular, of course, to try to DEFINE n using {1,...,n}.) In ZFC we have |{1,2,...,n}| = n for any natural n. And if we define the naturals due to von Neumann in addition we have |n| = n. :-) F. -- E-mail: infosimple-linede === Subject: Re: A consideration concerning the diagonal argument of G. Cantor In my assertions of infinite case induction, I made clear rules as to > what kind of properties could be derived > You haven't made a clear rule since the fouth grade when you said that > anything hit over Mr. Wilson's fence but not past his garden is a > ground rule double. > Good one > (pats self on back, injures shoulder) > But seriously, I stated that it applied to formulaic equalities, and > formulaic inequalities that were not based on a difference not with a > limit of 0 as n->oo. Can you unscramble that particular line of double talk? > Sure, though the request definitely seems to evidence the fact that you haven't listened to any of it in the past. But, hopefully, you'll try to make sense of what I'm saying this time, since you asked. Here goes. If we have a proof that f(n)=g(n), for all n in N (the finite cardinalities and set counts), then infinite-case induction allows us to assert that f(n)=g(n) for infinite set counts (positive whole numbers) as well. If we have a proof that f(n)oo: g(n)-f(n))>0, then we can assert that f(n) > It doesn't apply to is finite, has a last > element, likes chocolate eclairs, or has an attitude. So, you know, > I kinda don't appreciate it being characterized as, infinite sets are > the same as finite ones. That disingenuous. Not to get pissy or anything. You are saying precisely that there are contexts in which they are > alike, but fail to be specific about which contexts. I have said this before, though it's been a while, so you might not recall. Infinite case induction does not apply to all statements. As an example, it is quite easy to prove that for every n in N, 1-1/n < 1, since for every n in N 1/n>0. But, does this apply to the infinite case? No, because lim(n->oo: 1/n)=0, so in the infinite case, the difference between 1 and 1-1/n is zero or infinitesimal, and does not serve to distinguish between two real numbers. This proof for all n in N fails the criteria for infinite-case induction. I hope that was clear enough. Have a nice day. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > But seriously, I stated that it applied to formulaic equalities, and > formulaic inequalities that were not based on a difference not with a > limit of 0 as n->oo. Can you unscramble that particular line of double talk? > Sure, though the request definitely seems to evidence the fact that you > haven't listened to any of it in the past. But, hopefully, you'll try to > make sense of what I'm saying this time, since you asked. Here goes. If we have a proof that f(n)=g(n), for all n in N (the finite > cardinalities and set counts), then infinite-case induction allows us to > assert that f(n)=g(n) for infinite set counts (positive whole numbers) > as well. Given that f(n) and g(n) each have some meaning for each n in N, what guarantees that f(x) and g(x) both have meaning for an x not in N? If we have a proof that f(n) N, such that lim(n->oo: g(n)-f(n))>0, then we can assert that f(n) for infinite positive n as well. Given that f(n) and g(n) each have some meaning for each n in N, what guarantees that f(x) and g(x) both have meaning for some x not in N? Absent such a guarantee, you axioms are garbage. I think that's reasonable, and intuitive, and useful. You might, if you > give it a try. But, you have to put set theory aside temporarily, if you > can, or it will all seem wrong. You are requiring functions to have values at points outside their domains. Which doesn't work in any form of mathematics. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > In any case, the function WM addresses, f(x)=x, does not have any such > undefinedness at any defined x. If omega exists, f(omega)=omega. I don't > see how this identity function, a direct unit linear relationship, can > ever be considered discontinuous at any point that exists on that line. > Do you? > The function f has a certain DOMAIN. In Z set theories there is NO > function that has EVERYTHING in its domain. If w is not in the domain > of f, then, in ordinary mathematics, 'f(w)' is taken as undefined > (though, in more formal context, with the Fregean method, we can avoid > having anything undefined like that). And continuity applies to f > evaluated at points in the domain of f - such a domain as the set of > real numbers in which continuity can make sense as defined in terms of > some metric or whatever. Continuity doesn't apply to EVERYTHING you > want to shove in as an argument even though it is not in the domain of > the function. > MoeBlee > So, in other words, you don't want to consider the identity function to > apply to anything other than real numbers, > No, I've not suggested anything so ridiculous, and I don't appreciate > you suggesting that I have. > :0 Uhhhhh.... > What is your objection, then? Didn't it have something to do with the > domain of the identity function? Perhaps you were confused. Maybe it's me. > Let me reiterate: > f(x)=x -> f(w)=w > The nth natural is n, that is, N_n=n, thus index is equal to magnitude. A much better scheme is to use the von Neumann naturals, as otherwise > the empty set does not have a natural number cardinality. And with the > von Neumann naturals, {} = 0 is a natural first natural. That's fine, but doesn't change anything. You can say the 0th is 0 if you want to count from 0. > A set of size n has an nth element, that is, N_|N| e N. Then the set of the first n naturals does not contain n. I said we were using the naturals from 1, but if you want to say the first is 0, then it's the 0th. If you don't want to start the count from 0, then don't include 0 in the counting numbers. :) With the one-origin naturals, you have |{1,2,...,n}| = n, so that each n > has to be a member of the initial set having it as cardinal, which is > more than a bit circular. Oh no!! How irrelevant!!! :o === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > With the one-origin naturals, you have |{1,2,...,n}| = n, so that each n > has to be a member of the initial set having it as cardinal, which is > more than a bit circular. Oh no!! How irrelevant!!! Not to mathematicians. Anything circularly defined in mathematics is justifiably suspect. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor That wasn't my argument. Pay attention. There are only finite levels > in the infinite binary tree. At each finite level, there are twice as > many nodes in the tree as paths. Hence, there is no point in the tree, > even when fully complete, where paths exceed nodes. > I have no idea what this means, since I don't no what you mean by a > point in the tree [...] where paths exceed nodes. > Does that mean you yes it? Interesting. You're coming around. ;) No. It means I do not know what it means. How could you possibly > take that as agreement? I am playing with you. You said, I don't *no* what you mean, so I asked if you yessed it. (sigh) I guess that didn't compute... (bass thump and cymbal splash) So let's ask a simpler question. The full binary tree has paths > which > are infinitely long, yes? > As countably infinite as the number of levels in the tree, so yes. > Is the following true or false? > The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. > In other words: > There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. > There is a bijection between the set of rationals of the form > 1-m*(2^-n) for m and n in N and the set of all finite paths within the > tree. Since no node in the tree is ever infinitely distant from the > root, no irrational is ever specified fully within the tree, and > therefore doesn't exist within that tree. For every node in the tree, > there is a rational specified, and an infinite number of irrationals, > not to mention rationals not of the form 1-m*(2^-n) for m and n in N, > yet to be specified. There are no other nodes in the tree, and so many > reals are missed. The tree includes a number of levels equal to omega, > a number of nodes equal to 2^omega-1, and a number of paths equal to > 2^(omega-1), whatever value this phantom omega may represent. > You probably feel I didn't answer your question, but read again, and > see if you don't change your mind. Then rinse and repeat. They were true/false questions. You didn't even try to answer them. > Instead, you rambled a bit about nonsensical values such as omega - 1. I'll ask again. Are the above statements true or false? > They are true of a tree with uncountable levels. I don't accept that irrationals or even most rationals exists *within* that tree, but only as limits to the paths as the depth of the tree *approaches* some actual infinity. I have answered your question. Sometimes we want yes or no, but instead need to have the terms explained. I recommend you read my response again, and try to make sense of it. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor So let's ask a simpler question. The full binary tree has paths > which > are infinitely long, yes? > As countably infinite as the number of levels in the tree, so yes. > Is the following true or false? > The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. > In other words: > There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. > They were true/false questions. You didn't even try to answer them. > Instead, you rambled a bit about nonsensical values such as omega - 1. I'll ask again. Are the above statements true or false? > They are true of a tree with uncountable levels. Non-responsive. Are they true for the standard complete infinite binary tree with one level for each member of N and no other levbvels? I don't accept that > irrationals or even most rationals exists *within* that tree, but only > as limits to the paths as the depth of the tree *approaches* some actual > infinity. The infiniteness of the cardinality of N is as actual as it gets. > I have answered your question. Sometimes we want yes or no, > but instead need to have the terms explained. I recommend you read my > response again, and try to make sense of it. To make sense of it is a task even gods cannot achieve. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor I am playing with you. You said, I don't *no* what you mean, so I > asked if you yessed it. (sigh) I guess that didn't compute... (bass > thump and cymbal splash) Oh, I missed my own typo. Got it now. > Is the following true or false? > The set of infinite paths in the full binary tree has greater > cardinality than the set of nodes. > In other words: > There is no injection from the set of infinite paths in the full > binary tree to the set of nodes. There is an injection from the set > of nodes to the set of infinite paths. > There is a bijection between the set of rationals of the form > 1-m*(2^-n) for m and n in N and the set of all finite paths within the > tree. Since no node in the tree is ever infinitely distant from the > root, no irrational is ever specified fully within the tree, and > therefore doesn't exist within that tree. For every node in the tree, > there is a rational specified, and an infinite number of irrationals, > not to mention rationals not of the form 1-m*(2^-n) for m and n in N, > yet to be specified. There are no other nodes in the tree, and so many > reals are missed. The tree includes a number of levels equal to omega, > a number of nodes equal to 2^omega-1, and a number of paths equal to > 2^(omega-1), whatever value this phantom omega may represent. You probably feel I didn't answer your question, but read again, and > see if you don't change your mind. Then rinse and repeat. > They were true/false questions. You didn't even try to answer them. > Instead, you rambled a bit about nonsensical values such as omega - 1. > I'll ask again. Are the above statements true or false? They are true of a tree with uncountable levels. I don't accept that > irrationals or even most rationals exists *within* that tree, but only > as limits to the paths as the depth of the tree *approaches* some > actual infinity. I have answered your question. Sometimes we want yes > or no, but instead need to have the terms explained. I recommend you > read my response again, and try to make sense of it. Nah, I give up. No one mentioned a tree of uncountable depth. In fact, I'm not sure how to make sense of such a thing, but it's irrelevant to my question. And I don't see any reason to pursue this question any longer. In a fit of utter stupidity, I imagined that if I ask a simple question, Tony Orlow would give a clear, coherent response. I know, I know, it sounds really silly, but that's what I was thinking. That's the last time I drink Drano on an empty stomach. -- Jesse F. Hughes It's much better to live with my parents than with a wife. -- Quincy P. Hughes, age 4 1/2 === Subject: Re: A consideration concerning the diagonal argument of G. Cantor So, in other words, you don't want to consider the identity function > to apply to anything other than real numbers, so you're just not going > to do it. I am not allowed to say w=w or elephant=elephant, because > you don't want to talk about anything but reals. Gotcha. > How does saying w=w require an identity function with no domain or > range? How does it require any identity function at all? > f(x)=x. > f(w)=w? > w=w <-> f(w)=w? No. In any first order language with equality, we have the axiom (A x)(x=x). That doesn't depend on the existence of an identity function at all. > > While it is not true that w=w -> f(w)=w (though it might be under a > given rule that just occurred to me, intuitively), the converse is > true. Does w=w REQUIRE the identity function? No. Does the identity > function relate element count and value, thus shedding light on this > question? Yes. Shedding light on this trivial claim that w=w? No. > > Can a function have domain V? This one does. There is no function with domain V in ZFC. It's perfectly acceptable > to speak about an operation Set -> Set (or a functor), but that's not > a function in the set theoretic sense. > You're mightily confused. > About some things, but nothing so fundamental. > And Moe *never* said that the only legal identity function is the one > with domain and codomain R. For each set X, there is a corresponding > identity function id_X: X -> X. There is *not* a function id:Set - Set in ZFC (that is, a single function id such that id(x) is > defined for *every* set x). Your response is utterly irrelevant. > I'm obviously not restricting myself to ZFC. This thread isn't even > about ZFC per se, but about a more fundamental argument, the diagonal > argument. Like WM, I am appealing to more fundamental concepts than > ZFC for my argument, which is the same, under the hood, as his. Of > course he and I diverge in our choice of how to deal with the theory, > but it's pretty hard to accept any objection to the identity function > based on its domain. The objection has to do with whether the identity operator Set -> Set > is a set-theoretical function. It isn't. This is just a simple > terminological fact, not deep or controversial in the least. But your response was a lot stranger than you let on. First, you > claimed Moe disallowed any identity function other than id_R:R -> R. > Then you implied that equations like w=w depend on the existence of a > single identity *function* with domain V. This response was plainly > nutty, a lot nuttier than what you've said in your current post. Look, my point is simple. Everywhere within N (starting at 1) the nth element is n. There is infinite n within N, thus there is no infinite index for any element within N. We all agree on that. Here's the controversial part. In a set of size x, there exists an xth element. If there does not exist an xth element, we cannot say it has count x. Omega is declared the size of N, because the size cannot be a member of N for obvious reasons. But, there cannot be an omega-th element, because omega is defined to be larger than any n in N, and only nth elements exists for n in N. In other words, because of the identity function between element count and value, and the restriction that element value be finite, element count is finite everywhere within the set. There is no largest value, thus there is no largest count, thus there is no size. Please let me know, specifically, where you think this goes *logically* bad. Probably right after controversial part. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor So, in other words, you don't want to consider the identity function > to apply to anything other than real numbers, so you're just not going > to do it. I am not allowed to say w=w or elephant=elephant, because > you don't want to talk about anything but reals. Gotcha. > How does saying w=w require an identity function with no domain or > range? How does it require any identity function at all? > f(x)=x. > f(w)=w? > w=w <-> f(w)=w? No. In any first order language with equality, we have the axiom (A x)(x=x). That doesn't depend on the existence of an identity function at all. > > While it is not true that w=w -> f(w)=w (though it might be under a > given rule that just occurred to me, intuitively), the converse is > true. Does w=w REQUIRE the identity function? No. Does the identity > function relate element count and value, thus shedding light on this > question? Yes. Shedding light on this trivial claim that w=w? No. > > Can a function have domain V? This one does. There is no function with domain V in ZFC. It's perfectly acceptable > to speak about an operation Set -> Set (or a functor), but that's not > a function in the set theoretic sense. > You're mightily confused. > About some things, but nothing so fundamental. > And Moe *never* said that the only legal identity function is the one > with domain and codomain R. For each set X, there is a corresponding > identity function id_X: X -> X. There is *not* a function id:Set - Set in ZFC (that is, a single function id such that id(x) is > defined for *every* set x). Your response is utterly irrelevant. > I'm obviously not restricting myself to ZFC. This thread isn't even > about ZFC per se, but about a more fundamental argument, the diagonal > argument. Like WM, I am appealing to more fundamental concepts than > ZFC for my argument, which is the same, under the hood, as his. Of > course he and I diverge in our choice of how to deal with the theory, > but it's pretty hard to accept any objection to the identity function > based on its domain. The objection has to do with whether the identity operator Set -> Set > is a set-theoretical function. It isn't. This is just a simple > terminological fact, not deep or controversial in the least. But your response was a lot stranger than you let on. First, you > claimed Moe disallowed any identity function other than id_R:R -> R. > Then you implied that equations like w=w depend on the existence of a > single identity *function* with domain V. This response was plainly > nutty, a lot nuttier than what you've said in your current post. Look, my point is simple. Everywhere within N (starting at 1) the nth element is n. There is > infinite n within N, thus there is no infinite index for any element > within N. We all agree on that. Here's the controversial part. In a set of size x, there exists an xth > element. This only need hold for finite sets, i.e., sets having a natural number of members, where x can be a natural number. The set of natural numbers, whose size is not a natural number, for example, does not have an 'xth' element. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) Take a vase with no balls At time noon-1/n put 10 balls labelled > 10n, ... 10n+9 in the vase. > Remove the ball labelled n. Do this for every natural number n (and at no > other times, so only balls that have natural numbers > for labels are added to the vase). You argue at noon that there are an infinite number of balls > in the vase. You also agree that every ball labeled > with a natural number is removed from the vase. > The question is not where the balls with labels went, but where > did the balls without labels come from. - William Hughes within the domain of the experiment, since it would require infinite n. The Zeno machine is a primitively clever paradox factory, and marries well to the concept of a completed N with any definite end. But, I'm not impressed. sum(n=1->oo: (10-1)) diverges. Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > I also wonder where all those balls went. Maybe there is a hole in the > vase.... ;) Take a vase with no balls At time noon-1/n put 10 balls labelled > 10n, ... 10n+9 in the vase. > Remove the ball labelled n. Do this for every natural number n (and at no > other times, so only balls that have natural numbers > for labels are added to the vase). You argue at noon that there are an infinite number of balls > in the vase. You also agree that every ball labeled > with a natural number is removed from the vase. > The question is not where the balls with labels went, but where > did the balls without labels come from. - William Hughes within the domain of the experiment, since it would require infinite n. Infinitely many does not require infinitely large. For naturals, no largest is enough. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Consider the set X={0,1/2.3/4 ...} > Let P(r) be true if there is an element in X which is larger > than > r, > false otherwise. > There is no largest r for which P(r) is true, but P(1) is false > The difference occurs at r=1, > However difference and largest are not the same. > Yes the points at which P(r) is true have a supremum. > However, they do not have a maximum. > - William Hughes > At r=1, the function changes for obvious reasons. However, r<1, > because > r=1-2^(-n) for neN. Ne pas? The maximum does not exist, as we > agree, > again. Therefore, my consideration of such a set is largely > irrelevant. > No. Such a set shows that a change can occur by the supremum, but > not > before the supremum. Note the supremum of the finite binary trees > is the infinite tree. The change occurs by the supremum, but not > before > the supremum (only at the supremum are the infinite paths counted, > before > the supremum the infinite paths do not contribute to the count). > - William Hughes > I think WM's point is that the supremum doesn't exist, and that > consideration of such a point leads to contradictions. > I take it by the fact that you are trying to change the subject > that you now accept that the argument that something that holds > for every finite tree *must* hold for the infinite tree is worthless. > - William Hughes > I never made such a general statement. > However, you keep saying there is no step at which > things change as if this meant something. As steps > can only occur at finite points this is equivalent to > saying that something that holds for every finite tree > holds for an infinite tree. > - William Hughes > There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > At any step you have a finite tree. > Therefore, at no point in this set, > At every point you have a finite tree. > whether it includes every single one > of these finite steps or not, is this fact untrue. > The infinite tree is not one of those steps. Your > therefore is, as William stated, an attempt to jump > from something true for every finite tree to a > statement about the infinite tree. The set of things you get to, the set of things you > include in this sentence > There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > does not include the infinite tree. - Randy > No? Of what does the tree consist, besides finite levels with finitely > numbered nodes and edges? How do you define includes besides contains > all parts? There is no difference between the tree and all of its > finitely numbered nodes and edges. Nothing included in that totality > causes paths to outnumber nodes. > You argument that whatever holds for each finite tree must somehow hold > for the infinite tree would justify claiming that the infinite tree is > finite. > That wasn't my argument. Pay attention. There are only finite levels in > the infinite binary tree. At each finite level, there are twice as many > nodes in the tree as paths. Hence, there is no point in the tree, even > when fully complete, where paths exceed nodes. You are saying that what happens for all the finite subtrees must hold > for the infinite tree. By that argument the infinite tree must be finite. > In my assertions of infinite case induction, I made clear rules as to > what kind of properties could be derived, and is finite was explicitly > excluded. So, your generalization above has nothing to do with anything > I said, and is a straw man. You never justified you infinite case induction rule to the satisfaction > of anyone but yourself. See below. Now every path in the infinite tree corresponds to a unique endless > sequence of left or right branchings, from the root node onwards along > that path. If every bit position is finite and the levels of the tree countable, then the only numbers specified are of the form 1-(m*2^-n) for some m and n in N. 1/3 is not fully specified within the tree, despite the fact that one can specify a path within the tree that is arbitrarily close to 1/3. But Cantor's first diagonal proof shows that no enumeration (by > members of N) of those sequences can be complete, so they cannot be > counted. No surjection from N to the set of all such sequences (or all > such paths) can exist. Cantor's first diagonal argument is one of infinite-case induction, applying a consistent fact at each finite level to the infinite case. So, Georg, at least, sees eye to eye with me on this principle. > > If this does not correspond to the result demanded by your infinite > case induction rule, then it is your rule which loses, at least in > standard mathematics. That's ironic, since it's basically my rule that started this whole mess, and the experts in the field can't see that. The von Neumann ordinals are schlock. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > You never justified you infinite case induction rule to the satisfaction > of anyone but yourself. See below. > Now every path in the infinite tree corresponds to a unique endless > sequence of left or right branchings, from the root node onwards along > that path. If every bit position is finite and the levels of the tree countable, > then the only numbers specified are of the form 1-(m*2^-n) for some m > and n in N. 1/3 is not fully specified within the tree, despite the fact > that one can specify a path within the tree that is arbitrarily close to > 1/3. Since one is allowed infinite paths in an infinite tree, there will be a path that nails 1/3 exactly, i.e., 1/3 - pathvalue = 0. > But Cantor's first diagonal proof shows that no enumeration (by > members of N) of those sequences can be complete, so they cannot be > counted. No surjection from N to the set of all such sequences (or all > such paths) can exist. Cantor's first diagonal argument is one of infinite-case induction, > applying a consistent fact at each finite level to the infinite case. > So, Georg, at least, sees eye to eye with me on this principle. Cantor's first diagonal argument merely shows how to construct a binary sequence not in a given list of binary sequences. No induction beyond standard induction is needed. The von Neumann ordinals are schlock. Compared to the von Neumann naturals, everything that TO does is schlock. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > You argument that whatever holds for each finite tree must somehow hold > for the infinite tree would justify claiming that the infinite tree is > finite. > > That wasn't my argument. Pay attention. There are only finite levels in > the infinite binary tree. At each finite level, there are twice as many > nodes in the tree as paths. Hence, there is no point in the tree, even > when fully complete, where paths exceed nodes. Your argument is equivalent to saying that there are more > rationals than irrationals. In fact it's exactly the same argument > if you're considering binary fractions in the interval [0,1]. > Each rational is a terminating binary fraction (or node in a > binary tree), and each irrational is a non-terminating fraction > (or path within the tree). No irrational is ever completed within the tree of countable levels, since at each and every node in the tree, a rational is specified, and no irrational. Irrationals, to be fully specified, require infinite bit positions to be specified exactly. Where your argument fails (as has been pointed out to you > countless times) is that the rationals are denumerable but > the irrationals are not. Each rational binary fraction can be > labeled with its finite binary expansion, but none of the > irrationals can be labeled with any finite string of digits. I fully understand that. It's simply amazing that you persist in failing to comprehend > that what happens at every finite node has no bearing on what > happens to the set (or tree, or interval) as a whole. It's simply amazing that anyone thinks that anything but rationals are specified at any point within the tree, and not even ALL rationals, either. 1/3 is never fully specified at any finite bit position within the tree. Perhaps you would be willing to consider infinite bit positions? ;) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > No irrational is ever completed within the tree of countable levels, > since at each and every node in the tree, a rational is specified, and > no irrational. Irrationals, to be fully specified, require infinite bit > positions to be specified exactly. TO still conflates a set of infinitely many bit positions (indices) as necessarily having infinitely large ones. The existence of the former in any set does not require the existence of latter in that set in any axiom system of standard mathematics/set-theory. And TO has as yet to produce any axiom system in which it does. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > So what? Even if we have Y is made by steps one for each natural number > At every step P is true. We cannot conclude that P is true for Y. - William Hughes > We can conclude that P is true for all of Y, which consists of all the > steps you mention. We cannot conclude the property is true for Y as a > whole, but that's not what I'm arguing, anyway. > > If by all of Y Tony means each member of Y, he would be less > misunderstood if he used the latter form. > If he means something else, he is almost certainly wrong, but I await > his explanation before claiming certainty. Okay, yes. Indeed, I mean for every defining element of the structure/set Y, say, parametrically using n as the number of levels. At each level there are 2^n nodes and 2^n paths, and 2^(n+1)-1 nodes at that and all lower levels. There are clearly as many nodes as paths at each level, and that's not counting the nodes at all previous levels that have contributed to the formation of the paths. :) Tony === Subject: Re: A consideration concerning the diagonal argument of G. Cantor There are only finite steps in the sequence of adding levels, and at > none of those steps is the relationship nodes=2*paths-1 untrue. > So what? Even if we have Y is made by steps one for each natural number > At every step P is true. We cannot conclude that P is true for Y. - William Hughes > We can conclude that P is true for all of Y, which consists of all the > steps you mention. We cannot conclude the property is true for Y as a > whole, but that's not what I'm arguing, anyway. > > If by all of Y Tony means each member of Y, he would be less > misunderstood if he used the latter form. > If he means something else, he is almost certainly wrong, but I await > his explanation before claiming certainty. Okay, yes. Indeed, I mean for every defining element of the > structure/set Y, say, parametrically using n as the number of levels. At > each level there are 2^n nodes and 2^n paths, and 2^(n+1)-1 nodes at > that and all lower levels. There are clearly as many nodes as paths at > each level, and that's not counting the nodes at all previous levels > that have contributed to the formation of the paths. But TO, as with WM, is only counting those subpaths from the root that end at a given node, each such subpath representing the uncountable set of all endless paths that coincide on that subpath. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... > Hilarious!!! > Of course. As ususal. I'm perfectly willing to accept this as a statement > that the notation a1 + a2 + ... means limit of > the sequence of partial sums, if it exists. > But... (1) When using the definition symbol := we (definitely) have the following order of terms: Definiendum := Definiens So we MIGHT write a_1 + a_2 + ... := lim_{n -> oo} (a_1 + ... + a_n) Moreover... (2) On both sides of the := symbol we have expressions (formulas). For example 2 := 1 + 1 But we DON'T write 2 := 1 + 1. Comment: We MIGHT, on the other hand, formulate the definition in the following way: 2 denotes the number 1 + 1. Hence we MIGHT formulate: a_1 + a_2 + ... denotes the number lim_{n -> oo} (a_1 + ... + a_n) [if it exists]. I guess the latter is the way you interpreted WM's statement. VERY charitable. (Actually, NOT a good idea when arguing with WM, imho.) F. -- E-mail: infosimple-linede === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Definition: lim_[n --> oo] (a_1 + ...+ a_n) := a_1 + a_2 + .... > That's OK > Nope. That's NONSENSICAL - as usual. (1) When using the definition symbol := we have the following order of terms: Definiendum := Definiens So we MIGHT write a_1 + a_2 + ... := lim_{n -> oo} (a_1 + ... + a_n) (2) On both sides of the := symbol we have expressions (formulas). For example 2 := 1 + 1 But we (usually) DON'T write 2 := 1 + 1. Comment: We MIGHT, on the other hand, formulate the definition in the following way: 2 denotes the number 1 + 1. Hence we MIGHT formulate: a_1 + a_2 + ... denotes the number lim_{n -> oo} (a_1 + ... + a_n) [if it exists]. F. -- E-mail: infosimple-linede === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=bSICGQkAAADSbkxAJ5uMxFegr4rp0Qig AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) > IF actual infinity is a meaningful notion, THEN we can sum aleph 0 > terms. This is expressed by my definition. No it's not. You defined the notation > æ a 1 + a 2 + ... > to be another way of writing > æ lim(n->oo) (a 1 + a 2 + ...) > which we are all agreed is not a thing that involves > summing aleph 0 terms. It's just a limit of a sequence > æ æ{S 1, S 2, ... } > the sequence of partial sums. And that limit is defined > in terms of finite sums and their values at finite > values of n. As we all agree, including you. I agreed under the premise that aleph 0 is not a meaningful, i.e., > consistent notion. If aleph 0 elements exist, why then should the > limit be different from the sum over aleph 0 terms? Well, there's vague feelings about what should be according to the way you kinda sorta feel they should be when you squint at it, especially after a couple of beers... ... and then there's proof. You haven't even defined a sum over aleph 0 terms. As I said, you explicitly said that the notation a 1 + a 2 + ... refers to the same thing as the limit notation, which is nothing of the sort. - Randy === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) And then Isan, the cooking monk, tipped over the vase with his foot and became the new chief head dude of the monastery. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > (2^{n+1} - 1) / 2^n is Cauchy and converges to 2, and is always > greater than one, and is definitely not zero, but that is the ratio of > counts of nodes to paths, > in finite trees. æNot in the infinite tree, nor is > their any reason to assume that the count(limit) is > equal to the limit(count) (in this case it is false). > There is a good reason to state that the limit of the sequence > 1, 1, 1, ... is 1. Since we are not discussing the sequence 1,1,1 ... > this statement is not relevent. My arguing is based upon the fact that in mathematics the limit for n-- >oo can be obtained by Cauchy's criterion. In the binary tree we have > for level n number of nodes down to level n / number of paths which can be > distinguished down to level n : a_n = (2^(n+1)-1)/2^n = 2 - 1/2^n The sequence (a_n) has limit 2. This means, by Cauchy's convergence > criterion as well as by the simple delta-epsilon criterion, in the > infinite tree, there are not more paths than nodes. You must assert > that the limit is 0 in order to save set theory. Therefore set theory > is in contradiction with mathematics. Not at all. WM's set theory may be in contradiction to WM's mythematics, due to his inclusion of a variety of unstated and possibly mutually inconsistent axioms, but that is not the fault of either standard mathematics or any of the standard set theories. What WM says are being distinguished by a node are the two sets of all paths through one of its child nodes. The issue is not, as WM claims, whether there are only countably many such sets but whether any such set can contain uncountably many paths. In the infinite trees we are considering, each such set of paths as WM considers is equinumerous with the whole tree as a set of paths. For any path in the infinite binary tree, consider the set of levels at which it branches left. This is a subset of N, and there is unique path for every possible such set, so WM is claiming to be able to count the power set of N. And WM is again unable to see the forest for the trees. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor IF actual infinity is a meaningful notion, THEN we can sum aleph_0 > terms. This is expressed by my definition. No it's not. You defined the notation > æ a_1 + a_2 + ... > to be another way of writing > æ lim(n->oo) (a_1 + a_2 + ...) > which we are all agreed is not a thing that involves > summing aleph_0 terms. It's just a limit of a sequence > æ æ{S_1, S_2, ... } > the sequence of partial sums. And that limit is defined > in terms of finite sums and their values at finite > values of n. As we all agree, including you. I agreed under the premise that aleph_0 is not a meaningful, i.e., > consistent notion. If aleph_0 elements exist, why then should the > limit be different from the sum over aleph_0 terms? Would WM's alleged sum over aleph_0 many terms exist for every set of such terms? If so then ONE difference would be which ordered sets of terms which can be summed, WM allowing any such, but limits being limited them to convergent ones. And if WM's alleged sums need not always exist, what are his criteria for deciding when they do and his criteria for determining the value when they do? All of that is long since well settled for standard mathematics, but to yet for WM's mytheology. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <47be3d00@news2.lightlink.com> posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > 1. Potential > 2. Actual Aatu Koskensilta fed this fish to the penguins: 3. A deep fried banana. > Fried banana I can agree, but is it *potentially* fried or *actually* > fried. Big difference you know. > And is it to be fried in trans-finite fat? Not very healthful, > according to current beliefs. Like Rodriques, I'm still trying to see the difference between > potentially fried or actually fried. One would imagine that > potentially fried is exactly the same thing as not fried, but > that probably only indicates a lack of expertise with the > subtleties of frying. Presumably, there would be some kind of test we could apply > to tell the difference. Like the test we apply to see if a set is > potentially or actually infinite. I forget what that test is exactly, > though. The debate goes on. Its iterating a perfect example of the potential infinity. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor I said I'd stay out of this thread, but I just can't let this one go > through unchallenged. The statement every node at a finite level of the tree causes two > paths and only two paths to be recognized as being different does not > even work if the tree has finite depth. Every node at a finite level of the tree causes two paths and only two > paths to be recognized as being different. These paths represent > classes, as you call it. But in order to distinguish the members of a > class you need further nodes. Two paths like 0.000,... and 0.111,... (being representants of the > classes 0.0... and 0.1...) spring off from the root node (at level > zero), They can be recognized as being different at the first level. > 0.000... and 0.0111... and all the other path which begin with 0.0 > cannot be distinguished at the first level. In order to distinguish > them, you need a node at the second level. Assuming one is only concerned > with paths that start at the root of the tree and continue all the way > to the bottom, a finite binary tree of depth n has 2^n terminal nodes > (assuming that the root is at level 0), and so there are 2^(n-k) paths > through any node at level k. Going from level k to level k+1 one > divides these paths into two classes, each of which has 2^(n-k-1) > members. But in order to distinguish the members of one class, further nodes > are required. At level n you cannot distinguish the paths which begin > with n 1's. So if original class, as in an infinite tree, is uncountable, splitting it into parts of equal cardinality gives two uncountable parts The characteristic element of the tree is > | > o > / > where o is a node with value 0 or 1. This shows that there cannot be > more classes, as you call it, than nodes. It does not show it to anyone who is not already convinced of it, and we are not, at least for infinite trees. Things which you claim without proof, we require be proved, and you have not proved your claims. > One class with up to then > undistinguishable paths comes in, two distinct classes leave it. And in completed infinite binary trees all three such classes are necessarily uncountable. > As > there are only countably many nodes, there cannot be more than > countably many classes. It is not the number of such classes which we see as uncountable, but the number of members in each of them. And nothing in your argument speaks to that issue. This means, if there are uncountably many paths assumed, then most of > these paths cannot be distinguished by nodes. When a set of paths consists of all those paths passing through a given node, as in your construction, there can be a difference between the number of such sets and the size of any one set. You are looking, irrelevantly and irrationally, only at the number of sets, while countability versus uncountability only depends on the size of such sets. It is the size, of those sets, not the number of them, which we see as relavant. The only thing we care about with regard to the number of them is that there is at least one of them, as any one uncountable set is enough. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Assuming one is only concerned > with paths that start at the root of the tree and continue all the way > to the bottom, a finite binary tree of depth n has 2^n terminal nodes > (assuming that the root is at level 0), and so there are 2^(n-k) paths > through any node at level k. Going from level k to level k+1 one > divides these paths into two classes, each of which has 2^(n-k-1) > members. But in order to distinguish the members of one class, further nodes > are required. At level n you cannot distinguish the paths which begin > with n 1's. So if original class, as in an infinite ætree, is uncountable, splitting > it into parts of equal cardinality gives ætwo uncountable parts We vare not interested in the asserted number of paths in a class, but conclude to the complete infinite tree. The characteristic element of the tree is > æ | > æ o > / æ > where o is a node with value 0 or 1. This shows that there cannot be > more classes, as you call it, than nodes. It does not show it to anyone who is not already convinced of it, and we > are not, at least for infinite trees. Then start from the root node. There are two classes. After the two nodes of the first level have been passed, there are two further classes, in total 4. After the 4 nodes of the second level have been passed, there are 4 more classes, namely 8. Can yo follow down to this second level? If s, then try to find out how many nodes are on the third level and how many classes leave the third level. You will be surprised. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Assuming one is only concerned with paths that start at the > root of the tree and continue all the way to the bottom, a > finite binary tree of depth n has 2^n terminal nodes (assuming > that the root is at level 0), and so there are 2^(n-k) paths > through any node at level k. Going from level k to level k+1 > one divides these paths into two classes, each of which has > 2^(n-k-1) members. > But in order to distinguish the members of one class, further > nodes are required. At level n you cannot distinguish the paths > which begin with n 1's. So if original class, as in an infinite ætree, is uncountable, > splitting it into parts of equal cardinality gives ætwo uncountable > parts We vare not interested in the asserted number of paths in a class Then you disclaim interest in the number of paths in a tree? But that is precisely the whole point! > but only in the number of classes at a certain partial tree. An upper bound on the number of such classes places no upper bound on the number of paths unless there is an upper bound on the class SIZE as well. And for binary trees of the sort under discussion, the classes WM speaks of are sets of all paths through a given node. In our infinite binary trees, each such class contains as many paths as the whole tree, and that is uncountably many. Then start from the root node. There are two classes. Each of which, being the set of all infinite paths passing through a given node in our in infinite tree, contains uncountably many paths, just as many, in the sense of cardinality, as the entire tree. > After the two nodes of the first level have been passed, there are > two further classes, in total 4. Each of which contains uncountably many paths. > After the 4 nodes of the second level have been passed, there are 4 > more classes, namely 8. Each of which contains uncountably many paths. And so on ad infinitum. > Can yo follow down to this second level? If s, then try to find out > how many nodes are on the third level and how many classes leave the > third level. You will be surprised. Not at all. The number of such classes is admittedly countable, but when, as is the case here, each contains uncountably many paths, WE still get uncountably many paths in each, and in their union, whatever WM thinks he gets. WM misses the point that the union of a countable set of uncountable sets is itself uncountable. The number of sets in such a situation, as long as there is at least one, is irrelevant. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > (2^{n+1} - 1) / 2^n is Cauchy and converges to 2, and is always > greater than one, and is definitely not zero, but that is the ratio of > counts of nodes to paths, > in finite trees. Not in the infinite tree, nor is > their any reason to assume that the count(limit) is > equal to the limit(count) (in this case it is false). > There is a good reason to state that the limit of the sequence > 1, 1, 1, ... is 1. Since we are not discussing the sequence 1,1,1 ... > this statement is not relevent. > Since I agree that the limit(count nodes/count paths) = 1/2 this is not relevent. The questions is: does the fact that lim(count nodes / count paths) =1/2 mean that count(lim nodes) / count(lim paths) is greater than infinitesimal? The answer is no. - William Hughes === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Since I agree that the limit(count nodes/count paths) = 1/2 this is > not relevent. lim [n-->oo] (count nodes/count paths) = lim [n-->oo] (2^(n+1)-1)/2^n = lim [n-->oo] (2 - 1/2^n) = 2 > The questions is: does the fact that lim(count nodes / count paths) > =1/2 > mean that count(lim nodes) / count(lim paths) is greater than > infinitesimal? No, that is not the question. The question is whether in the complete infinite binary tree the ratio (count nodes/count paths) is larger than 1 or if it can happen to be 0. The answer is no. OK. If you have higher inspirations than such which can be drawn from mathematical considerations, then further discussion is meaningless. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor Since I agree that the limit(count nodes/count paths) = 1/2 this is > not relevent. lim [n-->oo] (count nodes/count paths) > = lim [n-->oo] (2^(n+1)-1)/2^n > = lim [n-->oo] (2 - 1/2^n) > = 2 The questions is: does the fact that lim(count nodes / count paths) > =1/2 > mean that count(lim nodes) / count(lim paths) is greater than > infinitesimal? No, that is not the question. That is certainly one of the questions to which WM has no satisfactory answer, and one of the questions he must answer satisfactorily in order to establish his thesis. > The question is whether in the complete > infinite binary tree the ratio (count nodes/count paths) is larger > than 1 or if it can happen to be 0. Relevant to that question, it has been shown, by a variety of methods, that the node count of the infinite binary trees under discussion is only countable while the path count is not. OK. If you have higher inspirations than such which can be drawn from > mathematical considerations, then further discussion is meaningless. Discussionwith WM, at least such as is based on accepting logical consequnces, is almost always meaningless, as he cannot neither follow the logic of others nor generate any on his own. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > For some sums like 1/2 + 1/4 + 1/8 + ... we need no Cauchy criterion > to prove that it is less than 100. No, we need some common sense. > We need something like a Cauchy criterion to prove that it is at all. Whether is is or not, we need no criterion to see that it is positive > and less than 100. With no criteria at all, it does not even have meaning, much less value. All we can see from your statement is that you do not understand that meaning. > Absent some proof of convergence, an infinite series has no value at > all. Inform yourself about summation of divergent series. Divergent series, by definition, do not converge to anything. Nevertheless I gave you a hint how some of them can be summed. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > For some sums like 1/2 + 1/4 + 1/8 + ... we need no Cauchy criterion > to prove that it is less than 100. > No, we need some common sense. > We need something like a Cauchy criterion to prove that it is at all. > Whether is is or not, we need no criterion to see that it is positive > and less than 100. With no criteria at all, it does not even have meaning, much less value. All we can see from your statement is that you do not understand that > meaning. The meaning to me is that there is a real number L such that for every positive real epsilon, all but finitely many of the partial sums, Sum_[1 <= k <= n] 1/2^k, are within epsilon of L. And that L has value 1. > Absent some proof of convergence, an infinite series has no value at > all. > Inform yourself about summation of divergent series. Divergent series, by definition, do not converge to anything. Nevertheless I gave you a hint how some of them can be summed. > === Subject: Picture of a 'real scheme' posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20071204 Ubuntu/7.10 (gutsy) Firefox/2.0.0.11,gzip(gfe),gzip(gfe) can anybody draw a picture of a scheme which is not affine? (for example a curve or something) At the first glance is a silly question (even for me), because the choice of the term 'affine' seems to be something which is like a piece of paper, so it should be impossible to draw a 'natural' picture of a non-affine scheme. However, it is not clear that every curve-like thing which I draw on a paper is defined by zeros of polynomials: Take for example the graph of x|-->abs(x) in the plane. Unfortunately this is not a scheme because there is no neighborhood of the point '0' which is defined by zeros of polynomials. Hence my question remains: Is there a way to glue affine schemes together (and get a non-affine scheme) such that the result remains 'drawable'? Sancho === Subject: Re: Picture of a 'real scheme' posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20080208 Fedora/2.0.0.12-1.fc8 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) can anybody draw a picture of a scheme which is not affine? (for > example a curve or something) At the first glance is a silly question (even for me), because the > choice of the term 'affine' seems to be something which is like a > piece of paper, so it should be impossible to draw a 'natural' picture > of a non-affine scheme. > However, it is not clear that every curve-like thing which I draw on a > paper is defined by zeros of polynomials: Take for example the graph > of x|-->abs(x) in the plane. Unfortunately this is not a scheme > because there is no neighborhood of the point '0' which is defined by > zeros of polynomials. > Hence my question remains: Is there a way to glue affine schemes > together (and get a non-affine scheme) such that the result remains > 'drawable'? What do you mean by drawable? You cannot expect any physical illustration to perfectly represent a geometrical object, but only represent it according to various conventions. If you make the appropriate convention, then, you can very much draw a projective line. -- m === posting-account=UM3jRwkAAADTHFmJ20qgwageu031CeWA 5.1),gzip(gfe),gzip(gfe) > Hi Google, > I love your services! > post thier garbage on usenet. > Can you please consider the following? > Adopt a human detector verification process when posting. > This process requires a human eye to read a word from a computer- > generated image, which is somewhat garbled, and then type it into a > submission box. > that is ruining the NGs! > For the life of usenet - man! > NGs: please help propogate this message - as many people will start > will be enough to catch their eye! > ~A They're already doing this.- Hide quoted text - - Show quoted text - Not when I post it isn't. I am using the web method - and it can be automated - that is, there > is only a post button. There is no human input based on a grphic. It is similar to what Craigslist has you do. Well, they don't do it on every post. Occasionally, I run into it, though. === > post thier garbage on usenet. Many people filter all posts from Google Groups. See, for example, the Usenet Improvment Project. Google's Freedom of speech model is weird. Chinese search results are filtered, YouTube comment boards are one of Dante's circles of HELL. I have no idea if Google knows what Breidbart is, or how they implement it === post thier garbage on usenet. Many people filter all posts from Google Groups. See, for example, the Usenet Improvment Project. Google's Freedom of speech model is weird. Chinese search results are > filtered, YouTube comment boards are one of Dante's circles of HELL. I have no idea if Google knows what Breidbart is, or how they implement it I have no idea if Google knows what usenet is! Phil -- -- Microsoft voice recognition live demonstration === > They're already doing this.- Hide quoted text - - Show quoted text - Not when I post it isn't. > What's this stuff about - hide/show quoted text-? Did you post it? How did it get there? Sure enought, you are posting from Google. inserting that into peoples posts. I've know of one poster in this newsgroup who blacks lists posts from Google. I'm considering doing the same my self. Be be a GooGoo Googler, use a different news server than Google! > I am using the web method - and it can be automated - that is, there > is only a post button. There is no human input based on a grphic. It is similar to what Craigslist has you do. > === > Hi Google, I love your services! > I hate you ever since you sold yourself to the stock market. Your quality has deteriorated and I find I use you less and less. Now again you abuse the web with your stupid ideas. Why do you when you quote a post in a reply use graphic characters where spaces should appear? In particular, when somebody posts hex 20 20, you quote it as hex 20 A0. The A0 appears as a graphic character on my news browser. How dare you change the contexts of a quoted post. In addition to that, you are inserting senseless Google formating tokens into the post like Hide quoted material Show quoted material. Why are you doing that? It doesn't belong in the post! More demonstration of software incompetence? Because of this and because you, as mentioned below, are a virulent vector easily review, tell the poster that I'm not replying, why I'm not replying, to repost their reply to me email and to use a different news server than Google. Indeed, I'm tired of your GooGoo Googlers and your GooGoo googy software. > post thier garbage on usenet. Can you please consider the following? Adopt a human detector verification process when posting. This process requires a human eye to read a word from a computer- > generated image, which is somewhat garbled, and then type it into a > submission box. > that is ruining the NGs! For the life of usenet - man! NGs: please help propogate this message - as many people will start > will be enough to catch their eye! > ~A === posting-account=W7RQ6gkAAACLDC5JWapQU1UV8ot8KkAS 1.0.3705; Dealio Toolbar 3.1.1),gzip(gfe),gzip(gfe) > Hi Google, I love your services! post thier garbage on usenet. Can you please consider the following? Adopt a human detector verification process when posting. This process requires a human eye to read a word from a computer- > generated image, which is somewhat garbled, and then type it into a > submission box. Well, the eye detectors have already been tried by the Quanutm wanks. But the problem was that the only difference between a virtual photon and a word is the virtual resize wanking. that is ruining the NGs! For the life of usenet - man! NGs: please help propogate this message - as many people will start > will be enough to catch their eye! > ~A === > Hi Google, I love your services! post thier garbage on usenet. Can you please consider the following? Adopt a human detector verification process when posting. This process requires a human eye to read a word from a computer- > generated image, which is somewhat garbled, and then type it into a > submission box. > fix that problem. However, you can. Simply killfile all readable group since I did that as it not only removed too. Sci.crypt's on a hair trigger currently... Phil -- -- Microsoft voice recognition live demonstration === <87d4qo7bn4.fsf@nonospaz.fatphil.org has actually become a readable group since I did that as it not only > Sci.crypt's on a hair trigger currently... > Hi Phil. You could put a weekly or monthly notice at sci.math -- Attention Google Users -- Important Notice to Google Users and explain that you are kill filing all GooGoo Googlers and that others here are also considering doing the same. In otherwords, go Google elsewhere. === has actually become a readable group since I did that as it not only > Sci.crypt's on a hair trigger currently... Hi Phil. You could put a weekly or monthly notice at sci.math > -- Attention Google Users > -- Important Notice to Google Users and explain that you are kill filing all GooGoo Googlers and that others > here are also considering doing the same. In otherwords, go Google > elsewhere. I generally find such spontanious broadcasts obnioxious. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: my theories posting-account=PCksUwoAAAAsb8KSbS36lmWOqee5wTyG CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0),gzip(gfe),gzip(gfe) all tachyons have spin one all forces are tachyons photons are tachyons - negative not order for universe order is increasing Kurt Stocklmeir Now that we know your theories while on the bottle, what are they when you are off the bottle? === Subject: Re: my theories posting-account=UM3jRwkAAADTHFmJ20qgwageu031CeWA 5.1),gzip(gfe),gzip(gfe) all tachyons have spin one all forces are tachyons photons are tachyons - negative not order for universe order is increasing Kurt Stocklmeir And all glibshins are narkwirsts, except on Saint Gribbowam's Day, when they are declared to be trondorffs. === Subject: Re: Finding k posting-account=no4jRwoAAADHjiy7leKH93A_GaYruAJ3 Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > (**note Picka = Guess**) Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. picka ? As consider 2x = k + 3r picka! http://upload.wikimedia.org/wikipedia/en/7/77/Pikachu.png === Subject: Re: JSH: Finding k posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Surprising answer with surrogate factoring that focuses on finding k, > and leverages a rather intriguingly simple little result to factor. > As consider > 2x = k + 3r > when > z^2 = y^2 + T > where T is the target to be factored, is odd and coprime to 3, and T > mod 3 = 2, as then z must have 3 as a factor, so > z = x+k > gives > x^2 + 2xk + k^2 = y^2 + T > which is x^2 = y^2 + T - 2xk - k^2, and I can substitute out 2x, to > get > x^2 = y^2 + T - 2k^2 - 3kr > and that's where a nifty thing pops in, as, you want r=0, but in > general, r will be NEGATIVE if you start with the optimal k when r=0 > and move about modulo 6, as that k will be even. That's because you'd > have > x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > and if you have positive k (no reason to use negative) and try to move > with positive j, then r will be negative, and even if you move with > negative j, the 36j^2 term will tend to dominate, forcing r to be > negative to compensate. > So r = 0 should be near the value at which abs(T-2k^2) is a minimum. > The value of k is going to be close to > sqrt(T/2). > So now suppose T has 200 digits or so. Then k is > going to have, say, 100 digits; k ~ 10^100. > So you are thinking you just need to search through > a range of values which are, you say, k/2p steps away > from k. > No, the algebra says so as > x^2 = y^2 + T - 2(k+6j)^2 - 3(k+6j)r > and if you multiply out the squared term you have > x^2 = y^2 + T - 2(k^2 + 12j + 36j^2) - 3(k+6j)r > and it's just trivial that as you move j around whether you do so > positively or negatively with x and y constant, then the -72j^2 term > will dominate, and to counterbalance it, r MUST be negative as j moves > out of a certain small range. > That's easy math. > Of course, for p = 3, k/2p = k/6 is also going to be > a large number: maybe 99 decimal digits. > How long do you think that kind of search is going > to take? > Well it's k/2p, so you don't have to use p=3 as the argument > generalizes, so you can use p, where p is an odd prime. > Are you really not getting this? If T has 200+ digits, > k/2p is going to have 99+ digits. That means you need to > search through something like > 1000000000000000000000000000000000000000000000000000000000000000000000000000 0 00 > 000000000000000000000 > different values. How long will it take you to do this? Ok, yeah, but what if p is approximately sqrt(public_key/2)? You just trade one problem for another. If you > choose p small, you have a huge range of k's to > check out. If you choose p large, you have a huge > range of p's to check out. Let's say T ~ 10^200. Nope. Any p will work for which k exists where k^2 = 2^{-1}(nT) mod p where n is 1 if T mod 3 = 1 or 5 if T mod 3 = 2. There is roughly a 50% chance that for any p in the desired range a solution will exist. > Then k ~ sqrt(T/2) ~ 10^50. RSA factors could > easily be in the range from 10^50 to 10^150. > Which would mean, you would need to search > for primes p in that range - again, a huge, huge > space. Yes, if the true factors were ***really*** > close to sqrt(T), you would have a smaller space Nope. Not from this theory. Seems you're still not bothering with the algebra. With 2x = k + pr, where p is the prime, you have z=x+k, when z^2 = y^2 + nT and z is being forced to have 3 as a factor, making the substitution you find that r=0--the desired value--will be near the maximal k such that abs(T - 2k^2) is a minimum. That is the easy algebra I keep repeating to you. If you need to multiply it out to see, note that x^2 = y^2 + nT - 2k^2 - kpr and if you change k's value, while x, y and nT remain constant, then r will tend to be negative as you move k around the maximal k mentioned above. It's actually kind of eery to watch the factorizations as yes I've been playing with factoring using the new theory. The math doesn't care much about p, except that k^2 = 2^{-1}(nT) mod p exists. James Harris === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=QtQiMAoAAABMwlM0ZvGpneo7VKGKjSsV Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Updated per January 15, 2008 I have the current edition of these comprehensive solutionsmanualfor > the following textbooks in electronic format (PDF/Word). The solutionsmanualare comprehensive with answers to both even & oddproblemsin > the text. The price is US$35 for each. The methods of payment is through PAYPAL (It is easy, safe, and you > can use debit or credit card to pay even if you don't have an account) Email me at sbooks4sale[at]hotmail[dot]com if you are interested. If > you could not find the book you are looking for, please let me know, I -------------------------- Data Abstraction & Problem Solving with C++ - Frank M. Carrano (5thed) (ISBN 0321433327) > Data and Computer Communications - William Stallings (8thed) (ISBN > 0132433109) > Data Structures and Algorithm Analysis in C++ - Mark Allen Weiss (3rded) (ISBN 032144146X) > Data Structures and Algorithm Analysis in Java - Mark Allen Weiss (2nded) (ISBN 0321370139) > Database Systems: A Practical Approach - Thomas M. Connolly (4thed) > (ISBN 0321294017) > Derivatives Markets - Robert L. McDonald (2nded) (ISBN 032128030X) > Detection and Estimation:Theory and Its Applications - Thomas > Schonhoff (1sted) (ISBN 0130894990)DifferentialEquations- John Polking (2nded) (ISBN 0131437380)DifferentialEquationsandBoundaryValueProblems: Computing and > Modeling - Henry Edwards (4thed) (ISBN 0131561073)DifferentialEquationsand Linear Algebra - Henry Edwards, David E. > Penney (2nded) (ISBN 0131481460)DifferentialEquationsand Linear Algebra - Jerry Farlow (2nded) > (ISBN 0131860615)DifferentialEquationsand Linear Algebra - Stephen W. Goode (3rded) > (ISBN 0130457949)DifferentialEquationsComputing and Modeling - Henry Edwards (4thed) > (ISBN 0136004385)DifferentialEquationsWithBoundaryValueProblems- John C. Polking > (2nded) (ISBN 0130911062) > Digital & Analog Communication Systems - Leon Couch (7thed) (ISBN > 0131424920) > Digital Communications - John Proakis (4thed) (ISBN 0072321113) > Digital Design - Morris Mano (4thed) (ISBN 0131989243) > Digital Electronics: A Practical Approach - William Kleitz (8thed) > (ISBN 0132435780) > Digital Signal Processing - John Proakis (4thed) (ISBN 0131873741) > Digital Signal Processing Using MATLAB -Vinay K. Ingle, John G. > Proakis (2nded) (ISBN 0495073113) > Digital Systems Design Using VHDL - Charles H. Roth (2nded) (ISBN > 0534384625) > Digital Systems: Principles and Applications - Ronald Tocci et al > (10thed) (ISBN 0131725793) > Discrete and Combinatorial Mathematics - Ralph P. Grimaldi (5thed) > (ISBN 0201726343) > Discrete Mathematics - Edgar G. Goodaire, Michael M Parmenter (3rded) > (ISBN 0131679953) > Discrete Mathematics - Otto, Eynden, Dossey, Spence (4thed) (ISBN > 0321079124) > Discrete Mathematics - Otto, Eynden, Dossey, Spence (5thed) (ISBN > 0321305159) > Discrete Mathematics - Richard Johnsonbaugh (6thed) (ISBN 0131176862) > Economic Development - Michael Todaro, Stephen Smith (9thed) (ISBN > 0321278887) > Economic Growth - David Weil (1sted) (ISBN 0201680262) > Economics - Richard Lipsey (13thed) (ISBN 0321369211) > Economics of Money, Banking, and Financial Markets, Update - Frederic > Mishkin (7thed) (ISBN 0321331850) > Economics Today: The Macro View - Roger Miller (13thed) (ISBN > 0321278992) > Economics Today: The Macro View - Roger Miller (14thed) (ISBN > 0321421442) > Economics Today: The Micro View - Roger Miller (13thed) (ISBN > 0321278984) > Economics Today: The Micro View - Roger Miller (14thed) (ISBN > 0321425065) > Economics: Principles and Policy - William J. Baumol (10thed) (ISBN > 0324537026) > Electric Circuits - James Nilsson (8thed) (ISBN 0131989251) > Electrical Engineering: Principles and Applications - Allan R. > Hambley (4thed) (ISBN 0131989227) > Electrical Machines, Drives and Power Systems - Theodore Wildi (6thed) (ISBN 0131776916) > Electronic Communications for Technicians - Tom Wheeler (2nded) (ISBN > 0131130498) > Electronics and Computer Math - Bill R. Deem (8thed) (ISBN > 0131711377) > Electronics Fundamentals: Circuits, Devices and Applications - Thomas > Floyd (7thed) (ISBN 013219709X) > ElementaryDifferentialEquations- Werner E. Kohler, Lee W.Johnson > (1sted) (ISBN 0201709260) > ElementaryDifferentialEquationsWithBoundaryValueProblems- Lee > Johnson et al (1sted) (ISBN 0321121643) > ElementaryDifferentialEquationsWithBoundaryValueProblems- Lee > Johnson et al (2nded) (ISBN 0321398505) > Elementary Linear Algebra with Applications - Bernard Kolman (9thed) > (ISBN 0132296543) > Elementary Number Theory - Kenneth H. Rosen (5thed) (ISBN 0321237072) > Elementary Statistics - Mario F. Triola (10thed) (ISBN 0321331834) > Elementary Statistics - Mario F. Triola (9thed) (ISBN 0201775700) > Elementary Statistics Using Excel - Mario Triola (3rded) (ISBN > 0321365135) > Elements of Forecasting - Francis X. Diebold (4thed) (ISBN > 032432359X) > Employment Law - John J. Moran (4thed) (ISBN 0136009964) > Engineering Economy - William G Sullivan (13thed) (ISBN 0131486497) > Engineering Economy and the Decision-Making Process - Joseph C. > Hartman (1sted) (ISBN 0131424017) > Engineering Fundamentals: An Introduction to Engineering - Saeed > Moaveni (3rded) (ISBN 0495082538) > Engineering Materials: Properties and Selection - Ken Budinski (8thed) (ISBN 0131837796) > Engineering Mechanics: Statics - Russell C. Hibbeler (11thed) (ISBN > 0132215004) > Engineering Mechanics: Statics Computational Edition - Robert W. > Soutas-Little (1sted) (ISBN 0534549217) > Engineering Vibration - Daniel Inman (3rded) (ISBN 0132281732) > Entrepreneurial Finance - Philip J. Adelman (4thed) (ISBN 0132434792) > Environmental and Natural Resource Economics - Tom Tietenberg (7thed) > (ISBN 0321305043) > Error Control Coding - Daniel J. Costello Jr., Shu Lin (2nded) (ISBN > 0130426725) > Essentials of Business Law - Jeffrey F. Beatty (3rded) (ISBN > 0324537123) > Essentials of Economics - Gregory Mankiw (4thed) (ISBN 0324236964) > Essentials of Logic - Irving Copi (2nded) (ISBN 013238034X) > Essentials of Managerial Finance - Scott Besley (13thed) (ISBN > 0324258755) > Essentials of Organizational Behavior - Stephen P Robbins (9thed) > (ISBN 0132431521) > Exploring Macroeconomics - Robert L. Sexton (4thed) (ISBN 0324395558) > Feedback Control of Dynamic Systems - Gene Franklin (5thed) (ISBN > 0131499300) > Financial & Managerial Accounting - Carl S. Warren (9thed) (ISBN > 0324401884) > Financial Accounting - Carl S. Warren, James M. Reeve (10thed) (ISBN > 0324380674) > Financial Accounting - Jane Reimers (1sted) (ISBN 0131492012) > Financial Accounting - Walter Harrison, Charles Horngren (6thed) > (ISBN 0131499459) > Financial Accounting and Financial Tips - Walter T. Harrison (7thed) > (ISBN 0135012848) > Financial Accounting: A Bridge to Decision Making - Robert Ingram (6thed) (ISBN 0324313357) > Financial Accounting: A Business Process Approach - Jane L. Reimers > (2nded) (ISBN 0131473867) > Financial Accounting: An Integrated Statements Approach - Jonathan > Duchac (2nded) (ISBN 0324312113) > Financial Accounting: An Introduction to Concepts, Methods and Uses - > Clyde P. Stickney (12thed) (ISBN 0324381980) > Financial and Managerial Accounting - Meg Pollard (1sted) (ISBN > 0136008984) > Financial Management For Public, Health, and Not-for-Profit > Organizations - Steven Finkler (2nded) (ISBN 0131471988) > Financial Management: Theory & Practice - Eugene Brigham (12thed) > (ISBN 0324422695) > Financial Markets and Institutions - Frederic S. Mishkin (5thed) > (ISBN 0321280296) > Financial Reporting and Analysis - Lawrence Revsine (3rded) (ISBN > 0131430211) > Financial Reporting and Analysis Using Financial Accounting > Information - Charles Gibson (10thed) (ISBN 0324304455) > Financial Reporting, Financial Statement Analysis, and Valuation - > Clyde P. Stickney (6thed) (ISBN 0324302959) > Financial/Managerial Accounting - Walter T. Harrison (1sted) (ISBN > 0131568779) > Finite Math and Its Application - Larry Goldstein (9thed) (ISBN > 0131873644) > Finite Mathematics - Margaret L. Lial et al (8thed) (ISBN 032122826X) > Finite Mathematics for Business, Economics, Life Sciences & Social > Sciences - Raymond Barnett (11thed) (ISBN 0132255707) > Finite Mathematics with Applications - Margaret L. Lial (9thed) (ISBN > 0321386728) > First Course in Abstract Algebra - John Fraleigh (7thed) (ISBN > 0201763907) > First Course in Abstract Algebra - Joseph Rotman (3rded) (ISBN > 0131862677) > First Course In Probability - Sheldon M. Ross (7thed) (ISBN > 0131856626) > Foundations of Finance - Arthur Keown, William Petty, John Martin, > David Scott (5thed) (ISBN 0131856057) > Foundations of Geometry - Gerard Venema (5thed) (ISBN 0131437003) > Foundations of MEMS - Chang Liu (1sted) (ISBN 0131472860) > Foundations of Microeconomics - Robin Bade (3rded) (ISBN 0321415957) > Fraud Examination - Steve Albrecht (2nded) (ISBN 0324651155) > Friendly Introduction to Analysis - Witold A.J. Kosmala (2nded) (ISBN > 0130457965) hello there i need to buy your solution manual Differential Equations and Boundary Value Problems: Computing and Modeling - Henry Edwards (4th ed) (ISBN 0131561073) even and odd problems?? so please contact me thank you. === Subject: Re: Number 47 posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) [snip a much better explanation than mine] To prove that 47 is the largest non-tileable number, we need to prove that > -48 x > -49 > -50 x > -51 > -52 x > -53 x > -54 > are all tileable. We also need to prove that 47 isn't. > All tileable numbers I could form I have marked. I hope one can find 49, 51 > and 54. Indeed, those are the same three numbers on which I'm stuck. This problem has been annoying me for the past few days, so much so in fact that last night I had a nightmare in which I was trying to assemble cubes into the shape of a bigger cube, until the cubes became angry and started to attack me. For the benefit of my sanity I wonder whether anybody can shed some light on the problem? The only progress I've made is this: given a tiling of the unit cube by smaller cubes, I believe that all the smaller cubes must have a rational side length. Since each cross section of such a tiling is a tiling of the unit square by squares it suffices to prove the corresponding statement for square tilings. My attempted proof of this is rather difficult to provide, since it involves considering various different cases which are best described by means of diagrams, but the idea is that, by removing some of the squares from such a tiling, we are left with a polygon which has an internal 90 degree vertex whose adjacent sides have lengths a and b such that a/b is irrational; and that inserting a square of any size into this vertex we necessarily create at least one new vertex with the same property, so that filling such a region with squares is impossible by induction on the number of squares required. Assuming that the above claim is correct, by taking the lowest common denominator h of the side lengths of every cube in a given tiling we can see that every tiling can be viewed as a tiling by h^3 cubes of equal size, in which some of the sets of cubes are fused into large cubes. Though whether this actually helps solve the problem I don't know. === Subject: Can n choose r * 2^(2(r-1)) ever be odd or twice odd for integers 1 < r <= [n/2] ? posting-account=Rkt6TwoAAACG_SqlrxmgPCl1Ozr0PWSD MathPlayer 2.10b; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) Are there integers r, n with 1 < r <= floor(n/2) for which n! * 2^(2(r-1)) / ((n-r)! * r!) is odd ? I'm hoping and expecting there aren't, and even none twice odd. John R Ramsden === Subject: Re: Can n choose r * 2^(2(r-1)) ever be odd or twice odd for integers 1 < r <= [n/2] ? Are there integers r, n with 1 < r <= floor(n/2) for which n! * >2^(2(r-1)) / ((n-r)! * r!) is odd ? Since ((n!) / ((n-r)! * r!)) is just (n choose r), your expression is therefore a multiple of 2^(2*(r-1)), hence always even, and in fact, always a multiple of 4. quasi === Subject: Exponential of a square matrix Hello! How can I calculate del E=e^A, where A is symmetric and positive definite? I know that is equal to the sum: I+A+(1/2)A+(1/3!)A^3... but I cannot execute infinite sums.... What is E^(-1)? Thansk! Eve === Subject: construction problem Can anybody solve (or give a reference for) the following problem? Given three parallel lines construct (with ruler and compass only) an equilateral triangle with a vertex on each of the three given lines. C. Wildhagen, The Netherlands === Subject: Re: Cardinals beta that satisfy beta = aleph_beta posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq Originator: israel@math.ubc.ca (Robert Israel) > Then can someone say what is known about cardinals > beta -- if any -- that solve the equation beta = aleph_beta ? Can someone at least please post references in the > literature to this equation? Despite how counterintuitive this seems, even wilder things happen, and it takes very little background to understand and prove these wilder things. The following is from a 20 July 2006 sci.math of mine (slightly edited). At its end is a beginner's level reference to the literature for what I discuss. --------------------------------------------------------- In fact, the cardinals that are greater than their ordinal subscripts show up for only relatively brief blips. As you get farther and further out along the cardinals, these blips blend together to form essentially the same number of cardinals (even when you count ordinally, and not just by cardinality) as those cardinals that are equal to their ordinal subscripts. The mapping beta |--> aleph_beta that takes the beta'th ordinal to the beta'th cardinal is a normal function (on a sufficiently large initial segment of the ordinals), and hence it has fixed points. In fact, its fixed points are order-isomorphic to every initial segment of the ordinals. Thus, relatively soon its fixed points start to look like the limit ordinals look like out around the first epsilon number. Moreover, these fixed points have fixed points as well, and these 2'nd order fixed points also have fixed points. Likewise for the 3'rd order fixed points, and so on -- even the omega'th order fixed points (use a countable intersection of clubs [*] to get them) and beyond. After a while, exotic landmarks such as the epsilon_0'th order fixed points will even start to look like the limit ordinals look like out around epsilon_0. And this is only the beginning, because we can jump past this pedestrian pace of iterating the fixed point operation by taking diagonal intersections (look it up), and on and on and on ... By the way, none of this requires us to come anywhere near the first inaccessible cardinal, to say nothing of the other Large Cardinals. [*] club = closed and unbounded set http://en.wikipedia.org/wiki/Club_set See the beginning of Section 4.33 in Azriel Levy's Basic Set Theory, Dover Publications, 1979/2002. Use 'Search in this book' = omitted by diagonal intersection --------------------------------------------------------- Dave L. Renfro dabve.renfrbo@bacbt.orbg [remove all b's] === Subject: Ten papers published by Geometry & Topology Monographs Geometry and Topology Publications is pleased to announce commencement of publication of G&T Monograph number 13: Groups, homotopy and configuration spaces (Tokyo 2005) Editors: Norio Iwase, Toake Kohno, Ran Levi, Dai Tamaki and Jie Wu This volume is the proceedings of the conference Groups, Homotopy and Configuration Spaces held at the University of Tokyo, July 5-11, 2005, in honor of the 60th birthday of Fred Cohen. The emphasis of the conference was on cohomology of groups, classical and modern homotopy theory, geometry and topology of configuration spaces and related topics. However, the conference was intended to have a broad scope, with talks on a variety of topics of current interests in topology. The organizing committee consisted of Norio Iwase, Toake Kohno, Ran Levi, Dai Tamaki and Jie Wu. The conference was supported by the COE program of the Graduate School of Mathematical Sciences, The University of Tokyo. The proceedings will consist of twenty-three papers, of which the first ten have now been published: (1) Geometry & Topology Monographs 13 (2008) 1-10 String topology of Poincare duality groups by Hossein Abbaspour, Ralph Cohen and Kate Gruher URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p001.xhtml DOI: 10.2140/gtm.2008.13.1 (2) Geometry & Topology Monographs 13 (2008) 11-40 Computation of the homotopy of the spectrum tmf by Tilman Bauer URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p002.xhtml DOI: 10.2140/gtm.2008.13.11 (3) Geometry & Topology Monographs 13 (2008) 41-83 A family of embedding spaces by Ryan Budney URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p003.xhtml DOI: 10.2140/gtm.2008.13.41 (4) Geometry & Topology Monographs 13 (2008) 85-104 Cohomology of Artin groups of type ~A_n, B_n and applications by Filippo Callegaro, Davide Moroni and Mario Salvetti URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p004.xhtml DOI: 10.2140/gtm.2008.13.85 (5) Geometry & Topology Monographs 13 (2008) 105-146 The boundary manifold of a complex line arrangement by Daniel C Cohen and Alexander I Suciu URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p005.xhtml DOI: 10.2140/gtm.2008.13.105 (6) Geometry & Topology Monographs 13 (2008) 147-168 Basis-conjugating automorphisms of a free group and associated Lie algebras by F R Cohen, J Pakianathan, V V Vershinin and J Wu URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p006.xhtml DOI: 10.2140/gtm.2008.13.147 (7) Geometry & Topology Monographs 13 (2008) 169-193 On braid groups and homotopy groups by F R Cohen and J Wu URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p007.xhtml DOI: 10.2140/gtm.2008.13.169 (8) Geometry & Topology Monographs 13 (2008) 195-201 Odd-primary homotopy exponents of compact simple Lie groups by Donald M Davis and Stephen D Theriault URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p008.xhtml DOI: 10.2140/gtm.2008.13.195 (9) Geometry & Topology Monographs 13 (2008) 203-227 Filtering the fiber of the pinch map by Brayton Gray URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p009.xhtml DOI: 10.2140/gtm.2008.13.203 (10) Geometry & Topology Monographs 13 (2008) 229-259 Homotopy algebra of open-closed strings by Hiroshige Kajiura and Jim Stasheff URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p010.xhtml DOI: 10.2140/gtm.2008.13.229 Abstracts follow (1) String topology of Poincare duality groups by Hossein Abbaspour, Ralph Cohen and Kate Gruher Let G be a Poincare duality group of dimension n. For a given element g in G, let C_g denote its centralizer subgroup. Let L_G be the graded abelian group defined by (L_G)_p = oplus_{[g]}H_p+n(C_g) where the sum is taken over conjugacy classes of elements in G. In this paper we construct a multiplication on L_G directly in terms of intersection products on the centralizers. This multiplication makes L_G a graded, associative, commutative algebra. When G is the fundamental group of an aspherical, closed oriented n-manifold M, then (L_G)_* = H_*+n(LM), where LM is the free loop space of M. We show that the product on L_G corresponds to the string topology loop product on H_*(LM) defined by Chas and Sullivan. (2) Computation of the homotopy of the spectrum tmf by Tilman Bauer This paper contains a complete computation of the homotopy ring of the spectrum of topological modular forms constructed by Hopkins and Miller. The computation is done away from 6, and at the (interesting) primes 2 and 3 separately, and in each of the latter two cases, a sequence of algebraic Bockstein spectral sequences is used to compute the E_2 term of the elliptic Adams-Novikov spectral sequence from the elliptic curve Hopf algebroid. In a further step, all the differentials in the latter spectral sequence are determined. The result of this computation is originally due to Hopkins and Mahowald (unpublished). (3) A family of embedding spaces by Ryan Budney Let Emb(S^j,S^n) denote the space of C^infty-smooth embeddings of the j-sphere in the n-sphere. This paper considers homotopy-theoretic properties of the family of spaces Emb(S^j,S^n) for n >= j > 0. There is a homotopy-equivalence of Emb(S^j,S^n) with SO_{n+1} times_{SO_{n-j}} K_{n,j} where K_{n,j} is the space of embeddings of R^j in R^n which are standard outside of a ball. The main results of this paper are that K_{n,j} is (2n-3j-4)-connected, the computation of pi_{2n-3j-3} (K_{n,j}) together with a geometric interpretation of the generators. A graphing construction Omega K_{n-1,j-1} --> K_{n,j} is shown to induce an epimorphism on homotopy groups up to dimension 2n-2j-5. This gives a new proof of Haefliger's theorem that pi_0 (Emb(S^j,S^n)) is a group for n-j>2. The proof given is analogous to the proof that the braid group has inverses. Relationship between the graphing construction and actions of operads of cubes on embedding spaces are developed. The paper ends with a brief survey of what is known about the spaces K_{n,j}, focusing on issues related to iterated loop-space structures. (4) Cohomology of Artin groups of type ~A_n, B_n and applications by Filippo Callegaro, Davide Moroni and Mario Salvetti We consider two natural embeddings between Artin groups: the group G_{tilde{A}_{n-1}} of type tilde{A}_{n-1} embeds into the group 1G_{B_n} of type B_n; G_{B_n} in turn embeds into the classical braid group Br_{n+1}:=G_{A_n} of type A_n. The cohomologies of these groups are related, by standard results, in a precise way. By using techniques developed in previous papers, we give precise formulas (sketching the proofs) for the cohomology of G_{B_n} with coefficients over the module Q[q^{+-1},t^{+-1}], where the action is (-q)-multiplication for the standard generators associated to the first n-1 nodes of the Dynkin diagram, while is (-t)-multiplication for the generator associated to the last node. As a corollary we obtain the rational cohomology for G_{tilde{A}_n} as well as the cohomology of Br_{n+1} with coefficients in the (n+1)-dimensional representation obtained by Tong, Yang and Ma. We stress the topological significance, recalling some constructions of explicit finite CW-complexes for orbit spaces of Artin groups. In case of groups of infinite type, we indicate the (few) variations to be done with respect to the finite type case. For affine groups, some of these orbit spaces are known to be K(pi,1) spaces (in particular, for type tilde{A}_n). We point out that the above cohomology of G_{B_n} gives (as a module over the monodromy operator) the rational cohomology of the fibre (analog to a Milnor fibre) of the natural fibration of K(G_{B_n},1) onto the 2-torus. (5) The boundary manifold of a complex line arrangement by Daniel C Cohen and Alexander I Suciu We study the topology of the boundary manifold of a line arrangement in CP^2, with emphasis on the fundamental group G and associated invariants. We determine the Alexander polynomial Delta(G), and more generally, the twisted Alexander polynomial associated to the abelianization of G and an arbitrary complex representation. We give an explicit description of the unit ball in the Alexander norm, and use it to analyze certain we also obtain a complete description of the first characteristic variety of G. Comparing this with the corresponding resonance variety of the cohomology ring of G enables us to characterize those arrangements for which the boundary manifold is formal. (6) Basis-conjugating automorphisms of a free group and associated Lie algebras by F R Cohen, J Pakianathan, V V Vershinin and J Wu Let F_n = denote the free group with generators {x_1,...,x_n}. Nielsen and Magnus described generators for the kernel of the canonical epimorphism from the automorphism group of F_n to the general linear group over the integers. In particular among them are the automorphisms chi_{k,i} which conjugate the generator x_k by the generator x_i leaving the x_j fixed for j not k. A computation of the cohomology ring as well as the Lie algebra obtained from the descending central series of the group generated by chi_{k,i} for i X union CA --> SA and this is used to study the boundary map in the fibration sequence of Cohen, Moore and Neisendorfer: Omega^2 S^{2n+1} --> Omega F_n --> Omega P^{2n+1} --> Omega S^{2n+1} The boundary map is shown to be compatible with the Hopf invariant and a filtration of the spliting is obtained. (10) Homotopy algebra of open-closed strings by Hiroshige Kajiura and Jim Stasheff This paper is a survey of our previous works on open-closed homotopy algebras, together with geometrical background, especially in terms of compactifications of configuration spaces (one of Fred's specialities) of Riemann surfaces, structures on loop spaces, etc. We newly present Merkulov's geometric A_infty-structure [Internat. Math. Res. Notices (1999) 153--164] as a special example of an OCHA. We also recall the relation of open-closed homotopy algebras to various aspects of deformation theory. === Subject: intersection theory, alg geometry According to Appendix A of Hartshorne's Algebraic Geometry, CH*(X) has a good intersection product if X is a smooth quasi-projective variety over an algebraically closed field k, but can we relax the condition on k? what if k is a perfect field, for example? Also, under what conditions does Serre's formula of intersection multiplicity make sense? I'm wondering this because when you define composition of finite correspondencees as in Voevodsky's Cor_k, you use Serre's formula. There is an imbedding of M_{rat}(k) into DM_{gm}(k), but if intersection product is defined only for sm proj vars over an algebraically closed field k, this imbedding makes sense only when k is alg closed.... My main interest if when k is finite or Z or Q. I will study intersection theory soon, but I really appreciate if you give me a few tips about these points. Do you know any nice introduction to intersection theory? === Subject: Re: intersection theory, alg geometry posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G According to Appendix A of Hartshorne's Algebraic Geometry, CH*(X) has a good intersection product if X is a smooth quasi-projective variety over an algebraically closed field k, but can we relax the condition on k? what if k is a perfect field, for example? Also, under what conditions does Serre's formula of intersection multiplicity make sense? I'm wondering this because when you define composition of finite correspondencees as in Voevodsky's Cor_k, you use Serre's formula. There is an imbedding of M_{rat}(k) into DM_{gm}(k), but if intersection product is defined only for sm proj vars over an algebraically closed field k, this imbedding makes sense only when k is alg closed.... My main interest if when k is finite or Z or Q. I will study intersection theory soon, but I really appreciate if you give me a few tips about these points. Do you know any nice introduction to intersection theory? > William Fulton, Intersection theory, Springer 1998. === Subject: Re: Exponential Function for Quaternions? Originator: baez@math.UUCP (John Baez) >On 2007-10-07 18:20:24 +0930, Harald Hanche-Olsen said: > exp(u theta) = cos(theta) + u sin(theta) > > when u is any imaginary quaternion of length 1. > I can't tell how he did it, but here is my own take on the whole > issue. Simply note that the linear span (over the reals) of 1 and u > is a subalgebra of the quaternions isomorphic to the complex numbers. > So everything that is true over the complex numbers transfers > automatically to this subalgebra. >Nothing so slick I am afraid ! I would have just expanded exp(u theta) in a power series and used the >fact that u^2 = -1 to get two power series one and rearrange using the >power series of cos and sin. I think that's essentially the same thing. You're basically saying u acts just like i, so all the usual stuff works. But that's just a less technical way of saying the linear span (over the reals) of 1 and u is isomorphic to the complex numbers, so all the usual stuff works. It's a good general principle! So we might as well state it even more simply: any quaternion formula that only involves *one* quaternion is secretly just a formula about the complex numbers. The tricky stuff only starts when you play with two quaternions at a time. I spend a lot of time using a similar principle one level up: any octonion formula that only involves *two* octonions is secretly just a formula about the quaternions. For example: the quaternions are associative, so the octonions satisfy rules like (xx)y = x(xy) (xy)x = x(yx) (yx)x = y(xx) even though they're not associative. Furthermore, exponentials of a single octonion are no harder to deal with than exponentials of a complex numbers. === Subject: Six papers published by Geometry & Topology Monographs Publication of GT Monographs Volume 13 in honour of Fred Cohen: Groups, homotopy and configuration spaces (Tokyo 2005) Editors: Norio Iwase, Toake Kohno, Ran Levi, Dai Tamaki and Jie Wu continues with six futher papers now published: (1) Geometry & Topology Monographs 13 (2008) 261-279 On the Rothenberg-Steenrod spectral sequence for the mod 2 cohomology of classifying spaces of spinor groups by Masaki Kameko and Mamoru Mimura URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p011.xhtml DOI: 10.2140/gtm.2008.13.261 (2) Geometry & Topology Monographs 13 (2008) 281-291 The homology of spaces of polynomials with roots of bounded multiplicity by Yasuhiko Kamiyama URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p012.xhtml DOI: 10.2140/gtm.2008.13.281 (3) Geometry & Topology Monographs 13 (2008) 293-306 Twisted Morita-Mumford classes on braid groups by Nariya Kawazumi URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p013.xhtml DOI: 10.2140/gtm.2008.13.293 (4) Geometry & Topology Monographs 13 (2008) 307-321 Twisted Alexander polynomials and a partial order on the set of prime knots by Teruaki Kitano and Masaaki Suzuki URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p014.xhtml DOI: 10.2140/gtm.2008.13.307 (5) Geometry & Topology Monographs 13 (2008) 323-334 On the Lusternik-Schnirelmann category of symmetric spaces of classical type by Mamoru Mimura and Kei Sugata URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p015.xhtml DOI: 10.2140/gtm.2008.13.323 (6) Geometry & Topology Monographs 13 (2008) 335-354 Lie algebras of symplectic derivations and cycles on the moduli spaces by Shigeyuki Morita URL: http://www.msp.warwick.ac.uk/gtm/2008/13/p016.xhtml DOI: 10.2140/gtm.2008.13.335 Abstracts follow (1) On the Rothenberg-Steenrod spectral sequence for the mod 2 cohomology of classifying spaces of spinor groups by Masaki Kameko and Mamoru Mimura We compute the cotorsion product of the mod 2 cohomology of spinor group spin(n), which is the E_2-term of the Rothenberg-Steenrod spectral sequence for the mod 2 cohomology of the classifying space of the spinor group spin(n). As a consequence of this computation, we show the non-collapsing of the Rothenberg-Steenrod spectral sequence for n > 16. (2) The homology of spaces of polynomials with roots of bounded multiplicity by Yasuhiko Kamiyama Let P_{k, n}^l be the space consisting of monic complex polynomials f(z) of degree k and such that the number of n-fold roots of f(z) is at most l. In this paper, we determine the integral homology groups of P_{k, n}^l. (3) Twisted Morita-Mumford classes on braid groups by Nariya Kawazumi Evaluating the twisted Morita-Mumford classes bar(h)_p on the Artin braid group B_n, we give the stable algebraic independence of the bar(h)_p's on the automorphism group of the free group, Aut(F_n). This is sharper than the results we obtained by restricting them to the mapping class group. (4) Twisted Alexander polynomials and a partial order on the set of prime knots by Teruaki Kitano and Masaaki Suzuki We give a survey of some recent papers by the authors and Masaaki Wada relating the twisted Alexander polynomial with a partial order on the set of prime knots. We also give examples and pose open problems. (5) On the Lusternik-Schnirelmann category of symmetric spaces of classical type by Mamoru Mimura and Kei Sugata We determine the Lusternik-Schnirelmann category of the irreducible, symmetric Riemann spaces SU(n)/SO(n) and SU(2n)/Sp(n) of type AI and AII respectively. (6) Lie algebras of symplectic derivations and cycles on the moduli spaces by Shigeyuki Morita We consider the Lie algebra consisting of all derivations on the free associative algebra, generated by the first homology group of a closed oriented surface, which kill the symplectic class. We find the first non-trivial abelianization of this Lie algebra and discuss its relation to unstable cohomology classes of the moduli space of curves via a theorem of Kontsevich. === Subject: Re: This Week's Finds in Mathematical Physics (Week 259) Originator: baez@math-lw-n07.math.ucr.edu (John Baez) >Another way to understand the analogy between sets and vector spaces >is via the language of matroids. Sets correspond to free matroids >and vector spaces correspond to representable matroids. How much >of what you say about the one-element field can be understood in >terms of matroids? I don't know, since I don't know much about matroids. But, from what little I know, a lot of projective geometry fits nicely into the matroid framework. That's promising, since the stuff I'm talking about is closely related to projective geometry. According to what you say, both for free matroids and representable matroids (whatever those are) over a given field we get one of these, up to isomorphism, for each natural number n - the cardinality or dimension. Are there other nice series of matroids, one for each natural number? That might lead to some fun stuff. === Subject: Macaulay 2, version 1.1, announcement Macaulay 2 is a computer program devoted to supporting research in algebraic geometry and commutative algebra. We've been working hard on the latest release of Macaulay 2, and now version 1.1 is ready for download from our web site. We hope you will try it out. See http://www.math.uiuc.edu/Macaulay2/Downloads/ Installation is easy. Versions have been compiled specifically for the following GNU/Linux systems: generic Linux, Ubuntu (32 bit and 64 bit), Debian (32 bit and 64 bit) both with *.deb files, Fedora 7, Fedora 8, and Red Hat Enterprise 4, with *.rpm files; for the following Macintosh OS X systems: 10.4 and 10.5 on Intel 32 bit, 10.5 on Intel 64 bit, and 10.4 on the Power PC; and on Microsoft Windows with the Cygwin compatibility package installed. Automatic installation and updating from our repositories is possible for Debian, Ubuntu, and Microsoft Windows with Cygwin. Documentation has been improved, with every function documented. Browse the latest version at our web site. It is easy to write, document, and distribute Macaulay 2 code. Packages have been contributed and included with Macaulay 2, including: NoetherNormalization, by Bart Snapp and Nathaniel Stapleton; GenericInitialIdeal and Regularity, by Alexandra Seceleanu and Nathaniel Stapleton; InvolutiveBases, by Daniel Robertz; ChainComplexExtras, by Frank Moore and Greg Smith; HyperplaneArrangements, by Graham Denham and Gregory G. Smith; LexIdeals, by Chris Francisco; ReesAlgebra, by David Eisenbud, Amelia Taylor, and Sorin Popescu; and TangentCone, by Craig Huneke and David Eisenbud. Packages can be posted on our web site for instant downloading and installation. Frequent updating by the author is possible. See http://www.math.uiuc.edu/Macaulay2/Packages We're hoping to help even more people get started with writing packages for Macaulay 2. Those interested should contact us and consider applying by March 1 for our workshop at the end of June (see the web site for details). http://www.math.uiuc.edu/Macaulay2/Events/Workshop2008 An interface with TeXmacs has been provided, so Macaulay 2 can be run with a good graphical user interface with beautiful formatting of Macaulay 2 output. A good implementation of real and complex numbers to arbitrary precision, based on the mpfr library from mpfr.org, has been implemented. The library is remarkable for the care taken to return correctly rounded results. It is hoped that this addition will form a good base for experimentation with algebraic algorithms that mix symbolic and numeric techniques. Basic transcendental functions are provided. An interface to lapack routines for singular value decomposition and eigenvectors is provided. A more complete list of improvements and bug fixes is available on the web site, see http://www.math.uiuc.edu/Macaulay2/Changes/1.1/ Let us know whether you have any problems getting started, and we'll do our best to help you. Dan Grayson Mike Stillman === Subject: Re: Maximum Average of Values drawn from normal distribution > .... > Given a normal distribution with mean 0.5 and standard deviation 0.25 and > given > a set X of values drawn from this distribution, what is the average of the > maximal value of set X.... You may be able to get some help in the news group. Ken Pledger. === Subject: Re: Probability Question posting-account=-jiXkQgAAAAZhbLPR1Fhvi1sNwKqv9WA FunWebProducts),gzip(gfe),gzip(gfe) > I have a Dice Probability Question. How many times will a 5 be rolled twice before a 7 in 1980 rolls of the > dice? I believe the equation would be p(5 twice before 7) * (Total # of 7's > rolled) = (# of times 5 is rolled before 7). That would be .16*330=52.8 A 5 would be rolled twice before a 7, 52.8 times. æCan someone confirm that > answer for me. Does not 5,11,3,5,4,3 9,4,4,7, etc. also count as 5 rolled twice before a 7? === Subject: Optimization PRoblem Here is a problem I've been trying to solve. A box with open top is to be made with a square base and a constant surface C. Determine the sides of the box if the volume is to be a maximum. ___ Answers are: Base = Square Root C/3, ____ height = Square Root C/I2 x = common side of base which is square v(x) = x(a - 2x)(b - 2x) = 4x^3 - 2(a + b)x^2 + abx. To find the local maximum we differentiate: v'(x) = 12x^2 - 4(a + b)x + AB. and then equate the derivative to 0. This leads to a quadratic equation 12x^2 - 4(a + b)x + AB = 0. Can't seem to get any further. TIA === Subject: Re: Optimization PRoblem Here is a problem I've been trying to solve. A box with open top is to be made with a square base and a constant > surface C. Determine the sides of the box if the volume is to be a > maximum. If you mean that the surface area of the box, C, is constant then C = b^2 + 4hb . . . . . . . . . . . . . (1) or 0 = b^2 + 4hb - C . . . . . . . . . . . (2) where b is the base side and h is the height. The volume is V = b^2 h . . . . . . . . . . . . . . . (3) which is to be maximized with constraint (2). One may use Lagrange multipliers. dV/db + L d(b^2 + 4hb - C)/db = 0 . . . . (4) dV/dh + L d(b^2 + 4hb - C)/dh = 0 . . . . (5) are to solved simultaneously and I've used L where people often write lambda. (4) is 2bh + L(2b + 4h) = 0 . . . . . . . . . . . (6) (5) is b^2 + L4b = 0 . . . . . . . . . . . . . . . (7) Now solve (2, 6 and 7). I get b = sqrt(C/3), h = (sqrt C/3)/2 which isn't a minimum, because h = 0 would give a minimum. > ___ Answers > are: Base = Square Root C/3, Right. ____ > height = Square Root C/I2 What's I? -- === Subject: Re: Optimization PRoblem original of the LaGrange factor. Also in working out the simultaneous equations. Could you expand on those items. TIA. I can see where you're right intuitively, but working it out logically is a little difficult. > > Here is a problem I've been trying to solve. > > A box with open top is to be made with a square base and a constant > surface C. Determine the sides of the box if the volume is to be a > maximum. If you mean that the surface area of the box, C, is constant then C = b^2 + 4hb . . . . . . . . . . . . . (1) >or > 0 = b^2 + 4hb - C . . . . . . . . . . . (2) where b is the base side and h is the height. The volume is V = b^2 h . . . . . . . . . . . . . . . (3) which is to be maximized with constraint (2). One may use Lagrange >multipliers. dV/db + L d(b^2 + 4hb - C)/db = 0 . . . . (4) dV/dh + L d(b^2 + 4hb - C)/dh = 0 . . . . (5) are to solved simultaneously and I've used L where people often write >lambda. (4) is 2bh + L(2b + 4h) = 0 . . . . . . . . . . . (6) (5) is b^2 + L4b = 0 . . . . . . . . . . . . . . . (7) Now solve (2, 6 and 7). I get b = sqrt(C/3), h = (sqrt C/3)/2 which isn't a minimum, because h = 0 would give a minimum. > ___ Answers > are: Base = Square Root C/3, Right. > ____ > height = Square Root C/I2 What's I? === Subject: Re: Optimization PRoblem original of the LaGrange factor. Also in working out the simultaneous > equations. Could you expand on those items. TIA. If you're not familiar with Lagrange multipliers, proceed as follows: Data: C = b^2 + 4hb . . . . . (i) V = hb^2 . . . . . . . (ii) Problem: find h, b to maximize V subject to C = const. (i) times b is Cb = b^3 + 4hb^2 in which (ii) gives Cb = b^3 + 4V . . . . . (iii) Differentiate (iii) w.r.t. b with a view to finding V's stationary points: C = 3b^2 + 4(dV/db) . . (iv) dV/db = 0 implies b = sqrt(C/3). considerations) b > 0. So V at b = sqrt(C/3) is a maximum. Put this b in (i): C = C/3 + 4h sqrt(C/3) to get h = (sqrt C/3)/2. -- === Subject: Re: Optimization PRoblem > > original of the LaGrange factor. Also in working out the simultaneous > equations. Could you expand on those items. TIA. If you're not familiar with Lagrange multipliers, proceed as follows: >Data: C = b^2 + 4hb . . . . . (i) V = hb^2 . . . . . . . (ii) Problem: find h, b to maximize V subject to C = const. (i) times b is Cb = b^3 + 4hb^2 in which (ii) gives Cb = b^3 + 4V . . . . . (iii) Differentiate (iii) w.r.t. b with a view to finding V's stationary >points: C = 3b^2 + 4(dV/db) . . (iv) > >dV/db = 0 implies b = sqrt(C/3). considerations) b > 0. So V at b = sqrt(C/3) > >is a maximum. Put this b in (i): C = C/3 + 4h sqrt(C/3) > >to get h = (sqrt C/3)/2. === Subject: Re: Optimization PRoblem original of the LaGrange factor. You can find out about Lagrange multipliers in any calculus book (Lang, Calculus of Several Variables, for example) or here: http://en.wikipedia.org/wiki/Lagrange_multiplier or here: http://mathworld.wolfram.com/LagrangeMultiplier.html. > Also in working out the simultaneous > equations. Simple algebra. Solve b^2 + 4hb - C = 0 2bh + L(2b + 4h) = 0 b^2 + L4b = 0. You don't want L. > Could you expand on those items. TIA. I can see where you're right intuitively, but working it out logically > is a little difficult. > Here is a problem I've been trying to solve. > A box with open top is to be made with a square base and a constant > surface C. Determine the sides of the box if the volume is to be a > maximum. If you mean that the surface area of the box, C, is constant then C = b^2 + 4hb . . . . . . . . . . . . . (1) >or > 0 = b^2 + 4hb - C . . . . . . . . . . . (2) where b is the base side and h is the height. The volume is V = b^2 h . . . . . . . . . . . . . . . (3) which is to be maximized with constraint (2). One may use Lagrange >multipliers. dV/db + L d(b^2 + 4hb - C)/db = 0 . . . . (4) dV/dh + L d(b^2 + 4hb - C)/dh = 0 . . . . (5) are to solved simultaneously and I've used L where people often write >lambda. (4) is 2bh + L(2b + 4h) = 0 . . . . . . . . . . . (6) (5) is b^2 + L4b = 0 . . . . . . . . . . . . . . . (7) Now solve (2, 6 and 7). I get b = sqrt(C/3), h = (sqrt C/3)/2 which isn't a minimum, because h = 0 would give a minimum. > ___ Answers > are: Base = Square Root C/3, Right. > ____ > height = Square Root C/I2 What's I? -- === Subject: Re: Optimization PRoblem > Here is a problem I've been trying to solve. A box with open top is to be made with a square base and a constant > surface C. Determine the sides of the box if the volume is to be a > maximum. > Answers > Base = Square Root C/3, > height = Square Root C/I2 x = common side of base which is square v(x) = x(a - 2x)(b - 2x) > Huh? v(x) = x^2 h(x) C = x^2 + 4x.h(x) === Subject: Re: Optimization PRoblem > A box with open top is to be made with a square base and a constant > surface C. Determine the sides of the box if the volume is to be a > maximum. > x = common side of base which is square v(x) = x(a - 2x)(b - 2x) How do you get this? If the base is square, and you're using x as the side of the base, then the volume should be (x^2)h where h is the height. What are a-2x and b-2x? > To find the local maximum we differentiate: v'(x) = 12x^2 - 4(a + b)x + AB. How did new variables A and B suddenly appear? > Can't seem to get any further. Have I used all the data? There are two quantities, volume and surface area. Have you written an equation for surface area in terms of the dimensions of the box? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Shikata ga nai... === Subject: Re: false, yet misleading proofs > Hello all, I am looking for false, yet misleading proofs. For > example, Claim 1 = 2 > [proof omitted] Here's a simple (false) proof of FLT. Assume x^n + y^n = z^n. Take derivatives: nx^(n-1) + ny^(n-1) = nz^(n-1). Cancel the n: x^(n-1) + y^(n-1) = z^(n-1). So we have lowered the degree by 1. Eventually we reach x^0 +y^0 =z^0, i.e.1+1=1. This is a contradictation. === Subject: Re: Fluid Mechanics by Cengel 1st Ed. solution manual + Ebook posting-account=OGuGzgoAAAAmBMlWlDF8HxLRktk5tOfu .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 1.1.4322; InfoPath.2),gzip(gfe),gzip(gfe) I have the solution manual + ebook for Fluid Mechanics by Cengel 1st > Ed. let me know if anybody wants it. my email is binbin...@hotmail.com ------------------ > Hi > I really want the solution manual + ebook of Fluid Mechanics by > Cengel > my email is khosravi....@gmail.com > plz email me. much if you could do that. My email is vela0050@gmail.com === Subject: Re: JSH: Assessing group opinion, survey 1. Do you believe that I may have valuable mathematical ideas? > No. Though one cannot be sure. I just didn't see any such ideas from your side so far. 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? > No, I don't think that this is possible. Since there are no valuable mathematical ideas (from your side), there actually is noting to block. 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? > Hardly. 4. Have you heard me talk of a z constraint? Does that mean > anything to you? > No. It mean nothing to me. (Nor to anyone else, I guess.) 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? > Post as much as you want -- like anyone else. But maybe it would be a good idea to read/study some textbooks instead, no? 6. Why do you think so many posters reply to me? > Masochism? 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? > Depends. Some (most of them, imho) will and some maybe won't, I guess. F. -- E-mail: infosimple-linede === Subject: Re: JSH: Assessing group opinion, survey posting-account=XBlrJAoAAADt98TEM65O-rVj9Zl2tTjM 1.1.4322),gzip(gfe),gzip(gfe) you are crazy! How often can we talk to a nut from afar? So you'd say that James Harris is NutsTM ;-) === Subject: Re: JSH: Assessing group opinion, survey > I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! > You can answer freely but I will give a few questions that reflect > areas that are of great interest to me, so responses to those question > would be appreciated: > 1. Do you believe that I may have valuable mathematical ideas? Unlikely. > At least, you doesn't appear to have shown them, and I doubt you may > do in the future. What does he know! You got a paper published! > 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? No. (don't belive him, he is one of them!!) > 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? Loaded question: efficient factor methods would not cause major > security breaches. (He is just saying that so you won't publish your results!!) > And anyway, analysis of security problems of widespread computation > and communication systems are regularly published despite the fact > that such knowledge actually causes security breaches. (Think about > the WEP flaws or the countless critical vulnerabilities in Internet > Explorer and Outlook). Research on efficient factoring is also published. See Shor's quantum > algorithm, for instance. > 4. Have you heard me talk of a z constraint? Does that mean > anything to you? I'm not following your posts closely. (sure, he's keeping an eye on you for sure....) > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? You can post as much as you want, of course. But what's the point of > doing so? JSH could and will discover new technology. > 6. Why do you think so many posters reply to me? Because the love poking at you, I suppose. JSH is right, and many don't like his sucess! > 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? What do you mean by 'regardless of the source' ? cut and paste ?? > You're welcome. > === Subject: Re: JSH: Assessing group opinion, survey Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! You can answer freely but I will give a few questions that reflect > areas that are of great interest to me, so responses to those question > would be appreciated: 1. Do you believe that I may have valuable mathematical ideas? > No. > 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? > No. > 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? > If a viable factoring algorithm was posted on the Internet, presented sufficiently clearly that any qualified person could understand it, then pretty soon RSA Laboratories would hear about it and would announce to their customers that the RSA cryptosystem was no longer secure. I don't know about government and security agencies getting involved. > 4. Have you heard me talk of a z constraint? Does that mean > anything to you? > No. > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? > Do whatever you feel like. If your goal is to do some original and interesting mathematical research, you'll have to make major changes in your strategy. If that's what you want to do, you really should listen to the people who are trying to help you. > 6. Why do you think so many posters reply to me? > You know, this is an interesting one. I have a weakness for trying to reason with cranks. But you attract more replies than other cranks. I don't know exactly why this is. There's something about you. > 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? > Yes. > Any time. > James Harris === Subject: Re: Assessing group opinion, survey But I have a reply here. Comments below... >I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! You can answer freely but I will give a few questions that reflect > areas that are of great interest to me, so responses to those question > would be appreciated: 1. Do you believe that I may have valuable mathematical ideas? yes, extending to larger numbers may be difficult Ok. 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? Deliberately yes, in the one case where your paper was withdrawn because > of > errors, it is their responsibility to prevent substandard technical > content > (any paper with errors) from being published. No fraud involved, that is > too time consuming and complicated. the goodness of its heart? Do you really think that one paper had the power to destroy and entire math journal and 9 years of other papers, because it was wrong? ***they were over extended, no time to do it right, you paper got in unreviewed, I saw it posted long ago, it needed too much work, your paper had no effect on the pblisher, they were folding anyway. Math guy are all honest, they are trying to solve problems tougher than yours, and need everybodies help. Being dishonest take too much time and effort. 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? yes, check out early days of PHP. Also there are several free complex > multi > level encription/decrytion SW on the internet that no Goveernment can > break > anyway. The Gov just want their stuff to stay secure, most of that are > very > very large Psudo Random number streams, and have nothing to do with > factoring. > I'm afraid that is the case, which is why I am somewhat paused as I consider what to do next. ****You misunderstand, breakers of the RSA show and tell how the did it freely and they are heros IF you do it you will be a hero too 4. Have you heard me talk of a z constraint? Does that mean > anything to you? yes, but how much does it improve anything in quantitave terms? > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? You may run into something that really works, and others do review what > you > have so far, and you have brought up interesting points in Abstract > Algebra > that are complicated. The whining is distracting and invites attacks. >But what if I've followed the rules--like with publication--and it >hasn't mattered? >Is my only hope factoring a large number? *****you should include large numbers in your formulas, if T is 100 bytes long, that means k is 25 bytes long and so on. I appreciate further feedback from others. I'm unlikely to reply, but I will try to read them all in this thread, within reason. James Harris === Subject: Re: JSH: Assessing group opinion, survey But I have a reply here. Comments below... fraudulently and deliberately blocking those ideas? Deliberately yes, in the one case where your paper was withdrawn because > of > errors, it is their responsibility to prevent substandard technical > content > (any paper with errors) from being published. No fraud involved, that is > too time consuming and complicated. >The journal had the paper for nine months and I was in continual >communication by email and told them that I was an amateur researcher. As a rhetorical question, why would the editors publish a paper from >an admitted amateur that had errors? One of my favorite later emails AFTER publication from one of the >editors in reply to me thinking them for publication noted that >mathematics was important regardless of the source, even if it came >from a janitor. I'm not a janitor. The cultural reference seemed obvious to me. They don't know what you do or look like, and they do not care. If you go to any major college se what nationalities are there even on a website, most are from India, China, Europe, etc, it is an international mix. Some amateurs are really good, but your paper did not get adequately reviewed before publication, they should have rejected it earlier in the process, and it was most likely the editor and crew did not have time to do it. And it is obvious they did not have time to do all the work required to publish a formal journal. >So then, if the editors were that all the way in, how does it make >sense to so many of you that the paper was actually flawed? Editors did not read it on first pass, they would have rejected it. A version of this paper was available to look at several years ago, and it was too short, lacked explination of your assumptions, missing dirivations, and lacked needed form and structure. If you worked wtih a Professor, he would have gotten it into shape for you. >Isn't a cover-up more likely? Especially considering that later the >journal DIED? And Cameron University removed all mention of it from >their websites so that EMIS had to save 9 years of math papers out of >the goodness of its heart? No cover-up at all. Your paper needed too much work, and just lack of time of the editors/sponcers to support a technical publication. Some bad papers do get published, and then they get rejected or modified, or corrected in a note in the next issue. Just go read a few of the journals at the library and you will see. >Do you really think that one paper had the power to destroy and entire >math journal and 9 years of other papers, because it was wrong? Your paper was not part of the decision to close the math journal, nor to remove papers from a website. It is all still public information anyone can look at it in the library. 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? > yes, check out early days of PHP. Also there are several free complex > multi > level encription/decrytion SW on the internet that no Goveernment can > break > anyway. The Gov just want their stuff to stay secure, most of that are > very > very large Psudo Random number streams, and have nothing to do with > factoring. I'm afraid that is the case, which is why I am somewhat paused as I >consider what to do next. You read my response wrong. What ever you come up with applicable to encription/decription, has already been exceeded by others and is freely available. You are looking at factoring, which applies to a very small subset of encryption. That has nothing to do with Psudo Random sequences, nor a majority of encryption systems, which are stronger than RSA. Most people that solve RSA publish how they do it, no problem 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? > You may run into something that really works, and others do review what > you > have so far, and you have brought up interesting points in Abstract > Algebra > that are complicated. The whining is distracting and invites attacks. >But what if I've followed the rules--like with publication--and it >hasn't mattered? >Is my only hope factoring a large number? include it in your research, something like that RSA100 because it will have an effect on what number sizes you choose for n, or z or the others, and it will show you the length of the subloops needed to support your approach James Harris === Subject: Re: JSH: Assessing group opinion, survey <47d0d9f4$1@news.x-privat.org> posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? Deliberately yes, in the one case where your paper was withdrawn because > of > errors, it is their responsibility to prevent substandard technical > content > (any paper with errors) from being published. No fraud involved, that is > too time consuming and complicated. > reviewed before publication, they should have rejected it earlier in the > process, and it was most likely the editor and crew did not have time to do > it. The very submission of the paper was fraudulent on the part of Mr. Harris. The ideas in the paper had been presented in this forum, and he had ALEADY been told by a number of professional mathematicians that they were erroneous. === Subject: Re: JSH: Assessing group opinion, survey > I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! You choose to feed the wrong part of yourself. You have also found a way to gorge by posting elementary mathematics accompanied by inflated and baseless claims; infantile fantasies of omnipotence; and vicious attacks. Some people know you engage in self destructive activities, and feed you anyway, because they despise you and want to see you, as the creature in the last reel of a science fiction movie, a decaying, foetid puddle of unimaginably foul slime on the floor. Free yourself. -- Michael Press === Subject: Re: Assessing group opinion, survey [...] > Do you really think that one paper had the power to destroy and entire > math journal and 9 years of other papers, because it was wrong? 21. Do you believe in The Hammer? I think this is the first invocation. === Subject: JSH: Hidden proof revealed, FLT for p>3 When I realized a couple of days ago that my latest attempt at a proof of FLT was still incomplete, I was a bit bummed. It took me about an hour to notice something interesting, which was that I had proven Case 1. ( Not so easy to see when you're playing with p=3 all the time which doesn't have a Case 1.) That heartened me a bit, so I made my ...self-pity posting which actually contained that portion of the proof within it, only slightly hidden. I thought it amusing. A bit later I realized the complete proof for p>3 and also put that in a reply to my ...self-pity posting. I still think it amusing. This one looks rock solid which is good because it's meant to do a teeny bit of bashing. The hammer is at the top of its arc but we have a while yet before the blow because...you folks will ignore the obvious. If I didn't think otherwise I wouldn't make this post, despite my growing ennui... ------------------------------------------------------------------- -- Michael Press === Subject: Re: JSH: Assessing group opinion, survey posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! You can answer freely but I will give a few questions that reflect > areas that are of great interest to me, so responses to those question > would be appreciated: 1. Do you believe that I may have valuable mathematical ideas? > Never have had any Don't have any now Never will have any (hope that's clear enough) > 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? > No. > 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? > No. > 4. Have you heard me talk of a z constraint? Does that mean > anything to you? > Yes. No. > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? > Difficult question. I guess it depends on why you are posting in the first place. I have a theory though that you are having one big joke at our expense. Sometimes, I think you are playing the motley fool intentionally and having one big joke on us. Part of me thinks you know you are wrong, and that you are in fact, inventing this up to create an Office like spoof. You are like the David Brent of mathematics (or Steve Carrell, if you like the US version better): the guy who can never get it right, who is always wrong, yet despite every situation in which he fails, never realizes it is *him*, and keeps on with his stupid ways. Although he knows nothing, that does not keep him from having an opinion on everything, and of course, he feels it his duty to share these deep, Jack Handy-like insights with everyone. Maybe I'm wrong about this, but you do have to admit that the coincidences are quite extraordinary! > 6. Why do you think so many posters reply to me? > Like Annie Lennox sings,Some of the want to abuse you, some of the want to be abused... I can't answer for other posters, but I *personally* find your failures very satisfying. They help validate my philosophy of what mathematics is. > 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? > Of course. Why wouldn't I? They certainly have exposed your hoaxes all too easily. > Why you are most welcome James. > James Harris M === Subject: Re: JSH: Assessing group opinion, survey posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > 1. Do you believe that I may have valuable mathematical ideas? No. > 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? No. > 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? No. > 4. Have you heard me talk of a z constraint? Does that mean > anything to you? Yes. No. > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? You should post much more than you do, especially after midnight. > 6. Why do you think so many posters reply to me? I can't speak for other, but my reason is simple. the only place you've found where you get the attention you need. I give you some attention, and you give me some comedy. > 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? Yes. -- I'm like a super janitor. -- James Harris === Subject: Re: Assessing group opinion, survey posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20071009 SeaMonkey/1.1.5,gzip(gfe),gzip(gfe) But I have a reply here. Comments below... >I am curious about the effectiveness of the approach I have taken in > communicating mathematical ideas and research I feel is important, so > I am making this post to ask your opinion! > You can answer freely but I will give a few questions that reflect > areas that are of great interest to me, so responses to those question > would be appreciated: > 1. Do you believe that I may have valuable mathematical ideas? yes, extending to larger numbers may be difficult Ok. > Mr Banks is being exceedingly generous. > 2. Do you believe that it is at all possible that mathematicians are > fraudulently and deliberately blocking those ideas? Deliberately yes, in the one case where your paper was withdrawn because of > errors, it is their responsibility to prevent substandard technical content > (any paper with errors) from being published. No fraud involved, that is > too time consuming and complicated. > It's not happening. You have an enormous free forum. Nothing is being blocked (but please try to expand your vocabulary a little). Papers you have submitted to journals are clearly without merit and have been evaluated appropriately and fairly. No, it is not a matter of style or scholarly references; it is a matter of content. You don't know what constitutes a valid proof. Worse, your papers (e.g. APF) assert incorrect statements; your incorrect proofs are irrelevant. > The journal had the paper for nine months and I was in continual > communication by email and told them that I was an amateur researcher. As a rhetorical question, why would the editors publish a paper from > an admitted amateur that had errors? > My guess: Sheer, outright clerical error: the paper was mistakenly put in the to-be-published file rather than the reject file. Easy to have happen with an electronic journal, almost impossible to happen with a print journal. > One of my favorite later emails AFTER publication from one of the > editors in reply to me thinking them for publication noted that > mathematics was important regardless of the source, even if it came > from a janitor. > True. > I'm not a janitor. You should consider it though. > The cultural reference seemed obvious to me. > What are you saying? What did the editors know about you that would cause them to make a 'cultural reference'? > So then, if the editors were that all the way in, how does it make > sense to so many of you that the paper was actually flawed? > Your premise is wrong. This was simply a filing error. We already know the editing of this journal was sloppy at best. With regard to your paper being flawed: more like, totally rotten to the core. There was one glaring, obvious, stupid error toward the end that you yourself overlooked for over a year. It was fixable, but it was NOT fixed in the version that was published for a few hours. Then there was the central error, based on the idea that if the constant term g(0) is divisible by a prime p, then g(x) is divisible by p for all x. Another glaringly obvious stupid error, but slightly more obscured by your unclear statement of it. And your main 'result' contradicts very well established important theorems. You think that that erroneous publication somehow validates your proof because it was peer- reviewed. If that were true, what about 150 years of papers in Galois theory and algebraic number theory, also peer reviewed, and many of them published in the best journals. How many of those were withdrawn? And what, incidentally, did the peer review of your paper actually say? > Isn't a cover-up more likely? There WAS a cover-up. The editor royally screwed up. Then he tried to hide it by (1) yanking your paper without your consent or any explanation to you, and (2) not publishing any kind of explanation in the journal. He treated both you and his readers with contempt. The root problem was that the paper was completely wrong and he saw that immediately when it was pointed out and did not want it to have it on the books. He covered his own tracks to save himself embarrassment. > Especially considering that later the > journal DIED? And Cameron University removed all mention of it from > their websites so that EMIS had to save 9 years of math papers out of > the goodness of its heart? > It is likely that the editor's incompetence was a factor in the death of the journal. It was a mediocre journal to start with and an unusually large proportion of the papers published in it were written by ... the editor! Your paper however was too obviously bad even for that journal. Your paper may have contributed a little to the journal's demise in that it was strong explicit evidence of incompetent editing. > Do you really think that one paper had the power to destroy and entire > math journal and 9 years of other papers, because it was wrong? > No. At best it was a small factor. > 3. Can important ideas in factoring be presented loudly without > anyone around the world noticing or caring, like governments or > security agencies if those factoring ideas could lead to major > breaches in security? yes, check out early days of PHP. Also there are several free complex multi > level encription/decrytion SW on the internet that no Goveernment can break > anyway. The Gov just want their stuff to stay secure, most of that are very > very large Psudo Random number streams, and have nothing to do with > factoring. > No, I doubt this is right. The government is not good at all at managing conspiracies. There are too many witnesses. The government could never pay them all to shut up, and anyway, a good many of them are honest. And anyway, there is no suppression of Harris's work. He gets all the air time he wants all the time. The government makes no effort to suppress it. Why should they? There is nothing worthwhile to suppress. > I'm afraid that is the case, which is why I am somewhat paused as I > consider what to do next. > 4. Have you heard me talk of a z constraint? Does that mean > anything to you? yes, but how much does it improve anything in quantitave terms? > Harris's constraint boils down to the fact that if T = r * s, where r and s are odd numbers and T is congruent to 2 mod 3, then if z - y = r and z + y = s, then z is divisible by 3. Trivial to prove and not a big deal and of minimal help in factoring. > It shows an equation that defines z when T mod 3 = 2, and z^2 = y^2 + T. Then z = (1 + 2.87^2)k/(2.87), where k^2 = (1 + .87^2)^{-1}(nT) mod p and p is an odd prime of your choice chosen such that k exists, and > since there are (p-1)/2 prime residues, it should exist about 50% of > the time. Then you also get factors mod p: f 1 = .87k mod p and f 2 = .87^{-1}(1 + .87^2)k mod p. But you'd still need to find k, but the research shows that k is near > the k such that abs(T - (1 + .87^2)k^2) is a minimum in that with positive k it is greater than or equal to > that value, and has the residue given by the equation above. > So which is it? Is k^2 = (1 + a^2)^{-1}nT mod p, or is k near to sqrt(T / (1 + a^2))? These are by no means necessarily close to each other. And what does near mean? If p is large, how much searching must you do to find a number whose quadratic residue is k? > So attacking an RSA number could be feasible if all that is true just > by picking a large prime p, and trying it. If it doesn't work, try > another, and 50% of them should work, and then you'd have z. > No. You would still have to search for k, from what you just said. That includes having to search for a quadratic residue modulo a large prime. How do you do that? > 5. Do you think I should just shut-up, or should I post as much as I > want--like anyone else? You may run into something that really works, and others do review what you > have so far, and you have brought up interesting points in Abstract Algebra > that are complicated. The whining is distracting and invites attacks. But what if I've followed the rules--like with publication--and it > hasn't mattered? > You HAVEN'T followed the rules. You have submitted papers that you KNEW were wrong (e.g. APF, the second time you submitted it). You have submitted a version of APF with the name of Andrew Beckwith (a published author) as a co-author, only in the hope of getting it accepted, even though Beckwith did not write one word of the paper. You have submitted papers when you knew there were substantive serious objections which you had not been able to refute. Far from following the rules, you have committed fraud. The papers were wrong and were duly rejected, and I doubt the editors ever knew that fraud was involved. They were rejected on their own lack of merit. Further: after APF was rejected (at least twice), you started claiming that the underlying ring in the paper was NOT the ring of algebraic numbers. This was a blatant obvious lie. Yet you clung to it for weeks, and probably still do. To say that you follow the rules must mean that the rules do not exclude lying. > Is my only hope factoring a large number? > That would be sufficient. Why don't you do it? > 6. Why do you think so many posters reply to me? Some are really trying to help you. Some attack the whining. Ok. > Most are just making fun of you. Many of these are hateful but you can just blow them off. The serious replies which demonstrate you are wrong are ignored or you dismiss them as lies. That is a mistake. > 7. Do you trust mathematicians of today to tell the truth as best > they know about mathematical ideas and research regardless of the > source? Most do tell the truth, it is an honor, and their truth is in the language > of Math and honorable traceable back to Newton, and before thousands of > years, etc Most are too busy working on hard problems, like you are, and > have no time to spend even thinking of being deceptive. > Mathematicians may be the most trustworthy group around. I appreciate further feedback from others. I'm unlikely to reply, but > I will try to read them all in this thread, within reason. > You may read them but you will not accept valid criticism. Marcus. > James Harris === Subject: permutation puzzle questions posting-account=emCEQwoAAAC2slHnMMHOpDGbgxTD6IYH MathPlayer 2.10b; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) I'm working on a solver for a permutation puzzle: Given a 3 x 3 grid; the task is to bring the fields of this grid into their natural order. You only can rotate each row / column of this grid left (right) / up (down). In other words: For some permutation in A_9, find a decomposition into cycles (1,2,3), (4,5,6), (7,8,9), (1,4,7), (2,5,8), (3,6,9). Experimentally, I found that there are permutations P such that the number of inversions always increases, no matter which of the permutations (1,2,3), (4,5,6), ... (or their inverses) I apply to P. An example is (5,9,8,7,6). So is there a way to characterize them, or am I only left with a brute-force search? Or does anybody have an idea why this is so? Thomas === Subject: Re: JSH: Thinking about equals posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20071009 SeaMonkey/1.1.5,gzip(gfe),gzip(gfe) I don't think you understand what I and others found so astounding about > that post was the theory that your critics are a parasitic genetic > subtype, and that evolution has produced you to prevent these parasites > from causing the demise of the human race. This is not the first incredible vision that you have presented to these > groups. For years you have been making the case that professional > mathematicians secretly know that your theories are true, but maintain a > conspiracy of lies to cover that up. You have also predicted that some > day, the public would discover the truth and fire all of your critics. > At times, you have taken this fantasy one step further and predicted > that the police would be sent in to forcibly remove Andrew Wiles and > others who disagree with you. This year's factoring threads have added even more outlandish theories. > We read of your fear of being assassinated for breaking RSA encryption. > You speculated that aliens may be in order to corrupt the study of pure > math. I had thought that your posts could not possibly become more extreme. > Then you came up with this story about the army of parasitic priests who > would have destroyed humanity if evolution had not produced you to stop > them. James, I've got to hand it to you: you still have the power to > astound even the most jaded of your readership. You are clearly uncomfortable with discussions of Narcissism, but when > you present outlandishly self-aggrandizing theories such as these, your > readers cannot help but think about Narcissism. Imagine a 400 pound man > who insists that his friends not discuss obesity in his presence. This > man may succeed in preventing his friends from talking about obesity, > but as his friends observe the fat man struggling to get through > doorways and avoid breaking furniture, they will certainly be *thinking* > about obesity. It is unavoidable. Narcissism can be controlled by therapy. You have clearly rejected this > option. Your decision to not treat this condition is extremely foolish, > and is bound to cause you much anguish in the coming years. But the > choice is yours. I'm not concerned about narcissism. I am concerned about my ability > to show major results with simple mathematics and still find myself > mired in dumb arguments. There just is no other research out there that will give you the > information that I can give in just a short space, like with an odd T, > coprime to 3, the target to be factored, and z^2 = y^2 + T, so you're > trying to find z and y to factor T non-trivially, if T mod 3 = 2, then > I have the explicit equation for z: z = (1 + 2.87^2)k/(2.87) which may seem trivial at first blush since that can easily be > satisfied with all integers when z has 3 as a factor, as it must if T > mod 3 = 2, and T is odd and coprime to 3, but I can also give factors > of T, modulo p where p is a picked prime where the condition on > picking p comes from the equation: k^2 = (1 + .87^2)^{-1}(T) mod p so you just find p and .87 such that k exists, as long as p is less than > both factors and p minus the smaller factor is less than that factor, > and now you can go further as you have: f 1 = .87k mod p and f 2 = .87^{-1}(1 + .87^2)k mod p where f 1*f 2 = T, and that is just so huge, as for any given > composite T where T mod 3 = 2, you will have k and .87 for each > factorization that will work with p. That is just mathematics. If the result is true then it is huge. And my narcissism is irrelevant, as most major figures in human > history have been narcissists anyway. You people would have treated Newton or Kant or Gauss. Forget whether or not I'm a narcissist and answer one question: 1. Is the result correct or not? > First, let's agree on what you are claiming to be true. Here is my version of that: 1. Assume T is the product of two odd primes, and T = 2 mod 3. 2. Choose a 'large' prime p - I am guessing you mean about as large as sqrt(T). 3. Choose a value for 'a'. Simplest choice, a = 1. 4. Compute k2 = (1 + a^2)^{-1} T mod p. 5. See whether k2 is a quadratic residue mod p - that is, whether k2 = k^2 mod p for some integer k. If not go back to step 2 and choose another large prime for p until you do find that k2 is a quadratic residue. 6. If k2 is a quadratic residue, then compute f 1 = ak mod p and f 2 = a^{-1}(1 + a^2)k mod p 7. Then f 1*f 2 = T and is a nontrivial factorization. Is this an accurate description of your method? Yes or no? Marcus. > James Harris === Subject: Re: JSH: Thinking about equals > A very well thought out and clearly articulated post, Mr. Lhota. > I would submit that the reason he has not sought out therapeutic help for > his Narcissism is because he is receiving all of the attention, and > more, that feeds his Narcissism. In other words, he doesn't feel like he > needs any help as long as he is getting all of the attention he desires. > I would guarantee that if everyone here, and that means every one, > including myself, stopped responding to him, (and I know, it's easier > said than done) he would show withdrawal symptoms like a heroin junkie > would, and basically he would go ballistic here in Usenet. You haven't > read anything from him to compare with what would be coming out of him > then, if we could only bite our lip, count to ten, walk around the block, > do whatever we have to do to not respond to him. > It's really just that simple. > --Fredrico Juarrez > For quite a while now, I've been pondering about what the appropriate > response to Mr. Harris. The increasingly bizarre, worrisome twists that > his recent writings have taken add a certain urgency to this issue. > Posters have tried serious critiques, sarcasm, supportive posts, and > candid confrontation in dealing with James Harris. Some of this makes > fascinating reading, but none of this has had any effect on his > self-destructive behavior. Your idea may be the one approach that could make a positive difference. > Granted, it would require that we refrain from posting absolutely any > reply, even a brief Begone, troll or a quick retort to an insult. Also, > we're not sure how much attention his NPD demands, so it may require near > unanimous agreement. OTOH the number of people who regularly post to the James Harris is small > enough that it might be possible to pull this off. I would also point out > that several posters who used to engage James Harris have stopped replying > to his posts. Tim Peters and Arturo Magidin come to mind. As for me, I'm willing to give it a try. I must confess that as he > endlessly recycles his materials, my rebuttals end up being repetitious as > well. It is time for me to take a break. I therefore take the following > pledge: I pledge to not reply to any Jame Harris post that demonstrates > narcissistic thinking, including conspiracy theories, plots for revenge, > and grandiose thinking. I will only reply to a James Harris post if it > presents ideas in a modest, respectable manner, and invites an open > discussion of these ideas. Who here will join me in taking this pledge? > I will pledge to review his math in like terms of his post === Subject: Re: JSH: Thinking about equals SV1),gzip(gfe),gzip(gfe) A very well thought out and clearly articulated post, Mr. Lhota. I would submit that the reason he has not sought out therapeutic help > for his Narcissism is because he is receiving all of the attention, and > more, that feeds his Narcissism. In other words, he doesn't feel like he > needs any help as long as he is getting all of the attention he desires. I would guarantee that if everyone here, and that means every one, > including myself, stopped responding to him, (and I know, it's easier > said than done) he would show withdrawal symptoms like a heroin junkie > would, and basically he would go ballistic here in Usenet. You haven't > read anything from him to compare with what would be coming out of him > then, if we could only bite our lip, count to ten, walk around the > block, do whatever we have to do to not respond to him. It's really just that simple. --Fredrico Juarrez > For quite a while now, I've been pondering about what the appropriate > response to Mr. Harris. The increasingly bizarre, worrisome twists that > his recent writings have taken add a certain urgency to this issue. > Posters have tried serious critiques, sarcasm, supportive posts, and > candid confrontation in dealing with James Harris. Some of this makes > fascinating reading, but none of this has had any effect on his > self-destructive behavior. Your idea may be the one approach that could make a positive difference. > Granted, it would require that we refrain from posting absolutely any > reply, even a brief Begone, troll or a quick retort to an insult. > Also, we're not sure how much attention his NPD demands, so it may > require near unanimous agreement. OTOH the number of people who regularly post to the James Harris is > small enough that it might be possible to pull this off. I would also > point out that several posters who used to engage James Harris have > stopped replying to his posts. Tim Peters and Arturo Magidin come to mind. As for me, I'm willing to give it a try. I must confess that as he > endlessly recycles his materials, my rebuttals end up being repetitious > as well. It is time for me to take a break. I therefore take the > following pledge: I pledge to not reply to any Jame Harris post that demonstrates > narcissistic thinking, including conspiracy theories, plots for revenge, > and grandiose thinking. I will only reply to a James Harris post if it > presents ideas in a modest, respectable manner, and invites an open > discussion of these ideas. Who here will join me in taking this pledge? > All right, I'll give it a go. > -- > All things extant in this world, > Gods of Heaven, gods of Earth, > Let everything be as it should be; > Thus shall it be! > - Magical chant from Magical Shopping Arcade Abenobashi Drizzle, Drazzle, Drozzle, Drome, > Time for this one to come home! > - Mr. Wizard from Tooter Turtle- Hide quoted text - - Show quoted text - === Subject: Re: JSH: Thinking about equals SV1),gzip(gfe),gzip(gfe) > A very well thought out and clearly articulated post, Mr. Lhota. > I would submit that the reason he has not sought out therapeutic help > for his Narcissism is because he is receiving all of the attention, and > more, that feeds his Narcissism. In other words, he doesn't feel like he > needs any help as long as he is getting all of the attention he desires. > I would guarantee that if everyone here, and that means every one, > including myself, stopped responding to him, (and I know, it's easier > said than done) he would show withdrawal symptoms like a heroin junkie > would, and basically he would go ballistic here in Usenet. You haven't > read anything from him to compare with what would be coming out of him > then, if we could only bite our lip, count to ten, walk around the > block, do whatever we have to do to not respond to him. > It's really just that simple. > --Fredrico Juarrez > For quite a while now, I've been pondering about what the appropriate > response to Mr. Harris. The increasingly bizarre, worrisome twists that > his recent writings have taken add a certain urgency to this issue. > Posters have tried serious critiques, sarcasm, supportive posts, and > candid confrontation in dealing with James Harris. Some of this makes > fascinating reading, but none of this has had any effect on his > self-destructive behavior. Your idea may be the one approach that could make a positive difference. > Granted, it would require that we refrain from posting absolutely any > reply, even a brief Begone, troll or a quick retort to an insult. > Also, we're not sure how much attention his NPD demands, so it may > require near unanimous agreement. OTOH the number of people who regularly post to the James Harris is > small enough that it might be possible to pull this off. I would also > point out that several posters who used to engage James Harris have > stopped replying to his posts. Tim Peters and Arturo Magidin come to mind. As for me, I'm willing to give it a try. I must confess that as he > endlessly recycles his materials, my rebuttals end up being repetitious > as well. It is time for me to take a break. I therefore take the > following pledge: I pledge to not reply to any Jame Harris post that demonstrates > narcissistic thinking, including conspiracy theories, plots for revenge, > and grandiose thinking. I will only reply to a James Harris post if it > presents ideas in a modest, respectable manner, and invites an open > discussion of these ideas. Who here will join me in taking this pledge? LOL. æI wish. You people stalk me. æIt's that simple. I've never wanted replies from people who avoid the mathematics like > you just did as I replied to you in this thread, addressing the issue > of narcissism. You stalk me because I'm right, as you know as I know that just > disagreeing even without merit has an impact on most people who > believe that a powerful and correct result would just be accepted. I challenge you and those of you who claim I just have NPD to refrain > from replying to my posts. I challenge you to show you believe what you claim you believe. As if I just have NPD then there is no reason to attack my > mathematical research at all, now is there. > You may or may not have NPD. There is certainly very strong evidence that you are completely impervious to reason so you have a point that there is not much point in trying to reason with you. On the other hand, why not reply to you, if it provides a few moments of diversion? The alternative is getting back to work on my lesson plans for next week. Would you prefer that people didn't make efforts to give you constructive feedback on your research? You're saying there's no rational motive for us to keep replying to you, well, fair enough, but may I ask what is your rational motive to keep posting? > And by replying by YOUR OWN ADMISSION you are simply aggravating the > disease, if I have it. So show what you truly believe--by your actions, if not your words. James Harris- Hide quoted text - - Show quoted text - === Subject: Re: JSH: Thinking about equals A very well thought out and clearly articulated post, Mr. Lhota. I would submit that the reason he has not sought out therapeutic help > for his Narcissism is because he is receiving all of the attention, and > more, that feeds his Narcissism. In other words, he doesn't feel like he > needs any help as long as he is getting all of the attention he desires. I would guarantee that if everyone here, and that means every one, > including myself, stopped responding to him, (and I know, it's easier > said than done) he would show withdrawal symptoms like a heroin junkie > would, and basically he would go ballistic here in Usenet. You haven't > read anything from him to compare with what would be coming out of him > then, if we could only bite our lip, count to ten, walk around the > block, do whatever we have to do to not respond to him. It's really just that simple. --Fredrico Juarrez > For quite a while now, I've been pondering about what the appropriate > response to Mr. Harris. The increasingly bizarre, worrisome twists that > his recent writings have taken add a certain urgency to this issue. > Posters have tried serious critiques, sarcasm, supportive posts, and > candid confrontation in dealing with James Harris. Some of this makes > fascinating reading, but none of this has had any effect on his > self-destructive behavior. Your idea may be the one approach that could make a positive difference. > Granted, it would require that we refrain from posting absolutely any > reply, even a brief Begone, troll or a quick retort to an insult. > Also, we're not sure how much attention his NPD demands, so it may > require near unanimous agreement. OTOH the number of people who regularly post to the James Harris is > small enough that it might be possible to pull this off. I would also > point out that several posters who used to engage James Harris have > stopped replying to his posts. Tim Peters and Arturo Magidin come to mind. As for me, I'm willing to give it a try. I must confess that as he > endlessly recycles his materials, my rebuttals end up being repetitious > as well. It is time for me to take a break. I therefore take the > following pledge: I pledge to not reply to any Jame Harris post that demonstrates > narcissistic thinking, including conspiracy theories, plots for revenge, > and grandiose thinking. I will only reply to a James Harris post if it > presents ideas in a modest, respectable manner, and invites an open > discussion of these ideas. Who here will join me in taking this pledge? 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It contains the solutions to the even problems as well, not just the odd ones. If so, would someone mind sending it to me? It would be a HUGE help! > Zang i do posess the manual that you require,pleae feel free to email me: santoshramani85@gmail.com,please do reply. === Subject: solution manuals posting-account=1LXcdwoAAAAOMsiNKAVJ2p8wxrEhk12V Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. MODERN CONTROL SYSTEM 4th Edition by OGATA Modern Control System 11th edition by Dorf and Bishop (11e) Fundamentals of Microelectronics. Author: Behzad Razavi Microelectronic Circuits. Author: Adel S. Sedra, Kenneth C. Smith Control Systems Engineering. Author: Norman S. Nise Probability and Random Processes for Electrical Engineering (3rd Edition)Chapters 1-6. Author: Albert Leon-Garcia Communication Systems Engineering (2nd Ed.). Author:Proakis and Salehi. 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HAYLER 's Probability and Statistics for Engineers and Scientists Manual Nanoengineering of Structural, Functional and Smart Materials If your interested do let me know at xplicitly666@yahoo.com Explicit Group === Subject: A First Course In Probability 7th Ed by Sheldon Ross posting-account=oaQOKgoAAACLUoTeYoU4afFgXLFVoS5b SV1),gzip(gfe),gzip(gfe) I need an ebook of the book - A First Course In Probability 7th Ed by Sheldon Ross Pls. send me a copy of this book === Subject: Re: A First Course In Probability 7th Ed by Sheldon Ross posting-account=lHNboAoAAACyasQ0uqX7OeM_tLuWGoQp CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) I need an ebook of the book - A First Course In Probability 7th Ed by > Sheldon Ross Pls. send me a copy of this book Please don't. This is just an attempt to rip off the publisher and author. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=LlRppgoAAAD1KDQAbEw51E0RIOUzJ0up Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > The two prime numbers were now telling me something James Harris Maybe to stop mathurbating? You are getting even more delusional. It isn't even April Fool's Day yet! Narcissist! No, but it's Valentine's day, and James is madly in love with himself. BTW, I motion the words STUMBLED and GUESSED be added to the list of James Harris BINGO words. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. So I expanded out a factorization with primes: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? And I realized that then you could use (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and reduce to a solution for d_1. d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. So suddenly I had it!!! Information from the intersection of the > primes. The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! And now you can go back to (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). That won't work. Turns out you need to check modulo p_1*p_2, as it is an intersection between BOTH primes. If you tried this and found that you always got matches then that's why. I finally tried it and freaked out when it wasn't working, so I went through a moment of extreme self-pity, hatred and anger. Then I figured it out. Funny how in math even the littlest things can make such a big difference. James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? > And I realized that then you could use > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and reduce to a solution for d_1. > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 > and then you also need d_2, so > d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. > So suddenly I had it!!! Information from the intersection of the > primes. > The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! > And now you can go back to > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to > k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 > so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since > k_2 = floor(T/p_2). That won't work. Turns out you need to check modulo p_1*p_2, as it is > an intersection between BOTH primes. If you tried this and found that you always got matches then that's > why. So you have to determine if they are factors before you determine that they might be factors. (???) I finally tried it and freaked out when it wasn't working, so I went > through a moment of extreme self-pity, hatred and anger. Then I figured it out. Funny how in math even the littlest things can make such a big > difference. > James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. So I expanded out a factorization with primes: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? And I realized that then you could use (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and reduce to a solution for d_1. d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 (1) and then you also need d_2, so d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. (2) Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. > But that's all you need since there are only so many numbers that fit. For instance, with 119, it factors as 7(17) or trivially as 1(119), so you only have 4 integers to fit, and for each you have f_1, as you have f_1 = 1 mod 11, f_1 = 7 mod 11, f_1 = 17 mod 11 = 6 mod 11, or f_1 = 119 mod 11 = 9 mod 11. So those are the ONLY values for f_1 that will fit with integers. > So suddenly I had it!!! Information from the intersection of the > primes. The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! And now you can go back to (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), I assume you mean to write m2 = floor(g1*g2/p2), otherwise the > following equation is wrong (since r2 has disappeared): Yup. and you can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 (3) so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since k_2 = floor(T/p_2). Now then, does this work? No. And here's why: as I mentioned above, you only know the values of > d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the > equation I have labelled (3). The trouble is that even if you have > incorrectly guessed f1 and g1 this equation will always hold mod p1. Nope, here's why. The relations (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) are explicit, as in exact and those are factors of T. Now mathematically there are only so many integers that can factor T, correct? So there are only so many values for which those explicit equations can be correct which is why this technique is a key in the lock. > This can be seen by directly substituting your formulae (1) and (2) k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and > this becomes k2 = T*p2^(-1) - r2*p2^(-1) mod p1, i.e. T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. Another tautology. That doesn't help you at all. > If you have those equations and they are true then, yes, you get equality on all sides. What you are doing by substituting is saying mathematically, these equations are true. But what if they're not for a specific value of the f's and g's? Then you're wrong and the substitutions are invalid. James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=BVr-MgkAAABE4LRE1rHDnN9heo0IZZTk .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) spider-ntc-te02.proxy.aol.com[CFC87082] (Prism/1.2.1), HTTP/1.1 cache-ntc-ac09.proxy.aol.com[CFC8748A] (Traffic-Server/6.1.5 [uScM]) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f 1 + c 1*p 1)(f 2 + c 2*p 1) = T = (r 1 + k 1*p 1) > and > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) > And I multiplied out and solved for k 1 and k 2 respectively and again > pondered, what if you GUESSED using f 1*f 2 = T mod p 1 and g 1*g 2 = > T > mod p 2, so you'd know the f's and the g's? > And I realized that then you could use > (f 1 + c 1*p 1) = (g 1 + d 1*p 2) > and > (f 2 + c 2*p 1) = (g 2 + d 2*p 2) > and reduce to a solution for d 1. > d 1 = (f 1 - g 1)*p 2^{-1} mod p 1 ? (1) > and then you also need d 2, so > d 2 = (f 2 - g 2)*p 2^{-1} mod p 1. ?(2) Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. But that's all you need since there are only so many numbers that fit. For instance, with 119, it factors as 7(17) or trivially as 1(119), so > you only have 4 integers to fit, and for each you have f 1, as you > have f 1 = 1 mod 11, f 1 = 7 mod 11, f 1 = 17 mod 11 = 6 mod 11, or f 1 = > 119 mod 11 = 9 mod 11. So those are the ONLY values for f 1 that will fit with integers. > So suddenly I had it!!! ?Information from the intersection of the > primes. > The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! > And now you can go back to > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) > multiply out, and divide by p 2, using m 2 = (g 1*g 2/p 2), I assume you mean to write m2 = floor(g1*g2/p2), otherwise the > following equation is wrong (since r2 has disappeared): Yup. > and you can get to > k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2 ? (3) > so now you just substitute modulo p 1 and compare what you get on the > right side with k 2 mod p 1 as you know what k 2 is, since > k 2 = floor(T/p 2). > Now then, does this work? No. And here's why: as I mentioned above, you only know the values of > d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the > equation I have labelled (3). The trouble is that even if you have > incorrectly guessed f1 and g1 this equation will always hold mod p1. Nope, here's why. The relations (f 1 + c 1*p 1) = (g 1 + d 1*p 2) and (f 2 + c 2*p 1) = (g 2 + d 2*p 2) are explicit, as in exact and those are factors of T. Now mathematically there are only so many integers that can factor T, > correct? So there are only so many values for which those explicit equations > can be correct which is why this technique is a key in the lock. This can be seen by directly substituting your formulae (1) and (2) k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and > this becomes k2 = T*p2^(-1) - r2*p2^(-1) mod p1, i.e. T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. Another tautology. That doesn't help you at all. If you have those equations and they are true then, yes, you get > equality on all sides. What you are doing by substituting is saying mathematically, these > equations are true. But what if they're not for a specific value of the f's and g's? Then you're wrong and the substitutions are invalid. James Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - For instance, with T=119, p1 = 11, and p2 = 13, I have f1f2 = 9 mod 11 and g1g2 = 2 mod 13 So I can use f1 = 2, as a bad guess along with f2 = 6, and g1 = 3 and g2 = 5. Sure enough! k 2 mod p 1 mismatch flags the error. Then you realize that f1 * f2 mod 11 does not equal T mod 11 = 9 Not knowing that f1 = 2 is the bad guess, you quickly change your last entry for f2 to 10 Now f1 * f2 mod 11 = 2 * 10 mod 11 = 9 = T mod 11 and k 2 mod p 1 matches, so you've got it! What's wrong with this picture? Enrico === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f 1 + c 1*p 1)(f 2 + c 2*p 1) = T = (r 1 + k 1*p 1) > and > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) > And I multiplied out and solved for k 1 and k 2 respectively and again > pondered, what if you GUESSED using f 1*f 2 = T mod p 1 and g 1*g 2 = > T > mod p 2, so you'd know the f's and the g's? > And I realized that then you could use > (f 1 + c 1*p 1) = (g 1 + d 1*p 2) > and > (f 2 + c 2*p 1) = (g 2 + d 2*p 2) > and reduce to a solution for d 1. > d 1 = (f 1 - g 1)*p 2^{-1} mod p 1 ? (1) > and then you also need d 2, so > d 2 = (f 2 - g 2)*p 2^{-1} mod p 1. ?(2) > Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. But that's all you need since there are only so many numbers that fit. For instance, with 119, it factors as 7(17) or trivially as 1(119), so > you only have 4 integers to fit, and for each you have f 1, as you > have f 1 = 1 mod 11, f 1 = 7 mod 11, f 1 = 17 mod 11 = 6 mod 11, or f 1 = > 119 mod 11 = 9 mod 11. So those are the ONLY values for f 1 that will fit with integers. > So suddenly I had it!!! ?Information from the intersection of the > primes. > The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! > And now you can go back to > (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) > multiply out, and divide by p 2, using m 2 = (g 1*g 2/p 2), > I assume you mean to write m2 = floor(g1*g2/p2), otherwise the > following equation is wrong (since r2 has disappeared): Yup. > and you can get to > k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2 ? (3) > so now you just substitute modulo p 1 and compare what you get on the > right side with k 2 mod p 1 as you know what k 2 is, since > k 2 = floor(T/p 2). > Now then, does this work? > No. And here's why: as I mentioned above, you only know the values of > d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the > equation I have labelled (3). The trouble is that even if you have > incorrectly guessed f1 and g1 this equation will always hold mod p1. Nope, here's why. The relations (f 1 + c 1*p 1) = (g 1 + d 1*p 2) and (f 2 + c 2*p 1) = (g 2 + d 2*p 2) are explicit, as in exact and those are factors of T. Now mathematically there are only so many integers that can factor T, > correct? So there are only so many values for which those explicit equations > can be correct which is why this technique is a key in the lock. > This can be seen by directly substituting your formulae (1) and (2) > k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 > Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and > this becomes > k2 = T*p2^(-1) - r2*p2^(-1) mod p1, > i.e. > T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. > Another tautology. That doesn't help you at all. If you have those equations and they are true then, yes, you get > equality on all sides. What you are doing by substituting is saying mathematically, these > equations are true. But what if they're not for a specific value of the f's and g's? Then you're wrong and the substitutions are invalid. James Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - g2 = 5. > Sure enough! k 2 mod p 1 mismatch flags the error. > Then you realize that f1 * f2 mod 11 does not equal T mod 11 = 9 > Not knowing that f1 = 2 is the bad guess, > you quickly change your last entry for f2 to 10 > Now f1 * f2 mod 11 = 2 * 10 mod 11 = 9 = T mod 11 > and k 2 mod p 1 matches, so you've got it! What's wrong with this picture? Enrico Oh yeah, I found my error, realized that they matched and freaked out!!! Then I went through a period of extreme mental pain and anguish. Ranted a bit in some posts, and started pondering some more. As I pondered I realized I'd forgotten to do something important. I'd forgotten to use BOTH primes and get the intersection of primes working for me, so I finally went back and took them modulo p 1*p 2 and it all worked. You know, that was kind of hard. I think it should have been easier. James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Sure enough! k_2 mod p_1 mismatch flags the error. > Then you realize that f1 * f2 mod 11 does not equal T mod 11 = 9 > Not knowing that f1 = 2 is the bad guess, > you quickly change your last entry for f2 to 10 > Now f1 * f2 mod 11 = 2 * 10 mod 11 = 9 = T mod 11 > and k_2 mod p_1 matches, so you've got it! What's wrong with this picture? Enrico Oh yeah, I found my error, *You* found your error? A quick read of this thread paints a different story. [...] > I'd forgotten to use BOTH primes and get the intersection of primes > working for me, so I finally went back and took them modulo p_1*p_2 > and it all worked. No, it still doesn't work. Allow me to quote from your latest blog entry, with a few additional comments of my own: *** begin quote **** > And now I can solve for the d's modulo p1: d1 = (f1 - g1)p2-1 mod p1 (1) and d2 = (f2 - g2)p2-1 mod p1. (2) Then I can use, but consider it modulo p1p2: Note that the equations I have labelled (1) and (2) only give you the residues of d1 and d2 mod p1. This narrows each of residues mod p1*p2 to p2 different possible values. In order to substitute d1 and d2 into an equation mod p1*p2 you therefore need to consider each of these pairs of values, not just one of them. k1 = c1f2 + c2f1 + c1c2p1 + m1 mod p1p2. (3) And see if k1 explicitly calculated corresponds with the > value given. If they do match you still have to check for the c's as > well, with c1 = (g1 - f1)p1-1 mod p2 and c2 = (g2 - f2)p1-1 mod p2 and k1 = c1f2 + c2f1 + c1c2p1 + m1 mod p1p2 to see if they match there as well. If they do, then you > have residues of a factorization, if not, then you can move > to the next set. For instance, with T=119, p1 = 11, and p2 = 13, I have f1f2 = 9 mod 11 and g1g2 = 2 mod 13 So I can use f1 = 2, as a hopefully bad guess along with > f2 = 10, and g1 = 3 and g2 = 5. Then d1 = (f1 - g1)p2-1 mod p1 = (2 - 3)13-1 mod 11 = 5 mod 11 and d2 = (f2 - g2)p2-1 mod p1 = (10 - 5)13-1 mod 11 = 8 mod 11. Then, m2 = floor(g1g2/p2) = floor(3(5)/13) = 1, and k2 = > (119-2)/13 = 9, so I have k2 = d1g2 + d2g1 + d1d2p2 + m2 mod p1p2 = 5(5) + 8(3) + 1 > = 50 mod 11(13). You have missed the term d1*d2*p2 from this sum. The correct answer for the RHS is 141 mod 143. So they do not match. Not for that particular choice of d1 and d2 mod p1*p2, no - but had you chosen different values (but ones which were still equal to the original values mod p1) you could have found that they did match. For example, try d1 = 5 and d2 = 107 = 8 mod 11. Then you find that d1*g2 + d2*g1 + d1*d2*p2 + m2 = 7302 = 9 mod 143. > If they did, I'd still have to check for the c's. So you loop through another set, and keep going until you > find matches. Notice though to complete the example, with f1 = 7 and f2 > = 6, and g1 = 7 and g2 = 4, I have that d1 = 0 mod mod 11, > d2 = 1 mod 11, and m2 = floor(7(4)/13) = 2, so k2 = d1g2 + d2g1 + d1d2p2 + m2 mod p1p2 = 0 + 7 + 2 = 9 > mod 143 as required. For that particular choice of d1 and d2, yes. But you could have made a different choice and you would have got a different answer. For example, try d1 = 0, d2 = 12 = 1 mod 11. Then d1*g2 + d2*g1 + d1*d2*p2 + m2 = 86 != 9 mod 143 *** end quote *** In fact, for *any* pair of non-zero residues f1 mod p1 and g1 mod p2, with d1 and d2 mod p1 defined by equations (1) and (2), there always exist values for d1 and d2 mod p1*p2 which have the right residues mod p1 and which satisfy equation (3). Proof: it was already shown in this thread that the values of d1 and d2 mod p1 given by equations (1) and (2) will automatically satisfy (3) mod p1. By the Chinese Remainder Theorem it therfore suffices to show that we can choose d1 and d2 to satisfy this equation mod p2 (since if a = b mod p1 and a = b mod p2 then a = b mod p1*p2). Suppose that for a given choice of d1 and d2 we have k2 - (d1*g2 + d2*g1 + d1*d2*p2 + m2) = n mod p2 Since p1*g1 is coprime to p2 it has an inverse mod p2 - denote this inverse by q. Then let d2' = d2 + n*q*p1, so that d2' = d2 mod p1. Replacing d2 by d2' in equation (3) mod p2 we find d1*g2 + d2'*g1 + d1*d2'*p2 + m2 = d1*g2 + d2*g1 + d1*d2*p2 + n*q*p1*g1 + d1*n*q*p1*p2 = k2 - n + n + 0 mod p2 as required. So in your above example you simply got lucky, in that the smallest possible values for d1 and d2 happened to follow your hypothesis; this will not be the case in general. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Sure enough! k_2 mod p_1 mismatch flags the error. > Then you realize that f1 * f2 mod 11 does not equal T mod 11 = 9 > Not knowing that f1 = 2 is the bad guess, > you quickly change your last entry for f2 to 10 > Now f1 * f2 mod 11 = 2 * 10 mod 11 = 9 = T mod 11 > and k_2 mod p_1 matches, so you've got it! > What's wrong with this picture? > Enrico Oh yeah, I found my error, *You* found your error? A quick read of this thread paints a different > story. [...] I'd forgotten to use BOTH primes and get the intersection of primes > working for me, so I finally went back and took them modulo p_1*p_2 > and it all worked. No, it still doesn't work. Allow me to quote from your latest blog > entry, with a few additional comments of my own: > Oh hey, you're right! Worse, this approach cannot lead to a viable factoring method. So I guess that's it with this idea. James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) Oh hey, you're right! Worse, this approach cannot lead to a viable factoring method. So I guess that's it with this idea. > James Harris- Where is the apology for thinking - yet again - that you dropped the Come on narcissist - you can do better than that. Grovel for us a bit - why don't you. What happened to all the chest pounding - your infinite one? As is typical with everything you have done in the past 12 years - you have NOTHING! NOTHING = James HARRIS! Have a nice day - delusional twit! ~A === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Oh hey, you're right! Worse, this approach cannot lead to a viable factoring method. So I guess that's it with this idea. > My pleasure. James Harris- Where is the apology for thinking - yet again - that you dropped the Come on narcissist - you can do better than that. Grovel for us a bit - why don't you. What happened to all the chest pounding - your infinite one? As is typical with everything you have done in the past 12 years - you > have NOTHING! NOTHING = James HARRIS! Have a nice day - delusional twit! Perhaps this will sound hypocritical, but I think you should lay off while James is being polite. Discussions are a lot more productive when they're civil, and it takes guts to be the one to step down from the animosity like he has done here. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=LlRppgoAAAD1KDQAbEw51E0RIOUzJ0up Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Have a nice day - delusional twit! Perhaps this will sound hypocritical, but I think you should lay off > while James is being polite. Discussions are a lot more productive > when they're civil, and it takes guts to be the one to step down from > the animosity like he has done here. No it doesn't sound hypocritical, just plumb stupid, that's all. Anyhow, go on and be PC all you want and feel good about yourself, but James's mixture of arrogance mixed with stupidity has earned him the the ridicule he so richly deserves. Besides, you miss the entire point, which is truly sad: there is no productive discussion with James. He knows math at (maybe) a sophomore level and all he does is guess (poorly). If you expect anything productive at the end of this, you're more of a wack job than he is. In point of fact, you are the one who fuel his hopes of finding a solution. If nobody took him seriously, he would simply give up. Your interest validates his hopes, and your suggestions give him new ideas to go on pondering. Maybe you are the one who needs to rethink things. Unless you like the status quo for some reason. M === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Have a nice day - delusional twit! Perhaps this will sound hypocritical, but I think you should lay off > while James is being polite. Discussions are a lot more productive > when they're civil, and it takes guts to be the one to step down from > the animosity like he has done here. No it doesn't sound hypocritical, just plumb stupid, that's all. I think that stupid is a bit strong, though you could certainly call me naive; the refutation that James readily accepted was actually more complicated than the ones he fought tooth and nail over last weekend, and his uncharacteristically polite reply made me think that perhaps he might turn over a new leaf. Though of course he was back to ranting about lying mathematicians within a couple of hours. > Anyhow, go on and be PC all you want and feel good about yourself, but > James's mixture of arrogance mixed with stupidity has earned him the > the ridicule he so richly deserves. I am well aware of his history. But I am inclined to think that insulting him even when he is being nice will only add to the paranoia which fuels his arrogance. Though that is just my opinion, of course I respect amzoti's right to post in whatever manner he pleases. > Besides, you miss the entire > point, which is truly sad: there is no productive discussion with > James. He knows math at (maybe) a sophomore level and all he does is > guess (poorly). If you expect anything productive at the end of this, > you're more of a wack job than he is. That all depends on what you mean by productive. Has it occurred to you that I might argue with Harris for reasons other than a belief that it will lead to a solution to the factoring problem? > In point of fact, you are the > one who fuel his hopes of finding a solution. If nobody took him > seriously, he would simply give up. Attempting to explain his errors hardly qualifies as taking him seriously. And there is ample evidence against your claim that he would give up if nobody took him seriously - as far as I can tell, nobody *does* take him seriously. > Your interest validates his hopes, > and your suggestions What suggestions? The only suggestion I recall having made is that he tests his algorithms before declaring the factoring problem solved, which will hardly give him new ideas - the only effect I can imagine is that it might make him post fewer ideas to usenet. > give him new ideas to go on pondering. Maybe you > are the one who needs to rethink things. Unless you like the status > quo for some reason. Yeah, that makes sense. I'm sure that the status quo would change if people were to just continuously insult James. Come to think of it, that's such a good idea I'm surprised that nobody has tried it before. === Subject: Re: JSH: Simple matching, factoring versus math politics > Have a nice day - delusional twit! > Perhaps this will sound hypocritical, but I think you should lay off > while James is being polite. Discussions are a lot more productive > when they're civil, and it takes guts to be the one to step down from > the animosity like he has done here. > No it doesn't sound hypocritical, just plumb stupid, that's all. I think that stupid is a bit strong, though you could certainly call > me naive; the refutation that James readily accepted was actually more > complicated than the ones he fought tooth and nail over last weekend, > and his uncharacteristically polite reply made me think that perhaps > he might turn over a new leaf. Though of course he was back to ranting > about lying mathematicians within a couple of hours. > > Anyhow, go on and be PC all you want and feel good about yourself, but > James's mixture of arrogance mixed with stupidity has earned him the > the ridicule he so richly deserves. I am well aware of his history. But I am inclined to think that > insulting him even when he is being nice will only add to the paranoia > which fuels his arrogance. Though that is just my opinion, of course I > respect amzoti's right to post in whatever manner he pleases. > > Besides, you miss the entire > point, which is truly sad: there is no productive discussion with > James. He knows math at (maybe) a sophomore level and all he does is > guess (poorly). If you expect anything productive at the end of this, > you're more of a wack job than he is. That all depends on what you mean by productive. Has it occurred to > you that I might argue with Harris for reasons other than a belief > that it will lead to a solution to the factoring problem? > > In point of fact, you are the > one who fuel his hopes of finding a solution. If nobody took him > seriously, he would simply give up. Attempting to explain his errors hardly qualifies as taking him > seriously. And there is ample evidence against your claim that he > would give up if nobody took him seriously - as far as I can tell, > nobody *does* take him seriously. > > Your interest validates his hopes, > and your suggestions What suggestions? The only suggestion I recall having made is that he > tests his algorithms before declaring the factoring problem solved, > which will hardly give him new ideas - the only effect I can imagine > is that it might make him post fewer ideas to usenet. > > give him new ideas to go on pondering. Maybe you > are the one who needs to rethink things. Unless you like the status > quo for some reason. Yeah, that makes sense. I'm sure that the status quo would change if > people were to just continuously insult James. Come to think of it, > that's such a good idea I'm surprised that nobody has tried it before. I submit to you that James is not at all interested in finding a solution to anything. He is using this 'factoring problem' of his to get people to respond to him. All James is interested in is being at the center of everyone's attention, thereby finding gratification for his Narcissistic Personality Disorder. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Then I went through a period of extreme mental pain and anguish. Ranted a bit in some posts, and started pondering some more. Hint, that's not normal. See a doctor. === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? > And I realized that then you could use > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and reduce to a solution for d_1. > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 (1) > and then you also need d_2, so > d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. (2) Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. But that's all you need since there are only so many numbers that fit. For instance, with 119, it factors as 7(17) or trivially as 1(119), so > you only have 4 integers to fit, and for each you have f_1, as you > have f_1 = 1 mod 11, f_1 = 7 mod 11, f_1 = 17 mod 11 = 6 mod 11, or f_1 = > 119 mod 11 = 9 mod 11. > This is not a good example. Here's why. You chose p1 = 11. That is larger than sqrt(T). In general you do not want to choose numbers that large. If, for example, T ~ 10^100, then sqrt(T) ~ 10^50. If p1 > 10^50, how many pairs (f1, f2) do you need to consider? What you want is that f1*f2 = T mod p1. Of course, since p1 is a prime, for ANY choice of f1 between 1 and p1 - 1, there exists f2 such that f1*f2 = T mod p1. Which means that the space of possible factors f1 has size on the order of 10^50. Many lifetimes of searching would be required. So this is not practical. You must choose p1 much smaller than sqrt(T). For example, if T = 119, then let p1 = 2. Note that T mod p1 = 1. This means that the only possible choice for f1 is f1 = 1. Of course any odd number is congruent to 1 mod 2, so the fact that f1 = 1 is the only possible choice does not tell you much. So you pick a different prime, say, p2 = 3. Note that T mod p2 = 2. Therefore you can choose g1 = 1 or 2. So you choose, perhaps, g1 = 2. So here is what you have: f1 mod p1 = 1 mod 2 = 1 g1 mod p2 = 1 mod 3 = 2. The Chinese Remainder Theorem then gives you a value x such that x mod 2 = 1 and x mod 3 = 2. The least value x for which this is true is x = 5. This is not a factor of T, so you choose a third prime less than sqrt(T), say, p3 = 5. Note that T mod 5 = 119 mod 5 = 4. Thus the possible choices for h1 are 1, 2, and 4. Perhaps you choose h1 = 1. Then you have 3 equations: f1 mod 2 = 1 g1 mod 3 = 2 h1 mod 5 = 1 Again using the Chinese Remainder Theorem, you can find x such that x mod 2 = 1 x mod 3 = 2 x mod 5 = 1 The smallest value for x is x = 11. Again, this is not a factor of T, and I have exhausted the supply of primes less than sqrt(T). Of course you can say, You fool, you should have chosen g1 = 1. You would then have found with the first two primes that x = 7 would be a solution. Instead you skipped over g1 = 1 and went to g1 = 2, which did not result in a factor. True, of course. But then maybe you begin to see the problem. There is only one choice for f1. There are two choices for g1. There are 4 choices for h1, and all together, for all combinations, there are 1 * 2 * 4 = 8 possible configurations. Most of these don't work! This is important. In general, for primes p1, p2, and p3, the number of combinations of f1, g1, and h1 is (p1 - 1)(p2 - 1)(p3 - 1), This can get to be a large number very fast. It grows somewhat like the factorial function. Suppose you need to factor a really big T. Perhaps you need to consider all the primes less than 1000. There are 168 such primes. The number of combinations which might work is (p1 - 1)(p2 - 1) ... (p168 - 1). An ENORMOUS number. You cannot expect to search all these possible combinations. True, you would eventually find the correct factors. But VERY inefficiently; far worse than random trial divisors or GCDs. > So those are the ONLY values for f_1 that will fit with integers. > You are choosing a bad example: much too small to reveal the problems with your method. Try something like T = 20413, and try it in a SYSTEMATIC way, not by factoring it first and then choosing f1, g1, h1, etc., after you know the answer. > So suddenly I had it!!! Information from the intersection of the > primes. > The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! > And now you can go back to > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), I assume you mean to write m2 = floor(g1*g2/p2), otherwise the > following equation is wrong (since r2 has disappeared): Yup. > and you can get to > k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 (3) > so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since > k_2 = floor(T/p_2). > Now then, does this work? No. And here's why: as I mentioned above, you only know the values of > d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the > equation I have labelled (3). The trouble is that even if you have > incorrectly guessed f1 and g1 this equation will always hold mod p1. Nope, here's why. The relations (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) are explicit, as in exact and those are factors of T. Now mathematically there are only so many integers that can factor T, > correct? > Definitely. But modulo a prime p1, there are (p1 - 1) ways. And unless p1 is bigger than sqrt(T), knowing all those possible modular factors does not tell you anything about the correct factors. Sure, you can choose p1 > sqrt(T), as you did in your little example, but you do NOT want to do that in general, for large T, as noted above. > So there are only so many values for which those explicit equations > can be correct which is why this technique is a key in the lock. > But you are grossly underestimating, for large T, how many combinations of factors you will need to examine. And, a key fact: the Chinese Remainder Theorem actually does not help you at all. In most cases it will eventually lead you to value that is larger than the largest factor of T and still is not itself a factor. Marcus. > This can be seen by directly substituting your formulae (1) and (2) k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and > this becomes k2 = T*p2^(-1) - r2*p2^(-1) mod p1, i.e. T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. Another tautology. That doesn't help you at all. If you have those equations and they are true then, yes, you get > equality on all sides. What you are doing by substituting is saying mathematically, these > equations are true. But what if they're not for a specific value of the f's and g's? Then you're wrong and the substitutions are invalid. James Harris === Subject: Re: JSH: Simple matching, factoring versus math politics posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20071204 Ubuntu/7.10 (gutsy) Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? > And I realized that then you could use > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and reduce to a solution for d_1. > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 (1) > and then you also need d_2, so > d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. (2) > Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. But that's all you need since there are only so many numbers that fit. For instance, with 119, it factors as 7(17) or trivially as 1(119), so > you only have 4 integers to fit, and for each you have f_1, as you > have f_1 = 1 mod 11, f_1 = 7 mod 11, f_1 = 17 mod 11 = 6 mod 11, or f_1 = > 119 mod 11 = 9 mod 11. This is not a good example. Here's why. You > chose p1 = 11. That is larger than sqrt(T). In > general you do not want to choose numbers that large. > If, for example, T ~ 10^100, then sqrt(T) ~ 10^50. > If p1 > 10^50, how many pairs (f1, f2) do you need > to consider? What you want is that f1*f2 = T mod p1. Of course, since p1 is a prime, for ANY choice of f1 > between 1 and p1 - 1, there exists f2 such that f1*f2 = T mod p1. Which means that the space of possible factors f1 > has size on the order of 10^50. Many lifetimes of > searching would be required. So this is not practical. > You must choose p1 much smaller than sqrt(T). For example, if T = 119, then let p1 = 2. Note that > T mod p1 = 1. This means that the only possible choice > for f1 is f1 = 1. Of course any odd number is > congruent to 1 mod 2, so the fact that f1 = 1 is > the only possible choice does not tell you much. > So you pick a different prime, say, p2 = 3. Note that > T mod p2 = 2. Therefore you can choose g1 = 1 or 2. > So you choose, perhaps, g1 = 2. So here is what you have: f1 mod p1 = 1 mod 2 = 1 g1 mod p2 = 1 mod 3 = 2. The Chinese Remainder Theorem then gives > you a value x such that x mod 2 = 1 and x mod 3 = 2. The least value x for which this is true is x = 5. This is not a factor of T, so you choose a > third prime less than sqrt(T), say, p3 = 5. > Note that T mod 5 = 119 mod 5 = 4. Thus the > possible choices for h1 are 1, 2, and 4. > Perhaps you choose h1 = 1. Then you have > 3 equations: f1 mod 2 = 1 g1 mod 3 = 2 h1 mod 5 = 1 Again using the Chinese Remainder Theorem, > you can find x such that x mod 2 = 1 > x mod 3 = 2 > x mod 5 = 1 The smallest value for x is x = 11. Again, > this is not a factor of T, and I have exhausted > the supply of primes less than sqrt(T). Of course you can say, You fool, you should > have chosen g1 = 1. You would then have found > with the first two primes that x = 7 would be > a solution. Instead you skipped over g1 = 1 > and went to g1 = 2, which did not result in a > factor. True, of course. But then maybe you begin to see > the problem. There is only one choice for f1. > There are two choices for g1. There are 4 choices > for h1, and all together, for all combinations, > there are 1 * 2 * 4 = 8 possible configurations. > Most of these don't work! This is important. > In general, for primes p1, p2, and p3, the > number of combinations of f1, g1, and h1 is (p1 - 1)(p2 - 1)(p3 - 1), This can get to be a large number very fast. > It grows somewhat like the factorial function. > Suppose you need to factor a really big T. > Perhaps you need to consider all the primes > less than 1000. There are 168 such primes. > The number of combinations which might work > is (p1 - 1)(p2 - 1) ... (p168 - 1). An ENORMOUS number. You cannot expect to > search all these possible combinations. True, > you would eventually find the correct factors. > But VERY inefficiently; far worse than random > trial divisors or GCDs. This shows very well what JSHs method amounts to and maybe more explicit than I have tried in a similar post many weeks ago. Essentially, it is trial division by a few small primes (say all primes <1000 as you propose) followed by finding all numbers Playing around with various approaches to the factoring problem I > noticed that if I did something as simple as consider T = 9 mod 11 and > T = 2 mod 13, I would find that the minimum positive number for T that > would work would be 119, which is a number I like to use in my > examples and it occurred to me that the primes were forcing something. > So I started thinking about what information prime numbers could give > about factors as if the two primes were forcing T to be 119 or > greater, then they were also forcing the factors to be certain values. > So I expanded out a factorization with primes: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered, what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = > T > mod p_2, so you'd know the f's and the g's? > And I realized that then you could use > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and reduce to a solution for d_1. > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 (1) > and then you also need d_2, so > d_2 = (f_2 - g_2)*p_2^{-1} mod p_1. (2) Note that these equations only give you the residues of the d's mod > p1, rather than exact values for the d's. But that's all you need since there are only so many numbers that fit. If you are searching for a factor g1 + d1*p2 of T which is less than sqrt(T) then knowing the value of d1 mod p1 narrows down the number of possible values of d1 to approximately sqrt(T)/(p1*p2). For T an RSA- sized composite this number is enormous. For instance, with 119, it factors as 7(17) or trivially as 1(119), so > you only have 4 integers to fit, and for each you have f_1, as you > have f_1 = 1 mod 11, f_1 = 7 mod 11, f_1 = 17 mod 11 = 6 mod 11, or f_1 = > 119 mod 11 = 9 mod 11. So those are the ONLY values for f_1 that will fit with integers. This is irrelevant; knowing that only certain values for f1 will fit is useless unless you have some efficient way of knowing *which* values, but you don't - more on this below. > So suddenly I had it!!! Information from the intersection of the > primes. > The two prime numbers were now telling me something about a key > variable in the expanded factorization!!! > And now you can go back to > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), I assume you mean to write m2 = floor(g1*g2/p2), otherwise the > following equation is wrong (since r2 has disappeared): Yup. > and you can get to > k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 (3) > so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since > k_2 = floor(T/p_2). > Now then, does this work? No. And here's why: as I mentioned above, you only know the values of > d1 mod p1 and d2 mod p1. You suggest substituting mod p1 into the > equation I have labelled (3). The trouble is that even if you have > incorrectly guessed f1 and g1 this equation will always hold mod p1. Nope, here's why. The relations (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) are explicit, as in exact and those are factors of T. Now mathematically there are only so many integers that can factor T, > correct? So there are only so many values for which those explicit equations > can be correct which is why this technique is a key in the lock. This can be seen by directly substituting your formulae (1) and (2) k2 = f1*f2*p2^(-1) - g1*g2*p2^(-1) + m2 mod p1 Now use the facts that f1*f2 = T mod p1 and g1*g2 = m2*p2 + r2 and > this becomes k2 = T*p2^(-1) - r2*p2^(-1) mod p1, i.e. T*p2^(-1) = (k2*p2 + r2)*p2^(-1) = T*p2^(-1) mod p1. Another tautology. That doesn't help you at all. If you have those equations and they are true then, yes, you get > equality on all sides. What you are doing by substituting is saying mathematically, these > equations are true. But what if they're not for a specific value of the f's and g's? Then you're wrong and the substitutions are invalid. Wrong. The substitutions I made did not assume that the f's and g's are equal mod pi to an actual factor of T. The only assumptions I made were your equations (1) and (2), which can be used to define d1 and d2 irrespective of whether you have the correct guesses for the f's and g's, and that f1*f2 = T mod p1 and g1*g2 = T mod p2; nowhere did I assume that you have the right f's and g's. And since the integers mod pi are a field, to any non-zero pair of residues f1 and g1 there exist f2 and g2 which satisfy f1*f2 = T mod p1 and g1*g2 = T mod p2. So checking whether equation (3) holds for a given choice of f1 and g1 will not tell you whether you guessed correctly, since it will always hold regardless. If you don't believe me and aren't up to the job of checking the algebra then try it with an example value for T and the wrong guesses for f1 and g1. Either that, or just start a load of new threads claiming I'm wrong without actually checking, like you did last weekend. Can you remember how that turned out? === Subject: JSH: Frustrated. posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Maybe I should say more after going on and on about an approach that I now accept was just useless for factoring, but I keep thinking about why I'm so desperate for something in the factoring area anyway, which is the blocking of my pure math research where math people don't follow their own rules. I NEED a practical math result because one dead journal shows how locked down math society has it now. Maybe we'll take each other down before all of this is over and I'll get my way of convincing the world that mathematicians routinely lie and your society will find a way to get me in return. And it is a sad testimony to the true reality of modern academia. It is a medieval system. And you end up in medieval crap with it, like what I call the Math Wars. Some of you seem to think the Math Wars are just fun and games or just some silly crackpot mouthing off, but it is about me finding a way to do things like end tenure, reduce funding for academics and convince the world that objective measures, rather than just letting academics say which of their buddies supposedly did something great, are necessary. So sit back. Think you have it all handled as I remain frustrated, and venting in futile anger at a feudal system of academia, because I need that practical mathematical result to break a broken system, but remember, it only takes one result for me to then go back and use every moment like this to emphasize to the public why it needs to do as much as necessary. Our modern world exists today because discovery was cherished. But parasites have turned things upside down for a few dollars and some empty accolades because they can't appreciate the value. If humanity loses then it loses down the line, and the extinction of our species whenever it occurs as we probably won't manage to get off this planet, may trace back to a tragic shift, when major problem solving was lost, and pretend took over. Call me crazy. But then nothing I do will work in the real world, right? But then you won't find funding drying up, and you won't find an increasingly skeptical public demanding more than just your say-so, right? Call me crazy, and if I am then the solutions I find to convince people that your society lies in the real world won't really be solutions, right? But I see the end of the Math Wars putting most of you in other areas of work outside of mathematics. But my vision is against your will to stop me. We'll see who wins down the line. James Harris === Subject: Re: JSH: Frustrated. > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. I NEED a practical math result because one dead journal shows how > locked down math society has it now. Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. And it is a sad testimony to the true reality of modern academia. It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. Our modern world exists today because discovery was cherished. But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. Call me crazy. But then nothing I do will work in the real world, > right? But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. We'll see who wins down the line. > James Harris I am deeply sorry to hear that your latest failure to solve the factoring problem has left you depressed and frustrated. As I indicated in the post nearly all of us could see this coming. You have been down this road before, many, many times, even this road always leads you to the sorrowful dead end. So what is the cause of your repeated failures? Once again, you attribute it to a powerful enemy, the math community. You are partially right: you are being held back by a powerful enemy. But your enemy is not the math community or academics. The enemy that has really been thwarting you at every turn, both in your math work and in other aspects of your life, is your Narcissism. The proof of this can be found in your latest threads. Granted, there were replies that consisted of nothing but insults. But you also got many replies with good, solid advice as to how to proceed. Had you followed the advice given in these posts, you would have discovered your errors sooner, and would be better equipped to try different approaches. Contrary to your perception, a lot of posters have been trying to help you. Your view of other mathematicians and posters has unfortunately been colored by your Narcissism. Because of your NPD, you end up battling with those who could be your allies, you shun suggestions that could help advance your ideas, and you inevitably end up alone and frustrated. You say We'll see who wins down the line. If history is any guide, you will surely be the loser, a big time loser. This is the course your life will take as long as you allow your NPD to run -- and ruin -- your life. James, it is time for you to take your life back. Seek therapy for your NPD. Taking this step will be tough, for your NPD will fight you on this one. But the rewards will be a much happier life. -- All things extant in this world, Gods of Heaven, gods of Earth, Let everything be as it should be; Thus shall it be! - Magical chant from Magical Shopping Arcade Abenobashi Drizzle, Drazzle, Drozzle, Drome, Time for this one to come home! - Mr. Wizard from Tooter Turtle === Subject: Re: JSH: Frustrated. >Maybe I should say more after going on and on about an approach that I >now accept was just useless for factoring, This is progress. Now delete the word Maybe, actually _say_ something about that, in particular say something about how all the people who lied about your result because they were too stupid to follow it or too evil to admit it was right were actually just stating the simple truth about the math, and we'll have even more progress. >but I keep thinking about >why I'm so desperate for something in the factoring area anyway, which >is the blocking of my pure math research where math people don't >follow their own rules. I NEED a practical math result because one dead journal shows how >locked down math society has it now. Maybe we'll take each other down before all of this is over and I'll >get my way of convincing the world that mathematicians routinely lie Maybe. HINT: This doesn't seem likely, since anyone who looks at the record can see literally hundreds of times when you've accused us of lying, when it turned out (surprise) we weren't. Really - if you want to convinve people that we're lying probably you should stop saying so many things that aren't true. Just a thought. >and your society will find a way to get me in return. And it is a sad testimony to the true reality of modern academia. It is a medieval system. And you end up in medieval crap with it, >like what I call the Math Wars. Some of you seem to think the Math Wars are just fun and games or just >some silly crackpot mouthing off, Some of us? Another hint: That applies to everyone reading these things, except maybe your mother, or other friends of yours who don't know any math. >but it is about me finding a way >to do things like end tenure, reduce funding for academics and Giggle. You don't see how funny this is: You say some of us may think you're just a crackpot, but... and then you say something that makes you sound like a raving lunatic. >convince the world that objective measures, rather than just letting >academics say which of their buddies supposedly did something great, >are necessary. So sit back. Think you have it all handled as I remain frustrated, >and venting in futile anger at a feudal system of academia, because I >need that practical mathematical result to break a broken system, but >remember, it only takes one result for me to then go back and use >every moment like this to emphasize to the public why it needs to do >as much as necessary. Our modern world exists today because discovery was cherished. But parasites have turned things upside down for a few dollars and >some empty accolades because they can't appreciate the value. If humanity loses then it loses down the line, and the extinction of >our species whenever it occurs as we probably won't manage to get off >this planet, may trace back to a tragic shift, when major problem >solving was lost, and pretend took over. Call me crazy. But then nothing I do will work in the real world, >right? But then you won't find funding drying up, and you won't find an >increasingly skeptical public demanding more than just your say-so, >right? Call me crazy, and if I am then the solutions I find to convince >people that your society lies in the real world won't really be >solutions, right? Uh, right. You haven't noticed that so far none of them _are_ really solutions? >But I see the end of the Math Wars putting most of you in other areas >of work outside of mathematics. But my vision is against your will to >stop me. Nobody has any will to stop you. When you say things about math that are not true people say that what you said is not true. Stop posting falsehoods about math and people will stop contradicting you. >We'll see who wins down the line. >James Harris David C. Ullrich === Subject: Re: JSH: Frustrated. >Some of you seem to think the Math Wars are just fun and games or just >some silly crackpot mouthing off, Some of us? Another hint: That applies to everyone reading these > things, except maybe your mother, or other friends of yours who > don't know any math. Ha! The Math Wars are a matter of utmost gravity. > Nobody has any will to stop you. When you say things about > math that are not true people say that what you said is not true. Well, they often have other things to say as well. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: JSH: Frustrated. 5.1),gzip(gfe),gzip(gfe) It's a shame you're frustrated. The reason you're frustrated is because your reality-testing skills are seriously lacking. This leads you to have goals which are completely unrealistic, and use strategies for achieving them which haven't got a hope of any degree of success. It's not totally out of the question that one day you might be able to produce some interesting mathematical research if you adopt a sensible approach and work really hard at it. But you'll have to let go of this idea that every time someone criticizes your efforts that's the mathematical establishment trying to hold you back. People who give you constructive feedback are your friends, not your enemies. I submitted my Ph.D. thesis last year and the examiner's reports were extremely harsh. They said my exposition was far too unclear and they doubted I'd proved what I said I'd proved. It's hardly going to get me anywhere to say they're just trying to block my research. If I want recognition from the mathematical community, I have to produce expositions of my arguments which will convince any qualified person. That's the standard we all have to meet. If I can't meet it, then mathematical research is not for me. Simple as that. This is true of you too and it's time you faced up to that reality. Plenty of people have told you this. You really should give some thought to the possibility that they might be right. You'd end up having a much happier life. === Subject: Re: JSH: Frustrated. posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring Maybe you shouldn't say more. Maybe you should *think* more about what happened (again). > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. Maybe you'll figure out that all these difficulties you perceive are due to your own faults, and not those of everyone else. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off Yes, that's exactly what it is. > but it is about me finding a way > to do things like end tenure, reduce funding for academics See, there it is! Some silly crackpot mouthing off. > So sit back. And enjoy the show. > Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. The show must go on! > Call me crazy. Crazy is a strong word. I'd prefer arrogant. Or deluded, insulting, ignorant, ineducable, or narcissistic. Especially narcissistic. > But then nothing I do will work in the real world, right? Nonsense. When you pay the fare, the bus takes you to work, right? That works. When you heat a microwave dinner in the oven, that works, right? But nothing you do will work in the real world and be *important. You're just ordinary, sorry. > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? What increasingly skeptical public? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? That's exactly right. You're catching on! > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. Is that what you see? I see you making an appointment with a mental health professional and beginning your recovery. > We'll see who wins down the line. The same people who won this time. -- You're right, I get it It all makes sense, you're the perfect person So right, so wrong Let's all live in your imaginary life -- Chevelle === Subject: Re: JSH: Frustrated. >Maybe I should say more after going on and on about an approach that I >now accept was just useless for factoring, but I keep thinking about >why I'm so desperate for something in the factoring area anyway, which >is the blocking of my pure math research where math people don't >follow their own rules. You are right to pick factoring as an area where it is easy to show the correctness of your results. If you say the two factors of 12827 are 101 and 127 then anyone can easily check that you are right. You could do better by actually factoring some numbers bigger than 119. By all means start with 119, but you need to show results for bigger integers. It is easy enough to generate random 40 or 50 bit semiprime numbers in Java. Factoring such numbers will let you test your methods and give you an idea of how fast they are. That will help you avoid making grandiose claims that are not in the end justified. Actually producing a working program will let you filter out any ideas that turn out to be non-starters. That will save you the embarassment of having other people do that for you. It will also tell you how fast you can factor, that will let you know if RSA (500 bits or more) is in any real danger. By checking things for yourself you can avoid having to retract quite so many times. I NEED a practical math result Again correct. I suggest that you have the practical math result actually available before you make any claims. I suspect that you work alone, so I can see that it is useful to bounce ideas of people here on usenet - they can catch typos and suggest improvements in the early stages of the development of one of your ideas. Try to stick to the this might be interesting level for a while until you have fully worked out the idea. You know by now that there are a great number of seemingly promising ideas that come to nothing. The chances are that any new idea will also come to nothing, so it is best to be cautious in the early stages. Try to curb your enthusiasm about your ideas. Your enthusiasm is understandable but it can lead you to react badly to valid criticism of your ideas and also to claim more than is warranted for them. Criticism of your ideas is not criticism of yourself. There is certainly direct criticism of you on usenet but you would do better to just ignore that. You should try to avoid making grandiose claims about any new idea until after you have actually coded it and got it working. One of the reasons you find it difficult to get people to listen to you is that too often you have made big claims, only to withdraw them a few weeks later. This is something you should try to avoid. >Maybe we'll take each other down before all of this is over and I'll >get my way of convincing the world that mathematicians routinely lie >and your society will find a way to get me in return. As has been pointed out, when mathematicians said that your latest idea was not as good as you wanted it to be, they were not lying. Better to tone down the antagonism, you will in time get less antagonism back. People do not like being called liars, especially whan they were correct in what they said. I hope that you think about this advice, however others have said similar things before, and you have been round this cycle many times before... rossum James Harris === Subject: Re: JSH: Frustrated. the correctness of your results. If you say the two factors of 12827 > are 101 and 127 then anyone can easily check that you are right. You could do better by actually factoring some numbers bigger than > 119. By all means start with 119, but you need to show results for > bigger integers. It is easy enough to generate random 40 or 50 bit > semiprime numbers in Java. Factoring such numbers will let you test > your methods and give you an idea of how fast they are. That will > help you avoid making grandiose claims that are not in the end > justified. Actually producing a working program will let you filter out any ideas > that turn out to be non-starters. That will save you the embarassment > of having other people do that for you. It will also tell you how > fast you can factor, that will let you know if RSA (500 bits or more) > is in any real danger. By checking things for yourself you can avoid > having to retract quite so many times. actually available before you make any claims. I suspect that you > work alone, so I can see that it is useful to bounce ideas of people > here on usenet - they can catch typos and suggest improvements in the > early stages of the development of one of your ideas. Try to stick to > the this might be interesting level for a while until you have fully > worked out the idea. You know by now that there are a great number of > seemingly promising ideas that come to nothing. The chances are that > any new idea will also come to nothing, so it is best to be cautious > in the early stages. Try to curb your enthusiasm about your ideas. > Your enthusiasm is understandable but it can lead you to react badly > to valid criticism of your ideas and also to claim more than is > warranted for them. Criticism of your ideas is not criticism of > yourself. There is certainly direct criticism of you on usenet but > you would do better to just ignore that. You should try to avoid making grandiose claims about any new idea > until after you have actually coded it and got it working. One of the > reasons you find it difficult to get people to listen to you is that > too often you have made big claims, only to withdraw them a few weeks > later. This is something you should try to avoid. idea was not as good as you wanted it to be, they were not lying. > Better to tone down the antagonism, you will in time get less > antagonism back. People do not like being called liars, especially > whan they were correct in what they said. I hope that you think about this advice, however others have said > similar things before, and you have been round this cycle many times > before... rossum > rossum, A simple yes or no question: Do you honestly think James is here in Usenet for anything other than finding satisfaction for his NPD? Fredrico Juarrez === Subject: Re: JSH: Frustrated. >rossum, A simple yes or no question: Do you honestly think James is here in Usenet for anything other than >finding satisfaction for his NPD? >Fredrico Juarrez I think that he is genuinely interested in maths, but for the most part it is his NPD talking. rossum === Subject: Re: JSH: Frustrated. > > rossum, > A simple yes or no question: > Do you honestly think James is here in Usenet for anything other than > finding satisfaction for his NPD? > Fredrico Juarrez > I think that he is genuinely interested in maths, but for the most > part it is his NPD talking. rossum > I appreciate the answer. I think most everyone here has a genuine interest in math. James knows this, and he uses it to gain access to a ready-made audience whereby he can find gratification for his NPD. It's just frustrating to see people trying to help him, when it's obvious he hasn't the slightest interest in finding an answer. All he wants is for people to respond to him. That's all he wants. Fredrico Juarrez === Subject: Re: JSH: Frustrated. posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless [.snip.] James Harris Can you please provide an example of anything you've done that isn't useless? As far as I know - EVERYTHING you have produced is useless! Coupled with your narcissism, this will lead you to drinking and drugs as it is a lost cause. Oh great one - have stock markets around the world rebounded after your acknowledment of yet another failure? Have we all been spared burning in hell? How many licks does it take for you to realize you are full of $hit? Well, I am 100% sure we will see another 12 years of rants, raves, accusations, fall of the empire, fall of humanity, fall of the planet and destruction of the galaxy using your new factor-truth-sqrt-prime- riemann-fermat-galois-ring - ray gun! === Subject: Re: JSH: Frustrated. >Can you please provide an example of anything you've done that isn't >useless? James' Prime Counting function works perfectly well. I have coded and tested it and it produces correct results. It is not of world shaking importance, but it is useful in its own small way. rossum === Subject: Re: Mechanics of Materials Solution manual posting-account=Xlph1QoAAADr9ktCfcKyOgWbmrqoAqHG .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > i have the solution of the textbook Mechanics of Materials > approach 4th edition by beer n jhonston Book: Mechanics of Materials > Author: Beer and Jhonston > Edition: 4th ( sixth 4 ) it is full and latest with all the solution worker out. > if you want contact me through mail me at nikhilpali...@yahoo.com and > i will send it to you BEWARE!! I purchaed this solutions manual from this seller and it is not the 4th edition. It is the 3rd edition. I contacted the seller and told him about the problem and he refused to refund my money. This seller is very untrustworthy. If you are looking for the 4th edition, look somewhere else beacuse you will be taken advantage of. === Subject: Re: complex numbers on a ti-89 anyone know how to do that? === Subject: Numerical Analysis 8th Edition - Faires/Burden posting-account=yw161QoAAADlj35bj3voOEcBw-D_xvsG Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Hi I'm looking for the Solutions manual for this book, anyone knows where i can find it? Numerical Analysis 8th Edition - Faires/Burden === Subject: Re: solutions.... loook hereeee.... posting-account=Xlph1QoAAADr9ktCfcKyOgWbmrqoAqHG .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > i have some solutions 1) Book: Engineering Mechanics - Statics And Dynamics > æ æ Edition: 11th ( it is the latest edition) > æ æ Author:Hibbeler 2) Thermodynamics: An engineering approach > æ æ Author: Cengel and Boles > æ æ Edition: 6th ( sixth 6 ) 3) Dynamics > æ æ 11th ed by hibbler 4) Fluid Mechanics Fundamentals and Application > æ æ 1st ed by Cenel and Cimbala 5) Physics > æ æ 8th ed by Halliday and resnick 6) Mechanics of Material > æ æ 4th ed by Beer and Jhonson 7) First course in diffrential equation > æ æ8st ed my Zill i may be having some more solution so please ask what you want, but > only engg n mathematics solution manual only. all the soluion are complete with worked solution of every questions > in the book. every solution is for 12$ through paypal, and 2 for 22$. > sorry the price is minimim n fixed. but you will get solution within > min of clearing of payment. > contact me at nikhilpali 88(at)yahoo.com , nikhilpali...@yahoo.com for > very fast reply. BEWARE!! I purchaed this solutions manual from this seller and it is not the 4th edition. It is the 3rd edition. I contacted the seller and told him about the problem and he refused to refund my money. This seller is very untrustworthy. If you are looking for the 4th edition, look somewhere else beacuse you will be taken advantage of. === Subject: Re: Proof Question forward proof that follows directly from the result and is obvious to an untrained reader too(someone like myself). I've put the following together. Does it seem valid? Let's assume that C(p,r) is an integer. PROOF: C(p,r) = p* [(p-1)(p-2)...(p-(r-1))]/r! First note that p is an integer since p is a prime number and that C(p,r) is an integer. Let k=[(p-1)(p-2)...(p-(r-1))]/r! k is a rational number since it consists of a quotient of products of natural numbers since 0 <6kJsj.95130$3m6.72163@fe2.news.blueyonder.co.uk>, Prove that C(n,m) is the number of distinct > m-element subsets of an > n-element set. Then it has to be an integer. this. It is the easiest way I know... You have n different ways of choosing the first > element of the subset; > then there are only n-1 elements left to choose the > second element; > n-2 for the third; ... ; and finally n-(m-1) ways > of choosing the mth > element. Since you are doing each of them in > series, and must do all > of them, that gives you n*(n-1)*...*(n-(m-1)) ways. But you've overcounted, because you will count each > m-element subset > as many ways as you have of listing them in order. > How many ways are > there to list an m-element subset? You should > divide your previous > answer by the number of times you repeatedly > counted each one. much simpler way of > doing this. I have worked out a couple of ways of > doing it but they all > amount to same thing. > -- It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes by Bill > Watterson) > Arturo Magidin > magidin-at-member-ams-org > Well, I guess having someone else prove it for you is easier! === Subject: Re: JSH: Factoring IS stupid simple posting-account=no4jRwoAAADHjiy7leKH93A_GaYruAJ3 Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > Ok, sorry, as I hate it to some extent when a problem turns out to be > harder to solve than I thought when some part of me must kind of know > the answer but it takes a while for the rest of me to get it. > Here's the correct answer to the factoring problem where I just kind > of diverged a bit before. > I was puzzling about the latest failed factoring idea wondering where > I went wrong, as I really felt that there was information wrapped up > in the intersection of the residues modulo T of two primes. > So I pondered some more. > Before too long I was writing out things along these lines: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = T > mod p_2, so you'd know the f's and the g's, and I realized that then > you could just solve for the c's by using > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and I realized you could solve for everything and reduce to a solution > for d_1. > That's from before, but after that point, before, I went in the wrong > direction as it is so OBVIOUS what you should do next, which is solve > d_1 modulo p_1 (maybe I was sleepy when I first typed that last). > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 > and then you also need d_2, so > d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 > and now you can substitute modulo p_1 into, oh, forgot something: > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to > k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 > so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since > k_2 = floor(T/p_2). > If you guessed correctly then you will match with k_2's residue modulo > p_1, but if you didn't, you shouldn't match unless this doesn't work > either!!! > So if you don't match you try another possible and try until you > match, and do that for a series of primes where the total is less than > m, such that T/m! < 1, or you can get an exact number by multiplying > primes together--assuming you start with the smallest and go up--until > you exceed T, and counting how many you have. > That is the solution to the factoring problem. > It's stupid simple. > I just took a little time to figure out exactly how the damn thing > worked. > But that's problem solving for you. Some part of me must have known > the answer but the rest of me was stupid, until simple won. > Why don't you just guess p1 such that T/p1 is integer? > Funny. Ok, yes, I admit it. I went on and on about having solved the > factoring problem through an intersection of primes when I hadn't > shown it. > So I had a strong feeling that there had to be a way to do it, and > then got waylaid in figuring out exactly how, but this latest result > is actually consistent with the original idea while what I had before > was not. > Here you DO test, and the guess just gets you SOME information, so you > need a series of primes. > The approach IS straightforward and it is stupid, simple, so get mad > and upset about the problem solving process but that just shows you > have no clue how it really works. > Mostly it's about hard work, lots of misses and being willing to keep > beating on that hunch until it pays off, and later let the freaking > historians re-write history and talk about what a genius you are. > It's a big mess until you're right. > During the process most of the time you feel like a freaking idiot. But the factoring problem has already been solved for at least 2300 > years. In fact, the method that I presented in one line does solve it. > The current issue is finding an efficient algorithm, since all known > algorithms are not efficient enough to be practically usable on large > numbers. > Instead of deriving new equations whose solutions are the prime > factors of a given number, which effectively amounts to rediscovering > the wheel, you should try to devise an efficient algorithm for > finding those solutions. If the approach holds i.e. if the values for d_1 mod p_1 and d_2 mod > p_1 are provisional values that work only when you guess correctly, > then this approach DOES lead to an efficient algorithm. It turns out that it is well-known that if you have f_1 mod p_1, and > f_1 mod p_2, ... f_1 mod p_n, where n is a sufficient number of primes > then you have f_1 explicitly. The problem though has been, how do you find f_1 modulo successive > primes? I contemplated this issue and hypothesized that when you have two > primes you have some kind of intersection that will give the answer, > and I went looking for that intersection. My first approaches went awry as I tried to actually solve for d_1 > exactly, but thinking about why those approaches didn't work, I looked > at the equations again and saw: d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and I had everything I needed where BOTH primes gave input, so I had > my intersection. So this approach IS a valuable one as long as the hypothesis holds > true and the value of d_1 modulo p_1 found above is invalid unless > you've guessed f_1 and g_1 correctly, as those are residues, where f_1*f_2 = T mod p_1 and g_1*g_2 = T mod p_2 and are NOT the full solutions. Maybe I could have picked other > variable letters. In any event, the point is an intersection between two primes with a > guess at residues and checking to see if the guess is right. If you can do that then factoring is trivial as you need less primes > than m, where T/m! < 1. So you have factorial working for you here, and that is a big over- > estimate for m. If this approach holds water then factoring an RSA public key of any > size possible on modern desktop computers would be trivially done in > minutes if not seconds. The factoring problem is solved by using the intersection of prime > numbers, in an elegant, precise, and remarkably short solution. James Harris In order for your method to be efficient, you need to prove that the number of possibilities for each guess is polynomial in the number of bits of T. I don't see such proof. === Subject: Re: JSH: Factoring IS stupid simple posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Ok, sorry, as I hate it to some extent when a problem turns out to be > harder to solve than I thought when some part of me must kind of know > the answer but it takes a while for the rest of me to get it. Here's the correct answer to the factoring problem where I just kind > of diverged a bit before. I was puzzling about the latest failed factoring idea wondering where > I went wrong, as I really felt that there was information wrapped up > in the intersection of the residues modulo T of two primes. So I pondered some more. Before too long I was writing out things along these lines: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again > pondered what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = T > mod p_2, so you'd know the f's and the g's, and I realized that then > you could just solve for the c's by using (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and I realized you could solve for everything and reduce to a solution > for d_1. That's from before, but after that point, before, I went in the wrong > direction as it is so OBVIOUS what you should do next, which is solve > d_1 modulo p_1 (maybe I was sleepy when I first typed that last). d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 and now you can substitute modulo p_1 into, oh, forgot something: (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 Now you have to take that modulo p_1*p_2, which is the crucial last step that I just figured out. so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since modulo p_1*p_2 k_2 = floor(T/p_2). If you guessed correctly then you will match with k_2's residue modulo > p_1, but if you didn't, you shouldn't match unless this doesn't work > either!!! That's correct, modulo p_1*p_2. Easy ultimate answer but still kind of subtle problem where I got lost on the intersection of two primes part, as I kept looking modulo a single prime at a time, which doesn't work. Problem solving is hard. Especially when you have mean people taunting you. No wonder so many mathematicians cheat! Doing the real thing is just so damn hard. Factoring problem is easy, but annoying as you have to get all the little details right, and I did it with obnoxious twerps trying to aggravate me the whole way. Now that is what I call a monster challenge! And I surmounted it. Guess no other human being in history can come close to an accomplishment like it, or would even want to--having nasty people being mean to you kind of takes away some of the fun, oh, along with the whole worrying about crashing the world economy thing... Come to think of it, how the hell did I do it? ___JSH === Subject: Re: JSH: Factoring IS stupid simple posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Ok, sorry, as I hate it to some extent when a problem turns out to be > harder to solve than I thought when some part of me must kind of know > the answer but it takes a while for the rest of me to get it. Here's the correct answer to the factoring problem where I just kind > of diverged a bit before. I was puzzling about the latest failed factoring idea wondering where > I went wrong, as I really felt that there was information wrapped up > in the intersection of the residues modulo T of two primes. So I pondered some more. Before too long I was writing out things along these lines: (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) and (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) And I multiplied out and solved for k_1 and k_2 respectively and again > pondered what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = T > mod p_2, so you'd know the f's and the g's, and I realized that then > you could just solve for the c's by using (f_1 + c_1*p_1) = (g_1 + d_1*p_2) and (f_2 + c_2*p_1) = (g_2 + d_2*p_2) and I realized you could solve for everything and reduce to a solution > for d_1. That's from before, but after that point, before, I went in the wrong > direction as it is so OBVIOUS what you should do next, which is solve > d_1 modulo p_1 (maybe I was sleepy when I first typed that last). d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 and then you also need d_2, so d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 and now you can substitute modulo p_1 into, oh, forgot something: (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 Now you have to take that modulo p_1*p_2, which is the crucial last > step that I just figured out. so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since modulo p_1*p_2 k_2 = floor(T/p_2). If you guessed correctly then you will match with k_2's residue modulo > p_1, but if you didn't, you shouldn't match unless this doesn't work > either!!! That's correct, modulo p_1*p_2. > Wrong. The idea can't lead to a factoring method. This approach fails. Can't be revived either as that's an absolute. James Harris === Subject: Re: JSH: Factoring IS stupid simple posting-account=mgs1FwoAAABD3j5T_RLZ06yrgt2dghDu Gecko/20050915,gzip(gfe),gzip(gfe) > Wrong. The idea can't lead to a factoring method. This approach fails. Don't sweat it, James. Everybody makes mistakes. The main thing to remember is that everyone who said you were mistaken for weeks and weeks was a LIAR or a FOOL! Another important thing to remember is that it's crucial to post and brag about every hare-brained idea you come up with while on the toilet. And never, ever test your ideas before declaring the end of the world. You can learn from your mistakes! -- I just thought up that idea while sitting on the toilet and I think it's a fairly good one. -- James Harris === Subject: Re: JSH: Factoring IS stupid simple posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Ok, sorry, as I hate it to some extent when a problem turns out to be > harder to solve than I thought when some part of me must kind of know > the answer but it takes a while for the rest of me to get it. > Here's the correct answer to the factoring problem where I just kind > of diverged a bit before. > I was puzzling about the latest failed factoring idea wondering where > I went wrong, as I really felt that there was information wrapped up > in the intersection of the residues modulo T of two primes. > So I pondered some more. > Before too long I was writing out things along these lines: > (f_1 + c_1*p_1)(f_2 + c_2*p_1) = T = (r_1 + k_1*p_1) > and > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > And I multiplied out and solved for k_1 and k_2 respectively and again > pondered what if you GUESSED using f_1*f_2 = T mod p_1 and g_1*g_2 = T > mod p_2, so you'd know the f's and the g's, and I realized that then > you could just solve for the c's by using > (f_1 + c_1*p_1) = (g_1 + d_1*p_2) > and > (f_2 + c_2*p_1) = (g_2 + d_2*p_2) > and I realized you could solve for everything and reduce to a solution > for d_1. > That's from before, but after that point, before, I went in the wrong > direction as it is so OBVIOUS what you should do next, which is solve > d_1 modulo p_1 (maybe I was sleepy when I first typed that last). > d_1 = (f_1 - g_1)*p_2^{-1} mod p_1 > and then you also need d_2, so > d_2 = (f_1 - g_2)*p_2^{-1} mod p_1 > and now you can substitute modulo p_1 into, oh, forgot something: > (g_1 + d_1*p_2)(g_2 + d_2*p_2) = T = (r_2 + k_2*p_2) > multiply out, and divide by p_2, using m_2 = (g_1*g_2/p_2), and you > can get to > k_2 = d_1*g_2 + d_2*g_1 + d_1*d_2*p_2 + m_2 Now you have to take that modulo p_1*p_2, which is the crucial last > step that I just figured out. > so now you just substitute modulo p_1 and compare what you get on the > right side with k_2 mod p_1 as you know what k_2 is, since modulo p_1*p_2 > k_2 = floor(T/p_2). > If you guessed correctly then you will match with k_2's residue modulo > p_1, but if you didn't, you shouldn't match unless this doesn't work > either!!! That's correct, modulo p_1*p_2. Wrong. The idea can't lead to a factoring method. This approach fails. Can't be revived either as that's an absolute. James Harris Nevertheless, you will be back within a day or two, or an hour or two, with a new variant, claiming it is easy algebra or stupid simple or some other condescending characterization, claiming that mathematicians who criticize your work are incompetent narrow-minded evil liars who are endangering the human race, etc., etc. And of course making claims without anything but overly simplistic numerical examples and without anything resembling a complete rigorous proof. I'm getting tired of this. One day you are puffing yourself up to be the super-genius and pissing on everyone else for daring to question your humanity-saving work; the next, you are admitting what you did was crap but NOT acknowledging that, far from being liars and morons, your critics were actually right and you yourself were wrong. Stop calling us liars and morons. The shoe has been found to be on the other foot, over and over and over again. Marcus. === Subject: Re: JSH: Factoring IS stupid simple posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) Wrong. æThe idea can't lead to a factoring method. This approach fails. Can't be revived either as that's an absolute. James Harris- Are you going to retract all of the bad things you said about mathemticians, the destruction of humanity and that our souls would rot in hell? You should really be more careful about your insults - what the hell could I be thinking. Narcissist - see you are wrong again. You have always been wrong. You will always be wrong. Having said that - you will be chest pounding with yet another incarnation of mathurbating/fuctoring becuase your delusions of grandeur can't stop you! You have been at this for 12+ years man - 12 years of your life .... WASTED! Go have a drink, smoke some crack, take some ludes and then mix in pain killers with prozac to relive the obvious pain you must be feeling! Have a nice day! ~A === Subject: Re: JSH: Factoring IS stupid simple posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) Ok, sorry, as I hate it to some extent when a problem turns out to be > harder to solve than I thought when some part of me must kind of know > the answer but it takes a while for the rest of me to get it. Here's the correct answer to the factoring problem where I just kind > of diverged a bit before. I was puzzling about the latest failed factoring idea wondering where > I went wrong, as I really felt that there was information wrapped up > in the intersection of the residues modulo T of two primes. So I pondered some more. Before too long I was writing out things along these lines: (f 1 + c 1*p 1)(f 2 + c 2*p 1) = T = (r 1 + k 1*p 1) and (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) And I multiplied out and solved for k 1 and k 2 respectively and again > pondered what if you GUESSED using f 1*f 2 = T mod p 1 and g 1*g 2 = T > mod p 2, so you'd know the f's and the g's, and I realized that then > you could just solve for the c's by using (f 1 + c 1*p 1) = (g 1 + d 1*p 2) and (f 2 + c 2*p 1) = (g 2 + d 2*p 2) and I realized you could solve for everything and reduce to a solution > for d 1. That's from before, but after that point, before, I went in the wrong > direction as it is so OBVIOUS what you should do next, which is solve > d 1 modulo p 1 (maybe I was sleepy when I first typed that last). d 1 = (f 1 - g 1)*p 2^{-1} mod p 1 and then you also need d 2, so d 2 = (f 1 - g 2)*p 2^{-1} mod p 1 and now you can substitute modulo p 1 into, oh, forgot something: (g 1 + d 1*p 2)(g 2 + d 2*p 2) = T = (r 2 + k 2*p 2) multiply out, and divide by p 2, using m 2 = (g 1*g 2/p 2), and you > can get to k 2 = d 1*g 2 + d 2*g 1 + d 1*d 2*p 2 + m 2 Now you have to take that modulo p 1*p 2, which is the crucial last > step that I just figured out. so now you just substitute modulo p 1 and compare what you get on the > right side with k 2 mod p 1 as you know what k 2 is, since modulo p 1*p 2 k 2 = floor(T/p 2). If you guessed correctly then you will match with k 2's residue modulo > p 1, but if you didn't, you shouldn't match unless this doesn't work > either!!! That's correct, modulo p 1*p 2. Easy ultimate answer but still kind of subtle problem where I got lost > on the intersection of two primes part, as I kept looking modulo a > single prime at a time, which doesn't work. Problem solving is hard. æEspecially when you have mean people > taunting you. No wonder so many mathematicians cheat! Doing the real thing is just so damn hard. Factoring problem is easy, but annoying as you have to get all the > little details right, and I did it with obnoxious twerps trying to > aggravate me the whole way. Now that is what I call a monster challenge! And I surmounted it. Guess no other human being in history can come close to an > accomplishment like it, or would even want to--having nasty people > being mean to you kind of takes away some of the fun, oh, along with > the whole worrying about crashing the world economy thing... Come to think of it, how the hell did I do it? JSH- Hide quoted text - - Show quoted text - Moron! Still wrong - narcissist! === Subject: solution manual posting-account=1LXcdwoAAAAOMsiNKAVJ2p8wxrEhk12V Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. MODERN CONTROL SYSTEM 4th Edition by OGATA Modern Control System 11th edition by Dorf and Bishop (11e) Fundamentals of Microelectronics. Author: Behzad Razavi Microelectronic Circuits. Author: Adel S. Sedra, Kenneth C. Smith Control Systems Engineering. Author: Norman S. Nise Probability and Random Processes for Electrical Engineering (3rd Edition)Chapters 1-6. Author: Albert Leon-Garcia Communication Systems Engineering (2nd Ed.). Author:Proakis and Salehi. ISBN 0130619746 Digital Signal Processing System Analysis and Design. Author: Paulo S.R. Diniz, Eduardo and Sergio Netto CMOS VLSI Design: A Circuits and Systems Perspective. Author: Neil Weste, David Harris. Fundamentals of Applied Electromagnetics (5th Ed., Fawwaz T. Ulaby) Signals and Systems (2nd edition) Author: Oppenheim and Willsky Principles and Applications of Electrical Engineering by Giorgio Rizzon Engineering Mechanics - Dynamics Anthony M. Bedford and Wallace Fowler 4th Edition Electric Machinery and Power System Fundamentals Stephen J. Chapman 1st Edition Fundamentals of Aircraft Structural Analysis Howard D. Curtis 1st Ed Electrical Engineering: Principles and Applications Allan R. Hambley 3rd Edition Fundamentals of Engineering Thermodynamics Michael J. Moran and Howard N. Shapiro 6th Edition Fluid Mechanics Yunus A. Cengel and John M. Cimbala 1st Edition Engineering Mechanics - Statics and Dynamics Russell C Hibbeler 11th Edition Process Dynamics and Control. Author: Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp Analytical Mechanics, 7th ed. Author: Fowles and Cassiday Electric Circuits. (6th edition). Author: Dorf and Svoboda Semiconductor Device Fundamentals , Robert Pierret Fundamentals of Electrical Engineering Bobrow 2nd ed. Signal Processing First-Mclellan, Schafer & Yoder Solution Manual Principles and Applications of Electrical Engineering:G. Rizzoni Electric Machinery and Power System Fundamentals Analysis Design of Analog IC Design Microwave And Rf Design Of Wireless Systems-Solution Digital Signal Processing--Principles, Algorithms & Applications Communication Systems 4th Edition FTP Link Design of Analog CMOS Integrated Circuit Digital signal Processing-Mitra FTP Link Introduction to Electrodynamics - Solution Manual By David J. Griffiths Signal Processing and Linear Systems - Solution Manual By B.P. Lathi Solutions Manual Signals and Systems 2nd Ed. Fundamentals of Semiconductor Devices Anderson Electric Circuits 8th edition Nilsson Riedel Engineering Circuit Analysis 7ed by Hayt TITAN- Solution Manual-Linear Circuit Analysis 2nd Edition by R.A. DeCarlo and P. Lin Classical Electrodynamics - 2nd Ed. John David Jackson Microwave Engineering 3E - David M Pozar RF Circuit Design Theory and Application by Ludwig bretchko Solution Manual-Electrical Engineering: Principles and Applications 3rd ed Fundamentals of Engineering Electromagnetics--Cheng Engineering Electromagnetics by Hayt 2001 Fundamentals of Electric Circuits 2nd Edition by Alexander Sadiku Microelectronic Circuits By Adel Sedra 5th Edition Introduction to Electric Circuits 6th Edition by Dorf, Svaboda Control Systems Engineering By Nise Electric Circuits 7th edition - Nilsson Selected Answers-Basic Engineering Circuit Analysis-7th Ed. by J. David Irwin Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Design with Operational Amplifiers & Analog Integrated Circuits Discrete-Time Signal Processing, 2nd ed-A.V. Oppenheim, R.W.Schafer & J. R. Buck Digital Communications 4th Edition Solution Manual Cheng - Field and Wave Electromagnetics 2nd edition Microwave and RF Design of Wireless Systems by David M. Pozar Introduction to Electrodynamics (3rd Edition) By David J. Griffiths classical electrodynamics - 2nd ed. john david jackson Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang Digital and Analog Communication Systems 7th Ed, Leon W. Couch - Solution Manual Device Electronics for Integrated Circuits - Muller, Kamins Communication Systems 4th edition - Simon Haykin Signals and Systems 2nd Edition Oppenheim, Willsky and Nawab Digital Signal Processing - Proakis & Manolakis Solution Manual Signal Processing and Linear Systems by B.P. Lathi Electric Machinery Fundamentals - 4th edition by Chapman Introduction to VLSI Circuits and Systems by John P. Uyemura DSP First: A Multimedia Approach-Mclellan, Schafer & Yoder Engineering Electromagnetics by William Hayt and Buck Electronic Devices-6th Edition by Thomas L. Floyd Electronic Devices-Electron Flow Version-4th Edition by Floyd Automatic Control Systems-Kuo and Golnaragh Solution Manual Modern Digital and Analog Communication Systems by Lathi 3rd Ed Feedback Control of Dynamic Systems Mechanical Design of Machine Elements and Machines Thermodynamics an Engineering Approach 6th edition Mechanics Of Materials Solution Manual (3Rd Ed , By Beer, Johnston, & Dewolf) vector mechanics for engineers and scientist by beer and johnstone for Fluids Mechanics 10th edition by finnemore Mechanics of Materials 6th Ed. by Riley, Sturges, and Morris Fundamentals of Thermal - Fluid Sciences 3rd Edition Engineering mechanics: Dynamics - 4th and 5th Ed. by Bedford and Fowler Heat Transfer-Fundamentals of Heat and Mass Transfer-Incropera & Dewitt Introduction to Fluid Mechanics 6th edition by Fox, McDonald and Pritchards Heat Transfer: A Practical Approach Solution Manual Cengel A. HAYLER 's Probability and Statistics for Engineers and Scientists Manual Nanoengineering of Structural, Functional and Smart Materials If your interested do let me know at xplicitly666@yahoo.com Explicit Group === Subject: solution manual posting-account=1LXcdwoAAAAOMsiNKAVJ2p8wxrEhk12V Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. I also have other manuals Feel free to contact me. MODERN CONTROL SYSTEM 4th Edition by OGATA Modern Control System 11th edition by Dorf and Bishop (11e) Fundamentals of Microelectronics. Author: Behzad Razavi Microelectronic Circuits. Author: Adel S. Sedra, Kenneth C. Smith Control Systems Engineering. Author: Norman S. Nise Probability and Random Processes for Electrical Engineering (3rd Edition)Chapters 1-6. Author: Albert Leon-Garcia Communication Systems Engineering (2nd Ed.). Author:Proakis and Salehi. ISBN 0130619746 Digital Signal Processing System Analysis and Design. Author: Paulo S.R. Diniz, Eduardo and Sergio Netto CMOS VLSI Design: A Circuits and Systems Perspective. Author: Neil Weste, David Harris. Fundamentals of Applied Electromagnetics (5th Ed., Fawwaz T. Ulaby) Signals and Systems (2nd edition) Author: Oppenheim and Willsky Principles and Applications of Electrical Engineering by Giorgio Rizzon Engineering Mechanics - Dynamics Anthony M. Bedford and Wallace Fowler 4th Edition Electric Machinery and Power System Fundamentals Stephen J. Chapman 1st Edition Fundamentals of Aircraft Structural Analysis Howard D. Curtis 1st Ed Electrical Engineering: Principles and Applications Allan R. Hambley 3rd Edition Fundamentals of Engineering Thermodynamics Michael J. Moran and Howard N. Shapiro 6th Edition Fluid Mechanics Yunus A. Cengel and John M. Cimbala 1st Edition Engineering Mechanics - Statics and Dynamics Russell C Hibbeler 11th Edition Process Dynamics and Control. Author: Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp Analytical Mechanics, 7th ed. Author: Fowles and Cassiday Electric Circuits. (6th edition). Author: Dorf and Svoboda Semiconductor Device Fundamentals , Robert Pierret Fundamentals of Electrical Engineering Bobrow 2nd ed. Signal Processing First-Mclellan, Schafer & Yoder Solution Manual Principles and Applications of Electrical Engineering:G. Rizzoni Electric Machinery and Power System Fundamentals Analysis Design of Analog IC Design Microwave And Rf Design Of Wireless Systems-Solution Digital Signal Processing--Principles, Algorithms & Applications Communication Systems 4th Edition FTP Link Design of Analog CMOS Integrated Circuit Digital signal Processing-Mitra FTP Link Introduction to Electrodynamics - Solution Manual By David J. Griffiths Signal Processing and Linear Systems - Solution Manual By B.P. Lathi Solutions Manual Signals and Systems 2nd Ed. Fundamentals of Semiconductor Devices Anderson Electric Circuits 8th edition Nilsson Riedel Engineering Circuit Analysis 7ed by Hayt TITAN- Solution Manual-Linear Circuit Analysis 2nd Edition by R.A. DeCarlo and P. Lin Classical Electrodynamics - 2nd Ed. John David Jackson Microwave Engineering 3E - David M Pozar RF Circuit Design Theory and Application by Ludwig bretchko Solution Manual-Electrical Engineering: Principles and Applications 3rd ed Fundamentals of Engineering Electromagnetics--Cheng Engineering Electromagnetics by Hayt 2001 Fundamentals of Electric Circuits 2nd Edition by Alexander Sadiku Microelectronic Circuits By Adel Sedra 5th Edition Introduction to Electric Circuits 6th Edition by Dorf, Svaboda Control Systems Engineering By Nise Electric Circuits 7th edition - Nilsson Selected Answers-Basic Engineering Circuit Analysis-7th Ed. by J. David Irwin Solution Manual-Microelectronics-Digital & Analog Circuits & Systems by Millman Electric Machinery 6Ed Fitzgerald, Kingsley, Uman Solution Manual Problems and Solutions on Electromagnetism (Major American Universities Ph.D. Engineering circuit analysis Solution Manual - William Hart Hayt Solution Manual-Design with Operational Amplifiers & Analog Integrated Circuits Discrete-Time Signal Processing, 2nd ed-A.V. Oppenheim, R.W.Schafer & J. R. Buck Digital Communications 4th Edition Solution Manual Cheng - Field and Wave Electromagnetics 2nd edition Microwave and RF Design of Wireless Systems by David M. Pozar Introduction to Electrodynamics (3rd Edition) By David J. Griffiths classical electrodynamics - 2nd ed. john david jackson Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang Digital and Analog Communication Systems 7th Ed, Leon W. Couch - Solution Manual Device Electronics for Integrated Circuits - Muller, Kamins Communication Systems 4th edition - Simon Haykin Signals and Systems 2nd Edition Oppenheim, Willsky and Nawab Digital Signal Processing - Proakis & Manolakis Solution Manual Signal Processing and Linear Systems by B.P. Lathi Electric Machinery Fundamentals - 4th edition by Chapman Introduction to VLSI Circuits and Systems by John P. Uyemura DSP First: A Multimedia Approach-Mclellan, Schafer & Yoder Engineering Electromagnetics by William Hayt and Buck Electronic Devices-6th Edition by Thomas L. Floyd Electronic Devices-Electron Flow Version-4th Edition by Floyd Automatic Control Systems-Kuo and Golnaragh Solution Manual Modern Digital and Analog Communication Systems by Lathi 3rd Ed Feedback Control of Dynamic Systems Mechanical Design of Machine Elements and Machines Thermodynamics an Engineering Approach 6th edition Mechanics Of Materials Solution Manual (3Rd Ed , By Beer, Johnston, & Dewolf) vector mechanics for engineers and scientist by beer and johnstone for Fluids Mechanics 10th edition by finnemore Mechanics of Materials 6th Ed. by Riley, Sturges, and Morris Fundamentals of Thermal - Fluid Sciences 3rd Edition Engineering mechanics: Dynamics - 4th and 5th Ed. by Bedford and Fowler Heat Transfer-Fundamentals of Heat and Mass Transfer-Incropera & Dewitt Introduction to Fluid Mechanics 6th edition by Fox, McDonald and Pritchards Heat Transfer: A Practical Approach Solution Manual Cengel A. HAYLER 's Probability and Statistics for Engineers and Scientists Manual Nanoengineering of Structural, Functional and Smart Materials If your interested do let me know at xplicitly666@yahoo.com Explicit Group === Subject: solution manual posting-account=x3Z6RgoAAACmYmbsf5fyQV3nm9mhYIq0 Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. I also have other manuals Feel free to contact me at xplicitly666@yahoo.com MODERN CONTROL SYSTEM 4th Edition by OGATA Modern Control System 11th edition by Dorf and Bishop (11e) Fundamentals of Microelectronics. Author: Behzad Razavi Microelectronic Circuits. Author: Adel S. Sedra, Kenneth C. Smith Control Systems Engineering. Author: Norman S. Nise Probability and Random Processes for Electrical Engineering (3rd Edition)Chapters 1-6. Author: Albert Leon-Garcia Communication Systems Engineering (2nd Ed.). Author:Proakis and Salehi. ISBN 0130619746 Digital Signal Processing System Analysis and Design. Author: Paulo S.R. Diniz, Eduardo and Sergio Netto CMOS VLSI Design: A Circuits and Systems Perspective. Author: Neil Weste, David Harris. Fundamentals of Applied Electromagnetics (5th Ed., Fawwaz T. Ulaby) Signals and Systems (2nd edition) Author: Oppenheim and Willsky Principles and Applications of Electrical Engineering by Giorgio Rizzon Engineering Mechanics - Dynamics Anthony M. Bedford and Wallace Fowler 4th Edition Electric Machinery and Power System Fundamentals Stephen J. Chapman 1st Edition Fundamentals of Aircraft Structural Analysis Howard D. Curtis 1st Ed Electrical Engineering: Principles and Applications Allan R. Hambley 3rd Edition Fundamentals of Engineering Thermodynamics Michael J. Moran and Howard N. Shapiro 6th Edition Fluid Mechanics Yunus A. Cengel and John M. Cimbala 1st Edition Engineering Mechanics - Statics and Dynamics Russell C Hibbeler 11th Edition Process Dynamics and Control. Author: Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp Analytical Mechanics, 7th ed. Author: Fowles and Cassiday Electric Circuits. (6th edition). Author: Dorf and Svoboda Semiconductor Device Fundamentals , Robert Pierret Fundamentals of Electrical Engineering Bobrow 2nd ed. Signal Processing First-Mclellan, Schafer & Yoder Solution Manual Principles and Applications of Electrical Engineering:G. Rizzoni Electric Machinery and Power System Fundamentals Analysis Design of Analog IC Design Microwave And Rf Design Of Wireless Systems-Solution Digital Signal Processing--Principles, Algorithms & Applications Communication Systems 4th Edition FTP Link Design of Analog CMOS Integrated Circuit Digital signal Processing-Mitra FTP Link Introduction to Electrodynamics - Solution Manual By David J. Griffiths Signal Processing and Linear Systems - Solution Manual By B.P. Lathi Solutions Manual Signals and Systems 2nd Ed. 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HAYLER 's Probability and Statistics for Engineers and Scientists Manual Nanoengineering of Structural, Functional and Smart Materials If your interested do let me know at xplicitly666@yahoo.com Explicit Group === Subject: Re: Sum of cumulative distribution functions <10350494.1202481112206.JavaMail.jakarta@nitrogen.mathforum.org>, > Hi > If you have a random variable X with cumulative distribution functinon > F(x,y)=G(x)+H(y) where G and H are two cumulative distributions functions. > How do you find mathematical expectation of X? .... You may have better luck with this question in the news group, or else perhaps . Ken Pledger. === Subject: Re: JSH: Frustrated. <1-qdneJV_4b5fivanZ2dnUVZ_sSlnZ2d@rcn.net> posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. I NEED a practical math result because one dead journal shows how > locked down math society has it now. Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. And it is a sad testimony to the true reality of modern academia. It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. Our modern world exists today because discovery was cherished. But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. Call me crazy. But then nothing I do will work in the real world, > right? But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. We'll see who wins down the line. James Harris I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. > You may be projecting. I would guess that Harris takes all this much less seriously than you expect, perhaps less seriously even than a number of other posters here. I think he is deliberately fantasizing, putting himself on, dreaming of how he would like things to be. He has indeed been down this road many times, and he is getting used to it. It disrupts the fantasy for a while, but then he comes up with another hare- brained idea - doesn't test it, because he knows it is likely wrong - doesn't spell out a complete proof because he knows he cannot - and the fantasy life takes over again. And he knows how to push a lot of people's buttons with his rapturous descriptions of the perfidy of mathematicians, world economic collapse, humanity escaping to the planet Venus, etc., because no one will recognize his genius. He knows he's not a genius and he knows how annoying it is for him to claim it with not a shred of evidence. He is basically a crank but he has the most masterful troll skills I have ever seen. > So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. > Oh, come on. NPD like a number of other pop-psych diagnoses is not even well-defined. It's an artificial construct, far more a product of human imagination than, say, sqrt(-1). Harris has an unusually rich array of character defects but they do not constitute any illness. They are HIS FAULT. Unlike a genuinely sick person he is responsible for what he does. Labelling him as NPD or a sociopath or whatever sheds no light on anything. Here you are saying he has a disease but at the same time you are saying he is to blame for it. It's like blaming him for having a heart defect. He acts like an arrogant spiteful mean-spirited prick because he IS an arrogant spiteful mean-spirited prick, and he is fully to blame for that condition. > The proof of this can be found in your latest threads. Granted, there > were replies that consisted of nothing but insults. But you also got > many replies with good, solid advice as to how to proceed. Had you > followed the advice given in these posts, you would have discovered your > errors sooner, and would be better equipped to try different approaches. Contrary to your perception, a lot of posters have been trying to help > you. Absolutely not the case. Everybody ultimately acts in their own interest, even Mother Theresa, even you. My own goal in trying to figure out where Harris goes off the rails is to convince him that he is wrong. This is not a generous altruistic activity - it is more like kicking someone in the ass while explaining to them why you are doing it as a way of rubbing salt and vinegar in their wounds. This guy is a and is undeserving of any kind of help. He NEEDS his ass kicked. > Your view of other mathematicians and posters has unfortunately > been colored by your Narcissism. Because of your NPD, you end up > battling with those who could be your allies, you shun suggestions that > could help advance your ideas, and you inevitably end up alone and > frustrated. > The same would happen if he listened carefully to his critics. He is not that talented and his mathematical intuition is feeble. If he paid attention to criticism it would kill his fantasies before they could even get started. He would quickly see that math is not the field for him and he would disappear. But he wants to fantasize about that 15 minutes of fame. So he is not about to take your advice. In fact he has no genuine interest in math anyway. He thinks of it as a tool, a way he can feed his fantasies without doing much real work. He dreams that he can do math without doing any work to learn current theory, that he is such a genius that he can create whatever he wants de novo. He gravitates toward famous problems because that is where the glory is, not because he is actually interested. Incidentally, he is already back on his blog with a variant of his old method, which was shot down repeatedly here but he turned a blind eye to the evidence and unlike this last episode, never admitted he was wrong. The identical behavior occurred with his purported FLT proof and with Advanced Polynomial Factorization. After a while he just quit trying to argue and declared victory. That is what he is moving toward here. > You say We'll see who wins down the line. If history is any guide, you > will surely be the loser, a big time loser. He will bounce back soon with yet another unbaked idea to solve factoring, and again the fantasy will take over and the cycle will begin again. This little episode will be forgotten - just a small pothole on the 8-lane interstate highway of his glorious fantasized destiny. Not that big a deal. He cares less about this than you do. He'll shrug it off and start in again. Really, these little setbacks do not mean that much to him. He is basically a pretty superficial guy. Marcus. > This is the course your life > will take as long as you allow your NPD to run -- and ruin -- your life. > James, it is time for you to take your life back. Seek therapy for your > NPD. Taking this step will be tough, for your NPD will fight you on this > one. But the rewards will be a much happier life. -- > All things extant in this world, > Gods of Heaven, gods of Earth, > Let everything be as it should be; > Thus shall it be! > - Magical chant from Magical Shopping Arcade Abenobashi Drizzle, Drazzle, Drozzle, Drome, > Time for this one to come home! > - Mr. Wizard from Tooter Turtle === Subject: Re: JSH: Frustrated. TBP_6.1_AC; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. æAnd you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. æThink you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. æBut then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. æBut my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. æ You may be projecting. æI would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. æI think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. æHe has indeed been down > this road many times, and he is getting used to it. æIt disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. æAnd he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. æHe knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. æHe is basically a crank but he has the > most masterful troll skills I have ever seen. So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. æ Oh, come on. æNPD like a number of other pop-psych diagnoses > is not even well-defined. It is in the DSM-IV, you know. It's got a very precise set of diagnostic criteria. You can't just call it pop-psych. Of course, some people say that all mental illnesses are social constructs but I don't think that's the argument you were making. You're right that we're not competent to diagnose any mental disorder James may have, of course. But I think it would be fair to say that for James to discuss his problems with a mental health professional or a counsellor of some kind might be a strategy worth trying. They might be able to persuade him that there are better ways to try to be happy. On the other hand, maybe you're right and maybe he's happy enough the way he is. > It's an artificial construct, far > more a product of human imagination than, say, sqrt(-1). > Harris has an unusually rich array of character defects but they > do not constitute any illness. æThey are HIS FAULT. æUnlike a > genuinely sick person he is responsible for what he does. > Labelling him as NPD or a sociopath or whatever sheds no light > on anything. æHere you are saying he has a disease but at the > same time you are saying he is to blame for it. æIt's like > blaming him for having a heart defect. æHe acts like an arrogant > spiteful mean-spirited prick because he IS an arrogant spiteful > mean-spirited prick, and he is fully to blame for that > condition. The proof of this can be found in your latest threads. Granted, there > were replies that consisted of nothing but insults. But you also got > many replies with good, solid advice as to how to proceed. Had you > followed the advice given in these posts, you would have discovered your > errors sooner, and would be better equipped to try different approaches. Contrary to your perception, a lot of posters have been trying to help > you. æ Absolutely not the case. æEverybody ultimately acts in their > own interest, even Mother Theresa, even you. æ Not on any meaningful definition of self-interest. You're basically saying that all preferences count as self-interested. That renders the word self-interested pretty meaningless. > My own goal in > trying to figure out where Harris goes off the rails is to > convince him that he is wrong. æThis is not a generous > altruistic activity - it is more like kicking someone in the ass > while explaining to them why you are doing it as a way of > rubbing salt and vinegar in their wounds. æThis guy is a > and is undeserving of any kind of help. æHe NEEDS his ass > kicked. > I don't feel that way about James, myself. I do feel that way about some of the other people I've met on usenet. James I see more as an object of pity. Maybe I shouldn't pity him, maybe you're right and he's perfectly happy. I believe that it is in James' interests to teach him how to evaluate mathematical arguments properly and I'm trying to teach him how to do that. I'm not particularly altruistic, when I don't get intellectual satisfaction out of discussing James' arguments I stop bothering. But I do have a goal other than self-interest. A silly waste of time? Maybe. But not that much sillier than investing your energies into trying to make him feel unhappy. === Subject: Re: JSH: Frustrated. posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. But then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris > I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. You may be projecting. I would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. I think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. He has indeed been down > this road many times, and he is getting used to it. It disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. And he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. He knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. He is basically a crank but he has the > most masterful troll skills I have ever seen. > So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. Oh, come on. NPD like a number of other pop-psych diagnoses > is not even well-defined. It is in the DSM-IV, you know. It's got a very precise set of > diagnostic criteria. You can't just call it pop-psych. Of course, > some people say that all mental illnesses are social constructs but > I don't think that's the argument you were making. You're right that we're not competent to diagnose any mental disorder > James may have, of course. But I think it would be fair to say that > for James to discuss his problems with a mental health professional or > a counsellor of some kind might be a strategy worth trying. They might > be able to persuade him that there are better ways to try to be happy. > On the other hand, maybe you're right and maybe he's happy enough the > way he is. > The accusations of mental illness are meant to be cruel AND they are hypocritical as math people go on and on against people (like me) trying to comment on mathematics when they are not expert, when they routinely make medical diagnoses. But being inconsistent is an ironic oddity of the modern math field. Diagnosing someone you're arguing with as being mentally ill is an old way to try and win the argument by just taking the other person out completely as someone to whom anyone would listen. The crazy diagnosis is just a way to say ignore everything that person says. Nothing more. Saying you're trying to help someone by accusing them publicly of being mentally ill is like saying you're fighting a fire by dumping gas on a flame. It's like, firefighters who are really arsonists and I think many of you are pretending to be mathematicians when you really don't believe in it, which is why you turn to unsavory tactics. No decent person ever accused someone else of being mentally ill to win an argument. James Harris === Subject: Re: JSH: Frustrated. 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) syd-pow-pr8.tpgi.com.au:3128 (squid) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. æAnd you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. æThink you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. æBut then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. æBut my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris > I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. > æ You may be projecting. æI would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. æI think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. æHe has indeed been down > this road many times, and he is getting used to it. æIt disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. æAnd he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. æHe knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. æHe is basically a crank but he has the > most masterful troll skills I have ever seen. > So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. > æ Oh, come on. æNPD like a number of other pop-psych diagnoses > is not even well-defined. It is in the DSM-IV, you know. It's got a very precise set of > diagnostic criteria. You can't just call it pop-psych. Of course, > some people say that all mental illnesses are social constructs but > I don't think that's the argument you were making. You're right that we're not competent to diagnose any mental disorder > James may have, of course. But I think it would be fair to say that > for James to discuss his problems with a mental health professional or > a counsellor of some kind might be a strategy worth trying. They might > be able to persuade him that there are better ways to try to be happy. > On the other hand, maybe you're right and maybe he's happy enough the > way he is. The accusations of mental illness are meant to be cruel AND they are > hypocritical as math people go on and on against people (like me) > trying to comment on mathematics when they are not expert, when they > routinely make medical diagnoses. But being inconsistent is an ironic oddity of the modern math field. Diagnosing someone you're arguing with as being mentally ill is an old > way to try and win the argument by just taking the other person out > completely as someone to whom anyone would listen. > Your craziness is not particularly subtle. These ideas you have that government agencies are observing the conversations you have here and are likely to kill you are not exactly sane. Of course we're not competent to make medical diagnoses. You should go and see a professional. But you do clearly have a very serious reality-testing problem. That's obvious. A mental health professional could probably help you. I've had psychosis myself, about six years ago. The medications they have these days are very good. We're not saying this to be cruel, it's just obvious that you could do with some help. It's nothing to be ashemed of. But if you're happy with what's been going on for the last ten years, don't worry about it, just enjoy life. But be aware that the next ten years will be more of the same unless you change your behaviour. > The crazy diagnosis is just a way to say ignore everything that > person says. > A lot of people here give you detailed feedback on what you say. We're not saying your arguments are wrong because you're mentally ill, we're saying they're wrong because we can see flaws in them. Anyway. Never mind. Any time you're interested in my views about how you could best go about achieving your goals, I'm happy to offer them in a constructive spirit. You're clearly not interested, so don't worry about it, just do what you think best. === Subject: Re: JSH: Frustrated. > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. But then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris > I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. > You may be projecting. I would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. I think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. He has indeed been down > this road many times, and he is getting used to it. It disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. And he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. He knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. He is basically a crank but he has the > most masterful troll skills I have ever seen. > So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. > Oh, come on. NPD like a number of other pop-psych diagnoses > is not even well-defined. > It is in the DSM-IV, you know. It's got a very precise set of > diagnostic criteria. You can't just call it pop-psych. Of course, > some people say that all mental illnesses are social constructs but > I don't think that's the argument you were making. > You're right that we're not competent to diagnose any mental disorder > James may have, of course. But I think it would be fair to say that > for James to discuss his problems with a mental health professional or > a counsellor of some kind might be a strategy worth trying. They might > be able to persuade him that there are better ways to try to be happy. > On the other hand, maybe you're right and maybe he's happy enough the > way he is. The accusations of mental illness are meant to be cruel AND they are > hypocritical as math people go on and on against people (like me) > trying to comment on mathematics when they are not expert, when they > routinely make medical diagnoses. It's funny that now you claim that you are not an expert on mathematics. I thought that you saw your self as one of the very few all time greatest mathematicians. On the other hand, if mathematicians around here should not post stuff about mental health, and more specifically about _your_ mental health, then why do you feel that you can? After all, you posted this: http://mathforum.org/kb/plaintext.jspa?messageID=444857 on the 2002 Christmas Day. Are *you* an expert on mental health? > But being inconsistent is an ironic oddity of the modern math field. Diagnosing someone you're arguing with as being mentally ill is an old > way to try and win the argument by just taking the other person out > completely as someone to whom anyone would listen. The crazy diagnosis is just a way to say ignore everything that > person says. Nothing more. Of course, if you actually did solve mathematical problems, you would not be ignored. > Saying you're trying to help someone by accusing them publicly of > being mentally ill is like saying you're fighting a fire by dumping > gas on a flame. It's like, firefighters who are really arsonists and I think many of > you are pretending to be mathematicians when you really don't believe > in it, which is why you turn to unsavory tactics. No decent person ever accused someone else of being mentally ill to > win an argument. I fully agree. Of course, several persons around here accused you of being mentally ill not to win an argument but because they actually feel (correctly, in my opinion) that you _are_ mentally ill and that you should get treatment. Jose Carlos Santos === Subject: Re: JSH: Frustrated. <61svkfF208eqhU1@mid.individual.net> posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. But then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris > I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. > You may be projecting. I would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. I think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. He has indeed been down > this road many times, and he is getting used to it. It disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. And he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. He knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. He is basically a crank but he has the > most masterful troll skills I have ever seen. > So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. > Oh, come on. NPD like a number of other pop-psych diagnoses > is not even well-defined. > It is in the DSM-IV, you know. It's got a very precise set of > diagnostic criteria. You can't just call it pop-psych. Of course, > some people say that all mental illnesses are social constructs but > I don't think that's the argument you were making. > You're right that we're not competent to diagnose any mental disorder > James may have, of course. But I think it would be fair to say that > for James to discuss his problems with a mental health professional or > a counsellor of some kind might be a strategy worth trying. They might > be able to persuade him that there are better ways to try to be happy. > On the other hand, maybe you're right and maybe he's happy enough the > way he is. The accusations of mental illness are meant to be cruel AND they are > hypocritical as math people go on and on against people (like me) > trying to comment on mathematics when they are not expert, when they > routinely make medical diagnoses. It's funny that now you claim that you are not an expert on mathematics. > I thought that you saw your self as one of the very few all time > greatest mathematicians. On the other hand, if mathematicians around here should not post stuff > about mental health, and more specifically about your mental health, > then why do you feel that you can? After all, you posted this: http://mathforum.org/kb/plaintext.jspa?messageID=444857 on the 2002 Christmas Day. Are *you* an expert on mental health? > I can talk about myself freely, as can you. I wouldn't try to abridge your right to call yourself mentally ill. But I would be wrong if *I* questioned your sanity. Understand the difference? It's about boundaries. > But being inconsistent is an ironic oddity of the modern math field. Diagnosing someone you're arguing with as being mentally ill is an old > way to try and win the argument by just taking the other person out > completely as someone to whom anyone would listen. The crazy diagnosis is just a way to say ignore everything that > person says. Nothing more. Of course, if you actually did solve mathematical problems, you would > not be ignored. I DO solve math problems and one of the more remarkable realities of discussions about my research is the continuing implication that I do not. For instance, I found a prime counting function. It counts prime numbers. There is no doubt that it does so, and it gives the correct count. Curious readers can see a page I had on Wikipedia which gives its full form which is only in the history now: http://en.wikipedia.org/w/index.php?title=Prime-counting function&oldid=8723634 Also with surrogate factoring I have had several methods which DID in fact, factor, but the problem was they were too slow to be practical. But that is like calling someone insane--pushing the idea that nothing they do works is a way to tell other people to ignore them. However, I like my example of Tiger Woods talking to an amateur golfer where some listening would say the amateur is inflating his own efforts. Would Tiger Woods mount a public campaign to discredit that person, including questioning his mental health, and justify it with all kinds of excuses including saying he was trying to HELP the amateur golfer by publicly calling him insane? If he did, what would people think of Mr. Woods? Why do math people expect different? James Harris === Subject: Re: JSH: Frustrated. > being mentally ill not to win an argument but because they actually feel > (correctly, in my opinion) that you _are_ mentally ill and that you > should get treatment. Treatment could also consist of reading math books on the subject. Treatment can consist of dragging *JSH the troll* round the internet totally humiliated. I don't think he is mental at all, just a troll that tries to keep the thread lengths as long as possible for the thrill of it. Just look at the titles to his posts. === Subject: Re: JSH: Frustrated. posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) > Maybe I should say more after going on and on about an approach that I > now accept was just useless for factoring, but I keep thinking about > why I'm so desperate for something in the factoring area anyway, which > is the blocking of my pure math research where math people don't > follow their own rules. > I NEED a practical math result because one dead journal shows how > locked down math society has it now. > Maybe we'll take each other down before all of this is over and I'll > get my way of convincing the world that mathematicians routinely lie > and your society will find a way to get me in return. > And it is a sad testimony to the true reality of modern academia. > It is a medieval system. And you end up in medieval crap with it, > like what I call the Math Wars. > Some of you seem to think the Math Wars are just fun and games or just > some silly crackpot mouthing off, but it is about me finding a way > to do things like end tenure, reduce funding for academics and > convince the world that objective measures, rather than just letting > academics say which of their buddies supposedly did something great, > are necessary. > So sit back. Think you have it all handled as I remain frustrated, > and venting in futile anger at a feudal system of academia, because I > need that practical mathematical result to break a broken system, but > remember, it only takes one result for me to then go back and use > every moment like this to emphasize to the public why it needs to do > as much as necessary. > Our modern world exists today because discovery was cherished. > But parasites have turned things upside down for a few dollars and > some empty accolades because they can't appreciate the value. > If humanity loses then it loses down the line, and the extinction of > our species whenever it occurs as we probably won't manage to get off > this planet, may trace back to a tragic shift, when major problem > solving was lost, and pretend took over. > Call me crazy. But then nothing I do will work in the real world, > right? > But then you won't find funding drying up, and you won't find an > increasingly skeptical public demanding more than just your say-so, > right? > Call me crazy, and if I am then the solutions I find to convince > people that your society lies in the real world won't really be > solutions, right? > But I see the end of the Math Wars putting most of you in other areas > of work outside of mathematics. But my vision is against your will to > stop me. > We'll see who wins down the line. > James Harris I am deeply sorry to hear that your latest failure to solve the > factoring problem has left you depressed and frustrated. As I indicated > in the post > nearly all of us could see this coming. You have been down this road > before, many, many times, even this road always leads you to the > sorrowful dead end. You may be projecting. I would guess that Harris takes all > this much less seriously than you expect, perhaps less > seriously even than a number of other posters here. I think > he is deliberately fantasizing, putting himself on, dreaming > of how he would like things to be. He has indeed been down > this road many times, and he is getting used to it. It disrupts > the fantasy for a while, but then he comes up with another hare- > brained idea - doesn't test it, because he knows it is likely > wrong - doesn't spell out a complete proof because he knows > he cannot - and the fantasy life takes over again. And he > knows how to push a lot of people's buttons with his rapturous > descriptions of the perfidy of mathematicians, world economic > collapse, humanity escaping to the planet Venus, etc., because > no one will recognize his genius. He knows he's not a genius > and he knows how annoying it is for him to claim it with not > a shred of evidence. He is basically a crank but he has the > most masterful troll skills I have ever seen. So what is the cause of your repeated failures? Once again, you > attribute it to a powerful enemy, the math community. You are > partially right: you are being held back by a powerful enemy. But your > enemy is not the math community or academics. The enemy that has really > been thwarting you at every turn, both in your math work and in other > aspects of your life, is your Narcissism. Oh, come on. NPD like a number of other pop-psych diagnoses > is not even well-defined. It's an artificial construct, far > more a product of human imagination than, say, sqrt(-1). > Harris has an unusually rich array of character defects but they > do not constitute any illness. They are HIS FAULT. Unlike a > genuinely sick person he is responsible for what he does. > Labelling him as NPD or a sociopath or whatever sheds no light > on anything. Here you are saying he has a disease but at the > same time you are saying he is to blame for it. It's like > blaming him for having a heart defect. He acts like an arrogant > spiteful mean-spirited prick because he IS an arrogant spiteful > mean-spirited prick, and he is fully to blame for that > condition. The proof of this can be found in your latest threads. Granted, there > were replies that consisted of nothing but insults. But you also got > many replies with good, solid advice as to how to proceed. Had you > followed the advice given in these posts, you would have discovered your > errors sooner, and would be better equipped to try different approaches. Contrary to your perception, a lot of posters have been trying to help > you. Absolutely not the case. Everybody ultimately acts in their > own interest, even Mother Theresa, even you. My own goal in > trying to figure out where Harris goes off the rails is to > convince him that he is wrong. This is not a generous > altruistic activity - it is more like kicking someone in the ass > while explaining to them why you are doing it as a way of > rubbing salt and vinegar in their wounds. This guy is a > and is undeserving of any kind of help. He NEEDS his ass > kicked. Your view of other mathematicians and posters has unfortunately > been colored by your Narcissism. Because of your NPD, you end up > battling with those who could be your allies, you shun suggestions that > could help advance your ideas, and you inevitably end up alone and > frustrated. The same would happen if he listened carefully to his critics. He > is not that talented and his mathematical intuition is feeble. If > he paid attention to criticism it would kill his fantasies before > they could even get started. He would quickly see that math is > not the field for him and he would disappear. But he wants to > fantasize about that 15 minutes of fame. So he is not about to > take your advice. In fact he has no genuine interest in math anyway. He thinks of > it as a tool, a way he can feed his fantasies without doing much > real work. He dreams that he can do math without doing any work to > learn current theory, that he is such a genius that he can > create whatever he wants de novo. He gravitates toward famous > problems because that is where the glory is, not because he > is actually interested. Incidentally, he is already back on his blog with a variant > of his old method, which was shot down repeatedly here but > he turned a blind eye to the evidence and unlike this last > episode, never admitted he was wrong. The identical behavior Wrong? Surrogate factoring is a concept I've worked on for I think it's over four years now. The concept is based on a question: can you factor one number by instead factoring another? > occurred with his purported FLT proof and with Advanced > Polynomial Factorization. After a while he just quit trying to I don't argue about settled results. And non-polynomial factorization is also a published result, though > argue and declared victory. That is what he is moving toward > here. > Nope. I don't argue about settled results. > You say We'll see who wins down the line. If history is any guide, you > will surely be the loser, a big time loser. He will bounce back soon with yet another unbaked idea > to solve factoring, and again the fantasy will take over and > the cycle will begin again. This little episode will be forgotten - > just a small pothole on the 8-lane interstate highway of his glorious > fantasized destiny. Not that big a deal. He cares less about > this than you do. He'll shrug it off and start in again. Really, > these little setbacks do not mean that much to him. He is > basically a pretty superficial guy. Marcus. LOL. Well I don't think there is anything left for me to say. James Harris === Subject: Re: JSH: Frustrated. > Sometimes I feel a little bit guilty about making fun of a lunatic. I always felt a little uncomfortable about that myself. The more I learned about NPD, and about how painful the down periods are for NPD sufferers, the more uncomfortable I became. That is why I choose to move in the direction of advocating treatment for his Narcissism. I do realize that it is highly unlikely that newsgroup posts will cause James to take the vital but difficult step of starting therapy. But stranger thing have happened. The very fact that James takes these groups so seriously indicates that there is some chance of this happening. I also take some comfort in promoting the right course of action. -- All things extant in this world, Gods of Heaven, gods of Earth, Let everything be as it should be; Thus shall it be! - Magical chant from Magical Shopping Arcade Abenobashi Drizzle, Drazzle, Drozzle, Drome, Time for this one to come home! - Mr. Wizard from Tooter Turtle === Subject: Re: JSH: Frustrated. <2lmdr35mo34bv9f52i83rc40ma4st2hrha@4ax.com> TBP_6.1_AC; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Sometimes I feel a little bit guilty about making fun of a lunatic. I always felt a little uncomfortable about that myself. The more I > learned about NPD, and about how painful the down periods are for NPD > sufferers, the more uncomfortable I became. That is why I choose to move > in the direction of advocating treatment for his Narcissism. I do realize that it is highly unlikely that newsgroup posts will cause > James to take the vital but difficult step of starting therapy. But > stranger thing have happened. The very fact that James takes these > groups so seriously indicates that there is some chance of this > happening. I also take some comfort in promoting the right course of action. -- > All things extant in this world, > Gods of Heaven, gods of Earth, > Let everything be as it should be; > Thus shall it be! > - Magical chant from Magical Shopping Arcade Abenobashi Drizzle, Drazzle, Drozzle, Drome, > Time for this one to come home! > - Mr. Wizard from Tooter Turtle Gee, James is giving a lot of 1 star ratings. Yeah, I agree with you, if one is going to respond to him at all the responsible thing to do is to try to encourage him to take a more constructive approach to his situation. Sometimes I try to rationalize that maybe there is a place for humour-as-therapy as well but that is probably just a rationalization. === Subject: Prove that (a^-1)(b^-1)(a)(b) is an even permutation. Let a and b belong to Sn. Prove that (a^-1)(b^-1)(a)(b) is an even permutation. This is what I know about permutations Every permutation in Sn, n>1 is a product of 2 cycles and A permutation that can be expressed as a product of an even number of 2 cycles is called an even permutation any help would be greatly appreciated. === Subject: Re: Prove that (a^-1)(b^-1)(a)(b) is an even permutation. <7908741.1203299294675.JavaMail.jakarta@nitrogen.mathforum.org>, allpro > Let a and b belong to Sn. Prove that (a^-1)(b^-1)(a)(b) is an even > permutation. This is what I know about permutations > Every permutation in Sn, n>1 is a product of 2 cycles > and > A permutation that can be expressed as a product of an even number of 2 > cycles is called an even permutation any help would be greatly appreciated. What is the inverse of a product of 2-cycles? -- Paul Sperry Columbia, SC (USA) === Subject: rao elements of engineering electromagnetics...solution manual posting-account=qjnP4goAAACC8cyxWczGWJFeF4-vvf32 CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) I need the solutions to elements of engineering electromagnetics by rao. 6th addition. Please send to christine.girolami@gmail.com if you have them! === Subject: Elements of engineering electromagnetics (6/e) by N.N.RAO posting-account=qjnP4goAAACC8cyxWczGWJFeF4-vvf32 CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Need solutions to Elements of engineering electromagnetics (6/e) by N.N.RAO christine.girolami@gmail.com === Subject: solution manual posting-account=1LXcdwoAAAAOMsiNKAVJ2p8wxrEhk12V Gecko/20080201 Firefox/2.0.0.12 Creative ZENcast v1.02.10,gzip(gfe),gzip(gfe) Hello I have solution manuals for a competitive price. 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HAYLER 's Probability and Statistics for Engineers and Scientists Manual Nanoengineering of Structural, Functional and Smart Materials If your interested do let me know at xplicitly666@yahoo.com Explicit Group === Subject: Dynamics posting-account=h1YwtAoAAACd7OcXtm3Lw5EytNvN81Vo AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) Could you please send the english dynamics solutions manual to my email: donough1@msu.edu === Subject: Dynamics posting-account=h1YwtAoAAACd7OcXtm3Lw5EytNvN81Vo AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) Could you please send me the english dynamics solutions manual to me email address? :donough1@msu.edu -Kaitlin === Subject: general equation of an ellipse & semi-axes I am following this site :- http://mathworld.wolfram.com/Ellipse.html which relates various attributes of an ellipse (such as its centre, angle of rotation, semi-axes etc.), to its general equation which is in the form :- ax^2 + 2bxy + cy^2 + 2d x + 2f y + g = 0 About 1/4 of the way down the page there are two large expressions which give both of the semi-axes in terms of the coefficients of the general ellipse equation. My question is, how would you find the semi-axes in the case that the x^2 and y^2 coefficients are the same, as both expressions in this case cannot be evaluated? (In this case the ellipse has an angle of rotation of 0 or 45degrees.) Anyone know what to do in this instance? Jeremy Watts === Subject: Re: general equation of an ellipse & semi-axes @hotmail.com>: > http://mathworld.wolfram.com/Ellipse.html a,b,c for parameters of the ellipse, and the second part uses them for coefficients of the general quadratic curve. It's particularly bad that, after introducing the general equation eccentricity in terms of the *previous* a, b, and c. http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node28.html at least uses consistent notation. > My question is, how would you find the semi-axes in the case that the x^2 > and y^2 coefficients are the same, as both expressions in this case cannot > be evaluated? (In this case the ellipse has an angle of rotation of 0 or > 45degrees.) If a = c, are you actually working with an ellipse any more, or can you make a simplification? > Anyone know what to do in this instance? Try dividing through by the common coefficient of x^2 and y^2, so that you have a new equation in the form x^2 + pxy + y^2 + qx + ry + s = 0 -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top posting such a bad thing? === Subject: Re: general equation of an ellipse & semi-axes > @hotmail.com>: > http://mathworld.wolfram.com/Ellipse.html a,b,c for parameters of the ellipse, and the second part uses them > for coefficients of the general quadratic curve. It's particularly bad that, after introducing the general equation > eccentricity in terms of the *previous* a, b, and c. http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node28.html > at least uses consistent notation. it is yes, i noticed that, later the same symbols used for the coefficients are used to denote the semi-major & minor axes. > My question is, how would you find the semi-axes in the case that the x^2 > and y^2 coefficients are the same, as both expressions in this case > cannot > be evaluated? (In this case the ellipse has an angle of rotation of 0 or > 45degrees.) If a = c, are you actually working with an ellipse any more, or can > you make a simplification? if the x^2 & y^2 coefficients are the same then yes this is still an ellipse, its that its orientation to the x-axis (the angle the major axis makes with the x-axis) is at either 0,45 or 135 degrees. i can either re-arrange the formula as david has suggested, or another idea is to translate the whole thing so that its centre is at the origin, and then rotate clockwise by the angle of rotation (both the centre and the angle of rotation are easy to find) , and then the axes should be obvious from the transformed equation. jeremy > Anyone know what to do in this instance? Try dividing through by the common coefficient of x^2 and y^2, so > that you have a new equation in the form > x^2 + pxy + y^2 + qx + ry + s = 0 > -- > Stan Brown, Oak Road Systems, Tompkins County, New York, USA > http://OakRoadSystems.com > A: Maybe because some people are too annoyed by top posting. > Q: Why do I not get an answer to my question(s)? > A: Because it messes up the order in which people normally read text. > Q: Why is top posting such a bad thing? === Subject: Re: general equation of an ellipse & semi-axes > I am following this site :- > http://mathworld.wolfram.com/Ellipse.html which relates various attributes of an ellipse (such as its centre, angle > of rotation, semi-axes etc.), to its general equation which is in the > form :- ax^2 + 2bxy + cy^2 + 2d x + 2f y + g = 0 About 1/4 of the way down the page there are two large expressions which > give both of the semi-axes in terms of the coefficients of the general > ellipse equation. I had planned to mention to you that it would have been more helpful if you had given the item numbers for those two expressions. Now, perhaps I see why you didn't! Those _two_ expressions seem to have _four_ item numbers, (22) - (25). > My question is, how would you find the semi-axes in the case that the x^2 > and y^2 coefficients are the same, as both expressions in this case > cannot be evaluated? (In this case the ellipse has an angle of rotation > of 0 or 45degrees.) Anyone know what to do in this instance? It's trivial, surely. What I don't understand is why MathWorld would have presented those expressions in that unfortunate way! Look at the first expression. In its denominator, you see the term (c - a) Sqrt(1 + 4b^2/(a - c)^2) and that's what's causing a problem when a = c. But if we merely replace that term by -sign(a - c) Sqrt((a - c)^2 + 4b^2) the problem vanishes. We can now let a = c. We can do similarly in the second expression, of course. David W. Cantrell === Subject: Re: general equation of an ellipse & semi-axes > I am following this site :- > http://mathworld.wolfram.com/Ellipse.html > which relates various attributes of an ellipse (such as its centre, angle > of rotation, semi-axes etc.), to its general equation which is in the > form :- ax^2 + 2bxy + cy^2 + 2d x + 2f y + g = 0 > About 1/4 of the way down the page there are two large expressions which > give both of the semi-axes in terms of the coefficients of the general > ellipse equation. I had planned to mention to you that it would have been more helpful if > you > had given the item numbers for those two expressions. Now, perhaps I see > why you didn't! Those _two_ expressions seem to have _four_ item numbers, > (22) - (25). > My question is, how would you find the semi-axes in the case that the x^2 > and y^2 coefficients are the same, as both expressions in this case > cannot be evaluated? (In this case the ellipse has an angle of rotation > of 0 or 45degrees.) > Anyone know what to do in this instance? It's trivial, surely. What I don't understand is why MathWorld would have > presented those expressions in that unfortunate way! Look at the first expression. In its denominator, you see the term (c - a) Sqrt(1 + 4b^2/(a - c)^2) and that's what's causing a problem when a = c. But if we merely replace > that term by -sign(a - c) Sqrt((a - c)^2 + 4b^2) the problem vanishes. We can now let a = c. We can do similarly in the second expression, of course. David W. Cantrell === Subject: Re: Solution Manual of Engineering Mechanics - Statics And Dynamics 11ed by Hibbeler in pdf form posting-account=WQOKlwoAAADBfCDl3lInh9MNGUnFNpRe Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) Could you send me the solutions for Dynamics 11e My email is steven.kerr@ndsu.edu This would be a big help! === Subject: Re: Balloon inflation (open problem... as far as I know) > Erm. The problem is just in P. Given your starting position, it is > straightforward to state a finite upper bound on the number of steps > than can answer the question. It is not in P unless it can be answered in a number of steps bounded by some polynomial in the size of the problem. That's what the P stands for. Merely finite isn't enough. - Tim === Subject: Rusty at rearranging! Hi! I've lost too many brain cells to work this out. Can anybody help me rearrange each of these equations to solve for N? X = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)+S-(2*F)) Y = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)-S) Mike -- === Subject: Re: Rusty at rearranging! > Hi! I've lost too many brain cells to work this out. Can anybody help me rearrange each of these equations to solve for N? X = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)+S-(2*F)) N = F^2 (X - S)/(C X (F - S)) > Y = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)-S) N = F^2 (Y - S)/(C Y (S - F)) David === Subject: Re: Rusty at rearranging! THANK YOU David! That did the trick beautifully! Folks like you are the very best of what the Internet has to offer, in my opinion. It's just wonderful to be able to ask a question like that and have a complete stranger come to the rescue. I really appreciate it. Mike Davis :> Hi! :> :> I've lost too many brain cells to work this out. :> :> Can anybody help me rearrange each of these equations to solve for N? :> :> X = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)+S-(2*F)) : N = F^2 (X - S)/(C X (F - S)) :> Y = S*((((F^2)/(N*C))+F)-F) / ((((F^2)/(N*C))+F)-S) : N = F^2 (Y - S)/(C Y (S - F)) : David -- === Subject: Maximum Average of Values drawn from normal distribution I am a bit stuck with the following -- I would appreciate if someone could assist me with this. Given a normal distribution with mean 0.5 and standard deviation 0.25 and given a set X of values drawn from this distribution, what is the average of the maximal value of set X. For example, if in the first drawing, we have 5 values, X_1 = {0.2, 0.6, 0.3, 0.4, and 0.5}, then the highest value x_1_max is 0.6. in the second draw we have, X_2 = {0.3, 0.1, 0.7, 0.5, and 0.3}, then the highest value x_2_max is 0.7. and so on. Now if I do this many times, what will the average of the x_max values converge to? It should be something like ln(|X|)*0.25, but I am not sure. So, for |X|=5 and standard deviation = 0.25 that would be ~0.40. Assistance is appreciated!! cheers, Chris === Subject: Re: Dedekind cuts > INTERVAL ARITHMETIC Ramon Moore defined the arithmetic operations on intervals as follows: [d,u] + [e,t] = [d+e, u+t] > - [d,u] = [-u, -d] > [d,u] x [e,t] = [min (d e,d t,u e,u t), max (d e,d t,u e,u t)] > [d,u]^-1 = [u^-1, d^-1] if 0 notin [d,u] > ie 0 in D or 0 in U > = [-oo, +oo] if 0 in [d,u] The above surprises me. (I'm not disputing that Moore used [d, u]^-1 = [-oo, +oo] if 0 in [d, u]; his book is not readily available for me to check.) In an extended interval arithmetic, it would be common nowadays to have [d, u]^-1 = [u^-1, +oo] union [-oo, d^-1]. That's not a single interval in the two-point extension of the reals. But using the one-point extension, we do get a single interval, and that interval can be expressed, using a generalized notation, as [u^-1, d^-1]. This allows us, neatly, to say that [d, u]^-1 = [u^-1, d^-1], regardless of whether 0 is in [d, u] or not. > The formula for multiplication is complicated by the need to consider > all possible combinations of signs. However, we have just said that supD and infU need not exist as real > numbers, so we have more work to do to define the arithmetic > operations for them. This extension can be defined, using methods from continuous lattices, > so long as each operation * is rounded in the sense that [d,u] * [e,t] << [a,z] => Some d' u' e' t'. > [d,u]<<[d',u'] & [e,t]<<[e',t'] & [d',u']*[e',t']<<[a',z'] where [d,u] << [d',u'] means d' addition, but rather messy for multiplication. It's neither difficult > nor deep, but the property is important, so is it stated anywhere? COMPLETENESS OF INTERVAL COMPUTATION Standard interval analysis evaluates a function with an interval as > its argument by replacing the usual arithmetic operations with Moore's > generalisation to intervals. This overestimates the range of a > function, often by a long way. However, we may compute the set-theoretic image of an interval [d,u] > to within any eps>0 by sub-dividing [d,u] into sufficiently but > finitely many parts, applying the function Moore-wise to each part, > and forming the union. Where is this theorem (first) proved in the > interval literature? This result is the basis of the construction of R > and the proof that [d,u] is compact in ASD. Perhaps I'm taking what you said too literally. But if we attempt to get the set-theoretic image under reciprocation of an interval [d, u] containing 0 as you indicate (that is, by sub-dividing it, applying the function Moore-wise to each part, and forming the union), then we would always get [-oo, +oo], which is not generally the set-theoretic image, [u^-1, +oo] union [-oo, d^-1]. David W. Cantrell === Subject: multiset question Can anybody give me an example of two finite sets A,B of different cardinality, with positive integer valued functions f:A->N, g:B->N such that sum_a{f(a)} = sum_b{f(b)} and sum_a{f(a)^2} = sum_b{f(b)^2} ? -- Gavin Wraith (gavin@wra1th.plus.com) Home page: http://www.wra1th.plus.com/ === Subject: Re: multiset question > Can anybody give me an example of two finite sets A,B > of different cardinality, with positive integer valued > functions f:A->N, g:B->N such that > sum_a{f(a)} = sum_b{f(b)} and > sum_a{f(a)^2} = sum_b{f(b)^2} ? > {1,2,6} and {4,5}. eegh === Subject: Re: Physics from Logic? Check the math please ABSTRACT: Starting from the premise that all facts are consistent > with each other, the Feynman path integral formulation of > quantum mechanics can be derived. You appear to be asserting that a universe without quantum mechanics is a logical impossibility. Since one can easily construct a deterministic universe (e.g., Conway's Game of Life), that assertion must be wrong, and any line of reasoning that appears to prove it must be flawed. -- === Subject: Re: Physics from Logic? Check the math please > derive physics from nothing more than logic. >This reminds me of the book > Information Mechanics > by Frederick William Kantor > 412 pages > John Wiley & Sons Inc (November 2, 1977) > ISBN: 0471029688 >which I spent some time looking at many years ago. The titel is interesting. What is the book about? -- Markus Redeker Hamburg, Germany === Subject: Inverse Limit Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm kind of confused with the meaning of inverse limits for groups. Could someone help me understand? Let X={a_1, a_2, a_3, ...} be a countably infinite set. For every 0=n the groups G_m and G_n are connected by the projection G_m->G_n, and for every n, there is a projection G_X->G_n. - Is G_X the inverse limit of the system (G_n) ? Clearly, G_X is the vertex of a cone. But is it the last one to be so? Namely, if G is defined to be the subgroup of the direct product of G_n's consisting of sequences (x_n) where for every m>=n, x_n is the projection x_m under G_m->G_n, I cannot see any homomorphism between G and G_X (in either direction) commuting with the projections. How can G be a factor of G_X, if it contains elements such as (a_1, a_1+a_2, a_1+a_2+a_3, ...) ? Have I misunderstood something? Or maybe there is a trivial answer? Siamak === Subject: Re: Inverse Limit Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >I'm kind of confused with the meaning of inverse limits for groups. >Could someone help me understand? Let X={a_1, a_2, a_3, ...} be a countably infinite set. >For every 0Let G_X be the free Abelian group generated by X, and G_n be the free >Abelian group generated by X_n. For m>=n the groups G_m and G_n are >connected by the projection G_m->G_n, and for every n, there is a >projection G_X->G_n. - Is G_X the inverse limit of the system (G_n) ? Clearly, G_X is the >vertex of a cone. But is it the last one to be so? Namely, if G is defined to be the subgroup of the direct product of >G_n's consisting of sequences (x_n) where for every m>=n, x_n is the >projection x_m under G_m->G_n, I cannot see any homomorphism between G >and G_X (in either direction) commuting with the projections. There is an obvious map from G_X to G. Consider an arbitrary element z of G_X, and let pi_m be the projection G_X -> G_m. By the universal property of the product the family of maps {pi_m} give you a map from G_X to the product of the G_n, namely z |-> (... pi_n(z), pi_{n-1}(z), ..., p_2(z), p_1(z)) and it is easy to verify that this element is in G. This gives you a map from G_X to G, and by construction it will commute with the projections. > How can >G be a factor of G_X, if it contains elements such as > (a_1, a_1+a_2, a_1+a_2+a_3, ...) >? Well, no, G is not a factor of G_X; G_X is not the beyond the last vertex of the cone of groups. G_X is a subgroup of G consisting of all the eventually constant inverse sequences. It is a similar relation as there is between the integers and the p-adic integers (where the former is a subgroup of the latter, given by all eventually constant sequences). >Have I misunderstood something? Or maybe there is a trivial answer? I'm not sure why you think the question asked has to do with checking if G is a factor of G_X; rather, the question is asking whether G_X and G are isomorphic. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Inverse Limit Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Let X={a_1, a_2, a_3, ...} be a countably infinite set. > For every 0 Let G_X be the free Abelian group generated by X, and G_n be the free > Abelian group generated by X_n. For m>=n the groups G_m and G_n are > connected by the projection G_m->G_n, and for every n, there is a > projection G_X->G_n. - Is G_X the inverse limit of the system (G_n) ? Clearly, G_X is the > vertex of a cone. But is it the last one to be so? Namely, if G is defined to be the subgroup of the direct product of > G_n's consisting of sequences (x_n) where for every m>=n, x_n is the > projection x_m under G_m->G_n, I cannot see any homomorphism between G > and G_X (in either direction) commuting with the projections. How can > G be a factor of G_X, if it contains elements such as > (a_1, a_1+a_2, a_1+a_2+a_3, ...) Your question contains the answer or am I misunderstanding something ? In the additive category of Abelian groups (=Z-modules), binary product = binary coproduct. But here you have something infinite. G_X does not contain the infinite sum a_1+a_2+a_3+..., the infinite product Za_1xZa_2x... does. And the latter group is the inverse limit (if I correctly understand the notations). pg. === Subject: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hardy and Wright mention that e.pi is irrational (4th ed, p 39), but I have heard that this is a misprint. I have further failed to find any reliable sources explaining the current status. So does anyone know: what is currently known about the rationality and/or transcendence of e.pi? (And as an aside, what about e+pi?) Julian Gilbey === Subject: Re: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) On Mar 4, 8:30æam, Julian Gilbey Hardy and Wright mention that e.pi is irrational (4th ed, p 39), but I > have heard that this is a misprint. æI have further failed to find any > reliable sources explaining the current status. æSo does anyone know: > what is currently known about the rationality and/or transcendence of > e.pi? æ(And as an aside, what about e+pi?) > Julian Gilbey This seems to be unknown. It is known that e^pi is transcendental; that is probably what the misprint is. It is also known that at least one of e.pi and e+pi is transcendental, if both were algebraic, then the roots of x^2-(e+pi)x+e.pi=0 would also be algebraic. === Subject: Re: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Hardy and Wright mention that e.pi is irrational (4th ed, p 39), but I > have heard that this is a misprint. I have further failed to find any > reliable sources explaining the current status. So does anyone know: > what is currently known about the rationality and/or transcendence of > e.pi? (And as an aside, what about e+pi?) At least one of e.pi and e+pi is transcendental, since e is a root of the polynomial x^2 - (e+pi) x + (e.pi), and e is known to be transcendental. As far as I am aware, nothing else is known about the status of these numbers. Schanuel's conjecture would imply that e and pi are algebraically independent, and therefore e.pi and e+pi are both transcendental. -- === Subject: Re: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > what is currently known about the rationality and/or transcendence of > e.pi? (And as an aside, what about e+pi?) At least one of these is transcendental (easy). Both are transcendental (true, but not yet proved). === Subject: Re: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Hardy and Wright mention that e.pi is irrational (4th ed, p 39), but I > have heard that this is a misprint. I have further failed to find any > reliable sources explaining the current status. So does anyone know: > what is currently known about the rationality and/or transcendence of > e.pi? (And as an aside, what about e+pi?) See the sci.math thread e*pi is irrational -- the title of which might, better, have ended in a question mark -- at I'd be surprised if the status of our knowledge has changed since then. David === Subject: Re: Current status of irrationality and transcendence of e.pi? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) * Julian Gilbey in sci.math.research: > Hardy and Wright mention that e.pi is irrational (4th ed, p 39), but > I have heard that this is a misprint. I have further failed to find > any reliable sources explaining the current status. So does anyone > know: what is currently known about the rationality and/or > transcendence of e.pi? (And as an aside, what about e+pi?) -- DW === Subject: Trojan Squares! Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Semi-Latin Squares could be found from :http://www.maths.qmw.ac.uk/ ~rab/slsopt.html I am searching for Optimal Semi Latin Squares , and apparently Trojan Squares were known to be optimal. Could i have a reference as to where i could find the complete Trojan Squares or any efficient code that could generate me such? === Subject: Looking for references on a set of expression transformation rules Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I'm wondering if the rules described below can be found in existing literature. In what follows we use the = sign to indicate that both sides of the equation can be beta-reduced to one and the same term. Classic definition of the beta reduction is well-known: (x. M A) -> M[x := A]. (*) Here substitution is determined by the following four rules: i) x[x := A] = A; ii) y[x := A] = y; iii) (y. M)[x := A] = y. M[x := A]; iv) (M N)[x := A] = (M[x := A] N[x := A]). One can observe that (x. x A) = A, (1) and (x. y A) = y. (2) Further, due to (*) (x. y. M A) -> (y. M)[x := A] = y. M[x := A] <- y. (x. M A), hence (x. y. M A) = y. (x. M A). (3) Finally, using (*) one can write (x. (M N) A) -> (M N)[x := A] = (M[x := A] N[x := A]) <- ((x. M A) (x. N A)), hence (x. (M N) A) = ((x. M A) (x. N A)). (4) Usage of the rules (1)-(4) could simplify construction of a machine for automated evaluation of lambda expressions: if each of the above rules is applied within one clock cycle, then after every cycle the memory will contain a valid expression. I will be very grateful for any information about where this or a similar approach to reduction of lambda expressions was considered in the literature. === Subject: Eigenvalue of Linear Partial Diff Operator Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I have a diff operator acting on real-valued functions (I give it in tex, the ascii is an impossible task for me, sorry) $PDEop:=frac{partial^2}{partialTheta^2}+cotThetafrac{partial } {partialTheta}+frac{1}{sin^2Theta}left(frac{partial^2} {partial phi^2}-2cosThetafrac{partial^2}{partialtaupartial phi}+frac{partial^2}{partial^2}{partialtau^2}right)$ Here 0<= tau < 2 pi, 0<= phi <2 pi, 0<= Theta < pi. How can we prove that its eigenvalues are real number, my conjecture is they are <=0 real numbers? I know and I have some books in the ordinary diff eg case, but my knowledge in the partial diff eq case is little. Sandor