mm-4429 === Subject: Re: A Question About Recursion > Is arithmetic or mathematics possible without it? Yep. Anything that can be done recursively can be done non-recursively. === Subject: Re: A Question About Recursion How can pi be iterated non-recursively? >> Is arithmetic or mathematics possible without it? Yep. Anything that can be done recursively can be done non-recursively. > === Subject: Re: A Question About Recursion > How can pi be iterated non-recursively? I'm not sure what you mean. Do you mean calculating pi to N digits without using recursion? Such as this sequence: pi/2 = 1 + 1/3 + (1*2)/(3*5) + (1*2*3)/(3*5*7) + ... which doesn't need to recurse to calculate. You simply build each term off the preceeding one and watching when the digits converge to the desired N. This is trivial with unlimited precision rationals and produces no accumulating roundoff errors. For example, my Python program that uses this technique can produce any number of digits in any base from 2 to 36: def pialso(n,b): # input number of digits (n) in requested base (b) # prints the digits and how many iterations it took > pialso(100,10) 3141592653589793238462643383279502884197169399375105820974944592307816406286 208998628034825342117067 327 > pialso(100,12) 3184809493b918664573a6211bb151551a05729290a7809a492742140a60a55256a0661a0375 3a3aa54805646880181a3682 353 > Is arithmetic or mathematics possible without it? Yep. Anything that can be done recursively can be done non-recursively === Subject: Re: Logarithm question Frederick Williams transported these words from his/hers keyboard to our eyes: > Further: even if it weren't a mathematical term that circumstance should > not be used to attack the op; he probably isn't a mathematician and is > just using a term he is familiar with. If the term is inappropriate Exactly ;) And I'm thinking to myself, who should I ask if not mathematicians? And I almost get my head chopped off while the rest of the folkes here try to make me use their vocabulary. And I thought I was nitpicking... > (and I don't think it is) by all means tell him what term he should > use. Actually, I think the important thing will turn out to be what > base of logarithms he is using. I think it's base ten, so antilog x =/= > exp x. Yap, base 10. So, basically, it can be done? I like hagman's rule of thumb. -- Where are you going? I am leaving, you are about to explode. === Subject: Re: Logarithm question > Frederick Williams transported these words > from his/hers keyboard to our eyes: Further: even if it weren't a mathematical term that circumstance should > not be used to attack the op; he probably isn't a mathematician and is > just using a term he is familiar with. If the term is inappropriate Exactly ;) And I'm thinking to myself, who should I ask if not > mathematicians? And I almost get my head chopped off while the rest of > the folkes here try to make me use their vocabulary. And I thought I was > nitpicking... > Yep - you were struck by a pedant. A pedant is a dull and inefficient weapon, most recognisable by the slow sucking sound it makes when striking an innocent body. Mike === Subject: Re: Logarithm question hagman transported these words from his/hers keyboard to our eyes: > Rule of thumb: If the difference is more than 10dB, you can safely > ignore the lower sound source. -- J: Teal'c, how about a little introduction of some sort? S: There is no need. Bra'tac has told me much of the Tauri. You are O'Neill, Teal'c's apprentice. J: Yeah... apprentice? S: Also a warrior of great skill and cunning. J: Apprentice? === Subject: Re: Space marker on the moon idea: Writing Big Letters on The Moon for Proof of our visit ;) How big is big enough ? reply-type=response How BIG do letters not to be written in the sand on the moon to be visible > from earth by telescopes ? So that we visitors of the moon can proof to people on the earth that we > or something has been there to anybody with a telescope ? ;) Good question for astronomers me thinks ? Need help from mathematicians ? Okdokie me include alt.math too :):):) (I call this idea a space marker ;) ) Bye, > Skybuck. > You don't need much maths, its a pretty simple calculation. If you let w = wavelength of the light you are using d = distance to moon a = aperture of the telescope t = thickness of the lines in the letters s = size of letter, say five times the line thickness then w/a = t/d t =dw/a s=5dw/a Now, we know 5dw= 5* (500)*10^-9 * 4*10^8 metres = 10^4 * 10^-9 * 10^8 = 10^3 So the size needed is s=10^3/a Here are the results for various aperture sizes: a = 5 mm (unaided eyesight), size = 200 kms a=50 mm (pair of binoculars), size = 20 kms a= 30 cms (reasonable amateur telescope) = 3 kms a = 10 metres (the world's largest telescope) = 100 metres. Peter Webb If you assume that the letters are five times as big as the width of the letters === Subject: Re: Is 2^(2^13466918-4)*(2^13466917-1) a big number? >> Define what you mean by big. That's not a standard math term!! Good question! > But that is exactly the purpose of the post. I am wondering what types of BIG NUMBERS > are being examined in the fields of mathematics > and how this one fits in the line-up. Can anyone assist? > Ta. Pete I have seen this question in several newsgroups so I have cross-posted this reply to three others. In the group of web sites called the Caldwell Prime Pages I found a Glossary with some terms used to categorize prime numbers as big. I suppose you could use the same terms to categorize all numbers as has been done for primes. The web site, http://primes.utm.edu/glossary/page.php?sort=Megaprime, has these meanings listed (I have paraphrased): a {titanic prime} has 1000 or more digits a {gigantic prime} has 10,000 or more digits a {megaprime} has 1,000,000 or more digits I suggest that what is meant by a big number depends on your context. For any particular purpose, to define a big number, first decide what is the {smallest big number}. A big number is then defined to be any number that big or larger. As suggested by the web site, most numbers are big. Note that most numbers are bigger than any number that is named. The symbols 2^(2^13466918 -4) * (2^13466917 -1) can be said to name a number. Most numbers are bigger than this number. To decide whether 2^(2^13466918 -4) * (2^13466917 -1) is a {big number}, any number you choose could be defined as the smallest {big number}. -- Dan in NY (for email, exchange y with g in dKlinkenbery at hvc dot rr dot com) === Subject: Re: Is 2^(2^13466918-4)*(2^13466917-1) a big number? > number is the number: 2^(2^13466918-4)*(2^13466917-1). > perfectly balanced, having precisely 2^(2^13466918-4)*(2^13466917-1) > ordered factorizations. If p is an odd prime then 2^(2p-2)*p is balanced. The record history for the largest known prime is at http://primes.utm.edu/notes/by_year.html The 6 latest records and still largest known: Prime Digits Month Year 2^13466917-1 4053946 November 2001 2^25964951-1 7816230 February 2005 2^30402457-1 9152052 December 2005 2^32582657-1 9808358 September 2006 All are Mersenne primes found by GIMPS. whether larger balanced numbers are known for other forms of numbers. > I am wondering what types of BIG NUMBERS are being examined in the fields > of mathematics and how this one fits in the line-up. Big number is not mathematically defined but without special context I would call 2^(2^13466918-4)*(2^13466917-1) big. http://en.wikipedia.org/wiki/Large_numbers mentions some numbers most people would probably call big. > The web site, http://primes.utm.edu/glossary/page.php?sort=Megaprime, has > these meanings listed (I have paraphrased): a {titanic prime} has 1000 or more digits > a {gigantic prime} has 10,000 or more digits > a {megaprime} has 1,000,000 or more digits I suggest that what is meant by a big number depends on your context. The terms titanic prime and gigantic prime were coined at a time where such primes were hard to find and few were known: http://primes.utm.edu/glossary/page.php?sort=TitanicPrime http://primes.utm.edu/glossary/page.php?sort=GiganticPrime Today a titanic prime can be found in seconds on a normal PC and a gigantic prime in less than an hour. The names for those sizes would probably not have been chosen today - at least not by the person who made them. Megaprime for 1,000,000 digits is of course because mega means 1,000,000 and not just because mega is sometimes used about big things. -- Jens Kruse Andersen === Subject: Re: Is 2^(2^13466918-4)*(2^13466917-1) a big number? <46f11b02$0$28858$4c368faf@roadrunner.com> perfectly balanced, having precisely 2^(2^13466918-4)*(2^13466917-1) > ordered factorizations. If p is an odd prime then 2^(2p-2)*p is balanced. > The record history for the largest known prime is athttp://primes.utm.edu/notes/by_year.html The 6 latest records and still largest known: > Prime Digits Month Year > 2^13466917-1 4053946 November 2001 > 2^25964951-1 7816230 February 2005 > 2^30402457-1 9152052 December 2005 > 2^32582657-1 9808358 September 2006 > All are Mersenne primes found by GIMPS. whether larger balanced numbers are known for other forms of numbers. I am wondering what types of BIG NUMBERS are being examined in the fields > of mathematics and how this one fits in the line-up. Big number is not mathematically defined but without special context I > would call 2^(2^13466918-4)*(2^13466917-1) big.http://en.wikipedia.org/wiki/Large_numbersmentions some numbers most > people would probably call big. The web site,http://primes.utm.edu/glossary/page.php?sort=Megaprime, has > these meanings listed (I have paraphrased): a {titanic prime} has 1000 or more digits > a {gigantic prime} has 10,000 or more digits > a {megaprime} has 1,000,000 or more digits I suggest that what is meant by a big number depends on your context. The terms titanic prime and gigantic prime were coined at a time where > such primes were hard to find and few were known:http://primes.utm.edu/glossary/page.php?sort=TitanicPrimehttp://primes . utm.edu/glossary/page.php?sort=GiganticPrime Today a titanic prime can be found in seconds on a normal PC and a > gigantic prime in less than an hour. The names for those sizes would > probably > not have been chosen today - at least not by the person who made them. Megaprime for 1,000,000 digits is of course because mega means > 1,000,000 and not just because mega is sometimes used about big things. So, if named today, titanic prime would be kiloprime and gigantic prime would be...decakiloprime? -- > Jens Kruse Andersen === Subject: wats interviewed referred to a bare interest rate figure as did the interviewer. Your complaint addressed Radio New Zealand 's coverage of the loan sharking issue in our Late Edition news bulletins on Friday the 17th of August. Your concern appears to be that the interest rates quoted were not referenced to a particular time period. Radio New Zealand notes that the Minister of the Crown interviewed referred to a bare interest rate figure as did the interviewer. no such word dictionary.com internet. 21/9/07. complaint. wat does it mean per annum week day etc. loan shark. they should tell me when i asked. don.lotto nz. It is considered that radio audiences are used to interest rates being quoted as bare figures without them being referenced to a time period. This is the case whether were are speaking of mortgage rates, inflation rates, or loan charges and avoids confusing our audience. In light of the above no inaccuracy was found and your complaint was not upheld. I am obliged by the Broadcasting Act to inform you of the reason why your complaint was not upheld; the letter above is intended to answer that obligation. I am also obliged to inform you of your right to refer this decision for review to the Broadcasting Standards Authority, P O Box 9213 , Wellington . You must do so within 20 working days if you wish to, otherwise the Authority will be time- barred by statute from accepting the reference. Yours sincerely George === Subject: RenContRes rencontresRenContRes rencontres rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes http://www.marocbreak.com http://www.cinebreak.com === Subject: rencontres RenContResrencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes http://www.marocbreak.com http://www.cinebreak.com === Subject: RenContRes rencontresRenContRes rencontres rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes rencontres RenContRes http://www.marocbreak.com http://www.cinebreak.com === Subject: Who Is Interested in Using Math to Help Conservation and Conservation Biology? Ben Fusaro tried to create an environmental math session at the joint there any interest out there? Here is my abstract. Please respond also via email. HOW MATHEMATICS CAN HELP CONSERVATION BIOLOGY (AND VICE VERSA) Mathematics enjoys the highest consensus of any science. Conservation biology, on the other hand, is fuzzy and suffers considerable controversy. They are the perfect couple! Mathematics can lend conservation biologists much-needed rigor and credibility. Conservation biology can supply mathematicians with a way to satisfy their need for ethical engagement in the real world -- to feel useful in a highly important way. Examples: (1) The Axiomatic Method: create a parsimonious set of basic assumptions, from which the principles of conservation can be deduced. E.g., we value genetic diversity. It follows that wildlife (at least in their reproductive years) should not be killed. (2) The concepts of Finiteness and Infinitude: when we are young, we feel, and act as if, we will live forever. As we get older, we come to realize just how finite our lifespan and energy are. Some policies implicitly assume that habitat and genetic resources are infinite (e.g. every development is okay, because the wildlife can always find someplace else to live), whereas in fact they are finite. (3) Probability: Explaining evolutionary gradualism vs. saltation: a mutation of one base pair can effect apparently gradual or quantum change, depending on the importance of the gene affected. -- I am working on creating wildlife habitat that is off-limits to humans (pure habitat). Want to help? (I spent the previous 8 years fighting auto dependence and road construction.) Please don't put a cell phone next to any part of your body that you are fond of! http://home.pacbell.net/mjvande 7694 === Subject: Modern Trigonometry 1) Domain and range for m(x)=sin^-1 (x - (x^2/2)) 2)Write sh x and ch x in mac-laurin series. 3)One projectil launched with inicial velocity (v0) and with inclination angle (b) makes an horizontal distance given by the formula r = (1/4.9).vo^2.(1/2)sin(2b) Determinate b in order to maximize r 4)Study f(x): R-E]-1,1[ x-E tgh x === Subject: Re: Modern Trigonometry Range y= sin ^-1 (x - x^2/2) <=> sin y = x- x^2/2 <=> -x^2+2x-2sin y =0 <=> x = 1 -+ sqrt (1-2sin y) <=> sin y C 0 {y E [-pi/2,pi/2] and sin y C 1/2 } D'm = [-pi/2,pi/6] Must a modern quality === Subject: Re: Modern Trigonometry >1) Domain and range for m(x)=sin^-1 (x - (x^2/2)) x - x^2/2 = -1 x^2 - 2x - 2 = 0 Domain: 1 - sqrt(3) <= x <= 1 + sqrt(3) Range: you find 2)Write sh x and ch x in mac-laurin series. Tedious. 3)One projectil launched with inicial velocity (v0) and >with inclination angle (b) makes an horizontal distance given by the formula r = (1/4.9).vo^2.(1/2)sin(2b) Determinate b in order to maximize r By inspection, sin(2b) = 1 2b = pi/2, b = pi/4 = 45 degrees > === Subject: Re: Modern Trigonometry 4. tanh x= ln (x+sqrt(1+x^2)/ln (x+sqrt(x^2-1) = ex-e(-x)/ex+e(-x),xER Positive if x>0,negative if x<0,if x=0,tanh x=0 Lim tanh x = 1 Lim tanh x = -1 Lim tanh x/x =0 1+tanh^2 x = sech^2 x d/dx [tanh v] = 1/cosh^2 (v) * d/dx[v] tanh(-x)=-tanh (x) === Subject: *~*~*Forex Your Hot Double Money*~*~* Cc: FOREX Forex Secretes INVESTIP = [I]nvestment [N]ews, [V]iews and [E]xpose for [S]tock [T]iming and [I]nsightful [P]rofits. The Investip(tm) group is dedicated to the exchange of practical, usable information and ideas relating to the stock market and individual stocks, including timing, strategies, and tactics, utilizing stocks, options and indices, both long and short http://forexonlineinvestment.blogspot.com === Subject: I need Electronic Devices & Electronic Principles Solutions I need Electronic Devices by Floyd Conventional Current Edition 8th Ed & Electronic Principles Solutions by Malvino 7th edition Solution's Manual === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I NEED THE SOLUTIONS TO 8TH EDITION!!!!!!!!!!!!! Fundamentals of Physics by Halliday, 8th Ed., Walker, Resnick === Subject: Looking For solution Manual I need the solution manual for Theory of Vibration with Applications (5th Edition). Can somebody please help? hmuchiri@mail.com === Subject: Advancing surrogate factoring I've noted that every integer factorization can be connected to some other factorization with simple congruences, and have talked about a method for factoring by going in one direction with that connection, but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an integer, such that all congruences are satisfied and that if S = p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of squares. Here is an example. Let the target composite to be factored be 77, so S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you have sqrt(S/2), while it can be much smaller than that, so you have progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, work, as it did with my example above, or can you have cases where abs(x)>T? If so that could be a problem with this method, otherwise it is guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection goes in BOTH DIRECTIONS, where I've focused on one direction until now. James Harris === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. James Harris Let's give this a shot, for example, factoring 13837. Then by your algorithm I should do the following: S = 13837, k = sqrt(floor(S/2)) = 83. T = S - 2k^2 = 13837 - 2(83^2) = 13837 - 13778 = 59. So then I factor 59, but 59 is prime! So then what do I do? x = (2)^(-1) k mod T. 2^-1 mod 59 is easily found as 30, so we get 30*k mod 59 = 30*83 mod 59 = 12. So how do I use that then to factor 13,837? How do I use this? You don't explain enough of your algorithm! :( Do you really want people to believe you have something really hot and awesome or not?! If this really goes somewhere then you've got to explain it ALL and show that it does. I'm stuck right on the above. === Subject: Re: Advancing surrogate factoring I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. Let's give this a shot, for example, factoring 13837. > Then by your algorithm I should do the following: S = 13837, k = sqrt(floor(S/2)) = 83. T = S - 2k^2 = 13837 - 2(83^2) = 13837 - 13778 = 59. So then I factor 59, but 59 is prime! So then > what do I do? x = (2)^(-1) k mod T. 2^-1 mod > 59 is easily found as 30, so we get 30*k mod 59 = > 30*83 mod 59 = 12. So how do I use that then to > factor 13,837? How do I use this? You don't explain > enough of your algorithm! :( Nimrod. He says what to do next, to calculate: > sqrt(((x+k)/2)^2 + 4*S) But it turns out this isn't an integer. (This formula might be a misprint, though. What you're supposed to do is to solve the equation m^2 + (x+k) m + S = 0 which is m^2 + 99 m + 13837 = 0 here, but there are no integer solutions to this equation, either. If there ARE integer solutions, they give you a factorization of S.) The algorithm dies here. --- Christopher Heckman > Do you really want people to believe you have > something really hot and awesome or not?! If this > really goes somewhere then you've got to explain > it ALL and show that it does. I'm stuck right on the > above. === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. Let's give this a shot, for example, factoring 13837. > Then by your algorithm I should do the following: S = 13837, k = sqrt(floor(S/2)) = 83. T = S - 2k^2 = 13837 - 2(83^2) = 13837 - 13778 = 59. So then I factor 59, but 59 is prime! So then > what do I do? x = (2)^(-1) k mod T. 2^-1 mod > 59 is easily found as 30, so we get 30*k mod 59 = > 30*83 mod 59 = 12. So how do I use that then to > factor 13,837? How do I use this? You don't explain > enough of your algorithm! :( Nimrod. He says what to do next, to calculate: > I didn't look at it enough, I guess. > sqrt(((x+k)/2)^2 + 4*S) But it turns out this isn't an integer. (This formula might be a > misprint, though. What you're supposed to do is to solve the equation m^2 + (x+k) m + S = 0 which is > m^2 + 99 m + 13837 = 0 here, > 99? The math I did gave x = 12, k = 83, so that should be 95. Regardless, she still don't have integer solutions. > but there are no integer solutions to this equation, either. If there > ARE integer solutions, they give you a factorization of S.) The > algorithm dies here. > *some* numbers (namely those for when there *are* such integer solutions), but is by no means a general purpose factoring algorithm, and hence I do not think James will be able to produce the factorization of an RSA challenge number with it. > --- Christopher Heckman Do you really want people to believe you have > something really hot and awesome or not?! If this > really goes somewhere then you've got to explain > it ALL and show that it does. I'm stuck right on the > above. === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, Do you need to factor anything? Why not just let T = S - 2k^2 ? If you do factor T, what do you do with the factors? > and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) This should be: sqrt((2*(x + k))^2 - 4 * S). to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. > It's clear from the errors noted above that you are again just carelessly throwing this out without any kind of systematic testing. Why don't you do that on a series of 100 or so small composites (for example, all the products of two primes which are both less than 50) and then report back the results in a table? Marcus. > And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. James Harris === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. > James Harris > It sounds to me like its more of an advanced case of retarditis.... === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). Probably ought to be m^2 + 18m + 77. I suspect JSH got lucky here, though. Try S = 32861419 (chosen at random). k = floor(sqrt(32861419/2)) = 4053. T = S - 2k^2 = 7801. x = 2^{-1}*k mod T = 3901 * 4053 = 5927. x+k = 9980, so I have to factor m^2 + 9980 m + 32861419 which has no real roots. (Oops!) m^2 + 9980 m - 32861419 has no integral roots, either. > So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, Oh, wait, I didn't do that. T = 29: x = 2^{-1}*k mod T = 11. x+k = 4064 so I have to factor m^2 + 4064 m + 32861419, which has no integer roots (neither does ... - 32861419). T = 269: x = 2^{-1}*k mod T = 9 x+k = 4062 so I have to factor m^2 + 4062 m + 32861419, which has no integer roots (neither does ... - 32861419). > and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. But there's no guarantee it works; see above. --- Christopher Heckman > Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist an x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. James Harris === Subject: Re: Advancing surrogate factoring I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). Probably ought to be m^2 + 18m + 77. I suspect JSH got lucky here, though. Try S = 32861419 (chosen at ( 3319 )( 9901 ) No, it's not about luck but about the size of the prime factors relative to k. That should have been obvious to you. > random). k = floor(sqrt(32861419/2)) = 4053. T = S - 2k^2 = 7801. x = 2^{-1}*k mod T = 3901 * 4053 = 5927. x+k = 9980, so I have to factor m^2 + 9980 m + 32861419 which has no real roots. (Oops!) There probably is one close though, if you loop through values for x, remember x = 2^{-1} k mod T so you're moving in multiples of 7801. Try it and see what happens. m^2 + 9980 m - 32861419 has no integral roots, either. So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, Oh, wait, I didn't do that. T = 29: x = 2^{-1}*k mod T = 11. > x+k = 4064 so I have to factor m^2 + 4064 m + 32861419, > which has no integer roots (neither does ... - 32861419). T = 269: x = 2^{-1}*k mod T = 9 > x+k = 4062 so I have to factor m^2 + 4062 m + 32861419, > which has no integer roots (neither does ... - 32861419). and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. But there's no guarantee it works; see above. --- Christopher Heckman > Yup there's a search involved and it's pointless to factor S-2k^2, but simply use it as is. And it's a mistake to use a small one, so the minimum I was looking for was a bad idea. Try a much bigger k, which gives a T greater than either of the prime factors of your target. Better yet, before you try, predict what it will do. James Harris === Subject: Re: Advancing surrogate factoring How did you factor that? > No, it's not about luck but about the size of the prime factors > relative to k. That should have been obvious to you. > simply use it as is. > How does the complexity of this search go, anyway? Is it polynomial time? If it's exponential time then your algorithm is no better than anything else out there. How large is the search for, say, a 1024-bit RSA product of strong primes? === Subject: Re: Advancing surrogate factoring relative to k. That should have been obvious to you. Yup there's a search involved and it's pointless to factor S-2k^2, but > simply use it as is. How does the complexity of this search go, anyway? > Is it polynomial time? It WAS exponential time, in the number of digits, but he decided to change repeat by factoring T with repeat by factoring _every factor of T_, and so the new algorithm might not be polynomial time any more. This is because T <= sqrt(S/2), so T will have half the digits of S. Recursion where you remove a constant number of digits of your numbers is enough to guarantee polynomial time, IIRC. --- Christopher Heckman > If it's exponential time > then your algorithm is no better than anything > else out there. How large is the search for, say, a 1024-bit RSA > product of strong primes? === Subject: Re: Advancing surrogate factoring > relative to k. That should have been obvious to you. Yup there's a search involved and it's pointless to factor S-2k^2, but > simply use it as is. How does the complexity of this search go, anyway? > Is it polynomial time? It WAS exponential time, That should be It was POLYNOMIAL time, of course. --- CCH > in the number of digits, but he decided to > change repeat by factoring T with repeat by factoring _every factor > of T_, and so the new algorithm might not be polynomial time any > more. This is because T <= sqrt(S/2), so T will have half the digits > of S. Recursion where you remove a constant number of digits of your numbers > is enough to guarantee polynomial time, IIRC. --- Christopher Heckman If it's exponential time > then your algorithm is no better than anything > else out there. How large is the search for, say, a 1024-bit RSA > product of strong primes? === Subject: Re: Advancing surrogate factoring relative to k. That should have been obvious to you. Yup there's a search involved and it's pointless to factor S-2k^2, but > simply use it as is. How does the complexity of this search go, anyway? > Is it polynomial time? It WAS exponential time, That should be It was POLYNOMIAL time, of course. --- CCH === Subject: Re: Advancing surrogate factoring > And it's a mistake to use a small one, so the minimum I was looking > for was a bad idea. Try a much bigger k, which gives a T greater than either of the prime > factors of your target. Better yet, before you try, predict what it will do. It is just Monkey Mathematics. > James Harris > === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). Probably ought to be m^2 + 18m + 77. I suspect JSH got lucky here, though. Try S = 32861419 (chosen at ( 3319 )( 9901 ) Yes, that is the answer, but I didn't get it from the algorithm that you posted. That was your claim, right? That the POSTED ALGORITHM factors integers? Or are we talking about a new algorithm here? > No, it's not about luck but about the size of the prime factors > relative to k. That should have been obvious to you. random). k = floor(sqrt(32861419/2)) = 4053. T = S - 2k^2 = 7801. x = 2^{-1}*k mod T = 3901 * 4053 = 5927. x+k = 9980, so I have to factor m^2 + 9980 m + 32861419 which has no real roots. (Oops!) There probably is one close though, if you loop through values for x, > remember x = 2^{-1} k mod T so you're moving in multiples of 7801. Try it and see what happens. Why? I've shown that the algorithm you posted is wrong. I shouldn't have to do your work and come up with variations of it. > m^2 + 9980 m - 32861419 has no integral roots, either. So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, Oh, wait, I didn't do that. T = 29: x = 2^{-1}*k mod T = 11. > x+k = 4064 so I have to factor m^2 + 4064 m + 32861419, > which has no integer roots (neither does ... - 32861419). T = 269: x = 2^{-1}*k mod T = 9 > x+k = 4062 so I have to factor m^2 + 4062 m + 32861419, > which has no integer roots (neither does ... - 32861419). and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. But there's no guarantee it works; see above. Yup there's a search involved and it's pointless to factor S-2k^2, but > simply use it as is. And it's a mistake to use a small one, so the minimum I was looking > for was a bad idea. I guess that's as close to I was wrong as we'll get out of you ... > Try a much bigger k, which gives a T greater than either of the prime > factors of your target. That's a stupid thing to say, because I don't *know* the factors ahead of time. If I want to choose a k which is greater than the factors of T, I'd have to rule out some small prime factors, less than M let's say, then choose k = T/M. If you choose k too big, though, you'll lose the polynomial running- time. > Better yet, before you try, predict what it will do. I predicted that your example was a lucky one, and that I could find a counterexample in the first three numbers I chose at random. I was right. (This was literally the first number I tried.) --- Christopher Heckman > James Harris === Subject: Re: Advancing surrogate factoring I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). Probably ought to be m^2 + 18m + 77. I suspect JSH got lucky here, though. Try S = 32861419 (chosen at ( 3319 )( 9901 ) Yes, that is the answer, but I didn't get it from the algorithm that > you posted. That was your claim, right? That the POSTED ALGORITHM > factors integers? Or are we talking about a new algorithm here? > You didn't read the original post carefully enough as I hedged on whether or not x would always be less than abs(T), and it turns out it's not. > No, it's not about luck but about the size of the prime factors > relative to k. That should have been obvious to you. random). k = floor(sqrt(32861419/2)) = 4053. T = S - 2k^2 = 7801. x = 2^{-1}*k mod T = 3901 * 4053 = 5927. x+k = 9980, so I have to factor m^2 + 9980 m + 32861419 which has no real roots. (Oops!) There probably is one close though, if you loop through values for x, > remember x = 2^{-1} k mod T so you're moving in multiples of 7801. Try it and see what happens. Why? I've shown that the algorithm you posted is wrong. I shouldn't > have to do your work and come up with variations of it. > Sigh. Then don't bother. James Harris === Subject: Re: Advancing surrogate factoring >> I've noted that every integer factorization can be connected to >> some >> other factorization with simple congruences, and have talked about >> a >> method for factoring by going in one direction with that >> connection, >> but why not go the other? >> Given >> x^2 = y^2 mod T >> where T is a non-zero integer, introducing k = 2x mod T, you can >> easily solve to find >> (x+k)^2 = y^2 + 2k^2 mod T >> so with S = 2k^2 mod T, you have a second difference of squares. >> But what if S is your target composite to be factored? >> Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a >> factor of S - 2k^2. >> Then x = 2^{-1} k mod T, gives you x. >> It can be shown that only for certain cases then will y exist as an >> integer, such that all congruences are satisfied and that if S = >> p_1*p_2 that case is equivalent to >> 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. >> And then it is trivial to factor S now through a difference of >> squares. >> Here is an example. Let the target composite to be factored be 77, >> so >> S = 77. >> Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so >> k >> = 6 gives me a minimum, so >> S - 2k^2 = 77 - 72 = 5. >> So T = 5. >> Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. >> Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you >> can >> non-trivially factor by solving the quadratic formula as >> m^2 + 18 + 77 = (m+7)(m+11). >> Probably ought to be m^2 + 18m + 77. >> I suspect JSH got lucky here, though. Try S = 32861419 (chosen at >> ( 3319 )( 9901 ) >> Yes, that is the answer, but I didn't get it from the algorithm that >> you posted. That was your claim, right? That the POSTED ALGORITHM >> factors integers? Or are we talking about a new algorithm here? You didn't read the original post carefully enough as I hedged on > whether or not x would always be less than abs(T), and it turns out > it's not. > you mean you were just guessing, hacking and haven't got a clue why? > No, it's not about luck but about the size of the prime factors >> relative to k. >> That should have been obvious to you. >> random). >> k = floor(sqrt(32861419/2)) = 4053. >> T = S - 2k^2 = 7801. >> x = 2^{-1}*k mod T = 3901 * 4053 = 5927. >> x+k = 9980, so I have to factor >> m^2 + 9980 m + 32861419 >> which has no real roots. (Oops!) >> There probably is one close though, if you loop through values for x, >> remember >> x = 2^{-1} k mod T >> so you're moving in multiples of 7801. >> Try it and see what happens. >> Why? I've shown that the algorithm you posted is wrong. I shouldn't >> have to do your work and come up with variations of it. Sigh. Then don't bother. Because, JSH cannot remember WTF he is doing or posting. > James Harris > === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to >> some >> other factorization with simple congruences, and have talked about >> a >> method for factoring by going in one direction with that >> connection, >> but why not go the other? >> Given >> x^2 = y^2 mod T >> where T is a non-zero integer, introducing k = 2x mod T, you can >> easily solve to find >> (x+k)^2 = y^2 + 2k^2 mod T >> so with S = 2k^2 mod T, you have a second difference of squares. >> But what if S is your target composite to be factored? >> Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a >> factor of S - 2k^2. >> Then x = 2^{-1} k mod T, gives you x. >> It can be shown that only for certain cases then will y exist as an >> integer, such that all congruences are satisfied and that if S = >> p_1*p_2 that case is equivalent to >> 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. >> And then it is trivial to factor S now through a difference of >> squares. >> Here is an example. Let the target composite to be factored be 77, >> so >> S = 77. >> Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so >> k >> = 6 gives me a minimum, so >> S - 2k^2 = 77 - 72 = 5. >> So T = 5. >> Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. >> Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you >> can >> non-trivially factor by solving the quadratic formula as >> m^2 + 18 + 77 = (m+7)(m+11). > Probably ought to be m^2 + 18m + 77. > I suspect JSH got lucky here, though. Try S = 32861419 (chosen at >> ( 3319 )( 9901 ) > Yes, that is the answer, but I didn't get it from the algorithm that > you posted. That was your claim, right? That the POSTED ALGORITHM > factors integers? Or are we talking about a new algorithm here? > You didn't read the original post carefully enough as I hedged on >> whether or not x would always be less than abs(T), and it turns out >> it's not. >> you mean you were just guessing, hacking and haven't got a clue why? >> No, it's not about luck but about the size of the prime factors >> relative to k. >> That should have been obvious to you. > random). > k = floor(sqrt(32861419/2)) = 4053. > T = S - 2k^2 = 7801. > x = 2^{-1}*k mod T = 3901 * 4053 = 5927. > x+k = 9980, so I have to factor > m^2 + 9980 m + 32861419 > which has no real roots. (Oops!) >> There probably is one close though, if you loop through values for x, >> remember >> x = 2^{-1} k mod T >> so you're moving in multiples of 7801. >> Try it and see what happens. > Why? I've shown that the algorithm you posted is wrong. I shouldn't > have to do your work and come up with variations of it. > Sigh. Then don't bother. Because, JSH cannot remember WTF he is doing or posting. (1/2)-Crackpot New Factoring Ideas: (1) Try neural networks (2) Try artificial intelligence (3) Try Bayesian inference, advanced statistics (4) ? >> James Harris >> === Subject: Re: Advancing surrogate factoring > There probably is one close though, if you loop through values for x, > remember x = 2^{-1} k mod T so you're moving in multiples of 7801. Try it and see what happens. Loop through the values of x? Then you have to increase k as well or else break your beautiful math. Are you doing that as well or just keeping k constant? > Try a much bigger k, which gives a T greater than either of the prime > factors of your target. Better yet, before you try, predict what it will do. AFAICT, your surrogate factoring algorithm essentially consists of pull a k out of thin air and it magically works. You have specified several different ways to generate k, and, upon being shown that the current method for k doesn't quite match up to your claims, you either change the mystical processes to generate your variables or you change your claims. I have yet to see a sufficiently coherent description of surrogate factoring to be able to write a program that can competently do it, let alone compare it to the current state-of-the-art factoring methods. === Subject: Re: Advancing surrogate factoring remember x = 2^{-1} k mod T so you're moving in multiples of 7801. Try it and see what happens. Loop through the values of x? Then you have to increase k as well or > else break your beautiful math. Are you doing that as well or just > keeping k constant? > k = 2x mod T so you can loop through values for x while keeping k constant. Intriguingly the answer that is coming in now from all my research paths with the primary congruences is that k should be LARGE. For instance to factor dudes target composite, I used my amp'd up program with a much larger than usual k, and here's the output. Begin program output: n_min=-32767 n_max=4095 n_max-n_min=36862 Surrogate factorization: factored? Yes. ( 5 )( 6297751 ) Product: 31488755 Surrogate factorization: factored? Yes. ( 2 )( 3 )( 7 )( 547 )( 2801 ) Product: 64350174 Number factored. k=+/-733758 n=-32766 n_max=4095 Total all combinations: 301 Time: 30 Time/combination: 0.09966777408637874 Surrogate: ( 2 )( 3 )( 7 )( 547 )( 2801 ) Product: 64350174 Surrogate combinations checked: 7 Initial Factorization: f_1=3319 f_2=9901 Now checking its factors... Success! Factors: ( 3319 )( 9901 ) Product: 32861419 In coming is 32861419 Surrogate factorization data for target: Surrogates factored : 2 Surrogates not factored : 0 Factored fuel percentage: 100% Data about all surrogates including those from recursions: Factored fuel : 39 Fuel not factored: 0 Factored fuel percentage: 100% Processing time: 562 Number of digits: 8 bitLength=25 -----------End program output---------- Notice that it took only 2 surrogates to factor!!! And looped through only 7 combinations or what other posters call probes. The difference I am increasingly certain is the size of k. This latest idea may be even more powerful though and not require any surrogate factoring at all, while still relying on the primary surrogate factoring congruence: (x+k)^2 = y^2 + 2k^2 mod T as simply going the other way and letting S = 2k^2 mod T, and solving for the difference of squares THAT way removes the need to factor anything as you can just let T = S - 2k^2 and maybe get a surprisingly simple answer about how to force a non- trivial factorization that totally blows my mind, so I find it impossible to believe even at this point. And renders any method based on factoring as a hard problem so sickeningly obsolete, you wonder how anyone could have thought they work. Or how they did for so long. Belief? Did RSA encryption work for as long as it did just because mathematicians were so convincing? James Harris === Subject: Re: Advancing surrogate factoring > There probably is one close though, if you loop through values for x, > remember > x = 2^{-1} k mod T > so you're moving in multiples of 7801. > Try it and see what happens. >> Loop through the values of x? Then you have to increase k as well or >> else break your beautiful math. Are you doing that as well or just >> keeping k constant? >> k = 2x mod T Suppose k = 1000 and T = 21. I derive that 2x must equal 13, which means x must equal 7.5 Unless you mean that k = 2x % T ? Additionally, you could also loop through k while keeping x constant. === Subject: Re: Advancing surrogate factoring > There probably is one close though, if you loop through values for x, > remember > x = 2^{-1} k mod T > so you're moving in multiples of 7801. > Try it and see what happens. >> Loop through the values of x? Then you have to increase k as well or >> else break your beautiful math. Are you doing that as well or just >> keeping k constant? k = 2x mod T Suppose k = 1000 and T = 21. I derive that 2x must equal 13, which means > x must equal 7.5 Unless you mean that k = 2x % T ? If T is odd, then there is a number y such that 2*y = 1 (mod T); in particular, y = (T + 1)/2 in normal (non-mod) arithmetic. This means x = 2^(-1)*k mod T = 11*1000 mod 21 = 17, and sure enough 2*17 = 13 mod 21. --- Christopher Heckman > Additionally, you could also loop through k while keeping x constant. === Subject: Re: Advancing surrogate factoring >> There probably is one close though, if you loop through values for x, >> remember >> GET OUT RACIST. GO HELP OJ STAY OUT OF PRIZON, BIGOT. === Subject: Re: Advancing surrogate factoring JSHs depth of understanding is like a monkey looking at coconuts. === Subject: Re: Advancing surrogate factoring > Notice that it took only 2 surrogates to factor!!! JSHs depth of understanding is like a monkey looking at coconuts. Amos, are you insuaten that Mr Feeble@Algebra is a flippin Chimp, a Monkey Boy? And he'b dosen't know what to do with a Coconuts?? Andy, well Is'been a thinken JSH be a throwback to the darkie ages, one o dem thar trolls, da de find at the bottom of a pool at midnight in the dark, blackness, no white at all, cause dey white boys got all the maths learnted alreadys. === Subject: JSH: Mixed result? I felt really good about my latest twist on surrogate factoring but thinking about it more I realize it may work best when prime factors are all roughly the same size and it is unknown to me how it would behave if one prime factor is very much greater than the others. Also, oddly enough, I think it may work best if you do NOT factor S - 2k^2 at all, but simply take it as it is, maybe even needing it to be rather large, especially if one prime factor is very much greater than sqrt(S). So the picture is muddled to me now when a little while before I was ready yet again to trumpet the demise of the impasse where mathematicians worldwide have so far managed to avoid the truth that my research is revolutionary and correct while they maintain the status quo--including teaching wrong info to trusting students. With that said, it is a new direction that I came up with hours ago, and experience shows that as time goes on the picture clears. So then, it's still about time. I pursue the demonstration that will end the impasse. And the math community lies. Eventually though I feel I will catch you, and the world will just see one more scandal to add to all the others. Most people probably won't even yawn, let alone be surprised to hear that the mathematical community is riddled with lying and corruption, with a lot of fake math on which plenty of professors depend, so they lied as long as they could. Too many liars in our world these days for most people to get too excited I'd think. You people will be just another soap opera. James Harris === Subject: Re: Mixed result? >I felt really good about my latest twist on surrogate factoring but Heard you joined the Black Panthers? Blaming WHITE people because you are too DUMB to understand simple algebra? Ever hear of the Bell Curve? Did you know that bigots are on the lower end of the bell curve too? You bigotry Proves you are *stupid*. === Subject: Re: Mixed result? >I felt really good about my latest twist on surrogate factoring but lying scumbag troll racial white hating bigot, GTF out. === Subject: Re: Mixed result? >I felt really good about my ....being a RACIST Just stay in the ter, dirtbag, racial bigot, post your hate spew in alt.HATE.ALL.WHITES. === Subject: Re: JSH: Mixed result? >I felt really good about my latest twist on surrogate factoring but >thinking about it more I realize it may work best when prime factors >are all roughly the same size and it is unknown to me how it would >behave if one prime factor is very much greater than the others. It may work best and it may not. All that means to me is that you need to do more work on it before posting it here. Also, oddly enough, I think it may work best if you do NOT factor S - >2k^2 at all, but simply take it as it is, maybe even needing it to be >rather large, especially if one prime factor is very much greater than >sqrt(S). It may and it may not. Again you are posting prematurely. So the picture is muddled to me now when a little while before I was >ready yet again to trumpet the demise of the impasse where >mathematicians worldwide have so far managed to avoid the truth that >my research is revolutionary and correct while they maintain the >status quo--including teaching wrong info to trusting students. With that said, it is a new direction that I came up with hours ago, >and experience shows that as time goes on the picture clears. Hours ago. You need to do days or weeks of work on this until you have answered all those may's and might's. You lose credibility by making grandiose claims that later prove to be false. Try doing more checking before making the claims. So then, it's still about time. I pursue the demonstration that will >end the impasse. That _may_ end the impasse. And the math community lies. So far the math community has been correct about your surrogate factoring methods. Eventually though I feel I will catch you, and the world will just see >one more scandal to add to all the others. Most people probably won't even yawn, let alone be surprised to hear >that the mathematical community is riddled with lying and corruption, >with a lot of fake math on which plenty of professors depend, so they >lied as long as they could. Too many liars in our world these days for most people to get too >excited I'd think. You people will be just another soap opera. Coronation Street or East Enders? rossum >James Harris === Subject: Re: JSH: Mixed result? making grandiose claims that later prove to be false. Note: Nothing that James does can possibly lower his credibility. - William Hughes === Subject: Re: JSH: Mixed result? >> ... You lose credibility by >> making grandiose claims that later prove to be false. Note: Nothing that James does can possibly lower his credibility. - William Hughes Sorry, I should have made it clear that credibility can take negative values. rossum === Subject: Re: JSH: Mixed result? Too many liars in our world these days for most people to get too > excited I'd think. James Harris Are you lying now were you lying then or are you lying both times? You a liar, charlatan, cheat, crank and troll. You start varying parameters - someone provides a counterexample before the cyberbits had time to settle - and then you start damage control. What gibberish! You have no clue and keep moving your moving target about into another flawed guess. I think it might be more fruitful for you to choose random factors of a challenge number and test it manually - that would be far superior to the absolute nonsense you keep trying to sell the math community. You have zero integrity and are a true example of scum - you angry racist!!! === Subject: Re: JSH: Mixed result? > I felt really good about my latest twist on surrogate factoring You shouldn't: It doesn't work. S = 32861419. Full details at the Advancing surrogate factoring thread. --- Christopher Heckman > but > thinking about it more I realize it may work best when prime factors > are all roughly the same size and it is unknown to me how it would > behave if one prime factor is very much greater than the others. Also, oddly enough, I think it may work best if you do NOT factor S - > 2k^2 at all, but simply take it as it is, maybe even needing it to be > rather large, especially if one prime factor is very much greater than > sqrt(S). So the picture is muddled to me now when a little while before I was > ready yet again to trumpet the demise of the impasse where > mathematicians worldwide have so far managed to avoid the truth that > my research is revolutionary and correct while they maintain the > status quo--including teaching wrong info to trusting students. With that said, it is a new direction that I came up with hours ago, > and experience shows that as time goes on the picture clears. So then, it's still about time. I pursue the demonstration that will > end the impasse. And the math community lies. Eventually though I feel I will catch you, and the world will just see > one more scandal to add to all the others. Most people probably won't even yawn, let alone be surprised to hear > that the mathematical community is riddled with lying and corruption, > with a lot of fake math on which plenty of professors depend, so they > lied as long as they could. Too many liars in our world these days for most people to get too > excited I'd think. You people will be just another soap opera. James Harris === Subject: solutions manual needed Anyone have the solutions manual to Moaveni, Finite Element Analysis, 3rd edition? Rob === Subject: Help with a Differential Equations problem please (probably easy) Hey all, Not sure if this is where I'm supposed to post this (tell me if I need to post it somewhere else), but I'd like some help on a beginner's differential equation problem. Anyway, the prob. is to just solve the diff. equation (using substitution): (y^2 + yx)dx - (x^2)(dy) = 0 so this is a homogeneous eq. to the 2nd order I started it out by substituting y = ux which gives me: (u^2 * x^2 + u*x^2)dx - x^2(udx + xdu) = 0 anyway, after using an identity for homogeneous equations, I eventually end up with: (dx/x) + du/(u^2 + 2u) which seems to always lead to the wrong answer no matter how I integrate from there. any help would be greatly appreciated. === Subject: Re: Help with a Differential Equations problem please (probably easy) > Anyway, the prob. is to just solve the diff. equation (using substitution): > (y^2 + yx)dx - (x^2)(dy) = 0 > so this is a homogeneous eq. to the 2nd order I started it out by substituting y = ux which gives me: > (u^2 * x^2 + u*x^2)dx - x^2(udx + xdu) = 0 x^2 y' = y^2 + xy y' = (y/x)^2 + y/x u = y/x; y = ux y' = u'x + u u'x + u = u^2 + u u'x = u^2 u'/u^2 = 1/x -1/u^3 = log x + c === Subject: Re: Help with a Differential Equations problem please (probably easy) <16139751.1190174914621.JavaMail.jakarta@nitrogen.mathforum.org>, > Hey all, > Not sure if this is where I'm supposed to post this (tell me if I need to > post it somewhere else), but I'd like some help on a beginner's differential > equation problem. Anyway, the prob. is to just solve the diff. equation (using substitution): > (y^2 + yx)dx - (x^2)(dy) = 0 > so this is a homogeneous eq. to the 2nd order I started it out by substituting y = ux which gives me: > (u^2 * x^2 + u*x^2)dx - x^2(udx + xdu) = 0 anyway, after using an identity for homogeneous equations, I eventually end > up with: (dx/x) + du/(u^2 + 2u) That + 2 u is wrong. > which seems to always lead to the wrong answer no matter how I integrate from > there. > any help would be greatly appreciated. === Subject: Re: Help with a Differential Equations problem please (probably easy) Hmmm... I don't know what I'm doing wrong to get there. Here's my steps... Starting with (u^2 * x^2 + u*x^2)dx - x^2(udx + xdu) = 0 I use the fact that it's a homogeneous 2nd order to re write in the form of x^n * M(1,u)dx + x^n * N(1,u)(udx + xdu) = 0 where n = the order M(1,u) = used in place of M(x,y), which is the left term of the equation N(1,u) = same as M(1,u) except for the right side After that I get (x^2)(u^2 + u)dx - (x^2)(udx + xdu) = 0 Then I get to the form dx/x + [N(1,u)du]/[M(1,u) + u*N(1,u)] Which gets me the (dx/x) + du/(u^2 + 2u) Where am I going wrong here? === Subject: Re: Help with a Differential Equations problem please (probably easy) >Hmmm... I don't know what I'm doing wrong to get there. Here's my steps... >Starting with (u^2 * x^2 + u*x^2)dx - x^2(udx + xdu) = 0 I use the fact that it's a homogeneous 2nd order to re write in the form of x^n * M(1,u)dx + x^n * N(1,u)(udx + xdu) = 0 >where n = the order >M(1,u) = used in place of M(x,y), which is the left term of the equation >N(1,u) = same as M(1,u) except for the right side After that I get >(x^2)(u^2 + u)dx - (x^2)(udx + xdu) = 0 Then I get to the form >dx/x + [N(1,u)du]/[M(1,u) + u*N(1,u)] Which gets me the >(dx/x) + du/(u^2 + 2u) Where am I going wrong here? N(1,u) = -1 ALso, crosspost, don't multipost (like in sci.math). === Subject: =?windows-1256?B?x83T5CDh2sjJIMrU2+Eg3e3lxyDH4czjzOPJINXNINHHyNgg5sfNzyDjyMf U 0SDmyM3M4yDV2+3RICjH4crH39PtIMfhw+XI4SAzKSDe1c/tIC4uLg==?= === Subject: Instructors' manuals for Engineering books I have the following solutions manuals in pdf. If you need any of them, send me email at modernbooks at hotmail dot com I accept paypal payments only * Solutions manual for Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray,Hurst, Lewis, Meyer * Solutions manual for Analytical Mechanics, 7th Edition, by Fowels, Cassiday * Solutions manual for An Interactive Introduction to Mathematical Analysis, by Jonathan Lewin * Solutions manual for An Introduction to Database Systems, 8th * Solutions manual for An Introduction to the Mathematics of Financial Derivatives, 2nd Ed.,by Neftci [ISBN:0125153929] * Solutions manual for Antenna Theory, 2nd Ed., by Balanis * Solutions manual for Antennas for all Applications, 3rd Edition, Kraus, Marhefka * Solutions manual for Applied Linear Statistical Models, 5th Ed., by Neter (Selected Sol.) * Solutions manual for Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley * Solutions manual for Applied Numerical Methods with MATLAB for Engineers and Scientists,1st Ed,. by Chapra * Solutions manual for Applied Statistics and Probability for Engineers, 3rd Ed., by Montgomery, Runger (Selected Solutions) * Solutions manual for Applied Strength of Materials, 4th Edition, by Mott * Solutions manual for A Transition to Advanced Mathematics, 5th Edition, by Smith, Eggen,Andre * Solutions manual for Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi * Solutions manual for A Course in Game Theory by Osborne, Rubinstein * Solutions manual for A Course in Algebraic Number Theory by Cohen * Solutions manual for Adaptive Filter Theory, 4th Edition, by Haykin * Solutions manual for Adaptive Control, 2nd. Ed., by Astrom, Wittenmark * Solutions manual for Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Macroeconomics, 1st Ed., by David Romer * Solutions manual for Advanced Mathematical Concepts Precalculus With Applications by Holliday [ISBN: 0028341759] * Solutions manual for Advanced Modern Engineering Mathematics, 3rd Ed., by G. James * Solutions manual for A First Course In Differential Equations, 7th Edition, by Zill, Cullen * Solutions manual for Analog Integrated Circuit Design, 1st Ed., by Johns, Martin (text ebook and solution manual) * Solutions manual for Basic Business Statistics: Concepts and Applications, 10th Ed., by Berenson, Krehbiel, Levine (chap1-18) * Solutions manual for Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin * Solutions manual for Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms (Missing a chapter or 2) * Solutions manual for Bioprocess Engineering Principles by Doran * Solutions manual for Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson,Hostetler, Edwards * Solutions manual for Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler * Solutions manual for Chemical Engineering Volume 1, 6th Edition, by Richardson, Coulson,Backhurst, Harker 5th Ed, by Marion, Thornton * Solutions manual for College Physics, Volume 1: 7th Edition, by Serway, Faugh * Solutions manual for College Physics, Volume 2: 7th Edition, by Serway, Faughn * Solutions manual for Communications Systems, 4th Ed., by Haykin * Solutions manual for Communications Systems Engineering, 2nd Edition, by Proakis * Solutions manual for Computational Techniques for Fluid Dynamics by Srinivas, Fletcher * Solutions manual for Computer Networks, 4th Ed., by Andrew S. Tanenbaum * Solutions manual for Computer Networks: A Systems Approach, 3rd Edition, by Davie * Solutions manual for Control Systems Engineering, 4th Ed., by Norman Nise * Solutions manual for Corporate Finance, 6th Edition, by Ross * Solutions manual for C++ How to Program: Intro Object-Oriented Design with the UML, 3rd Ed., by Deitel, Nieto * Solutions manual for Calculus Early Transcendental, 5th Ed., by James Stewart * Solutions manual for Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis * Solutions manual for Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney,Demana, Kennedy * Solutions manual for Calculus: Multivariable, 5th Edition, by James Stewart * Solutions manual for Calculus: Single Variable, Early Transcendental, 5th Edition, by James Stewart * Solutions manual for Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett,McCallum * Solutions manual for Device Electronics for * Solutions manual for Integrated Circuits 3rd Edition by Muller * Solutions manual for Differential Equations with Boundary Value Problems, 2nd Ed., by Polking, Arnold * Solutions manual for Digital And Analog Communication Systems 7th Ed., Leon W. Couch * Solutions manual for Digital Communications, 4th Edition, by Proakis * Solutions manual for Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar * Solutions manual for Digital Design, 4th Edition, by Mano, Ciletti * Solutions manual for Digital Image Processing, 2nd Edition, by Gonzalez, Woods * Solutions manual for Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for Chapters 3, 5, 6, 10) * Solutions manual for Digital Signal Processing: A Computer Based Approach, 1st Ed., by Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: Principles, Algorithms and Applications,3rd Edition, by Proakis * Solutions manual for Discrete Time Signal Processing, 2nd Edition, Oppenheim * Solutions manual for Dynamics of Mechanical Systems by C.T.F. Ross * Solutions manual for Data and Computer Communications, 8th Edition by Stallings * Solutions manual for Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol.for Chapters 2-21, odd only) * Solutions manual for Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi Design of Analysis of Experiments, 6th Edition, Montgomery (missing chapter 6-8) * Solutions manual for Design of Machinery, 3rd Ed by Robert L. Norton * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 2nd Ed., by Sergio Franco * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 3rd Ed., by Sergio Franco * Solutions manual for Elementary Principles of Chemical Processes, 3rd Ed., by Felder,Rousseau * Solutions manual for Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler * Solutions manual for Engineering and Chemical Thermodynamics, by Koretsky [ISBN:0471385867] (No sol. for chapt 6) * Solutions manual for Engineering Circuit Analysis, 6th Edition, Hyat * Solutions manual for Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck * Solutions manual for Engineering Electromagnetics, 7th Ed., Hayt, Buck * Solutions manual for Engineering Fluids Mechanics 7th Edition by Crowe * Solutions manual for Engineering Fluids Mechanics 8th Edition by Crowe * Solutions manual for Engineering Mathematics, 4th Ed., by John Bird * Solutions manual for Engineer Mechanics: Dynamics, 4th Ed., by Bedford * Solutions manual for Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 11th Ed. by Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 5th Ed. by Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace Fowler * Solutions manual for Engineering Mechanics: Statics, 5th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 6th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Statics 11th Ed. by Hibbeler * Solutions manual for Experiments with Economic Principles by Bergstrom, Miller * Solutions manual for Econometric Analysis, 5th Edition, by Greene * Solutions manual for Econometric Analysis of Cross Section and Panel * Solutions manual for Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, Viceira * Solutions manual for Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar * Solutions manual for Electric Circuits 6th Ed. by Nilsson * Solutions manual for Electric Circuits 7th Ed. by Nilsson * Solutions manual for Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans * Solutions manual for Electric Machinery Fundamentals, 4th Ed by Chapman * Solutions manual for Electromagnetic Fields and Waves by Iskander * Solutions manual for Electronic Circuit Analysis, 2nd Ed., by Donald Neamen * Solutions manual for Electronics, 2nd Ed., by Allan R. Hambley * Solutions manual for Elementary Differential Equations, 8th Edition, by Boyce, DiPrima(some odd/even) * Solutions manual for Fundamentals of Applied Electromagnetics, 5th Ed., 2008 Media Edition,by Ulaby * Solutions manual for Fundamentals of Digital Logic with Verilog Design, 1st Edition, by Brown, Vranesic * Solutions manual for Fundamentals of Electric Circuits, 2nd Edition, by Alexander * Solutions manual for Fundamentals of Electromagnetics with Engineering Appls by Wentworth * Solutions manual for Fundamentals of Fluid Mechanics, 5th Ed. by Munson, Young.. * Solutions manual for Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 5th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... * Solutions manual for Fundamentals of Logic Design, 5th Ed., by Roth Jr. * Solutions manual for Fundamentals of Machine Component Design, 3rd Ed., by Juvinall * Solutions manual for Fundamentals of Machine Component Design, 4th Ed., by Juvinall * Solutions manual for Fundamentals of Machine Elements, 2nd Ed., Hamrock, Jacobson, Schmid * Solutions manual for Fundamentals of Physics by Halliday, 7th Ed., Walker, Resnick * Solutions manual for Fundamentals of Semiconductor Devices, 1st Edition by Anderson * Solutions manual for Fundamentals of Structural Analysis, 2nd Ed., Chia-Ming Uang, Kenneth Leet * Solutions manual for Fundamentals of Thermal-Fluid Sciences, 2nd Ed. by Cengel * Solutions manual for Fundamentals of Thermal-fluid Sciences, Int'l 2nd Ed. by Cengel * Solutions manual for Fundamentals of Engineering Thermodynamics, 5th Ed. by Shapiro * Solutions manual for Fundamentals of Thermodynamics, 5th Ed., by Sonntag, Borgnakke... * Solutions manual for Fundamentals of Thermodynamics, 6th Ed., by Sonntag * Solutions manual for Facilities Planning, 3rd Edition, by Tompkins, White, Bozer, Tanchoco * Solutions manual for Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- Naeini * Solutions manual for Financial Accounting, 4th Ed., by Libby, Short (Chap1-14) * Solutions manual for Financial Accounting: An International Introduction, 2nd Ed., by Alexander, Nobes * Solutions manual for Finite Element Techniques in Structural Mechanics by Ross * Solutions manual for Fluid Mechanics - 5th Edition by Frank M. White * Solutions manual for Fluid Mechanics and Thermodynamics of Turbomachinery, 5th Ed., by S. L. Dixon [ISBN: 0750678704] * Solutions manual for Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., by Cengel & Cimbala * Solutions manual for Fluid Mechanics with Engineering Applications, 10th Edition, by Finnemore * Solutions manual for Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. * Solutions manual for Fundamentals of Applied Electromagnetics, 1st Ed., 2001 Media Edition, by Ulaby * Solutions manual for Geometry, 04 Edition, by McGraw-Hill [ISBN: 0078296374] * Solutions manual for Guide to Energy Management, 5th Edition, by Pawlik * Solutions manual for Heat Transfer: A Practical Approach - 2nd Edition by Cengel * Solutions manual for Hydraulics in Civil and Environmental Engineering, 4th Ed., by Andrew Chadwick Applications by Oz Shy Introduction to Algorithms, 2nd Ed by Cormen, Leiserson (Selected Sol.) * Solutions manual for Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van Ness, Smith, Abbott * Solutions manual for Introduction to Electric Circuits, 6th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electric Circuits, 7th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electrodynamics, 3rd Ed. by David Griffiths * Solutions manual for Introduction to Fluid Mechanics - 5th Ed. by Fox.. * Solutions manual for Introduction to Fluid Mechanics - 6th Ed by Fox, McDonald... * Solutions manual for Introduction to Linear Algebra, 3rd Ed., by Gilbert Strang * Solutions manual for Introduction to Linear Algebra, 5th Ed., Arnold, Johnson, Riess * Solutions manual for Introduction to Probability by Grinstead, Snell (odd solutions only, not just answers but step by step solutions) * Solutions manual for Introduction to Quantum Mechanics, 2nd Ed. by Griffiths * Solutions manual for Introdution to Solid State Physics, 8th Edition by Kittel * Solutions manual for Introduction to Statistical Quality Control, 4th Edition, by Montgomery * Solutions manual for Introduction to Thermal Systems Engineering: Thermodynamics, Fluid Mechanics, and Heat Transfer by Moran, Shapiro, Munson, DeWitt * Solutions manual for Introduction to Thermal Systems Engineering, by Moran, Shapiro * Solutions manual for Linear Algebra, by J. Hefferon Linear Algebra And Its Applications, 3rd Ed., by David C. Lay * Solutions manual for Linear Algebra with Applications, 2nd Edition - by Otto Bretscher * Solutions manual for Linear Algebra with Applications, 3rd Edition - by Otto Bretscher * Solutions manual for Linear Circuit Analysis: Time Domain, Phasor and Laplace.., 2nd Ed, Lin * Solutions manual for Machine Design: An Integrated Approach, 2nd Ed., by Robert L. Norton * Solutions manual for Machine Design: An Integrated Approach, 3rd Ed., by Robert L. Norton * Solutions manual for Managerial Accounting, 11th Ed., by Noreen, Brewer, Garrison * Solutions manual for Materials Science and Engineering: An Introduction, 6th Ed. by Callister * Solutions manual for Matrix Analysis and Applied Linear Algebra by Carl Meyer * Solutions manual for MC68HC11: An Introduction: Software/Hardware Interf, 2nd Ed, by Huang * Solutions manual for Mechanical Engineering Design, 7th Ed. by Mischke, Shigley * Solutions manual for Mechanical Vibrations, 3rd Edition, by S. S. Rao (99% same as 4th Edition, No Solutions for Chapters 6, 9, and 12) * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Fluids, 4th Ed., Irving H. Shames * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Materials - 3rd Ed. by Beer, Johnston, Dewolf * Solutions manual for Mechanics of Materials - 6th Ed. by Hibbeler * Solutions manual for Mechanics of Materials, 6th Edition by James M. Gere (missing small portion, section 8.5) * Solutions manual for Mechanics of Materials, 6th Ed., by Sturges, Morris, Riley (part of Chapt 2 is missing but only #1 thru #60) * Solutions manual for Mechanics of Solids by C.T.F. Ross Microeconomic Analysis, 3rd Ed., by H. Varian (Ans. to Exercises: Ch. 1- Ch.25) * Solutions manual for Microeconomic Theory, by Mas-Colell, Whinston, Green * Solutions manual for Microelectronic Circuit Analysis and Design, 3rd Edition, by D. Neamen * Solutions manual for Microelctronic Circuits, 5th Ed. by Sedra and Smith * Solutions manual for Microelectronic Circuit Design, 2nd Edition by Jaeger, Blalock * Solutions manual for Microelectronic Circuit Design, 3rd Edition by Jaeger, Blalock * Solutions manual for Microelectronics: Digital and Analog Circuits and Systems by Millman * Solutions manual for Microwave and Rf Design of Wireless Systems, 1st Edition, by Pozar * Solutions manual for Microwave Engineering, 3rd Ed., by David M. Pozar * Solutions manual for Microwave Transistor Amplifiers: Analysis and Design, 2nd Ed., by Guillermo Gonzalez Miller & Freund's Probability and Statistics for Engineers, 7th Edition, Johnson, Miller * Solutions manual for Modern Compressible Flow, 3rd Edition, by Anderson * Solutions manual for Modern Control Engineering, 3rd Edition, by Ogata * Solutions manual for Modern Control Engineering, 4th Edition, by Ogata * Solutions manual for Modern Digital and Analog Communication Systems, 3rd Ed., by Lathi * Solutions manual for Modern Control Systems, 9th Ed., by Richard C. Dorf, Robert H Bishop (98% same as the 10th Ed.) * Solutions manual for Modern Operating Systems,2nd Ed., by Andrew Tanenbaum * Solutions manual for Modern Physics 4th Edition by Tipler * Solutions manual for Monetary Theory and Policy, 2nd Edition, by Walsh * Solutions manual for Multivariable Calculus, 5th Edition, by James Stewart * Solutions manual for Operating Systems: Internals and Design Principles, 4th Edition, by Stallings * Solutions manual for Operating System Concepts, 7th Ed., Silberschatz, Galvin, Gagne * Solutions manual for Options, Futures and Other Derivatives, 4th Ed., by John Hull * Solutions manual for Options, Futures and Other Derivatives, 5th Ed., by John Hull (Chapters 1 thru 18 ONLY) * Solutions manual for Orbital Mechanics: For Engineering Students by Howard Curtis (includes matlab scripts) * Solutions manual for Organic Chemistry, 4th Ed., by Carey, Atkins (Student Study Guide and Sol. Man.) * Solutions manual for Partial Differential Equations with Fourier Series and Boundary Value * Solutions manual for Problems, 2nd Ed., by Asmar (Student Solutions Manual) * Solutions manual for Physical Chemistry - 7th Edition - by Julio de Paula, Peter Atkins * Solutions manual for Physics, 6th Edition, by John Cutnell * Solutions manual for Physics, 5th Edition, Vol 2 by Halliday, Resnick, Krane (Chap 25-52) * Solutions manual for Physics for Scientist and Engineers by Knight (No Chapt 36-42) * Solutions manual for Physics for Scientist and Engineers, 6th Ed., by Serway * Solutions manual for Physics for Scientists and Engineers-Vol 1, 5th Edition, Serway, Beichner (Chap. 1 - 22) * Solutions manual for Physics for Scientists and Engineers-Vol 2, 5th Edition, Serway, Beichner (Chap. 23 - 46) * Solutions manual for Physics for Scientists and Engineers, 3rd Ed., by Douglas C. Giancoli * Solutions manual for Physics for Scientist and Engineers, 5th Edition, by Tipler, Mosca * Solutions manual for Physics: Principles with Applications, 6th Ed. by Giancoli * Solutions manual for Power System Analysis and Design, 3rd Ed., by Glover, Sarma * Solutions manual for Principles and Applications of Electrical Engineering 4th (Revised) Edition by Rizzoni * Solutions manual for Principles and Practices of Automatic Process Control, 3rd Edition by Smith, Corripio [ISBN: 0471431907] * Solutions manual for Principles of Communication: Systems, Modulation Noise, 5th Ed., Ziemer * Solutions manual for Principles of Physics, 3rd Edition, by Serway * Solutions manual for Principles of Physics, 4th Edition, by Serway * Solutions manual for Principles of Statics, 10th Ed., by Russell C. Hibbeler [ISBN: 0131866745] * Solutions manual for Probability and Statistics for Engineers and Scientists, 3rd Edition, Hayter * Solutions manual for Probability and Statistics for Engineering and the Sciences, 6th Ed., by Jay L. Devore * Solutions manual for Probability Random Variables, and Stochastic Processes, 4th Ed., by Papoulis, Pillai * Solutions manual for Quantum Mechanics: An Accessible Introduction, 1st Ed., by Robert Scherrer * Solutions manual for Recursive Macroeconomic Theory, 1st Ed., by Ljungqvist, Sargent * Solutions manual for Recursive Methods in Economic Dynamics, (2002) by Irigoyen, Rossi- Hansberg, Wright * Solutions manual for RF Circuit Design: Theory & Applications, by Bretchko, Ludwig * Solutions manual for Sears and Zemansky's University Physics 11th Edition by Young.. * Solutions manual for Semiconductor Device Fundamentals by Pierret * Solutions manual for Semiconductor Devices: Physics and Technology, 2nd Ed., S.M. Sze * Solutions manual for Semiconductor Physics And Devices -3rd Ed. by D. Neamen * Solutions manual for Separation Process Principles, 2nd Ed., Seader, Henley * Solutions manual for Signal Processing and Linear Systems by Lathi * Solutions manual for Signals and Systems, 2nd Edition, by Haykin, Van Veen * Solutions manual for Signals and Systems, 2nd Edition, Oppenheim, Willsky, Hamid, Nawab * Solutions manual for Signals and Systems: Analysis Using Transform Methods and MATLAB, 1st Ed., by M. J. Roberts * Solutions manual for Signals, Systems, and Transforms, 3rd Ed., by Charles L. Phillips, Eve A. Riskin, John M. Parr * Solutions manual for Shigley's Mechanical Engineering Design, 8th Ed. by Budynas, Nisbett (No Sol. for Chapt 18 & 19) * Solutions manual for Simply C#: An Application-Driven Tutorial Approach, by Deitel, Hoey (Chapters 1-32) * Solutions manual for Soil Mechanics: Concepts and Applications, 2nd Ed., by Powrie * Solutions manual for Solid State Electronic Devices - 5th Ed by Streetman * Solutions manual for Solid State Electronic Devices - 6th Ed by Streetman * Solutions manual for Statics and Mechanics of Materials: An Integrated Approach, 2nd Ed., by Riley, Sturges, Morris * Solutions manual for Structural Analysis, 5th Edition, by Hibbeler * Solutions manual for University Physics 11th Edition by Young.. * Solutions manual for Vector Mechanics: Statics 7th Edition by Beer * Solutions manual for Vector Mechanics: Dynamics, 7th Ed., by Beer, Johnston, Staab, Clausen * Solutions manual for Vibrations and Stability: Advanced Theory, Analysis, and Tools, 7th Ed., by Thomsen * Solutions manual for Wireless Communications: Principles and Practice, 2nd Ed, by Rappaport * Solutions manual for Theory and Design for Mechanical Measurements, 4th Ed., Beasley, Figliola * Solutions manual for Thermal Physics, 2nd Edition, by Charles Kittel * Solutions manual for Thermal Physics, by Ralph Baierlein * Solutions manual for Thermodynamics: An Engineering Approach, 5th Ed., by Cengel, Boles (Missing solutions #118-149 of Chapter 7) * Solutions manual for Thermodynamics: An Engineering Approach, 6th Ed., by Cengel, Boles * Solutions manual for The Science and Engineering of Materials, 4th Ed., by Donald R. Askeland, Pradeep P. Phule * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 10th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 10th Ed. (Multivariable, chs. 8-13), by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 11th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 11th Ed. (Multivariable, chs. 11-16), by * Solutions manual for Thomas, Weir, Hass, Giordano * Solutions manual for Transport Phenomena, 1st Edition, by R. Byron Bird * Solutions manual for Transport Phenomena, 2nd Ed., by Bird. Contact me at modernbooks@hotmail.com === Subject: Instructors' manuals for Engineering books I have the following solutions manuals in pdf. If you need any of them, send me email at modernbooks at hotmail dot com I accept paypal payments only * Solutions manual for Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray,Hurst, Lewis, Meyer * Solutions manual for Analytical Mechanics, 7th Edition, by Fowels, Cassiday * Solutions manual for An Interactive Introduction to Mathematical Analysis, by Jonathan Lewin * Solutions manual for An Introduction to Database Systems, 8th * Solutions manual for An Introduction to the Mathematics of Financial Derivatives, 2nd Ed.,by Neftci [ISBN:0125153929] * Solutions manual for Antenna Theory, 2nd Ed., by Balanis * Solutions manual for Antennas for all Applications, 3rd Edition, Kraus, Marhefka * Solutions manual for Applied Linear Statistical Models, 5th Ed., by Neter (Selected Sol.) * Solutions manual for Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley * Solutions manual for Applied Numerical Methods with MATLAB for Engineers and Scientists,1st Ed,. by Chapra * Solutions manual for Applied Statistics and Probability for Engineers, 3rd Ed., by Montgomery, Runger (Selected Solutions) * Solutions manual for Applied Strength of Materials, 4th Edition, by Mott * Solutions manual for A Transition to Advanced Mathematics, 5th Edition, by Smith, Eggen,Andre * Solutions manual for Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi * Solutions manual for A Course in Game Theory by Osborne, Rubinstein * Solutions manual for A Course in Algebraic Number Theory by Cohen * Solutions manual for Adaptive Filter Theory, 4th Edition, by Haykin * Solutions manual for Adaptive Control, 2nd. Ed., by Astrom, Wittenmark * Solutions manual for Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Macroeconomics, 1st Ed., by David Romer * Solutions manual for Advanced Mathematical Concepts Precalculus With Applications by Holliday [ISBN: 0028341759] * Solutions manual for Advanced Modern Engineering Mathematics, 3rd Ed., by G. James * Solutions manual for A First Course In Differential Equations, 7th Edition, by Zill, Cullen * Solutions manual for Analog Integrated Circuit Design, 1st Ed., by Johns, Martin (text ebook and solution manual) * Solutions manual for Basic Business Statistics: Concepts and Applications, 10th Ed., by Berenson, Krehbiel, Levine (chap1-18) * Solutions manual for Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin * Solutions manual for Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms (Missing a chapter or 2) * Solutions manual for Bioprocess Engineering Principles by Doran * Solutions manual for Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson,Hostetler, Edwards * Solutions manual for Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler * Solutions manual for Chemical Engineering Volume 1, 6th Edition, by Richardson, Coulson,Backhurst, Harker 5th Ed, by Marion, Thornton * Solutions manual for College Physics, Volume 1: 7th Edition, by Serway, Faugh * Solutions manual for College Physics, Volume 2: 7th Edition, by Serway, Faughn * Solutions manual for Communications Systems, 4th Ed., by Haykin * Solutions manual for Communications Systems Engineering, 2nd Edition, by Proakis * Solutions manual for Computational Techniques for Fluid Dynamics by Srinivas, Fletcher * Solutions manual for Computer Networks, 4th Ed., by Andrew S. Tanenbaum * Solutions manual for Computer Networks: A Systems Approach, 3rd Edition, by Davie * Solutions manual for Control Systems Engineering, 4th Ed., by Norman Nise * Solutions manual for Corporate Finance, 6th Edition, by Ross * Solutions manual for C++ How to Program: Intro Object-Oriented Design with the UML, 3rd Ed., by Deitel, Nieto * Solutions manual for Calculus Early Transcendental, 5th Ed., by James Stewart * Solutions manual for Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis * Solutions manual for Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney,Demana, Kennedy * Solutions manual for Calculus: Multivariable, 5th Edition, by James Stewart * Solutions manual for Calculus: Single Variable, Early Transcendental, 5th Edition, by James Stewart * Solutions manual for Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett,McCallum * Solutions manual for Device Electronics for * Solutions manual for Integrated Circuits 3rd Edition by Muller * Solutions manual for Differential Equations with Boundary Value Problems, 2nd Ed., by Polking, Arnold * Solutions manual for Digital And Analog Communication Systems 7th Ed., Leon W. Couch * Solutions manual for Digital Communications, 4th Edition, by Proakis * Solutions manual for Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar * Solutions manual for Digital Design, 4th Edition, by Mano, Ciletti * Solutions manual for Digital Image Processing, 2nd Edition, by Gonzalez, Woods * Solutions manual for Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for Chapters 3, 5, 6, 10) * Solutions manual for Digital Signal Processing: A Computer Based Approach, 1st Ed., by Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: Principles, Algorithms and Applications,3rd Edition, by Proakis * Solutions manual for Discrete Time Signal Processing, 2nd Edition, Oppenheim * Solutions manual for Dynamics of Mechanical Systems by C.T.F. Ross * Solutions manual for Data and Computer Communications, 8th Edition by Stallings * Solutions manual for Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol.for Chapters 2-21, odd only) * Solutions manual for Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi Design of Analysis of Experiments, 6th Edition, Montgomery (missing chapter 6-8) * Solutions manual for Design of Machinery, 3rd Ed by Robert L. Norton * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 2nd Ed., by Sergio Franco * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 3rd Ed., by Sergio Franco * Solutions manual for Elementary Principles of Chemical Processes, 3rd Ed., by Felder,Rousseau * Solutions manual for Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler * Solutions manual for Engineering and Chemical Thermodynamics, by Koretsky [ISBN:0471385867] (No sol. for chapt 6) * Solutions manual for Engineering Circuit Analysis, 6th Edition, Hyat * Solutions manual for Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck * Solutions manual for Engineering Electromagnetics, 7th Ed., Hayt, Buck * Solutions manual for Engineering Fluids Mechanics 7th Edition by Crowe * Solutions manual for Engineering Fluids Mechanics 8th Edition by Crowe * Solutions manual for Engineering Mathematics, 4th Ed., by John Bird * Solutions manual for Engineer Mechanics: Dynamics, 4th Ed., by Bedford * Solutions manual for Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 11th Ed. by Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 5th Ed. by Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace Fowler * Solutions manual for Engineering Mechanics: Statics, 5th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 6th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Statics 11th Ed. by Hibbeler * Solutions manual for Experiments with Economic Principles by Bergstrom, Miller * Solutions manual for Econometric Analysis, 5th Edition, by Greene * Solutions manual for Econometric Analysis of Cross Section and Panel * Solutions manual for Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, Viceira * Solutions manual for Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar * Solutions manual for Electric Circuits 6th Ed. by Nilsson * Solutions manual for Electric Circuits 7th Ed. by Nilsson * Solutions manual for Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans * Solutions manual for Electric Machinery Fundamentals, 4th Ed by Chapman * Solutions manual for Electromagnetic Fields and Waves by Iskander * Solutions manual for Electronic Circuit Analysis, 2nd Ed., by Donald Neamen * Solutions manual for Electronics, 2nd Ed., by Allan R. Hambley * Solutions manual for Elementary Differential Equations, 8th Edition, by Boyce, DiPrima(some odd/even) * Solutions manual for Fundamentals of Applied Electromagnetics, 5th Ed., 2008 Media Edition,by Ulaby * Solutions manual for Fundamentals of Digital Logic with Verilog Design, 1st Edition, by Brown, Vranesic * Solutions manual for Fundamentals of Electric Circuits, 2nd Edition, by Alexander * Solutions manual for Fundamentals of Electromagnetics with Engineering Appls by Wentworth * Solutions manual for Fundamentals of Fluid Mechanics, 5th Ed. by Munson, Young.. * Solutions manual for Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 5th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... * Solutions manual for Fundamentals of Logic Design, 5th Ed., by Roth Jr. * Solutions manual for Fundamentals of Machine Component Design, 3rd Ed., by Juvinall * Solutions manual for Fundamentals of Machine Component Design, 4th Ed., by Juvinall * Solutions manual for Fundamentals of Machine Elements, 2nd Ed., Hamrock, Jacobson, Schmid * Solutions manual for Fundamentals of Physics by Halliday, 7th Ed., Walker, Resnick * Solutions manual for Fundamentals of Semiconductor Devices, 1st Edition by Anderson * Solutions manual for Fundamentals of Structural Analysis, 2nd Ed., Chia-Ming Uang, Kenneth Leet * Solutions manual for Fundamentals of Thermal-Fluid Sciences, 2nd Ed. by Cengel * Solutions manual for Fundamentals of Thermal-fluid Sciences, Int'l 2nd Ed. by Cengel * Solutions manual for Fundamentals of Engineering Thermodynamics, 5th Ed. by Shapiro * Solutions manual for Fundamentals of Thermodynamics, 5th Ed., by Sonntag, Borgnakke... * Solutions manual for Fundamentals of Thermodynamics, 6th Ed., by Sonntag * Solutions manual for Facilities Planning, 3rd Edition, by Tompkins, White, Bozer, Tanchoco * Solutions manual for Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- Naeini * Solutions manual for Financial Accounting, 4th Ed., by Libby, Short (Chap1-14) * Solutions manual for Financial Accounting: An International Introduction, 2nd Ed., by Alexander, Nobes * Solutions manual for Finite Element Techniques in Structural Mechanics by Ross * Solutions manual for Fluid Mechanics - 5th Edition by Frank M. White * Solutions manual for Fluid Mechanics and Thermodynamics of Turbomachinery, 5th Ed., by S. L. Dixon [ISBN: 0750678704] * Solutions manual for Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., by Cengel & Cimbala * Solutions manual for Fluid Mechanics with Engineering Applications, 10th Edition, by Finnemore * Solutions manual for Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. * Solutions manual for Fundamentals of Applied Electromagnetics, 1st Ed., 2001 Media Edition, by Ulaby * Solutions manual for Geometry, 04 Edition, by McGraw-Hill [ISBN: 0078296374] * Solutions manual for Guide to Energy Management, 5th Edition, by Pawlik * Solutions manual for Heat Transfer: A Practical Approach - 2nd Edition by Cengel * Solutions manual for Hydraulics in Civil and Environmental Engineering, 4th Ed., by Andrew Chadwick Applications by Oz Shy Introduction to Algorithms, 2nd Ed by Cormen, Leiserson (Selected Sol.) * Solutions manual for Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van Ness, Smith, Abbott * Solutions manual for Introduction to Electric Circuits, 6th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electric Circuits, 7th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electrodynamics, 3rd Ed. by David Griffiths * Solutions manual for Introduction to Fluid Mechanics - 5th Ed. by Fox.. * Solutions manual for Introduction to Fluid Mechanics - 6th Ed by Fox, McDonald... * Solutions manual for Introduction to Linear Algebra, 3rd Ed., by Gilbert Strang * Solutions manual for Introduction to Linear Algebra, 5th Ed., Arnold, Johnson, Riess * Solutions manual for Introduction to Probability by Grinstead, Snell (odd solutions only, not just answers but step by step solutions) * Solutions manual for Introduction to Quantum Mechanics, 2nd Ed. by Griffiths * Solutions manual for Introdution to Solid State Physics, 8th Edition by Kittel * Solutions manual for Introduction to Statistical Quality Control, 4th Edition, by Montgomery * Solutions manual for Introduction to Thermal Systems Engineering: Thermodynamics, Fluid Mechanics, and Heat Transfer by Moran, Shapiro, Munson, DeWitt * Solutions manual for Introduction to Thermal Systems Engineering, by Moran, Shapiro * Solutions manual for Linear Algebra, by J. Hefferon Linear Algebra And Its Applications, 3rd Ed., by David C. Lay * Solutions manual for Linear Algebra with Applications, 2nd Edition - by Otto Bretscher * Solutions manual for Linear Algebra with Applications, 3rd Edition - by Otto Bretscher * Solutions manual for Linear Circuit Analysis: Time Domain, Phasor and Laplace.., 2nd Ed, Lin * Solutions manual for Machine Design: An Integrated Approach, 2nd Ed., by Robert L. Norton * Solutions manual for Machine Design: An Integrated Approach, 3rd Ed., by Robert L. Norton * Solutions manual for Managerial Accounting, 11th Ed., by Noreen, Brewer, Garrison * Solutions manual for Materials Science and Engineering: An Introduction, 6th Ed. by Callister * Solutions manual for Matrix Analysis and Applied Linear Algebra by Carl Meyer * Solutions manual for MC68HC11: An Introduction: Software/Hardware Interf, 2nd Ed, by Huang * Solutions manual for Mechanical Engineering Design, 7th Ed. by Mischke, Shigley * Solutions manual for Mechanical Vibrations, 3rd Edition, by S. S. Rao (99% same as 4th Edition, No Solutions for Chapters 6, 9, and 12) * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Fluids, 4th Ed., Irving H. Shames * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Materials - 3rd Ed. by Beer, Johnston, Dewolf * Solutions manual for Mechanics of Materials - 6th Ed. by Hibbeler * Solutions manual for Mechanics of Materials, 6th Edition by James M. Gere (missing small portion, section 8.5) * Solutions manual for Mechanics of Materials, 6th Ed., by Sturges, Morris, Riley (part of Chapt 2 is missing but only #1 thru #60) * Solutions manual for Mechanics of Solids by C.T.F. Ross Microeconomic Analysis, 3rd Ed., by H. Varian (Ans. to Exercises: Ch. 1- Ch.25) * Solutions manual for Microeconomic Theory, by Mas-Colell, Whinston, Green * Solutions manual for Microelectronic Circuit Analysis and Design, 3rd Edition, by D. Neamen * Solutions manual for Microelctronic Circuits, 5th Ed. by Sedra and Smith * Solutions manual for Microelectronic Circuit Design, 2nd Edition by Jaeger, Blalock * Solutions manual for Microelectronic Circuit Design, 3rd Edition by Jaeger, Blalock * Solutions manual for Microelectronics: Digital and Analog Circuits and Systems by Millman * Solutions manual for Microwave and Rf Design of Wireless Systems, 1st Edition, by Pozar * Solutions manual for Microwave Engineering, 3rd Ed., by David M. Pozar * Solutions manual for Microwave Transistor Amplifiers: Analysis and Design, 2nd Ed., by Guillermo Gonzalez Miller & Freund's Probability and Statistics for Engineers, 7th Edition, Johnson, Miller * Solutions manual for Modern Compressible Flow, 3rd Edition, by Anderson * Solutions manual for Modern Control Engineering, 3rd Edition, by Ogata * Solutions manual for Modern Control Engineering, 4th Edition, by Ogata * Solutions manual for Modern Digital and Analog Communication Systems, 3rd Ed., by Lathi * Solutions manual for Modern Control Systems, 9th Ed., by Richard C. Dorf, Robert H Bishop (98% same as the 10th Ed.) * Solutions manual for Modern Operating Systems,2nd Ed., by Andrew Tanenbaum * Solutions manual for Modern Physics 4th Edition by Tipler * Solutions manual for Monetary Theory and Policy, 2nd Edition, by Walsh * Solutions manual for Multivariable Calculus, 5th Edition, by James Stewart * Solutions manual for Operating Systems: Internals and Design Principles, 4th Edition, by Stallings * Solutions manual for Operating System Concepts, 7th Ed., Silberschatz, Galvin, Gagne * Solutions manual for Options, Futures and Other Derivatives, 4th Ed., by John Hull * Solutions manual for Options, Futures and Other Derivatives, 5th Ed., by John Hull (Chapters 1 thru 18 ONLY) * Solutions manual for Orbital Mechanics: For Engineering Students by Howard Curtis (includes matlab scripts) * Solutions manual for Organic Chemistry, 4th Ed., by Carey, Atkins (Student Study Guide and Sol. Man.) * Solutions manual for Partial Differential Equations with Fourier Series and Boundary Value * Solutions manual for Problems, 2nd Ed., by Asmar (Student Solutions Manual) * Solutions manual for Physical Chemistry - 7th Edition - by Julio de Paula, Peter Atkins * Solutions manual for Physics, 6th Edition, by John Cutnell * Solutions manual for Physics, 5th Edition, Vol 2 by Halliday, Resnick, Krane (Chap 25-52) * Solutions manual for Physics for Scientist and Engineers by Knight (No Chapt 36-42) * Solutions manual for Physics for Scientist and Engineers, 6th Ed., by Serway * Solutions manual for Physics for Scientists and Engineers-Vol 1, 5th Edition, Serway, Beichner (Chap. 1 - 22) * Solutions manual for Physics for Scientists and Engineers-Vol 2, 5th Edition, Serway, Beichner (Chap. 23 - 46) * Solutions manual for Physics for Scientists and Engineers, 3rd Ed., by Douglas C. Giancoli * Solutions manual for Physics for Scientist and Engineers, 5th Edition, by Tipler, Mosca * Solutions manual for Physics: Principles with Applications, 6th Ed. by Giancoli * Solutions manual for Power System Analysis and Design, 3rd Ed., by Glover, Sarma * Solutions manual for Principles and Applications of Electrical Engineering 4th (Revised) Edition by Rizzoni * Solutions manual for Principles and Practices of Automatic Process Control, 3rd Edition by Smith, Corripio [ISBN: 0471431907] * Solutions manual for Principles of Communication: Systems, Modulation Noise, 5th Ed., Ziemer * Solutions manual for Principles of Physics, 3rd Edition, by Serway * Solutions manual for Principles of Physics, 4th Edition, by Serway * Solutions manual for Principles of Statics, 10th Ed., by Russell C. Hibbeler [ISBN: 0131866745] * Solutions manual for Probability and Statistics for Engineers and Scientists, 3rd Edition, Hayter * Solutions manual for Probability and Statistics for Engineering and the Sciences, 6th Ed., by Jay L. Devore * Solutions manual for Probability Random Variables, and Stochastic Processes, 4th Ed., by Papoulis, Pillai * Solutions manual for Quantum Mechanics: An Accessible Introduction, 1st Ed., by Robert Scherrer * Solutions manual for Recursive Macroeconomic Theory, 1st Ed., by Ljungqvist, Sargent * Solutions manual for Recursive Methods in Economic Dynamics, (2002) by Irigoyen, Rossi- Hansberg, Wright * Solutions manual for RF Circuit Design: Theory & Applications, by Bretchko, Ludwig * Solutions manual for Sears and Zemansky's University Physics 11th Edition by Young.. * Solutions manual for Semiconductor Device Fundamentals by Pierret * Solutions manual for Semiconductor Devices: Physics and Technology, 2nd Ed., S.M. Sze * Solutions manual for Semiconductor Physics And Devices -3rd Ed. by D. Neamen * Solutions manual for Separation Process Principles, 2nd Ed., Seader, Henley * Solutions manual for Signal Processing and Linear Systems by Lathi * Solutions manual for Signals and Systems, 2nd Edition, by Haykin, Van Veen * Solutions manual for Signals and Systems, 2nd Edition, Oppenheim, Willsky, Hamid, Nawab * Solutions manual for Signals and Systems: Analysis Using Transform Methods and MATLAB, 1st Ed., by M. J. Roberts * Solutions manual for Signals, Systems, and Transforms, 3rd Ed., by Charles L. Phillips, Eve A. Riskin, John M. Parr * Solutions manual for Shigley's Mechanical Engineering Design, 8th Ed. by Budynas, Nisbett (No Sol. for Chapt 18 & 19) * Solutions manual for Simply C#: An Application-Driven Tutorial Approach, by Deitel, Hoey (Chapters 1-32) * Solutions manual for Soil Mechanics: Concepts and Applications, 2nd Ed., by Powrie * Solutions manual for Solid State Electronic Devices - 5th Ed by Streetman * Solutions manual for Solid State Electronic Devices - 6th Ed by Streetman * Solutions manual for Statics and Mechanics of Materials: An Integrated Approach, 2nd Ed., by Riley, Sturges, Morris * Solutions manual for Structural Analysis, 5th Edition, by Hibbeler * Solutions manual for University Physics 11th Edition by Young.. * Solutions manual for Vector Mechanics: Statics 7th Edition by Beer * Solutions manual for Vector Mechanics: Dynamics, 7th Ed., by Beer, Johnston, Staab, Clausen * Solutions manual for Vibrations and Stability: Advanced Theory, Analysis, and Tools, 7th Ed., by Thomsen * Solutions manual for Wireless Communications: Principles and Practice, 2nd Ed, by Rappaport * Solutions manual for Theory and Design for Mechanical Measurements, 4th Ed., Beasley, Figliola * Solutions manual for Thermal Physics, 2nd Edition, by Charles Kittel * Solutions manual for Thermal Physics, by Ralph Baierlein * Solutions manual for Thermodynamics: An Engineering Approach, 5th Ed., by Cengel, Boles (Missing solutions #118-149 of Chapter 7) * Solutions manual for Thermodynamics: An Engineering Approach, 6th Ed., by Cengel, Boles * Solutions manual for The Science and Engineering of Materials, 4th Ed., by Donald R. Askeland, Pradeep P. Phule * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 10th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 10th Ed. (Multivariable, chs. 8-13), by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 11th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 11th Ed. (Multivariable, chs. 11-16), by * Solutions manual for Thomas, Weir, Hass, Giordano * Solutions manual for Transport Phenomena, 1st Edition, by R. Byron Bird * Solutions manual for Transport Phenomena, 2nd Ed., by Bird. Contact me at modernbooks@hotmail.com === Subject: Instructors' manuals for Engineering books I have the following solutions manuals in pdf. If you need any of them, send me email at modernbooks at hotmail dot com I accept paypal payments only * Solutions manual for Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray,Hurst, Lewis, Meyer * Solutions manual for Analytical Mechanics, 7th Edition, by Fowels, Cassiday * Solutions manual for An Interactive Introduction to Mathematical Analysis, by Jonathan Lewin * Solutions manual for An Introduction to Database Systems, 8th * Solutions manual for An Introduction to the Mathematics of Financial Derivatives, 2nd Ed.,by Neftci [ISBN:0125153929] * Solutions manual for Antenna Theory, 2nd Ed., by Balanis * Solutions manual for Antennas for all Applications, 3rd Edition, Kraus, Marhefka * Solutions manual for Applied Linear Statistical Models, 5th Ed., by Neter (Selected Sol.) * Solutions manual for Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley * Solutions manual for Applied Numerical Methods with MATLAB for Engineers and Scientists,1st Ed,. by Chapra * Solutions manual for Applied Statistics and Probability for Engineers, 3rd Ed., by Montgomery, Runger (Selected Solutions) * Solutions manual for Applied Strength of Materials, 4th Edition, by Mott * Solutions manual for A Transition to Advanced Mathematics, 5th Edition, by Smith, Eggen,Andre * Solutions manual for Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi * Solutions manual for A Course in Game Theory by Osborne, Rubinstein * Solutions manual for A Course in Algebraic Number Theory by Cohen * Solutions manual for Adaptive Filter Theory, 4th Edition, by Haykin * Solutions manual for Adaptive Control, 2nd. Ed., by Astrom, Wittenmark * Solutions manual for Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even solutions) * Solutions manual for Advanced Macroeconomics, 1st Ed., by David Romer * Solutions manual for Advanced Mathematical Concepts Precalculus With Applications by Holliday [ISBN: 0028341759] * Solutions manual for Advanced Modern Engineering Mathematics, 3rd Ed., by G. James * Solutions manual for A First Course In Differential Equations, 7th Edition, by Zill, Cullen * Solutions manual for Analog Integrated Circuit Design, 1st Ed., by Johns, Martin (text ebook and solution manual) * Solutions manual for Basic Business Statistics: Concepts and Applications, 10th Ed., by Berenson, Krehbiel, Levine (chap1-18) * Solutions manual for Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin * Solutions manual for Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms (Missing a chapter or 2) * Solutions manual for Bioprocess Engineering Principles by Doran * Solutions manual for Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson,Hostetler, Edwards * Solutions manual for Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler * Solutions manual for Chemical Engineering Volume 1, 6th Edition, by Richardson, Coulson,Backhurst, Harker 5th Ed, by Marion, Thornton * Solutions manual for College Physics, Volume 1: 7th Edition, by Serway, Faugh * Solutions manual for College Physics, Volume 2: 7th Edition, by Serway, Faughn * Solutions manual for Communications Systems, 4th Ed., by Haykin * Solutions manual for Communications Systems Engineering, 2nd Edition, by Proakis * Solutions manual for Computational Techniques for Fluid Dynamics by Srinivas, Fletcher * Solutions manual for Computer Networks, 4th Ed., by Andrew S. Tanenbaum * Solutions manual for Computer Networks: A Systems Approach, 3rd Edition, by Davie * Solutions manual for Control Systems Engineering, 4th Ed., by Norman Nise * Solutions manual for Corporate Finance, 6th Edition, by Ross * Solutions manual for C++ How to Program: Intro Object-Oriented Design with the UML, 3rd Ed., by Deitel, Nieto * Solutions manual for Calculus Early Transcendental, 5th Ed., by James Stewart * Solutions manual for Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis * Solutions manual for Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney,Demana, Kennedy * Solutions manual for Calculus: Multivariable, 5th Edition, by James Stewart * Solutions manual for Calculus: Single Variable, Early Transcendental, 5th Edition, by James Stewart * Solutions manual for Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett,McCallum * Solutions manual for Device Electronics for * Solutions manual for Integrated Circuits 3rd Edition by Muller * Solutions manual for Differential Equations with Boundary Value Problems, 2nd Ed., by Polking, Arnold * Solutions manual for Digital And Analog Communication Systems 7th Ed., Leon W. Couch * Solutions manual for Digital Communications, 4th Edition, by Proakis * Solutions manual for Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar * Solutions manual for Digital Design, 4th Edition, by Mano, Ciletti * Solutions manual for Digital Image Processing, 2nd Edition, by Gonzalez, Woods * Solutions manual for Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for Chapters 3, 5, 6, 10) * Solutions manual for Digital Signal Processing: A Computer Based Approach, 1st Ed., by Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S.Mitra * Solutions manual for Digital Signal Processing: Principles, Algorithms and Applications,3rd Edition, by Proakis * Solutions manual for Discrete Time Signal Processing, 2nd Edition, Oppenheim * Solutions manual for Dynamics of Mechanical Systems by C.T.F. Ross * Solutions manual for Data and Computer Communications, 8th Edition by Stallings * Solutions manual for Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol.for Chapters 2-21, odd only) * Solutions manual for Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi Design of Analysis of Experiments, 6th Edition, Montgomery (missing chapter 6-8) * Solutions manual for Design of Machinery, 3rd Ed by Robert L. Norton * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 2nd Ed., by Sergio Franco * Solutions manual for Design With Operational Amplifiers and Analog Integrated Circuits, 3rd Ed., by Sergio Franco * Solutions manual for Elementary Principles of Chemical Processes, 3rd Ed., by Felder,Rousseau * Solutions manual for Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler * Solutions manual for Engineering and Chemical Thermodynamics, by Koretsky [ISBN:0471385867] (No sol. for chapt 6) * Solutions manual for Engineering Circuit Analysis, 6th Edition, Hyat * Solutions manual for Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck * Solutions manual for Engineering Electromagnetics, 7th Ed., Hayt, Buck * Solutions manual for Engineering Fluids Mechanics 7th Edition by Crowe * Solutions manual for Engineering Fluids Mechanics 8th Edition by Crowe * Solutions manual for Engineering Mathematics, 4th Ed., by John Bird * Solutions manual for Engineer Mechanics: Dynamics, 4th Ed., by Bedford * Solutions manual for Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 11th Ed. by Hibbeler * Solutions manual for Engineering Mechanics: Dynamics 5th Ed. by Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace Fowler * Solutions manual for Engineering Mechanics: Statics, 5th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 6th Ed., Meriam, Kraige * Solutions manual for Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler * Solutions manual for Engineering Mechanics: Statics 11th Ed. by Hibbeler * Solutions manual for Experiments with Economic Principles by Bergstrom, Miller * Solutions manual for Econometric Analysis, 5th Edition, by Greene * Solutions manual for Econometric Analysis of Cross Section and Panel * Solutions manual for Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, Viceira * Solutions manual for Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar * Solutions manual for Electric Circuits 6th Ed. by Nilsson * Solutions manual for Electric Circuits 7th Ed. by Nilsson * Solutions manual for Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans * Solutions manual for Electric Machinery Fundamentals, 4th Ed by Chapman * Solutions manual for Electromagnetic Fields and Waves by Iskander * Solutions manual for Electronic Circuit Analysis, 2nd Ed., by Donald Neamen * Solutions manual for Electronics, 2nd Ed., by Allan R. Hambley * Solutions manual for Elementary Differential Equations, 8th Edition, by Boyce, DiPrima(some odd/even) * Solutions manual for Fundamentals of Applied Electromagnetics, 5th Ed., 2008 Media Edition,by Ulaby * Solutions manual for Fundamentals of Digital Logic with Verilog Design, 1st Edition, by Brown, Vranesic * Solutions manual for Fundamentals of Electric Circuits, 2nd Edition, by Alexander * Solutions manual for Fundamentals of Electromagnetics with Engineering Appls by Wentworth * Solutions manual for Fundamentals of Fluid Mechanics, 5th Ed. by Munson, Young.. * Solutions manual for Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 5th Ed by Incropera... * Solutions manual for Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... * Solutions manual for Fundamentals of Logic Design, 5th Ed., by Roth Jr. * Solutions manual for Fundamentals of Machine Component Design, 3rd Ed., by Juvinall * Solutions manual for Fundamentals of Machine Component Design, 4th Ed., by Juvinall * Solutions manual for Fundamentals of Machine Elements, 2nd Ed., Hamrock, Jacobson, Schmid * Solutions manual for Fundamentals of Physics by Halliday, 7th Ed., Walker, Resnick * Solutions manual for Fundamentals of Semiconductor Devices, 1st Edition by Anderson * Solutions manual for Fundamentals of Structural Analysis, 2nd Ed., Chia-Ming Uang, Kenneth Leet * Solutions manual for Fundamentals of Thermal-Fluid Sciences, 2nd Ed. by Cengel * Solutions manual for Fundamentals of Thermal-fluid Sciences, Int'l 2nd Ed. by Cengel * Solutions manual for Fundamentals of Engineering Thermodynamics, 5th Ed. by Shapiro * Solutions manual for Fundamentals of Thermodynamics, 5th Ed., by Sonntag, Borgnakke... * Solutions manual for Fundamentals of Thermodynamics, 6th Ed., by Sonntag * Solutions manual for Facilities Planning, 3rd Edition, by Tompkins, White, Bozer, Tanchoco * Solutions manual for Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- Naeini * Solutions manual for Financial Accounting, 4th Ed., by Libby, Short (Chap1-14) * Solutions manual for Financial Accounting: An International Introduction, 2nd Ed., by Alexander, Nobes * Solutions manual for Finite Element Techniques in Structural Mechanics by Ross * Solutions manual for Fluid Mechanics - 5th Edition by Frank M. White * Solutions manual for Fluid Mechanics and Thermodynamics of Turbomachinery, 5th Ed., by S. L. Dixon [ISBN: 0750678704] * Solutions manual for Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., by Cengel & Cimbala * Solutions manual for Fluid Mechanics with Engineering Applications, 10th Edition, by Finnemore * Solutions manual for Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. * Solutions manual for Fundamentals of Applied Electromagnetics, 1st Ed., 2001 Media Edition, by Ulaby * Solutions manual for Geometry, 04 Edition, by McGraw-Hill [ISBN: 0078296374] * Solutions manual for Guide to Energy Management, 5th Edition, by Pawlik * Solutions manual for Heat Transfer: A Practical Approach - 2nd Edition by Cengel * Solutions manual for Hydraulics in Civil and Environmental Engineering, 4th Ed., by Andrew Chadwick Applications by Oz Shy Introduction to Algorithms, 2nd Ed by Cormen, Leiserson (Selected Sol.) * Solutions manual for Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van Ness, Smith, Abbott * Solutions manual for Introduction to Electric Circuits, 6th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electric Circuits, 7th Ed., by Dorf, Svoboda * Solutions manual for Introduction to Electrodynamics, 3rd Ed. by David Griffiths * Solutions manual for Introduction to Fluid Mechanics - 5th Ed. by Fox.. * Solutions manual for Introduction to Fluid Mechanics - 6th Ed by Fox, McDonald... * Solutions manual for Introduction to Linear Algebra, 3rd Ed., by Gilbert Strang * Solutions manual for Introduction to Linear Algebra, 5th Ed., Arnold, Johnson, Riess * Solutions manual for Introduction to Probability by Grinstead, Snell (odd solutions only, not just answers but step by step solutions) * Solutions manual for Introduction to Quantum Mechanics, 2nd Ed. by Griffiths * Solutions manual for Introdution to Solid State Physics, 8th Edition by Kittel * Solutions manual for Introduction to Statistical Quality Control, 4th Edition, by Montgomery * Solutions manual for Introduction to Thermal Systems Engineering: Thermodynamics, Fluid Mechanics, and Heat Transfer by Moran, Shapiro, Munson, DeWitt * Solutions manual for Introduction to Thermal Systems Engineering, by Moran, Shapiro * Solutions manual for Linear Algebra, by J. Hefferon Linear Algebra And Its Applications, 3rd Ed., by David C. Lay * Solutions manual for Linear Algebra with Applications, 2nd Edition - by Otto Bretscher * Solutions manual for Linear Algebra with Applications, 3rd Edition - by Otto Bretscher * Solutions manual for Linear Circuit Analysis: Time Domain, Phasor and Laplace.., 2nd Ed, Lin * Solutions manual for Machine Design: An Integrated Approach, 2nd Ed., by Robert L. Norton * Solutions manual for Machine Design: An Integrated Approach, 3rd Ed., by Robert L. Norton * Solutions manual for Managerial Accounting, 11th Ed., by Noreen, Brewer, Garrison * Solutions manual for Materials Science and Engineering: An Introduction, 6th Ed. by Callister * Solutions manual for Matrix Analysis and Applied Linear Algebra by Carl Meyer * Solutions manual for MC68HC11: An Introduction: Software/Hardware Interf, 2nd Ed, by Huang * Solutions manual for Mechanical Engineering Design, 7th Ed. by Mischke, Shigley * Solutions manual for Mechanical Vibrations, 3rd Edition, by S. S. Rao (99% same as 4th Edition, No Solutions for Chapters 6, 9, and 12) * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Fluids, 4th Ed., Irving H. Shames * Solutions manual for Mechanics of Fluids, 8th Ed., by Bernard Massey * Solutions manual for Mechanics of Materials - 3rd Ed. by Beer, Johnston, Dewolf * Solutions manual for Mechanics of Materials - 6th Ed. by Hibbeler * Solutions manual for Mechanics of Materials, 6th Edition by James M. Gere (missing small portion, section 8.5) * Solutions manual for Mechanics of Materials, 6th Ed., by Sturges, Morris, Riley (part of Chapt 2 is missing but only #1 thru #60) * Solutions manual for Mechanics of Solids by C.T.F. Ross Microeconomic Analysis, 3rd Ed., by H. Varian (Ans. to Exercises: Ch. 1- Ch.25) * Solutions manual for Microeconomic Theory, by Mas-Colell, Whinston, Green * Solutions manual for Microelectronic Circuit Analysis and Design, 3rd Edition, by D. Neamen * Solutions manual for Microelctronic Circuits, 5th Ed. by Sedra and Smith * Solutions manual for Microelectronic Circuit Design, 2nd Edition by Jaeger, Blalock * Solutions manual for Microelectronic Circuit Design, 3rd Edition by Jaeger, Blalock * Solutions manual for Microelectronics: Digital and Analog Circuits and Systems by Millman * Solutions manual for Microwave and Rf Design of Wireless Systems, 1st Edition, by Pozar * Solutions manual for Microwave Engineering, 3rd Ed., by David M. Pozar * Solutions manual for Microwave Transistor Amplifiers: Analysis and Design, 2nd Ed., by Guillermo Gonzalez Miller & Freund's Probability and Statistics for Engineers, 7th Edition, Johnson, Miller * Solutions manual for Modern Compressible Flow, 3rd Edition, by Anderson * Solutions manual for Modern Control Engineering, 3rd Edition, by Ogata * Solutions manual for Modern Control Engineering, 4th Edition, by Ogata * Solutions manual for Modern Digital and Analog Communication Systems, 3rd Ed., by Lathi * Solutions manual for Modern Control Systems, 9th Ed., by Richard C. Dorf, Robert H Bishop (98% same as the 10th Ed.) * Solutions manual for Modern Operating Systems,2nd Ed., by Andrew Tanenbaum * Solutions manual for Modern Physics 4th Edition by Tipler * Solutions manual for Monetary Theory and Policy, 2nd Edition, by Walsh * Solutions manual for Multivariable Calculus, 5th Edition, by James Stewart * Solutions manual for Operating Systems: Internals and Design Principles, 4th Edition, by Stallings * Solutions manual for Operating System Concepts, 7th Ed., Silberschatz, Galvin, Gagne * Solutions manual for Options, Futures and Other Derivatives, 4th Ed., by John Hull * Solutions manual for Options, Futures and Other Derivatives, 5th Ed., by John Hull (Chapters 1 thru 18 ONLY) * Solutions manual for Orbital Mechanics: For Engineering Students by Howard Curtis (includes matlab scripts) * Solutions manual for Organic Chemistry, 4th Ed., by Carey, Atkins (Student Study Guide and Sol. Man.) * Solutions manual for Partial Differential Equations with Fourier Series and Boundary Value * Solutions manual for Problems, 2nd Ed., by Asmar (Student Solutions Manual) * Solutions manual for Physical Chemistry - 7th Edition - by Julio de Paula, Peter Atkins * Solutions manual for Physics, 6th Edition, by John Cutnell * Solutions manual for Physics, 5th Edition, Vol 2 by Halliday, Resnick, Krane (Chap 25-52) * Solutions manual for Physics for Scientist and Engineers by Knight (No Chapt 36-42) * Solutions manual for Physics for Scientist and Engineers, 6th Ed., by Serway * Solutions manual for Physics for Scientists and Engineers-Vol 1, 5th Edition, Serway, Beichner (Chap. 1 - 22) * Solutions manual for Physics for Scientists and Engineers-Vol 2, 5th Edition, Serway, Beichner (Chap. 23 - 46) * Solutions manual for Physics for Scientists and Engineers, 3rd Ed., by Douglas C. Giancoli * Solutions manual for Physics for Scientist and Engineers, 5th Edition, by Tipler, Mosca * Solutions manual for Physics: Principles with Applications, 6th Ed. by Giancoli * Solutions manual for Power System Analysis and Design, 3rd Ed., by Glover, Sarma * Solutions manual for Principles and Applications of Electrical Engineering 4th (Revised) Edition by Rizzoni * Solutions manual for Principles and Practices of Automatic Process Control, 3rd Edition by Smith, Corripio [ISBN: 0471431907] * Solutions manual for Principles of Communication: Systems, Modulation Noise, 5th Ed., Ziemer * Solutions manual for Principles of Physics, 3rd Edition, by Serway * Solutions manual for Principles of Physics, 4th Edition, by Serway * Solutions manual for Principles of Statics, 10th Ed., by Russell C. 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Askeland, Pradeep P. Phule * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 10th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 10th Ed. (Multivariable, chs. 8-13), by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus, Early Trans., Part 1, 11th Ed. by Thomas, Weir, Hass, Giordano * Solutions manual for Thomas' Calculus: Part 2, 11th Ed. (Multivariable, chs. 11-16), by * Solutions manual for Thomas, Weir, Hass, Giordano * Solutions manual for Transport Phenomena, 1st Edition, by R. Byron Bird * Solutions manual for Transport Phenomena, 2nd Ed., by Bird. Contact me at modernbooks@hotmail.com === Subject: Re: Matrix/Vector differentiation > .... > Let g(x) = x'Mx, where M is a n-by-n real constant matrix and x' > denotes the transpose of vector x. Then the derivative of g(x) = (M + > M')x.... You need to be very clear about what's meant. Your x is a vector with n components x_1, x_2, .... x_n so what do you mean by the derivative of g(x)? I think you probably mean the gradient vector. Anyway, a good way to attack many mathematical problems is to look carefully at a special case. If n = 2, you have a general 2-by-2 matrix and two scalar variables to deal with. Expand the expression x'Mx in full, and think about the derivatives etc. until you thoroughly understand that simple special case. Then the general case should be a lot more manageable. Ken Pledger. === Subject: Advancing surrogate factoring I've noted that every integer factorization can be connected to some other factorization with simple congruences, and have talked about a method for factoring by going in one direction with that connection, but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an integer, such that all congruences are satisfied and that if S = p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of squares. Here is an example. Let the target composite to be factored be 77, so S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you have sqrt(S/2), while it can be much smaller than that, so you have progressively smaller numbers to factor than your target. There must exist and x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, work, as it did with my example above, or can you have cases where abs(x)>T? If so that could be a problem with this method, otherwise it is guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection goes in BOTH DIRECTIONS, where I've focused on one direction until now. James Harris === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist and x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. James Harris JSH must be hunting snarks here. Just because you post it three times doesn't make it true, James! --- Christopher Heckman === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) is that the postive square root, or the negitive square root ? to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. how can you check if it is integer? it can still have a remainder close to zero. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist and x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. > James Harris > === Subject: Re: Advancing surrogate factoring <46f17697$0$47133$892e7fe2@authen.yellow.readfreenews.net I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) is that the postive square root, or the negitive square root ? to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. how can you check if it is integer? it can still have a remainder close to > zero. calculate floor of the sqrt and check if the right thing comes out when squaring again. Actually, any decent algorithm to calulate floo(sqrt(n)) will produce the answer as a spin-off. === Subject: Re: Advancing surrogate factoring <46f17697$0$47133$892e7fe2@authen.yellow.readfreenews.net > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) is that the postive square root, or the negitive square root ? to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. how can you check if it is integer? it can still have a remainder close to > zero. calculate floor of the sqrt and check if the right thing comes out > when squaring again. > Actually, any decent algorithm to calculate floor(sqrt(n)) will produce > the answer as a spin-off. There's also that classical algorithm, which starts off by grouping the digits by 2's. --- Christopher Heckman === Subject: Advancing surrogate factoring I've noted that every integer factorization can be connected to some other factorization with simple congruences, and have talked about a method for factoring by going in one direction with that connection, but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an integer, such that all congruences are satisfied and that if S = p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of squares. Here is an example. Let the target composite to be factored be 77, so S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you have sqrt(S/2), while it can be much smaller than that, so you have progressively smaller numbers to factor than your target. There must exist and x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, work, as it did with my example above, or can you have cases where abs(x)>T? If so that could be a problem with this method, otherwise it is guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection goes in BOTH DIRECTIONS, where I've focused on one direction until now. James Harris === Subject: Re: Advancing surrogate factoring > I've noted that every integer factorization can be connected to some > other factorization with simple congruences, and have talked about a > method for factoring by going in one direction with that connection, > but why not go the other? Given x^2 = y^2 mod T where T is a non-zero integer, introducing k = 2x mod T, you can > easily solve to find (x+k)^2 = y^2 + 2k^2 mod T so with S = 2k^2 mod T, you have a second difference of squares. But what if S is your target composite to be factored? Then S - 2k^2 = 0 mod T, so by some choice of k, you have T as a > factor of S - 2k^2. Then x = 2^{-1} k mod T, gives you x. It can be shown that only for certain cases then will y exist as an > integer, such that all congruences are satisfied and that if S = > p_1*p_2 that case is equivalent to 2(x+k) = p_1 + p_2 or 2(x+k) = p_1 - p_2. And then it is trivial to factor S now through a difference of > squares. Here is an example. Let the target composite to be factored be 77, so > S = 77. Looking for the smallest S - 2k^2, I take floor(sqrt(77/2)) = 6, so k > = 6 gives me a minimum, so S - 2k^2 = 77 - 72 = 5. So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. Then x+k = 9, and p_1 + p_2 = 18, so 2(x+k) = p_1 + p_2, and you can > non-trivially factor by solving the quadratic formula as m^2 + 18 + 77 = (m+7)(m+11). So the algorithm that follows naturally is with a target S, let k = > floor(sqrt(S/2)), then you factor S - 2(floor(sqrt(S/2))^2 to get T, looping through its factors, and solve for x, using x = > 2^{-1} k mod T, and then check sqrt(((x+k)/2)^2 + 4*S) to see if it is an integer and if it is, check to see if you have a > non-trivial factorization of your target composite S. Notice this technique can be recursive, as for a MAXIMUM for T you > have sqrt(S/2), while it can be much smaller than that, so you have > progressively smaller numbers to factor than your target. There must exist and x for the given k that will work as x = (p_1+p_2)/2 - k while the only question then that remains is, how often will x < T, > work, as it did with my example above, or can you have cases where > abs(x)>T? If so that could be a problem with this method, otherwise it is > guaranteed to non-trivially factor. And that extension of surrogate factoring comes because the connection > goes in BOTH DIRECTIONS, where I've focused on one direction until > now. James Harris Not much point, really, in posting this a 3rd time, and with the same careless errors, wouldn't you say? It just means having to delete 3 times as many messages in the long run. Marcus. === Subject: Re: Advancing surrogate factoring ... > Then x = 2^{-1} k mod T, gives you x. ... > So T = 5. Now x = 2^{1}*6 mod 5 = 3*6 mod 5 = 3. ... Q1 ) - Is it {-1} or {1} ? Q2 ) - Does this mean that 2^{1} = 3 ? Mike === Subject: Re: Advancing surrogate factoring Serragoat Factoring quoting JSH, slower than random guessing What else you got, Monkey-Boy? === Subject: Re: Cute elementary algebra problem <46EC83CD.7090208@netscape.net> On Sep 15, 6:15 pm, Stephen J. Herschkorn Came across this problem in the first section of the first >>chapter of the intro calculus text from which I am >>teaching this semester. I thought it was kind of cute. >Solve x^4 - 4x = 1 exactly. Let s = sqrt(2) and write x^4 - 4x - 1 as the difference of >two squares: x^4 - 4x - 1 = > (x^4 + 2x^2 + 1) - (2x^2 + 4x + 2) = > (x^2 + 1)^2 - 2(x + 1)^2 = > (x^2 + 1 + s(x + 1))(x^2 + 1 - s(x + 1)) = > (x^2 + sx + (1 + s)) (x^2 - sx + (1 - s)) Clearly either x^2 + sx + (1+s) = 0 or x^2 - sx + (1-s) = 0. >But s^2 - 4(1+s) = 2 - 4 - 4s < 0, so only the second factor >produces real roots. They are (1/2)(s +/- sqrt(-2 + 4s)) = >(s/2)(1 +/- sqrt(-1 + 2s)). (Barring algebraic errors at >some point, of course.) Why do you exclude the complex roots, easily found from your derivation? I got there this way, which is pretty close to Brian's route. x^4 - 4x = 1 > x^4 = 4x + 1 > x^4 + 2x^2 + 1 = 2 x^2 + 4x + 2 > (x^2 + 1)^2 = 2 (x + 1)^2 > x^2 + 1 = +/- sqrt(2) (x+1) We now have two quadratic equations to solve. Any idea if we can express sqrt(sqrt(2) - 1) as a rational linear > combination of roots of integers? I suspect not. -- > Stephen J. Herschkorn sjhersc...@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan I get (x^2 + 2*sqrt(x)) * (x^2 - 2*sqrt(x)) = 1 by simple factoring. Doesn't solve the problem but raises the question; Does this preclude negative values for x? Bill J === Subject: Re: Cute elementary algebra problem <46EC83CD.7090208@netscape.net >alt.math.recreational,alt.math.undergrad: >Came across this problem in the first section of the first >>chapter of the intro calculus text from which I am >>teaching this semester. I thought it was kind of cute. >Solve x^4 - 4x = 1 exactly. Let s = sqrt(2) and write x^4 - 4x - 1 as the difference of >two squares: x^4 - 4x - 1 = > (x^4 + 2x^2 + 1) - (2x^2 + 4x + 2) = > (x^2 + 1)^2 - 2(x + 1)^2 = > (x^2 + 1 + s(x + 1))(x^2 + 1 - s(x + 1)) = > (x^2 + sx + (1 + s)) (x^2 - sx + (1 - s)) Clearly either x^2 + sx + (1+s) = 0 or x^2 - sx + (1-s) = 0. >But s^2 - 4(1+s) = 2 - 4 - 4s < 0, so only the second factor >produces real roots. They are (1/2)(s +/- sqrt(-2 + 4s)) = >(s/2)(1 +/- sqrt(-1 + 2s)). (Barring algebraic errors at >some point, of course.) Why do you exclude the complex roots, easily found from your derivation? I got there this way, which is pretty close to Brian's route. x^4 - 4x = 1 > x^4 = 4x + 1 > x^4 + 2x^2 + 1 = 2 x^2 + 4x + 2 > (x^2 + 1)^2 = 2 (x + 1)^2 > x^2 + 1 = +/- sqrt(2) (x+1) We now have two quadratic equations to solve. Any idea if we can express sqrt(sqrt(2) - 1) as a rational linear > combination of roots of integers? I suspect not. -- > Stephen J. Herschkorn sjhersc...@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan I get (x^2 + 2*sqrt(x)) * (x^2 - 2*sqrt(x)) = 1 by simple > factoring. That doesn't get you anywhere, since you don't have a 0 on the right- hand side. > Doesn't solve the problem but raises the question; Does this preclude negative values for x? Not necessarily; if we're working with complex numbers, square roots of negative numbers are okay. --- Christopher Heckman === Subject: Re: Introduction to heat transfer I need the solutions manual for introduction to heat transfer > incropera/dewitt/bergmann/lavine > Fight edition. i think i got teh fifth edition solution mankunak in pdf form. hey, > is it possible for you to send me the solution manual? > or tell me where to find it? > thank you so much. hi, > i also need the solutions manual for the 5th edition of intro to heat > transfer by incropera/dewitt/bergmann/lavine. > can you help? I would like a copy of the solutions manual as well if possible. === Subject: Re: Introduction to heat transfer > I need the solutions manual for introduction to heat transfer > incropera/dewitt/bergmann/lavine > Fight edition. i think i got teh fifth edition solution mankunak in pdf form. hey, > is it possible for you to send me the solution manual? > or tell me where to find it? > thank you so much. hi, > i also need the solutions manual for the 5th edition of intro to heat > transfer by incropera/dewitt/bergmann/lavine. > can you help? I would like a copy of the solutions manual as well if possible. === Subject: Engineering Mech Stat 10 edition Hibbeler HI i need a solutions manual for engineering mech stat 10th edition (Hibbeler) if it is possible. baranceandragos@yahoo.com === Subject: solution manual I need Fundamentals of Communication Systems (John G. Proakis, Masoud Salehi) . Can Anyone help? web976@aol.com === Subject: Cosine pre-calc problem Doing an exercise in sine/cosine relationships. Problem from the text is as follows: The perimeter and the area of an oblique triangle are 20 centimeters and 1O(sqrt3) square centimeters, respectively. Find side a and side b. The figure shown is pretty close to a right angle. Sides a and b are connected at a vertex of 60 degrees. Side a sits on the x axis. Side b goes off at a 60 degree angle. (Assume a < b.) According to the text the answer is a = 5 b = 8 I stared at this problem for a long time. Finally I simply used the formula for area of an oblique triangle: Area = [abSineC (60 degrees)]/2 That is 17.32 sq centimeters = et ceterea I brought all knowns over to the left hand side of formula which came out to 40 Thus 40 = ab Then I simply factored the number 40 and came up with 5 & 8. Seems to me the problem calls for a more precise manner of solution. Is there a more precise way to solve this problem? --Senior citizen math undergrad here === Subject: Re: Cosine pre-calc problem >Doing an exercise in sine/cosine relationships. Problem from the text is as follows: The perimeter and the area of an oblique triangle >are 20 centimeters and 1O(sqrt3) square centimeters, respectively. >Find side a and side b. The figure shown is pretty close to a right angle. Sides a and b are connected at a vertex of 60 degrees. Side a sits >on the x axis. Side b goes off at a 60 degree angle. (Assume a < b.) According to the text the answer is a = 5 b = 8 I stared at this problem for a long time. Finally I simply used the >formula for area of an oblique triangle: Area = [abSineC (60 degrees)]/2 That is 17.32 sq centimeters = et ceterea I brought all knowns over to the left hand side of formula which came >out to 40 Thus 40 = ab Then I simply factored the number 40 and came up with 5 & 8. No, that's not valid. So far, you have one equation in two unknowns, ab = 40. You need another equation to take advantage of the fact that you know the perimeter, 20 = a + b + c. By the law of cosines, you have c^2 = a^2 + b^2 - 2 a b cos(60 degrees) so c^2 = a^2 + b^2 - ab Substitute (20 - a - b) for c, and 40/a for b . Solve the resulting quadratic equation for a. Then, given a < b, you can easily find b. Seems to me the problem calls for a more precise manner of solution. >Is there a more precise way to solve this problem? --Senior citizen math undergrad here === Subject: Re: Cosine pre-calc problem >Doing an exercise in sine/cosine relationships. Problem from the text is as follows: The perimeter and the area of an oblique triangle >are 20 centimeters and 1O(sqrt3) square centimeters, respectively. >Find side a and side b. The figure shown is pretty close to a right angle. Sides a and b are connected at a vertex of 60 degrees. Side a sits >on the x axis. Side b goes off at a 60 degree angle. (Assume a < b.) According to the text the answer is a = 5 b = 8 I stared at this problem for a long time. Finally I simply used the >formula for area of an oblique triangle: Area = [abSineC (60 degrees)]/2 That is 17.32 sq centimeters = et ceterea I brought all knowns over to the left hand side of formula which came >out to 40 Thus 40 = ab Then I simply factored the number 40 and came up with 5 & 8. Seems to me the problem calls for a more precise manner of solution. >Is there a more precise way to solve this problem? --Senior citizen math undergrad here Well, 40 = ab is one equation in two unknowns. You need another equation in order to get the unknowns. I suspect that the perimeter will provide you with another equation. Hope this helps, Brian === Subject: Solutions for Intro to Heat 5th ed === Subject: Recurrence Relations Problem Hello all, I hope I'm posting to the right forum. This is a question from my instructor as a prep to the first midterm that's coming up on Tuesday. I just need someone to help me work through this concept of finding a recurrence relation...I don't necessarily want the answer but maybe a yea or nay if I'm going in the right direction? (and maybe even a hint!! :)) The problem: Instead of the conventional algorithm in finding the minimum of a set S of size N, consider the findmin algorithm that returns the minimum of S using the divide and conquer as follows: findmin(S,N) = min(findmin(left half of S,N/2),findmin(right half of S,N/2)) where min(a,b) returns the minimum of its 2 arguments using one binary compare operation. Let C(N) denote the complexity of findmin(in terms of compare operations) Define a recurrence relation and initial condition C(1) whose solution is C(N). (I don't need to solve it, but just state the recurrence relation.) What I know so far is, the number of compares to find a minimum in a conventional minimum function would be n-1 compares. (This is a linear search algorithm, I think.) So, for this problem, since N is really N/2, the reccurence relation for the left side would be: C(N) = C(N/2-1) so if I were to include the right side, the equation is now: C(N) = C(2(N/2-1)) + 1 where the '+ 1' refers to the final compare. And then, I need to define C(1). Hrmm, do I just plug in 1 into the equation to get that value? Or do I say if the set S was a size 1, then C(1) = 0? since if the set S was a size 1 then there are zero compares to do. Am I working through this problem in the right manner? I'm having such a hard time with this so any sort of comments or input would be appreciated. Marion P.S. I apologize for any stupid sounding questions that I may be asking. :x === Subject: Re: Recurrence Relations Problem > The problem: > Instead of the conventional algorithm in finding the minimum of a set > S of size N, consider the findmin algorithm that returns the minimum > of S using the divide and conquer as follows: > The usual method? Look at each individually? > findmin(S,N) = min(findmin(left half of S,N/2),findmin(right half of > S,N/2)) > where min(a,b) returns the minimum of its 2 arguments using one binary > compare operation. > Let C(N) denote the complexity of findmin(in terms of compare > operations) > This looks much more complicated. Say you have 8 elements. a b c d e f g h. You have to make 7 comparisons ab, cd, ef, gh; ab.cd, ef.gh; ab.cd:ef.gh which is no improvement and puts strain on the stack which the usual doesn't. That method is used to find in a sorted list of size n, the nearest match to a given element in log_2 n searches. > Define a recurrence relation and initial condition C(1) whose solution > is C(N). > (I don't need to solve it, but just state the recurrence relation.) What I know so far is, the number of compares to find a minimum in a > conventional minimum function would be n-1 compares. (This is a > linear search algorithm, I think.) So, for this problem, since N is > really N/2, the reccurence relation for the left side would be: > C(N) = C(N/2-1) so if I were to include the right side, the equation > is now: > C(N) = C(2(N/2-1)) + 1 where the '+ 1' refers to the final compare. And then, I need to define C(1). Hrmm, do I just plug in 1 into the > equation to get that value? Or do I say if the set S was a size 1, > then C(1) = 0? since if the set S was a size 1 then there are zero > compares to do. Am I working through this problem in the right manner? I'm having > such a hard time with this so any sort of comments or input would be > appreciated. Marion P.S. I apologize for any stupid sounding questions that I may be > asking. :x === Subject: Re: Recurrence Relations Problem > The problem: >> Instead of the conventional algorithm in finding the minimum of a set >> S of size N, consider the findmin algorithm that returns the minimum >> of S using the divide and conquer as follows: >The usual method? Look at each individually? > findmin(S,N) = min(findmin(left half of S,N/2),findmin(right half of >> S,N/2)) >> where min(a,b) returns the minimum of its 2 arguments using one binary >> compare operation. >> Let C(N) denote the complexity of findmin(in terms of compare >> operations) >This looks much more complicated. Say you have 8 elements. > a b c d e f g h. >You have to make 7 comparisons > ab, cd, ef, gh; ab.cd, ef.gh; ab.cd:ef.gh >which is no improvement and puts strain on the stack which the >usual doesn't. Ignoring the fact that this really has no bearing on the question that was asked: Say you start with a list of 2^n elements. How large a stack does that require? >That method is used to find in a sorted list of size n, the nearest match >to a given element in log_2 n searches. > Define a recurrence relation and initial condition C(1) whose solution >> is C(N). >> (I don't need to solve it, but just state the recurrence relation.) >> What I know so far is, the number of compares to find a minimum in a >> conventional minimum function would be n-1 compares. (This is a >> linear search algorithm, I think.) So, for this problem, since N is >> really N/2, the reccurence relation for the left side would be: >> C(N) = C(N/2-1) so if I were to include the right side, the equation >> is now: >> C(N) = C(2(N/2-1)) + 1 where the '+ 1' refers to the final compare. >> And then, I need to define C(1). Hrmm, do I just plug in 1 into the >> equation to get that value? Or do I say if the set S was a size 1, >> then C(1) = 0? since if the set S was a size 1 then there are zero >> compares to do. >> Am I working through this problem in the right manner? I'm having >> such a hard time with this so any sort of comments or input would be >> appreciated. >> Marion >> P.S. I apologize for any stupid sounding questions that I may be >> asking. :x >> ************************ David C. Ullrich === Subject: Re: Recurrence Relations Problem Considering an array of size 2^N confuses me further. So I chose a smaller number like 6. I'm having trouble with the small number much less having to think about an abstract number. The algorithm that is given to me is not sorted, as I understand it. It can be in the worse possible order (backwards sorted) because it seems the worst possible case is what is needed to be considered. So, consider the list as: 6 5 4 3 2 1, (N = 6) Left half = 6, 5, 4 = 2 compares (comparing 6,5 & 6,4) = C(N) = N/2 - 1 Right half = 3, 2, 1 = 2 compares (comparing 3,2 & 2,1) = N/2 -1 Then add 1 because of the last initial compare of the right and left side. The reccurence relation must be stated using the D/C method which I understand in this case is in the form of this: C(N) = C(min result of left side) + C(min result of the right side) + 1 and then I need to figure out C(1) and right now I don't know why. (A hint given by the professor) Thus my conclusion is C(N) = 2(C(N/2 - 1)) + 1, C(1) = 0 Which brings another question: is 2(C(N/2-1)) equivalent to C(2(N/ 2-1))? To me it's the same, but all this uncertainty has created doubt in my judgement. Again, I don't know if I'm making any sense, am I solving for the answer the wrong way? or? Any comments are appreciated. === Subject: Re: Recurrence Relations Problem smaller number like 6. I'm having trouble with the small number much > less having to think about an abstract number. The algorithm that is given to me is not sorted, as I understand it. > It can be in the worse possible order (backwards sorted) because it > seems the worst possible case is what is needed to be considered. So, consider the list as: 6 5 4 3 2 1, (N = 6) Left half = 6, 5, 4 = 2 compares (comparing 6,5 & 6,4) = C(N) = N/2 - > 1 > Right half = 3, 2, 1 = 2 compares (comparing 3,2 & 2,1) = N/2 -1 > Then add 1 because of the last initial compare of the right and left > side. The reccurence relation must be stated using the D/C method which I > understand in this case is in the form of this: C(N) = C(min result of left side) + C(min result of the right side) + > 1 > and then I need to figure out C(1) and right now I don't know why. (A > hint given by the professor) Thus my conclusion is C(N) = 2(C(N/2 - 1)) + 1, C(1) = 0 Which brings another question: is 2(C(N/2-1)) equivalent to C(2(N/ > 2-1))? To me it's the same, but all this uncertainty has created > doubt in my judgement. You can check this directly: C(2(N/2-1)) = C(N-2). For N-2 elements, you need C(N/2-1) for the first half, and C(N/2-1) for the right half, and one for the final comparison, so C(2(N/2-1)) = 2*C(N/2 -1) + 1. They are not equivalent. Back to your original questions: for any Recurrence, you need to give the starting values. Thus, you should define C(1). Defining it as 0 makes sense for the reason you stated. You have two problems in your proposed answer of C(N)=2*C(N/2-1)+1. First, consider finding C(4), which should be 3. You have: C(4) = 2 C(4/2-1) +1 = 2 C(1) + 1 = 0 + 1 = 1. There must be an error there. Or, even worse: C(2) = 2 C(2/2-1) + 1 = 2 C(0) +1. You haven't defined C(0). (although, given you know C(2) should be 1, you could define that as 0. Second, note that you also need to refine your recurrence relation, since if N is odd, then N/2-1 is not an integer; unless you have N=2^m, for some integer m, you will have and odd number at some step in the recursion. For example: C(5) = 2 C(5/2-1) + 1 = 2 C(3/2) + 1. Can you find a set with 3/2 elements? I'm sure you see the problem. I would suggest using the floor and ceiling operation to solve this problem. (one for each half) === Subject: Re: Recurrence Relations Problem Hello QD, for letting me bounce ideas back and forth like this. There is virtually no one that I can comfortably ask to help solve this without feeling stupid. > You can check this directly: > C(2(N/2-1)) = C(N-2). For N-2 elements, you need C(N/2-1) for the > first half, and C(N/2-1) for the right half, and one for the final > comparison, so C(2(N/2-1)) = 2*C(N/2 -1) + 1. They are not equivalent. > After looking at your answer, there are a couple of things that are bothering me. First, the issue I had with whether 2(C(N/2-1)) equivalent to C(2(N/2-1)). I'm remembering in my classes last semester and this that a constant can always be pulled out an equation at any time. In summation equations and the like. why would it matter in that equation? To me, if I were to use the logic you presented me, I check directly by substituting N with 4. Orig equation: C(2(N/2-1)) = C(N-2). Substituting C(2(4/2 - 1) = C(4-2) I am assuming when you substitute a variable (N) with a real value, you evaluate the expressions then stick them into the functions...so what the value in the functions has no meaning right now..I'm just at the evaluating before processing them into the functions. I am unconvinced still that they are not equal. What is wrong with my logic? Or what am I missing here? > Back to your original questions: for any Recurrence, you need to give > the starting values. Thus, you should define C(1). Defining it as 0 > makes sense for the reason you stated. You have two problems in your proposed answer of C(N)=2*C(N/2-1)+1. First, consider finding C(4), which should be 3. You have: > C(4) = 2 C(4/2-1) +1 = 2 C(1) + 1 = 0 + 1 = 1. There must be an error > there. Or, even worse: > C(2) = 2 C(2/2-1) + 1 = 2 C(0) +1. You haven't defined C(0). > (although, given you know C(2) should be 1, you could define that as > 0. For your proposed question for C(4) = 1. Hrm...ok, I see what you're saying now. Is it possible for me to define C(1) = 1 instead of 0? Then my rec. relation would work then or is that a bad thing to do? I see why now defining the base cases is important now, too. How many base cases should I define? C(0), C(1) and C(2)? C(0) = 0, C(1) = 1, C(2) = 1. Alternatively, changing C(1) = 1 bothers me too since technically, no compares are done when we have one element. We already know that it is zero compares if N = 1. I'm going to have to think about that one. For that part about a non integer resulting from (N/2 - 1), I like that solution of using ceiling or floor function. That part I understand with ease. Ok, I need to get ready for school....thank you much for helping me work on this exercise. I know the issues that need to be addressed and can go into class knowing what I need to be cleared up on. Marion === Subject: Re: Recurrence Relations Problem in alt.math.undergrad: [...] >> You can check this directly: >> C(2(N/2-1)) = C(N-2). For N-2 elements, you need C(N/2-1) >> for the first half, and C(N/2-1) for the right half, and >> one for the final comparison, so C(2(N/2-1)) = 2*C(N/2-1) + 1. >> They are not equivalent. > After looking at your answer, there are a couple of things > that are bothering me. First, the issue I had with > whether 2(C(N/2-1)) equivalent to C(2(N/2-1)). > I'm remembering in my classes last semester and this that > a constant can always be pulled out an equation at any > time. It cannot: it is most definitely *not* true that f(2x) = 2f(x) for any old function f. For instance sin(2x) is not in general equal to 2 sin x, as you can easily see by substituting x = pi/2: sin(2*pi/2) = sin pi = 0, but 2 sin (pi/2) = 2 * 1 = 2 (which isn't even a possible value for the sine function!). > In summation equations and the like. why would it matter > in that equation? To me, if I were to use the logic you > presented me, I check directly by substituting N with 4. > Orig equation: C(2(N/2-1)) = C(N-2). > Substituting C(2(4/2 - 1) = C(4-2) But this isn't the equivalence about which you were asking: you wanted to know whether C(2(N/2 - 1)) = 2C(N/2 - 1), and all that you've calculated here is the first of these expressions. Now calculate 2C(N/2 - 1) with N = 4: you get 2C(2 - 1) = 2C(1). C(2) = 1, and 2C(1) = 0, so the two are *not* the same. [...] >> Back to your original questions: for any Recurrence, you >> need to give the starting values. Thus, you should >> define C(1). Defining it as 0 makes sense for the reason >> you stated. >> You have two problems in your proposed answer of >> C(N)=2*C(N/2-1)+1. >> First, consider finding C(4), which should be 3. You have: >> C(4) = 2 C(4/2-1) +1 = 2 C(1) + 1 = 0 + 1 = 1. There must be an error >> there. Or, even worse: >> C(2) = 2 C(2/2-1) + 1 = 2 C(0) +1. You haven't defined C(0). >> (although, given you know C(2) should be 1, you could define that as >> 0. > For your proposed question for C(4) = 1. Hrm...ok, I see what you're > saying now. Is it possible for me to define C(1) = 1 instead of 0? Not a good idea, since C(1) really is 0: you don't perform any comparisons when N = 1. Think about what your recurrence represents: C(N) should be the number of comparisons made in getting the minimum of the left half plus the number made in getting the minimum of the right half plus the one that's needed to pick the smaller of those two partial results. When N = 4, the left and right halves both have 2 elements, so in this case the formula should boil down to C(4) = C(2) + C(2) + 1. Unfortunately, your formula boils down to C(4) = C(1) + C(1) + 1. Clearly the problem isn't an incorrect value of C(1), because you shouldn't be evaluating C(1) here in the first place; the problem is with your N/2 - 1. You need to think a bit more about how big the left and right halves will be; this is where the floor and/or ceiling function(s) will be useful. [...] Brian === Subject: Statics and Mechanics of Materials (2nd Edition ) R.C.Hibbeler hey i wanted the solution manual for Statics and Mechanics of Materials (2nd Edition ) R.C.Hibbeler if you have it let me know and contact me on kushsura@hotmail.com or kushura@gmail.com Thnak you Kush Sura === Subject: Norton's Design of Machinery I would like the solution manual to Norton's Design of Machinery kalens99@wpi.edu === Subject: Primes conjecture with surrogate factoring Oddly enough looking at going in a different direction with what I call the surrogate factoring congruence lead me to a fascinating little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime for any non-zero k, unless S is a product of two primes, and they have to be differing primes. Is that already known? It is kind of an odd little result. Here's an example of when a prime is allowed, S = 77, k = 6, then 77 - 2*36 = 5 but in contrast, say if S = 105, which has 3 prime factors, by this conjecture no non-zero k exists such that abs(105 - 2k^2) is a prime number. The proof is oddly easy but I'll call it a conjecture in case I made a mistake in the argument which can be found at my math blog. James Harris === Subject: Re: Primes conjecture with surrogate factoring > The proof is oddly easy but I'll call it a conjecture in case I made a > mistake in the argument which can be found at my math blog. > Evil White Mathematicians made you make mistakes in your own racist blog, Monkey Boy. James Harris > === Subject: Re: Primes conjecture with surrogate factoring > Oddly enough looking at going in a different direction with what I > call the surrogate factoring congruence lead me to a fascinating > little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime > for any non-zero k, unless S is a product of two primes, and they have > to be differing primes. Is that already known? It is kind of an odd little result. Here's an example of when a prime > is allowed, S = 77, k = 6, then 77 - 2*36 = 5 but in contrast, say if S = 105, which has 3 prime factors, by this > conjecture no non-zero k exists such that abs(105 - 2k^2) is a prime number. The proof is oddly easy but I'll call it a conjecture in case I made a > mistake in the argument which can be found at my math blog. James Harris Wow, amazing! I'd love to see the proof! I wonder how it applies to S = 49, k = 1 ... Or S = 45, k = 2 ... Or S = 105, k = 4 ... You do prove the most amazing things. And since it's a proof, and an oddly easy proof at that, it must be right. After all, as you have often said, a proof cannot be wrong, and you are the master who defined 'proof' in the first place. And you never make any mistakes of course. And no, I don't think it's already known. Do you ever bother to check ANYTHING ? You must think that every hare-brained idea that comes into your head is a stroke of genius. Marcus. === Subject: Re: Primes conjecture with surrogate factoring > Oddly enough looking at going in a different direction with what I > call the surrogate factoring congruence lead me to a fascinating > little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime > for any non-zero k, unless S is a product of two primes, and they have > to be differing primes. second term is 2, 8, 18, 32, 50, ............. why subtracting that should result in a prime ? Is that already known? It is kind of an odd little result. Here's an example of when a prime > is allowed, S = 77, k = 6, then 77 - 2*36 = 5 but in contrast, say if S = 105, which has 3 prime factors, by this > conjecture no non-zero k exists such that abs(105 - 2k^2) abs is meaningless in this case, look up another function, like erf is a prime number. The proof is oddly easy but I'll call it a conjecture in case I made a > mistake in the argument which can be found at my math blog. > James Harris > === Subject: Re: Primes conjecture with surrogate factoring >but in contrast, say if S = 105, which has 3 prime factors, by this >conjecture no non-zero k exists such that abs(105 - 2k^2) is a prime number. For S =105, your claim is false for k = 1, 2, 4, 7. quasi === Subject: Re: Primes conjecture with surrogate factoring >but in contrast, say if S = 105, which has 3 prime factors, by this >>conjecture no non-zero k exists such that >>abs(105 - 2k^2) >>is a prime number. For S =105, your claim is false for k = 1, 2, 4, 7. quasi In fact, I just noticed that you also allow abs, hence for S=105, there lots more counterexamples, probably infinitely many. For example, in the range 1, 2, 3, ... 1000, there are 19 values of k such that abs(105 - 2k^2) is prime. quasi === Subject: Re: Primes conjecture with surrogate factoring > Oddly enough looking at going in a different direction with what I > call the surrogate factoring congruence lead me to a fascinating > little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime > for any non-zero k, unless S is a product of two primes, and they have > to be differing primes. you should try to prove it, instead of stateing it as fact. no need for abs at all. Is that already known? It is kind of an odd little result. Here's an example of when a prime > is allowed, S = 77, k = 6, then 77 - 2*36 = 5 but in contrast, say if S = 105, which has 3 prime factors, by this > conjecture no non-zero k exists such that abs(105 - 2k^2) is a prime number. If you make broad conjecture by looking at a few numbers,you will always be wrong. Try a simple algebraic proof. You can post it here and I will correct it for you. - chow The proof is oddly easy but I'll call it a conjecture in case I made a > mistake in the argument which can be found at my math blog. > James Harris > === Subject: Re: Primes conjecture with surrogate factoring > Oddly enough looking at going in a different > direction with what I > call the surrogate factoring congruence lead me to a > fascinating > little result! Given a composite S, it is not possible for abs(S - > 2k^2) to be prime > for any non-zero k, unless S is a product of two > primes, and they have > to be differing primes. Is that already known? Let S = 45 = 3^2 x 5 (S does not factor into two primes) Let k = 1. Then: S - 2(1)^2 = 45 - 2 = 43 43 is prime. Is this a counter-example to your conjecture? M> Oddly enough looking at going in a different direction with what I > call the surrogate factoring congruence lead me to a fascinating > little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime > for any non-zero k, unless S is a product of two primes, and they have > to be differing primes. Counterexample: S = 27, k=2 === Subject: Re: Primes conjecture with surrogate factoring Oddly enough looking at going in a different direction with what I > call the surrogate factoring congruence lead me to a fascinating > little result! Given a composite S, it is not possible for abs(S - 2k^2) to be prime > for any non-zero k, unless S is a product of two primes, and they have > to be differing primes. Counterexample: S = 27, k=2 > I admit error. [...] My mistakes prove my dedication to getting the answer. And my honesty about them shows my humanity. So what do your attempts to remove evidence of your mistakes by deleting posts from Google (such as the opening post of this thread, for example) show? === Subject: Re: +1600 Solutions Manual to low cost Hey, Did he respond to any of you guys? Nope. I'm guessing its just some scam. Heaven forbid. Just because the vendor > * students who want to cheat, by offering > * pirated copyright works > doesn't necessarily mean that he also cheats his > marks^H^H^H^H^Hcustomers. And here's a clue for all you potential marks^H^H^H^H^Hcustomers: > require that the vendor first send you the first few pages (title page > and table of contents) before you part with your money. -- > --------------------------- > | BBB b Barbara at LivingHistory stop co stop uk > | B B aa rrr b | > | BBB a a r bbb | Quidquid latine dictum sit, > | B B a a r b b | altum viditur. > | BBB aa a r bbb | > ----------------------------- pliss Mechanical Engineering Design (6th Ed., Shigley) mail:david_westcoast@yahoo.es === Subject: Re: +1600 Solutions Manual to low cost > My List of Solutions Manual > (Available on print-original copy, Cd - for heavy files & by e-mail) contact me to : bergh...@yahoo.com If your solutions manual wanted ins't on this list, also can make the request. These are some only. 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Marghitu) + Ebook > Kinematics and Dynamics of Machinery (3rd Ed., Wilson & Sadler) > Kinematics, Dynamics, and Design of Machinery (2nd Ed., Waldron & Kinzel) > Mechanism Design: Analysis and Synthesis-Volume 1 (4th Ed., Erdman & Sandor) > Machines and Mechanisms: Applied Kinematic Analysis (3rd Ed., Myszka) > Mechanical Design: A Components Approach (Peter Childs) > Mechanical Design of Machine Elements and Machines: A Failure Prevention Perspective (Collins) > Fundamentals of Machine Component Design (3rd Ed., Juvinall) > Fundamentals of Machine Component Design (4th Ed., Juvinall) > Design of Machine Elements (8th Ed., Spotts) > Machine Design (Wentzell) > Solutions to problems at the text : Problems on the Design of Machine Elements (Faires) > Machine Elements in Mechanical Design (4th Ed., Mott) > Mechanical Design: An Integrated Approach (1st Ed., Ugural) > Design of Machinery (3rd Ed., Norton) > Machine Design (2nd Ed., Norton) > Machine Design : An Integrated Approach (3rd Ed., Norton) > Mechanical Engineering Design (6th Ed., Shigley) > Mechanical Engineering Design (7th Ed., Shigley) > Shigley's Mechanical Engineering Design (8th Ed., Budynas) > Fundamentals of Machine Elements (1st Ed., Hamrock) > Fundamentals of Machine Elements (2nd Ed., Hamrock) > Metal Fatigue in Engineering (2nd Ed., Stephens, Fatemi & Fuchs) > Principles of Metal Manufacturing Processes (Beddoes & Bibby) > Materials Selection in Mechanical Design (3rd Ed., Michael Ashby) > Introduction to Manufacturing Processes (3rd Ed., Schey) > Manufacturing, Engineering & Technology (4th Ed. Kalpakjian & Smith) > Manufacturing, Engineering & Technology (5th Ed. Kalpakjian & Smith) > Automation, Production Systems, and Computer-Integrated Manufacturing (2nd Ed., Groover) > Introduction to Robotics: Mechanics and Control (3rd Ed, Craig) > Applied Manufacturing Process Planning: With Emphasis on Metal Forming and Machining (Nelson, Schneider) > Mechanics of Materials: A Modern Integration of Mechanics and Materials in Structural Design (Christopher Jenkins & Sanjeev Khanna) > Mechanics of Materials (3th Ed., Beer) > Mechanics of Materials (5th Ed., Gere) > Mechanics of Materials (6th Ed., Gere) > Vector Mechanics for Engineers: Statics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (7th Ed., Ferdinand P. Beer) > Vector Mechanics for Engineers: Dynamics (8th Ed., Ferdinand P. Beer) > Statics: Analysis and Design of Systems in Equilibrium (Sheppard & Tongue) > Statics and Mechanics of Materials: An Integrated Approach (2nd Ed., Riley, Sturges & Morris) > Mechanics of Materials (6th Ed., Riley, Sturges & Morris) > Deformable Bodies and Their Material Behavior (Haslach & Armstrong) > Strength of Materials - Volume 1 : Elementary Theory and Problems (Timoshenko) > Intermediate Mechanics of Materials, (1st Ed., Barber) > Elasticity (2nd Ed., J.R. Barber) + Ebook > Elasticity: Theory, Applications, and Numerics (Martin Sadd) + Ebook > Elasticity in Engineering Mechanics (2nd Ed., Boresi) > Advanced Mechanics of Materials (6th Ed., Boresi) + Ebook > Engineering Mechanics: Dynamics (Boresi) > Flight Performance of Fixed and Rotary Wing Aircraft (Antonio Filippone) > Aircraft Structures for Engineering Students (4th Ed., T.H.G. 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Stronge) > Statics and Strengths of Materials (6th Ed., Morrow & Kokernak) > Engineering Mechanics : Statics (11th Ed., Hibbeler)-Not mathcad files converted to pdf, real instructor sol. manual > Principles of Statics (10th Ed., Hibbeler) > Engineering Mechanics : Dynamics (11th Ed., Hibbeler)-Not mathcad files converted to pdf, real instructor sol. manual > Principles of Dynamics (10th Ed., Hibbeler) > Mechanics of Materials (6th Ed, Hibbeler) > Statics and Mechanics of Materials (2nd Ed., Hibbeler) > Energy Principles and Variational Methods in Applied Mechanics (2nd Ed., Reddy) > Theory of Vibrations with Applications (5th Ed., Thomson & Dahleh) > Engineering Vibrations (2nd Ed., Inman) > Engineering Vibrations (3rd Ed., Inman) > Introduction to Finite Element Vibration Analysis (Maurice Petyt) > Vibrations and Stability: Advanced Theory, Analysis, and Tools (2nd Ed., Jon J. Thomsen) > Mechanical Vibrations (4th Ed., Rao) > The Finite Element Method and Applications in Engineering Using ANSYS (Erdogan Madenci, Ibrahim Guven) > Finite Element Analysis Theory and Application with ANSYS (2nd Ed., Moaveni) > Finite Element Analysis Theory and Application with ANSYS (3rd Ed., Moaveni) > The Finite Element Method and Applications in Engineering Using ANSYS (Madenci & Guven) + Ebook > Modeling and Analysis of Dynamic Systems (3rd Ed, Close, Frederick & Newell) > System Dynamics: Modeling and Simulation of Mechatronic Systems (4th Ed., Karnopp, Margolis & Rosenberg) > Concepts and Applications of Finite Element Analysis (4th Ed., Cook, Malkus, Plesha & Witt) > Advanced Strength and Applied Elasticity (4th Ed., Ugural) > Fracture Mechanics: An Introduction (2nd Ed., by E.E. Gdoutos) > Fracture Mechanics (2nd Ed., Anderson) > Mechanical Behavior of Materials (2nd Ed. Dowling) > Mechanical Behavior of Materials (3rd ... read more E hello > I need the solution manual for > Advanced Strength and Applied Elasticity (2 ,or 3th, or 4th Ed., > Ugural) hey I need the solution manual to this book > Vector Mechanics for > Engineers: Dynamics (7th Ed., Ferdinand P. Beer), if you have the one > for the 8th edition it would be better....what do i have to do to get? > either one will work can I have the Shigley's Mechanical Engineering Design (8th ed.) solutions? katie === Subject: Re: +1600 Solutions Manual to low cost Aite Frederick, I don't know if you have a stick up your butt all the time. But if you read the title it does say solution manual thread. Also, i'm a graduate student and having solutions to the book will help in my studies and where I don't do any homework or assignments. Please don't make a full of yourself and think you are the man on forums on the internet. Post to others: Any of you guys send the money and actually received the solution manuals? I'm definitely not parting my money until I know he is legit. === Subject: solutions manual Cc: bergh...@yahoo.com I need the solutions manual to this book. Structural Steel design 3rd edition. Mccormac and nelson by Prentice Hall. e-mail me at ycastel2@mail.usf.edu I need it fast. please send me your price quick I will pay! === Subject: Re: Solutions manual to low cost hey bergh@yahoo.com how do I get in touch with you to get Fundamentals of Communication Systems (John G. Proakis, Masoud Salehi)? Web976@aol.com === Subject: Engineering Mechanics - Statics (11th ) Engineering Mechanics - Statics (11th ) by R.C.HIBBELER Anyone have solutions to this? Please email me if you know where to get them!!! === Subject: solution Hi i need solution of Microwave Engineering 3rd Ed. David M Pozar === Subject: Millman, Grabel I would like to get a pdf copy of Solution Manual for Microelectronics by Jacob Millman and Arvin Grabel. Wilbur === Subject: Re: wmacias22@yahoo.com solutions manual trade,sell,etc just talk to me.. Introduction.to.Quantum.Mechanics.Solutions.Griffiths.2nd can you send this to me sabya_altius@rediffmail.com === Subject: Re: wmacias22@yahoo.com solutions manual trade,sell,etc just talk to me.. I need a solutions manual even problems prefer....for Discrete > Mathematics and its applic 6th edition by Kenneth Rosen. If you have > it email me back as soon as possible kakarotstyle@hotmail.com > === Subject: Re: wmacias22@yahoo.com solutions manual trade,sell,etc just talk to me.. I need a solutions (even) problems manual for Discrete Mathematics 6th edition by Kenneth Rosen. If any 1 has it email me back as soon as possible : kakarotstyle@hotmail.com === Subject: Can someone tell me what type of problem this is? Can someone tell me what this problem is called, and what I would be looking for? I am taking a pre algebra course and one of the questions in the book is this. however, they don't cover the problem in the text. http://img2.freeimagehosting.net/uploads/eb7dba0a4c.jpg Email: 3dheaven@comcast.net === Subject: Re: Can someone tell me what type of problem this is? days. My association with the Department is that of an alumnus. >Can someone tell me what this problem is called, It's not a problem, it is an expression. It reads the fifth root of 145 (or of 1.45, if that blurry dot two thirds of the way up the line between the 1 and the 4 is meant to be a period). > and what I would be looking for? A major that does not require you to know any basic arithmetic? -- Arturo Magidin magidin-at-member-ams-org === Subject: solution manual of A Systems Approach 3rd Ed. by larry L. Peterson & Bruce S. Davie Hi rakthik@gmail.com i need the solution manual to the following book in pdf format. Computer Networks: A Systems Approach 3rd Ed. by larry L. Peterson and Bruce S. Davie your help is appreciated. Also if possible could anyone upload the pdf version of the book Computer Networks: A Systems Approach 3rd Ed. by larry L. Peterson and Bruce S. Davie rak === Subject: relatively prime? 1 = a (m) + b (n) 1 = 17 ( ) + 20 ( ) what would be in place of m & n in the above example === Subject: Re: relatively prime? > 1 = 17 ( ) + 20 ( ) Quick solution using Math program... ExtendedGCD[17, 20] {1, {-7, 6}} Hence... 17*-7 + 20*6 = 1 -- Dana >1 = a (m) + b (n) > 1 = 17 ( ) + 20 ( ) what would be in place of m & n in the above example > === Subject: Re: relatively prime? days. My association with the Department is that of an alumnus. >1 = a (m) + b (n) >1 = 17 ( ) + 20 ( ) what would be in place of m & n in the above example http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm -- Arturo Magidin magidin-at-member-ams-org === Subject: modular arithmetic also, i am stuck on this problem. 2^1000 mod 7. how would i go about solving this? i know i need to break down the power, what is the easiest (and least painful) way to do it. === Subject: Re: modular arithmetic days. My association with the Department is that of an alumnus. >also, i am stuck on this problem. 2^1000 mod 7. >how would i go about solving this? i know i need to break down the >power, what is the easiest (and least painful) way to do it. 1. Compute the first few powers of 2 modulo 7: 2^0 = 1 (mod 7) 2^1 = 2 (mod 7) 2^2 = 4 (mod 7) 2^3 = 8 = 1 (mod 7) 2^4 = 16 = 2 (mod 7) . . . 2. Try to see the pattern. 3. Use the pattern. -- Arturo Magidin magidin-at-member-ams-org === Subject: Re: JSH: Teaching false math hello all, >i have a question. i need to find the least positive integer that is >congruent modulo 7 to the given product >(4)(9)(15)(59). >if anyone could show me step by step how to get it, i would really >appreciate..even hints and pointers would do! :) Okay, hint: it will be either 1, 2, 3, 4, 5, 6, or 7. Another hint? If a is congruent to b modulo 7, and x is congruent to y modulo 7, then ax is congruent to by modulo 7. -- Arturo Magidin magidin-at-member-ams-org === Subject: Surgery Obstruction Groups for Z[Z] I am considering a surgery problem, but I am a newbie to surgery. I have read the expository paper The Surgery-Theoretic Classification of High-Dimensional Smooth and Piecewise Linear Simply Connected Manifolds by Kelner (2002) and The Surgery Obstruction Groups of C.T.C. Wall by Lees (1973) Is it known what the surgery obstruction groups for the integral group ring of the integers, Ln(Z[Z]), are for n >= 5? If so, could you give me pointers to where the groups are enumerated? -- Jeffrey Rolland === Subject: Re: Space and anti-space > In principle, can two mathematical spaces, having dimensionality, > be constructed, where each is in some rigorous sense the other's > anti-space, in the sense that an appropriate combination of both [ Moderator's Note. Perhaps some situation in topology, where > (equivalence classes of) spaces, with an appropriate operation, form a > group? ] > Of interest may be paper by A.K. Steiner, April 1966 The Lattice of Topologies: Structure and Complementation Theorem 7.8. The lattice of topologies on any set is complemented. Moreover, each topology has a principal complement. === Subject: Re: Essential extensions of reduced rings. Originator: israel@math.ubc.ca (Robert Israel) > Let C be a category. Let x,y be in ob(C), then a monomorphism f : x -- >> y is said to be an essential extension > iff > for any g : y --> z such that fg is a monomorphism ==> g is a > monomorphism Ok, what did you do here to have a rewording in the category of > commutative rings using properties which can be immediately verified (at > least in principle)? > And later on... > Are you sure that you mean f o g? The morphisms don't seem to be > composable as you defined g. Sorry for the confusion I caused.. of course I meant (g o f) so again.. Let C be a category and x,y be in Ob(C). Then a monomorphism f : x -> y is said to be an essential extension iff [ If g : y -> z such that (g o f) : x->y->z is a monomorphism ==> g : y -> z is a monomorphism ] You asked what I did to have a rewording in the category of commutative unitary rings (Crings).. Recall that originally I defined in Crings that a Cring B is an essential extension of a Cring A iff for any non-null ideal I of B, I / A is nonull.. Thus the canonical injection f : A --> B is claimed to be essential in the category Crings.. one can easily show that this is equivalent with the category theoretical definition I just gave. Proof: ==> Suppose by contradiction that A --> B is not essential in the ring theoretical definition. Let I be an ideal of B that is non-null and yet I/A is null. Then the composition of the canonical maps A --> B --> B/I is a monomorphism in Crings.. . and thus B --> B/I is a monomorphism in Crings.. but this is only possible if I=(0) .. a contradiction <== Let B --> C be a Cring morphism such that A --> B --> C be a monomorphism. If B --> C is NOT a monomorphism, then I:=ker(B --> C) is a non-null ideal in B and has I / A = (0) , a contradiction. Jose Capco === Subject: Properties of a Limit of a Sequence of Sets Originator: israel@math.ubc.ca (Robert Israel) What properties of a set pass through the limit? For example, given a sequence of measurable sets {A_n} with measurable limit set A, and a set X, is it true that 1) lim_{n-->inf} d(A_n, X) = d(A,X); 2) If A_n are all convex, the limit is convex; 3) lim_{n-->inf} mu(A_n) = mu(A), assuming that {A_n} is contained in a compact subset of R_n; 4) Other possible properties? Also, does the answer depend on whether the convergence is analytic (lim sup=lim inf) or topological (Hausdorff)? === Subject: Re: Secondary group cohomology Originator: israel@math.ubc.ca (Robert Israel) > Ordinary group cohomology is derived from the invariant functor H^0. > Now let G be a group and M a G-module. > Let S(M) be the subgroup of M consisting of all m such that > g.m-m is in H^0(M) for every g in G. > Then S is a left exact functor from G-modules to abelian groups. > Let's call the derivatives the secondary cohomology functors. > Anythin interesting about them? Do they describe obstructions? > Are they connected to topological constructions on the classifying space > of G? Do they occur in literature? I haven't come across this secondary cohomology before, but there has been recent interest in second-order modular forms which are more-or-less functions f on the upper half-plane such that (f|_k A) - f is a modular form for G for all A in G (where G is your favourite subgroup of the modular group). See for instance: Diamantis, N.(4-NOTT-SM); Knopp, M.(1-TMPL); Mason, G.(1-UCSC); O'Sullivan, C.(1-CUNYX) $L$-functions of second-order cusp forms. (English summary) Ramanujan J. 12 (2006), no. 3, 327--347. 11F12 (11F66) Some easy observations about the functor S: S(M) consists of those m in M with (h-1)(g-1)m = 0 for all g, h in G. Thus S(M) is the set of elements annihilated by the square of the augmentation ideal I of ZG. Thus S(M) = Hom_G(ZG/I^2,M). Thus the derived functors of S are simply the Ext-functors Ext_{ZG}^*(ZG/I^2, -). Analogously one could define secondary homology by deriving the functor ZG/I^2 (x) -. These are the Tor-functors Tor^{ZG}_*(ZG/I^2, -). Both these homology and cohomology theories can be derived from a free resolution of ZG/I^2. I can't at the moment see a simple resolution of this analogous to the standard resolution of ZG/I, alas. It's easy to see that gh-hg lies in I^2 for all g, h in G. Thus in the case where G is a perfect group. I^2 = I so that secondary (and higher!) (co)homology coincides with standard (co)homology. In general a Z-basis of Z/I^2 is given by the g + I^2 where g runs over a system of coset representatives in G for the derived subgroup G'. Robin Chapman === Subject: Re: Measure-preserving words Originator: israel@math.ubc.ca (Robert Israel) Hi! Indeed there is some combinatorial group theory underneath this Benjamin === Subject: Primitive words Originator: israel@math.ubc.ca (Robert Israel) Hi all, I have currently read the following statement in a paper: Let u be a primitive word of F_n (free group in n letters x_1,...,x_n). By definition of primitive words, there is an automorphism A in Aut(F_n), such that A(x_1) = u. Can this be true? I mean, the word u = (x_1)^2 (x_2)^2 is primitive in F_2, but how can there be an automorphism A: F_2 -> F_2 such that A(x_1) = (x_1)^2 (x_2)^2? I would guess that there can be no inverse to A, since having an inverse of A would mean to have two words w_1, w_2 in the two letters x_1, x_2 such that (w_1)^2(w_2)^2 = x_1, right? But simply by counting exponents, this cannot be true: in (w_1)^2(w_2)^2, both x_1 and x_2 appear with even powers, contrary to x_1. Or am I missing something? Benjamin === Subject: Re: Primitive words Originator: israel@math.ubc.ca (Robert Israel) > Hi all, I have currently read the following statement in a paper: Let u be a primitive word of F_n (free group in n letters > x_1,...,x_n). By definition of primitive words, there is an > automorphism A in Aut(F_n), such that A(x_1) = u. When asking questions like this, it would be helpful if you told us what your definition of primitive word is. I believe that the standard definition is a word (or element) in a free group is primitive if it forms part of a free basis of the group. Since for any two free bases of the group there is an automorphism mapping one to the other, it follows immediately that there is an automorphism mapping x_1 to the primitive element. So this property does indeed follow directly from the definition. > Can this be true? I mean, the word u = (x_1)^2 (x_2)^2 is primitive But it isn't primitive according to the standard definition! Derek Holt. > in > F_2, but how can there be an automorphism A: F_2 -> F_2 such that > A(x_1) = (x_1)^2 (x_2)^2? I would guess that there can be no inverse > to A, since having an inverse of A would mean to have two words w_1, > w_2 in the two letters x_1, x_2 such that (w_1)^2(w_2)^2 = x_1, right? > But simply by counting exponents, this cannot be true: in > (w_1)^2(w_2)^2, both x_1 and x_2 appear with even powers, contrary to > x_1. Or am I missing something? > Benjamin === Subject: Re: Primitive words Originator: israel@math.ubc.ca (Robert Israel) >> Hi all, >> I have currently read the following statement in a paper: >> Let u be a primitive word of F_n (free group in n letters >> x_1,...,x_n). By definition of primitive words, there is an >> automorphism A in Aut(F_n), such that A(x_1) = u. When asking questions like this, it would be helpful if you told us >what your definition of primitive word is. I believe that the >standard definition is a word (or element) in a free group is >primitive if it forms part of a free basis of the group. I wondered about that myself... [...] >> Can this be true? I mean, the word u = (x_1)^2 (x_2)^2 is primitive But it isn't primitive according to the standard definition! Benjamin may have been misled by something along the following lines: taken from Combinatorics on Words by M. Lothaire, Cambridge Mathematical Library. Let A be an alphabet, and let A* be the monoid of all words in A. On p 7, section 1.3, Lothaire defines: A word x in A* is said to be primitive if it is not a power of another word; that is, if x is not the trivial word, and for all z in A*, if w is in {z}*, then w = z. Under that interpretation, his u would be primitive. -- Arturo Magidin magidin-at-member-ams-org === Subject: Re: Primitive words Originator: israel@math.ubc.ca (Robert Israel) Hi! Yes, in fact the definition of primitive that I have read about says not the proper power of another word. But if the definition in this elements in F_n such that they together freely generate F_n, then I would understand it, the above quote would make much more sense to me! It is just that (since I don't know much about group theory) I haven't heard this definition before. By the way: Given the latter definition of primitive, is there a good way of deciding, whether a given word is primitive? With the exmaple (x_1)^n(x_2)^n this is probably easy to see, but what about generic words? Is there a good reference for this? Benjamin === Subject: Re: Primitive words Originator: israel@math.ubc.ca (Robert Israel) > Hi! Yes, in fact the definition of primitive that I have read about says > not the proper power of another word. But if the definition in this > elements in F_n such that they together freely generate F_n, then I > would understand it, the above quote would make much more sense to me! > It is just that (since I don't know much about group theory) I haven't > heard this definition before. > By the way: Given the latter definition of primitive, is there > a good way of deciding, whether a given word is primitive? With the > exmaple (x_1)^n(x_2)^n this is probably easy to see, but what about > generic words? Is there a good reference for this? > Yes, there is an algorithm due to Whitehead for deciding whether an element of a free group is primitive. Let F be free on x1,..,xr. Then an automorphism f of F is called a Whitehead automorphism if either (i) f permutes the set X := { x1, ..., xr, x1^-1, ..., xr^-1 }; or (ii) There is some fixed x in X such that f(x)=x and, for each y in X, f(y) is one of y, yx, x^-1y, or x^-1yx. Then Whitehead's Theorem says that in any equivalence class of elements of F under Aut(F), it is possible to reduce any (reduced) word in F to a word of minimal length in the class by applying a sequence of Whitehead automorphisms (of Type (ii)) each of which strictly reduces the length of the word. So if the element is primitive then it can be reduced to a word of length 1. On the other hand, if the word has length more than one and no Whitehead automorphism reduces its length, then it is not primitive. Of course there is a large number of Whitehead automorphisms, particularly when r is not small, but the algortihm can easily be implemented on a computer. There is a generalization to test whether any subset of F can be extended to a free basis. Derek Holt. === Subject: impact factor Originator: israel@math.ubc.ca (Robert Israel) hi friends I am vikash. I am confused about impact factor of journals. So please let me know whether a journal with high impact factor is better or a journal with low impact factor is better.Also i want to know, how to calculate impact factor of a journal. Vikash === Subject: Re: impact factor Originator: tchow@lebesgue.mit.edu.mit.edu (Timothy Chow) Originator: israel@math.ubc.ca (Robert Israel) >So please let me know whether a journal with high impact >factor is better or a journal with low impact factor is better. Journals with high impact factors are *supposedly* better. The importance of the impact factor is controversial. For some skeptical such as the following: http://www.ams.org/notices/200603/comm-milman.pdf http://www.ams.org/ewing/Documents/Judgingjournals.pdf http://www.ams.org/notices/200707/tx070700821p.pdf -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: impact factor Originator: israel@math.ubc.ca (Robert Israel) > hi friends > I am vikash. I am confused about impact factor of > journals. So please let me know whether a journal with high impact > factor is better or a journal with low impact factor is better.Also i > want to know, how to calculate impact factor of a journal. Vikash Hi; I have just taken the definition of impact factor from help page of Thomson Corporation and it is better to choose journal having higher impact factor in order to publish your manuscript. Impact Factor: the journal published in the past two years have been cited in the JCR year. The impact factor is calculated by dividing the number of citations in previous years. An impact factor of 1.0 means that, on average, the one or two year ago have been cited two and a half times. Citing different journals. The aggregate impact factor for a subject category is calcuted the same way as the impact factor for a journal, but it takes into account the number of citations to all journals in the category and the number category published one or two years ago have been cited one time. The median impact factor is the median value of all journal impact factors in the subject category. The impact factor mitigates the importance of absolute citation frequencies. It tends to discount the advantage of large journals over small journals because large journals produce a larger body of citable literature. For the same reason, it tends to discount the advantage of frequently issued journals over less frequently issued ones and of older journals over newer ones. Because the journal impact factor offsets the advantages of size and age, it is a valuable tool for journal evaluation. The impact factor trend graph shows the impact factor for a five-year period. To view the graph, click the Impact Factor Trend button at the top of the journal page. web address : http://admin.isiknowledge.com/JCR/help/h_impfact.htm#impact_factor === Subject: Re: impact factor Originator: israel@math.ubc.ca (Robert Israel) > hi friends > I am vikash. I am confused about impact factor of > journals. So please let me know whether a journal with high impact > factor is better or a journal with low impact factor is better.Also i > want to know, how to calculate impact factor of a journal. Vikash See http://en.wikipedia.org/wiki/Impact_factor === Subject: Re: impact factor Originator: israel@math.ubc.ca (Robert Israel) + vicky : > I am confused about impact factor of journals. So > please let me know whether a journal with high impact factor is > better or a journal with low impact factor is better.Also i want to > know, how to calculate impact factor of a journal. calculate the impact factor and a discussion of its usefulness. Yes, one generally expexts higher quality journals to have a higher impact factor, and no, the correspondence is not exact. Far from it. -- * Harald Hanche-Olsen - It is undesirable to believe a proposition when there is no ground whatsoever for supposing it is true. -- Bertrand Russell === Subject: Re: impact factor Originator: israel@math.ubc.ca (Robert Israel) > hi friends > I am vikash. I am confused about impact factor of > journals. So please let me know whether a journal with high impact > factor is better or a journal with low impact factor is better. High is better. (Among journals in the same field.) > Also i > want to know, how to calculate impact factor of a journal. Vikash > http://scientific.thomson.com/free/essays/journalcitationreports/impactf actor/ === Subject: Re: impact factor Originator: israel@math.ubc.ca (Robert Israel) >hi friends > I am vikash. I am confused about impact factor of >journals. So please let me know whether a journal with high impact >factor is better or a journal with low impact factor is better.Also i >want to know, how to calculate impact factor of a journal. High is better. Impact factor is essentially an average number You can find impact factors as computed by Thomson ISI at http://portal.isiknowledge.com/portal.cgi and a description of the methodology at http://scientific.thomson.com/free/essays/journalcitationreports/impactfacto r/ -Ted -- [E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] === Subject: odd perfect number Originator: israel@math.ubc.ca (Robert Israel) The requirement for an odd perfect number N = p1^k1 * p2^k2 * ..... is that 2 * p1^k1 * p2^k2 * ..... = (p1^(k1+1) / (p1-1)) * (p2^(k2+1) / (p2-1)) * ..... Of course, no odd perfect number has ever been found. Suppose we relax the requirement that each p be prime, that is, we require each p to be merely a distinct odd integer. Obviously, N will not be a perfect number, but are there any solutions to the equation? If so, what? === Subject: Re: odd perfect number Originator: israel@math.ubc.ca (Robert Israel) > The requirement for an odd perfect number N = p1^k1 * p2^k2 * ..... is > that > 2 * p1^k1 * p2^k2 * ..... = (p1^(k1+1) / (p1-1)) * (p2^(k2+1) / > (p2-1)) * ..... Of course, no odd perfect number has ever been found. Suppose we relax the requirement that each p be prime, that is, we > require each p to be merely a distinct odd integer. Obviously, N will > not be a perfect number, but are there any solutions to the equation? > If so, what? Descartes found that if you make believe 22021 is prime then 3^2 7^2 11^2 13^2 22021 is an odd perfect number. See B1 in Guy, Unsolved Problems in Number Theory. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: odd perfect number Originator: israel@math.ubc.ca (Robert Israel) On Sep 20, 1:59 pm, Gerry Myerson The requirement for an odd perfect number N = p1^k1 * p2^k2 * ..... is > that > 2 * p1^k1 * p2^k2 * ..... = (p1^(k1+1) / (p1-1)) * (p2^(k2+1) / > (p2-1)) * ..... Of course, no odd perfect number has ever been found. Suppose we relax the requirement that each p be prime, that is, we > require each p to be merely a distinct odd integer. Obviously, N will > not be a perfect number, but are there any solutions to the equation? > If so, what? Descartes found that if you make believe 22021 is prime > then 3^2 7^2 11^2 13^2 22021 is an odd perfect number. > See B1 in Guy, Unsolved Problems in Number Theory. -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email) Hi Gerry, result... Tim === Subject: Asymptotic evaluation of a stochastic integral Originator: israel@math.ubc.ca (Robert Israel) Is it possible to evaluate the integral int(g(x)*exp(if(x)L),x=a..b) with L a large parameter and f(x) a stochastic function with =0 =C(x1-x2) with an asymptotic series in 1/L? Also references are good as well. Jon === Subject: second-order Gauss-Markov process Importance: Normal Originator: israel@math.ubc.ca (Robert Israel) Let X(t) denote a continuous-time random process. If X(t) is stationary, Gaussian, Markovian and continuous in probability, then X(t) satisfies the following linear stochastic differential equation [1,2,3]: dX(t) = -rho*(X_t - mu)*dt + sigma*dW(t) where mu, sigma & rho are constants and W(t) is Brownian motion. X(t) is often called an Ornstein-Uhlenbeck process or a first-order Gauss-Markov process. A process Y(t) is second-order Markovian if, for every k and every set of timepoints t_1 < t_2 < ... < t_k, it is true that P( Y(t_k) < y | Y(t_{k-1}), Y(t_{k-2}), ..., Y(t_2), Y(t_1) ) = P( Y(t_k) < y | Y(t_{k-1}), Y(t_{k-2}) ) That is, the probability distribution of Y(t) depends only on the two points immediately in the past. If Y(t) is stationary, Gaussian, second-order Markovian and continuous in probability, then Y(t) would seem to satisfy a certain second-order linear SDE [4]. Does anyone know a reference for a rigorous proof of Steve Finch http://algo.inria.fr/bsolve/ References 1. J. L. Doob, The Brownian movement and stochastic equations, Annals of Math. 43 (1942) 351-369. 2. L. Breiman, Probability, Addison-Wesley, 1968, pp. 347-351. 3. I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, Springer-Verlag, 1988, p. 358. 4. A. Gelb et al, Applied Optimal Estimation, MIT Press, 1974. _________________________________________________________________ Kick back and relax with hot games and cool activities at the Messenger Caf.8e. http://www.cafemessenger.com?ocid=TXT_TAGLM_SeptWLtagline === Subject: Independent sets problem Originator: israel@math.ubc.ca (Robert Israel) Hello everyone, Let A be a nonempty finite set. Let I be a nonempty system of subsets of A such that every pair of distinct sets in I is incomparable with respect to inclusion (in other words, I is an antichain in the poset 2^A). Problem: Characterize systems (A,I) such that there is a mapping f:A- >N such that * f is not everywhere zero and * for every B in I, the sum of all f(x), where x runs through all elements of B is always the same. This a generalization of a problem from finite orthomodular lattices; A is the set of all atoms and I is the system of all maximal compatible subsets of A. The existence of f is equivalent to the existence of a state on the orthomodular lattice. Initial observations: * If the intersection of the system I is nonempty, then the answer is yes: put f(x)=1 for all elements x of the intersection and f(x)=0 otherwise. * If all sets in the system I are of the same cardinality, the answer is yes: any nonzero constant mapping f will do. * If there is a subset S of A such that, for all B in I, (S intersection B) is a singleton, then the answer is yes: put f(x)=1 for all x in S and f(x)=0 otherwise. A sufficient condition for nonexistence of such f can be the following: if there are subsets I1, I2 of I such that * I1 and I2 are disjoint, * the sets in I1 are pairwise disjoint, * the sets in I2 sre pairwise disjoint, * card(I1) is not equal to card(I2), * (union I1)=(union I2)=A then f does not exist. Questions: 1) Has anyone seen anything like this before? Any references appreciated. 2) Consider the abstract simplicial complex (A,I), where A are points and I are maximal simplices. Is there any connection to homology groups of this complex? My intuition says: If f does not exist, the Betti number of (A,I) must be big with repect to card(A). 3) Is the sufficient condition for nonexistence of f (mentioned above) necessary? Gejza Jenca Dept. of Mathematics Slovak Technical University Bratislava === Subject: Two papers published by Geometry & Topology Monographs Originator: israel@math.ubc.ca (Robert Israel) Geometry and Topology Publications are pleased to announce completion of the following monograph: Editors: David Auckly and Jim Bryan The final two papers are: (1) Geometry & Topology Monographs 8 (2006) 167-194 Hodge-type integrals on moduli spaces of admissible covers by Renzo Cavalieri URL: http://www.msp.warwick.ac.uk/gtm/2006/08/p008.xhtml DOI: 10.2140/gtm.2006.8.167 (2) Geometry & Topology Monographs 8 (2006) 195-456 Introduction to the Gopakumar-Vafa Large N Duality by Dave Auckly and Sergiy Koshkin URL: http://www.msp.warwick.ac.uk/gtm/2006/08/p009.xhtml DOI: 10.2140/gtm.2006.8.195 Abstracts follow: (1) Hodge-type integrals on moduli spaces of admissible covers by Renzo Cavalieri In this paper we study a natural class of intersection numbers on moduli spaces of degree d admissible covers from genus g curves to P^1, using techniques of localization. These intersection numbers involve tautological lambda and psi classes, and are in some sense analogous to Hodge Integrals on moduli spaces of stable curves. We compute explicitly these numbers for all genera in degrees 2 and 3 and express the result in generating function form; we provide a conjecture for the general degree d case. (2) Introduction to the Gopakumar-Vafa Large N Duality by Dave Auckly and Sergiy Koshkin Gopakumar-Vafa Large N Duality is a correspondence between Chern-Simons invariants of a link in a 3-manifold and relative Gromov-Witten invariants of a 6-dimensional symplectic manifold relative to a Lagrangian submanifold. We address the correspondence between the Chern-Simons free energy of S^3 with no link and the Gromov-Witten invariant of the resolved conifold in great detail. This case avoids mathematical difficulties in formulating a definition of relative Gromov-Witten invariants, but includes all of the important ideas. There is a vast amount of background material related to this duality. We make a point of collecting all of the background material required to check this duality in the case of the 3-sphere, and we have tried to present the material in a way complementary to the existing literature. This paper contains a large section on Gromov-Witten theory and a large section on quantum invariants of 3-manifolds. It also includes some physical motivation, but for the most part it avoids physical terminology. === Subject: normal form of matrices under congruence transformations Originator: israel@math.ubc.ca (Robert Israel) I am looking for information on the normal form of matrices with real entries under congruence transformations. A equiv B <=> B=G^TAG for some invertible G Symmetric matrices are classified by their signature, and antisymmetric matrices by their rank; what is the classification in general? Can one always choose G such that B is block diagonal with diagonal blocks of size 1 or 2, with blocks of size 2 of the form diagonal+antidsymmetric? Surely this problem has been solved completely many, many years ago, but I don't know how to find out where... Arnold Neumaier === Subject: The universal algebra of quandles Originator: israel@math.ubc.ca (Robert Israel) In order to do some work with welded knots and surface knots I have run into questions involving the universal algebra of quandles. However, I have had trouble tracking down sources for this. Does anyone know of any papers studying this problem? From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! 22g2000hsm.googlegroups.com!not-for-mail From: JSH Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Surrogate factoring algorithm Date: Sat, 22 Sep 2007 02:04:37 -0000 Organization: http://groups.google.com Lines: 75 Message-ID: <1190426677.656410.200610@22g2000hsm.googlegroups.com> NNTP-Posting-Host: 76.102.237.111 Mime-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 X-Trace: posting.google.com 1190426677 7395 127.0.0.1 (22 Sep 2007 02:04:37 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 22 Sep 2007 02:04:37 +0000 (UTC) User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.7) Gecko/20070914 Firefox/2.0.0.7,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: 22g2000hsm.googlegroups.com; posting-host=76.102.237.111; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109479 alt.math.undergrad:57772 alt.math:17947 X-Received-Date: Fri, 21 Sep 2007 22:04:37 EDT (newsdbm02.news.prodigy.net) Simple relations connect every integer factorization to an infinity of others as I have shown with simple congruences: In the ring of integers given x^2 = y^2 mod T and k = 2x mod T it must be true that (x+k)^2 = y^2 + S where S = 2k^2 mod T so one congruence of squares is connected to an infinity of others. That can be used to factor in either direction, but I will focus on letting S be the target composite to be factored in this post. If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a non-trivial factorization is given by x + k - y = f_1 and x + k + y = f_2 so 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. So the algorithm is that given a target S to be factored, choose a non- integer k, and find T, from S - 2k^2 = 0 mod T, so if factor_pool = S - 2k^2, then you factor factor_pool, and loop through combinations of factors of factor_pool, using each combination as T. Then you find x from x = 2^{-1} k mod T, and can then check sqrt((x+k)^2 - 4S) to see if it is an integer and if it is then f_1 = (-(x+k) + sqrt((x+k)^2 - 4S))/2 but how do you pick k? Well it can be shown that 3k - f_1 - f_2 = 0 mod T so if S is odd, then k cannot be even, unless you get lucky and 3k - f_1 - f_2 = 0. So k should be odd. Also, if k has 3 as a factor, and f_1 + f_2 does as well, then it is blocked from working unless S has 3 as a factor, so k should be coprime to 3. Other than that any k can be chosen so it would make sense to choose k such that S - 2k^2 is a minimum with k even and not divisible by 3. Oddly enough this may be a perfect factoring algorithm which would end considering the factoring problem being a hard problem, but that depends on there not being some other conditions which would stop a particular k from working, or some other reason that looping through the combination of factors of the factor_pool would not give a non- trivial factorization. This method is more specific than my previous surrogate factoring methods going the other way where T is a target as with those it was not possible to prove that only a single k with some minimal qualifications should work. James Harris From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!newspeer.monmouth.com!newsfeed.icl.net! newsfeed.fjserv.net!oleane.net!oleane!news.ecp.fr!club-internet.fr! feedme-small.clubint.net!aioe.org!not-for-mail From: biggus Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sat, 22 Sep 2007 12:02:33 -0500 Organization: dd34e@yahoo.com Lines: 20 Message-ID: References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> Reply-To: biggus NNTP-Posting-Host: G8ScRq70w4hSPI3vK9mB4Q.user.aioe.org X-Complaints-To: abuse@aioe.org X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-RFC2646: Format=Flowed; Original X-Newsreader: Microsoft Outlook Express 6.00.2900.3138 X-Priority: 3 X-MSMail-Priority: Normal Xref: prodigy.net sci.math:1109596 alt.math.undergrad:57783 alt.math:17954 X-Received-Date: Sat, 22 Sep 2007 13:01:37 EDT (newsdbm02.news.prodigy.net) JSH wrote in message news:1190426677.656410.200610@22g2000hsm.googlegroups.com... sqrt((x+k)^2 - 4S) > this is wrong, numbers are complex, Who would ever put complex numbers in a simplistic factoring search ? OBW, do you mean + or - sqrt() ? > James Harris > => moron troll <= From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! o80g2000hse.googlegroups.com!not-for-mail From: marcus_b Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sat, 22 Sep 2007 09:17:02 -0700 Organization: http://groups.google.com Lines: 50 Message-ID: <1190477822.973052.182610@o80g2000hse.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> NNTP-Posting-Host: 128.101.116.194 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190477823 23886 127.0.0.1 (22 Sep 2007 16:17:03 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 22 Sep 2007 16:17:03 +0000 (UTC) In-Reply-To: <1190426677.656410.200610@22g2000hsm.googlegroups.com> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.0.8) Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: o80g2000hse.googlegroups.com; posting-host=128.101.116.194; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109586 alt.math.undergrad:57782 alt.math:17953 X-Received-Date: Sat, 22 Sep 2007 12:17:03 EDT (newsdbm02.news.prodigy.net) On Sep 21, 9:04 pm, JSH wrote: > Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: In the ring of integers given x^2 = y^2 mod T and k = 2x mod T it must be true that (x+k)^2 = y^2 + S where S = 2k^2 mod T so one congruence of squares is connected to an infinity of others. That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a > non-trivial factorization is given by x + k - y = f_1 and x + k + y = f_2 so 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. So the algorithm is that given a target S to be factored, choose a non- > integer k, and find T, from S - 2k^2 = 0 mod T, so if factor_pool = S - 2k^2, then you factor > factor_pool, and loop through combinations of factors of factor_pool, > using each combination as T. Then you find x from x = 2^{-1} k mod T, and can then check sqrt((x+k)^2 - 4S) > Why can't you get this right? Marcus. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!border1.nntp.dca.giganews.com!nntp.giganews.com ! postnews.google.com!r29g2000hsg.googlegroups.com!not-for-mail From: mike3 Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 11:08:18 -0700 Organization: http://groups.google.com Lines: 56 Message-ID: <1190570898.806448.149060@r29g2000hsg.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> NNTP-Posting-Host: 74.38.33.242 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190570899 27823 127.0.0.1 (23 Sep 2007 18:08:19 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 18:08:19 +0000 (UTC) In-Reply-To: <1190477822.973052.182610@o80g2000hse.googlegroups.com> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.1) Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: r29g2000hsg.googlegroups.com; posting-host=74.38.33.242; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Bytes: 2735 Xref: prodigy.net sci.math:1109864 alt.math.undergrad:57801 alt.math:17961 X-Received-Date: Sun, 23 Sep 2007 14:08:20 EDT (newsdbm02.news.prodigy.net) On Sep 22, 10:17 am, marcus_b wrote: > On Sep 21, 9:04 pm, JSH wrote: Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: In the ring of integers given x^2 = y^2 mod T and k = 2x mod T it must be true that (x+k)^2 = y^2 + S where S = 2k^2 mod T so one congruence of squares is connected to an infinity of others. That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a > non-trivial factorization is given by x + k - y = f_1 and x + k + y = f_2 so 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. So the algorithm is that given a target S to be factored, choose a non- > integer k, and find T, from S - 2k^2 = 0 mod T, so if factor_pool = S - 2k^2, then you factor > factor_pool, and loop through combinations of factors of factor_pool, > using each combination as T. Then you find x from x = 2^{-1} k mod T, and can then check sqrt((x+k)^2 - 4S) Why can't you get this right? Marcus. Because James likes to think that his way must be right, and nobody else's can be. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! 57g2000hsv.googlegroups.com!not-for-mail From: marcus_b Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 11:44:12 -0700 Organization: http://groups.google.com Lines: 96 Message-ID: <1190573052.288861.89410@57g2000hsv.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> <1190570898.806448.149060@r29g2000hsg.googlegroups.com> NNTP-Posting-Host: 128.101.116.194 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190573052 3129 127.0.0.1 (23 Sep 2007 18:44:12 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 18:44:12 +0000 (UTC) In-Reply-To: <1190570898.806448.149060@r29g2000hsg.googlegroups.com> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.0.8) Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: 57g2000hsv.googlegroups.com; posting-host=128.101.116.194; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109879 alt.math.undergrad:57803 alt.math:17962 X-Received-Date: Sun, 23 Sep 2007 14:44:14 EDT (newsdbm02.news.prodigy.net) On Sep 23, 1:08 pm, mike3 wrote: > On Sep 22, 10:17 am, marcus_b wrote: On Sep 21, 9:04 pm, JSH wrote: Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: In the ring of integers given x^2 = y^2 mod T and k = 2x mod T it must be true that (x+k)^2 = y^2 + S where S = 2k^2 mod T so one congruence of squares is connected to an infinity of others. That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a > non-trivial factorization is given by x + k - y = f_1 and x + k + y = f_2 so 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. So the algorithm is that given a target S to be factored, choose a non- > integer k, and find T, from S - 2k^2 = 0 mod T, so if factor_pool = S - 2k^2, then you factor > factor_pool, and loop through combinations of factors of factor_pool, > using each combination as T. Then you find x from x = 2^{-1} k mod T, and can then check sqrt((x+k)^2 - 4S) Why can't you get this right? Marcus. Because James likes to think that his way must be right, > and nobody else's can be. Actually I think that's not the reason in this case. I was referring not to his general approach, but specifically to the expression sqrt((x + k)^2 - 4S), which should have been sqrt((2(x + k))^2 - 4S), which is somewhat corrected from his first version of this, which was sqrt(((x + k)/2)^2 + 4S). The real reason he is not getting this right I think is that he is rushing headlong in an attempt to find something, ANYTHING that might work, after his previous approach [using S as the surrogate, T as the target] was shown to perform inefficiently and contrary to his 'theory'. He is flailing about frantically, hoping that some kind of magic will enable him to stumble on an algorithm that other people have overlooked. He is the Las Vegas gamble sticking quarter after quarter in the $1M slot machine, hoping to strike it rich. Like the gambler and unlike, e.g., Dixon or Pomerance, there is no rationale behind his attempts. Factors of the target and factors of the surrogate, whichever is which, have no useful relationship to each other. Ironically, using the wrong formula for the expression above is just about as likely to produce the result he wants as using the right one. He might as well compute random GCDs. Marcus. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!border1.nntp.dca.giganews.com!nntp.giganews.com ! local01.nntp.dca.giganews.com!nntp.bt.com!news.bt.com.POSTED!not-for-mail NNTP-Posting-Date: Sun, 23 Sep 2007 14:42:44 -0500 From: rossum Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 20:41:45 +0100 Message-ID: References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> <1190570898.806448.149060@r29g2000hsg.googlegroups.com> <1190573052.288861.89410@57g2000hsv.googlegroups.com> X-Newsreader: Forte Agent 1.93/32.576 English (American) MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Lines: 29 X-Usenet-Provider: http://www.giganews.com NNTP-Posting-Host: 86.142.182.237 X-AuthenticatedUsername: NoAuthUser X-Trace: sv3-csPzPYJlrXhh5c/39cM0HBxirhuqWBS4bBikAS1s3AtaIKtWv1/ R3ZzcsZv95sCuYPJRTo2qx3XlUCX!3qZwxrH2eA5b5Qc+W7sMDXseo3tKyK1jcHwUrPhaRA0NaVG / 8KVLiYsD/HtMxpR8RFAyvxgmFsTG!lAy/ X-Complaints-To: abuse@btinternet.com X-DMCA-Complaints-To: abuse@btinternet.com X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.36 Bytes: 2674 Xref: prodigy.net sci.math:1109900 alt.math.undergrad:57805 alt.math:17963 X-Received-Date: Sun, 23 Sep 2007 15:42:41 EDT (newsdbm02.news.prodigy.net) On Sun, 23 Sep 2007 11:44:12 -0700, marcus_b wrote: >He is the Las Vegas gamble sticking quarter after >quarter in the $1M slot machine, hoping to strike it >rich. Like the gambler and unlike, e.g., Dixon or >Pomerance, there is no rationale behind his attempts. >Factors of the target and factors of the surrogate, >whichever is which, have no useful relationship to each >other. Ironically, using the wrong formula for the >expression above is just about as likely to produce >the result he wants as using the right one. He >might as well compute random GCDs. Marcus. When coding up James' many different versions of his method I sometimes make mistakes. It is not unusual for the mistakes to perform as well as or better than James' actual method. As you say, to a first approximation his various methods are inefficient PRNGs. It may be that James is actually developing a meta-method. He creates numerous random versions of his factoring algorithm and throws them to us for some ruthless selection. Unfortunately he does not then go to the next stage and develop those versions that fare least badly. Eventually he might find a genetic algorithm to develop new mathematical algorithms using sci.math for the selection part. rossum From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! n39g2000hsh.googlegroups.com!not-for-mail From: marcus_b Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 14:31:22 -0700 Organization: http://groups.google.com Lines: 40 Message-ID: <1190583082.068491.238720@n39g2000hsh.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> <1190570898.806448.149060@r29g2000hsg.googlegroups.com> <1190573052.288861.89410@57g2000hsv.googlegroups.com> NNTP-Posting-Host: 128.101.116.194 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190583082 7000 127.0.0.1 (23 Sep 2007 21:31:22 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 21:31:22 +0000 (UTC) In-Reply-To: User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.0.8) Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: n39g2000hsh.googlegroups.com; posting-host=128.101.116.194; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109925 alt.math.undergrad:57807 alt.math:17965 X-Received-Date: Sun, 23 Sep 2007 17:31:24 EDT (newsdbm02.news.prodigy.net) On Sep 23, 2:41 pm, rossum wrote: > On Sun, 23 Sep 2007 11:44:12 -0700, marcus_b wrote: >He is the Las Vegas gamble sticking quarter after >quarter in the $1M slot machine, hoping to strike it >rich. Like the gambler and unlike, e.g., Dixon or >Pomerance, there is no rationale behind his attempts. >Factors of the target and factors of the surrogate, >whichever is which, have no useful relationship to each >other. Ironically, using the wrong formula for the >expression above is just about as likely to produce >the result he wants as using the right one. He >might as well compute random GCDs. Marcus. When coding up James' many different versions of his method I > sometimes make mistakes. It is not unusual for the mistakes to > perform as well as or better than James' actual method. As you say, > to a first approximation his various methods are inefficient PRNGs. It may be that James is actually developing a meta-method. He creates > numerous random versions of his factoring algorithm and throws them to > us for some ruthless selection. Unfortunately he does not then go to > the next stage and develop those versions that fare least badly. > Eventually he might find a genetic algorithm to develop new > mathematical algorithms using sci.math for the selection part. > This is just about as likely to work as mixing up chemicals at random would produce a cure-all for cancer. Funny, isn't it, that real progress in mathematics is made by deep rational and logical thinkers, not by know-nothings tossing out random ideas? Marcus. > rossum From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!border1.nntp.dca.giganews.com! nntp.giganews.com!local01.nntp.dca.giganews.com!nntp.comcast.com! news.comcast.com.POSTED!not-for-mail NNTP-Posting-Date: Sun, 23 Sep 2007 16:30:12 -0500 Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: JSH: Surrogate factoring algorithm From: Tim Peters References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> <1190570898.806448.149060@r29g2000hsg.googlegroups.com> <1190573052.288861.89410@57g2000hsv.googlegroups.com> Message-ID: User-Agent: Xnews/5.04.25 Date: Sun, 23 Sep 2007 16:30:12 -0500 Lines: 38 X-Usenet-Provider: http://www.giganews.com NNTP-Posting-Host: 69.255.16.201 X-Trace: sv3-wC2481EaAD1bH0nd1l2gWaVOHE1VjCPiXQIfcBV3OvBir6NjzkeUdiG58Jz4+ mWR1ejpNWcE54FHFbg!+oPUi6mITb9wrO5VUFQ9jjgr5AW8+ H7SpL0jfPOBKSFJhJcpKKXkKEqkFK2t6n/IR1753sdoVPI9!LaDbCPg2FMX94MnjElFJpA7g X-Complaints-To: abuse@comcast.net X-DMCA-Complaints-To: dmca@comcast.net X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.35 Xref: prodigy.net sci.math:1109924 alt.math.undergrad:57806 alt.math:17964 X-Received-Date: Sun, 23 Sep 2007 17:30:16 EDT (newsdbm02.news.prodigy.net) [added JSH: to subject] [marcus_b, on JSH] > ... He is the Las Vegas gamble sticking quarter after > quarter in the $1M slot machine, hoping to strike it > rich. Like the gambler and unlike, e.g., Dixon or > Pomerance, there is no rationale behind his attempts. Spoken like a mathematician who's never been to Vegas ;-) Really, the analogy is /better/ than you think: many gamblers follow a system (well, a succession of systems) they're sure is going to pay off. There are even systems for determining when a slot machine has gone cold (unlikely to pay off soon) so can you move on to a hot machine (more likely to pay off soon). All gibberish, but like JSH math sometimes using technical phrases more-or-less appropriately. Of course casinos love systems for the obvious reason -- which may be related to why sci.math loves JSH-math so much ;-) > Factors of the target and factors of the surrogate, > whichever is which, have no useful relationship to each > other. Ah, but can you /prove/ that they don't? James's inability to discover a useful relationship doesn't bother him a bit, not in the face of his unreasoning intuition that he /must/ be on a brilliant (heck, beyond brilliant) track. After a year of thrashing around the same basic approach, he isn't even impressed by that the equations he starts from /still/ aren't satisfied by the algorithms he makes up. > Ironically, using the wrong formula for the > expression above is just about as likely to produce > the result he wants as using the right one. He > might as well compute random GCDs. More, he'd be better off: random-GCD is efficient compared to most of his methods. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! w3g2000hsg.googlegroups.com!not-for-mail From: marcus_b Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: JSH: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 15:17:04 -0700 Organization: http://groups.google.com Lines: 80 Message-ID: <1190585824.314188.54830@w3g2000hsg.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <1190477822.973052.182610@o80g2000hse.googlegroups.com> <1190570898.806448.149060@r29g2000hsg.googlegroups.com> <1190573052.288861.89410@57g2000hsv.googlegroups.com> NNTP-Posting-Host: 128.101.116.194 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190585824 16721 127.0.0.1 (23 Sep 2007 22:17:04 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 22:17:04 +0000 (UTC) In-Reply-To: User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.0.8) Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: w3g2000hsg.googlegroups.com; posting-host=128.101.116.194; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109928 alt.math.undergrad:57808 alt.math:17966 X-Received-Date: Sun, 23 Sep 2007 18:17:04 EDT (newsdbm02.news.prodigy.net) On Sep 23, 4:30 pm, Tim Peters wrote: > [added JSH: to subject] [marcus_b, on JSH] ... He is the Las Vegas gamble sticking quarter after > quarter in the $1M slot machine, hoping to strike it > rich. Like the gambler and unlike, e.g., Dixon or > Pomerance, there is no rationale behind his attempts. Spoken like a mathematician who's never been to Vegas ;-) Aha, but I *have* been to Vegas. > Really, the > analogy is /better/ than you think: many gamblers follow a system > (well, a succession of systems) they're sure is going to pay off. There > are even systems for determining when a slot machine has gone cold > (unlikely to pay off soon) so can you move on to a hot machine (more > likely to pay off soon). That's exactly what I observed on one of my trips there (strictly on business, you understand). I met a young guy who was completely convinced he was on a hot streak and he was intending to play for all it was worth in the next day or two. I never found out if he actually came out ahead. > All gibberish, but like JSH math sometimes > using technical phrases more-or-less appropriately. Of course casinos > love systems for the obvious reason -- which may be related to why > sci.math loves JSH-math so much ;-) > Love, hate. Two sides of the same coin? I think not. But people do love to hate the guy. Without him, lots of jokes here would have no butt. > Factors of the target and factors of the surrogate, > whichever is which, have no useful relationship to each > other. Ah, but can you /prove/ that they don't? The burden of proof has to be on the other side. Otherwise we waste all our time on crackpottery. Which is what we are doing right now, BTW. > James's inability to discover > a useful relationship doesn't bother him a bit, not in the face of his > unreasoning intuition that he /must/ be on a brilliant (heck, beyond > brilliant) track. Way beyond brilliant. HUGE. Think of the world economy. Think of your own bank account, right now being systematically looted of its last nickel by Harrisian factorers. Think of Paris Hilton's cell phone. Think of Britney Spears. Think of Dolly Parton. Yes, we're talking BEYOND HUGE. > After a year of thrashing around the same basic > approach, he isn't even impressed by that the equations he starts from > /still/ aren't satisfied by the algorithms he makes up. Ironically, using the wrong formula for the > expression above is just about as likely to produce > the result he wants as using the right one. He > might as well compute random GCDs. More, he'd be better off: random-GCD is efficient compared to most of > his methods. Surely not. How could you write something that does worse than random? By having it waste lots of time in factoring useless surrogates, perhaps? But yeah. It's way beyond brilliant. It's HUGE. Marcus. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!news-out.octanews.net! indigo.octanews.net!authen.yellow.readfreenews.net.POSTED!not-for-mail From: Mas Plak Newsgroups: sci.math,alt.math.undergrad,alt.math References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> Subject: Re: Surrogate factoring algorithm Date: Sat, 22 Sep 2007 10:54:23 -0500 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 6.00.2900.3138 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-RFC2646: Format=Flowed; Original Lines: 70 Message-ID: <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> Organization: Read Free News NNTP-Posting-Date: 22 Sep 2007 10:53:40 CDT X-Trace: DXC=UK>W8 wrote in message news:1190426677.656410.200610@22g2000hsm.googlegroups.com... WHY ARE YOU STILL POSTING HERE RACIST ? > Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: WRONG. You have shown NOTHING. You are too DUMB to do simple algebra. You were born retarded in math. In the ring of integers given =>Which Ring? 'O Ringless Monkey Boy, troll for brains ? x^2 = y^2 mod T and k = 2x mod T *meaningless drivel*, snotty noodles thrown on a wall > it must be true that (x+k)^2 = y^2 + S must be ? WHY DON'T YOU DERIVE IT ? can you PROOVE it ? where S = 2k^2 mod T How do you pick (GUESS) k ? T ? S? x? y ? > so one congruence of squares is connected to an infinity of others. Prove it, retard. > That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. Factor your ass back to Africa, racist. methods going the other way where T is a target as with those it was > not possible to prove that only a single k with some minimal > qualifications should work. But the DON'T WORK. STFU Racist slime ball troll > James ( J.B.) Harris > From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! d55g2000hsg.googlegroups.com!not-for-mail From: mike3 Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sat, 22 Sep 2007 11:36:50 -0700 Organization: http://groups.google.com Lines: 24 Message-ID: <1190486210.239799.110600@d55g2000hsg.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> NNTP-Posting-Host: 74.38.33.242 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190486210 21325 127.0.0.1 (22 Sep 2007 18:36:50 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 22 Sep 2007 18:36:50 +0000 (UTC) In-Reply-To: <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.1) Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: d55g2000hsg.googlegroups.com; posting-host=74.38.33.242; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109628 alt.math.undergrad:57786 alt.math:17955 X-Received-Date: Sat, 22 Sep 2007 14:36:53 EDT (newsdbm02.news.prodigy.net) On Sep 22, 9:54 am, Mas Plak wrote: > JSH wrote in message news:1190426677.656410.200610@22g2000hsm.googlegroups.com... WHY ARE YOU STILL POSTING HERE RACIST ? Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: WRONG. You have shown NOTHING. You are too DUMB to do simple algebra. You were born retarded in math. > And he obviously doesn't seem interested in overcoming it, either. Is that monkey thing racism on your part TOO? From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!news-out.octanews.net! indigo.octanews.net!authen.yellow.readfreenews.net.POSTED!not-for-mail From: Mas Plak Newsgroups: sci.math,alt.math.undergrad,alt.math References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> <1190486210.239799.110600@d55g2000hsg.googlegroups.com> Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 12:38:46 -0500 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 6.00.2900.3138 X-MIMEOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-RFC2646: Format=Flowed; Original Lines: 32 Message-ID: <46f6a47b$0$47155$892e7fe2@authen.yellow.readfreenews.net> Organization: Read Free News NNTP-Posting-Date: 23 Sep 2007 12:38:03 CDT X-Trace: DXC=P2_9 wrote in message news:1190486210.239799.110600@d55g2000hsg.googlegroups.com... > On Sep 22, 9:54 am, Mas Plak wrote: >> JSH wrote in message >> news:1190426677.656410.200610@22g2000hsm.googlegroups.com... >> WHY ARE YOU STILL POSTING HERE RACIST ? >> Simple relations connect every integer factorization to an infinity of >> others as I have shown with simple congruences: >> WRONG. You have shown NOTHING. You are too DUMB to do simple algebra. >> You were born retarded in math. And he obviously doesn't seem interested in overcoming it, > either. > Is that monkey thing racism on your part TOO? > I was going to say stinking cabbage head, but I have been harassed by Vegetable Rights people. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! d55g2000hsg.googlegroups.com!not-for-mail From: mike3 Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 11:08:09 -0700 Organization: http://groups.google.com Lines: 38 Message-ID: <1190570889.177928.288650@d55g2000hsg.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> <1190486210.239799.110600@d55g2000hsg.googlegroups.com> <46f6a47b$0$47155$892e7fe2@authen.yellow.readfreenews.net> NNTP-Posting-Host: 74.38.33.242 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190570889 27801 127.0.0.1 (23 Sep 2007 18:08:09 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 18:08:09 +0000 (UTC) In-Reply-To: <46f6a47b$0$47155$892e7fe2@authen.yellow.readfreenews.net> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.1) Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: d55g2000hsg.googlegroups.com; posting-host=74.38.33.242; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109863 alt.math.undergrad:57800 alt.math:17960 X-Received-Date: Sun, 23 Sep 2007 14:08:10 EDT (newsdbm02.news.prodigy.net) On Sep 23, 11:38 am, Mas Plak wrote: > mike3 wrote in message news:1190486210.239799.110600@d55g2000hsg.googlegroups.com... On Sep 22, 9:54 am, Mas Plak wrote: >> JSH wrote in message >news:1190426677.656410.200610@22g2000hsm.googlegroups.com... > WHY ARE YOU STILL POSTING HERE RACIST ? > Simple relations connect every integer factorization to an infinity of >> others as I have shown with simple congruences: > WRONG. You have shown NOTHING. You are too DUMB to do simple algebra. > You were born retarded in math. And he obviously doesn't seem interested in overcoming it, > either. > Is that monkey thing racism on your part TOO? I was going to say stinking cabbage head, but I have been harassed by > Vegetable Rights people. You could have just said, . If you were complaining that James was racist, then you should not also practice racism yourself. Otherwise isn't it then hypocrisy? From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! d55g2000hsg.googlegroups.com!not-for-mail From: Rotwang Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Sun, 23 Sep 2007 10:53:59 -0700 Organization: http://groups.google.com Lines: 14 Message-ID: <1190570039.910172.214410@d55g2000hsg.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> <46f53a84$0$47140$892e7fe2@authen.yellow.readfreenews.net> <1190486210.239799.110600@d55g2000hsg.googlegroups.com> <46f6a47b$0$47155$892e7fe2@authen.yellow.readfreenews.net> NNTP-Posting-Host: 81.96.205.181 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190570040 24677 127.0.0.1 (23 Sep 2007 17:54:00 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 17:54:00 +0000 (UTC) In-Reply-To: <46f6a47b$0$47155$892e7fe2@authen.yellow.readfreenews.net> User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-GB; rv:1.8.1.7) Gecko/20070914 Firefox/2.0.0.7,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: d55g2000hsg.googlegroups.com; posting-host=81.96.205.181; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109854 alt.math.undergrad:57797 alt.math:17959 X-Received-Date: Sun, 23 Sep 2007 13:54:02 EDT (newsdbm02.news.prodigy.net) On 23 Sep, 18:38, Mas Plak wrote: > mike3 wrote in message >> Is that monkey thing racism on your part TOO? I was going to say stinking cabbage head, but I have been harassed by > Vegetable Rights people. What about the Factor your ass back to Africa, racist comment? Was that racism? From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!wns14feed!worldnet.att.net!204.71.34.3! newsfeed.cwix.com!newsfeed.vmunix.org!news.cs.univ-paris8.fr!feeder1-2. proxad.net!proxad.net!feeder1-1.proxad.net!club-internet.fr! feedme-small.clubint.net!aioe.org!not-for-mail From: Jane Winston Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Fri, 21 Sep 2007 19:18:25 -0700 Organization: Aioe.org NNTP Server Lines: 67 Message-ID: References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> Reply-To: Jane Winston NNTP-Posting-Host: S8BgaoXylOPtNwvb+9ciXA.user.aioe.org X-Complaints-To: abuse@aioe.org X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-RFC2646: Format=Flowed; Original X-Newsreader: Microsoft Outlook Express 6.00.2900.3138 X-Priority: 3 X-MSMail-Priority: Normal Xref: prodigy.net sci.math:1109483 alt.math.undergrad:57775 alt.math:17948 X-Received-Date: Fri, 21 Sep 2007 22:17:47 EDT (newsdbm02.news.prodigy.net) JSH wrote in message news:1190426677.656410.200610@22g2000hsm.googlegroups.com... > Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: Idiot. > In the ring of integers given Moron. > x^2 = y^2 mod T and k = 2x mod T Imbecile. > it must be true that Cretin. > (x+k)^2 = y^2 + S 'tard. > where S = 2k^2 mod T Douche-nozzle. > so one congruence of squares is connected to an infinity of others. Ass-tunnel. > That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. -wit. > If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a > non-trivial factorization is given by Copraphageous twat. > x + k - y = f_1 and Impotent Puke. > x + k + y = f_2 so > Dog sodomizer. > 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. Baby raper. > So the algorithm is that given a target S to be factored, choose a non- > integer k, and find T, from Snip. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! y42g2000hsy.googlegroups.com!not-for-mail From: mike3 Newsgroups: sci.math,alt.math.undergrad,alt.math Subject: Re: Surrogate factoring algorithm Date: Fri, 21 Sep 2007 21:44:07 -0700 Organization: http://groups.google.com Lines: 77 Message-ID: <1190436247.212770.67090@y42g2000hsy.googlegroups.com> References: <1190426677.656410.200610@22g2000hsm.googlegroups.com> NNTP-Posting-Host: 74.38.33.242 Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii X-Trace: posting.google.com 1190436247 8253 127.0.0.1 (22 Sep 2007 04:44:07 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 22 Sep 2007 04:44:07 +0000 (UTC) In-Reply-To: User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.8.1) Gecko/20061023 SUSE/2.0-30 Firefox/2.0,gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: y42g2000hsy.googlegroups.com; posting-host=74.38.33.242; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net sci.math:1109509 alt.math.undergrad:57776 alt.math:17949 X-Received-Date: Sat, 22 Sep 2007 00:44:07 EDT (newsdbm02.news.prodigy.net) On Sep 21, 8:18 pm, Jane Winston wrote: > JSH wrote in message news:1190426677.656410.200610@22g2000hsm.googlegroups.com... Simple relations connect every integer factorization to an infinity of > others as I have shown with simple congruences: Idiot. In the ring of integers given Moron. x^2 = y^2 mod T and k = 2x mod T Imbecile. it must be true that Cretin. (x+k)^2 = y^2 + S 'tard. where S = 2k^2 mod T Douche-nozzle. so one congruence of squares is connected to an infinity of others. Ass-tunnel. That can be used to factor in either direction, but I will focus on > letting S be the target composite to be factored in this post. -wit. If S has non-unit factors f_1 and f_2, such that S = f_1*f_2, then a > non-trivial factorization is given by Copraphageous twat. x + k - y = f_1 and Impotent Puke. x + k + y = f_2 so Dog sodomizer. 2(x+k) = f_1 + f_2 and all you need do is determine x and y, and choose k. Baby raper. So the algorithm is that given a target S to be factored, choose a non- > integer k, and find T, from Snip. Does any of that explain to people like me who want to learn something why exactly James is wrong here? I've noticed that he just says choose a non-integer k, but then it seems to turn into a wild guessing game. Now that's a good reason to dismiss the algorithm. He hasn't produced the RSA challenge factors yet, just more of these unworkable algorithms that magically fail to solve jack. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!wn13feed!worldnet.att.net!208.48.142.85! newsfeed.news2me.com!sn-xt-sjc-05!sn-xt-sjc-06!sn-post-sjc-01!supernews.com! corp.supernews.com!hevanet.com!marsh From: William Elliot Newsgroups: alt.math Subject: Re: Intersections of circles Date: Fri, 21 Sep 2007 22:06:35 -0700 Organization: Posted via Supernews, http://www.supernews.com Message-ID: In-Reply-To: <1190000903.483568.224200@g4g2000hsf.googlegroups.com> References: <1190000903.483568.224200@g4g2000hsf.googlegroups.com> MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII X-Complaints-To: abuse@supernews.com Lines: 39 Xref: prodigy.net alt.math:17950 X-Received-Date: Sat, 22 Sep 2007 01:06:47 EDT (newsdbm02.news.prodigy.net) On Sun, 16 Sep 2007, Nicholas Sherlock wrote: > I'm stuck on a geometry problem that I don't know how to solve > efficiently. I have a set of circles on an x,y plane. All the circles > have the same radius. No matter where I place a point on this plane, I > need the number of circles which contain this point to be no larger > than a given maximum. Actually, I want to find this maximum number. You need to clarify the unusual use of contain. Do you mean on the circle, inside the circle or, on or inside the circle? > So, if none of my circles intersect, the answer will be 1 (An > arbitrary point on the plane is contained by at most one circle). If a > pair of circles intersect at two points, the answer would be two. If What if two tangent circles? > four circles all enclose one area, the answer will be four. Four > circles could also intersect such that any area is enclosed by at most > two circles, so the answer would be two, or a couple of other ways. At the moment, I'm constructing a graph where the nodes represent the > circles, and I add a bidirectional edge between two nodes when the > circles at the two nodes overlap each other (There is an area that > they both enclose). Then, I find the largest clique in the graph. I > think this will give me an upper bound on the maximum number of times > a point could be contained by my circles. I can see that I could compute the points of intersections between all > the circles in the clique. If one of these points was contained by or > was on the boundary of all the circles in the clique, then I would > know that there is a point enclosed the same number of times as the > number of circles in the clique. But I'm hoping that somebody knows a > faster solution. Thanks, > Nicholas Sherlock From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!news2.arglkargh.de!noris.net! newsfeed.velia.net!newsfeed01.sul.t-online.de!t-online.de!newsfeeder.wxs.nl! divide.nova.planet.nl!posting.nova.planet.nl!not-for-mail From: Peter Hendriks Newsgroups: sci.logic,sci.math,alt.math,alt.brain.teasers Subject: MathPuzzle 195: Easy arithmetic Date: Sat, 22 Sep 2007 09:40:05 +0200 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 6.00.2900.3138 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 X-RFC2646: Format=Flowed; Original X-Antivirus: avast! (VPS 000775-6, 22-09-2007), Outbound message X-Antivirus-Status: Clean Lines: 23 Message-ID: <46f4c6d1$0$25487$ba620dc5@text.nova.planet.nl> NNTP-Posting-Host: 86.81.128.227 X-Trace: 1190446801 text.nova.planet.nl 25487 86.81.128.227:3911 X-Complaints-To: abuse@planet.nl Xref: prodigy.net sci.logic:202342 sci.math:1109516 alt.math:17951 alt.brain.teasers:62157 X-Received-Date: Sat, 22 Sep 2007 03:42:02 EDT (newsdbm02.news.prodigy.net) A good Saturday morning to all of you, from sunny Ruurlo ! This time my new puzzle is really simple. Just easy arithmetics. To be honest, the puzzle I had in mind for this week was a tiny bit more difficult. But when I woke up this Thursday morning, I realised that the puzzle had an error I could not repair easily. Therefore this puzzle from my little stock. Puzzle 188 (Prime cross) has been closed now. This week I will try to close a few more puzzles, but I seem to have a time constraint. Nevertheless I will try the best I can. Have (a little?) fun with the new puzzle! Solutions are more than welcome. Best regards, Peter direct link: http://home.planet.nl/~p.j.hendriks/p195e.htm Please answer by email and not in this newsgroup. From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! 19g2000hsx.googlegroups.com!not-for-mail From: Deadspeak2 Newsgroups: alt.fiction.interactive,alt.books,alt.writing,alt.math,alt.technology Subject: about irrational numbers and Ockam's razor - by sEung kim Date: Sat, 22 Sep 2007 12:34:44 -0700 Organization: http://groups.google.com Lines: 16 Message-ID: <1190489684.029846.279690@19g2000hsx.googlegroups.com> NNTP-Posting-Host: 66.8.158.190 Mime-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 X-Trace: posting.google.com 1190489684 1377 127.0.0.1 (22 Sep 2007 19:34:44 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sat, 22 Sep 2007 19:34:44 +0000 (UTC) User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 1.1.4322; Hotbar 10.0.356.0),gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: 19g2000hsx.googlegroups.com; posting-host=66.8.158.190; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net alt.books:89495 alt.writing:200862 alt.math:17956 X-Received-Date: Sat, 22 Sep 2007 15:34:46 EDT (newsdbm02.news.prodigy.net) I have long discovered that relativity exists in size and distance, where the size of the object in proportion to the distance covered is of ratioistic equivalence to smaller or greater distance covered in proportion to the size of the object. It is clear that Ockam's Razor proves that irrational numbers are not merely conceptual but as zooming indefinately proves through Ockam's Razor that it can regress in decimal point indefinately where there is no resolution reached since the distance travelled regresses indefinately. But according to proportional distance we realize also that there is a leap in space travelled which resolves logically this problem that there may be infinite microcosms but motion requires that an object leap at a uniform speed..where infinite micrcosms are traversed in continuum by decline of speed at an accelerated rate of decline. Conceptually True and Proven by Ockam's Razor (by Seung Kim) From ???@??? Fri Jan 01 00:00:00 1999 Path: newsdbm02.news.prodigy.net!newsdst02.news.prodigy.net!prodigy.com!newscon02. news.prodigy.net!prodigy.net!news.glorb.com!postnews.google.com! d55g2000hsg.googlegroups.com!not-for-mail From: don.lotto@paradise.net.nz Newsgroups: alt.bible,nz.general,alt.math Subject: GODSLV > FFF FF(8) Date: Sun, 23 Sep 2007 00:20:57 -0700 Organization: http://groups.google.com Lines: 24 Message-ID: <1190532057.352166.114960@d55g2000hsg.googlegroups.com> NNTP-Posting-Host: 121.73.65.225 Mime-Version: 1.0 Content-Type: text/plain; charset=iso-8859-1 X-Trace: posting.google.com 1190532057 21012 127.0.0.1 (23 Sep 2007 07:20:57 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Sun, 23 Sep 2007 07:20:57 +0000 (UTC) User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Complaints-To: groups-abuse@google.com Injection-Info: d55g2000hsg.googlegroups.com; posting-host=121.73.65.225; posting-account=ps2QrAMAAAA6_jCuRt2JEIpn5Otqf_w0 Xref: prodigy.net alt.bible:1340748 nz.general:1299967 alt.math:17957 X-Received-Date: Sun, 23 Sep 2007 03:20:57 EDT (newsdbm02.news.prodigy.net) God's love for us: God's love for me and God's love for you. Love ...The Ephesians did not fully grasp God's love for them. .... deep is the love of Christ, and to know this love that surpasses knowledge--that I may be filled ... www.dougbrittonbooks.com/onlinebiblestudies-selfworthandrespect/ godsloveforusandgodsloveforme.asp - 19k - Cached - Similar pages Ephesians 3:19 And to know the love of 8th Fermat number F(8) = 2 ^ (2^8) +1. = 2 ^ 256 +1 ~ = 1E 77. FF FF F8 incomprehensible large. GD LV > F 8. number plate. 23/9/07. don.lotto nz. gods love surpasses knowledge.